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Exercise 9A

Page No 139:

Question 1:

Write each of the following statements as an equation:
(i) 5 times a number equals 40.
(ii) A number increased by 8 equals 15.
(iii) 25 exceeds a number by 7.
(iv) A number exceeds 5 by 3.
(v) 5 subtracted from thrice  a number is 16.
(vi) If 12 is subtracted from a number, the result is 24.
(vii) Twice a number subtracted from 19 is 11.
(viii) A number divided by 8 gives 7.
(ix) 3 less than 4 times a number is 17.
(x) 6 times a number is 5 more than the number.

ANSWER:

(i) Let the required number be x.
  So, five times the number will be 5x.
  ∴ 5x = 40

(ii) Let the required number be x.
     So, when it is increased by 8, we get x + 8.
     ∴ x + 8 = 15

(iii) Let the required number be x.
     So, when 25 exceeds the number, we get 25 −- x.
 ∴ 25 −- x  = 7

(iv) Let the required number be x.
    So, when the number exceeds 5, we get x −- 5.
   ∴ x −- 5  = 3

(v) Let the required number be x.
     So, thrice the number will be 3x.
  ∴ 3x −- 5 = 16

(vi) Let the required number be x.
     So, 12 subtracted from the number will be x −- 12.
   ∴ x −- 12 = 24

(vii) Let the required number be x.
     So, twice the number will be 2x.
   ∴ 19 −- 2x = 11

(viii) Let the required number be x.
     So, the number when divided by 8 will be x8x8.
   ∴ x8x8 = 7

(ix) Let the required number be x.
     So, four times the number will be 4x.
 ∴ 4x −- 3 = 17

(x) Let the required number be x.
     So, 6 times the number will be 6x.
    ∴ 6x = x + 5

Page No 140:

Question 2:

Write a statement for each of the equations, given below:
(i) x − 7 = 14
(ii) 2y = 18
(iii) 11 + 3x = 17
(iv) 2x − 3 = 13
(v) 12y − 30 = 6
(vi) 2z3=82z3=8

ANSWER:

(i) 7 less than the number x equals 14.
(ii) Twice the number y equals 18.
(iii) 11 more than thrice the number x equals 17.
(iv) 3 less than twice the number x equals 13.
(v) 30 less than 12 times the number y equals 6.
(vi) When twice the number z is divided by 3, it equals 8.

Page No 140:

Question 3:

Verify by substitution that
(i) the root of 3x − 5 = 7 is x = 4
(ii) the root of 3 + 2x = 9 is x = 3
(iii) the root of 5x − 8 = 2x − 2 is x = 2
(iv) the root of 8 − 7y = 1 is y = 1
(v) the root of z7=8z7=8 is z = 56

ANSWER:

(i)

   3x − 5 = 7Substituting x = 4  in the given equation:L.H.S. : 3×4 −5or, 12 − 5 = 7 = R.H.S.L.H.S. = R.H.S. Hence,  x = 4  is the root of the given equation. 3x – 5 = 7Substituting x = 4  in the given equation:L.H.S. : 3×4 -5or, 12 – 5 = 7 = R.H.S.L.H.S. = R.H.S. Hence,  x = 4  is the root of the given equation. 

(ii)

    3 + 2x= 9Substituting x = 3 in the given equation:L.H.S. : 3 + 2×3or,  3 + 6 = 9 = R.H.S. L.H.S. = R.H.S. Hence,  x = 3  is the root of the given equation. 3 + 2x= 9Substituting x = 3 in the given equation:L.H.S. : 3 + 2×3or,  3 + 6 = 9 = R.H.S. L.H.S. = R.H.S. Hence,  x = 3  is the root of the given equation. 

(iii)

    5x − 8 = 2x −2 Substituting x = 2 in the given equation:L.H.S. :                                   R.H.S. :5×2− 8                           = 2×2−2or, 10 − 8 = 2                      = 4 −2 = 2     L.H.S. = R.H.S. Hence, x =2  is the root of the given equation. 5x – 8 = 2x -2 Substituting x = 2 in the given equation:L.H.S. :                                   R.H.S. :5×2- 8                           = 2×2-2or, 10 – 8 = 2                      = 4 -2 = 2     L.H.S. = R.H.S. Hence, x =2  is the root of the given equation. 

(iv)

     8 − 7y = 1 Substituting y = 1  in the given equation:L.H.S. : 8 −7×1or, 8 − 7 = 1 = R.H.S. L.H.S. = R.H.S. Hence, y =1  is the root of the given equation. 8 – 7y = 1 Substituting y = 1  in the given equation:L.H.S. : 8 -7×1or, 8 – 7 = 1 = R.H.S. L.H.S. = R.H.S. Hence, y =1  is the root of the given equation. 

(v)

     z7 = 8 Substituting z = 56 in the given equation:L.H.S. : 567 = 8 =R.H.S.L.H.S. = R.H.S. Hence, z =56  is the root of the given equation.z7 = 8 Substituting z = 56 in the given equation:L.H.S. : 567 = 8 =R.H.S.L.H.S. = R.H.S. Hence, z =56  is the root of the given equation.

Page No 140:

Question 4:

Solve each of the following equations by the trial-and-error method:
(i) y + 9 = 13
(ii) x − 7 = 10
(iii) 4x = 28
(iv) 3y = 36
(v) 11 + x = 19
(vi) x3=4×3=4
(vii) 2x − 3 = 9
(viii) 12x + 7 = 1112x + 7 = 11
(ix) 2y + 4 = 3y
(x) z − 3 = 2z − 5

ANSWER:

(i) y + 9 = 13
    We try several values of y until we get the  L.H.S. equal to the R.H.S.

y	L.H.S.	R.H.S.	Is LHS =RHS ?
1	1 + 9 = 10	13	No
2	2 + 9 = 11	13	No
3	3 + 9 = 12	13	No
4	4 + 9 = 13	13	Yes

 ∴ y = 4

(ii) x − 7= 10
We try several values of x until we get the  L.H.S. equal to the R.H.S.

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
10	10 − 7 = 3	10	No
11	11 − 7 = 4	10	No
12	12 − 7 = 5	10	No
13	13 − 7 = 6	10	No
14	14 − 7 = 7	10	No
15	15 − 7 = 8	10	No
16	16 − 7 = 9	10	No
17	17 − 7 = 10	10	Yes

∴ x = 17

(iii) 4x = 28
     We try several values of x until we get the  L.H.S. equal to the R.H.S.

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
1	4 ×× 1 = 4	28	No
2	4 ×× 2 = 8	28	No
3	4 ×× 3 = 12	28	No
4	4 ×× 4 = 16	28	No
5	4 ×× 5 = 20	28	No
6	4 ×× 6 = 24	28	No
7	4 ×× 7 = 28	28	Yes

  ∴ x = 7

(iv) 3y = 36
    We try several values of x until we get the L.H.S. equal to the R.H.S.

y	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
6	3 ×× 6 = 18	36	No
7	3 ×× 7 = 21	36	No
8	3 ×× 8 = 24	36	No
9	3 ×× 9 = 27	36	No
10	3 ×× 10 = 30	36	No
11	3 ××11 = 33	36	No
12	3 ×× 12 = 36	36	Yes

 ∴ y = 12

(v) 11 + x = 19
     We try several values of x until we get the L.H.S. equal to the R.H.S.

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
1	11 + 1 = 12	19	No
2	11 + 2 = 13	19	No
3	11 + 3 = 14	19	No
4	11 + 4 = 15	19	No
5	11 + 5 = 16	19	No
6	11 + 6 = 17	19	No
7	11 + 7 = 18	19	No
8	11 + 8 = 19	19	Yes

   ∴ x = 8

(vi) x3 = 4×3 = 4
    Since R.H.S. is an natural number so L.H.S. must also be a natural number. Thus, x has to be a multiple of 3.

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
3	33=133=1	4	No
6	63=263=2	4	No
9	93=393=3	4	No
12	123=4123=4	4	Yes

  ∴ x = 12

(vii) 2x − 3 = 9

  We try several values of x until we get the L.H.S. equal to the R.H.S.

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
1	2 ×× 1 − 3 = −1	9	No
2	2 ×× 2 − 3 = 1	9	No
3	2 ×× 3 − 3 = 3	9	No
4	2 ×× 4 − 3 = 5	9	No
5	2 ×× 5 − 3 = 7	9	No
6	2 ×× 6 − 3 = 9	9	Yes

  ∴  x = 6

(viii) 12x + 7 = 1112x + 7 = 11
     Since, R.H.S. is a natural number so L.H.S. must be a natural number Thus, we will try values if x which are multiples of ‘x’

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
2	2/2 + 7 = 8	11	No
4	4/2 + 7 = 9	11	No
6	6/2 + 7 = 10	11	No
8	8/2 + 7 = 11	11	Yes

    ∴ x = 8

(ix) 2y + 4 = 3y
      We try several values of y until we get the L.H.S. equal to the R.H.S.

y	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
1	2 ×× 1 + 4 = 6	3 ×× 1 = 3	No
2	2 ×× 2 + 4 = 8	3 ×× 2 = 6	No
3	2 ×× 3 + 4 = 10	3 ×× 3 = 9	No
4	2 ×× 4 + 4 = 12	3 ×× 4 = 12	Yes

  ∴ y = 4

(x) z − 3 = 2z − 5
 We try several values of z till we get the L.H.S. equal to the R.H.S.

z	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.?
1	1 − 3 = −2	2 ×× 1 − 5 = −3	No
2	2 − 3 = −1	2 ×× 2 − 5 = −1	Yes

∴ z = 2

Page No 143:

Exercise 9B

Question 1:

Solve each of the following equations and verify the answer in each case:
x + 5 = 12

ANSWER:

 x + 5 = 12

Subtracting 5 from both the sides:
⇒ x + 5 − 5 = 12 − 5              
⇒ x = 7
Verification:
Substituting x = 7 in the L.H.S.:
⇒ 7 + 5 = 12 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 2:

Solve each of the following equations and verify the answer in each case:
x + 3 = −2

ANSWER:

 x + 3 = −2

Subtracting 3 from both the sides:
⇒ x + 3 − 3 = −2 − 3             
⇒ x = −5

Verification:
Substituting x = −5 in the L.H.S.:
⇒ −5 + 3 =  −2 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 3:

Solve each of the following equations and verify the answer in each case:
x − 7 = 6

ANSWER:

 x − 7 = 6
Adding 7 on both the sides:
⇒ x − 7 + 7 = 6 + 7               
⇒ x = 13

Verification:
Substituting x = 13 in the L.H.S.:
⇒ 13 − 7 =  6 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 4:

Solve each of the following equations and verify the answer in each case:
x − 2 = −5

ANSWER:

 x − 2 = −5

Adding 2 on both sides:
⇒ x − 2 + 2 = −5 + 2            
⇒ x = −3
Verification:
Substituting x = −3 in the L.H.S.:
⇒  −3 − 2 = −5 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 5:

Solve each of the following equations and verify the answer in each case:
3x − 5 = 13

ANSWER:

3x − 5 = 13
⇒ 3x − 5 + 5 = 13 + 5             [Adding 5 on both the sides]
⇒ 3x = 18
⇒ 3×3 = 1833×3 = 183                        [Dividing both the sides by 3]
⇒ x = 6
Verification:
Substituting x = 6 in the L.H.S.:
⇒  3 ×× 6 − 5 = 18 − 5 = 13 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 6:

Solve each of the following equations and verify the answer in each case:
4x + 7 = 15

ANSWER:

4x + 7 = 15
⇒ 4x + 7 − 7 = 15 − 7              [Subtracting 7 from both the sides]
⇒ 4x = 8
⇒ 4×4 = 844×4 = 84                        [Dividing both the sides by 4]
⇒ x = 2
Verification:
Substituting x = 2 in the L.H.S.:
⇒  4××2 + 7 = 8 + 7 = 15 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 7:

Solve each of the following equations and verify the answer in each case:
x5=12×5=12

ANSWER:

x5 = 12×5 = 12
⇒ x5×5  = 12×5×5×5  = 12×5                                  [Multiplying both the sides by 5]
⇒ x = 60
Verification:
Substituting x = 60 in the L.H.S.:
⇒ 605605 = 12 = R.H.S.
⇒ L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 8:

Solve each of the following equations and verify the answer in each case:
3×5=153×5=15

ANSWER:

     3×5 = 153×5 = 15
⇒ 3×5× 5  = 15 × 53×5× 5  = 15 × 5                                  [Multiplying both the sides by 5]
⇒ 3x = 75
⇒ 3×3 = 7533×3 = 753
⇒ x = 25
Verification:
Substituting x = 25 in the L.H.S.:
⇒ 3 × 2553 × 255 = 15 = R.H.S.
⇒ L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 9:

Solve each of the following equations and verify the answer in each case:
5x − 3 = x + 17

ANSWER:

5x − 3 = x + 17
⇒ 5x − x = 17 + 3                  [Transposing x to the L.H.S. and 3 to the R.H.S.]
⇒ 4x = 20
⇒ 4×4 = 2044×4 = 204                        [Dividing both the sides by 4]
⇒ x = 5
Verification:
Substituting x = 5 on both the sides:
L.H.S.:  5(5) − 3
⇒ 25 − 3
⇒ 22

R.H.S.:  5 + 17 = 22
⇒ L.H.S. = R.H.S.
 Hence, verified.

Page No 143:

Question 10:

Solve each of the following equations and verify the answer in each case:
2x−12=32x-12=3

ANSWER:

2x−12 = 32x-12 = 3
⇒ 2x −-1212 + 1212 = 3 + 1212                              [Adding 1212 on both the sides]
⇒ 2x  = 6 + 126 + 12
⇒ 2x = 7272
⇒ 2×2 = 72 × 22×2 = 72 × 2                                            [Dividing both the sides by 3]
⇒ x = 7474
Verification:
Substituting  x = 7474 in the  L.H.S.:
2(74) − 12= 72 − 12 = 62 = 3 = R.H.S.274 – 12= 72 – 12 = 62 = 3 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 11:

Solve each of the following equations and verify the answer in each case:
3(x + 6) = 24

ANSWER:

3(x + 6) = 24
⇒ 3×x + 3×6 = 243×x + 3×6 = 24                     [On expanding the brackets]
⇒  3x + 18 = 24
⇒ 3x + 18 −- 18 = 24 −- 18            [Subtracting 18 from both the sides]
⇒ 3x = 6
⇒ 3×3 = 633×3 = 63                                [Dividing both the sides by 3]
⇒ x = 2
Verification:
Substituting x = 2 in the L.H.S.:
 3(2 + 6) = 3 ××8 = 24  = R.H.S.
 L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 12:

Solve each of the following equations and verify the answer in each case:
6x + 5 = 2x + 17

ANSWER:

6x + 5 = 2x + 17
⇒⇒6x  −- 2x = 17 −- 5                          [Transposing 2x to the L.H.S. and 5 to the R.H.S.]
⇒⇒4x = 12
⇒⇒4×4= 1244×4= 124                                      [Dividing both the sides by 4]
⇒⇒x = 3
Verification:
Substituting x = 3 on both the sides:
L.H.S.: 6(3) + 5
=18 + 5
=23
R.H.S.:   2(3) + 17
= 6 + 17
= 23
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 13:

Solve each of the following equations and verify the answer in each case:
x4−8=1×4-8=1

ANSWER:

x4− 8 = 1×4- 8 = 1
⇒x4− 8 + 8 = 1 + 8⇒x4- 8 + 8 = 1 + 8                  [Adding 8 on both the sides]
⇒x4 = 9⇒x4 = 9
⇒x4 × 4 = 9 × 4⇒x4 × 4 = 9 × 4                        [Multiplying both the sides by 4]
or, x = 36
Verification:
Substituting x = 36 in the L.H.S.:
or, 364 − 8364 – 8 = 9 − 8 = 1 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 14:

Solve each of the following equations and verify the answer in each case:
x2=x3+1×2=x3+1

ANSWER:

x2 = x3 + 1×2 = x3 + 1
⇒x2 − x3 = 1⇒x2 – x3 = 1                                      [Transposing x3x3 to the L.H.S.]
⇒3x − 2×6 = 1⇒3x – 2×6 = 1
⇒x6 = 1⇒x6 = 1
⇒x6 × 6 = 1 × 6⇒x6 × 6 = 1 × 6                                    [Multiplying both the sides by 6]
or, x = 6
Verification:
Substituting x = 6 on both the sides:
L.H.S.: 62 62 = 3
R.H.S.: 63 + 163 + 1 =  2 + 1 =  3
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 15:

Solve each of the following equations and verify the answer in each case:
3(x + 2) − 2(x − 1) = 7

ANSWER:

3(x + 2) − 2(x − 1) = 7
⇒3×x + 3×2 − 2×x −2×(−1) = 7⇒3×x + 3×2 – 2×x -2×(-1) = 7              [On expanding the brackets]
or, 3x + 6 −-2x + 2 = 7
or, x + 8 = 7
or, x + 8 −- 8 = 7 −- 8                                        [Subtracting 8 from both the sides]
or, x = −-1
Verification:
Substituting x = −-1 in the L.H.S.:
3(−1+2) −2(−1−1)or, 3(1) −2(−2)or, 3 + 4 = 7 = R.H.S.3(-1+2) -2(-1-1)or, 3(1) -2(-2)or, 3 + 4 = 7 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 16:

Solve each of the following equations and verify the answer in each case:

ANSWER:

5(x-1) +2(x+3) + 6 = 0
⇒⇒5x -5 +2x +6 +6 = 0        (Expanding within the brackets)
⇒⇒7x +7 = 0
⇒⇒x +1 = 0       (Dividing by 7)
⇒⇒x = -1

Verification:
Putting x = -1 in the L.H.S.:
L.H.S.: 5(-1 -1) + 2(-1 + 3) + 6
          = 5(-2) + 2(2) + 6
          = -10 + 4 + 6  = 0 = R.H.S.

Hence, verified.

Page No 143:

Question 17:

Solve each of the following equations and verify the answer in each case:
6(1 − 4x) + 7(2 + 5x) = 53

ANSWER:

6(1 − 4x) + 7(2 + 5x) = 53
or, 6 × 1 − 6 × 4x + 7 × 2 + 7 × 5x = 536 × 1 – 6 × 4x + 7 × 2 + 7 × 5x = 53              [On expanding the brackets]
or, 6 −- 24x + 14 + 35x = 53
or, 11x + 20 = 53
or, 11x + 20 −- 20 = 53 −- 20                                        [Subtracting 20 from both the sides]
or, 11x = 33
or, 11×11= 331111×11= 3311                                                      [Dividing both the sides by 11]
or, x = 3
Verification:
Substituting x = 3 in the L.H.S.:
6(1 − 4 × 3) + 7(2 + 5 × 3)⇒6(1 − 12) + 7(2 + 15)⇒6(−11) + 7(17)⇒−66 + 119  = 53 = R.H.S.6(1 – 4 × 3) + 7(2 + 5 × 3)⇒6(1 – 12) + 7(2 + 15)⇒6(-11) + 7(17)⇒-66 + 119  = 53 = R.H.S.

L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 18:

Solve each of the following equations and verify the answer in each case:
16(3x − 5) − 10(4x − 8) = 40

ANSWER:

16(3x − 5) − 10(4x − 8) = 40
or, 16 × 3x − 16 × 5 −10 × 4x − 10 × (−8) = 4016 × 3x – 16 × 5 -10 × 4x – 10 × (-8) = 40              [On expanding the brackets]
or, 48x −- 80 −- 40x + 80 = 40
or, 8x  = 40
or, 8×8=4088×8=408                                                      [Dividing both the sides by 8]
or, x = 5
Verification:
Substituting x = 5 in the L.H.S.:

16(3 × 5 − 5) − 10( 4 × 5 − 8)⇒16(15 − 5) − 10(20 − 8)⇒16(10) −10(12)⇒160 − 120  = 40 = R.H.S.16(3 × 5 – 5) – 10( 4 × 5 – 8)⇒16(15 – 5) – 10(20 – 8)⇒16(10) -10(12)⇒160 – 120  = 40 = R.H.S.

L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 19:

Solve each of the following equations and verify the answer in each case:
3(x + 6) + 2(x + 3) = 64

ANSWER:

3(x + 6) + 2(x + 3) = 64
⇒⇒3 × x   +   3 × 6 + 2 × x  + 2 × 3   = 64            [On expanding the brackets]
⇒⇒3x + 18 +  2x + 6 = 64
⇒5x + 24 = 64
⇒5x + 24 −- 24 = 64 −- 24                                       [Subtracting 24 from both the sides]
⇒5x = 40
⇒5×5 = 4055×5 = 405                                                           [Dividing both the sides by 5]
⇒x = 8
Verification:
Substituting x = 8 in the L.H.S.:
3(8 + 6) + 2(8 + 3)3(14) + 2(11)42 + 22 = 64 = R.H.S.3(8 + 6) + 2(8 + 3)3(14) + 2(11)42 + 22 = 64 = R.H.S.

L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 20:

Solve each of the following equations and verify the answer in each case:
3(2 − 5x) − 2(1 − 6x) = 1

ANSWER:

3(2 − 5x) − 2(1 − 6x) = 1
or, 3 × 2  + 3 × (−5x) − 2 × 1 − 2 × (−6x) = 1           [On expanding the brackets]
or, 6 − 15x −  2 + 12x = 1
or, 4 – 3x = 1
or,  3  =3x                                                        
or, x = 1

Verification:
Substituting x = 1 in the L.H.S.:
3(2 − 5 × 1) − 2(1 − 6 × 1)⇒3(2 − 5) − 2(1− 6)⇒3(−3) −2(−5)⇒−9 + 10 = 1 = R.H.S.3(2 – 5 × 1) – 2(1 – 6 × 1)⇒3(2 – 5) – 2(1- 6)⇒3(-3) -2(-5)⇒-9 + 10 = 1 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 21:

Solve each of the following equations and verify the answer in each case:
n4−5=n6+12n4-5=n6+12

ANSWER:

n4−5 = n6 + 12n4-5 = n6 + 12
or, n4 − n6 = 12 + 5   n4 – n6 = 12 + 5                                     [Transposing n/6 to the L.H.S. and 5 to the R.H.S.]
or, 3n−2n12 = 1+1023n-2n12 = 1+102
or, n12 = 112n12 = 112
or, n12×12 = 112×12n12×12 = 112×12                                    [Dividing both the sides by 12]
or, n = 66
Verification:
Substituting n = 66 on both the sides:

L.H.S.:
664−5 =332 − 5  =33 − 102 =232 = 232 R.H.S.: 666+12= 11 + 12 = 22+12= 232 664-5 =332 – 5  =33 – 102 =232 = 232 R.H.S.: 666+12= 11 + 12 = 22+12= 232 
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 22:

Solve each of the following equations and verify the answer in each case:
2m3+ 8=m2−12m3+ 8=m2-1

ANSWER:

2m3 + 8 = m2− 12m3 + 8 = m2- 1
or, 2m3 − m2 = −1 −82m3 – m2 = -1 -8                      [Transposing m/2 to the L.H.S. and 8 to the R.H.S.]
or, 4m−3m6 = −9or, m6 = −9or, 4m-3m6 = -9or, m6 = -9
or, m6×6 = −9×6m6×6 = -9×6                            [Multiplying both the sides by 6]
or, m = −-54
Verification:
Substituting x = −54 on both the sides:

L.H.S.: 2(−54)3 +  8 = −542−1= −1083 + 8 = −36+ 8 = −28  R.H.S.:−542−1 = −27 − 1= −28L.H.S.: 2(-54)3 +  8 = -542-1= -1083 + 8 = -36+ 8 = -28  R.H.S.:-542-1 = -27 – 1= -28
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 23:

Solve each of the following equations and verify the answer in each case:
2×5−32=x2+12×5-32=x2+1

ANSWER:

2×5 −32 = x2 + 12×5 -32 = x2 + 1
or, 2×5− x2 = 1+ 322×5- x2 = 1+ 32                        [Transposing x/2 to the L.H.S. and 3/2 to R.H.S.]
or, 4x−5×10= 2+32or, −x10 = 52or, 4x-5×10= 2+32or, -x10 = 52
or, −x10(−10) =52 ×(−10)or, -x10(-10) =52 ×(-10)               [Multiplying both the sides by −10]
or, x = −25
Verification:
Substituting x = −25 on both the sides:
L.H.S.: 2(−25)5 − 32 = −505 − 32  = −10 − 32 = −232R.H.S.: −252+ 1= −25 + 22 = −232L.H.S.: 2(-25)5 – 32 = -505 – 32  = -10 – 32 = -232R.H.S.: -252+ 1= -25 + 22 = -232
L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 24:

Solve each of the following equations and verify the answer in each case:
x−35−2=2x5x-35-2=2×5

ANSWER:

x−35 − 2 = 2×5 x-35 – 2 = 2×5 
or, x5− 35 −2 = 2x5x5- 35 -2 = 2×5
or, − 35− 2 = 2×5−x5- 35- 2 = 2×5-x5                        [Transposing x/5 to the R.H.S.]
or, −3− 105 = x5-3- 105 = x5
or, −135 = x5-135 = x5
or, −135(5) =x5 ×(5)-135(5) =x5 ×(5)               [Multiplying both the sides by 5]
or, x = −13
Verification:
Substituting x = −13 on both the sides:
L.H.S.: −13 − 35 − 2 =−165 − 2= −16 − 105 = −265R.H.S.: 2×(−13)5  = −265  L.H.S.: -13 – 35 – 2 =-165 – 2= -16 – 105 = -265R.H.S.: 2×(-13)5  = -265  

L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 25:

Solve each of the following equations and verify the answer in each case:
3×10−4=143×10-4=14

ANSWER:

3×10 − 4 = 14 3×10 – 4 = 14 
or, 3×10− 4  + 4= 14 + 43×10- 4  + 4= 14 + 4                        [Adding 4 on both the sides]
or, 3×10 = 183×10 = 18
or, 3×10×10 = 18×103×10×10 = 18×10                              [Multiplying both the sides by 10]
or, 3x = 1803x = 180              
or, 3×3 = 18033×3 = 1803                  [Dividing both the sides by 3]
or, x = 60
Verification:
Substituting x = 60 on both the sides:
3×6010 − 4 =18010 − 4 = 18 − 4 = 14 = R.H.S.3×6010 – 4 =18010 – 4 = 18 – 4 = 14 = R.H.S.

L.H.S. = R.H.S.
Hence, verified.

Page No 143:

Question 26:

Solve each of the following equations and verify the answer in each case:
34 (x − 1) = x − 334 (x – 1) = x – 3

ANSWER:

34(x−1) = x − 334x-1 = x – 3
 ⇒34×x  − 34 × 1= x − 3 ⇒34×x  – 34 × 1= x – 3                   [On expanding the brackets]
⇒3×4− 34  = x − 3⇒3×4- 34  = x – 3                      
⇒3×4− x = −3 + 34⇒3×4- x = -3 + 34                           [Transposing x to the L.H.S. and −34-34 to the R.H.S.]
⇒3x−4×4=  −12+34⇒3x-4×4=  -12+34
⇒−x4 =  −94⇒-x4 =  -94
⇒−x4×(−4) =  −94×(−4)⇒-x4×-4 =  -94×-4                              [Multiplying both the sides by -4]
or, x = 9             

Verification:
Substituting x = 9 on both the sides:
L.H.S. : 34(9−1) = 34(8) = 6  R.H.S.: 9 − 3 = 6L.H.S. : 349-1 = 34(8) = 6  R.H.S.: 9 – 3 = 6

L.H.S. = R.H.S.
Hence, verified.

Page No 144:

Exercise 9C

Question 1:

If 9 is added to a certain number, the result is 36. Find the number.

ANSWER:

Let the required number be x.
According to the question:
9 + x = 36
or, x + 9 −- 9 = 36 −- 9                        [Subtracting 9 from both the sides]
or, x = 27
Thus, the required number is 27.

Page No 144:

Question 2:

If 11 is subtracted from 4 times a number, the result is 89. Find the number.

ANSWER:

Let the required number be x.
According to the question:
4x −-11 = 89
or, 4x −- 11 +11 = 89 + 11                        [Adding 11 on both the sides]
or, 4x = 100
or, 4×4 = 10044×4 = 1004                                       [Dividing both the sides by 4]
or, x = 25
Thus, the required number is 25.

Page No 144:

Question 3:

Find a number which when multiplied by 5 is increased by 80.

ANSWER:

Let the required number be x.
According to the question:
or, 5x = x + 80
or, 5x −- x = 80                       [Transposing x to the L.H.S.]
or, 4x = 80
or, 4×4 = 8044×4 = 804                                        [Dividing both the sides by 4]
or, x = 20
Thus, the required number is 20.

Page No 144:

Question 4:

The sum of three consecutive natural numbers is 114. Find the numbers.

ANSWER:

Let the three consecutive natural numbers be x, (x+1), (x+2).
According to the question:
x + (x + 1) + (x + 2) = 114
or, x + x + 1 + x + 2 = 114
or, 3x + 3 = 114
or, 3x + 3 −- 3 = 114 −- 3                     [Subtracting 3 from both the sides]
or, 3x = 111
or, 3×3 = 11133×3 = 1113                                   [Dividing both the sides by 3]
or, x = 37
Required numbers are:
x = 37
or, x + 1 = 37 + 1 = 38
or ,x + 2 = 37 + 2 = 39
Thus, the required numbers are 37, 38 and 39.

Page No 144:

Question 5:

When Raju multiplies a certain number by 17 and adds 4 to the product, he gets 225. Find that number.

ANSWER:

Let the required number be x.
When Raju multiplies it with 17, the number becomes 17x.
According to the question :
17x + 4 = 225
or, 17x + 4 −- 4 = 225 −- 4                          [Subtracting 4 from both the sides]
or, 17x = 221
or, 17×17 = 2211717×17 = 22117                                        [Dividing both the sides by 17]
or, x = 13
Thus, the required number is 13.

Page No 144:

Question 6:

If a number is tripled and the result is increased by 5, we get 50. Find the number.

ANSWER:

Let the required number be x.
According to the question, the number is tripled and 5 is added to it
∴ 3x + 5
or, 3x  + 5 = 50
or, 3x + 5 −- 5 = 50 −- 5                         [Subtracting 5 from both the sides]
or, 3x = 45
or, 3×3 =4533×3 =453                              [Dividing both the sides by 3]
or, x = 15
Thus, the required number is 15.

Page No 144:

Question 7:

Find two numbers such that one of them exceeds the other by 18 and their sum is 92.

ANSWER:

Let one of the number be x.
∴ The other number = (x + 18)
According to the question:
x + (x + 18) = 92
or, 2x + 18 −- 18 = 92 −- 18                         [Subtracting 18 from both the sides]
or, 2x =74
or, 2×2 = 7422×2 = 742                                           [Dividing both the sides by 2]
or, x = 37
Required numbers are:
x = 37
or, x + 18 = 37 + 18 = 55

Page No 144:

Question 8:

One out of two numbers is thrice the other. If their sum is 124, find the numbers.

ANSWER:

Let one of the number be ‘x’
∴ Second number = 3x
According to the question:
x + 3x = 124
or, 4x  = 124                         
or, 4×4 = 12444×4 = 1244                      [Dividing both the sides by 4]
or, x = 31
Thus, the required number is x = 31 and 3x = 3××31 = 93.

Page No 144:

Question 9:

Find two numbers such that one of them is five times the other and their difference is 132.

ANSWER:

Let one of the number be x.
∴ Second number = 5x
According to the question:
5x −- x = 132
or, 4x = 132
or, 4×4 = 13244×4 = 1324                        [Dividing both the sides by 4]
or, x = 33
Thus, the required numbers are x = 33 and 5x = 5××33 = 165.

Page No 144:

Question 10:

The sum of two consecutive even numbers is 74. Find the numbers.

ANSWER:

Let one of the even number be x.
Then, the other consecutive even number is (x + 2).
According to the question:
x + (x + 2)  = 74
or, 2x + 2  = 74
or, 2x + 2 −- 2 = 74 −- 2           [Subtracting 2 from both the sides]
or, 2x = 72
or, 2×2 = 7222×2 = 722                         [Dividing both the sides by 2]
or, x = 36
Thus, the required numbers are x = 36 and x+ 2 = 38.

Page No 144:

Question 11:

The sum of three consecutive odd numbers is 21. Find the numbers.

ANSWER:

Let the first odd number be x.
Then, the next consecutive odd numbers will be (x + 2) and (x + 4).
According to the question:
x + (x + 2) + (x + 4)  = 21
or, 3x + 6  = 21
or, 3x + 6 −- 6 = 21 −- 6           [Subtracting 6 from both the sides]
or, 3x = 15
or, 3×3 = 1533×3 = 153                         [Dividing both the sides by 3]
or, x = 5
∴ Required numbers are:
x = 5
x + 2 = 5 + 2 = 7
x + 4 = 5 + 4 = 9

Page No 144:

Question 12:

Reena is 6 years older than her brother Ajay. If the sum of their ages is 28 years, what are their present ages?

ANSWER:

Let the present age of Ajay be x years.
Since Reena is 6 years older than Ajay, the present age of Reena will be (x+ 6) years.
According to the question:
x + (x + 6) = 28
or, 2x + 6 = 28
or, 2x + 6 −- 6 = 28 −- 6                 [Subtracting 6 from both the sides]
or, 2x = 22
or, 2×2 = 2222×2 = 222                              [Dividing both the sides by 2]
or, x = 11
∴ Present age of Ajay = 11 years
Present age of Reena  = x +6 = 11 + 6
                                 = 17 years

Page No 144:

Question 13:

Deepak is twice as old as his brother Vikas. If the difference of their ages be 11 years, find their present ages.

ANSWER:

Let the present age of Vikas be x years.
Since Deepak is twice as old as Vikas, the present age of Deepak will be 2x years.
According to the question:
2x −- x = 11
x = 11
∴ Present age of Vikas = 11 years
Present age of Deepak  = 2x = 2××11
                                   = 22 years

Page No 144:

Question 14:

Mrs Goel is 27 years older than her daughter Rekha. After 8 years she will be twice as old as Rekha. Find their present ages.

ANSWER:

Let the present age of Rekha be x years.
As Mrs. Goel is 27 years older than Rekha, the present age of Mrs. Goel will be (x + 27) years.
After 8 years:
Rekha’s age = (x + 8) years
Mrs. Goel’s age = (x + 27 + 8)
                      = (x + 35) years

According to the question:
(x + 35) = 2(x + 8)
or, x + 35 = 2××x + 2××8                [On expanding the brackets]
or, x + 35 = 2x + 16
or, 35 −- 16 = 2x −- x                [Transposing 16 to the L.H.S. and x to the R.H.S.]
or, x = 19
∴ Present age of Rekha = 19 years
Present age of Mrs. Goel = x + 27
                                     = 19 + 27
                                     = 46 years

Page No 145:

Question 15:

A man is 4 times as old as his son. After 16 years he will be only twice as old as his son. Find their present ages.

ANSWER:

Let the present age of the son be x years.
As the man is 4 times as old as his son, the present age of the man will be (4x) years.
After 16 years:
Son’s age  = (x + 16) years
Man’s age = (4x + 16) years

According to the question:
(4x + 16) = 2(x + 16)
or, 4x + 16 = 2××x + 2××16                [On expanding the brackets]
or, 4x + 16 = 2x + 32
or, 4x −- 2x = 32 −- 16                [Transposing 16 to the R.H.S. and 2x to the L.H.S.]
or, 2x = 16
or, 2×2 = 1622×2 = 162                           [Dividing both the sides by 2]
or, x = 8
∴ Present age of the son = 8 years
Present age of the man  = 4x = 4××8
                                    = 32 years

Page No 145:

Question 16:

A man is thrice as old as his son. Five years ago the man was four times as old as his son. Find their present ages.

ANSWER:

Let the present age of the son be x years.
As the man is 3 times as old as his son, the present age of the man will be (3x) years.

5 years ago:
Son’s age = (x −- 5) years
Man’s age = (3x −- 5) years

According to the question:
(3x −- 5) = 4(x −- 5)
or, 3x −- 5 = 4××x −- 4××5                [On expanding the brackets]
or, 3x −- 5 = 4x −- 20
or, 20 −- 5 = 4x −- 3x                [Transposing 3x to the R.H.S. and 20 to the L.H.S.]
or, x = 15
∴ Present age of the son = 15 years
Present age of the man  = 3x = 3××15
                                    = 45 years

Page No 145:

Question 17:

After 16 years, Fatima will be three times as old as she is now. Find her present age.

ANSWER:

Let the present age of Fatima be x years.

After 16 years:
Fatima’s age = (x + 16) years

According to the question:
x + 16 = 3(x)
or, 16 = 3x −- x               [Transposing x to the R.H.S.]
or, 16 = 2x
or, 2×2 = 1622×2 = 162                [Dividing both the sides by 2]
or, x = 8
∴ Present age of Fatima = 8 years

Page No 145:

Question 18:

After 32 years, Rahim will be 5 times as old as he was 8 years ago. How old is Rahim today?

ANSWER:

Let the present age of Rahim be x years.
After 32 years:
Rahim’s age = (x + 32) years
8 years ago:
Rahim’s age = (x −- 8) years
According to the question:
x + 32 = 5(x −- 8)
or, x + 32  = 5x −- 5××8              
or, x + 32 = 5x −- 40
or, 40 + 32 = 5x −- x                      [Transposing ‘x’ to the R.H.S. and 40 to the L.H.S.]
or, 72 = 4x
or, 4×4 = 7244×4 = 724                               [Dividing both the sides by 4]
or, x = 18
Thus, the present age of Rahim is 18 years.

Page No 145:

Question 19:

A bag contains 25-paisa and 50-paisa coins whose total value is Rs 30. If the number of 25-paisa coins is four times that of 50-paisa coins, find the number of each type of coins.

ANSWER:

Let the number of 50 paisa coins be x.
Then, the number of 25 paisa coins will be 4x.
According to the question:
0.50(x) + 0.25(4x) = 30
or, 0.5x + x = 30
or, 1.5x = 30
or, 1.5×1.5 = 301.51.5×1.5 = 301.5          [Dividing both the sides by 1.5]
or, x = 20
Thus, the number of 50 paisa coins is 20.
Number of 25 paisa coins = 4x = 4××20 = 80

Page No 145:

Question 20:

Five times the price of a pen is Rs 17 more than three times its price. Find the price of the pen.

ANSWER:

Let the price of one pen be Rs x.
According to the question:
5x = 3x + 17
or, 5x −- 3x = 17                    [Transposing 3x to the L.H.S.]
or, 2x = 17
or, 2×2 = 1722×2 = 172                        [Dividing both the sides by 2]
or, x = 8.50
∴ Price of one pen = Rs 8.50

Page No 145:

Question 21:

The number of boys in a school is 334 more than the number of girls. If the total strength of the school is 572, find the number of girls in the school.

ANSWER:

Let the number of girls in the school be x.
Then, the number of boys in the school will be (x + 334).
Total strength of the school = 572

∴ x + (x + 334) = 572
or, 2x + 334 = 572
or, 2x + 334 −- 334 = 572 −- 334                {Subtracting 334 from both the sides]
or, 2x = 238
or, 2×2 = 23822×2 = 2382                                          [Dividing both the sides by 2]
or, x = 119
∴ Number of girls in the school = 119

Page No 145:

Question 22:

The length of a rectangular park is thrice its breadth. If the perimeter of the park is 168 metres, fund its dimensions.

ANSWER:

Let the breadth of the park be x metres.
Then, the length of the park will be 3x metres.
Perimeter of the park = 2 (Length + Breadth) = 2 ( 3x + x ) m
Given perimeter = 168 m

∴ 2(3x + x) = 168                           
or, 2 ( 4x ) = 168
or, 8x = 168                                      [On expanding the brackets]
or, 8×8 = 16888×8 = 1688                                [Dividing both the sides by 8]
or, x = 21 m
∴ Breadth of the park = x = 21 m
Length of the park = 3x = 3××21 = 63 m

Page No 145:

Question 23:

The length of a rectangular hall is 5 metres more than its breadth. If the perimeter of the hall is 74 metres, find its length and breadth.

ANSWER:

Let the breadth of the hall be x metres.
Then, the length of the hall will be (x + 5) metres.
Perimeter of the hall = 2(Length + Breadth) = 2( x + 5 + x) metres
Given perimeter of the rectangular hall = 74 metres

∴ 2( x + 5 + x) = 74                             
or, 2 ( 2x + 5 ) = 74
or, 2 ×2x + 2 ×5 = 74                              [On expanding the brackets]
or, 4x + 10 = 74
or, 4x + 10 −- 10 = 74 −- 10                     [Subtracting 10 from both the sides]
or, 4x = 64
or, 4×4 = 6444×4 = 644                                        [Dividing both the sides by 4]
or, x = 16 metres
∴ Breadth of the park = x
                                  = 16 metres
Length of the park = x + 5 = 16 + 5
                            = 21 metres

Page No 145:

Question 24:

A wire of length 86 cm is bent in the form of a rectangle such that its length is 7 cm more than its breadth. Find the length and the breadth of the rectangle so formed.

ANSWER:

Let the breadth of the rectangle be x cm.
Then, the length of the rectangle will be (x + 7) cm.
Perimeter of the rectangle = 2(Length + Breadth) = 2( x + 7 + x) cm
Given perimeter of the rectangle = Length of the wire = 86 cm

∴ 2( x + 7 + x) = 86                            
or, 2 ( 2x + 7 ) = 86
or, 2 ×2x + 2 × 7 = 86                              [On expanding the brackets]
or, 4x + 14 = 86
or, 4x + 14 – 14 = 86 – 14                   [Subtracting 14 from both the sides]
or, 4x = 72
or, 4×4 = 7244×4 = 724                                   [Dividing by 4 on both the sides]
or, x = 18 metres
Breadth of the hall = x
                             = 18 metres
Length of the hall = x + 7
                           = 18 + 7
                           = 25 metres

Page No 146:

Exercise 9D

Question 1:

A man earns Rs 25 per hour. How much does he earn in x hours?

ANSWER:

Earning of the man per hour = Rs 25

Earning of the man in x hours = Rs (25××x)
                                             = Rs 25x

Page No 146:

Question 2:

The cost of 1 pen is Rs 16 and the cost of 1 pencil is Rs 5. What is the total cost of x pens and y pencils.

ANSWER:

Cost of 1 pen = Rs 16
∴ Cost of ‘x‘ pens = Rs 16 ×× x
                           = Rs 16x
Similarly, cost of 1 pencil = Rs 5
∴ Cost of ‘y’ pencils = Rs 5××y
                               = Rs 5y
∴ Total cost of x pens and y pencils = Rs (16x + 5y)

Page No 146:

Question 3:

Lalit earns Rs x per day and spends Rs y per day. How much does he save in 30 days?

ANSWER:

Lalit’s earning per day = Rs x
∴ Lalit’s earning in 30 days = Rs 30 ××x
                                          = Rs 30x

Similarly, Lalit’s expenditure per day = Rs y
∴ Lalit’s expenditure in 30 days = Rs 30 ×× y
                                                = Rs 30y
∴ In 30 days, Lalit saves = (Total earnings −- Total expenditure)
                                      = Rs (30x −- 30y)
                                      = Rs 30(x – y)

Page No 146:

Question 4:

Three times a number added to 8 gives 20. Find the number.

ANSWER:

Let the required number be x.
Three times this number is 3x.
On adding 8, the number becomes 3x + 8.
3x + 8 = 20
or, 3x + 8 −- 8 = 20 −- 8                [Subtracting 8 from both the sides]
or, 3x = 12
or, 3×3 = 1233×3 = 123                             [Dividing both the sides by 3]
or, x = 4
∴ Required number = 4

Page No 146:

Question 5:

If x = 1, y = 2 and z = 3, find the value of x² + y² + 2xyz.

ANSWER:

Given:
x =1
y = 2
z = 3

Substituting x = 1, y = 2 and z = 3 in the given equation (x² + y² + 2xyz):

(1)2 +( 2)2  + 2(1)(2)(3)⇒1 + 4 + 12 = 1712 + 22  + 2123⇒1 + 4 + 12 = 17

Page No 146:

Question 6:

Solve: 4x + 9 = 17.

ANSWER:

4x + 9 = 17
or, 4x + 9 −- 9 = 17 −- 9                          [Subtracting 9 from both the sides]
or, 4x = 8
or, 4×4 = 844×4 = 84                                          [Dividing both the sides with 4]
or, x = 2

Page No 146:

Question 7:

Solve: 3(x + 2) − 2(x − 1) = 7.

ANSWER:

3(x + 2) − 2(x − 1) = 7.
or, 3 × x + 3 × 2 − 2 × x − 2 × (−1) = 73 × x + 3 × 2 – 2 × x – 2 × (-1) = 7                    [On expanding the brackets]
or, 3x + 6 − 2x + 2 = 7
or, x + 8 = 7
or, x + 8 − 8 = 7 − 8                                            [Subtracting 8 from both the sides]
or, x = −1

Page No 146:

Question 8:

Solve: 2×5−x2=522×5-x2=52.

ANSWER:

2×5 −x2 = 522×5 -x2 = 52
or, 4x − 5×10 = 524x – 5×10 = 52                               [Taking the L.C.M. as 10]
or, −x10 = 52-x10 = 52
or, −x10×(−10) = 52×(−10)-x10×-10 = 52×-10                          [Multiplying both the sides by (−10)]
or, x = −25

Page No 146:

Question 9:

The sum of three consecutive natural numbers is 51. Find the numbers.

ANSWER:

Let the three consecutive natural numbers be x, (x + 1) and (x + 2).

∴ x + (x + 1) + (x + 2) = 51
3x + 3 = 51
3x + 3 −- 3 = 51 −- 3                   [Subtracting 3 from both the sides]
3x = 48
3×3 = 4833×3 = 483                                [Dividing both the sides by 3]
x = 16
Thus, the three natural numbers are x = 16, x+1 = 17 and x+2 = 18.

Page No 146:

Question 10:

After 16 years, Seema will be three times as old as she is now. Find her present age.

ANSWER:

Let the present age of Seema be x years.
After 16 years:
Seema’s age = x + 16

After 16 years, her age becomes thrice of her age now
∴ x + 16 = 3x
or, 16 = 3x −- x                     [Transposing x to the R.H.S.]
or, 2x = 16
or, 2×2 = 1622×2 = 162                      [Dividing both the sides by 2]
or, x = 8 years

Page No 146:

Question 11:

By how much does I exceed 2x − 3y − 4?
(a) 2x − 3y − 5
(b) 2x − 3y − 3
(c) 5 − 2x + 3y
(d) none of these

ANSWER:

(c) 5 − 2x + 3y

1 exceeds 2x − 3y − 4.

∴1 −- (2x − 3y − 4) = 1 −-2x + 3y + 4
                              = 5 −- 2x + 3y

∴ 1 exceeds 2x − 3y − 4 by 5 −- 2x + 3y.

Page No 146:

Question 12:

What must be added to 5x³ − 2x² + 6x + 7 to make the sum x³ + 3x² − x + 1?
(a) 4x³ − 5x² + 7x + 6
(b) −4x³ + 5x² − 7x − 6
(c) 4x³ + 5x² − 7x + 6
(d) none of these

ANSWER:

  (b) −4x³ + 5x² − 7x − 6
In order to find what must be added, we subtract (5x³ − 2x² + 6x + 7) from (x³ + 3x² − x + 1).

    (x³ + 3x² − x + 1)  − ( 5x³ − 2x² + 6x + 7)
or,  x³ + 3x² − x + 1 − 5x³  + 2x² − 6x − 7
or,  x³ − 5x³+ 3x²+ 2x²− x − 6x+ 1 − 7
or, −4x³ + 5x² − 7x − 6

Page No 146:

Question 13:

2x − [3y − {2x − (y − x)}] = ?
(a) 5x − 4y
(b) 4y − 5x
(c) 5y − 4x
(d) 4x − 5y

ANSWER:

(a) 5x − 4y

2x − [3y − {2x − (y − x)}]
= 2x − [3y − {2x −  y + x}]
= 2x − [3y − {3x −  y}]
= 2x −  [3y −  3x + y]
= 2x −  [4y −  3x]
= 2x −  4y + 3x
= 5x −  4y

Page No 146:

Question 14:

The coefficient of x in −5xyz is
(a) −5
(b) 5yz
(c) −5yz
(d) yz

ANSWER:

(c) −5yz

All the terms in the expression −5xyz barring x will be the coefficient of x, i.e. −5yz.

Page No 146:

Question 15:

13(x + 7 + z)13(x + 7 + z) is a
(a) monomial
(b) binomial
(c) trinomial
(d) quadrinomial

ANSWER:

(b) trinomial
Since it contains three variables, i.e. ‘x’, ‘y’ and ‘z’, it is a trinomial.

Page No 146:

Question 16:

If x5=1×5=1, then
(a) x=15x=15
(b) x = 5
(c) x = (5 + 1)
(d) none of these

ANSWER:

(b) x = 5

x5 = 1or, x5 × 5= 1 × 5   [Multiplying both the sides by 5]x5 = 1or, x5 × 5= 1 × 5   [Multiplying both the sides by 5]
or, x = 5

Page No 146:

Question 17:

If x = 1, y = 2 and z = 3 then (x² + y² + z²) = ?
(a) 6
(b) 12
(c) 14
(d) 15

ANSWER:

(c) 14
Substituting x = 1, y = 2 and z = 3 in (x² + y² + z²):
(1)2  + (2)2 + (3)212  + 22 + 32
or, 1 + 4 + 9 = 14

Page No 146:

Question 18:

If 13 x + 5 = 813 x + 5 = 8, then x = ?
(a) 3
(b) 6
(c) 9
(d) 12

ANSWER:

(c) 9

13x + 5 = 813x + 5 = 8
or, 13x + 5 − 5 = 8−513x + 5 – 5 = 8-5                                  [Subtracting 5 from both the sides]
or, 13x = 313x = 3
or, 13x × 3 = 3×313x × 3 = 3×3                                        [Multiplying both the sides by 3]
or, x = 9

Page No 146:

Question 19:

Fill in the blanks.
(i) An expression having one term is called a …… .
(ii) An expression having two term is called a …… .
(i) An expression having three term is called a …… .
(iv) 3x − 5 = 7 − x ⇒ x = …… .3x – 5 = 7 – x ⇒ x = …… .
(v) (b2 − a2) − (a2 − b2) = …… .(b2 – a2) – (a2 – b2) = …… .

ANSWER:

(i) monomial
(ii) binomial
(iii) trinomial
(iv) x = 3
3x −- 5 = 7 −- x
or, 3x + x = 7 + 5                   [Transposing x to the L.H.S. and 5 to the R.H.S.]
or, 4x = 12
or, 4×4 = 1244×4 = 124
or, x = 3
(v) 2b² −- 2a²
or, b² −- a² −- a² + b²
or, 2b² −- 2a²
( b2 − a2) − (a2 − b2)
(b2 − a2) − (a2 − b2
(b2 − a2) − (a2 − b2)
(b2 − a2) − (a2 − b2)

Page No 147:

Question 20:

Write ‘T’ for true and ‘F’ for false for each of the statements given below:
(i) −3xy²z is a monomial.
(ii) x=23x=23 is solution of 2x + 5 = 8.
(iii) 2x + 3 = 5 is a linear equation.
(iv) The coefficient of x in 5xy is 5.
(v) 8 − x = 5 ⇒ x = 38 – x = 5 ⇒ x = 3.

ANSWER:

(i) True
Since it has one term, it is a monomial.

(ii) False
2x + 5 = 8
or, 2x + 5 −- 5 = 8 −- 5               [Subtracting 5 from both the sides]
or, 2x = 3
or, x = 3/2 and not x = 2/3

(iii) True
This is because the maximum power of the variable x is 1.

(iv) False
The coefficient of x in 5xy would be 5y.

(v) True
    8 − x = 5
  or, 8 − 5 = x
    or, 3 = x

Exercise 10A

Page No 152:

Question 1:

Find each of the following ratios in the simplest form:
(i) 24 to 56
(ii) 84 paise to Rs 3
(iii) 4 kg to 750 g
(iv) 1.8 kg to 6 kg
(v) 48 minutes to 1 hour
(vi) 2.4 km to 900 m

ANSWER:

(i) 24:56 =     24  =     24 ​÷ 8     =    3     
                       56       56 ​÷ 8            7
As the H.C.F. of 3 and 7 is 1, the simplest form of 24:56 is 3:7.

(ii) 84 paise to Rs 3 = Rs 0.84 to R. 3 =  0.84  =  0.84​ ÷ 3    =  0.28   =   28    =  28 ​÷  4   =   7  
                                                                   3            3 ​÷ 3             1          100      100 ​÷ 4        25
As the H.C.F. of 7 and 25 is 1, the simplest form of  0.84:3 is 7:25.

(iii) 4 kg:750 g = 4000 g:750 g =   4000 ​÷ 250   =    16   
                                                            750 ​÷ 250            3
As the H.C.F. of 16 and 3 is 1, the simplest form of 4000:750 is 16:3.

(iv) 1.8 kg:6 kg  =   1.8   =   18   =  18 ​÷ 6  =     3  
                               6           60      60 ​÷ 6        10 
As the H.C.F. of 3 and 10 is 1, the simplest form of 1.8:6 is 3:1.

(v) 48 minutes to 1 hour = 48 minutes to 60 minutes = 48:60 =  48 ​÷ 12   =  4   
                                                                                                 60 ​÷ 12        5
As the H.C.F. of 4 and 5 is 1, the simplest form of 48:60 is 4:5.
 
(vi) 2.4 km to 900 m = 2400m:900m =    2400   =   24    =   24 ​÷ 3   =    8  
                                                               900          9           9 ​÷ 3          3
 As the H.C.F. of 8 and 3 is 1, the simplest form of 2400:900 is 8:3.

                    

Page No 152:

Question 2:

Express each of the following ratios in the simplest form:
(i) 36 : 90
(ii) 324 : 144
(iii) 85 : 561
(iv) 480 : 384
(v) 186 : 403
(vi) 777 : 1147

ANSWER:

(i) 36:90 =   36    =    36 ​÷ 18    =     2          (As the H.C.F. of 36 and 90 is 18.)
                   90          90 ​÷ 18            5
Since the H.C.F. of 2 and 5 is 1, the simplest form of 36:90 is 2:5.

(ii) 324:144 =   324   =    324 ​÷ 36    =    9       (As the H.C.F. of  324 and 144 is 36.)
                          144          144 ​÷ 36           4
 Since the H.C.F. of 9 and 4 is 1, the simplest form of 324:144 is 9:4.

(iii) 85:561 =  85   =   85 ​÷ 17   =      5         (As the H.C.F. of 85 and 561 is 17.)
                        561      561 ​​÷ 17           33
Since the H.C.F. of 5 and 33 is 1, the simplest form of 85:561 is 5:33.

(iv) 480:384 =    480    =    480 ​÷ 96    =   5       (As the H.C.F. of 480 and 384 is 96.)
                            384          384 ​​÷ 96          4
Since the H.C.F. of 5 and 4 is 1, the simplest form of 480:384 is 5:4.

(v) 186:403 =     186   =   186 ÷ 31    =   6        (As the H.C.F. of 186 and 403 is 31.)
                             403        403 ÷ 31        13
Since the H.C.F. of 6 and 13 is 1, the simplest form of 186:403 is 6:13.

(vi) 777:1147 =  777  ​÷ 37      =   21           (As the H.C.F. of 777 and 1147 is 37.)
                            1147  ​÷  37         31
Since the H.C.F. of 21 and 31 is 1, the simplest form of 777:1147 is 21:31.

Page No 152:

Question 3:

Write each of the following ratios in the simplest form:
(i) Rs 6.30 : Rs 16.80
(ii) 3 weeks : 30 days
(iii) 3 m 5 cm : 35 cm
(iv) 48 min : 2 hours 40 min
(v) 1 L 35 mL : 270 mL
(vi) 4 kg : 2 kg 500 g

ANSWER:

(i) Rs 6.30:Rs 16.80
             6.30     =    63    =   63 ​÷ 21     =   3        (H.C.F. of 63 and 168 is 21.)
            16.80         168        168  ​÷ 21        8
    Ratio = 3 : 8
(ii)3 weeks:30 days = 21days:30 days          (1 week = 7 days)
          21     =   21 ​÷ 3   =   7          (H.C.F. of 21 and 30 is 3.)
          30          30 ​ ​÷ 3       10
     Ratio = 7 : 10
 (iii) 3 m 5 cm:35 cm = 305 cm:35 cm      (1 m = 100 cm)
            305  =   305  ​÷ 5   =   61        (H.C.F. of 305 and 35 is 5.)
             35         35  ​÷ 5          7
      Ratio = 61:7
(iv) 48 min:2 hours 40 min = 48 min:160 min        (1 hour = 60 mins)
             48   =    48  ​÷ 16    =    3         (H.C.F. of 48 and 160 is 16.)
            160        160  ​÷ 16       10
      Ratio = 3:10
(v) 1 L 35 mL:270 mL = 1035 mL:270 mL         (1 L = 1000 mL)
             1035   =   1035  ​÷ 45    =   23        (H.C.F. of 1035 and 270 is 45.)
              270           270  ​÷ 45          6
       Ratio = 23:6
(vi) 4 kg:2 kg 500 g = 4000 g:2500 g        (1 kg= 1000 g)
            4000   =   40  =  40  ​÷ 5   =   8     (H.C.F. of 40 and 25 is 5.)
            2500         25      25  ​÷ 5        5
      Ratio = 8:5

Page No 152:

Question 4:

Mr Sahai and his wife are both school teachers and earn Rs 16800 and Rs 10500 per month respectively. Find the ratio of
(i) Mr Sahai’s income to his wife’s income;
(ii) Mrs Sahai’s income to her husband’s income;
(iii) Mr Sahai’s income to the total income of the two.

ANSWER:

Mr Sahai’s earning = Rs 16800
Mrs Sahai’s earning = Rs 10500
(i) Ratio = 16800:10500 = 168:105 =  168  ​÷ 21  =   8          (H.C.F. of 168 and 105 is 21.)
                                                           105 ​ ​÷ 21           5
Mr Sahai’s income:Mrs Sahai’s income = 8:5
(ii)Ratio = 10500:16800 = 105:168 =  105  ​÷ 21   =    5        (H.C.F. of 168 and 105 is 21.)
                                                         168 ​ ​÷ 21          8
Mrs Sahai’s income:Mr Sahai’s income = 5:8

(iii) Total income = 16800 + 10500 = Rs  27300
   Ratio = 16800:27300 = 168:273 =  168   =  168  ​÷ 21   =  8  (H.C.F. of 168 and 273 is 21.)
                                                        273        273  ​÷ 21     13
Mrs Sahai’s income:Total income = 8:13

Page No 152:

Question 5:

Rohit earns Rs 15300 and saves Rs 1224 per month. Find the ratio of
(i) his income and savings;
(ii) his income and expenditure;
(iii) his expenditure and savings.

ANSWER:

Rohit’s income = Rs 15300
Rohit’s savings = Rs 1224
(i) Income:Savings = 15300:1224 = 15300 ​÷ 612   =  25        (H.C.F. of 15300 and 1224 is 612.)
                                                                  1224  ​÷ 612         2 
    Income:Savings = 25:2
(ii) Monthly expenditure = Rs (15300 −- 1224) = Rs 14076
    Income:Expenditure = 15300:14076 =  15300  ÷ 612  =   25       (H.C.F. of 15300 and 14076 is 612.)
                                                                       14076  ​÷ 612        23
     Income:Expenditure = 25:23
(iii) Expenditure : Savings = 14076:1224 =  14076  ÷ 612  =  23       (H.C.F. of 14076 and 1224 is 612.)
                                                                           1224  ​÷ 612         2
      Expenditure:Savings = 23:2
                    

Page No 152:

Question 6:

The ratio of the number of male and female workers in a textile mill is 5 : 3. If there are 115 male workers, what is the numkber of female workers in the mill?

ANSWER:

Number of male:Number of female = 5:3
Let the number be x.
Number of male = 5x
​Number of female = 3x
Number of male workers = 115
   Now, 5x = 115
         ⇒  x =   115   = 23
                       5
Number of female workers in the mill = 3x = 3 × 23 = 69

Page No 152:

Question 7:

The bosys and the girls in a school are in the ratio 9 : 5. If the total strength of the school is 448, find the number of girls.

ANSWER:

Boys:Girls = 9:5
Let the number of boys = 9x 
Let the number of girls = 5x
Total strength of the school = 448
According to given condition, we have:
                                                      9x + 5x = 448
                                              ⇒         14x = 448
                                               ⇒            x =   448   = 32
                                                                      14
Number of boys = 9x = 9 × 32 = 288
Number of girls = 5x = 5 ​× 32 = 160

Page No 152:

Question 8:

Divide Rs 1575 between Kamal and Madhu in the ratio 7 : 2.

ANSWER:

Kamal:Madhu = 7:2
Sum of the ratio terms = 7 + 2 = 9
Kamal’s share =   7   × 1575 =   11025   = Rs 1225
                            9                         9
Madhu’s share =   2   × 1575 =   3150   = Rs 350
                            9                       9

Page No 152:

Question 9:

Divide Rs 3450 among A, B and C in the ratio 3 : 5 : 7.

ANSWER:

A:B:C = 3:5:7
Sum of the ratio terms = 3 + 5 +7 = 15
A’s share =   3    × 3450 =   10350   = Rs 690
                   15                       15

B’s share =   5   × 3450 =   17250   = Rs 1150
                   15                        15

 C’s share =     7   × 3450 =   24150  = Rs 1610      
                      15                       15

Page No 152:

Question 10:

Two numbers are in the ratio 11 : 12 and their sum is 460. Find the numbers.

ANSWER:

Two number are in the ratio 11:12.
Let the numbers be 11x and 12x.
Given:     11x + 12x = 460
         ⇒          23x = 460
        ⇒               x =   460   =  20
                                    23
First number = 11x = 11 × 20 = 220
Second number = 12x = 12 × 20 = 240
Hence, the numbers are 220 and 240.

Page No 152:

Question 11:

A 35-cm line segment is divided into two parts in the ratio 4 : 3. Find the length of each part.

ANSWER:

Ratio of the two parts of line segment = 4:3
Sum of the ratio terms = 4 + 3 = 7
First part =   4   × 35 cm = 4 × 5 cm = 20 cm
                    7
Second part =    3   × 35 cm = 3 × 5 cm = 15 cm
                         7

Page No 152:

Question 12:

A factory produces electric bulbs. If 1 out of every 10 bulbs is defective and the factory produces 630 bulbs per day, find the number of defective bulbs produced each day.

ANSWER:

Number of bulbs produced each day = 630
Out of 10 bulbs, 1 is defective.
Number of defective bulbs =  630  = 63
                                             10

∴∴ Number of defective bulbs produced each day = 63

Page No 152:

Question 13:

Find the ratio of the price of a pencil to that of a ball pen if pencils cost Rs 96 per score and ball pens cost Rs 50.40 per dozen.

ANSWER:

Price of pencil = Rs 96 per score 
Price of ball pen = Rs 50.40 per dozen
Price per unit of pencil =  96  = 4.8
                                       20
Price per unit of ball pen =  50.40  =  4.2
                                            12
   Ratio =    4.8   =   48   =   48  ​÷ 6    =    8  
                   4.2        42        42  ​÷  6         7
Price of a pencil:Price of a ball pen = 8:7

Page No 152:

Question 14:

The ratio of the length of a field to its width is 5 : 3. Find its length if the width is 42 metres.

ANSWER:

Length:Width = 5:3
Let the length and the width of the field be 5x m and 3x m, respectively.
Width = 42 m
3x = 42
 x =   42   = 14
          3
∴∴ Length = 5x = 5 × 14 = 70 metres

Page No 152:

Question 15:

The ratio of income to savings of a family is 11 : 2. Find the expenditure if the savings is Rs 1520.

ANSWER:

Income:Savings = 11:2
Let the income and the saving be Rs 11x and Rs 2x, respectively.
Saving = Rs 1520
  2x = 1520
   x =   1520   = 760
              2
∴∴ Income = Rs 11x =Rs (11 × 760) = Rs 8360
   Expenditure = Income −- Saving
                      = Rs (8360 −- 1520 )
                      = Rs 6840

Page No 152:

Question 16:

The ratio of income to expenditure of a family is 7 : 6. Find the savings if the income is Rs 14000.

ANSWER:

Income:Expenditure = 7:6
Let the income and the expenditure be Rs 7x and Rs 6x, respectively.
Income = Rs 14000
7x = 14000
x =    14000  =  2000
             7
Expenditure = Rs 6x = Rs 6 × 2000 = Rs 12000
∴∴ Saving = Income −- Expenditure
          = Rs (14000 −- 12000)
         = Rs 2000

Page No 152:

Question 17:

The ratio of zinc and copper in an alloy is 7 : 9. If the weight of copper in the alloy is 11.7 kg find the weight of zinc in it.

ANSWER:

Let the weight of zinc be x kg.
Ratio of zinc and copper = 7:9
Weight of copper in the alloy = 11.7 kg
     7   =    x     
     9       11.7
⇒  x =  11.7 × 7   =  81.9   = 9.1
                9                9
Weight of zinc = 9.1 kg

Page No 152:

Question 18:

A bus covers 128 km in 2 hours and a train covers 240 km in 3 hours. Find the ratio of their speeds.

ANSWER:

A bus covers 128 km in 2 hours.
Speed of the bus =   Distance  =  128 km   = 64 km/ hr
                               Time               2 hr

A train covers 240 km in 3 hours.
Speed of the train =  Distance   =  240   = 80 km /hr
                                 Time             3

Ratio of their speeds = 64:80 =  64  =   64 ÷ 16   =   4  
                                                 80       80 ÷ 16        5
∴∴ Ratio of the speeds of the bus and the train = 4:5

Page No 153:

Question 19:

From each of the given pairs, find which ratio is larger:
(i) (3 : 4) or (9 : 16)
(ii) (5 : 12) or (17 : 30)
(iii) (3 : 7) or (4 : 9)
(iv) (1 : 2) or (13 : 27)

ANSWER:

(i) (3:4) or (9:16) 

Making the denominator equal:

        3 × 4   =  12  and 12   >   9 
        4 × 4       16         16       16

 ∴∴ (3:4) > (9:16)

(ii) (5:12) or (17:30) 

Making the denominator equal:

        5 × 5   =    25   and   17 × 2    =    34     
      12 × 5         60            30 × 2          60
⇒    25   <    34         
      60         60
 ∴∴ (5:12) < (17:30)

(iii) (3:7) or (4:9) 

Making the denominator equal:

          3 × 9  =   27   and   4 × 7   =  28  
          7 × 9       63            9 ​× 7       63
⇒       27  <   28      
         63       63

(3:7) < (4:9)

(iv) (1:2) or (13:27)

Making the denominator equal:

        1× 27   =   27   and   13 × 2   =   26  
        2 × 27       54            27 ​× 2        54

⇒   27  >  26          
     54       54

(1:2) > (13:27)

Page No 153:

Question 20:

Fill in the place holders:
(i) 2440=   5=12   2440=   5=12   
(ii) 3663=4   =   213663=4   =   21
(iii) 57=   28=35   57=   28=35   

ANSWER:

(i)   24   =   24 ​÷ 8  =   3   =    3 × 4  =  12      
      40         40 ​÷ 8      5          5  × 4      20

(ii)    36  =   36  ​÷ 9  =  4   =   4 × 3  =   12       
        63        63  ​÷ 9      7        7 × 3        21

(iii)   5   =   5 × 4   =  20  =   5 × 7   =   35  
        7        7 × 4       28        7 × 7         49

Page No 155:

Exercise 10B

Question 1:

Determine if the following numbers are in proportion:
(i) 4, 6, 8, 12
(ii) 7, 42, 13, 78
(iii) 33, 121, 9, 96
(iv) 22, 33, 42, 63
(v) 32, 48, 70, 210
(vi) 150, 200, 250, 300

ANSWER:

(i) 4, 6, 8, 12
    4  =    4 ​÷ 2   =   2 ;     8  =    8  ​÷ 4   =   2  
    6        6  ​÷ 2       3       12       12  ​÷ 4       3
Hence, 4:9::8:12 are in proportion.

(ii) 7, 42, 13, 78
      7    =  7  ​÷ 7   =   1 ;     13   =  13  ​÷ 13     =    1  
      42      42  ​÷ 7      6        78        78  ​÷ 13           6
Hence, 7:42::13:78 are in proportion.

(iii) 33, 121, 9, 96
      33   =   33  ​÷ 11   =   3   ;     9   =   9  ​÷ 3   =   3  
     121      121  ​÷ 11      11         96       96  ​÷ 3      32
 Hence, 33:121::9:96 are not in proportion.

(iv) 22, 33, 42, 63
     
2233=22÷1133÷11=23 and 4263=42÷2163÷21=232233=22÷1133÷11=23 and 4263=42÷2163÷21=23

Hence, 22:33 :: 42 : 63 are not in proportion.
 
(v) 32, 48, 70, 210
       32   =   32  ​÷ 6  =   7 ;    70   =   70  ​÷ 70   =  1 
       48        48  ​÷ 6       8     210      210  ​÷ 70      3
    Hence, 32:48::70:210 are not in proportion.


 (vi) 150, 200, 250, 300
       150   =   150  ​÷ 50   =  3;   250  =  250  ​÷ 50  =   5  
      200         200  ​÷ 50        4   300      300  ​÷ 50       6
    Hence, 150:200::250:300 are not in proportion.

Page No 155:

Question 2:

Verify the following:
(i) 60 : 105 : : 84 : 147
(ii) 91 : 104 : : 119 : 136
(iii) 108 : 72 : : 129 : 86
(iv) 39 : 65 : : 141 : 235

ANSWER:

(i) 60:105::84:147
      60   =   60  ​÷ 15   =   4         (H.C.F. of 60 and 105 is 15.)
     105       105  ​÷ 15       7
       84   =   84  ​÷ 21  =   4         (H.C.F. of 84 and 147 is 21.)
      147     147  ​÷ 21        7
    Hence, 60:105::84:147 are in proportion.
(ii) 91:104::119:136
      91   =  91  ​÷ 13   =   7         (H.C.F. of 91 and 104 is 13.)
     104       104  ​÷ 13      8  
      119  =   119  ​÷ 17  =   7      (H.C.F. of 11 and 136 is 17.)
     136        136  ​÷ 17       8
  Hence, 91:104::119:136 are in proportion.
(iii) 108:72::129:86
      108  =   108  ​÷ 36   =   3        (H.C.F. of 108 and 72 is 36.)
      72           72 ​ ​÷ 36        2
       129   =   ​129  ​÷  43  =  3       (H.C.F. of 129 and 86 is 43.)
        86          86   ​÷ 43        2
   Hence, 108:72::129:86 are in proportion.
(iv) 39:65::141:235
       39   =   39  ​÷ 13  =   3        (H.C.F. of 39 and 65 is 13.)
       65         65  ​÷ 13      5
      141   =   141  ​÷  47   =   3     (H.C.F. of 141 and 235 is 47.)
      235         235  ​÷ 47        5
   Hence, 39:65::141:235 are in proportion.

Page No 155:

Question 3:

Find the value of x in each of the following proportions:
(i) 55 : 11 : : x : 6
(ii) 27 : x : : 63 : 84
(iii) 51 : 85 : : 57 : x
(iv) x : 92 : : 87 : 116

ANSWER:

(i) 55:11::x:6
      Product of extremes = Product of means
                          55 × 6 = 11 × x
⇒                            11x = 330
⇒                               x =  330   = 30
                                          11
(ii) 27:x::63:84
      Product of extremes = Product of means
                      27 ​× 84 = x ​× 63
 ⇒                         63x = 2268
 ⇒                             x =  2268  = 36
                                           63
(iii) 51:85::57:x 
      Product of extremes = Product of means
                           51 × x = 85 × 57
 ⇒                           51x = 4845
 ⇒                               x =   4845   = 95
                                             51
(iv) x:92::87:116
     Product of extremes = Product of means
                       x ×  116 = 92 ​× 87
 ⇒                      116x = 8004
 ⇒                           x  =   8004  = 69
                                       116

Page No 155:

Question 4:

Write (T) for true and (F) for false in case of each of the following:
(i) 51 : 68 : : 85 : 102
(ii) 36 : 45 : : 80 : 100
(iii) 30 bags : 18 bags : : Rs 450 : Rs 270
(iv) 81 kg : 45 kg : : 18 men : 10 men
(v) 45 km : 60 km : : 12 h : 15 h
(vi) 32 kg : Rs 36 : : 8 kg : Rs 9

ANSWER:

(i) 51:68::85:102
   Product of means = 68 × 85 = 5780
   Product of extremes = 51 × 102 = 5202
   Product of means ≠ Product of extremes
    Hence, (F).
(ii) 36:45::80:100
  Product of means = 45 ​× 80 = 3600
  Product of extremes = 36 × 100 = 3600
  Product of means = Product of extremes 
   Hence, (T).
(iii) 30 bags:18 bags::Rs 450:Rs 270
       or 30:18::450:270
     Product of means = 18 × 450 = 8100
     Product of extremes = 30 ​× 270 = 8100
     Product of means = Product of extremes 
     Hence, (T).
(iv) 81 kg:45 kg::18 men:10 men
      or 81:45::18:10
     Product of means = 45 × 18 = 810
     Product of extremes = 81 × 10 = 810
     Product of means = Product of extremes
      Hence, (T).
(v) 45 km:60 km::12 h:15 h
     or,45:60::12:15
     Product of means = 60 × 12 = 720
     Product of extremes = 45 × 15 = 675
     Product of means ≠ Product of extremes 
      Hence, (F).
(vi) 32 kg:Rs 36::8 kg:Rs 9
     Product of means = 36 × 8 = 288
     Product of extremes = 32 × 9 = 288
     Product of means = Product of extremes
     Hence, (T).
    

Page No 155:

Question 5:

Determine if the following ratios form a proportion:
(i) 25 cm : 1 m and Rs 40 : Rs 160
(ii) 39 litres : 65 litres and 6 bottles : 10 bottles
(iii) 200 mL : 2.5 L and Rs 4 : Rs 50
(iv) 2 kg : 80 kg and 25 g : 625 kg

ANSWER:

(i) 25 cm:1 m and Rs 40:Rs 160 (or) 25 cm:100 cm and Rs 40:Rs 160
      25  =  25 ​÷ 25  =  1  and  40  =  40 ÷ 40  =  1
    100      100 ​​÷ 25     4         160    160 ​÷ 40      4 
       Hence, they are in proportion.

(ii) 39 litres:65 litres and 6 bottles:10 bottles
       39   =  39 ​÷ 13   =   3    and   6   =   6 ​÷ 2   =   3 
       65        65 ​​÷ 13       5            10      10 ​÷ 2        5
      Hence they are  in proportion.

(iii) 200 mL:2.5 L and Rs 4:Rs 50 (or) 200 mL:2500 mL and Rs 4:Rs 50
        200   =   2   and   4   =   4 ​÷ 2   =    2  
       2500      25           50       50 ÷ 2       25
     Hence, they are in proportion.

(iv) 2 kg:80 kg and 25 g:625 kg  (or)  2 kg:80 kg and 25 g:625000 g
        2   =  2 ​÷ 2    =   1    and   25      =   25 ​÷ 25     =     1   
       80      80 ​÷ 2       40         625000    625000 ​​÷ 25     25000
        Hence, they are not in proportion.

Page No 155:

Question 6:

In a proportion, the 1st, 2nd and 4th terms are 51, 68 and 108 respectively. Find the 3rd term.

ANSWER:

Let the 3rd term be x.
Thus, 51:68::x:108
 We know:
                      Product of extremes = Product of means
                               51 × 108 = 68 × x
             ⇒                     5508 = 68x
             ⇒                          x =  5508  = 81
                                                   68
Hence, the third term is 81.

Page No 155:

Question 7:

The 1st, 3rd and 4th terms of a proportion are 12, 8 and 14 respectively. Find the 2nd term.

ANSWER:

Let the second term be x.
Then. 12:x::8:14
We know:
  Product of extremes = Product of means
                                 12 × 14 = 8x
            ⇒                       168 = 8x
​            ⇒                           x =   168  = 21
                                                    8
 Hence, the second term is 21.

Page No 155:

Question 8:

Show that the following numbers are in continued proportion:
(i) 48, 60, 75
(ii) 36, 90, 225
(iii) 16, 84, 441

ANSWER:

(i) 48:60, 60:75
      Product of means = 60 × 60 = 3600
      Product of extremes = 48 × 75 = 3600
Product of means = Product of extremes
       Hence, 48:60::60:75 are in continued proportion.

(ii) 36:90, 90:225
     Product of means = 90 × 90 = 8100
     Product of extremes = 36 × 225 = 8100
Product of means = Product of extremes
      Hence, 36:90::90:225 are in continued proportion.

(iii) 16:84, 84:441
    Product of means = 84 × 84 = 7056
    Product of extremes = 16 × 441 = 7056
Product of means = Product of extremes
    Hence, 16:84::84:441 are in continued proportion.
                             

Page No 155:

Question 9:

If 9, x, x 49 are in proportion, find the value of x.

ANSWER:

Given: 9:x::x:49
We know:
  Product of means = Product of extremes
                                     x × x = 9 × 49
                       ⇒               x² = 441
                       ⇒               x² = (21)²
                       ⇒                x = 21

Page No 155:

Question 10:

An electric pole casts a shadow of length 20 m at a time when a tree 6 m high casts a shadow of length 8 m. Find the height of the pole.

ANSWER:

Let the height of the pole = x m
Then, we have:
      x:20::6:8
Now, we know:
        Product of extremes = Product of means
                                   8x = 20​ × 6
                                     x =  120  = 15
                                              8
​Hence, the height of the pole is 15 m.

Page No 155:

Question 11:

Find the value of x if 5 : 3 : : x : 6.

ANSWER:

5:3::x:6
We know:
   Product of means = Product of extremes
                               3x = 5 ​× 6
                            ⇒ x =  30  = 10
                                        3
∴∴ x = 10

Page No 157:

Exercise 10C

Question 1:

If the cost of 14 m of cloth is Rs 1890, find the cost of 6 m of cloth.

ANSWER:

Cost of 14 m of cloth = Rs 1890
Cost of 1 m of cloth =  1890  = Rs 135
                                     14
Cost of 6 m of cloth = 6​ × 135 = Rs 810

Page No 157:

Question 2:

If the cost of a dozen soaps is Rs 285.60, what wil be the cost of 15 such soaps?

ANSWER:

Cost of dozen soaps = Rs 285.60
Cost of 1 soap =  285.60 
                            12
Cost of 15 soaps = 15​ ×  285.60  =  4284  = Rs 357
                                         12            12

Page No 157:

Question 3:

If 9 kg of rice costs Rs 327.60, what will be the cost of 50 kg of rice?

ANSWER:

Cost of 9 kg of rice = Rs 327.60
Cost of 1 kg of rice =  327.60  
                                     9 
Cost of 50 kg of rice = 50​ ×  327.60  =  16380  = Rs 1820
                                               9              9
Hence, the cost of 50 kg of rice is Rs 1820.

Page No 157:

Question 4:

If 22.5 m of a uniform iron rod weighs 85.5 kg, what will be the weight of 5 m of the  same rod?

ANSWER:

Weight of 22.5 m of uniform iron rod = 85.5 kg
Weight of 1 m of uniform iron rod =  85.5  kg
                                                        22.5
Weight of 5 m of uniform iron rod = 5​ ×  85.5  =  427.5  = 19 kg
                                                              22.5       22.5
Thus, the weight of 5 m of iron rod is 19 kg.

Page No 157:

Question 5:

If 15 tins of the same size contain 234 kg of oil, how much oil will there be in 10 such tins?

ANSWER:

Oil contained by 15 tins = 234 kg
Oil contained by 1 tin =  234  kg
                                       15
Oil contained by 10 tins = 10 ×  234  =  2340  = 156 kg
                                                  15         15

Page No 157:

Question 6:

If 12 L of diesel is consumed by a car in covering a distance of 222 km, how many kilometres will it go in 22 L of diesel?

ANSWER:

Distance covered by a car in 12 L diesel = 222 km
Distance covered by it in 1 L diesel =  222  km
                                                          12
Distance covered by it in 22 L diesel = 22 ×  222  =  4884  = 407 km
                                                                   12          12

Page No 157:

Question 7:

A transport company charges Rs 540 to carry 25 tonnes of weight. What will it charge to carry 35 tonnes?

ANSWER:

Cost of transporting 25 tonnes of weight = Rs 540
Cost of transporting 1 tone of weight =  540 
                                                             25
Cost of transporting 35 tonnes of weight = 35​ ×  540  =  18900  = Rs 756
                                                                          25          25

Page No 158:

Question 8:

4.5 g of an alloy of copper and zinc contains 3.5 g of copper. What weight of copper will there be in 18.9 g of the alloy?

ANSWER:

Let the weight of copper be x g.
​Then, 4.5:3.5::18.9:x
        Product of extremes = Product of means
              4.5 × x  = 3.5 × 18.9
          ⇒ x =  66.15  = 14.7
                       4.5
So, the weight of copper is 14.7 g.

Page No 158:

Question 9:

35 inland letters cost Rs 87.50. How many such letters can we buy for 315?

ANSWER:

Number of inland letters whose total cost is Rs 87.50 = 35
Number of inland letters of whose cost is Re 1 =   35    
                                                                         87.50
Number of inland letters whose cost is Rs 315 = 315​ ×   35    =  11025  = 126
                                                                                  87.50      87.50
Hence, we can buy 126 inland letters for Rs 315.

Page No 158:

Question 10:

Cost of 4 dozen bananas is Rs 104. How many bananas can be purchased for Rs 6.50?

ANSWER:

Number of bananas that can be purchased for Rs 104 = 48 (4 dozen)
Number of bananas that can be purchased for Re 1 =  48 
                                                                               104
Number of bananas that can be purchased for Rs 6.50 = 6.50 ×  48   =  312  = 3
                                                                                               104       104
Hence, 3 bananas can be purchased for Rs 6.50.

Page No 158:

Question 11:

The cost of 18 chairs is Rs 22770. How many such chairs can be bought for Rs 10120?

ANSWER:

Number of chairs that can be bought for Rs 22770 = 18
Number of chairs that can be bought for Re 1 =    18   
                                                                        22770
Number of chairs that can be bought for Rs 10120 = 10120 ×    18     =  182160  = 8
                                                                                             22770       22770

Page No 158:

Question 12:

A car travels 195 km in 3 hours.
(i) How long will it take to travel 520 km?
(ii) How far will it travel in 7 hours with the same speed?

ANSWER:

(i) Time taken by the car to travel 195 km = 3 hours
   Time taken by it to travel 1 km =   3   hours
                                                      195
  Time taken by it to travel 520 km = 520 ×   3   =  1560  = 8 hours
                                                                  195      195

(ii) Distance covered by the car in 3 hours = 195 km
    Distance covered by it in 1 hour =  195  = 65 km
                                                          3
   Distance covered by it in 7 hours = 7 × 65 = 455 km

Page No 158:

Question 13:

A labourer earns Rs 1980 in 12 days.
(i) How much does he earn in 7 days?
(ii) In how many days will he earn Rs 2640?

ANSWER:

(i) Earning of a labourer in 12 days = Rs 1980
    Earning of the labourer in 1 day =  1980  = Rs 165
                                                        12
    Earning of the labourer in 7 days = 7​ × 165 = Rs 1155
(ii) Number of days taken by the labourer to earn Rs 1980 = 12 days
     Number of days taken by him to earn Re 1 =  12  days
                                                                       1980
     Number of days taken by him to earn Rs 2640 = 2640 ×  12   =  31680  = 16 days
                                                                                         1980      1980

Page No 158:

Question 14:

The weight of 65 books is 13 kg.
(i) What is the weight of 80 such books?
(ii) How many such books weigh 6.4 kg?

ANSWER:

Weight of 65 books = 13 kg
(i) Weight of 1 book =  13  kg
                                   65
    Weight of 80 books = 80 ×  13   =  1040   = 16 kg
                                              65         65

(ii) Number of books weighing 13 kg = 65
    Number of books weighing 1 kg =  65  = 5 
                                                        13
   Number of books weighing 6.4 kg = 6.4 × 5 = 32

Page No 158:

Question 15:

If 48 boxes contain 6000 pens, how many such boxes will be needed for 1875 pens?

ANSWER:

Number of boxes containing 6000 pens = 48
Number of boxes containing 1 pen =   48  
                                                       6000
Number of boxes containing 1875 pens = 1875 ×   48    =   90000  = 15
                                                                          6000        6000
15 boxes are needed for 1875 pens.

Page No 158:

Question 16:

24 workers can build a wall in 15 days. How many days will 9 workers take to build a similar wall?

ANSWER:

Number of days taken by 24 workers to build a wall = 15 days
Number of days taken by 1 worker to build the wall = 15 × 24 = 360 days         (less worker means more days)
Number of days taken by 9 workers to build the wall =  360  = 40 days
                                                                                     9

Page No 158:

Question 17:

40 men can finish a piece of work in 26 days. How many men will be needed to finish it in 16 days?

ANSWER:

Number of men required to complete the work in 26 days = 40
Number of men required to complete the work in 1 day = 40 × 26 = 1040 men  (less men more days)
Number of men required to complete the work in 16 days =  1040  = 65
                                                                                                   16

Page No 158:

Question 18:

In an army capm, there were provisions for 550 men for 28 days. But, 700 men attended the camp. How long did the provisions last?

ANSWER:

Number of days the provisions will last for 550 men = 28 days
Number of days the provisions will last for 1 man = 28 × 550 = 15400 days  (less men means more days)
Number of days the provisions will last for 700 men =  15400  = 22 days
                                                                                    700
The provision will last for 22 days. 

Page No 158:

Question 19:

A given quantity of rice is sufficient for 60 persons for 3 days. How many days would the rice last for 18 persons?

ANSWER:

Number of days for which the given quantity of rice is sufficient for 60 persons = 3 days
Number of days for which it is sufficient for 1 person = 3 × 60 = 180 days      (less men means more days )
Number of days for which it is sufficient for 18 persons =  180  = 10 days
                                                                                       18

Page No 158:

Exercise 10D

Question 1:

The ratio 92 : 115 in its simplest for is
(a) 23 : 25
(b) 18 : 23
(c) 3 : 5
(d) 4 : 5

ANSWER:

(d) 4 : 5
92:115 =   92 ​÷ 23   =  4  (As H.C.F. of 92 and 115 is 23.)
                115 ​÷ 23       5

Page No 158:

Question 2:

If 57 : x : : 51 : 85, then the value of x is
(a) 95
(b) 76
(c) 114
(d) none of these

ANSWER:

(a) 95
57:x::51:85
    57  =   51 
     x        85
⇒ x =  57 × 85  
               51
⇒ x =  4845  = 95
            51

Page No 158:

Question 3:

If 25 : 35 : : 45 : x, then the value of x is
(a) 63
(b) 72
(c) 54
(d) none of these

ANSWER:

(a) 63
25:35::45:x
          25  =  45 
          35       x
⇒ x =  35 × 45  =  1575  = 63
              25             25

Page No 158:

Question 4:

If 4 : 5 : : x : 35, then the value of x is
(a) 42
(b) 32
(c) 28
(d) none of these

ANSWER:

(c) 28
4:5::x:35
⇒  4  =  x  
     5      35
⇒ x =  4 × 35  = 4 × 7 = 28
              5

Page No 158:

Question 5:

If a, b, c, d are in proportion, then
(a) ac = bd
(b) ad = bc
(c) ab = cd
(d) none of these

ANSWER:

(b) ad = bc
Given:
a, b, c, d are in proportion.
a:b::c:d
     a  =  c  
     b      d
⇒ ad = bc

Page No 158:

Question 6:

If a, b, c are in proportion, then
(a) a² = bc
(b) b² = ac
(c) c² = ab
(d) none of these

ANSWER:

(b) b² = ac
Given:
a, b, c are in proportion.
a:b::b:c
    Product of means = Product of extremes
⇒​ b^​2 = ac

Page No 158:

Question 7:

Choose the correct statement:
(a) (5 : 8) > (3 : 4)
(b) (5 : 8) < (3 : 4)
(c) two ratios cannot be compared

ANSWER:

(b) (5 : 8) < (3 : 4)

We can write
(5:8) = 58 and (3:4) = 34(5:8) = 58 and (3:4) = 34
Making the denominator equal:
 5   and   3 × 2  =   6       
 8           4 × 2        8
As 6 > 5,   5   <   3    
                 8        4

Page No 159:

Question 8:

If Rs 760 is divided between A and B in the ratio 8 : 11, then B’s share is
(a) Rs 440
(b) Rs 320
(c) Rs 430
(d) Rs 330

ANSWER:

(a) Rs 440
A:B = 8:11
Sum of ratio terms = 8 + 11 = 19
B’s share =  11  × 760 =  8360  = Rs 440
                   19                  19

Page No 159:

Question 9:

Two numbers are in the ratio 5 : 7 and the sum of these numbers is 252. The larger of these numbers is
(a) 85
(b) 119
(c) 105
(d) 147

ANSWER:

(d) 147
Ratio = 5:7
Let x be any number such that we have:
   5x + 7x = 252
⇒ 12x = 252
⇒ x =  252  = 21
           12
Now, 5x = 5 × 21= 105
7x = 7 × 21 = 147

The largest number is 147.

Page No 159:

Question 10:

The sides of a triangle are in the ratio 1 : 3 : 5 and its perimeter is 90 cm. The length of its largest side is
(a) 40 cm
(b) 50 cm
(c) 36 cm
(d) 54 cm

ANSWER:

(b) 50 cm
The sides of the triangle are in the ratio 1:3:5.
Let x be any number such that the sides are 1x cm, 3x cm and 5x cm.
          1x + 3x + 5x = 90
       ⇒ 9x = 90
       ​⇒ x =  90  = 10
                   9
First side = 1x = 1 ​× 10 = 10 cm
Second side = 3x = 3 ​× 10 = 30 cm
Third side = 5x = 5 × 10 = 50 cm
The length of the largest side is 50 cm.

Page No 159:

Question 11:

The ratio of boys and girls in a school is 12 : 5. If the number of girls is 840, the total strength of the school is
(a) 1190
(b) 2380
(c) 2856
(d) 2142

ANSWER:

(c) 2856
Ratio of boys and girls = 12:5
Let x be any number such that the number of boys and girls are 12x and 5x, respectively.
Number of girls = 840
      5x = 840
 ⇒ x =  840  = 168
             5
Number of boys = 12x = 12 × 168 = 2016
Number of girls = 840
Total strength of the school = 2016 + 840 = 2856

Page No 159:

Question 12:

If the cost of 12 pens is Rs 138, then the cost of 14 such pens is
(a) Rs 164
(b) Rs 161
(c) Rs 118.30
(d) Rs 123.50

ANSWER:

(b) Rs 161
Cost of 12 pens = Rs 138
Cost of 1 pen = Rs  138  
                              12
Cost of 14 pens = Rs  138  × 14 = Rs 1932  = Rs 161
                                     12                     12

Page No 159:

Question 13:

If 24 workers can build a wall in 15 days, how many days will 8 workers take to build a similar wall?
(a) 42 days
(b) 45 days
(c) 48 days
(d) none of these

ANSWER:

(b) 45 days
Time taken by 24 workers to build a wall = 15 days
Time taken by 1 worker to build a wall = 24 × 15 = 360 days        (clearly less workers will take more time to build a wall)
Time taken by 8 workers to build a wall =  360  = 45 days
                                                                       8

Page No 159:

Question 14:

If 40 men can finish a piece of work in 26 days, how many men will be required to finish it in 20 days?
(a) 52
(b) 31
(c) 13
(d) 65

ANSWER:

(a) 52
Number of men required to finish the work in 26 days = 40
Number of men required to finish it in 1 day = 40 × 26 = 1040 men          (More men means less days)
Number of men required to finish it in 20 days =  1040  = 52
                                                                                  20

Page No 159:

Question 15:

In covering 111 km, a car consumes 6 L of petrol. How many kilometres will it go in 10 L of petrol?
(a) 172 km
(b) 185 km
(c) 205 km
(d) 266.4 km

ANSWER:

(b) 185 km
Distance covered in 6 L of petrol = 111 km
Distance covered in 1 L of  petrol =  111 km
                                                       6
Distance covered in 10 L of petrol =  111  × 10 =  1110  = 185 km
                                                              6                    6

Page No 159:

Question 16:

In a fort, 550 men had provisions for 28 days. How many days will it last for 700 men?
(a) 22 days
(b) 3571135711 days
(c) 34 days
(d) none of these

ANSWER:

(a) 22 days
Number of days for which 550 men had provisions = 28 days
Number of days for which 1 man had provisions = 28 × 550 = 15400 days (more men means less days)
Number of days for which 700 men had provisions =  15400  = 22 days
                                                                                        700

Page No 159:

Question 17:

The angles of a triangle are in the ratio 3 : 1 : 2. The measure of the largest angle is
(a) 30°
(b) 60°
(c) 90°
(d) 120°

ANSWER:

(c) 90°
Ratio of the angles of a triangle is 3:1: 2
Let x be any number such that the three angles are (3x)°°, (1x)°° and (2x)°°.
We know, the sum of the angles of a triangle is 180°°.
        3x + 1x + 2x = 180
      ⇒ 6x = 180
​      ⇒ x =  180  = 30
                  6
∴∴  (3x )°° = (3 ​× 30)°° = 90^o
​     (1x)°° = (1​ × 30)°° = 30^o
​     (2x)°° = (2 × 30)°° = 60^o
The measure of the largest angle is 90^o​.

Page No 159:

Question 18:

Length and breadth of a rectangular field are in the ratio 5 : 4. If the width of the field is 36 m, what is its length?
(a) 40 m
(b) 45 m
(c) 54 m
(d) 50 m

ANSWER:

(b) 45 m
Length:Breadth = 5:4
Let x be any number such that the length and the breadth are 5x and 4x, respectively.
Now , 4x = 36
            x =  36  = 9
                    4
Length = 5x = 5 × 9 = 45 m

Page No 159:

Question 19:

If a bus covers 195 km in 3 hours and a train covers 300 km in 4 hours, then the ratio of their speeds is
(a) 13 : 15
(b) 15 : 13
(c) 13 : 12
(d) 12 : 13

ANSWER:

(a) 13 : 15

Speed =  Distance  
                Time 
Speed of the bus =  195 km = 65 km/hr
                                  3 hr
Speed of the train =  300 km  = 75 km/hr
                                    4 hr
Ratio =  65  =  65 ÷ 5  =  13  = 13:15
             75       75 ÷ 5      15

Page No 159:

Question 20:

If the cost of 5 bars of soap is Rs 82.50, then the cost of one dozen such bars is
(a) Rs 208
(b) Rs 192
(c) Rs 198
(d) Rs 204

ANSWER:

(c) Rs 198
Cost of 5 bars of soap = Rs 82.50
Cost of 1 bar of soap =  82.50  = Rs 16.5
                                         5
Cost of 12 (1 dozen) bars of soap = 16.5 × 12 = Rs 198

Page No 159:

Question 21:

If the cost of 30 packets of 8 pencils each is Rs 600, what is the cost of 25 packets of 12 pencils each?
(a) Rs 725
(b) Rs 750
(c) Rs 480
(d) Rs 720

ANSWER:

(b) Rs 750
Cost of 30 packets of 8 pencils each = Rs 600
Cost of 1 packet of 8 pencils =  600   = Rs 20
                                                 30
Cost of  1 pencil = Rs  20   
                                      8
Cost of 1 packet of 12 pencils = 12​ ×  20  =  240  = Rs 30
                                                              8         8
Cost of 25 packets of 12 pencils each = 25 × 30 = Rs 750

Page No 159:

Question 22:

A rail journey of 75 km costs Rs 215. How much will a journey of 120 km cost?
(a) Rs 344
(b) Rs 324
(c) Rs 268.75
(d) none of these

ANSWER:

(a) Rs 344
Cost of rail journey of 75 km = Rs 215
Cost of rail journey of 1 km = Rs  215 
                                                   75
Cost of rail journey of 120 km = 120​ ×  215   = 25800 = Rs 344
                                                              75          75

Page No 159:

Question 23:

The 1st, 2nd and 4th terms of a proportion are 12, 21 and 14 respectively. Its third term is
(a) 16
(b) 18
(c) 21
(d) 8

ANSWER:

(d) 8
Let the third term be x.
Then, we have:
12:21::x:14
We know:
    Product of means = Product of extremes
      21x = 12 × 14
  ⇒ 21x = 168
  ⇒ x =  168  = 8
              21
The third term is 8

Page No 159:

Question 24:

10 boys can dig a pitch in 12 hours. How long will 8 boys take to do it?
(a) 9 h 36 min
(b) 15 h
(c) 6 h 40 min
(d) 13 h 20 min

ANSWER:

(b) 15 h
Time taken by 10 boys to dig a pitch = 12 hours
Time taken by 1 boy to dig a pitch = 12 × 10 = 120 hours        (less boys means more time)
Time taken by 8 boys to dig a pitch =  120 = 15 hours
                                                                8

Page No 161:

Exercise 10E

Question 1:

Find the ratio of:
(a) 90 cm to 1.05 m
(b) 35 minutes to an hour
(c) 150 mL to 2 L
(d) 2 dozens to a score

ANSWER:

(a) 90 cm:1.05 m (or) 90 cm:105 cm             (1 m = 100 cm)
        90   =  90 ​÷ 15  =   6         (H.C.F. of 90 and 105 is 15.) 
       105      105 ​÷ 15      7
     ∴∴ 6:7

(b) 35 minutes to an hour (or) 35 minutes:60 minutes      (1 hour = 60 minutes)
        35  =  35 ​÷ 5  =  7       (H.C.F. of 35 and 60 is 5.)
        60       60 ​÷ 5     12  
   ∴∴  7:12

(c) 150 mL to 2 L (or) 150 L:2000 L                (1 L= 1000 mL)
         150  =   150 ​÷ 50  =   3    (HCF of 150 and 2000 is 50)
        2000      2000​ ​÷50       40
    ∴∴ 3:40

(d) 2 dozens to a score (or) 24:20       (1 dozen = 12 and 1 score = 20)
        24  =   24 ​÷ 4  =  6     (H.C.F. of 24 and 20 is 4)
        20        20​ ​÷ 4      5
   ∴∴ 6:5

Page No 161:

Question 2:

The ratio of zinc and copper in an alloy is 7 : 9. If the weight of copper in the alloy is 12.6 kg, find the weight of zinc in it.

ANSWER:

Ratio of zinc and copper in an alloy is = 7:9
Let the weight of zinc and copper in it be (7x) and (9x), respectively.
Now, the weight of a copper = 12.6 kg   (given)
∴ 9x = 12.6
⇒  x = 12.6  = 1.4
             9
∴ Weight of zinc = 7x = 7​ × 1.4 = 9.8 kg

Page No 161:

Question 3:

Divide Rs 1400 among A. B and C in the ratio 2 : 3 : 5.

ANSWER:

Given:
  A:B:C = 2:3:5
Sum of ratio = 2 + 3 + 5 = 10
Total money = Rs 1400
Then, share of A =   2   × Rs 1400 = Rs  2800  = Rs 280
                             10                             10
Share of B =  3  ​× Rs 1400 = Rs  4200  = Rs 420
                    10                             10
Share of C =  5  ​× Rs 1400 = Rs  7000  = Rs 700
                    10                            10

Page No 161:

Question 4:

Prove that (5 : 6) > (3 : 4).

ANSWER:

We can write:
(5:6) = 56 and (3:4) = 34(5:6) = 56 and (3:4) = 34
By making their denominators same: (Taking the L.C.M. of 6 and 4, which is 24.)
Consider, 5:6
         5 ​× 4  =  20 
         6 ​× 4      24
              
And,  3 ​× 6  =  18 
         4 ​× 6       24
As 20 > 18
Clearly, (5:6) > (3:4)

Page No 161:

Question 5:

40 men can finish a piece of work in 26 days. How many men will be needed to finish it in 16 days?

ANSWER:

Number of men needed to finish a piece of work in 26 days = 40
Number of men needed to finish it in 1 day = 26 × 40 = 1040    (less days means more men)
Number of men needed to finish it in 16 days =  1040  = 65
                                                                          16 

Page No 161:

Question 6:

In an army camp, there were provisions for 425 men for 30 days. How long did the provisions last for 375 men?

ANSWER:

Number of days for which provisions last for 425 men = 30 days
Number of days for which provisions last for 1 men = 30 × 425 = 12750 days. (less men means more days)
Number of days for which provisions last for 375 men = 12750  = 34 days
                                                                                   375
Hence, provisions will last for 34 days for 375 men.

Page No 161:

Question 7:

Find the value of x when 36 : x : : x : 16.

ANSWER:

Given:
36:x::x:16
We know:
Product of means = Product of extremes 
  x × x = 36 × 16
⇒ x² = 576
⇒ x² = 24²
⇒ x = 24

Page No 161:

Question 8:

Show that 48, 60, 75 are in continued proportion.

ANSWER:

Consider 48:60::60:75

Product of means = 60 × 60 = 3600
Product of extremes = 48 × 75 = 3600
So product of means = Product of extremes
Hence, 48, 60, 75 are in continued proportion.

Page No 161:

Question 9:

Two numbers are in the ratio 3 : 5 and their sum is 96. The larger number is
(a) 36
(b) 42
(c) 60
(d) 70

ANSWER:

(c) 60
Ratio = 3:5
Let x be any number such that we have:
    3x + 5x = 96
 ⇒ 8x = 96
 ⇒ x =  96  = 12
             8
The numbers are:
     3x = 3 ​× 12 = 36
     5x = 5 ​× 12 = 60
The largest number = 60

Page No 161:

Question 10:

A car travels 288 km is 4 hours and a train travels 540 km in 6 hours. The ratio of their speeds is
(a) 5 : 4
(b) 4 : 5
(c) 5 : 6
(d) 3 : 5

ANSWER:

(b) 4 : 5
Speed of the car =  Distance  =  288 km  = 72 km/hr
                                 Time            4 hr
 
Speed of the train =  Distance  =  540 km  = 90 km/hr
                                   Time             6 hr
 
Ratio of their speeds = 72:90
where, 72  = 72 ​÷ 18  =  4      (H.C.F. of 72 and 90 is 18.)
           90      90 ​÷ 18      5

Page No 161:

Question 11:

The first three terms of a proportion are 12, 21 and 8 respectively. The 4th term is
(a) 18
(b) 16
(c) 14
(d) 20

ANSWER:

(c) 14
Let the 4th term be x, such that we have:
     12:21::8:x
 Now, we know:
     Product of extremes = Product of means
                                    12x = 21 × 8 
                                       x =  168  = 14
                                               12

Page No 161:

Question 12:

The ratio 92 : 115 in simplest form is
(a) 23 : 25
(b) 18 : 23
(c) 3 : 5
(d) 4 : 5

ANSWER:

(d) 4 : 5
92:115
 92  =  92 ​÷ 23   =  4         (H.C.F. of 92 and 115 is 23)
115     115 ​÷ 23      5

Page No 161:

Question 13:

If 57 : x : : 51 : 85, then the value of x is
(a) 95
(b) 76
(c) 114
(d) none of these

ANSWER:

(a) 95
Given :  
57:x::51:85
We know:
Product of means = Product of extremes
                                      51x = 57 × 85
                                          x =   4845  = 95
                                                     51

Page No 161:

Question 14:

If 4 : 5 : : x : 45, then the value of x is
(a) 54
(b) 60
(c) 36
(d) 30

ANSWER:

(c) 36
Given:
4:5::x:45
We know:
Product of mean = Product of extremes
                                      5x = 4 ​× 45
                                       x =    180   = 36
                                                  5

Page No 161:

Question 15:

If a, b, c are in proportion, then
(a) a² = bc
(b) b² = ac
(c) c² = ab
(d) none of these

ANSWER:

(b) b² = ac
Given:
a, b, c are in proportion, such that we have:
  a:b::b:c
Now, we know:
  Product of means = Product of extremes
                         b ​× b = a ​× c
                              b² = ac

Page No 161:

Question 16:

10 boys can dig a pitch in 12 hours. How long will 8 boys take to do it?
(a) 9 hrs 36 min
(b) 15 hrs
(c) 6 hrs 40 min
(d) 13 hrs 10 min

ANSWER:

(b) 15 hrs
Time taken by 10 boys to dig a pitch = 12 hours
Time taken by 1 boy to dig a pitch = 12 × 10 = 120 hours   (Less boys would take more hours.)
Time taken by 8 boys to dig a pitch =  120  = 15 hours
                                                               8

Page No 161:

Question 17:

In covering 148 km, a car consumes 8 litres of petrol. How many kilometres will it go in 10 litres of petrol?
(a) 172 km
(b) 185 km
(c) 205 km
(d) 266.4 km

ANSWER:

 (b) 185 km
Distance covered by a car in 8 litres of petrol = 148 km
Distance covered by it in 1 litre of petrol =  148  km
                                                                    8 
Distance covered by it in 10 litres of petrol = 10 × 148 = 1480 = 185 km
                                                                              8         8

Page No 161:

Question 18:

Fill in the blanks.
(i) 1421=   3=6   1421=   3=6   
(ii) 90 cm : 1.5 m = …… .
(iii) If 36 : 81 : : x : 63, then x = …… .
(iv) If 25, 35, x are in proportion, then x = …… .
(v) If 9, x, x, 49 are in proportion, then x = …… .

ANSWER:

(i)
           Let 1421 = x3Thus, we have: 21x = 14 × 3 ⇒ x = 14 × 321 = 2∴ 1421 = 23Again,  let 23=6yThus, we have: 2y = 6 × 3 ⇒ y = 6 × 32 = 9∴ 23 = 69∴  1421 = 23 = 69Let 1421 = x3Thus, we have: 21x = 14 × 3 ⇒ x = 14 × 321 = 2∴ 1421 = 23Again,  let 23=6yThus, we have: 2y = 6 × 3 ⇒ y = 6 × 32 = 9∴ 23 = 69∴  1421 = 23 = 69

(ii) 90 cm:1.5 m (or) 90 cm:150 cm          (1 m = 100 cm)
      90  =  9  =  9 ÷ 3  =  3              (H.C.F. of 9 and 15 is 3.)
     150     15    15 ​​÷ 3      5
  
(iii) If 36:81::x:63
      Product of means = Product of extremes
                          81x = 36 × 63
                             x =  2268 
                                      81
                             x = 28

(iv) Given:
      25, 35, x are in proportion.
       25:35::35:x
     Now, we know:
     Product of extremes = Product of means
                          25 × x = 35 ​× 35
                            25x = 1225
                              x =  1225  = 49
                                       25

(v) Given:
     9, x, x, 49 are in proportion.
         9:x::x:49
     Now, we know:
     Product of means = Product of extremes
                             x ​× x = 9 ​× 49
                                   x² = 441
                                   x² = 21²
                                    x = 21

Page No 162:

Question 19:

Write ‘T’ for true and ‘F’ for false for each of the statements given below:
(i) 30, 40, 45, 60 are in proportion.
(ii) 6 : 8 and 9 : 12 are equivalent ratios of 3 : 4.
(iii) a dozen : a score = 5 : 3.
(iv) 60 p : Rs 3 = 1 : 5.

ANSWER:

(i) 30, 40, 45, 60 
      30  =    3 ,    45  =   45 ​÷ 15  =     3      They are in proportion.
      40        4      60        60 ​÷ 15         4
  Hence, true.

(ii)  6  =  6 ​÷ 2  =  3 ,  9  =  9 ​÷ 3  =  3     Hence, they are equivalent to 3:4.
      8       8 ​÷ 2      4    2      12 ​÷ 3     4
  Hence, true.
(iii) 1 dozen:1 score = 12:20
      12  =  12 ​÷ 4  =  3       
      20      20 ​÷ 4       5
Hence, false.
(iv) 60p:Rs 3 = 60p:300p                        (1 Re = 100 p)
       60  =  6  =  6 ​÷ 6  =  1 
      300     30    30 ​÷ 6      5

 Hence, true.

Exercise 8A

Page No 130:

Question 1:

Write the following using literals, numbers and signs of basic operations:
(i) x increased by 12
(ii) y decreased by 7
(iii) The difference of a and b, when a > b
(iv) The product of x and y added to their sum
(v) One-third of x multiplied by the sum of a and b
(vi) 5 times x added to 7 times y
(vii) Sum of x and the quotient of y by 5
(viii) x taken away from 4
(ix) 2 less than the quotient of x by y
(x) x multiplied by itself
(xi) Twice x increased by y
(xii) Thrice x added to y squared
(xiii) x minus twice y
(xiv) x cubed less than y cubed
(xv) The quotient of x by 8 is multiplied by y

ANSWER:

(i) x increased by 12 is (x + 12).
(ii) y decreased by 7 is (y – 7).
(iii) The difference of a and b, when a>b is (a – b).
(iv) The product of x and y is xy.
The sum of x and y is (x + y).
      So, product of x and y added to their sum is xy + (x + y).
(v) One third of x is x3x3.
The sum of a and b is (a + b).
      ∴∴ One-third of x multiplied by the sum of a and b = x3×(a+b)=x(a+b)3 x3×(a+b)=x(a+b)3 
(vi) 5 times x added to 7 times y = (5×x)+(7×y),  which is equal to 5x+7y.(5×x)+(7×y),  which is equal to 5x+7y.

(vii) Sum of x and the quotient of y by 5 is x+y5x+y5.
(viii) x taken away from 4 is (4-x).
(ix) 2 less than the quotient of x by y is xy−2xy-2.
(x) x multiplied by itself is x×x=x2x×x=x2.
(xi) Twice x increased by y is (2×x)+y = 2x+y(2×x)+y = 2x+y.
(xii) Thrice x added to y squared is (3×x)+(y×y)=3x+y2(3×x)+(y×y)=3x+y2.
(xiii) x minus twice y is x−(2×y)=x−2yx-(2×y)=x-2y.
(xiv) x cubed less than y cubed is (y×y×y)−(x×x×x)=y3−x3.(y×y×y)-(x×x×x)=y3-x3.
(xv) The quotient of x by 8 is multiplied by y is x8×y=xy8x8×y=xy8.

Page No 130:

Question 2:

Ranjit scores 80 marks in English and x marks in Hindi. What is his total score in the two subjects?

ANSWER:

Ranjit’s score in English = 80 marks
Ranjit’s score in Hindi = x marks
Total score in the two subjects = (Ranjit’s score in English + Ranjit’s score in Hindi)
∴ Total score in the two subjects = (80 + x) marks

Page No 130:

Question 3:

Write the following in the exponential form:
(i) b × b × b × … 15 times
(ii) y × y × y × … 20 times
(iii) 14 × a × a × a × a × b × b × b
(iv) 6 × x × x × y × y
(v) 3 × z × z × z × y × y × x

ANSWER:

(i) b × b × b × … 15 times = b15b15
(ii) y × y × y × … 20 times = y20y20
(iii) 14 × a × a × a × a × b × b × b = 14×(a×a×a×a)×(b×b×b) =14a4b314×(a×a×a×a)×(b×b×b) =14a4b3
(iv) 6 × x × x × y × y = 6×(x×x)×(y×y)=6x2y26×(x×x)×(y×y)=6x2y2
(v) 3 × z × z × z × y × y × x = 3×(z×z×z)×(y×y)×x=3z3y2x3×(z×z×z)×(y×y)×x=3z3y2x

Page No 130:

Question 4:

Write down the following in the product form:
(i) x²y⁴
(ii) 6y⁵
(iii) 9xy²z
(iv) 10a³b³c³

ANSWER:

(i) x2y4=(x×x)×(y×y×y×y)=x×x×y×y×y×yx2y4=(x×x)×(y×y×y×y)=x×x×y×y×y×y
(ii) 6y5=6×(y×y×y×y×y)=6×y×y×y×y×y6y5=6×(y×y×y×y×y)=6×y×y×y×y×y
(iii) 9xy2z=9×x×(y×y)×z=9×x×y×y×z9xy2z=9×x×(y×y)×z=9×x×y×y×z
(iv) 10a3b3c3=10×(a×a×a)×(b×b×b)×(c×c×c)=10×a×a×a×b×b×b×c×c×c10a3b3c3=10×(a×a×a)×(b×b×b)×(c×c×c)=10×a×a×a×b×b×b×c×c×c

Page No 132:

Question 1:

If a = 2 and b = 3, find the value of
(i) a + b
(ii) a² + ab
(ii) ab − a²
(iv) 2a − 3b
(v) 5a² − 2ab
(vi) a³ − b³

ANSWER:

(i) a+b
    Substituting a = 2 and b = 3 in the given expression:
    2+3 = 5

(ii) a2+aba2+ab
    Substituting a = 2 and b = 3 in the given expression:
    (2)2+(2×3)=4+6=10(2)2+(2×3)=4+6=10

(iii) ab−a2ab-a2
    Substituting a = 2 and b = 3 in the given expression:
     (2×3)−(2)2=6−4=2(2×3)-(2)2=6-4=2

(iv) 2a-3b
    Substituting a = 2 and b = 3 in the given expression:
    (2×2)−(3×3)=4−9=−5(2×2)-(3×3)=4-9=-5

(v) 5a2−2ab5a2-2ab
    Substituting a=2 and b=3 in the given expression:
    5×(2)2−2×2×3=5×4−12=20−12=85×(2)2-2×2×3=5×4-12=20-12=8

(vi) a3−b3a3-b3
    Substituting a=2 and b=3 in the given expression:
    23−33=2×2×2−3×3×3=8−27=−1923-33=2×2×2-3×3×3=8-27=-19

Page No 132:

Question 2:

If x = 1, y = 2 and z = 5, find the value of
(i) 3x − 2y + 4z
(ii) x² + y² + z²
(iii) 2x² − 3y² + z²
(iv) xy + yz − zx
(v) 2x² y − 5yz + xy²
(vi) x³ − y³ − z³

ANSWER:

(i) 3x-2y+4z
    Substituting x = 1, y = 2 and z = 5 in the given expression:
    3×(1)−2×(2)+4×(5)=3−4+20=193×(1)-2×(2)+4×(5)=3-4+20=19

(ii) x2+y2+z2 x2+y2+z2 
    Substituting x = 1, y = 2 and  z = 5 in the given expression:
    12+22+52=(1×1)+(2×2)+(5×5)=1+4+25=3012+22+52=(1×1)+(2×2)+(5×5)=1+4+25=30

(iii) 2×2−3y2+z22x2-3y2+z2
      Substituting x = 1, y = 2 and z = 5 in the given expression:
     2×(1)2−3×(2)2+52=2×(1×1)−3×(2×2)+(5×5)=2−12+25=152×(1)2-3×(2)2+52=2×(1×1)-3×(2×2)+(5×5)=2-12+25=15

(iv) xy+yz−zxxy+yz-zx
     Substituting x = 1, y = 2 and z = 5 in the given expression:
      (1×2)+(2×5)−(5×1)=2+10−5=7(1×2)+(2×5)-(5×1)=2+10-5=7

(v) 2x2y−5yz+xy22x2y-5yz+xy2
      Substituting x = 1, y = 2 and z = 5 in the given expression:
     2×(1)2×2−5×2×5+1×(2)2=4−50+4=−422×(1)2×2-5×2×5+1×(2)2=4-50+4=-42

(vi) x3−y3−z3x3-y3-z3
    Substituting x = 1, y = 2 and z = 5 in the given expression:
    13−23−53=(1×1×1)−(2×2×2)−(5×5×5)=1−8−125=−13213-23-53=(1×1×1)-(2×2×2)-(5×5×5)=1-8-125=-132

Page No 132:

Question 3:

If p = −2, q = −1 and r = 3, find the value of
(i) p² + q² − r²
(ii) 2p² − q² + 3r²
(iii) p − q − r
(iv) p³ + q³ + r³ + 3pqr
(v) 3p²q + 5pq² + 2pqr
(vi) p⁴ + q⁴ − r⁴

ANSWER:

(i) p2+q2−r2p2+q2-r2
     Substituting p = -2, q = -1 and r = 3 in the given expression:
    (−2)2+(−1)2−(3)2=(−2×−2)+(−1×−1)−(3×3)⇒4+1−9=−4(-2)2+(-1)2-(3)2=(-2×-2)+(-1×-1)-(3×3)⇒4+1-9=-4

(ii) 2p2−q2+3r22p2-q2+3r2
     Substituting p = -2, q = -1 and r = 3 in the given expression:
   
2×(−2)2−(−1)2+3×(3)2=2×(−2×−2)−(−1×−1)+3×(3×3)⇒8−1+27=342×(-2)2-(-1)2+3×(3)2=2×(-2×-2)-(-1×-1)+3×(3×3)⇒8-1+27=34

(iii) p−q−rp-q-r
      Substituting p = -2, q = -1 and r = 3 in the given expression:
     (−2)−(−1)−(3)=−2+1−3=−4(-2)-(-1)-(3)=-2+1-3=-4

(iv) p3+q3+r3+3pqrp3+q3+r3+3pqr
    Substituting p = -2, q = -1 and r = 3 in the given expression:

   (−2)3+(−1)3+(3)3+3×(−2×−1×3)=(−2×−2×−2)+(−1×−1×−1)+(3×3×3)+3×(6)=(−8)+(−1)+(27)+18=36(-2)3+(-1)3+(3)3+3×(-2×-1×3)=(-2×-2×-2)+(-1×-1×-1)+(3×3×3)+3×(6)=(-8)+(-1)+(27)+18=36
   
(v) 3p2q+5pq2+2pqr3p2q+5pq2+2pqr
     Substituting p = -2, q = -1 and r = 3 in the given expression:

     3×(−2)2×(−1)+5×(−2)×(−1)2+2×(−2×−1×3)=3×(−2×−2)×(−1)+5×(−2)×(−1×−1)+2×(−2×−1×3)=−12−10+12=−103×(-2)2×(-1)+5×(-2)×(-1)2+2×(-2×-1×3)=3×(-2×-2)×(-1)+5×(-2)×(-1×-1)+2×(-2×-1×3)=-12-10+12=-10

(vi) p4+q4−r4p4+q4-r4
     Substituting p = -2, q = -1 and r = 3 in the given expression:
    (−2)4+(−1)4−(3)4=(−2×−2×−2×−2)+(−1×−1×−1×−1)−(3×3×3×3)=16+1−81=−64(-2)4+(-1)4-(3)4=(-2×-2×-2×-2)+(-1×-1×-1×-1)-(3×3×3×3)=16+1-81=-64
  

Page No 132:

Question 4:

Write the coefficient of
(i) x in 13x
(ii) y in −5y
(iii) a in 6ab
(iv) z in −7xz
(v) p in −2pqr
(vi) y² in 8xy²z
(vii) x³ in x³
(viii) x² in −x²

ANSWER:

(i) Coefficient of x in 13x is 13.
(ii) Coefficient of y in -5y is -5.
(iii) Coefficient of a in 6ab is 6b.
(iv) Coefficient of z in -7xz is -7x.
(v) Coefficient of p in -2pqr is -2qr.
(vi) Coefficient of y² in 8xy²z is 8xz.
(vii) Coefficient of x³ in  x³ is 1.
(viii) Coefficient of x² in -x² is -1.

Page No 132:

Question 5:

Write the numerical coefficient of
(i) ab
(ii) −6bc
(iii) 7xyz
(iv) −2x³y²z

ANSWER:

(i) Numerical coefficient of ab is 1.
(ii) Numerical coefficient of -6bc is -6.
(iii) Numerical coefficient of 7xyz is 7.
(iv) Numerical coefficient of −2x³y²z is -2.

Page No 132:

Question 6:

Write the constant term of
(i) 3x² + 5x + 8
(ii) 2x² − 9
(iii) 4y2−5y+354y2-5y+35
(iv) z3−2z2+z−83z3-2z2+z-83

ANSWER:

A term of expression having no literal factors is called a constant term.
(i) In the expression 3x² + 5x + 8, the constant term is 8.
(ii) In the expression 2x² − 9, the constant term is -9.
(iii) In the expression  4y2−5y+354y2−5y+35, the constant term is 3535.
(iv) In the expression z3−2z2+z−83z3−2z2+z−83 , the constant term is −83-83.

Page No 132:

Question 7:

Identify the monomials, binomials and trinomials in the following:
(i) −2xyz
(ii) 5 + 7x³y³z³
(iii) −5x³
(iv) a + b − 2c
(v) xy + yz − zx
(vi) x⁵
(vii) ax³ + bx² + cx + d
(viii) −14
(ix) 2x + 1

ANSWER:

The expressions given in (i), (iii), (vi) and (viii) contain only one term. So, each one of them is monomial.
The expressions given in (ii) and (ix) contain two terms. So, both of them are binomial.
The expressions given in (iv) and (v) contain  three terms. So, both of them are trinomial.
The expression given in (vii) contains four terms. So, it does not represents any of the given types.

Page No 133:

Question 8:

Write all the terms of the algebraic expressions:
(i) 4x⁵ − 6y⁴ + 7x²y − 9
(ii) 9x³ − 5z⁴ + 7z³ y − xyz

ANSWER:

(i) Expression 4x⁵ − 6y⁴ + 7x²y − 9 has four terms, namely 4x⁵ ,-6y⁴ , 7x²y and -9.
(ii) Expression 9x³ − 5z⁴ + 7z³ y − xyz has four terms, namely 9x³ , -5z⁴ , 7z³ y and -xyz.

Page No 133:

Question 9:

Identify the like terms in the following:
(i) a² , b², −2a², c², 4a
(ii) 3x, 4xy, −yz,12zy3x, 4xy, -yz,12zy
(iii) −2xy², x²y, 5y²x, x²z
(iv) abc, ab²c, acb², c²ab, b²ac, a²bc, cab²

ANSWER:

The terms that have same literals are called like terms.
(i)  a² and 2a² are like terms.
(ii) −yz and 12zy-yz and 12zy are like terms.
(iii) −2xy² and 5y²x are like terms.
(iv) ab²c , acb² , b²ac and cab² are like terms.

Page No 134:

Exercise 8B

Question 1:

Add:
(i) 3x, 7x
(ii) 7y, −9y
(iii) 2xy, 5xy, −xy
(iv) 3x, 2y
(v) 2x², − 3x², 7x²
(vi) 7xyz, − 5xyz, 9xyz, −8xyz
(vii) 6a³, − 4a³, 10a³, −8a³
(viii) x² − a², −5x² + 2a², −4x² + 4a²

ANSWER:

  (i) Required sum = 3x + 7x
     = (3+7)x = 10x

(ii) Required sum = 7y +(−9y)
     = (7-9)y = -2y

(iii) Required sum = 2xy +5xy + (−xy)
    = (2+5-1)xy = 6xy

(iv) Required sum = 3x+2y

(v) Required sum = 2x² + (− 3x²) + 7x²
    =(2-3+7)x² = 6x²

(vi)Required sum =  7xyz + (− 5xyz) + 9xyz + (−8xyz)
       = (7-5+9-8)xyz = 3xyz

(vii) Required sum = 6a³ +(− 4a³) + 10a³ +( −8a³)
     =(6-4+10-8)a³ = 4a³
 
(viii) Required sum = x² − a² + (−5x² + 2a²) +( −4x² + 4a² )
       Rearranging and collecting the like terms =  x² -5x² -4x² -a² + 2a² +4a²
      = (1-5-4)x² +(-1+2+4)a²
     = -8x² + 5a²

Page No 134:

Question 2:

Add the following:
(i)    x − 3y − 2z  5x + 7y − z− 7x − 2y + 4z                            x – 3y – 2z  5x + 7y – z- 7x – 2y + 4z                         

(ii)    m2 − 4m + 5− 2m2 + 6m − 6  − m2 − 2m − 7                               m2 – 4m + 5- 2m2 + 6m – 6  – m2 – 2m – 7                            

(iii)   2×2 − 3xy + y2− 7×2 − 5xy − 2y2  4×2 + xy − 6y2                                     2×2 – 3xy + y2- 7×2 – 5xy – 2y2  4×2 + xy – 6y2                                   

(iv)   4xy − 5yz − 7zx− 5xy + 2yz + zx− 2xy − 3yz + 3zx                               4xy – 5yz – 7zx- 5xy + 2yz + zx- 2xy – 3yz + 3zx                             

ANSWER:

(i)
     x −  3y    −2z  5 x + 7y    −  z−7x − 2y + 4z−x   +2y   +  z     x –  3y    -2z  5 x + 7y    –  z-7x – 2y + 4z-x   +2y   +  z

(ii)
          m2 − 4m + 5   − 2m2 + 6m − 6     − m2 − 2m − 7       −2m2  +0×m−8= −2m2  + 0 −8 = −2m2−8          m2 – 4m + 5   – 2m2 + 6m – 6     – m2 – 2m – 7       -2m2  +0×m-8= -2m2  + 0 -8 = -2m2-8

(iii)
     2×2 − 3xy   + y2− 7×2 − 5xy − 2y2     4×2 + xy  − 6y2                                      −x2−7xy  −7y2     2×2 – 3xy   + y2- 7×2 – 5xy – 2y2     4×2 + xy  – 6y2                                      -x2-7xy  -7y2

(iv)
     4xy − 5yz − 7zx− 5xy + 2yz + zx− 2xy − 3yz + 3zx−3xy   −6yz −3zx     4xy – 5yz – 7zx- 5xy + 2yz + zx- 2xy – 3yz + 3zx-3xy   -6yz -3zx

Page No 134:

Question 3:

Add:
(i) 3a − 2b + 5c, 2a + 5b − 7c, − a − b + c
(ii) 8a − 6ab + 5b, −6a − ab − 8b, −4a + 2ab + 3b
(iii) 2x³ − 3x² + 7x − 8, −5x³ + 2x² − 4x + 1, 3 − 6x + 5x² − x³
(iv) 2x² − 8xy + 7y² − 8xy², 2xy² + 6xy − y² + 3x², 4y² − xy − x² + xy²
(v) x³ + y³ − z³ + 3xyz, − x³ + y³ + z³ − 6xyz, x³ − y³ − z³ − 8xyz
(vi) 2 + x − x² + 6x³, −6 −2x + 4x² −3x³, 2 + x², 3 − x³ + 4x − 2x²

ANSWER:

(i) Sum of the given expressions
 = (3a − 2b + 5c)+(2a + 5b − 7c)+ (− a − b + c)
Rearranging and collecting the like terms
= 3a+2a-a-2b+5b-b+5c-7c+c
= (3+2-1)a + (-2+5-1)b + (5-7+1)c
= 4a+2b-c


(ii) Sum of the given expressions
 = (8a − 6ab + 5b) + (−6a − ab − 8b) + (−4a + 2ab + 3b)
Rearranging and collecting the like terms
  =(8−6−4)a + (- 6 −1+2)ab + (5− 8+ 3)b
  = -2a-5ab+0 = -2a – 5ab

(iii) Sum of the given expressions
    = (2x³ − 3x² + 7x − 8) + (−5x³ + 2x² − 4x + 1) + ( 3 − 6x + 5x² − x³ )
      Rearranging and collecting the like terms
     =2x³−5x³ − x³ − 3x² + 2x² + 5x² +7x-4x-6x-8+1+3
     = (2-5-1)x³ +(-3+2+5)x²+(7-4-6)x-4
     = -4x³ +4x²-3x-4


(iv) Sum of the given expressions
= (2x² − 8xy + 7y² − 8xy²)+( 2xy² + 6xy − y² + 3x²)+( 4y² − xy − x² + xy² )
     Rearranging and collecting the like terms
  = 2x² +3x² − x²   + 7y² − y² +4y² − 8xy + 6xy − xy− 8xy² +2xy² + xy²
  = (2 +3− 1)x²   + (7 − 1 +4)y² + (-8 + 6 −1)xy + (− 8 +2 +1)xy²
 = 4x²   + 10y² − 3xy -5xy²

(v) Sum of the given expressions
 = (x³ + y³ − z³ + 3xyz)+(− x³ + y³ + z³ − 6xyz)+(x³ − y³ − z³ − 8xyz)
      Rearranging and collecting the like terms
   = x³ -x³ + x³ + y³ + y³ − y³ -z³ + z³ − z³ + 3xyz-6xyz-8xyz
  = (1-1+1)x³ + (1+1-1)y³ + (-1+1-1)z³ +(3-6-8)xyz
  = x³ + y³ − z³ -11xyz

(vi) Sum of the given expressions
  = (2 + x − x² + 6x³)+(−6 −2x + 4x² −3x³)+( 2 + x²)+( 3 − x³ + 4x − 2x² )
   Rearranging and collecting the like terms
  = 6x³ −3x³− x³− x² +4x²+ x²− 2x²+ x −2x+ 4x+2-6+2+3
=  (6-3-1)x³+(-1+4+1-2)x²+(1-2+4)x+1
= 2x³+2x²+3x+1

Page No 135:

Question 4:

Subtract:
(i) 5x from 2x
(ii) −xy from 6xy
(iii) 3a from 5b
(iv) −7x from 9y
(v) 10x² from −7x²
(vi) a² − b² from b² − a²

ANSWER:

Change the sign of each term of the expression that is to be subtracted and then add.

(i) Term to be subtracted = 5x
 Changing the sign of each term of the expression gives -5x.
 On adding:

   2x+(-5x) = 2x-5x
   = (2-5)x
   = -3x

(ii)  Term to be subtracted = -xy
 Changing the sign of each term of the expression gives xy.
 On adding:

   6xy+xy
   = (6+1)xy
   = 7xy

(iii) Term to be subtracted = 3a
 Changing the sign of each term of the expression gives -3a.
 On adding:

   5b+(-3a)
   = 5b-3a

(iv) Term to be subtracted = -7x
 Changing the sign of each term of the expression gives 7x.
On adding:
9y+7x


(v) Term to be subtracted = 10x²
 Changing the sign of each term of the expression gives -10x².
On adding:
−7x² + (-10x²) = −7x² −10x²
   = (−7−10)x²
   = −17x²

(vi) Term to be subtracted = a² − b²
 Changing the sign of each term of the expression gives -a² + b².
On adding:
   b² − a² + (-a² + b²) = b² − a² -a² + b²
   = (1+1)b²  +(−1-1) a²
   = 2b² − 2a²

Page No 135:

Question 5:

Subtract:
(i) 5a + 7b − 2c from 3a − 7b + 4c
(ii) a − 2b − 3c from −2a + 5b − 4c
(iii) 5x² − 3xy + y² from 7x² − 2xy − 4y²
(iv) 6x³ − 7x² + 5x − 3 from 4 − 5x + 6x² − 8x³
(v) x³ + 2x²y + 6xy² − y³ from y³ − 3xy² − 4x²y
(vi) −11x²y² + 7xy −6 from 9x²y² −6xy + 9
(vii) −2a + b + 6d from 5a − 2b − 3c

ANSWER:

Change the sign of each term of the expression that is to be subtracted and then add.

(i) Term to be subtracted = 5a + 7b − 2c
 Changing the sign of each term of the expression gives -5a -7b + 2c.
 On adding:
  (3a − 7b + 4c)+(-5a -7b + 2c ) = 3a − 7b + 4c-5a -7b + 2c
   = (3-5)a+( − 7-7)b + (4+2)c
   = -2a − 14b + 6c

(ii) Term to be subtracted = a − 2b − 3c
 Changing the sign of each term of the expression gives -a +2b + 3c.
  On adding:
  (−2a + 5b − 4c)+(-a +2b + 3c ) = −2a + 5b − 4c-a +2b + 3c
   = (−2-1)a + (5+2)b +(−4+3)c
   = −3a + 7b − c

(iii) Term to be subtracted = 5x² − 3xy + y²
 Changing the sign of each term of the expression gives  -5x² + 3xy – y².

   On adding:
(7x² − 2xy − 4y²)+(-5x² + 3xy – y²) = 7x² − 2xy − 4y²-5x² + 3xy – y²
   = (7-5)x² +(−2+3)xy +(−4-1)y²
   = 2x² +xy − 5y²


(iv) Term to be subtracted = 6x³ − 7x² + 5x − 3
 Changing the sign of each term of the expression gives  -6x³ + 7x² – 5x + 3.
  On adding:
  (4 − 5x + 6x² − 8x³)+(-6x³ + 7x² – 5x + 3) = 4 − 5x + 6x² − 8x³-6x³ + 7x² – 5x + 3
   = (-8-6)x³ +(6+7)x² +(-5- 5)x + 7
   = -14x³ + 13x² – 10x + 7

(v) Term to be subtracted = x³ + 2x²y + 6xy² − y³
 Changing the sign of each term of the expression gives  -x³ – 2x²y – 6xy² + y³.
  On adding:
(y³ − 3xy² − 4x²y)+(-x³ – 2x²y – 6xy² + y³) = y³ − 3xy² − 4x²y-x³ – 2x²y – 6xy² + y³
   = -x³ +(- 2-4)x²y +(-6-3)xy² + (1+1)y³
   = -x³ – 6x²y – 9xy² + 2y³

(vi) Term to be subtracted = −11x²y² + 7xy −6
 Changing the sign of each term of the expression gives  11x²y² -7xy +6.
  On adding:
(9x²y² −6xy + 9)+(11x²y² -7xy +6) = 9x²y² −6xy + 9+11x²y² -7xy +6
   = (9+11)x²y² (-7−6)xy + 15
   = 20x²y² −13xy +15

(vii) Term to be subtracted = −2a + b + 6d
 Changing the sign of each term of the expression gives 2a-b-6d.
 On adding:
(5a − 2b -3c)+(2a-b-6d ) = 5a − 2b -3c +2a-b-6d
   = (5+2)a+(− 2-1)b -3c -6d
   = 7a − 3b-3c -6d

Page No 135:

Question 6:

Simplify:
(i) 2p³ − 3p² + 4p − 5 − 6p³ + 2p² − 8p − 2 + 6p + 8
(ii) 2x² − xy + 6x − 4y + 5xy − 4x + 6x² + 3y
(iii) x⁴ − 6x³ + 2x − 7 + 7x³ − x + 5x² + 2 − x⁴

ANSWER:

(i) 2p³ − 3p² + 4p − 5 − 6p³ + 2p² − 8p − 2 + 6p + 8
   Rearranging and collecting the like terms
 = (2-6)p³ +(−3+2)p² + (4-8+6)p − 5-2+8
 = -4p³ −p² +2p +1

(ii) 2x² − xy + 6x − 4y + 5xy − 4x + 6x² + 3y
     Rearranging and collecting the like terms
  = (2+6)x² +(−1+5) xy + (6-4)x +(− 4+3)y
 = 8x² + 4xy + 2x − y

(iii) x⁴ − 6x³ + 2x − 7 + 7x³ − x + 5x² + 2 − x⁴
       Rearranging and collectingthe like terms
    = (1-1)x⁴ +(− 6+7)x³ + 5x² +(2-1)x-7+ 2
    = 0 +  x³ + 5x² +x-5
    = x³ + 5x² +x-5

Page No 135:

Question 7:

From the sum of 3x² − 5x + 2 and −5x² − 8x + 6, subtract 4x² − 9x + 7.

ANSWER:

Adding:
(3x² − 5x + 2) + (−5x² − 8x + 6)
Rearranging and collecting the like terms:
(3-5)x² +(− 5-8)x + 2 +6
= -2x² − 13x + 8

Subtract 4x² − 9x + 7 from -2x² − 13x + 8.

Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 4x² − 9x + 7
Changing the sign of each term of the expression gives -4x² + 9x – 7.
On adding:
   ( -2x² − 13x + 8 )+(-4x² + 9x – 7 )    = -2x² − 13x + 8 -4x² + 9x – 7
   = ( -2-4)x² +(−13+9)x + 8 -7
      = -6x² − 4x + 1

Page No 135:

Question 8:

If A = 7x² + 5xy − 9y², B = −4x² + xy + 5y² and C = 4y² − 3x² − 6xy then show that A + B + C = 0.

ANSWER:

A = 7x² + 5xy − 9y²
B = −4x² + xy + 5y² 
C = 4y² − 3x² − 6xy

Substituting the values of A, B and C in A+B+C:
= (7x² + 5xy − 9y²)+(−4x² + xy + 5y²)+(4y² − 3x² − 6xy)
= 7x² + 5xy − 9y²−4x² + xy + 5y²+4y² − 3x² − 6xy

Rearranging and collecting the like terms:
(7-4-3)x² + (5+1-6)xy +(−9+5+4)y²
= (0)x² + (0)xy + (0)y²
= 0
⇒A+B+C=0⇒A+B+C=0

Page No 135:

Question 9:

What must be added to 5x³ − 2x² + 6x + 7 to make the sum x³ + 3x² − x + 1?

ANSWER:

Let the expression to be added be X.
(5x³ − 2x² + 6x + 7)+X = (x³ + 3x² − x + 1)
X = (x³ + 3x² − x + 1) – (5x³ − 2x² + 6x + 7)
Changing the sign of each term of the expression that is to be subtracted and then adding:
X = (x³ + 3x² − x + 1) + (-5x³ + 2x² – 6x – 7)
X = x³ + 3x² − x + 1-5x³ + 2x² – 6x – 7

Rearranging and collecting the like terms:
X = (1-5)x³ + (3+2)x² +(−1-6) x + 1-7
X = -4x³ + 5x² − 7x -6

So, -4x³ + 5x² − 7x -6 must be added to 5x³ − 2x² + 6x + 7 to get the sum as x³ + 3x² − x + 1.

Page No 135:

Question 10:

Let      P = a² − b² + 2ab,      Q = a² + 4b² − 6ab,      R = b² + 6,      S = a² − 4ab      and      T = −2a² + b² − ab + a. Find P + Q + R + S − T.

ANSWER:

P = a² − b² + 2ab
Q = a² + 4b² − 6ab
R = b² + 6
S = a² − 4ab 
T = −2a² + b² − ab + a

Adding P, Q, R and S:
P+Q+R+S
= (a² − b² + 2ab)+(a² + 4b² − 6ab)+(b² + 6)+(a² − 4ab )
= a² − b² + 2ab+a² + 4b² − 6ab+b² + 6+a² − 4ab

Rearranging and collecting the like terms:
= (1+1+1)a² +(−1+4+1) b² + (2-6-4)ab+6
P+Q+R+S = 3a² +4b² – 8ab+6

To find P + Q + R + S − T, subtract T = (−2a² + b² − ab + a) from P+Q+R+S = (3a² +4b² – 8ab+6).

On changing the sign of each term of the expression that is to be subtracted and then adding:
Term to be subtracted = −2a² + b² − ab + a
Changing the sign of each term of the expression gives 2a² – b² + ab – a.
 Now add:
(3a² +4b² – 8ab+6)+(2a² – b² + ab – a) = 3a² +4b² – 8ab+6+2a² – b² + ab – a
= (3+2)a² +(4-1) b² +(-8+1) ab – a+6
 
P + Q + R + S − T = 5a² +3b² -7 ab – a+6

Page No 135:

Question 11:

What must be subtracted from a³ − 4a² + 5a − 6 to obtain a² − 2a + 1?

ANSWER:

Let the expression to be subtracted be X.
(a³ − 4a² + 5a − 6)-X = (a² − 2a + 1)
X = (a³ − 4a² + 5a − 6)- (a² − 2a + 1)
Since ‘-‘ sign precedes the parenthesis, we remove it and change the sign of each term within the parenthesis.
X = a³ − 4a² + 5a − 6- a² + 2a – 1
Rearranging and collecting the like terms:
X = a³ +(− 4-1)a² + (5+2)a − 6 – 1
X = a³ −5a² + 7a − 7
So, a³ −5a² + 7a − 7 must be subtracted from a³ − 4a² + 5a − 6 to obtain a² − 2a + 1.

Page No 135:

Question 12:

How much is a + 2a − 3c greater than 2a − 3b + c?

ANSWER:

To calculate how much is a + 2b − 3c greater than 2a − 3b + c, we have to subtract 2a − 3b + c from a + 2b − 3c.

Change the sign of each term of the expression that is to be subtracted and then add.

 Term to be subtracted = 2a − 3b + c
Changing the sign of each term of the expression gives -2a + 3b – c.
On adding:
   (a + 2b − 3c )+(-2a + 3b – c )   
   = a + 2b − 3c -2a + 3b – c
   = (1-2)a + (2+3)b +(− 3-1)c
   = -a + 5b − 4c

Page No 135:

Question 13:

How much less than x − 2y + 3z is 2x − 4y − z?

ANSWER:

To calculate how much less than x − 2y + 3z is 2x − 4y − z, we have to subtract 2x − 4y − z from x − 2y + 3z.

Change the sign of each term of the expression that is to be subtracted and then add.

 Term to be subtracted = 2x − 4y − z
Changing the sign of each term of the expression gives -2x + 4y + z.
On adding:
    (x − 2y + 3z)+(-2x + 4y + z )   
   = x − 2y + 3z-2x + 4y + z 
   = (1-2)x +(−2+4)y + (3+1)z
   = -x + 2y + 4z

Page No 135:

Question 14:

By how much does 3x² − 5x + 6 exceed x³ − x² + 4x − 1?

ANSWER:

To calculate how much does 3x² − 5x + 6 exceed x³ − x² + 4x − 1, we have to subtract x³ − x² + 4x − 1 from 3x² − 5x + 6.
Change the sign of each term of the expression that is to be subtracted and then add.

 Term to be subtracted = x³ − x² + 4x − 1
Changing the sign of each term of the expression gives -x³ + x² – 4x + 1.
On adding:
(3x² − 5x + 6)+(-x³ + x² – 4x + 1 )   
   = 3x² − 5x + 6-x³ + x² – 4x + 1
   = -x³ + (3+1)x² +(-5-4)x+6 + 1
   = -x³ +4 x² – 9x + 7

Page No 135:

Question 15:

Subtract the sum of 5x − 4y + 6z and −8x + y − 2z from the sum of 12x − y + 3z and −3x + 5y − 8z.

ANSWER:

Add 5x − 4y + 6z and −8x + y − 2z.

(5x − 4y + 6z )+(−8x + y − 2z)
= 5x − 4y + 6z −8x + y − 2z
= (5-8)x +(−4+1)y + (6-2)z
= -3x − 3y + 4z

Adding 12x − y + 3z and −3x + 5y − 8z:
(12x − y + 3z )+(−3x + 5y − 8z)
= 12x − y + 3z −3x + 5y − 8z
= (12-3)x +(−1+5)y + (3-8)z
= 9x +4y -5z

Subtract -3x − 3y + 4z from 9x +4y -5z.
Change the sign of each term of the expression that is to be subtracted and then add.

 Term to be subtracted = -3x − 3y + 4z
Changing the sign of each term of the expression gives 3x + 3y – 4z.
On adding:
  (9x +4y -5z)+(3x + 3y – 4z )   
   = 9x +4y -5z+3x + 3y – 4z   
   = (9+3)x +(4+3)y + (-5-4)z
  = 12x +7y -9z

Page No 135:

Question 16:

By how much is 2x − 3y + 4z greater than 2x + 5y − 6z + 2?

ANSWER:

To calculate how much is 2x − 3y + 4z greater than 2x + 5y − 6z + 2, we have to subtract 2x + 5y − 6z + 2 from 2x − 3y + 4z.
Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 2x + 5y − 6z + 2
Changing the sign of each term of the expression gives -2x – 5y + 6z – 2.
On adding:
    (2x − 3y + 4z )+(-2x – 5y + 6z – 2 )   
   = 2x − 3y + 4z-2x – 5y + 6z – 2   
   = (2-2)x + (-3-5)y +(4+6)z-2
   = 0-8y+10z-2
   = -8y+10z-2

Page No 135:

Question 17:

By how much does 1 exceed 2x − 3y − 4?

ANSWER:

To calculate how much does 1 exceed 2x-3y-4, we have to subtract 2x-3y-4 from 1.
Change the sign of each term of the expression to be subtracted and then add.

 Term to be subtracted = 2x-3y-4
Changing the sign of each term of the expression gives -2x+3y+4.
On adding:
   (1)+(-2x+3y+4 )   
   = 1-2x+3y+4
   = 5-2x+3y

Page No 136:

Exercise 8C

Question 1:

Simplify:
a − (b − 2a)

ANSWER:

a – (b – 2a)
Here,  ‘-‘ sign precedes the parenthesis. So, we will remove it and change the sign of each term within the parenthesis.
=a – b + 2a
=3a – b

Page No 136:

Question 2:

Simplify:
4x − (3y − x + 2z)

ANSWER:

4x − (3y − x + 2z)
Here, ‘−’ sign precedes the parenthesis. So, we will remove it and change the sign of each term within the parenthesis.
= 4x − 3y + x − 2z
= 5x − 3y − 2z

Page No 136:

Question 3:

Simplify:
(a² + b² + 2ab) − (a² + b² −2ab)

ANSWER:

(a² + b² + 2ab) − (a² + b² − 2ab)
Here, ‘−’ sign precedes the second parenthesis. So, we will remove it and change the sign of each term within the parenthesis.
 a² + b² + 2ab − a² − b² +2ab

Rearranging and collecting the like terms:
a² − a² +b² − b² + 2ab + 2ab
=(1 − 1)a² + (1− 1)b² + (2 + 2)ab
=0 + 0 + 4ab
= 4ab

Page No 136:

Question 4:

Simplify:
−3(a + b) + 4(2a − 3b) − (2a − b)

ANSWER:

−3(a + b) + 4(2a − 3b) − (2a − b)
Here, ‘−’ sign precedes the first and the third parenthesis. So, we will remove them and change the sign of each term within the two parenthesis.
= −3a − 3b + (4××2a )−(4××3b) − 2a + b
= − 3a − 3b + 8a − 12b − 2a + b

Rearranging and collecting the like terms:
 −3a + 8a − 2a − 3b − 12b + b
= (−3 + 8 − 2)a + (−3 − 12 + 1)b
= 3a −14b

Page No 136:

Question 5:

Simplify:
−4x² + {(2x² − 3) − (4 − 3x²)}

ANSWER:

−4x² + {(2x² − 3) − (4 − 3x²)}

We will first remove the innermost grouping symbol (  ) and then {  }.

∴ −4x² + {(2x² − 3) − (4 − 3x²)}
= −4x² + {2x² − 3 − 4 + 3x²}
= −4x² + {5x² − 7}
= −4x² + 5x² − 7
= x² − 7

Page No 136:

Question 6:

Simplify:
−2(x² − y² + xy) −3(x² + y² − xy)

ANSWER:

−2(x² − y² + xy) −3(x² + y² − xy)
Here a ‘−’ sign precedes both the parenthesis. So, we will remove them and change the sign of each term within the two parenthesis.
= −2x² +2y² − 2xy −3x² − 3y² + 3xy
= (−2 − 3)x² +(2 − 3)y² + (− 2 + 3)xy
= −5x² − y² + xy

Page No 136:

Question 7:

Simplify:
a − [2b − {3a − (2b − 3c)}]

ANSWER:

a − [2b − {3a − (2b − 3c)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ a − [2b − {3a − (2b − 3c)}]
= a − [2b − {3a − 2b + 3c}]
= a − [2b − 3a + 2b − 3c]
= a − [4b − 3a − 3c]
= a − 4b + 3a + 3c
= 4a − 4b + 3c

Page No 136:

Question 8:

Simplify:
−x + [5y − {x − (5y − 2x)}]

ANSWER:

−x + [5y − {x − (5y − 2x)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ −x + [5y − {x − (5y − 2x)}]
= −x + [5y − {x − 5y + 2x}]
= −x + [5y − {3x − 5y}]
= −x + [5y − 3x + 5y]
= −x + [10y − 3x]
= −x + 10y − 3x
=  − 4x + 10y

Page No 137:

Question 9:

Simplify:
86 − [15x − 7(6x − 9) −2{10x − 5(2 − 3x)}]

ANSWER:

86 − [15x − 7(6x − 9) −2{10x − 5(2 − 3x)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 86 − [15x − 7(6x − 9) −2{10x − 5(2 − 3x)}]
= 86 − [15x − 42x + 63 −2{10x − 10 + 15x}]
= 86 − [15x − 42x + 63 −2{25x − 10}]
= 86 − [15x − 42x + 63 −50x + 20]
= 86 − [− 77x + 83 ]
= 86 + 77x − 83
= 77x + 3

Page No 137:

Question 10:

Simplify:
12x − [3x³ + 5x² − {7x² − (4 − 3x − x³) + 6x³} − 3x]

ANSWER:

12x − [3x³ + 5x² − {7x² − (4 − 3x − x³) + 6x³} − 3x]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 12x − [3x³ + 5x² − {7x² − (4 − 3x − x³) + 6x³} − 3x]
= 12x − [3x³ + 5x² − {7x² − 4 + 3x + x³+ 6x³} − 3x]
= 12x − [3x³ + 5x² − {7x² − 4 + 3x + 7x³} − 3x]
= 12x − [3x³ + 5x² − 7x² + 4 − 3x − 7x³ − 3x]
= 12x − [ − 2x² + 4 − 4x³ − 6x]
= 12x + 2x² − 4 + 4x³ + 6x
= 4x³ + 2x² +18x-4

Page No 137:

Question 11:

Simplify:
5a − [a² − {2a(1 − a + 4a²) − 3a(a² − 5a − 3)}] −8a

ANSWER:

5a − [a² − {2a(1 − a + 4a²) − 3a(a² − 5a − 3)}] −8a
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 5a − [a² − {2a(1 − a + 4a²) − 3a(a² − 5a − 3)}] −8a
= 5a − [a² − {2a − 2a² + 8a³ − 3a³ + 15a² + 9a}] −8a
= 5a − [a² − {5a³ + 13a² + 11a}] − 8a
= 5a − [a² − 5a³ − 13a² −11a] − 8a
= 5a − [ − 5a³ − 12a² − 11a] − 8a
= 5a + 5a³ + 12a² + 11a − 8a
= 5a³ + 12a² + 8a

Page No 137:

Question 12:

Simplify:
3 − [x − {2y − (5x + y − 3) + 2x²} − (x² − 3y)]

ANSWER:

3 − [x − {2y − (5x + y − 3) + 2x²} − (x² − 3y)]

We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 3 − [x − {2y − (5x + y − 3) + 2x²} − (x² − 3y)]
= 3 − [x − {2y − 5x – y + 3 + 2x²} − x² + 3y]
= 3 − [x − {y − 5x + 3 + 2x²} − x² + 3y]
= 3 − [x − y + 5x − 3 − 2x² − x² + 3y]
= 3 − [ 6x − 3 − 3x² + 2y]
= 3 − 6x + 3 + 3x² − 2y
= 3x² − 2y − 6x+6

Page No 137:

Question 13:

Simplify:
xy − [yz − zx − {yx − (3y − xz) − (xy − zy)}]

ANSWER:

xy − [yz − zx − {yx − (3y − xz) − (xy − zy)}]

We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ xy − [yz − zx − {yx − (3y − xz) − (xy − zy)}]
= xy − [yz − zx − {yx − 3y + xz − xy + zy}]
= xy − [yz − zx − {− 3y + xz + zy}]  (∵xy=yx)(∵xy=yx)
= xy − [yz − zx + 3y – xz – zy] 
= xy − [− 2zx + 3y ] (∵ yz=zy, zx=xz)(∵ yz=zy, zx=xz)
= xy + 2zx − 3y 

Page No 137:

Question 14:

Simplify:
2a − 3b − [3a − 2b − {a − c − (a − 2b)}]

ANSWER:

2a − 3b − [3a − 2b − {a − c − (a − 2b)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 2a − 3b − [3a − 2b − {a − c − (a − 2b)}]
= 2a − 3b − [3a − 2b − {a − c − a + 2b}]
= 2a − 3b − [3a − 2b − {− c  + 2b}]
= 2a − 3b − [3a − 2b + c  − 2b]
= 2a − 3b − [3a − 4b + c ]
= 2a − 3b − 3a + 4b − c
= − a + b − c

Page No 137:

Question 15:

Simplify:
−a − [a + {a + b − 2a − (a − 2b)} − b]

ANSWER:

-a − [a + {a + b − 2a − (a − 2b)} − b]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].


∴ −a − [a + {a + b − 2a − (a − 2b)} − b]
= −a − [a + {a + b − 2a − a + 2b} − b]
= −a − [a + {3b − 2a } − b]
= −a − [a + 3b − 2a  − b]
= −a − [2b − a ]
= −a − 2b + a
= −2b

Page No 137:

Question 16:

Simplify:
2a − [4b − {4a − (3b − 2a + 2b)}]2a – [4b – {4a – (3b – 2a + 2b)}]

ANSWER:

2a-[4b-{4a-(3b-2a+2b¯¯¯¯¯¯¯¯¯)}]2a-[4b-{4a-(3b-2a+2b¯)}]
We will first remove the innermost grouping symbol bar bracket. Next, we will remove (  ), followed by {  } and then [   ].

∴ 2a-[4b-{4a-(3b-2a+2b¯¯¯¯¯¯¯¯¯)}]2a-[4b-{4a-(3b-2a+2b¯)}]
= 2a-[4b-{4a-(3b-2a-2b)}]
= 2a-[4b-{4a-(b-2a)}]
= 2a-[4b-{4a-b+2a}]
=2a-[4b-{6a-b}]
= 2a-[4b-6a+b]
= 2a-[5b-6a]
= 2a-5b+6a
= 8a-5b

Page No 137:

Question 17:

Simplify:
5x − [4y − {7x − (3z − 2y) + 4z − 3(x + 3y − 2z)}]

ANSWER:

5x − [4y − {7x − (3z − 2y) + 4z − 3(x + 3y − 2z)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 5x − [4y − {7x − (3z − 2y) + 4z − 3(x + 3y − 2z)}]
= 5x − [4y − {7x − 3z + 2y + 4z − 3x − 9y + 6z}]
= 5x − [4y − {4x + 7z − 7y}]
= 5x − [4y − 4x − 7z + 7y]
= 5x − [11y − 4x − 7z ]
= 5x − 11y + 4x + 7z
= 9x − 11y + 7z

Chapter 12 – Some Applications of Trigonometry Exercise Ex. 12.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, find the height of the wall.Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21A 1.6 m tall girl stands at a distance of 3.2 m from a lamp-post and casts a shadow of 4.8 m on the ground. Find the height of the lamp-post by using (i) trigonometric ratios (ii) property of similar triangles.Solution 21

Question 22A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30^o to 60^o as he walks towards the building. Find the distance he walked towards the building.Solution 22

Question 23The shadow of a tower standing on a level ground is found to be 40 m longer when Sun’saltitude is 30^o. Find the height of the tower.Solution 23

Question 24From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45^o and 60^o respectively. Find the height of the tower.Solution 24

Let BC be the building, AB be the transmission tower, and D be the point on ground from where elevation angles are to be measured.

Question 25The angles of depression of the top and bottom of 8 m tall building from the top of a multistoried building are 30^o and 45^o respectively. Find the height of the multistoried building and the distance between the two buildings.Solution 25

Question 26A statue, 1.6 m tall, stands on a top of pedestal, from a point on the ground, the angle of elevation of the top of statue is 60^o and from the same point the angle of elevation of the top of the pedestal is 45^o. Find the height of the pedestal.Solution 26

Let AB be the statue, BC be the pedestal and D be the point on ground from where elevation angles are to be measured.

Question 27A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower the angle of elevation of the top of the tower is 60^o. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30^o. Find the height of the tower and the width of the canal.

Solution 27

Question 28From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60^o and the angle of depression of its foot is 45^o. Determine the height of the tower.Solution 28

Question 29As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30^o and 45^o. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.Solution 29

Let AB be the lighthouse and the two ships be at point C and D respectively.

Question 30The angle of elevation of the top of a building from the foot of the tower is 30^o and the angle of elevation of the top of the tower from the foot of the building is 60^o. If the tower is 50 m high, find the height of the building.Solution 30

Question 31From a point on a bridge across a river the angles of depression of the banks on opposite side of the river are 30^o and 45^o respectively. If bridge is at the height of 30 m from the banks, find the width of the river.Solution 31

Question 32Two poles of equal heights are standing opposite each other an either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60^o and 30^o, respectively. Find the height of poles and the distance of the point from the poles.Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

An aeroplane is flying at a height of 210 m. Flying at this height at some instant the angles of depression of two points in a line in opposite directions on both the banks of the river are 45° and 60°. Find the width of the river.

Solution 40

Question 41

The angle of elevation of the top of a chimney from the top of the tower is 60° and the angle of depression of the foot of the chimney from the top of the tower is 30°. If the height of the tower is 40, find the height of the chimney. According to pollution control norms, the minimum height of a smoke emitting chimney should be 100m. State if the height of the above mentioned chimney meets the pollution norms. What value is discussed in this question?Solution 41

Let AC = h be the height of the chimney.

Height of the tower = DE = BC = 40 m

In ∆ABE,

∴AB = BE√3….(i)

In ∆CBE,

tan 30° = 

Substituting BE in (i),

AB = 40√3 × √3

= 120 m

Height of the chimney = AB + BC = 120 + 40 = 160 m

Yes, the height of the chimney meets the pollution control norms.Question 42

Two ships are there in the sea on either side of a light house in such away that the ships and the light house are in the same straight line. The angles of depression of two ships are observed from the top of the light house are 60° and 45°  respectively. If the height of the light house is 200 m, find the distance between the two ships.

Solution 42

Let the ships be at B and C.

In D ABD,

∴ BD = 200 m

In D ADC,

Distance between the two ships = BC = BD + DC

Question 43

The horizontal distance between two poles is 15 m. The angle of depression of the top of the first pole as seen from the top of the second pole is 30°. If the height of the second pole is 24 m, find the height of the first pole.

Solution 43

Here m∠CAB = m∠FEB = 30°.

Let BC = h m, AC = x m

In D ADE,

In D BAC,

Height of the second pole is 15.34 mQuestion 44

The angles of depression of two ships from the top of a light house and on the same side of it are found to be 45^o and 30^o respectively. If the ships are 200 m apart, find the height of the light house.Solution 44

Question 45

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.Solution 45

Let AQ be the tower and R, S respectively be the points which are 4m, 9m away from base of tower.


As the height can not be negative, the height of the tower is 6 m.Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

From the top of building AB, 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30^o and 60^o respectively. Find

(i) the horizontal distance between AB and CD.

(ii) the height of the lamp post.

(iii) the difference between the heights of the building and the lamp post.Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 56 (i)

ΔA moving boat is observed from the top of a 150m high cliff moving away from the cliff. The angle of depression of the boat changes from 60˚ to 45˚ in 2 minutes. Find the speed of the boat in m /h.Solution 56 (i)

Let AB be the cliff, so AB=150m.

C and D are positions of the boat.

DC is the distance covered in 2 min.

∠ACB = 60^o and ∠ADB = 45^o

∠ABC = 90^o

In ΔABC,

tan(∠ACB)= 

In ΔABD,

tan(∠ADB)= 

So, DC=BD – BC

  =

Now,

Question 56 (ii)

A man in a boat rowing away from a light house 100 m high takes 2 minutes to change the angle of elevation of the top of the light house from 60^o to 30^o. Find the speed of the boat in metres per minute. (use ) Solution 56 (ii)

Let AB be the lighthouse and C be the position of man initially.

Suppose, a man changes his position from C to D.

As per the question, we obtain the following figure

Let speed of the boat be x metres per minute.

Therefore, CD = 2x

Using trigonometry, we have

Also,

Hence, speed of the boat is 57.8 m.Question 57

From the top of a 120 m high tower, a man observes two cars on the opposite sides of the tower and in straight line with the base of tower with angles of depression as 60˚ and 45˚. Find the distance between the cars. Solution 57

AB is the tower.

DC is the distance between cars.

AB=120m

In ΔABC,

tan(∠ACB) = 

In ΔABD,

tan(∠ADB) = 

So, DC=BD+BC

Question 58

Two points A and B are on the same side of a tower and in the same straight line with its base. The angles of depression of these points from the top of the tower are 60˚ and 45˚ respectively. If the height of the tower is 15 m, then find the distance between these points.Solution 58

Let CD be the tower.

So CD =15m

AB is the distance between the points.

∠CAD = 60^o and ∠CBD = 45^o

∠ADC = 90^o

In ΔADC,

tan(∠CAD)= 

In ΔCBD,

tan(∠CBD)= 

So AB=BD – AD

Question 59

A fire in a building B is reported on telephone to two fire stations P and Q, 20 km apart from each other on a straight road. P observes that the fire is at an angle of 60^o to the road and Q observes that it is at an angle of 45^o to the road. Which station should send its team and how much will this team have to travel?Solution 59

Now, in triangle APB,

sin 60^o = AB/ BP

√3/2 = h/ BP

This gives

h = 14.64 kmQuestion 60

Solution 60

Question 61

A man standing on the deck of a ship, which is 8 m above water level. He observes the angle of elevation of the top of a hill as 60^o and the angle of depression of the base of the hill as 30^o. Calculate the distance of the hill from the ship and the height of the hill.Solution 61

Question 62

Solution 62

Question 63

The angle of elevation of an aeroplane from a point on the ground is 45^o. After a flight of 15 seconds, the elevation changes to 30^o. If the aeroplane is flying at a height of 3000 metres, find the speed of the aeroplane.Solution 63

Question 64

Solution 64

Question 65

Solution 65

Question 66

The angle of elevation of a stationery cloud from a point 2500 m above a lake is 15^o and the angle of depression of its reflection in lake is 45^o. What is the height of the cloud above the lake level? (Use tan 15^o = 0.268)Solution 66

Question 67

Solution 67

Question 68

Solution 68

Question 69

Solution 69

Question 70

Solution 70

Question 71

Solution 71

Question 72

Solution 72

Question 73

Solution 73

Question 74

Solution 74

Question 75

Solution 75

Question 76

From the top of a tower h metre high, the angles of depression of two objects, which are in the line with the foot of the tower are α and β (β > α). Find the distance between the two objects.Solution 76

Question 77

A window of a house is h metre above the ground. From the window, the angles of elevation and depression of the top and bottom of another house situated on the opposite side of the lane are found to be a and b respectively. Prove that the height of the house is h (1 + tan α cot β) metres.Solution 77

Question 78

The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these window are observed to be 60° and 30° respectively. Find the height of the balloon above the ground.Solution 78

Chapter 4 – Triangles Exercise Ex. 4.1

Question 1

Fill in the blanks using correct word given in the brackets:-
(i) All circles are __________. (congruent, similar)
(ii) All squares are __________. (similar, congruent)
(iii) All __________ triangles are similar. (isosceles, equilateral)

(iv) Two triangles are similar, if their corresponding angles are __________. (proportional, equal)

(v) Two triangles are similar, if their corresponding sides are __________. (proportional, equal)
(vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)

Solution 1

(i) All circles are similar.
(ii) All squares are similar.
(iii) All equilateral triangles are similar.

(iv) Two triangles are similar, if their corresponding angles are equal.

(v) Two triangles are similar, if their corresponding sides are proportional.
(vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.Question 2

Write the truth value (T/F) of each of the following statements:

(i) Any two similar figures are congruent.

(ii) Any two congruent figures are similar.

(iii) Two polygons are similar, if their corresponding sides are proportional.

(iv) Two polygons are similar, if their corresponding angles are proportional.

(v) Two triangles are similar if their corresponding sides are proportional.

(vi) Two triangles are similar if their corresponding angles are proportional

Solution 2

(i) False

(ii) True

(iii) False

(iv) False

(v) True

(vi) True

Chapter 4 – Triangles Exercise Ex. 4.2

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

Question 1(x)

Solution 1(x)

Question 1(xi)

Solution 1(xi)

Question 1(xii)

Solution 1(xii)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 3

Solution 3

Question 4

Solution 4

Question 5

In Fig 7.35, state if PQ || EF.

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Chapter 4 – Triangles Exercise Ex. 4.3

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 2

in Fig. 7.57, AE is the AE is the bisector of the exterior CAD Meeting BC produced in E. If AB = 10 cm, AC = 6 cm and BC = 12 cm, find CE.

Solution 2

Question 3

Solution 3

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 4(v)

Solution 4(v)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Chapter 4 – Triangles Exercise Ex. 4.4

Question 1(i)

In fig., if AB||CD, find the value of x.

Solution 1(i)

Question 1(ii)

In fig., if AB || CD, find the value of x.

Solution 1(ii)

Question 1(iii)

In fig., AB||CD. If OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4, find x.

Solution 1(iii)

Chapter 4 – Triangles Exercise Ex. 4.5

Question 1

Solution 1

Question 2

In Fig. 7.137, AB || QR. Find the length of PB.

Solution 2

Question 3

In Fig. 7.138, XY || BC. Find the length of XY.

Solution 3

Question 4

In a right angled triangle with sides a and b and hypotenuse c, the altitude drawn on the hypotenuse is x. Prove that ab = cx.Solution 4

We have: 

Question 5

In Fig. 7.140, ABC = 90^o and BD AC. If BD = 8 cm and AD = 4 cm, find CD.

Solution 5

Question 6

In Fig. 7.140, ABC = 90^o and BD AC> If AB = 5.7 cm , BD = 3.8 cm and CD = 5.4 cm, find BC.

Solution 6

Question 7

In fig. 7.141, DE || BC such that AE = (1/4) AC. If AB = 6 cm, find AD

Solution 7

Question 8

Solution 8

Question 9

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using similarity criterion for two triangles, show that  Solution 9

Question 10

If ABC and  AMP are two right triangles, right angled at B and M respectively such that  MAP =  BAC. Prove that

Solution 10

Question 11

A vertical stick 10 cm long casts a shadow 8 cm long. At the same time a tower casts a shadow 30 m long. Determine the height of the tower.Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q. Prove that the rectangle obtained by BP and DQ is equal to the rectangle contained by AB and BC.Solution 17

Question 18

In ABC, AL and CM are the perpendiculars from the vertices A anf C to BC and AB respectively. If AL and CM intersect at O, prove that:

(i) 

(ii) Solution 18

Question 19

ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the mid-points of AB, AC, CD and BD respectively, show that PQRS is a rhombus.Solution 19

Question 20

In an isosceles ABC, the base AB is produced both the ways to P and Q such that AP  BQ = AC². Prove that .Solution 20

Question 21

A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2m/sec. If the lamp is 3.6m above the ground, find the length of her shadow after 4 seconds.Solution 21

Question 22

A vertical stick of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.Solution 22

Question 23

In fig. 7.144, ΔABC is right angled at C and DE AB. prove that ΔABC  ΔADE and hence find the lengths of AE and DE.

Solution 23

Question 24

Solution 24

Question 25

In Fig. 7.144, We have AB||CD||EF, if AB = 6 cm, CD = x cm, EF = 10 cm, BD = 4 cm and DE = y cm, calculate the values of x and y.

Solution 25

Chapter 4 – Triangles Exercise Ex. 4.6

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 2

Solution 2

Question 3

The areas of two similar traingles are 81 cm² and 49 cm² respectively. Find the ratio of their corresponding heights. What is the ratio of their corresponding medians?Solution 3

Question 4

The areas of two similar triangles are 169 cm² and 121 cm² respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.?Solution 4

Question 5

The areas of two similar triangles are 25 cm² and 36 cm² respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.Solution 5

Question 6

The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.Solution 6

Question 7

Solution 7

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 9

Solution 9

Question 10

Solution 10

Question 11

The areas of two similar triangles are 121 cm² and 64 cm² respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other.Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 19

In Fig.7.180, ΔABC and ΔDBC are two triangles on the same base BC. If AD intersects BC at O,

show that    

Solution 19

Since ABC and DBC are one same base,
Therefore ratio between their areas will be as ratio of their heights.
Let us draw two perpendiculars AP and DM on line BC.


In APO and DMO,
APO = DMO    (Each is90^o)
AOP = DOM          (vertically opposite angles)
OAP = ODM         (remaining angle)
Therefore APO ~  DMO    (By AAA rule)
Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 18

Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3RS. Find the ratio of the areas of triangles POQ and ROS.Solution 18

In trapezium PQRS, PQ || RS and PQ = 3RS.

 … (i)

In ∆POQ and ∆ROS,

∠SOR = ∠QOP … [Vertically opposite angles]

∠SRP = ∠RPQ … [Alternate angles]

∴ ∆POQ ∼ ∆ROS … [By AA similarity criteria]

Using the property of area of areas of similar triangles, we have

Hence, the ratio of the areas of triangles POQ and ROS is 9:1. 

Chapter 4 – Triangles Exercise Ex. 4.7

Question 1

Solution 1

Question 2(i)

The sides of a triangle are a = 7 cm, b = 24 cm and c = 25 cm. Determine whether it is a right triangle.Solution 2(i)

Question 2(ii)

The sides of a triangle are a = 9 cm, b = 16 cm and c = 18 cm. Determine whether it is a right triangle.Solution 2(ii)

Question 2(iii)

The sides of a triangle are a = 1.6 cm, b = 3.8 cm and c = 4 cm. Determine whether it is a right triangle.Solution 2(iii)

Question 2(iv)

The sides of a triangle are a = 8 cm, b = 10 cm and c = 6 cm. Determine whether it is a right triangle.Solution 2(iv)

Question 3

A man goes 15 metres due west and then 8 metres due north. How far is he from the starting point?Solution 3

Question 4

A ladder 17 m long reaches a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building. Solution 4

Question 5

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.Solution 5

Let CD and AB be the poles of height 11 and 6 m.
Therefore CP = 11 – 6 = 5 m
From the figure we may observe that AP = 12m
In triangle APC, by applying Pythagoras theorem

Therefore distance between their tops = 13 m.Question 6

In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC.Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Using pythagoras theorem determine the length of AD terms of b and c shown in Fig. 7.221.

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

In Fig. 7.222, B<90^o and segment AD BC, show that

Solution 17

(i)

Question 18

Solution 18

Question 19

ABD is a right triangle right angled at A and AC  BD. Show that
(i)    AB² = BC . BD
(ii)    AC² = BC . DC
(iii)    AD² = BD . CD

(iv) AB²/ AC² = BD/ DCSolution 19

Question 20

A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?Solution 20

Question 21

Determine whether the triangle having sides (a – 1) cm,  cm and (a + 1) cm is a right angled triangle.Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

In Fig. 7.223, D is the mid-point of side BC and AE  BC. If

Solution 24

Question 25

Solution 25

(i)

Question 26

Solution 26

Question 27

Solution 27

Question 28

An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after Solution 28