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
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