RD SHARMA SOLUTION CHAPTER – 6 Algebraic Expressions and Identities | CLASS 8TH MATHEMATICS-EDUGROWN

Exercise 6.1

Question 1.
Identify the terms, their co-efficients for each of the following expressions.
(i) 7x2yz – 5xy
(ii) x2 + x + 1
(iii)3x2y2 – 5x2y2 + z2 + z2
(iv) 9 – ab + be-ca
(v) a2+b2-ab
(vi)2x – 0.3xy + 0.5y
Solution:
(i) Co-efficient of 7x2yz = 7
co-efficient of -5xy = -5
(ii) Co-efficient of x1 = 1
co-efficient of x = 1
co-efficient of 1 = 1
(iii) Co-efficient of 3x2_y2 = 3
co-efficient of -5x2y2z2 = -5
co-efficient of z2 – 1
(iv) Co-efficient of 9 = 9
co-efficient of -ab = -1
co-efficient of be = 1
co-efficient of -ca = -1
(v) Co-efficient of a2=12
Co-efficient of b2=12
co-efficient of -ab = -1
(vi) co-efficient of 0.2x = 0.2
co-efficient of-0.3xy = -0.3
co-efficient of 0.5y = 0.5

Question 2.
Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any category ?
(i) x+y
(ii) 1000
(iii) x + x2 + x3 + x4
(iv) 7 + a + 5b
(v) 2b – 3 b2    
(vi) 2y – 3y2 +4y3
(vii) 5x – 4y + 3x
(viii) 4a – 15a2
(ix) xy+yz + zt + tx

(x)   pqr
(xi) p2q + pq2       
(xii)  2p + 2 q
Solution:
Monomials are (ii), (x)
Binomials are (i), (v), (viii), (xi), (xii)
Trinomials are (iv), (vi) and (vii)
None of these are (iii) and (ix)

Exercise 6.2

Question 1.
Add the following algebraic expressions
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 1
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 2
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 3
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 4
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 5
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 6

Question 2.
Subtract:
(i) -5xy from 12xy
(ii) 2a2 from -7a2
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 7
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 8
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 9
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 10
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Question 3.
Take away :
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 12
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 13
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 14
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RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 16
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RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 18
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 19

Question 4.
Subtract 3x – 4y – 7z from the sum of x – 3y + 2z and – 4X + 9y- 11z.
Solution:
Sum of x – 3y + 2z and – 4x + 9y – 11z
= x – 3y + 2z + (- 4x + 9y – 11z)
= x – 3y + 2z – 4x + 9y – 11z
= x – 4x – 3y + 9y + 2z – 11z
= – 3x + 6y – 9z
Now (-3x + 6y – 9z) – (3x – 4y – 7z)
= -3x + 6y – 9z – 3x + 4y + 7z
= -3x – 3x + 6y + 4y -9z +7z
= -6x + 10y – 2z

Question 5.
Subtract the sum of 3l- 4m – 7n2 and 2l + 3m – 4n2 from the sum of 9l + 2m – 3nand -3l + m + 4n2.
Solution:
Sum of 9l + 2m – 3n2 and -3l + m + 4n2
= 9l + 2m – 3 n2 + (-3l) + m + 4n2
= 9l + 2m – 3n2 – 3l + m + 4n2
= 9l- 3l+ 2m + m – 3 n2 + 4n2
= 6l + 3m + n2
and sum of 3l – 4m – 7n2 and 2l +3m- 4n2
= 3l- 4m – 7n2 + 2l+ 3m- 4n2
= 3l + 2l – 4m + 3m- 7n2 – 4n2
= 5l -m- 11n2
Now (6l + 3m + n2) – (5l – m – 11n2)
= 6l + 3m + n2 – 5l + m + 11n2
= 6l – 5l + 3m + m + n2 + 11n2
= l + 4m+ 12n2

Question 6.
Subtract the sum of 2x – x2 + 5 and -4x – 3 + 7x2 from 5.
Solution:
5 – (2x-x2 + 5-4x-3 + 7x2)
= 5 – (2x – 4x- x2 + 7x2 + 5-3)
= 5 – (-2x + 6x2 + 2)
= 5 + 2x – 6x2 – 2
= – 6x2+2x+3
= 3 + 2x – 6x2

Question 7.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 20
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 21
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 22
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 23

Exercise 6.3

Find each of the following products (1-8)
Question 1.
5x2 x 4x3
Solution:
5x2 x 4x3 = 5 x 4 x x2 x x3
= 20x2 + 3 = 20xs

Question 2.
3a2 x 4b4
Solution:
-3a2 x 4b4 = -3 x 4 x a2b4
= -12a2b4

Question 3.
(-5xy) x (-3x2yz)
Solution:
(-5xy) x (-3x2yz)
= (-5) x (-3)xy x x2yz
= 15x1 + 2xy1+ 1z= 15x3y2z

Question 4.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 1
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 2
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 3

Question 5.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 4
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 5

Question 6.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 6
Solution:

Question 7.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 7
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 8
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 9

Question 8.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 10
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 11

Find each of the following products : (9-17)

Question 9.
(7ab) x (-5ab2c) x (6abc2)
Solution:
(7ab) x (-5ab2c) x (6abc2)
= 7 x (-5) x 6 x a x a x a x b x b2 x b x c x c2
=-210 x a1+1+1 x b1+2+1x c1+2
=-210 x a3b4c3

Question 10.
(-5a) x (-10a2) x (-2a3)
Solution:
(-5a) x (-10a2) x (-2a3)
= (-5) (-10) (-2) x a x a2 x a3
= -100a1 + 2 + 3 = -100a6

Question 11.
(-4x2) x (-6xy2) x (-3yz2)
Solution:
(-4x2) x (-6xy2) x (-3yz2)
= (-4) x (-6) x (-3) x2 x x x y2 x y xz2
= -72x2+1 x y2+1 x z2
= 72x3y3z3

Question 12.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 12
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 13

Question 13.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 14
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 15

Question 14.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 16
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 17

Question 15.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 18
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 19

Question 16.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 20
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 21
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 22

Question 17.
(2.3xy) x (0.1x) x (0.16)
Solution:
(2.3xy) x (0.1x) x (0.16)
= 2.3 x 0.1 x 0.16 x x x x x y
= 0.0368x1 +1 x y = 0.0368x2y

Express each of the following products as a monomials and verify the result in each case for x = 1 : (18 -26)

Question 18.
(3x) x (4x) x (-5x)
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 23

Question 19.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 24
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 25
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 26

Question 20.
(5x4) x (x2)3 x (2x)2
Solution:
(5x4) x (x2)3 x (2x)2
= 5x4 x x2 x 3 x 2x x 2x
= 5x4 * x6 x 4x2 = 5 x 4 x x4 + 6 + 2
= 20x12
Verification:
L.H.S. = (5x4) x (x2)3 x (2x)2
= 5 x (1)4 x [(1)2]3 x (2 x 1)2
= 5 x 1 x (1)2 x 3x (2)2
= 5 x 16 x 22 = 5 x 1 x 4 = 20
R.H.S. = 20x12 = 20 (1)12 = 20 x 1 = 20
∴ L.H.S. = R.H.S.

Question 21.
(x2)3 x (2x) x (-4x) x 5
Solution:
(x2)3 x (2x) x (-4x) x (5)
= x2 x 3 X 2x X (-4x) X 5
= x6 x 2x x (-4x) x 5 = 2 x (-4) x 5x6+1 +1
= -40x8
Verification
L.H.S. = (x2)3 x (2x) x (-4x) x (5)
= (12)3 x (2 x 1) x (-4 x 1) x 5
= 12 x x 2 x (- 4) x 5 = 16 x 2 x (-4) x 5
= 1 x 2 x (-4) x 5 = -40
R.H.S. = -40x8 = -40 x (1)8
= -40 x 1 = -40
∴ L.H.S. = R.H.S.

Question 22.
Write down the product of -8x2y6 and – 20xy Verify the product for x = 2.5, y = 1.
Solution:
Product of -8x2y6 and -20xy
= (-8x2y6) x (-20xy)
= -8 x (-20) x2 x x x y6 x y = 160x2 + 1 x y6 + 1
= 160x3y3
Verification.
L.H.S. = (-8x2y6) x (-20xy)
= -8 x (2.5)2 x (1) x (-20 x 2.5 x 1)
= -8 x 6.25 x 1 x -20 x 2.5
= (-50) x (-50) = 2500
R.H.S. = 160 x = 160 (2.5)3 x (1)7
= 160 x 15.625 x 1 =2500
∴ L.H.S. = R.H.S.

Question 23.
Evaluate : (3.2x6y3) x (2.1x2y2) when x = 1 and y = 0.5.
Solution:
3.2x6y3 x 2.1x2y2
= 3.2 x 2.1 x x6+2 x y3+2
= 6.72x8y5 = 6.72 x (1)8 x (0.5)5
= 6.72 x 1 x 0.03125
= 0.21

Question 24.
Find the value of (5x6) x (-1.5x2y3) x (-12xy2) when x = 1, y = 0.5.
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 27

Question 25.
Evaluate : (2.3a5b2) x (1.2a2b2) when a = 1, b = 0.5.
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 28

Question 26.
Evaluate : (-8x2y6) x (-20xy) for x = 2.5 and y = 1.
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 29

Express each of the following products as a monomials and verify the result for x = 1,y = 2: (27-31)

Question 27.
(-xy3) x (yx3) x (xy)
Solution:
(-xy3) x (yx3) x (xy)
= -x x xx x x yx y x y = -x3 + 1 x y3 + 1 + = -x5y5
Verification:
L.H.S. = (-xy3) x (yx3) x (xy)
= (-1 x 23) x [2 x (1)3] x (1 X 2)
= (-1 x 8) x (2 x 1) x (1 x 2)
= -8 x 2 x 2 = -32
R.H.S. =-x5y5  = -(1)5 (2)5
= -1 x 32 =-32
∴ L.H.S. = R.H.S.

Question 28.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 30
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 31
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 32

Question 29.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 33
Solution:
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Question 30.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 35
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 36

Question 31.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 37
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 38
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 39

Question 32.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 40
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 41
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 42

Question 33.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 43
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 44

Exercise 6.4

Find the following products 

Question 1.
2a3 (3a + 5b)
Solution:
2a3 (3a + 5b) = 2a3 x 3a + 2a3 x 5b
= 6a3 +1 + 10a3b
= 6a4 + 10a3b

Question 2.
-11a (3a + 2b)
Solution:
-11a (3a + 2b) = -11a x 3a – 11a x 2b
= -33a2– 22ab

Question 3.
-5a (7a – 2b)
Solution:
-5a (7a – 2b) = -5a x 7a- 5a x (-2b)
= -35a2 + 10ab

Question 4.
-11y2 (3y + 7)
Solution:
-11y2 (3y + 7) = -11y2 x 3y – 11y2 x 7
= -33y2+1-77y2
= 33y3-77y2

Question 5.
6×5 (x3+y3)
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 1

Question 6.
xy (x3-y3)
Solution:
xy (x3 – y3) =xy x x3 – xy x y3
= x1 + 3 x y – x x y1+3
= x4y – xy4

Question 7.
0.1y (0.1x5 + 0.1y)
Solution:
0.1y (0.1x5 + 0.1y) = 0.1y x 0.1x5 + 0.1y x 0.1y
= 0.01x5y + 0.01y2

Question 8.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 2
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 3

Question 9.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 4
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 5

Question 10.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 6
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 7

Question 11.
5x (10x2y – 100xy2)
Solution:
5x (10x2y – 100xy2)
= 1.5x x 10x2y – 1.5x x 100xy2
= 15x1 + 2y- 150x1+1 x y2
15 x3y- 150xy2

Question 12.
4.1xy (1.1x-y)
Solution:
4.1xy (1.1x-y) = 4.1xy x 1.1x – 4.1xy x y
= 4.51x2y-4.1xy2

Question 13.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 8
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 9

Question 14.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 10
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 11

Question 15.
43 a (a2 + 62 – 3c2)
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 12
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 13

Question 16.
Find the product 24x2 (1 – 2x) and evaluate its value for x = 3.
Solution:
24x2 (1 – 2x) = 24x2 x 1 + 24x2 x (-2x)
= 24x2 + (-48x2+1)
= 24x2 – 48x3
If x = 3, then
= 24 (3)2 – 48 (3)3
= 24 x 9-48 x 27 = 216- 1296
= -1080

Question 17.
Find the product of -3y (xy +y2) and find its value for x = 4, and y = 5.
Solution:
-3y (xy + y2) = -3y x xy – 3y x y2
= -3xy2 -3y2 +1  = -3xy2 – 3y3
If x = 4, y = 5, then
= -3 x 4 (5)2 – 3 (5)3 = -12 x 25 – 3 x 125
= -300 – 375 = – 675

Question 18.
Multiply – 32 x2y3 by (2x-y) and verify the answer for x = 1 and y = 2.
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 14
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 15

Question 19.
Multiply the monomial by the binomial and find the value of each for x = -1, y = 25 and z =05 :
(i) 15y2 (2 – 3x)
(ii) -3x (y2 + z2)
(iii) z2 (x – y)
(iv) xz (x2 + y2)
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 16
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 17

Question 20.
Simplify :
(i) 2x2 (at1 – x) – 3x (x4 + 2x) -2 (x4 – 3x2)
(ii) x3y (x2 – 2x) + 2xy (x3 – x4)
(iii) 3a2 + 2 (a + 2) – 3a (2a + 1)
(iv) x (x + 4) + 3x (2x2 – 1) + 4x2 + 4
(v) a (b-c) – b (c – a) – c (a – b)
(vi) a (b – c) + b (c – a) + c (a – b)
(vii) 4ab (a – b) – 6a2 (b – b2) -3b2 (2a2 – a) + 2ab (b-a)
(viii) x2 (x2 + 1) – x3 (x + 1) – x (x3 – x)
(ix) 2a2 + 3a (1 – 2a3) + a (a + 1)
(x) a2 (2a – 1) + 3a + a3 – 8
(xi) 32-x2 (x2 – 1) + 14-x2 (x2 + x) – 34x (x3 – 1)
(xii) a2b (a – b2) + ab2 (4ab – 2a2) – a3b (1 – 2b)
(xiii )a2b (a3 – a + 1) – ab (a4 – 2a2 + 2a) – b (a3– a2 -1)
Solution:
(i) 2x2 (x3 -x) – 3x (x4 + 2x) -2 (x4 – 3x2)
= 2xx x3-2x2x x-3x x x4-3x x 2x-2x+ 6x2
= 2x2 + 3– 2x2 +1 – 3x,1+ 4-6x,1+1 -2x4 + 6x2
= 2x5 – 2x3 – 3x5 — 6x2 – 2x4 + 6x2
= 2x5 – 3x5 – 2a4 – 2x3 + 6x2 – 6x2
= -x5 – 2x4 – 2x3 + 0
= -x5-2x4-2x3
(ii) x3y (x2 – 2x) + 2xy (x3 – x4)
= x3y x x2 – x3y x 2x + 2ay x ac3 – 2xy x x4
= x3 + 2y-2x3 + 1 y + 2x1 + 3y – 2yx4+1
= x5y – 2x4y + 2x4y – 2yx5
= -x5y
(iii) 3a2 + 2 (a + 2) – 3a (2a + 1)
= 3a2 + 2a + 4 – 6a2 – 3a
= 3a2 – 6a2 + 2a – 3a + 4
= -3a2 – a + 4
(iv) x (x + 4) + 3x (2x2 – 1) + 4x2 + 4
= x2 + 4x + 3x x 2x2 – 3x x 1 + 4x2 + 4
= x2 + 4x + 6x2 +1 – 3x + 4x2 + 4
= x2 + 4x + 6x3 – 3x + 4x2 + 4
= 6a3 + 4x2 + x2 + 4x – 3x + 4
= 6x3 + 5x2 + x + 4
(v) a (b – c)-b (c – a) – c (a – b)
= ab – ac – be + ab – ac + bc
= 2ab – 2ac
(vi) a (b – c) + b (c – a) + c (a – b)
= ab – ac + bc – ab + ac – bc
= ab – ab + bc – be + ac – ac
= 0
(vii) 4ab (a – b) – 6a2 (b – b2) -3b2 (2a2 – a) + 2ab (b – a)
= 4a2b – 4ab2 – 6a2b + 6a2b2 – 6a2b2 + 3ab+ 2ab2 – 2a2b
= 4a2b- 6a2b – 2 a2b – 4ab2 + 3 ab2 + 2ab2 + 6a2b2 – 6a2b2
= 4a2b – 8a2b – 4ab2 + 5 ab2 + 0
= – 4a2b + ab2
(viii) x2 (x2 + 1) – x3 (x + 1) – x (x3 – x)
= x2 + 2 + x2 – x3 + 1 – x3 – x1 + 3 + x1 + 1
= x4 + x2-x4-x3-x4 + x2
= x4-x4-x4-x3 + x2 + x2
= -x4 – x3 + 2x2
(ix) 2a2 + 3a (1 – 2a3) + a (a + 1)
= 2a2 + 3 a – 3 a x 2a3 + a2 + a
= 2a2 + 3a – 6a1 + 3 + a2 + a
= 2a2 + 3a – 6a4 + a2 + a
= -6a4 + 3a2 + 4a
(x) a2 (2a – 1) + 3a + a3 – 8
= 2 a2 x a – a2 x 1+3a + a3-8
= 2a3 – a2 + 3a + a3 – 8
= 2a3 + a3 – a2 + 3a – 8
= 3a3 – a2 + 3a – 8
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.4 18
(xii) a2b (a – b2) + ab2 (4ab – 2a2) – a3b (1 – 2b)
= a2b x a – a2b x b1 + ab2 x 4ab – ab1 x2a-a3b x 1 + a3b x 2b
= a2+1 b-a2b2 +1+ 4a1 +1 b2 +1 -2a2+1 b2-a3b + 2a3b1 +1
= a’b – a2b3 + 4a2b3 – 2a3b2 – a3b + 2a3b2
= a3b – a3b – a2b3 + 4a2b3 – 2a3b2 + 2a3b2
= 0 + 3a2b3 + 0 = 3 a2b3
(xiii) a2b (a3 – a + 1) – ab (a4 – 2a2 + 2a) – b (a-a2– 1)
= a2b x a3 – a2b x a + a2b – ab x a2 + ab x 2a2 – ab x 2a- ba3 + ba2 + b
= a2+ 3b – a2+1 b + a2b -a1 + 4b + 2a1 + 2b- 2a1+1 b- a3b + a2b + b
= a5b – a3b + a26 – a5b + 2a3b – 2a2b – a3b + a2b + b
= a5b – a3b + 2a3b – a36 – a3b + a2b – 2a2b + a2b + b
= a3b – a5b + 2a3b – 2a3b + 2a2b-2a2b + b
= 0 + 0 + 0 + b = b

Exercise 6.5

Question 1.
(5x + 3) by (7x + 2)
Solution:
(5x + 3) x (7x + 2)
= 5x (7x + 2) + 3 (7x + 2)
= 35x2 + 10x + 21x + 6
= 35x2 + 31x + 6

Question 2.
(2x + 8) by (x – 3)
Solution:
(2x + 8) x (x – 3)
= 2x (x – 3) + 8 (x – 3)
= 2x2 – 6x + 8x – 24
= 2x2 + 2x – 24

Question 3.
(7x +y) by (x + 5y)
Solution:
(7x + y) x (x + 5y)
= 7x (x + 5y) + y (x + 5y)
= 7x2 + 35xy + xy + 5y2
=7x2 + 36xy + 5y2

Question 4.
(a – 1) by (0.1a2 + 3)
Solution:
(a – 1) x (0.1a2 + 3)
= a (0.1a2 + 3) – 1 (0.1a2+ 3)
= 0.1a3 + 3a-0.1a2-3
= 0.1a3 – 0.1a2 + 3a-3

Question 5.
(3x2 +y2) by (2x2 + 3y2)
Solution:
(3x2+y2) x (2x2 + 3y2)
= 3x2 (2x2 + 3y2) + y2(2x2 + 3y2)
= 6x2 +2 + 9x2y2 + 2x2y2 + 3y2 + 2
= 6x4 + 11 x2y2 + 3y4

Question 6.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 1
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 2

Question 7.
(x6-y6) by (x2+y2)
Solution:
(x6 – y6) x (x2 + y2)
= x6 (x2 + y2) – y6 (x2 + y2)
= x6 x x2 + x6y2 – x2y6 -y6 x y2
= x6 + 2 + x6y2 – x2y6 – y+2
= x  + x6y2 – x2y6 – y8

Question 8.
(x2 + y2) by (3a+2b)
Solution:
(x2 + y2) x (3a + 2b)
= x2 (3a + 2b) + y2 (3a + 2b)
= 3x2a + 2x2b + 3y2a + 2y2b
3ax2 + 3av2 + 2bx2 + 2by2

Question 9.
[-3d + (-7ƒ)] by (5d +ƒ)
Solution:
[-3d + (-7ƒ)] x (5d +ƒ)
= -3d x (5d +ƒ) + (-7ƒ) x (5d +ƒ)
= -15d2-3dƒ- 35dƒ- 7ƒ2
= -15d2 – 38dƒ- 7ƒ2

Question 10.
(0.8a – 0.5b) by (1.5a -3b)
Solution:
(0.8a – 0.5b) x (1.5a-3b)
= 0.8a x (1.5a – 36) – 0.56 (1.5a -3b)
= 1.2a2 – 2.4ab – 0.75ab + 1.5b2
= 1.2a2-3.15ab+ 1.5b2

Question 11.
(2x2 y2 – 5xy2) by (x2 -y2)
Solution:
(2x2 y2 – 5xy2) x (x2 -y2)
= 2x2y2 (x2 – y2) – 5x_y2 (x2 – y2)
= 2x2y2 x x2 – 2x2y2 xy2– 5xy2 x x2 + 5x2 xy2
= 2x2 + 2 y2– 2x2 x y2 + 2– 5x1+2 y2+5xy2 + 2
= 2x4y2– 2x2y4 – 5x3y2+ 5xy4

Question 12.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5-q12
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 3

Question 13.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 4
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 5
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 6

Question 14.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 7
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 8

Question 15.
(2x2-1) by (4x3 + 5x2)
Solution:
(2x2-1)x(4x3 + 5x2)
= 2x2 x (4x3 + 5x2) – 1 (4x3 + 5x2)
= 2x2 x 4x3 + 2x2 x 5x2 – 4x3 – 5x2
= 8x2 + 3 + 10x2 + 2-4x3-5x2
= 8x5 + 10x4 – 4x3 – 5x2

Question 16.
(2xy + 3y2) (3y2 – 2)
Solution:
(2xy + 3y2) (3y2 – 2)
= 2xy x (3y2-2) + 3y2 x (3y2-2)
= 2xy x Zy2+ 2xy x (-2) + Zy2 x Zy2 – Zy2 x 2
= 6xy1 + 2– 4xy + 9y2 + 2– 6y2
= 6xy3 – 4xy + 9y4– 6y2
Find the following products and verify the result for x = -1, y = -2 :

Question 17.
(3x-5y)(x+y)
Solution:
(3x-5y)(x+y)
= 3x x (x + y) – 5y x (x + y)
= 3x x x + 3x x y-5y x x-5y x y
= 3x2 + 3xy – 5xy – 5y2
= 3x2 – 2xy – 5y2
Verfification:
x = -1,y = -2
L.H.S. = (3x-5y)(x+y)
= [3 (-1) -5 (-2)] [-1 – 2]
= (-3 + 10) (-3) = 7 x (-3) = -21
R.H.S. = 3x2 – 2xy – 5y2
= 3 (-1)2 – 2 (-1) (-2) -5 (-2)2
=3×1-4-5×4=3-4-20
= 3-24 = -21
∴ L.H.S. = R.H.S.

Question 18.
(x2y-1) (3-2x2y)
Solution:
(x2y-1) (3-2x2y)
= x2y (3 – 2x2y) -1(3-2x2y)
= x2y x 3 – x2y x 2x2y – 1 x 3 + 1 x 2x2y
= 3x2y-2x2 + 2x y1 +1-3 + 2x2y
= 3x2y – 2x4y2– 3 + 2x2y
= 3x2y + 2x2y – 2x4y2 – 3
= 5x2y – 2x4y2 – 3
Verification : (x = -1, y = -2)
L.H.S. = (x2y – 1) (3 – 2x2y)
= [(-1)2 x (-2) -1] [3 – 2 x (-1)2 x (-2)]
= [1 x (-2) -1) [3 – 2 x 1 x (-2)]
= (-2 – 1) (3 + 4) = -3 x 7 = -21
R.H.S. = 5x2y – 2x4y2 – 3
= 5 (-1)2 (-2) -2 (-1)4 (-2)2 -3
5 x 1 (-2) – 2 (1 x 4) -3
= -10-8-3 = -21
∴ L.H.S. = R.H.S

Question 19.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 9
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.5 10

Simplify :

Question 20.
x2 (x + 2y) (x – 3y)
Solution:
x2 (x + 2y) (x – 3y)
= x2 [x (x – 3y) + 2y (x – 3y)]
= x2 [x2 – 3xy + 2xy – 6y2]
= x2 [x2 – xy – 6y2)
= x2 x x2 – x2 x xy – x26y2
= x4 – x3y – 6x2y2

Question 21.
(x2 – 2y2) (x + 4y)
Solution:
(x2 – 2y2) (x + 4y) x2y2
= [x2 (x + 4y) -2y2 (x + 4y)] x2y2
= (x3 + 4x2y – 2xy2 – 8y3) x2y2
= x2y2 x x3 + x2y2 x 4x2y – 2x2y2 x xy2 – 8x2y2 x y3
= x2 +3 y2 + 4x2 + 2 y2 +1 – 2x2 +1 y2+ 2 – 8x2y2+3
= xy + 44xy3 – 2x3y4 – 8x2y5

Question 22.
a2b2 (a + 2b) (3a + b)
Solution:
a2b2 (a + 2b) (3a + b)
= a2b2 [a (3a + b).+ 2b (3a + b)]
= a2b2 [3a2 + ab + 6ab + 2b2]
= a2b2 [3a2 + lab + 2b2]
= a2b2 x 3a2 + a2b2 x 7ab + a2b2 x 2b2
= 3a2 + 2b2 + 7a2+1 b2+1+ 2a2b2 + 2
= 3a4b2 + 7a3b3 + 2a2b4

Question 23.
x2 (x-y) y2 (x + 2y)
Solution:
x2 (x -y) y2 (x + 2y)
= [x2 x x – x2 x y] [y2 x x + y2 x 2y]
= (x3 – x2y) (xy2 + 2y3)
= x3 (xy2 + 2y3) – x2y (xy2 + 2y3)
= x3 x xy2 + x3 x 2y3 – x2y x xy2 – x2y x 2y3
= x3 +1 y2 + 2x3y3 – x2 +1 y1+ 2 – 2x2y1 + 3
= x4y2 + 2x3y3 – x3y3 – 2x2y4
= x4y2 + x3y3 – 2x2y4

Question 24.
(x3 – 2x2 + 5x-7) (2x-3)
Solution:
(x3 – 2x2 + 5x – 7) (2x – 3)
= (2x – 3) (x3 – 2x2 + 5x – 7)
= 2x (x3 – 2x2 + 5x – 7) -3 (x3 – 2x2 + 5x – 7)
= 2x x x3 – 2x x 2x2 + 2x x 5x – 2x x 7 -3 x x3 – 3 x (-2x2) – 3 x 5x – 3 x (-7)
= 2x4-4x3 + 10x2– 14x-3x3 + 6x2– 15x + 21
= 2x4 – 4x3 – 3x3 + 10x2 + 6x2– 14x- 15x + 21
= 2x4-7x3 + 16x2-29x+ 21

Question 25.
(5x + 3) (x – 1) (3x – 2)
Solution:
(5x + 3) (x – 1) (3x – 2)
= (5x + 3) [x (3x – 2) -1 (3x – 2)]
= (5x + 3) [3x2 – 2x – 3x + 2]
= (5x + 3) [3x2 – 5x + 2]
= 5x (3x2 – 5x + 2) + 3 (3x2 – 5x + 2)
= (5x x 3x2 – 5x x Sx + 5x x 2)+ [3 x 3x2 + 3 x (-5x) + 3×2]
= 15x3 – 25x2 + 10x + 9x2 – 15x + 6
= 15x3 – 25x2 + 9x2 + 10x – 15x + 6
= 15x3 – 16x2 – 5x + 6

Question 26.
(5-x) (6-5x) (2-x)
Solution:
(5-x) (6-5x) (2-x)
= [5 (6 – 5x) -x (6 – 5x)] (2 – x)
= [30 – 2$x – 6x + 5x2] (2 – x)
= (30 – 3 1x + 5x2) (2-x)
= 2 (30 – 31x + 5x2) – x (30 – 31x + 5x2)
= 60 – 62x + 10x2 – 30x + 3 1x2 – 5x3
= 60 – 62x – 30x + 10x2 + 3 1x2 – 5x3
= 60 – 92x + 41x2 – 5x3

Question 27.
(2x2 + 3x – 5) (3x2 – 5x + 4)
Solution:
(2x2 + 3x – 5) (3x2 – 5x + 4)
= 2x2 (3x2 – 5x + 4) + 3x (3x2 – 5x + 4) -5 (3x2 – 5x + 4)
= 2x2 x 3x2 – 2x2 x 5x + 2x2 x 4 + 3x x 3x2 – 3x x 5x + 3x x 4 – 5 x 3x2 – 5 (-5x) -5×4
= 6x4 – 10x3 + 8x2 + 9x3 – 15x2 + 12x – 15x+ 25x-20
= 6x4 – 10x3 + 9x3 + 8x2 – 15x2 – 15x2 + 12x + 25x – 20
= 6x4 – x3 – 22x2 + 37x – 20

Question 28.
(3x – 2) (2x – 3) + (5x – 3) (x + 1)
Solution:
(3x – 2) (2x – 3) + (5x – 3) (x + 1)
= 3x (2x – 3) -2 (2x – 3) + 5x (x + 1) – 3 (x + 1)
= 6x2 – 9x – 4x + 6 + 5x2 + 5x – 3x – 3
= 6x2 + 5x2 – 9x – 4x + 5x – 3x + 6 – 3
= 11x2– 11x + 3

Question 29.
(5x – 3) (x + 2) – (2x + 5) (4x – 3)
Solution:
(5x – 3) (x + 2) – (2x + 5) (4x – 3)
= [5x (x + 2) -3 (x + 2)] – [2x (4x – 3) + 5 (4x – 3)]
= [5x2 + 1 0x – 3x – 6] – [8x2 – 6x + 20x -15]
= (5x2 + 7x – 6) – (8x2 + 14x – 15)
= 5x2 + lx – 6 – 8x2 – 14x + 15
= 5x2 – 8x2 + 7x – 14x – 6 + 15
= -3x2 – 7x + 9

Question 30.
(3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)
Solution:
(3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)
= [3x (4x + 3y) + 2y (4x + 3y)]-[2x (7x-3y)-y(7x-3y)]
= (12x2 + 9xy + 8xy + 6y2) – (14x2 – 6xy – 7xy + 3y2)
= (12x2 + 17xy + 6y2) – (14x2 – 13xy + 3y2)
= 12x2 + 17xy + 6y2 – 14x2 + 13xy – 3y2
= 12x2 – 14x2 + 17xy + 13xy + 6y2 – 3y2
= -2x2 + 30xy + 3y2
= -2x2 + 3y+ 30xy

Question 31.
(x2-3x + 2) (5x- 2) – (3x2 + 4x-5) (2x- 1)
Solution:
(x2-3x + 2) (5x- 2) – (3x2 + 4x-5) (2x- 1)
= [5x (x2 – 3x + 2) -2 (x2 – 3x + 2)] – [2x (3x2 + 4x – 5) -1 (3x2 + 4x – 5)]
= [5x3 – 15x2 + 10x – 2x2 + 6x – 4] – [6x3 + 8x2 – 10x – 3x2 – 4x + 5]
= [5x3 – 15x2 – 2x2 + 10xc + 6x – 4] – [6x3 + 8x2 – 3x2 – 10x – 4x + 5]
= (5x3 – 17x2 + 16x-4) – (6x3 + 5x2 – 14x + 5)
= 5x3 – 17x2 + 16x – 4 – 6x3 – 5x2 + 14x – 5
= 5x3 – 6x3 – 17x2 – 5x2 + 16x + 14x – 4 – 5
= -x3 – 22x2 + 30x – 9

Question 32.
x3 – 2x2 + 3x – 4) (x – 1) – (2x – 3) (x2 – x + 1)
Solution:
(x3 – 2x2 + 3x – 4) (x – 1) – (2x – 3) (x2 – x + 1)
= [x (x3 – 2x2 + 3x – 4) – 1 (x3 – 2x2 + 3x – 4)] – [2x (x2 – x + 1) – 3 (x2 – x + 1)]
= [x4 – 2x3 + 3x2 – 4x – x3 + 2x2 – 3x + 4] [2x3 – 2x2 + 2x – 3x2 + 3x – 3]
= (x4 – 2x3 – x3 + 3x2 + 2x2 – 4x – 3x + 4) (2x3 – 2x2 – 3x2 + 2x + 3x – 3)
= (x4 – 3x3 + 5x2 – 7x + 4) – (2x3 – 5x2 + 5x – 3)
= x4 – 3x3 + 5x2 – 7x + 4 – 2x3 + 5x2 – 5x + 3
= x4 – 3x3 – 2x3 + 5x2 + 5x2 – 7x – 5x + 4 + 3
= x4 – 5x3 + 10x2 – 12x + 7

Exercise 6.6

Question 1.
Write the following squares of bionomials as trinomials :
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 1
Solution:
Using the formulas
(a + b)2 = a2 + 2ab + band (a – b)2 = a2 – 2ab + b2
(i) (a + 2)2 = (a)2 + 2 x a x 2 + (2)2
{(a + b)2 = a2 + 2ab + b2}
= a2 + 4a + 4
(ii) (8a + 3b)2 = (8a)2 + 2 x 8a * 3b + (3b)= 642 + 48ab + 9 b2
(iii) (2m+ 1)2 = (2m)2 + 2 x 2m x1 + (1)2
= 4m2 + 4m + 1
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 2
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 3

Question 2.
Find the product of the following binomials :
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 4
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 5
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 6
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 7

Question 3.
Using the formula for squaring a binomial, evaluate the following :
(i) (102)2
(ii) (99)2
(iii) (1001)2

(iv) (999)2
(v) (703)
2
Solution:
(i) (102)2 = (100 + 2)2
= (100)2 + 2 x 100 x 2 + (2)2
{(a + b)2 = a2 + 2ab + b2}
= 10000 + 400 + 4 = 10404
(ii) (99)2 = (100 – 1)2
= (100)2 – 2 x 100 X 1 +(1)2
{(a – b)2 = a2 – 2ab + b2}
= 10000 -200+1
= 10001 -200 =9801
(iii) (1001 )2 = (1000 + 1)2
{(a + b)2 = a2 + 2ab + b2}
= (1000)2 + 2 x 1000 x 1 + (1)2
= 1000000 + 2000 + 1 = 1002001
(iv) (999)2 = (1000 – 1)2
{(a – b)2 = a2 – 2ab + b2}
= (1000)2 – 2 x 1000 x 1 + (1)2
= 1000000 – 2000 + 1
= 1000001 -2000 = 998001

Question 4.
Simplify the following using the formula:
(a – b) (a + b) = a2 – b2 :
(i) (82)2 (18)2
(ii) (467)2 (33)2
(iii) (79)2 (69)2
(iv) 197 x 203
(v) 113 x 87
(vi) 95 x 105
(vii) 1.8 x 2.2
(viii) 9.8 x 10.2
Solution:
(i) (82)2 – (18)2 = (82 + 18) (82 – 18)
{(a + b)(a- b) = a2 – b2} = 100 x 64 = 6400
(ii) (467)2 – (33)2 = (467 + 33) (467 – 33)
= 500 x 434 = 217000
(ii) (79)2 – (69)2 = (79 + 69) (79 – 69)
148 x 10= 1480
(iv) 197 x 203 = (200 – 3) (200 + 3)
= (200)2 – (3)2
= 40000-9 = 39991
(v) 113 x 87 = (100 + 13) (100- 13)
= (100)2 – (13)2
= 10000- 169 = 9831
(vi) 95 x 105 = (100 – 5) (100 + 5)
= (100)2 – (5)2
= 10000 – 25 = 9975
(vii) 8 x 2.2 = (2.0 – 0.2) (2.0 + 0.2)
= (2.0)2 – (0.2)2
= 4.00 – 0.04 = 3.96
(viii)9.8 x 10.2 = (10.0 – 0.2) (10.0 + 0.2)
(10.0)2 – (0.2)2
= 100.00 – 0.04 = 99.96

Question 5.
Simplify the following using the identities :
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 8
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 9
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 10

Question 6.
Find the value of x, if
(i)  4x = (52)2 – (48)2
(ii) 14x = (47)2 – (33)2
(iii)  5x = (50)2 – (40)2
Solution:
(i) 4x = (52)2 – (48)2
⇒ 4x = (52 + 48) (52 – 48)
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 11

Question 7.
If x + 1x= 20, find the value of x2+ 1×2

Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 12

Question 8.
If x – 1x = 3, find the values of x2 + 1×2 and x4 + 1×4

Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 13

Question 9.
If x2 – 1×2= 18, find the values of x+ 1x  and x– 1x
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 14

Question 10.
Ifx+y = 4 and xy = 2, find the value of x2+y2.
Solution:
x + y = 4
Squaring on both sides,
(x + y)2 = (4)2
⇒ x2 +y2 + 2xy = 16
⇒ x2+y2 + 2 x 2 = 16                       (∵ xy = 2)
⇒ x2 + y2 + 4 = 16
⇒ x2+y2 = 16 – 4= 12           ‘
∴ x2+y2 = 12

Question 11.
If x-y = 7 and xy = 9, find the value of x2+y2.
Solution:
x-y = 7
Squaring on both sides,
(x-y)2 = (7)2
⇒ x2+y2-2xy = 49
⇒ x2 + y2 – 2 x 9 = 49                    (∵ xy = 9)
⇒ x2 +y2 – 18 = 49
⇒ x2 + y2 = 49 + 18 = 67
∴ x2+y2 = 67

Question 12.
If 3x + 5y = 11 and xy = 2, find the value of 9x2 + 25y2
Solution:
3 x + 5y = 11, xy = 2
Squaring on both sides,
(3x + 5y)2 = (11)2
⇒ (3x)2 + (5y)2 + 2 x 3x x 5y = 121
⇒ 9x2 + 25y2 + 30 x 7 = 121
⇒ 9x2 + 25y2+ 30 x 2 = 121           (∵ xy = 2)
⇒ 9x2 + 25y2 + 60 = 121
⇒ 9x2 + 25y2 = 121 – 60 = 61
∴ 9x2 + 25y2 = 61

Question 13.
Find the values of the following expressions :
(i)16x2 + 24x + 9, when X’ = 745
(ii) 64x2 + 81y2 + 144xy when x = 11 and y = 43
(iii) 81x2 + 16y2-72xy, whenx= 23 andy= 34
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 15
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 16

Question 14.
If x + 1x = 9, find the values of x+ 1×4.
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 17
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 18

Question 15.
If x + 1x = 12, find the values of x–  1x.
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 19

Question 16.
If 2x + 3y = 14 and 2x – 3y = 2, find the value of xy.
Solution:
2x + 3y = 14, 2x – 3y= 2
We know that
(a + b)2 – (a – b)2 = 4ab
∴ (2x + 3y)2 – (2x – 3y)2 = 4 x 2x x 3y = 24xy
⇒ (14)2 – (2)2 = 24xy
⇒ 24xj= 196-4= 192
⇒ xy = 19224 = 8
∴  xy = 8

Question 17.
If x2 + y2 = 29 and xy = 2, find the value of
(i) x+y
(ii) x-y
(iii) x4 +y4
Solution:
x2 + y2 = 29, xy = 2
(i) (x + y)2 = x2 + y2 + 2xy
= 29 + 2×2 = 29+ 4 = 33
∴  x + y= ±√33
(ii) (x – y)2 = x2 + y2 – 2xy
= 29- 2×2 = 29- 4 = 25
∴ x-y= ±√25= ±5
(iii) x2 + y2 = 29
Squaring on both sides,
(x2 + y2)2 = (29)2
⇒ (x2)2 + (y2)2 + 2x2y2 = 841
⇒ x4 +y + 2 (xy)2 = 841
⇒ x4 + y + 2 (2)2 = 841          (∵ xy = 2)
⇒ x4 + y + 2×4 = 841
⇒ x4 + y + 8 = 841
⇒ x4 + y = 841 – 8 = 833
∴ x4 +y = 833

Question 18.
What must be added to each of the following expressions to make it a whole square ?’
(i) 4x2 – 12x + 7
(ii) 4x2 – 20x + 20
Solution:
(i) 4x2 – 12x + 7 = (2x)2 – 2x 2x x 3 + 7
In order to complete the square,
we have to add  32 – 7 = 9 – 7 = 2
∴ (2x)2 – 2 x 2x x 3 + (3)2
= (2x-3)2
∴ Number to be added = 2
(ii) 4x2 – 20x + 20
⇒ (2x)2 – 2 x 2x x 5 + 20
In order to complete the square,
we have to add (5)2 – 20 = 25 – 20 = 5
∴ (2x)2 – 2 x 2x x 5 + (5)2
= (2x – 5)2
∴ Number to be added = 5

Question 19.
Simplify :
(i) (x-y) (x + y) (x2 + y2) (x4 + y4)
(ii) (2x – 1) (2x + 1) (4x2 + 1) (16x4 + 1)
(iii) (7m – 8m)2 + (7m + 8m)2
(iv) (2.5p -5q)2 – (1.5p – 2.5q)2
(v) (m2 – n2m)2 + 2m3n2

Solution:
(i) (x – y) (x + y) (x2 + y2) (x4 +y)
= (x2 – y2) (x2 + y) (x4 + y4)
= [(x2)2 – (y2)2] (x4+y4)
= (x4-y4) (x4+y4)
= (x4)2 – (y4)2 = x8 – y8
(ii) (2x – 1) (2x + 1) (4x2 + 1) (16x4 + 1)
= [(2x)2 – (1)2] (4x2 + 1) (16x4 + 1)
= (4x2 – 1) (4x2 + 1) (16x4 + 1)
= [(4x2)2-(1)2] (16x4+ 1)
= (16x4-1) (16x4+ 1)
= (16x4)2– (1)= 256x8 – 1
(iii) (7m – 8m)2 + (7m + 8n)2
= (7m)2 + (8n)2 – 2 x 7m x 8n + (7m)2 + (8n)2 + 2 x 7m x 8n
= 49m2 + 64m2 – 112mn + 49m2 + 64m2 + 112mn
= 98 m2 + 128n2
(iv) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
= (2.5p)2 + (1.5q)2 – 2 x 2.5p x 1.5q
= [(1.5p)2 + (1.5q)2 – 2 x 1.5 p x 2.5q]
= (6.25p2 + 2.25q2 – 7.5 pq) – (2.25p2 + 6.25q2-7.5pq)
= 6.25p2 + 2.25q2 – 7.5pq – 2.25p2 – 6.25q+ 7.5pq
= 6.25p2 – 2.25p2 + 2.25g2 – 6.25q2
= 4.00P2 – 4.00q2
= 4p2 – 4q2 = 4 (p2 – q2)
(v) (m2 – n2m)2 + 2m3M2
= (m2)2 + (n2m)2 -2 x m2 x n2m + 2;m3m2
= m4 + n4m2 – 2m3n2 + 2m3n2
= m4 + n4m2 = m4 + m2n4

Question 20.
Show that :
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 20
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 21
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 22

Exercise 6.7

Question 1.
Find the following products :
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7 1
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7 2
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7 3
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.7 4

Question 2.
Evaluate the following :
(i) 102 x 106
(ii) 109 x 107
(iii) 35 x 37
(iv) 53 x 55
(v) 103 x 96
(vi) 34 x 36
(vii) 994 x 1006
Solution:
(i) 102 x 106 = (100 + 2) (100 + 6)
= (100)2 + (2 + 6) x 100 + 2 x 6
= 10000 + 800 + 12 = 10812

(ii) 109 x 107 = (100 + 9) (100 + 7)
= (100)2 + (9 + 7) x 100 + 9 x 7
=10000 + 1600 + 63 = 11663

(iii) 35 x 37 = (30 + 5) (30 + 7)
= (30)2 + (5 + 7) x 30 + 5 x 7
= 900 + 12 x 30 + 35
= 900 + 360 + 35 = 1295

(iv) 53 x 55 = (50 + 3) (50 + 5)
= (50)2 + (3 + 5) x 50 + 3 x 5
= 2500 + 8 x 50 + 15
= 2500 + 400+ 15 = 2915

(v)103 x 96 = (100 + 3) (100-4)
= (100)2 + (3 – 4) x 100 + 3 x (-4)
= 10000+ (-1) x 100-12
= 10000 – 100 – 12 = 10000 – 112 = 9888

(vi) 34 x 36 = (30 + 4) (30 + 6)
= (30)2 + (4 + 6) x 30 + 4 x 6
= 900 + 10 x 30 + 24
= 900 + 300 + 24 = 1224

(vii) 994 x 1006 = (1000 – 6) (1000 + 6)
= (1000)2 + (-6 + 6) x 1000 + (-6) x 6
= 1000000 + 0-36
= 1000000-36 = 999964

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Chapter -1 हम पंछी उन्मुक्त गगन के | Class 7th | NCERT Hindi Vasant 2 Solutions | Edugrown

NCERT Solutions for Class 7th Hindi Vasant Part 2

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Chapter -1 हम पंछी उन्मुक्त गगन के

पाठ्यपुस्तक के प्रश्न-अभ्यास

कविता से

प्रश्न 1.
हर तरह की सुख सुविधाएँ पाकर भी पक्षी पिंजरे में बंद क्यों नहीं रहना चाहते ?
उत्तर-
हर प्रकार की सुख सुविधाएँ पाकर भी पक्षी पिंजरे में बंद नहीं रहना चाहते, क्योंकि उन्हें वहाँ उड़ने की आजादी नहीं है। वे तो खुले आसमान में ऊँची उड़ान भरना, नदी-झरनों का बहता जल पीना, कड़वी निबौरियाँ खाना, पेड़ की ऊँची डाली पर झूलना, कूदना, फुदकना अपनी पसंद के अनुसार अलग-अलग ऋतुओं में फलों के दाने चुगना और क्षितिज मिलन करना ही पसंद है। यही कारण है कि हर तरह की सुख-सुविधाओं को पाकर भी पक्षी पिंजरे में बंद नहीं रहना चाहते।

प्रश्न 2.
पक्षी उन्मुक्त रहकर अपनी कौन-कौन सी इच्छाएँ पूरी करना चाहते हैं?
उत्तर-
पक्षी उन्मुक्त रहकर अपनी इन इच्छाओं को पूरा करना चाहते हैं
(क) वे खुले आसमान में उड़ना चाहते हैं।
(ख) वे अपनी गति से उड़ान भरना चाहते हैं।
(ग) नदी-झरनों का बहता जल पीना चाहते हैं।
(घ) नीम के पेड़ की कड़वी निबौरियाँ खाना चाहते हैं।
(ङ) पेड़ की सब ऊँची फुनगी पर झूलना चाहते हैं।
वे आसमान में ऊँची उड़ान भरकर अनार के दानों रूपी तारों को चुगना चाहते हैं। क्षितिज मिलन करना चाहते हैं।

प्रश्न 3.
भाव स्पष्ट कीजिए-
या तो क्षितिज मिलन बन जाता या तनती साँसों की डोरी।
उत्तर
इस पंक्ति में कवि पक्षी के माध्यम से कहना चाहता है कि यदि मैं स्वतंत्र होता तो उस असीम क्षितिज से मेरी होड़ हो जाती। मैं इन छोटे-छोटे पंखों से उड़कर या तो उस क्षितिज से जाकर मिल जाता या फिर मेरा प्राणांत हो जाता।

कविता से आगे

प्रश्न 1.
कई लोग पक्षी पालते हैं
(क) पक्षियों को पालना उचित है अथवा नहीं? अपने विचार लिखिए।
(ख) क्या आपने या आपकी जानकारी में किसी ने कभी कोई पक्षी पाला है? उसकी देखरेख किस प्रकार की जाती होगी, लिखिए।
उत्तर-
(क) हमारे दृष्टिकोण से पक्षियों को पालना उचित नहीं है, इससे हम उनकी आजादी पर रोक लगा देते हैं। उनकी इच्छाओं, सपनों तथा अरमानों पर पाबंदी लग जाता है। अतः पक्षियों को पालना सही नहीं है। उन्हें प्रकृति में स्वच्छंद विचरण करने देना चाहिए। उन्हें वहीं प्रसन्नता मिलती है।
(ख) हमारे एक पड़ोसी ने तोता पाला था। उस पड़ोसी ने उसे मेले से खरीदकर लाया था। उसके परिवार के सभी सदस्य मन से उसकी देखरेख किया करते थे। प्रतिदिन उसके पिंजरे की सफ़ाई किया करते थे। एक कटोरी में पानी पीने के लिए तथा खाने के लिए चना दिया जाता था। इसके अलावे तोते को मौसमी फल तथा मिर्च भी खाने को दिया जाता था। मेरा पडोसी घंटों उस तोते से बातें किया करता था और उसे लेकर उसे घुमाने पार्क में जाया करता था। तोते ने घर के सभी सदस्यों के नाम रट लिए थे, लेकिन तोता खाना भारी मन से खाता था। जब मैं पड़ोसी के घर पिंजरे के पास जाता था तो वह हमारी ओर आशा भरी दृष्टि से देखता था।

प्रश्न 2.
पक्षियों को पिंजरे में बंद करने से केवल उनकी आज़ादी का हनन ही नहीं होता, अपितु पर्यावरण भी प्रभावित होता है। इस विषय पर दस पंक्तियों में अपने विचार लिखिए।
उत्तर
पक्षियों को पिंजरे में बंद करके उनकी आजादी का हनन होता ही है क्योंकि उनकी प्रकृति है ‘उड़ना। पिंजरे में बंद करके हम उन्हें पराधीन बना लेते हैं। जिससे उनकी आज़ादी तो समाप्त हो ही जाती है साथ ही पर्यावरण भी प्रभावित होता है क्योंकि पर्यावरण को संतुलित करने में भी पक्षियों का सहयोग रहता है। पक्षी आहार श्रृंखला को नियमित करते हैं। जैसे-घास को टिड्डा खाता है, टिड्डे को पक्षी खाते हैं और यदि पक्षी न हों तो टिड्डों की संख्या अत्यधिक हो जाएगी जो फसलों को नष्ट कर देंगे। यदि टिड्डे न हों तो घास इतनी बढ़ जाएगी कि मनुष्य परेशान हो जाएगा।

अनुमान और कल्पना

प्रश्न 1.
क्या आपको लगता है कि मानव की वर्तमान जीवन-शैली और शहरीकरण से जुड़ी योजनाएँ पक्षियों के लिए घातक हैं? पक्षियों से रहित वातावरण में अनेक समस्याएँ उत्पन्न हो सकती हैं। इन समस्याओं से बचने के लिए हमें क्या करना चाहिए? उक्त विषय पर वाद-विवाद प्रतियोगिता का आयोजन कीजिए।
उत्तर
यह कहना गलत नहीं कि मानव की वर्तमान जीवन-शैली और शहरीकरण से जुड़ी योजनाएँ पक्षियों के लिए घातक हैं क्योंकि शहरों में औद्योगीकरण के कारण विषैली गैसें और प्रदूषित जल पक्षियों के लिए हानिकारक होता है। दूसरी ओर अधिक-से-अधिक भवन निर्माण के कारण वनों व हरियाली वाले इलाकों को काटकर बड़े-बड़े भवन बना दिए जाते हैं, जिससे पक्षियों का आश्रय स्थल समाप्त हो जाता है। साथ ही वृक्षों से प्राप्त खाद्य पदार्थ, फल-फूल आदि उन्हें नहीं मिल पाते। ऐसा होने पर उन्हें बहुत मुश्किलों का सामना करना पड़ता है।

पक्षियों से रहित वातावरण में आहार श्रृंखला प्रभावित हो जाएगी। पर्यावरण संतुलित नहीं रहेगा। इसके लिए हमें अधिक-से-अधिक वृक्ष लगाने चाहिए व बाग-बगीचों का निर्माण करना चाहिए। फैक्टरियों को भी शहरों से दूर लगाकर धुएँ व प्रदूषित जल हेतु उचित प्रबंध करने चाहिए। (नोट-इन्हीं विचारों के आधार में वाद-विवाद कीजिए)।

प्रश्न 2.
यदि आपके घर के किसी स्थान पर किसी पक्षी ने अपना आवास बनाया है और किसी कारणवश आपको अपना घर बदलना पड़ रहा है तो आप उस पक्षी के लिए किस तरह के प्रबंध करना आवश्यक समझेंगे? लिखिए।
उत्तर-
यदि हमारे घर में किसी पक्षी ने अपना घोंसला बनाया हो और किसी कारणवश हमें घर बदलना पड़ रहा हो, तो हम प्रयास करेंगे कि जब तक घोंसलों में रखे अंडों से बच्चे न निकल जाएँ और पक्षी उन्हें उड़ना न सिखा ले तब तब घोसलों को न छेड़ा जाए। यदि फिर भी घर छोड़ना अनिवार्य हुआ तो उस घर में जाने वाले नए परिवार से मिलकर यह अनुरोध करेंगे कि वे घोसलों को यथावत रहने दें और न छेड़े तथा उनका ध्यान रखें।

भाषा की बात

प्रश्न 1.
स्वर्ण-श्रृंखला और लाल किरण-सी में रेखांकित शब्द गुणवाचक विशेषण हैं। कविता से हूँढ़कर इस प्रकार के तीन और उदाहरण लिखिए।
उत्तर-
(क) कनक-तिलियाँ,
(ख) कटुक-निबौरी,
(ग) तारक-अनार

प्रश्न 2.
‘भूखे-प्यासे’ में द्वंद्व समास है। इन दोनों शब्दों के बीच लगे चिह्न को सामासिक चिह्न (-) कहते हैं। इस चिह्न से ‘और’ का संकेत मिलता है, जैसे-भूखे-प्यासे = भूखे और प्यासे।
इस प्रकार के दस अन्य उदाहरण खोजकर लिखिए।
उत्तर-
दाल-रोटी – दाल और रोटी
अन्न-जल – अन्न और जल
सुबह-शाम – सुबह और शाम
पाप-पुण्य – पाप और पुण्य
राम-लक्ष्मण – राम और लक्ष्मण
सुख-दुख – सुख और दुख
तन-मन – तन और मन
दिन-रात – दिन और रात
दूध-दही – दूध और दही
कच्चा-पक्का – कच्चा और पक्का

अन्य पाठेतर है हल प्रश्न

बहुविकल्पी प्रश्नोत्तर
(क) ‘हम पंछी उन्मुक्त गगन के’ पाठ के रचयिता हैं
(i) भवानी प्रसाद मिश्र
(ii) सर्वेश्वर दयाल सक्सेना
(iii) शिवमंगल सिंह ‘सुमन’
(iv) महादेवी वर्मा

(ख) पक्षी कहाँ का जल पीना पसंद करते हैं?
(i) नल का जल
(ii) वर्षा का जल
(iii) नदी-झरनों का जल
(iv) पिंजरे में रखी कटोरी का जल

(ग) बंधन किसका है?
(i) स्वर्ण का
(ii) श्रृंखला का
(iii) स्वर्ण श्रृंखला का
(iv) मनुष्य का

(घ) लंबी उड़ान में क्या-क्या संभावनाएँ हो सकती थीं?
(i) क्षितिज की सीमा मिल जाती
(ii) साँसों की डोरी तन जाती
(iii) ये दोनों बातें हो सकती थीं
(iv) कुछ नहीं होता

(ङ) पक्षी क्यों व्यथित हैं?
(i) क्योंकि वे बंधन में हैं।
(ii) क्योंकि वे आसमान की ऊँचाइयाँ छूने में असमर्थ हैं।
(iii) क्योंकि वे अनार के दानों रूपी तारों को चुगने में असमर्थ हैं।
(iv) उपर्युक्त सभी

उत्तर-
(क) (iii)
(ख) (iii)
(ग) (iii)
(घ) (iii)
(ङ) (iv)

अतिलघु उत्तरीय प्रश्न

(क) इस कविता तथा कवि का नाम लिखिए।
उत्तर-
कविता का नाम- ‘हम पंछी उन्मुक्त गगन के
कवि का नाम- शिवमंगल सिंह ‘सुमन’

(ख) पक्षी कैसा जीवन जीना चाहते हैं?
उत्तर-
पक्षी एक स्वतंत्र जीवन जीना चाहते हैं।

(ग) पक्षी ऊँची उड़ान के लिए क्या-क्या बलिदान देते हैं?
उत्तर-
पक्षी ऊँची उड़ान के लिए अपना घोंसला, डाली का सहारा आदि सब कुछ न्योछावर करने को तैयार हैं। उनका मानना है। कि ईश्वर ने उन्हें सुंदर पंख दिए हैं इसलिए उनकी उड़ान में कोई बाधक न बनें।

(घ) अपनी किन इच्छाओं को पूरा करने के लिए पिंजरे से आजाद होने के लिए व्याकुल हैं।
उत्तर-
पक्षी नदी-झरनों का बहता जल पीने, तेज़ गति से उड़ान भरने नीले आसमान की सीमा तक उड़ने, पेड़ की फुनगी पर झूलने, कड़वी निबौरियाँ खाने और अनार रूपी दाने चुगने के लिए पिंजरे के बाहर निकलने के लिए व्याकुल होते हैं।

लघु उत्तरीय प्रश्न

(क) पिंजरे में पक्षियों को क्या-क्या कष्ट है?
उत्तर-
पिंजरे में पक्षी खुले आसमान में उड़ान नहीं भर सकते, नदी-झरनों का बहता जल नहीं पी सकते, कड़वी निबौरियाँ नहीं खा सकते, फुदक नहीं सकते, अपने पंख नहीं फैला सकते, अनार के दानों रूपी तारों को चुग नहीं सकते। इसके अतिरिक्त पिंजरे में पक्षियों को वह वातावरण नहीं मिलता, जिसमें रहने के वे आदी हैं।

(ख) पक्षियों के सपने और अरमान क्या हैं?
उत्तर-
पक्षियों का सपना है कि वह वृक्ष की सबसे ऊँची फुनगी पर बैठकर झूला झूलें उनका अरमान है कि वे नीले आसमान में दूर-दूर तक उड़ते हुए आकाश की सीमा तक पहुँच जाएँ। इस कोशिश में क्षितिज से मुकाबला करते हुए उसका अंतिम छोर ढूंढ़ निकालें या अपने प्राण त्याग दें।

(ग) पक्षी मनुष्यों से क्या चाहते हैं?
उत्तर-
पक्षी मनुष्यों से चाहते हैं कि उसे स्वतंत्र होकर उड़ान भरने दें। वह इसके बदले अपना घोंसला और टहनी का अपना आश्रय भी देने को तैयार हैं। वे हम लोगों से यह प्रार्थना करते हैं कि उन्हें ईश्वर ने जब उड़ने के लिए पंख दिए हैं तो मानव उनकी उड़ान में विघ्न न डालें और उन्हें स्वतंत्र रूप से उड़ने दें।

(घ) यह कविता हमें किस बात के लिए प्रेरित करती है?
उत्तर-
यह कविता हमें इस बात के लिए प्रेरित करती है कि बंधन में रखकर हमें कितनी भी सुविधाएँ क्यों न दी जाएँ, सभी व्यर्थ होती हैं। स्वतंत्र जीवन में ही हम अपनी इच्छा से सभी काम कर सकते हैं, जबकि पराधीनता में दूसरों की इच्छाओं को मानना पड़ता है।

दीर्घ उत्तरीय प्रश्न

(क) पक्षी को मैदा से भरी सोने की कटोरी से कड़वी निबौरी क्यों अच्छी लगती है?
उत्तर-
परतंत्र जीवन सदैव कष्टमय होता है। ऐसे समय में मन की स्वतंत्रता समाप्त हो जाती है। स्वतंत्र जीवन में कठिनाइयाँ भी कितनी अधिक क्यों न हों, वह गुलामी के जीवन से अच्छा होता है। अतः पक्षी भी खुले में रहकर मैदा से भरी सोने की कटोरी की अपेक्षा नीम के कड़वे फल खाना अधिक पसंद करते हैं।

(ख) कवि ने इस कविता के माध्यम से हमें क्या संदेश देना चाहा है?
उत्तर-
कवि ने इस कविता के माध्यम से संदेश देना चाहा है कि पराधीन सपनेहुँ सुख नाहीं। यानी स्वतंत्रता सबसे अच्छी है। स्वतंत्र रहकर ही अपने सपने और अरमान पूरे किए जा सकते हैं। पराधीनता में सारी इच्छाएँ खत्म हो जाती हैं। पराधीन रहने से हमें अपनी मूलभूत आवश्यकताओं के लिए भी दूसँरों पर निर्भर हो जाना पड़ता है। अतः कवि ने इस कविता के माध्यम से स्वतंत्रता के महत्त्व को दर्शाया है। अतः हमें पक्षियों को बंदी बनाकर नहीं रखना चाहिए। उन्हें आजाद कर आसमान में उड़ान भरने देना चाहिए।

मूल्यपरक प्रश्न

(क) स्वतंत्रता के महत्व को लिखिए?
उत्तर-
स्वतंत्रता सर्वोपरि होता है। स्वतंत्र व्यक्ति अपनी इच्छा से अपनी आवश्यकताओं की पूर्ति कर सकता है, खा-पी सकता है,

कहीं घूम – फिर सकता है तथा विचारों को अभिव्यक्त कर सकता है। गुलामी का जीवन कष्टमय होता है। हमें अंग्रेजों ने दो सौ वर्षों तक गुलाम बनाकर रखा जिसमें हमें काफ़ी यातनाएँ झेलनी पड़ी। हमें काफ़ी संघर्ष के बाद आजादी मिली। अतः स्वतंत्रता को सँभालकर रखना हम सभी का दायित्व है। इसी प्रकार की स्वतंत्रता पक्षियों पर भी लागू होती है।

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Chapter -9 Garden Snake | Class 7th | NCERT English Honeycomb Poem Solutions | Edugrown

NCERT Solutions for Class 7 English Honeycomb Poem

Here our subject experts provided the Chapter-wise NCERT Solutions for Class 7 English as per the latest cbse guidelines. Students of class 7 can grab this opportunity. Access the links prevailing on this page and dip deep into the chapterwise NCERT English subject solutions and get knowledge on how to present the answer perfectly in the exams. Go ahead and understand how important referring to the NCERT Class 7 English solutions during preparation.

Chapter - 9 Garden Snake

Question 1.
Answer the following questions :
(i) Pick out the line that suggests that the child is afraid of snakes.
(ii) Which line shows a complete change of the child’s attitude towards snakes?
Read it aloud.
(iii) But mother says that kind is good…….” What is mother referring to?
Answer:
(i) ‘Some snakes are dangerous, they say’
(ii) ‘It’s just a harmless garden snake!’
(iii) Mother is referring to the snake that was seen by the child.

Question 2.
Find the word that refers to the snake’s movements in the grass. ….
Answer:
The word is ‘wiggles’.

Question 3.
There are four pairs of rhyming words in the poem. Say them aloud.
Answer:
Away-say; Good-food; Grass-pass; Mistake-snake.

Question 4.
A snake has no legs or feet, but it moves very fast. Can you guess how? Discuss in the group.
Answer:
The snake wiggles on its body with the help of scales.

Question 5.
Can you’recall the word used for a cobra’s long sharp teeth ? Where did you come across this word first ?
Answer:
The word ‘fang’ is used for cobra’s long, sharp teeth. I read this word in a story when I was in sixth class.

Garden Snake Introduction

The poet feels that snakes are dangerous creatures. So, he ran away on seeing the snake. But on the basis of mother’s observation he concludes that “that kind is good.” Watching from a distance, he feels that it was just a harmless garden snake.

Garden Snake Word notes
NCERT Solutions for Class 7 English Honeycomb Poem 9 Garden Snake 1
Garden Snake Complete hindi translation

Have you………. ………….snake. (Page 137)

क्या आपने किसी नेवले से साँप को लड़ते हुए या बिल में जाते हुए या नदी में तैरते हुए देखा है? अधिकतर हर व्यक्ति यह विश्वास करता है कि साँप खतरनाक होते हैं। कुछ होते हैं, कुछ नहीं होते। एक हानिरहित बगीचे के साँप पर लिखी यह कविता पढ़ो।

1. I saw…… ……………. his food. (Page 137)

मैंने साँप देखा और दौड़ा…… कुछ साँप खतरनाक होते हैं, वे कहते हैं; पर माँ कहती है उस किस्म के अच्छे होते हैं, और भोजन के लिए कीड़े-मकीड़े खाते हैं।

2. So when. ….snake! (Page 137)
इसलिए जब वह घास में टेढ़ा-मेढ़ा रेंगता है मैं एक तरफ खड़ा होऊँगा और उसे गुजरते हुए देखूगा और स्वयं को बताऊँगा, “इसमें कोई गलती नहीं है, यह एक हानिरहित बगीचे का साँप है।”

 
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Chapter -8 Meadow Surprises | Class 7th | NCERT English Honeycomb Poem Solutions | Edugrown

NCERT Solutions for Class 7 English Honeycomb Poem

Here our subject experts provided the Chapter-wise NCERT Solutions for Class 7 English as per the latest cbse guidelines. Students of class 7 can grab this opportunity. Access the links prevailing on this page and dip deep into the chapterwise NCERT English subject solutions and get knowledge on how to present the answer perfectly in the exams. Go ahead and understand how important referring to the NCERT Class 7 English solutions during preparation.

Chapter - 8 Meadow Surprises

Question 1.
Read the lines in which the following phrases occur. Then discuss with your partner the meaning of each phrase in its context.
(i) velvet grass
(ii) drinking straws
(iii) meadow houses
(iv) amazing mound
(v) fuzzy head.
Answer:
(i) The meadows are covered with velvet grass which is very soft to walk in.
(ii) The butterfly has long probosics with help of which it sucks nector.
(iii) Meadow house means the open place covered with grass where creatures like rabbits, snakes, ants etc. make burrows or mounds.
(iv) Mound is a small hill. Ants have collected soil particles which looks like a wonderful mound.
(v) Fuzzy head is the puffy tuft of the dandelion which floats in the air.

Question 2.
Which line in the poem suggests that you need a keen eye and a sharp ear to enjoy a meadow ? Read aloud the stánza that contains this line.
Answer:
The line is : “If you look and listen well.” The stanza is given below which you may read aloud: Oh! Meadows have surprises And many things to tell; You may discover these yourself, If you look and listen well.

Question 3.
Find pictures of the kinds of birds, insects and scenes mentioned in the poem.
Answer:
For self-attempt. You yourself have to collect pictures of the birds, insects and scenes mentioned in the poem.

Question 4.
Watch a tree or a plant, or walk across a field or park at the same time everyday for a week. Keep a diary of what you see and hear. At the end of the week, write a short paragraph or poem about your experiences. Put your writing up on the class bulletin board.
Answer:
For self-attempt and a class-room activity.

Meadow Surprises Introduction

The poet says the meadows offer surprises. We see a butterfly resting upon a butter cup, the rabbit when it hops and dandelion flutters when air blows. Poet wishes to explore the meadows to find about the burrows, nest and ant’s mound. But for this we have to “look and listen well.”

Meadow Surprises Word notes

NCERT Solutions for Class 7 English Honeycomb Poem 8 Meadow Surprises
Meadow Surprises Complete hindi translation

Walk across. ……………. ear.(Page 123)

एक हरे मैदान से, पार्क अथवा वृक्षों के झुरमुट से गुजरो और आप कई आश्चर्यजनक चीजें पायेंगे। एक तीक्ष्ण दृष्टि . तथा तेज कान के व्यक्ति के लिए चरागाह में कितने आश्चर्य हैं?

1. Meadows.. ……. the brook? (Page 123)

चरागाहों में आश्चर्य हैं
तुम अगर देखो तो पाओगे;
मखमली घास पर आराम से चलो
और जलधारा (की आवाज) सुनो।

2. You may……………..(Page 123)

तुम्हें शायद तितली दिखाई दे
बटरकप पर आराम करती हई
और अपनी पीने वाली नलिका को खोलती हुई
ताकि वह रस को पी सके।

3. You may…………. …………will. (Page 123)

तुम खरगोश को डरा सकते हो
जो बहुत शांत बैठा है
जबकि पहले तुम उसे न देख पाओ
जब वह कूदेगा, तुम देखोगे।

4. A dandelion……….. ……you blow (Page 123)

एक छोटा पीला फूल जिसका अस्पष्ट कोंपल
कुछ दिनों पहले सुनहरा था
वह हवा से भरपूर छाते में बदल गया है
जो हवा चलने से फड़फड़ाता है।

5. Explore the…. ……..mound. (Page 124)

चरागाहों की खोज करो,
धरती में बिल है, जो
लम्बी घास के नीचे एक घोंसला है,
चीटी का अद्भुत टीला है!

6. Oh! Meadows…….. ………………….will. (Page 124)

ओह! चरागाह में आश्चर्य हैं
और कई चीजें बताने को हैं;
तुम्हें इन्हें स्वयं खोजना है,
अगर तुम देखो और ध्यान से सुनो।

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Chapter -7 Dad and the Cat and the Tree | Class 7th | NCERT English Honeycomb Poem Solutions | Edugrown

NCERT Solutions for Class 7 English Honeycomb Poem

Here our subject experts provided the Chapter-wise NCERT Solutions for Class 7 English as per the latest cbse guidelines. Students of class 7 can grab this opportunity. Access the links prevailing on this page and dip deep into the chapterwise NCERT English subject solutions and get knowledge on how to present the answer perfectly in the exams. Go ahead and understand how important referring to the NCERT Class 7 English solutions during preparation.

Chapter -7 Dad and the Cat and the Tree

Question 1.
Why was Dad sure he wouldn’t fall ?
Answer:
Dad thought himself as a great climber. That is why he was sure that he would not fall.

Question 2.
Which phrase in the poem expresses Dad’s self-confidence best?
Answer:
It is : “A climber like me ?/ Child’s play, this is !”

Question 3.
Describe plan A and its consequences.
Answer:
Plan A was to climb the tree with the help of a ladder. Dad could not climb successfully as the ladder fell. Then he landed on the flower bed.

Question 4.
Plan C was a success. What went wrong then ?
Answer:
Plan C of Dad was a success as he was able to climb on the garden wall. But he landed upon the cat. As a result of it, he was stuck up on the tree.

Question 5.
The cat was very happy to be on the ground. Pick out the phrase used to express this idea.
Answer:
The phrase is : “Pleased as Punch to be/safe and sound.”

Question 6.
Describe the Cat and Dad situation in the beginning and at the end of the poem.
Answer:
In the beginning of the poem the cat is stuck up on the tree. But Dad is proudly standing on the ground.. While at the end of the poem, Dad is stuck up on the tree and cat is smiling and ‘smirking on the ground.

Question 7.
Why and when did Dad say each of the following ?
(i) Fail ?
(ii) Never mind
(iii)Funny joke
(iv) Rubbish
Answer:
(i) This word is uttered by Dad when mother shows fear about falling of Dad. Father is very confident of climbing. That is why he is scoffing at mother.
(ii) Dad uttered Never mind to hide his embarrassment. He said so when he fell on the flower bed from the ladder. :!!
(iii) Mother said to Dad, “Don’t fall again.” This seemed quite absurd to him. Then he said what a funny joke it was.
(iv) Dad got angry when mother asked him to be careful. She was afraid that he might not break his neck. So, Dad said “Rubbish’ in an angry mood.

Question 8.
Do you find the poem humorous? Read aloud lines which make you laugh.
Answer:
I liked the poem very much because it is so humorous. These lines made me laugh.
The cat gave a yell
And sprang to the ground,
Pleased as punch to be Safe and sound.
But poor old Dad’s
Still
Stuck
Up
The
Tree!

Dad and the Cat and the Tree Introduction

This is a humorous poem about a cat which got stuck upon the tree. Dad thought that it is a “child’s play” to free the cat. He tries but falls again and again. Ultimately he is able to be on the tree. But he lands flat on the cat. The cat jumps on the ground. But Dad is now stuck on the tree.

Dad and the Cat and the Tree Word notes

NCERT Solutions for Class 7 English Honeycomb Poem 7 Dad and the Cat and the Tree

Dad and the Cat and the Tree Complete hindi translation

Have you ever……. ……….find out. (Page 107)

क्या आपने कभी पेड़ पर चढ़ती एक बिल्ली देखी है ? कभी कभी बिल्ली इतना ऊपर चढ़ जाती है और पेड़ में फंस जाती है। बेचारी बिना सहायता के नीचे नहीं आ सकती। आप उसकी सहायता किस प्रकार करेंगे? निश्चित रूप से, कविता में पिताजी की तरह नहीं। क्या पिता जी एक अच्छे आरोही हैं ? उनकी योजनाएं क्या थी? कविता को पढ़िये और जानिये।

1. This morning…. …………..me.(Page 107)

इस सुबह एक बिल्ली ।
हमारे पेड़ में अटक गई।
पिताजी ने कहा,
“ठीक है यह मुझ पर छोड़ दो।”

2. The tree…. ………………fall (Page 107)

पेड़ हिल रहा था
पेड़ लम्बा था माँ ने कहा,
“ईश्वर के लिए आप मत गिरना।”

3. “Fall?”…….. ………………wait and see. (Page 107)

“गिरना?” पिता जी ने डाँटा,
मेरे जैसा आरोही ?
यह बच्चों का खेल है!
इंतजार करो और देखो।”

4. He got. ………bed. (Page 107)

वह सीढ़ी लाये
बगीचे के गोदाम से।
वह फिसल गई।
वे फूलों की क्यारी में गिरे।

5. “Never………..shirt. (page 108)

“कोई बात नहीं,” पिता जी ने कहा
अपने बालों और चेहरे से अपनी पतलून और कमीज से,
गन्दगी को साफ करते हुए।

6. “We’ll………O.K? (page 108)

हम उपाय ‘ब’ की कोशिश करेंगे
रास्ते से दूर हट जाओ!
माँ ने कहा, “गिरना मत
दुबारा, ठीक है न?”

7. “Fell……………….It broke.(page 108)

“दुबारा गिरूँगा?” पिता जी बोले
“मजाकिया चुटकला है!”
तब उन्होंने अपने को टहनी पर झुलाया।
वह टूट गई।

8. Dad. landed…………me! (page 108)

पिता जी जोर से गिरे
डैक के ऊपर।
माँ के कहा, “बस करो,
आप अपनी गर्दन तुड़वा लोगे!”

9. “Rubbish!”………….. me (page 109)

“क्या बकवास है!” पिताजी ने कहा,
“अब उपाय सी (c) पर काम करेंगे।”
मेरे जैसे चढ़ने वाले के लिए
आँख झपकने की तरह आसान है।”

10. Then he………………. fall (page 109)

तब वह ऊँचे चढ़े
बगीचे की दीवार पर।
अनुमान लगाओ कि क्या हुआ?
वह गिरे नहीं!

11. He gave………..cat! (page 109)

वह जोर से उछले
और सीधे पेड़ के
तने के झुके हिस्से में
सीधे बिल्ली पर जा गिरे!

12. The cat……sound (page 109)

बिल्ली जोर से चिल्लाई
और मैदान पर कूदी,
सुरक्षित व स्वस्थ ।
बहुत खुश होते हुए।

13. So it’s………….Tree (page 109)

इसलिए वह व्यंग्यात्मक हँसी से मुस्कुराई
अपनी उपलब्धि पर संतुष्ट थी,
पर बेचारे पिता जी
शांति
अटके रहे
पेड़ पर ऊपर!

 

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Chapter -6 Mystery of the Talking Fan | Class 7th | NCERT English Honeycomb Poem Solutions | Edugrown

NCERT Solutions for Class 7 English Honeycomb Poem

Here our subject experts provided the Chapter-wise NCERT Solutions for Class 7 English as per the latest cbse guidelines. Students of class 7 can grab this opportunity. Access the links prevailing on this page and dip deep into the chapterwise NCERT English subject solutions and get knowledge on how to present the answer perfectly in the exams. Go ahead and understand how important referring to the NCERT Class 7 English solutions during preparation.

Chapter -6 Mystery of the Talking Fan

Question 1.
Fans don’t talk, but it is possible to imagine that they do. What is it, then, that sounds like the fan’s chatter ?
Answer:
When the motor of the fan is not oiled, a peculiar sound is made by rotation of the blades. Poet calls it as the chatter of the fan. We too can imagine the sound created by a fan as their “talking” to us. We can interpret the sound in any way we like. It all depends on the working of our mind at that time.

Question 2.
Completed the following sentences:
(i) The chatter is electrical because…..
(ii) It is mysterious because …………….
Answer:
(i) The chatter is electrical because the fan is electrically operated or works with the passage of electricity.
(ii) It is mysterious because the poet cannot exactly interpret what the sound is saying.

Question 3.
What do you think the talking fan was demanding ?
Answer:
I think the talking fan was demanding oil. I think so because the poor fan must have been fed up of chattering continuously.

Question 4.
How does an electric fan manage to throw so much air when it is switched on?
Answer:
When electric fan is switched on, it throws so much air. It is because the blades rotate very fast and strike against the passing air.

Question 5.
Is there a “talking fan’ in your house ? Create a dialogue between the fan and the mechanic.
Answer:
Fan : At last you have come!
Mechanic : I came when I was called.
Fan : I think you were called late. I am fed up of producing meaningless noises. My prestige has also been lowered-because of my old-age ailments.
Mechanic : Don’t worry, I shall set you perfectly alright. You will feel young again.
Fan : Thanks a lot!

Mystery of the Talking Fan Introduction

The poet tells us about a fan that would chatter or make noise because of some electrical fault. The poet imagines about what the fan said. But after its motor was oiled, it ran “as still as water.” Thus the mystery of what the fan said was “spoiled” for the poet.

Mystery of the Talking Fan Word notes

NCERT Solutions for Class 7 English Honeycomb Poem 6 Mystery of the Talking Fan
Mystery of the Talking Fan Complete hindi translation

Is there …………..to read. (Page 97)

जिस कमरे में आप इस समय बैठे हो, क्या वहाँ कोई छत का पंखा है ? क्या पंखा खामोश है या शोर कर रहा है? यदि वह शोर कर रहा है, तुम्हें यह जान लेना चाहिए कि वह इस कविता में, जिसे आप पढ़ने जा रहे हैं, ‘बोलते पंखे’ का कोई दूर का रिश्तेदार है।

1. Once there……………..matter (Page 97)

एक बार एक बोलने वाला पंखा थाबिजली से उसकी चटर-पटर (आवाजें) थीं। जो वह कहता था मुझे स्पष्ट सुनाई नहीं दे रहा था। और मुझे आशा है कि इससे कोई फर्क नहीं पड़ता।

2. Because one………..water

क्योंकि एक दिन किसी ने । उसकी छोटी घूमती मोटर को तेल दिया और सारा रहस्य बेकार हो गया। वह पानी की तरह शांत चलने लगा।

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Chapter -5 Trees | Class 7th | NCERT English Honeycomb Poem Solutions | Edugrown

NCERT Solutions for Class 7 English Honeycomb Poem

Here our subject experts provided the Chapter-wise NCERT Solutions for Class 7 English as per the latest cbse guidelines. Students of class 7 can grab this opportunity. Access the links prevailing on this page and dip deep into the chapterwise NCERT English subject solutions and get knowledge on how to present the answer perfectly in the exams. Go ahead and understand how important referring to the NCERT Class 7 English solutions during preparation.

Chapter - 5 Trees

Question 1.
What are the games or human activities which use trees, or in which trees also “participate’?
Answer:
We can play many games by using trees. We can swing swings on tree. One can hide behind a tree while playing the game of hide and seek. Our kites can get caught in the trées. There are a lot of human activities in which trees also participate. Human beings can make tree-houses in trees. We can have tea parties under the trees. Human beings can enjoy cool shade under the trees in summer. Trees also provide us with various kinds of fruits. Trees provide us clean air. We get timber from trees to build our houses and furniture etc. Wood is also used for burning.

Question 2.
(i) “Trees are to make no shade in winter.” What does this mean ?
(Contrast this line with the line immediately before it.)
Answer:
As compared to summer season, sunshine in winter is less brighter. Moreover shade is not needed in winter. The people enjoy basking in the direct sunshine in winter.

(ii) “Trees are for apples to grow on, or pears.” Do you agree that one purpose of a tree is to have fruit on it ? Or do you think this line is humorous ?
Answer:
One of the most important functions of the trees is to bear and provide us fruits. It is a fact. So, this line cannot be called humorous.

Question 3.
With the help of your partner, try to rewrite some lines in the poem, or add new ones of your own as in the following examples.
Trees are for birds to build nests in.
Trees are for people to sit under.
Now try to compose a similar poem about water, or air.
Answer:
Water is the elixir of life for us.
Water is essential for our survival
Water is for fish to live in
Water is a boon for the birds
Water is for plants to trees to grow
Water is a means of transportation
Water is for clothes to be cleaned.

Trees Introduction

The poet tells us how important the trees are for all of us. In one way or the other trees are used to make tree houses, tree homes and children play various games under them. Trees provide us with shade and fruits. Trees give us timber and joy also.

Trees Word notes

NCERT Solutions for Class 7 English Honeycomb Poem 5 Trees

Trees Complete hindi translation

Part-I

Take a few ………. ……tree. (Page 83)

एक दूसरे को उन वृक्षों के नाम बताने के लिए कुछ मिनट लो जिन्हें तुम जानते हो या जिनके बारे में तुमने सुना है। उन चीजों के नाम बताओ जो वृक्ष हमें देते हैं। तब वृक्षों के बारे में यह कविता पढ़ो।

1. Trees are …………… on. (Page 83)

पेड़ पक्षियों के लिए हैं।
पेड़ बच्चों के लिए हैं।
पेड़ वृक्ष-घर बनाने के लिए हैं।
पेड़ झूले झूलने के लिए हैं।

2. Trees are…… ……..caught in. (Page 83)

पेड़ उनमें से हवा बहने के लिए हैं।
पेड़ छुपा छुपाई के खेल में छिपने के लिए हैं।
पेड़ उनके नीचे इकट्ठे चाय पीने के लिए हैं।
पेड़ पतंगें अटकने के लिए हैं।

3. Trees are to.. ……”Timber (Page 83)

पेड़ गर्मियों में शीतल छाया देने के लिए हैं।’
पेड़ सर्दियों में छाया न देने के लिए हैं।
पेड़ सेब और नाशपातियाँ उगाने के लिए हैं;
पेड़ काटकर गिराने और ‘लकड़ी’ पुकारने के लिए हैं।

4. Trees make…. ……………this fall! (Page 83)

पेड़ माताओं को यह कहने के लिए हैं,
“वाह! कितना सुन्दर चित्र चित्रण के लिए है!” .
पेड़ पिताओं को यह कहने के लिए हैं,
“इस पतझड़ में कितने पत्ते बटोरने हैं!”

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Chapter -4 Chivvy | Class 7th | NCERT English Honeycomb Poem Solutions | Edugrown

NCERT Solutions for Class 7 English Honeycomb Poem

Here our subject experts provided the Chapter-wise NCERT Solutions for Class 7 English as per the latest cbse guidelines. Students of class 7 can grab this opportunity. Access the links prevailing on this page and dip deep into the chapterwise NCERT English subject solutions and get knowledge on how to present the answer perfectly in the exams. Go ahead and understand how important referring to the NCERT Class 7 English solutions during preparation.

Chapter - 4 Chivvy

Question 1.
Discuss these questions in small groups before you answer them.
(i) When is a grown-up likely to say this?
Don’t talk with your mouth full.
Answer:
A grown up is likely to say this when a child is 2 or 3 years old. This will be said when a child is talking while eating.

(ii) When are you likely to be told this? Say thank you.
Answer:
I will be told to say thank you’ when I am about five to six years old.

(iii) When do you think an adult would say this ? No one thinks you are funny.
Answer:
I think this will be told when one starts understanding things.

Question 2.
The last two lines of the poem are not prohibitions or instructions. What is the adult now asking the child to do? Do you think the poet is suggesting that this is unreasonable ? Why ?
Answer:
The child is now growing up. He is about to enter his teens. So, he is being told to make out choices and take his own decision. I think the poet is suggesting that it is unreasonable because all the time he had been told to do this or don’t do that. So how he can think independently now.

Question 3.
Why do you think grown-ups say the kind of things mentioned in the poem ? Is it important that they teach children good manners, and how to behave in public?
Answer:
I think the grown-ups give such instructions because they want to make their children more sensible and better citizens. It is very important to teach children good manners and how to behave in public. This will make children well-behaved and useful members of society.

Question 4.
If you had to make some rules for grown-ups to follow, what would you say? Make at least five such rules. Arrange the lines as in a poem.
Answer:
Don’t nag us.
Let us think independently.
Let us take our own decisions.
Give us clean and green environment.
Guide us but don’t goad us.

Chivvy Introduction

In this poem the poet wishes to highlight the irritating effect of nagging. All the time children are given instructions as do this or don’t do that. The children can’t develop their independent thinking if too much instructions are given to them. They get fed up also.

Chivvy Word notes

NCERT Solutions for Class 7 English Honeycomb Poem 4 Chivvy

Chivvy Complete hindi translation

Part-I

Ask yourself ……… ………poem now. (Page 69)

स्वयं से तथा अपने सहपाठी से पूछो : क्या तुम्हें यह पसंद आता है हर समय यह कहा जाए कि यह करो या यह ना करो? आपके अनुभव में क्या बड़े लोग ऐसा करते हैं? जब माइकल 5 वर्ष का था, उसकी माँ उसे एक पास के विद्यालय में प्रवेश दिलाने ले गई। अध्यापक ने पूछा, “बच्चे, तुम्हारी माँ तुम्हें घर पर क्या कहकर बुलाती है?” “माइकल नहीं,” विश्वासपूर्वक उत्तर आया।
नोट : छिवी का अर्थ है आलोचना करना, “हमारे कुछ भी करने की कोशिश करते रहना,
प्रायः नाराज करने के तरीके में”, शब्दकोश के अनुसार। अब कविता पढ़ें।

1. Grown-ups say …… ….your nose.(Page 69)

बड़े लोग ऐसी बातें (चीजें) कहते हैं;
बोलो
अपना मुँह भर कर मत बोलो,
मत घूरो,
इशारा मत करो,
नाक में अंगुली मत डालो।

2. Sit up……. ……… your pockets. (Page 69)

बैठ जाओ, विनम्रता से बोलो
शोर कम करो।
अपने पीछे दरवाजा बंद करो
अपने पैर मत घसीटो
क्या तुम्हारे पास रूमाल नहीं है?
अपनी जेब से हाथ बाहर निकालो।

3. Pull your socks up.. …..anything? (Page 70)

अपनी जुराबें ऊपर करो
सीधे खड़े हो जाओ
धन्यवाद कहो
बीच में हस्तक्षेप मत करो
कोई नहीं सोचता कि तुम मज़ाकिया हो ।
अपनी कोहनियाँ मेज़ से हटाओ
क्या तुम अपना मन किसी भी चीज के बारे में नहीं बना सकते ?

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Chapter -3 The Shed | Class 7th | NCERT English Honeycomb Poem Solutions | Edugrown

NCERT Solutions for Class 7 English Honeycomb Poem

Here our subject experts provided the Chapter-wise NCERT Solutions for Class 7 English as per the latest cbse guidelines. Students of class 7 can grab this opportunity. Access the links prevailing on this page and dip deep into the chapterwise NCERT English subject solutions and get knowledge on how to present the answer perfectly in the exams. Go ahead and understand how important referring to the NCERT Class 7 English solutions during preparation.

Chapter - 3 The Shed

Question 1.
Answer the following questions :

(i) Who is the speaker in the poem ?
Answer:
The poet is the speaker.

(ii) Is she/he afraid or curious, or both ?
Answer:
He is both afraid and curious as he wants to step in the shed.

(iii) What is she/he planning to do soon ?
Answer:
The poet is planning to take a peep one day soon.

(iv) “But not just yet…..” suggests doubt, fear, hesitation, laziness or something else. Choose the word which seems right to you. Tell others why you chose it.
Answer:
The poet shows fear in his mind as well as he is hesitant to go in. His brother has made him doubtful that there is a ghost in the shed.

Question 2.
Is there a room in your house or a house in your neighbourhood/locality where you would rather not go alone, and never at night? If there is such a place and a story to go with it, let others hear all about it.
Answer:
In our locality there is a primary school in which a basement has been made to gather the waste things like blackboard, table, chair, bricks and even the torn charts etc. Nobody is allowed to enter there as it is supposed to be haunted by some reptiles. The basement has some strange kind of smell and a few rats can be found running here and there. Getting in such a place is an adventure. I’ll go there one day but not today.

The Shed Introduction

This simple poem explains what a shed is. It is a small room, away from the main house, for keeping things, tools, vehicles and even animals. However, it is a must for every big house to keep its unused belongings. Read the poem and enjoy it with your partner.

The Shed Word notes

NCERT Solutions for Class 7 English Honeycomb Poem 3 The Shed

The Shed Complete hindi translation

Do you ……….. …………..poem. (Page 48)

क्या आप जानते हो कि छप्पर क्या होता है? जैसे गाय के लिए छप्पर, औजार-शाला, लकड़ी के लिए छप्पर, यह एक छोटा सा कमरा होता है जो घर से दूर चीजें, जानवर, औजार, वाहन आदि रखने के लिए प्रयोग किया जाता है। अपने सहपाठी से पूछो कि क्या उसने कोई छप्पर देखा है। उसे कहिए कि वह उसका वर्णन कक्षा में करे। आइये कविता पढ़ें।

1. There’s a shed. …………………………….one day.(Page 48)

हमारे बगीचे में एक ओर एक छप्पर बना है जिसमें द्वार पर मकड़ी के जाले लटके हुए हैं, जिसके कब्जे जंग लगे हैं और हवा के चलने भर से चरचराहट की आवाज करते हैं। जब मैं अपने बिस्तर पर लेटता हूँ, और उसे सुनता हूँ। मैं अवश्य . एक दिन इसका दरवाजा खोलूंगा।

2. There’s a dusty…. ….one day. (Page 49)

उसकी एक ओर एक मैली पुरानी खिड़की है जिसमें तीन टूटे-फूटे शीशे हैं। मैं प्रायः सोचता हूँ कि वहाँ ऐसा कोई है जो मुझे घूर रहा है जब भी मैं उसके पास से गुजरता हूँ। मैं अवश्य एक दिन इसकी खिड़की से अन्दर झांकूगा।

3. My brother….. …………. one day. (Page 49)

मेरा भाई कहता है कि छप्पर में एक भूत है जो गले हुए फर्श के तख्तों के नीचे छिपकर बैठा है। और यदि मैंने कभी भी उनके अन्दर कदम रखने की कोशिश की, वह कूद कर बाहर आयेगा और मेरा सिर काट देगा। परंतु एक दिन अवश्य मैं अन्दर जाऊंगा।

4. I know that,…. …….. just yet. (Page 49)

मैं जानता हूँ कि वास्तव में कोई भूत नहीं होता। मेरा भाई झूठ बोलता है क्योंकि वह उसे अपनी गुफा बनाना चाहता _है; वहाँ कोई भी नहीं है जो मुझे घूर रहा है या अजीब-अजीब आवाजें निकालता है और अब मकड़ी भी अपने जाल से दूर जा चुकी है, मैं नहीं जानता कब से गई है। मैं शीघ्र ही एक दिन इस छप्पर में जाऊंगा परंतु अभी नहीं……..

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Chapter -2 The Rebel | Class 7th | NCERT English Honeycomb Poem Solutions | Edugrown

NCERT Solutions for Class 7 English Honeycomb

Here our subject experts provided the Chapter-wise NCERT Solutions for Class 7 English as per the latest cbse guidelines. Students of class 7 can grab this opportunity. Access the links prevailing on this page and dip deep into the chapterwise NCERT English subject solutions and get knowledge on how to present the answer perfectly in the exams. Go ahead and understand how important referring to the NCERT Class 7 English solutions during preparation.

Chapter - 2 The Rebel

Question 1.

Answer the following questions :
(i) If someone doesn’t wear a uniform to school, what do you think thee teacher will say ?
Answer:
The teacher will say that one should not do it again in future. He may punish the student for not wearing the uniform.

(ii) When everyone wants a clear sky, what does the rebel want most ?
Answer:
The rebel wishes for the rain.

(iii) If the rebel has a dog for a pet, what is everyone else likely to have ?
Answer:
Everyone is likely to have a kitten.

(iv) Why is it good to have rebels ?
Answer:
The rebels act as critics who always keep things under control.

(v) Why is it not good to be a rebel oneself ?
Answer:
One will always live in tense atmosphere.

(vi) Would you like to be a rebel ? If yes, why ? If not, why not?
Answer:
No, I do not want to be a rebel in general. A rebel would not get regard and respect in the surrounding.

Question 2.
Find in the poem an antoynm (a word opposite in meaning) for each of the following words.
(i) long ………………
(ii) grow …………
(iii) quietness ………….
(iv) sober. ……………..
(v) lost ………
Answer:
(i) long – short
(ii) grow – die
(iii) quietness – noisy
(iv) sober – fantastic
(v) lost – found

Question 3.
Find in the poem lines that match the following. Read both one after the other.
(i) The rebel refuses to cut his hair.
(ii) He says cats are better.
(iii) He recommends dogs.
(iv) He is unhappy because there is no sun.
(v) He is noisy on purpose.
Answer:
(i) The rebel refuses to cut his hair
when everyone has short hair.

(ii) He says cats are better
When everyone loves dog.

(iii) He recommends dog
When everyone loves cat

(iv) He is unhappy because there is no sun
When everyone wishes for sun

(v) He is noisy on purpose
When everyone wishes to be quiet.

The Rebel Introduction

A rebel is a person who fights against injustice in general. It is good to have ‘real rebels They act as critics who always keep the chain of the society move well. Of course, generally the rebels do not get regard and respect in their contemporary world but praised by the coming generation, if they give something better and fruitful.

The Rebel Word notes

NCERT Solutions for Class 7 English Honeycomb Poem 2 The Rebel

The Rebel Complete hindi translation

Do you know (Page 33)

क्या आप किसी ऐसे व्यक्ति को जानते हैं जो सदैव तुमसे या तुम्हारे दोस्तों से असहमत रहता है, या हर व्यक्ति की सोच का उलटा करना पसंद करता है? ऐसे व्यक्ति को परिभाषित करने के लिए एक शब्द सोचिये। ऐसे व्यक्ति द्वारा की गई कुछ बातों के बारे में अपने सहपाठी के विचार-विमर्श कीजिए।

1. When everybody …… ………….clothes. (Page 33)

जब सब के बाल छोटे होते हैं, विरोधी अपने बालों को लम्बा करने की सोचता है। जब सबके बाल लम्बे होते हैं, तो विरोधी (विद्रोही) अपने बाल छोटे कर लेता है। जब हर कोई पाठ के बीच बोलता है तो विद्रोही एक शब्द भी नहीं बोलता। जब पाठ के बीच में कोई नहीं बोल रहा होता तो विद्रोही परेशानी उत्पन्न कर देता है। जब प्रत्येक वर्दी पहनता है तो विद्रोही भड़कीले वस्त्रों में सजकर आता है।

2. When everybody ……………. (Pages 33-34)

जब हर कोई भड़कीले वस्त्रों को पहनता है तो विद्रोही सादे कपड़े पहनता है। कुत्तों से प्यार रखने वालों के बीच, विद्रोही बिल्लियों के लिए अपनी प्राथमिकता दिखाता है। बिल्ली को प्यार करने वालों के बीच विद्रोही कुत्तों के लिए अच्छी शब्दावली का प्रयोग करता है। जब प्रत्येक व्यक्ति सूर्य की प्रशंसा करता है तब विद्रोही वर्षा की आवश्यकता पर टिप्पणी करता है। जब प्रत्येक वर्षा के आगमन पर खुश होता है तो विद्रोही सूर्य की अनुपस्थिति पर खेद व्यक्त करता है।

3. When everybody……………… ………….one. (Page 34)

जब हर कोई सभा के लिए जाता है तो विद्रोही घर पर ठहरता है और पुस्तक पढ़ता है। जब प्रत्येक घर पर ठहरता है और पुस्तक पढ़ता है तो विद्रोही सभा पर चला जाता है। जब हर कोई ‘हाँ जी कहिए’ कहता है तो विद्रोही ‘न धन्यवाद’ कहता है। जब हर कोई ‘न धन्यवाद’ कहता है तो विद्रोही ‘हाँ जनाब’ कहता है। यह अच्छा है कि हमारे पास विद्रोही हैं। तुम इनमें से एक होकर अच्छा महसूस नहीं करोगे!

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