CBSE NCERT Solutions for Class 7 Chapter 12

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CBSE NCERT Solutions for Class 7 Maths Chapter 12:

Question-1:

Identify like terms in the following:
(a)x y^2, 4yx^2, 8 x^2, 2xy^2, 7y, 11x^2, 100x, 11yx, 20x^2y, 6x^2, y, 2xy, 3x
(b)10pq, 7p, 8q, p^2 q^2, 7qp, 100q, 23, 12q^2p^2, 5p^2, 41, 2405p, 78qp, 13p^2q, qp^2, 701 p^2

Solution:

(a) Like terms are:
(i)x y^2 , 2x y^2
(ii) 4 y x^2, 20x^2 y
(iii) 8x^2 , 11x^2, 6x^2
(iv) 7 y, y
(v) 100x, 3x
(vi)11yx, 2xy
(b)
Like terms are:
(i) 10pq, 7pq, 78pq
(ii) 7p, 2405p
(iii) 8q,100q
(iv) p^2 q^2 , 12 p^2 q^2
(v) 12, 41
(vi) 5p^2 ,701 p^2
(vii) 13 p^2 q , q p^2

Question-2:

(a) What should be added to x^2 + xy+ y^2 to obtain 2 x^2 + 3xy ?
(b) What should be subtracted from 2a + 8b+ 10 to get 3a + 7b+ 16 ?

Solution:

(a) Let p should be added.
Then according to question,
x^2 +xy+ y^2 +p =2 x^2 + 3xy
p =2 x^2 + 3xy - x^2 - xy - y^2
p= x^2- y^2 +2xy
Hence, x^2 -y^2 + 2xy should be added.

(b) Let q should be subtracted.
Then according to question,
2a + 8b+ 10 - q = 3a + 7b+ 16
2a + 8b+ 10 -3a - 7b- 16 = q
-a +b – 6 = q
q = -a +b – 6

Question-3:

If p=2, find the value of:
(i) 4p-7
(ii) 3 p^2 +4p-7
(iii) 2p^3 + 3 p^2 -4p -17

Solution:

(i) 4p-7=42- 7 [Putting p= 2]
= 8 - 7 =1
(ii) 3p^2 +4p-7 =3(2^2) +4(2) -7 [Putting p =2]
= 12+8 -7
= 20 - 7 = 13
(iii) 2 p^3 +3 p^2 -4p -17= 2
2^3 +3* 2^2 - 4 * 2- 17 [Putting p =2]
=16 +12 - 8 -1 7
= 20 - 17 =3