NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2
NCERT
Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 are the part of NCERT Solutions
for Class 7 Maths. Here you can find the NCERT Solutions for Class 7 Maths
Chapter 12 Algebraic Expressions Ex 12.2.
- NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1
- NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2
- NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3
- NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4
Ex 12.2 Class 7 Maths Question 1.
Simplify combining like terms:(i) 21b – 32 + 7b – 20b
(ii) -z2 + 13z2 – 5z + 7z3 – 15z
(iii) p – (p – q) – q – (q – p)
(iv) 3a – 2b — ab – (a – b + ab) + 3ab + b – a
(v) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2
(vi) (3y2 + 5y – 4) – (8y – y2 – 4)
Solution:
(i) 21b – 32 + 7b – 20b
Rearranging the like terms, we get
21b + 7b – 20b – 32
= (21 + 7 – 20)b – 32
= 8b – 32
(ii) -z2 + 13z2 – 5z
+ 7z3 – 15z
Rearranging the like terms, we get
7z3 – z2 + 13z2 – 5z – 15z
= 7z3 + (-1 + 13)z2 + (-5 – 15)z
= 7z3 + 12z2 – 20z
(iii) p – (p – q) – q – (q – p)
= p – p + q – q – q + p
Rearranging the like terms, we get
(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3ab + b – a
Rearranging the like terms, we get
= 3a – a – a – 2b + b + b – ab – ab + 3ab
(v) 5x2y – 5x2 + 3yx2 –
3y2 + x2 – y2 + 8xy2 –
3y2
Rearranging the like terms, we get
5x2y + 3x2y + 8xy2 – 5x2 +
x2 – 3y2 – y2 – 3y2
= 8x2y + 8xy2 – 4x2 – 7y2
(vi) (3y2 + 5y – 4) – (8y – y2 –
4)
= 3y2 + 5y – 4 – 8y + y2 + 4 (Solving the brackets)
Rearranging the like terms, we get
= 3y2 + y2 + 5y – 8y – 4 + 4
= 4y2 – 3y
Ex 12.2 Class 7 Maths Question 2.
Add:(i) 3mn, -5mn, 8mn, -4mn
(ii) t – 8tz, 3tz – z, z – t
(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
(iv) a + b – 3, b – a + 3, a – b + 3
(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(vii) 4x2y, -3xy2, -5xy2, 5x2y
(viii) 3p2q2 – 4pq + 5, -10p2q2, 15 + 9pq + 7p2q2
(ix) ab – 4a, 4b – ab, 4a – 4b
(x) x2 – y2 – 1, y2 – 1 – x2, 1 – x2 – y2
Solution:
(i) 3mn, -5mn, 8mn, -4mn
= (3mn) + (-5mn) + (8mn) + (-4mn)
= (3 – 5 + 8 – 4)mn
= 2mn
(ii) t – 8tz, 3tz – z, z – t
= t – 8tz + 3tz – z + z – t
Rearranging the like terms, we get
= t – t – 8tz + 3tz – z + z
= 0 – 5tz + 0
= -5tz
(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
= -7mn + 5 + 12mn + 2 + 9mn – 8 + (-2mn) – 3
Rearranging the like terms, we get
= -7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3
= -7mn – 2mn + 12mn + 9mn + 7 – 8 – 3
(iv) a + b – 3, b – a + 3, a – b + 3
= a + b – 3 + b – a + 3 + a – b + 3
Rearranging the like terms, we get
= a – a + a + b + b – b – 3 + 3 + 3
= a + b + 3
(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
= 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
Rearranging the like terms, we get
= -12xy + 8xy + 4xy + 14x – 7x + 10y – 10y – 13 + 18
= 7x + 5
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
= 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5
Rearranging the like terms, we get
= 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5
= 3m – 4n – 3mn – 3
(vii) 4x2y, -3xy2, -5xy2,
5x2y
= 4x2y – 3xy2 – 5xy2
+ 5x2y
Rearranging the like terms, we get
= 4x2y + 5x2y – 3xy2 – 5xy2
= 9x2y — 8xy2
(viii) 3p2q2 – 4pq + 5,
-10p2q2, 15 + 9pq + 7p2q2
= (3p2q2 – 4pq + 5) + (-10p2q2)
+ (15 + 9pq + 7p2q2)
= 3p2q2 – 4pq + 5 – 10p2q2 +
15 + 9pq + 7p2q2
Rearranging the like terms, we get
= 3p2q2 + 7p2q2 – 10p2q2 –
4pq + 9pq + 5 + 15
= 10p2q2 – 10p2q2 + 5pq +
20
= 0 + 5pq + 20
= 5pq + 20
(ix) ab – 4a, 4b – ab, 4a – 4b
= ab – 4a + 4b – ab + 4a – 4b
Rearranging the like terms, we get
= 0 + 0 + 0 = 0
(x) x2 – y2 – 1, y2 –
1 – x2, 1 – x2 – y2
= x2 – y2 – 1 + y2 – 1 – x2 +
1 – x2 – y2
Rearranging the like terms, we get
= -(x2 + y2 + 1)
Ex 12.2 Class 7 Maths Question 3.
Subtract:(i) -5y2 from y2
(ii) 6xy from -12xy
(iii) (a – b) from (a + b)
(iv) a(b – 5) from b(5 – a)
(v) -m2 + 5mn from 4m2 – 3mn + 8
(vi) -x2 + 10x – 5 from 5x – 10
(vii) 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2
(viii) 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq
Solution:
(i) -5y2 from y2
y2 – (-5y2) = y2 +
5y2 = 6y2
(ii) 6xy from -12xy
-12xy – 6xy = -18xy
(iii) (a – b) from (a + b)
(a + b) – (a – b)
= a + b – a + b = 2b
(iv) a(b – 5) from b(5 – a)
b(5 – a) – a(b – 5)
= 5b – ab – ab + 5a
= 5a + 5b – 2ab
(v) -m2 + 5mn from 4m2 –
3mn + 8
(4m2 – 3mn + 8) – (-m2 + 5mn)
= 4m2 – 3mn + 8 + m2 – 5mn
= 4m2 + m2 – 3mn – 5mn + 8
= 5m2 – 8mn + 8
(vi) -x2 + 10x – 5 from 5x – 10
(5x – 10) – (-x2 + 10x – 5)
= 5x – 10 + x2 – 10x + 5
= x2 + 5x – 10x – 10 + 5
= x2 – 5x – 5
(vii) 5a2 – 7ab + 5b2 from
3ab – 2a2 – 2b2
(3ab – 2a2 – 2b2) – (5a2 – 7ab + 5b2)
= 3ab – 2a2 – 2b2 – 5a2 + 7ab –
5b2
= 3ab + 7ab – 2a2 – 5a2 – 2b2 –
5b2
= 10ab – 7a2 – 7b2
(viii) 4pq – 5q2 – 3p2 from
5p2 + 3q2 – pq
(5p2 + 3q2 – pq) – (4pq – 5q2 –
3p2)
= 5p2 + 3q2 – pq – 4pq + 5q2 +
3p2
= 5p2 + 3p2 + 3q2 + 5q2 –
pq – 4pq
= 8p2 + 8q2 – 5pq
Ex 12.2 Class 7 Maths Question 4.
(a) What should be added to x2 + xy + y2 to obtain 2x2 + 3xy?(b) What should be subtracted from 2a + 8b + 10 to get -3a + 7b + 16?
Solution:
(a) (2x2 + 3xy) – (x2 +
xy + y2)
= 2x2 + 3xy – x2 – xy – y2
= 2x2 – x2 + 3xy – xy – y2
= x2 + 2xy – y2
Thus, x2 + 2xy – y2 is
required expression to be added.
(b) (2a + 8b + 10) – (-3a + 7b + 16)
= 2a + 8b + 10 + 3a – 7b – 16
= 2a + 3a + 8b – 7b + 10 – 16
= 5a + b – 6
Thus, 5a + b – 6 is required expression to be
subtracted.
Ex 12.2 Class 7 Maths Question 5.
What should be taken away from 3x2 – 4y2 + 5xy + 20 to obtain -x2 – y2 + 6xy + 20?Solution:
Let A should be taken away. Then,
(3x2 – 4y2 + 5xy + 20) – A = -x2 –
y2 + 6xy + 20
⇒ A = (3x2 – 4y2 + 5xy + 20) –
(-x2 – y2 + 6xy + 20)
A = 3x2 – 4y2 + 5xy + 20 + x2 +
y2 – 6xy – 20
A = 3x2 + x2 – 4y2 + y2 +
5xy – 6xy + 20 – 20
A = 4x2 – 3y2 – xy
Thus, 4x2 – 3y2 – xy
is required expression to be taken away.
Ex 12.2 Class 7 Maths Question 6.
(a) From the sum of 3x – y + 11 and -y – 11, subtract 3x – y – 11.(b) From the sum of 4 + 3x and 5 – 4x + 2x2, subtract the sum of 3x2 – 5x and -x2 + 2x + 5.
Solution:
(a) Sum of 3x – y + 11 and -y – 11
= (3x – y + 11) + (-y – 11)
= 3x – y + 11 – y – 11
= 3x – 2y
∴ (3x – 2y) – (3x – y – 11)
= 3x – 2y – 3x + y + 11
= 3x – 3x – 2y + y + 11
= -y + 11
(b) Sum of (4 + 3x) and (5 – 4x + 2x2)
= 4 + 3x + 5 – 4x + 2x2
= 2x2 – 4x + 3x + 4 + 5 = 2x2 – x + 9
Sum of (3x2 – 5x) and (-x2 + 2x + 5)
= (3x2 – 5x) + (-x2 + 2x + 5)
= 3x2 – 5x – x2 + 2x + 5 = 2x2 –
3x + 5
Now, (2x2 – x + 9) – (2x2 – 3x + 5)
= 2x2 – x + 9 – 2x2 + 3x – 5
= 2x2 – 2x2 + 3x – x + 9 – 5
= 2x + 4
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