NCERT Solutions For Class 7 Maths Chapter 12

NCERT Solutions Class 7 Maths Algebraic Expressions

NCERT Solutions For Class 7 Maths Chapter 12 Algebraic Expressions are given here in a simple and detailed way. These NCERT Solutions for class 7 maths chapter 12 can be extremely helpful for the students to clear all their doubts easily and understand the basics of this chapter in a better and detailed way.

NCERT class 7 maths chapter 12 Algebraic Expressions solutions given here are very easily understandable so that students does not face any difficulties regarding any of the solutions. The NCERT solutions for class 7 maths chapter 12 PDF is also available here that the students can download and study.

NCERT Solutions For Class 7 Maths Chapter 12 Exercises

Exercise 12.1

Q1: Using arithmetic operations, constants and variables find the algebraic expressions of the cases given below:

(i) Numbers a and b both squared and added.

(ii) Number 5 added to three times the product of s and t.

(iii) One-fourth of the product of numbers m and n.

(iv) One-half of the sum of numbers a and b.

(v) Product of numbers e and f subtracted from 10.

(vi) Subtraction of v from u.

(vii) Sum of numbers s and t subtracted from their product

(viii) The number x multiplied by itself.

 

Sol:

(i) a2+b2

(ii) 3st+5

(iii) mn4

(iv) a+b2

(v) 10ef

(vi) u-v

(vii) st(s+t)

(viii) x2

 

Q2:  

(a) Figure out the terms and their factors in the expression given below and show them by the help of tree diagram

(i) a3

(ii) 1+a+a2

(iii) yy3  

(iv)5ab2+7x2y  

(v) xy+2y23x2

 

(b) Figure out the terms and factors in the expressions below:

(i)  4a+5

(ii) 4a+5b

(iii) 5a+3a2  

(iv) ab+2a2b2

(v) ab+b  

(vi) 1.2xy2.4y+3.6x  

(vii) 34x+14

(viii) 0.1a2+0.2b2

Also show the terms and factors by tree diagram.

 

Sol:

(a)

(i) a3

(ii)   1+a+a2

(iii) yy3

(iv)  5ab2+7x2y

(v)   xy+2y23x2

 

(b)-

(i)   4a+5

Terms:  4a,5

Factors:  4,a;5

 

(ii) 4a+5b

Terms:  4a,5b

Factors:   4,a;5,b

 

(iii) 5a+3a2

Terms:  5a,3a2

Factors:   5,a;3,aa

 

(iv) ab+2a2b2

Terms:  ab,2a2b2

Factors:   a,b;2,a,a;b,b

 

(v) ab+b

Terms:  ab,b

Factors:   a,b;b

 

(vi) 1.2xy2.4y+3.6x

Terms:  1.2xy,2.4y,3.6x

Factors:   1.2,x,y;2.4,y;3.6x

 

(vii) 34x+14

Terms:  34x,14

Factors:   34,x;14

 

(viii) 0.1a2+0.2b2

Terms:  0.1a2,0.2b2

Factors:   0.1,a,a;0.2,bb

 

Q3: Other than the constants figure out the numerical coefficients of the given expressions:

(i) 53a2

(ii)  1=a+a2+a3

(iii) a+2ab+3b

(iv) 100x+100y

(v) x2y2+7xy

(vi) 1.2x+0.8y

(vii) 3.14x2

(viii) 2(a+b)

(ix) 0.1x+0.01x2

 

S.no Expression Terms Numerical Coefficient
(i) 53a2 3a2
(ii) 1=a+a2+a3  

a
a2
a3
 

1
1
1
(iii) a+2ab+3b  

a
2ab
3b
 

1
2
3
(iv) 100x+100y  

100m
100n
 

100
100
(v) x2y2+7xy  

x2y2
7xy
 

-1
7
(vi) 1.2x+0.8y  

1.2x
0.8y
 

1.2
0.8
(vii) 3.14x2 3.14x2 3.14
(viii) 2(a+b)  

2a
2b
 

2
2
(ix) 0.1x+0.01x2  

0.1x
0.01x2
 

0.1
0.01

 

Q4:

(a) Identify the terms which contain ‘a’ and give the coefficient of a.

 

(i) b2a+b

(ii) 13b28ab

(iii) a+b+15

(iv) 5+m+ma

(v) 1+a+ab

(vi) 12ab2+10

(vii) 7a+am2

 

(b) Figure out the terms which contain b2 and also give the coefficient of the same term.

(i) 8ab2

(ii) 5b2+10a

(iii) 2a2b5ab2+15b2

 

Sol:

 

S.no Expression Terms with factor a Coefficient of a
(i) b2a+b b2a b2
(ii) 13b28ab 8ab 8b
(iii) a+b+15 a 1
(iv) 5+m+ma ma m
(v) 1+a+ab  

a
ab
 

1
b
(vi) 12ab2+10 12ab2 12b2
(vii) 7a+am2  

am2
7a
 

m2
7

 

(b)

 

S.no Expression Terms containing b2 Coefficient of b2
(i) 8ab2 ab2 a
(ii) 5b2+10a 5b2 5
(iii) 2a2b5ab2+15b2  

5ab2
15b2
 

5a
15

 

Q5: Classify into monomials, binomials and trinomials:

(i) 4b7a

(ii) b2

(iii) a+bab

(iv) 50

(v) ab+b+a

(vi) 5+10x

(vii) 15a2b10ab2

(viii) 10yz

(ix) x2+10x5

(x) x2+y2

(xi) x2+y

(xii) a2+a+50

 

Sol:

 

S.no Expression Type of Polynomial
(i) 4b7a Binomial
(ii) b2 Monomial
(iii) a+bab Trinomial
(iv) 50 Monomial
(v) ab+b+a Trinomial
(vi) 5+10x Binomial
(vii) 15a2b10ab2 Binomial
(viii) 10yz Monomial
(ix) x2+10x5 Trinomial
(x) x2+y2 Binomial
(xi) x2+y Binomial
(xii) a2+a+50 Trinomial

 

Q6: State whether a given pair of term is of like or unlike terms:

(i) 1,100

(ii) 20x,12x

(iii) 10x,10y

(iv) 50ab,30ba

(v) 2a2b,8ab2

(vi) 10ab,20a2b

 

Sol:

 

S.no Pair of terms Like/Unlike terms
(i) 1,100 Like terms
(ii) 20x,12x Like terms
(iii) 10x,10y Unlike terms
(iv) 50ab,30ba Like terms
(v) 2a2b,8ab2 Unlike terms
(vi) 10ab,20a2b Unlike terms

 

Q7: Identify the like terms in the following:

(a) a2b,4ab2,9a2,2ab2,10a,20a2,30a,5a2b,2ab,35a

(b) 10pq,10p,5q,2p2q2,5pq,50q,30,18p2q2,55,100p,30pq,105p2q,200

 

Sol:

(a) Like terms are:

(i) a2b,5a2b

(ii) 4ab2,2ab2

(iii) 9a2,20a2

(iv) 10a,30a,35a

(v) 2ab

 

(b) Like terms are:

(i)  10pq,5pq,30pq

(ii)  10p,100p,

(iii) 5q,50q

(iv) 2p2q2,18p2q2

(v) 30,55,200

(vi) 105p2q

 

Exercise 12.2 

 

Q1: Simplify the terms:

(i) 21a32+7a20a

(ii) x2+13x25x+7x315x

(iii) a(ab)b(ba)

(iv) 3x2yxy(xy+xy)+3xy+yx

(v) 5a2b5a2+3a2b3b2+a2b2+8ab23b2

(vi) (3b2+5b4)(8bb24)

Sol:

(i) 21a32+7a20a=21a+7a20b32

8b32

 

(ii) x2+13x25x+7x315x=7x3+13x2x25x15x

=7x3+12x220x

 

(iii) a(ab)b(ba)=aa+bbb+a

=ab

 

(iv) 3x2yxy(xy+xy)+3xy+yx=3x2yxyx+yxy+3xy+yx

=3xxx+y+y2yxyxy+3xy =x2xy+3xy

 

(v)  5a2b5a2+3a2b3b2+a2b2+8ab23b2

5a2b+3a2b+8ab25a2+a23b2b23b2=8a2b+8ab24a27b2

 

(vi) (3b2+5b4)(8bb24)

3b2+5b48b+b2+4=3b2+b2+5b8b+44 =4b23b

 

Q2: Add:

(i) 3mn,5mn,8mn,4mn

(ii) a8ab,3abb,ba

(iii) 7mn+5,12mn+2,9mn8,2mn3

(iv) a+b3,ba+3,ab+3

(v) 14x+10y12xy13,187x10y+8xy,4xy

(vi) 5m7n,3n4m+2,2m3mn5

(vii) 4x2y,3xy2,5xy2,5x2y

(viii) 3p2q24pq+5,10p2q2,15+9pq+7p2q2

(ix) ab4a,4bab,4a4b

(x) x2y21,y21x2,1x2y2

Sol:

(i) 3mn,5mn,8mn,4mn

3mn+(5mn)+8mn(4mn)=(35+84)mn =(2)mn

 

(ii) a8ab,3abb,ba

a8ab+3abb+ba=aa+bb8ab+3ab =5ab

 

(iii) 7mn+5,12mn+2,9mn8,2mn3

7mn+5+12mn+2+9mn8+2mn3=7mn+12mn+9mn+5+283 =(7+12+9)mn+(5+283)=14mn+2

 

(iv) a+b3,ba+3,ab+3

a+b3+ba+3+ab+3=a+aa+b+bb+3+33 =(1+11)a+(1+11)b+(3+33)=a+b+3

 

(v) 14x+10y12xy13,187x10y+8xy,4xy

=14x7x+10y10y+8xy+4xy12xy+1813=7x+18 =7x+18

 

(vi) 5m7n,3n4m+2,2m3mn5

5m4m+2m7n+3n+253mn=3m4n3mn3

 

(vii) 4x2y,3xy2,5xy2,5x2y

4x2y+(3xy2)+(5xy2)+5x2y=4x2y+5x2y3xy25xy2 =9x2y8xy2

 

(viii) 3p2q24pq+5,10p2q2,15+9pq+7p2q2

3p2q24pq+5+(10p2q2)+15+9pq+7p2q2=3p2q2+7p2q210p2q2+9pq4pq+155 =5pq+10

 

(ix) ab4a,4bab,4a4b

ab4a+4bab+4a4b=4a4a+4b4b+abab =0

 

(x) x2y21,y21x2,1x2y2

x2y21+y21x2+1x2y2=x2x2x2+y2y2y2+111 =x2y21

 

Q3: Subtract:

(i)  5y2 from y2

(ii) 6xy from 12xy

(iii) (ab) from (a+b)

(iv) a(b5) from b(5a)

(v) m2+5mn from 4m23mn+8

(vi) x2+10x5 from 5x10

(vii) 5a27ab+5b2 from 3ab2a22b2

(viii) 4pq5q23p2 from 5p2+3q2pq

 

Sol:

(i) y2(5y2)

    =y2+5y2

=6y2

 

(ii) 12xy6xy

=18xy

 

(iii) (a+b)(ab)

=a+ba+b =2b

 

(iv) b(5a)a(b5)

=5babab+5a =5a+5b2ab

 

(v)  4m23mn+8(m2+5mn)

=4m23mn+8+m25mn =5m28mn+8

 

(vi) 5x10(x2+10x5)

=5x10+x210x+5 =x25x5

 

(vii) 3ab2a22b2(5a27ab+5b2)

=3ab2a22b25a2+7ab5b2 =3ab+7ab2a25a22b25b2 =10ab7a27b2

 

(viii) 5p2+3q2pq(4pq5q23p2)

=5p2+3q2pq4pq+5q2+3p2 =5p2+3p2+3q2+5q2pq4pq =8p2+8q25pq

 

Q4: (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?

 

Sol:

(a) Let a should be added

Then according to the question

x2+xy+y2+a=2x2+3xy

 

a=2x2+3xy(x2+xy+y2)

 

a=2x2+3xyx2xyy2

 

a=2x2x2y2+3xyxy

 

a=x2y2+2xy

Hence the value of a comes out to be x2y2+2xy.

Hence  x2y2+2xy should be added.

 

(b) Let b should be subtracted

Then according to the question,

2a+8b+10q=3a+7b+16

 

2a+8b+10q=3a+7b+16

 

q=2a+8b+10(3a+7b+16)

 

q=2a+8b+10+3a7b16

 

q=2a+3a+8b7b+1016

 

q=5a+b6

 

Q5: What should be taken from 3x2-4y2+5xy+20 to obtain –x2-y2+6xy+20 ?

Sol:

Let a be subtracted

Then according to the question,

3x-4y2+5xy+20 – q= –x2-y2+6xy+20

q=  3x-4y2+5xy+20 -(–x2-y2+6xy+20)

q= 3x2-4y2+5xy+20+x2+y2-6xy-20

q=3x2+x2-4y2+y2+5xy-6xy +20 -20

q=4x2-3y2-xy

Hence, 4x2-3y2-xy should be subtracted in the given equation.

 

Q6:

(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 + 2x, subtract the sum of 3x2 – 5x and –x2 + 2x + 5.

Sol:

(a)According to the question

(3x – y + 11)+( – y – 11)-( 3x – y – 11)= 3x – y + 11 – y – 11- 3x + y + 11

= 3x-3x+y-y+11+11-11

=11

(b) According to question,

(4 + 3x)+( 5 – 4x + 2x2)-( 3x2 – 5x)-( –x2 + 2x + 5)

= 4 + 3x+ 5 – 4x + 2x2– 3x2 + 5x +x2 – 2x – 5

= 3x-4x+5x-2x +2x2– 3x2+x2+4+5-5

= 2x+4

 

Exercise 12.3

 

Q1: If a=2, find the values of:

(i) a-2

(ii) 3a-5

(iii) 9-5a

(iv) 3a22a7

(v) 5m24

 

Sol:

(i) a-2 =2-2  (Putting a=2)

=0

 

(ii)  3a-5= 3×25   (Putting a=2)

=1

 

(iii)   9-5a=95×2 (Putting a=2)

= -1

 

(iv)  3a22a7=3×222×27  (Putting a=2)

=12-4-7

=1

 

(v)  5m24=5×224=54   (Putting a=2)

=1

 

Q2: If x=-2, find

(i)  4x+7

(ii)  3x2+4x+7

(iii) 2x33x2+4x+7

 

Sol:

(i)  4x+7=4(-2)+7   (Putting x= -2)

= -8+7=-1

 

(ii)  3x2+4x+7=3(2)2+4(2)+7

= -3(4)-8+7=-12-8+7

= -13

 

(iii) 2x33x2+4x+7=2(2)33(2)2+4(2)+7  (Putting x= -2)

= -2(-8)-3(4)+4(-2)+7

=  16-12-8+7

=3

 

Q3: Find the value of the following expressions, when x= -1:

(i) 5x-35

(ii) -2x+4

(iii) 3x2+6x+3

(iv) 6x23x6  

Sol:

(i) 5x-35 = 5(-1)-35 =-5-35              [Putting x= -1 ]

= -40

(ii)  -2x+4  = -2(-1)+4            [Putting x= -1 ]

= 2 + 4 = 6

(iii) 3x2+6x+3 = 3(1)2+6(1)+3      [Putting x= -1 ]

= 3-6+3 =0

(iv) 6x23x6  = 6(1)23(1)6        [Putting x= -1 ]

= 6+1-6 =1

 

Q 4: If x=2, y= -2, find the value of:

(i) x2+y2  

(ii) x2+xy+y2  

(iii) x2y2

 

Sol:

(i) x2+y2 = 22+(2)2                        [Putting a=2,  b= -2 ]

= 4 + 4 = 8

(ii) x2+xy+y2  = 22+2(2)+(2)2        [Putting a=2,b= -2 ]

= 4 – 4 + 4 = 4

(iii) x2y2 = (2)2(2)2                        [Putting a=2, b= -2]

= 4 – 4 = 0

 

Q5: When x=0,y= -1, find the value of the given expressions:

(i) 2x+2y

(ii) 2x2+y2+1  

(iii) 2x2y+2xy2+xy

(iv) x2+xy+2

Sol:

(i) 2x+2y = 2(0)+2(-1)     [Putting x=0,y= -1 ]

= 0 – 2 = -2

(ii) 2x2+y2+1 = 2(0)2+(1)2+1   [Putting x=0, y=-1 ]

= 0 + 1 + 1 = 2

(iii) 2x2y+2xy2+xy =  2(0)2(1)+2(0)(1)2+0(1)      [Putting x=0, y= -1]

= 0 + 0 + 0 = 0

(iv) x2+xy+2 = (0)2+(0)(1)+2    [Putting x=0, y= -1 ]

= 0 + 0 + 2 = 2

 

Q6: Simplify the following expressions and find the value at a= 2:

(i) a+7+4(a-5)

 (ii) 3(a+2)+5a-7  

(iii) 10a+4(a-2)

(iv) 5(3a-2)+4a+8    

 

Sol:

(i) a+7+4(a-5) = a+7+4a-20

=4a+a+7-20 =5a-13

= 5(2)-13 =10-13                                                                                   [Putting a=2 ]

= -3

 

(ii) 3(a+2)+5a-7 = 3a+6+5a-7

= 3a+5a+6-7  = 8a-1

= 8( 2) – 1                                                                                               [Putting a=2 ]

= 16 – 1 = 15

 

(iii) 10a+4(a-2) = 10a+4a-8

= 14a-8

= 14( 2) – 8                                                                                            [Putting a= 2 ]

= 28 – 8 = 20

 

(iv) 5(3a-2)+4a+8 = 15a-10+4a+8

=15a+4a-10+8 = 19a-2                                                                           [Putting =2  ]

= 19(2)-2 = 38-2

= 36

 

Q7: Simplify the expression given below and find the value at x=3, y= -1, z= -2  :

(i) 8x-10-3x+5

(ii) 10-5x+3x+6

(iii) 5y+3-2y+6

(iv) 5-8z-12-4z

(v) 3y-5z-6x+15

Sol:

(i) 8x-10-3x+5 = 8x-3x-10+5

=5x-5 = 5(3)-5                                                                                                [Putting x=3 ]

= 15-5 = 0

(ii) 10-5x+3x+6 = 10+6-5x+3x

= 16-2x = 16-2(3)                                                                                           [Putting x= 3 ]

=  16-6 =10

(iii) 5y+3-2y+6 = 5y-2y+3+6

= 3y+9 = 3(-1)+9                                                                                             [Putting y= -1 ]

= -3 + 9 = 6

(iv) 5-8z-12-4z = 5-12-8z-4z

= -7-12 z                                                                                                         [Putting z= -2 ]

= -7 -12(-2) = -7+24

= 17

(v) 3y-5z-6x+15

= 3(-1)-5(-2)-6(3)+15                                                                       [Putting x=3, y=-1, z=-2]

= -3+10-18+15

= 25-21

= 4

 

Q8:

(i) If x= 10, find the value of x33x25x+15 .

(ii) If y= -10, find the value of 2y23y+50

 

Sol:

(i) x33x25x+15

= 1033(10)25(10)+15                                                   [Putting x=10 ]

=1000-300-50+15

= 665

 

(ii) 2y23y+50

=2(10)23(10)+50                                                            [Putting y= -10 ]

=200+30+50

=280

 

Q9: What should be the value of p if the value of 2a2+ap=5 equals to 5, when a=0 ?

Sol:

2a2+ap=5

2(0)2+0p=5                                                                         [ Putting x= 0 ]

p=5

Hence, the value of p is -5.

 

Q10: Simplify the expression and find its value when x= 5 and y= -3:   

2(x2+xy)+3xy.

Sol:

 

Given:

2(x2+xy)+3xy

 

2x2+2xy+3xy

 

2x2+2xyxy+3

 

2x2+xy+3

 

2(5)2+(5)(3)+3             [Putting x=5, y= -3 ]

 

2(25)+(15)+3

 

5015+3 38

 

Exercise 12.4

Q1: Observe the pattern made from the line segment, which are of equal length which are found in display of calculators and digital speedometer: If n is the number of digits, and the number of required segments to form the digit n is given by the algebraic expression on the right of the digit. So how many segments are required to form 5,10,100 digits of the kind .

Sol:

 

S.no Symbols Digit’s number Pattern Formulae No. of segments
(i)  

5
10
100
5n+1  

26
51
501
(ii)  

5
10
100
3n+1  

16
31
301
(iii)  

5
10
100
5n+2  

27
52
502

 

(i) 5n+1

Putting n=5,          5×5+1=25+1=26

Putting n=10,        5×10+1=50+1=51

Putting n=100,       5×100+1=500+1=501

 

(ii) 3n+1

Putting n=5,          3×5+1=15+1=16

Putting n=10,        3×10+1=30+1=31

Putting n=100,        3×100+1=300+1=301

 

(iii) 5n+2

Putting n=5,          5×5+2=25+2=27

Putting n=10,        5×10+2=50+1=52