NCERT Solutions for Class 7 Maths Chapter 12 (Algebraic Expressions), in this chapter we learn how to make algebraic expressions and how to solve the basic problem of algebraic expressions.
The student can learn online and download the solution in pdf format, solution in pdf is very helpful for the students.Direct access the solution of the all questions of this chapter by exercise wise.
- Solution Exercise 12.1
- Solution Exercise 12.2
- Solution Exercise 12.3
- Download the pdf of Exercise 12.1, Exercise 12.2 and Exercise 12.3
Class 7th Solution Exercise 12.1
Direct access the NCERT Solutions for Class 7 Maths Chapter 12 , Exercise 12.2 and Exercise 12.3
Question 1:
Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.
(i) Subtraction of z from y.
(ii) One-half of the sum of numbers x and y.
(iii) The number z multiplied by itself.
(iv) One-fourth of the product of numbers p and q.
(v) Numbers x and y both squared and added.
(vi) Number 5 added to three times the product of numbers m and n.
(vii) Product of numbers y and z subtracted from 10 .
(viii) Sum of numbers a and b subtracted from their product.
Answer 1:
(i) y-z
(ii) 
(iii) z×z = z2
(iv) 
(v) x2 + y2
(vi) 3mn+5
(vii) 10 – yz
(viii) ab – (a+b)
Question 2:
(i) Identify the terms and their factors in the following expressions
Show the terms and factors by tree diagrams.
(a) x-3 (b) 1+x+ x2 (c) y- y2
(d) 5xy2 +7x2y (e) -ab+2b2 -3a2
(ii) Identify terms and factors in the expressions given below:
(a) – 4x+5 (b) – 4x+5y (c) 5y+3y2
(d) xy+2x2 y2 (e) pq+q (f) 1.2ab -2.4b+3.6a
Answer 2: (i)
(ii)
Question 4:
(a) Identify terms which contain x and give the coefficient of x.
(i) y2x+y (ii) 13y2 -8yx (iii) x+y+2
(iv) 5+z+zx (v) 1+x+xy (vi) 12xy2 +25
(vii) 7x+xy2
(b) Identify terms which contain y2 and give the coefficient of y2 .
(i) 8-xy2 (ii) 5y2 +7x (iii) 2x2 y – 15xy2 +7y2
Answer 4:
Question 5:
Classify into monomials, binomials and trinomials.
(i) 4y -7z (ii) y2 (iii) x+y-x
(iv) 100 (v) ab-a-b (vi) 5-3t
(vii) 4p2 q- 4pq2 (viii) 7mn (ix) z2 -3z+8
(x) a2 + b2 (xi) z2 +z (xii) 1+x+x2
Answer 5:
(i) Binomial
(ii) Monomial
(iii) Trinomial
(iv) Monomial
(v) Trinomial
(vi) Binomial
(vii) Binomial
(viii) Monomial
(ix) Trinomial
(x) Binomial
(xi) Binomial
(xii) Trinomial
Question 6:
State whether a given pair of terms is of like or unlike terms.
(i) 1,100 (ii) 
(iii) -29x, -29y
(iv) 14xy, 42yx (v) 4m2 p, 4mp2 (vi) 12xz, 12x2z2
Answer 6:
(i) like
(ii) like
(iii) unlike
(iv) like
(v) unlike
(vi) unlike
Question 7:
Identify like terms in the following:
Solution Exercise 12.2
Student can direct access the NCERT Solutions for Class 7 Maths Chapter 12 Exercise 12.1 and Exercise 12.3
Question 1:
Simplify combining like terms:
Answer 1:
Question 2:
Answer 2:
Question 3:
Answer 2:
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?
Answer 4:
(a) We get the required result by subtracting x2+xy+y2 from 2x2 +3xy
Now, Difference
⇒ 2x2 +3xy-(x2 +xy+y2 )
⇒2x2 +3xy-x2 -xy-y2
⇒ x2 – y2 +2xy
Hence we shoud x2 – y2 +2xy add to get desired result.
(b) Subtracting -3a+7b+16? from 2a+8b+10
We get, (2a+8b+10)-(-3a+7b+16? )
⇒2a+8b+10+3a-7b-16
⇒5a+b -6
Hence we should subtract 5a+b -6 for getting the required result.
Question 5:
What should be taken away from 3x2 -4y2 +5xy+20
to obtain – x2 – y2 +6xy+20?
Answer 5:
Subtract -x2 – y2 +6xy+20 from 3x2 – 4y2 +5xy+20
We get, (3x2 -4y2 +5xy+20)-(- x2 – y2 +6xy+20)
⇒3x2 – 4y2 +5xy+20+x2 +y2 – 6xy -20
⇒4x2 -3y2 – xy
Hence we should subtract 2x2 -3y2 -xy to get required result.
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
Answer 6:
Solution Exercise 12.3 Maths Class 7
Access here the NCERT Solutions for Class 7 Maths Chapter 12 Exercise 12.1 and Exercise 12.2
Question 1:
Question 7:
Simplify these expressions and find their values
if x = 3, a = – 1, b = – 2.
(i) 3x – 5 – x + 9 (ii) 2 – 8x + 4x + 4
(iii) 3a + 5 – 8a + 1 (iv) 10 – 3b – 4 – 5b
(v) 2a – 2b – 4 – 5 + a
