CBSE students can refer to NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions InText Questions and Answers are provided by experts in order to help students secure good marks in exams.

## Class 7 Maths NCERT Solutions Chapter 12 Algebraic Expressions InText Questions

Try These (Page 230)

Question 1.

Describe how the following expressions are obtained :

Solution:

In 7xy + 5, we first obtain XY, multiply it by 7 to get 7xy, and add 5 to 7xy to get the expression 7xy + 5.

In x^{2}y, we obtain x^{2} and multiply it by y to get x^{2}y.

In 4x^{2} – 5x, we first obtain x^{2}, and multiply it by 4 to get 4x^{2} and then obtain the second term by multiplying x by the constant 5. From 4x^{2}, we subtract 5x to finally arrive at 4x^{2} – 5x.

Try These (Page 231)

Question 1.

Let us draw a tree diagram for the expression 5xy + 10.

The expression + 10 consists of two terms 5xy and 10. The term 5xy is a product of 5, x, and y i.e., 5, x and y are factors of the term 5xy. But the term, 10 has only one factor, that is, 10.

These terms and factors of the terms of an expression can be represented by a tree diagram as shown in the figure.

Try These (Page 231)

Question 1.

What are the terms in the following expressions? Show how the terms are formed. Draw a tree diagram for each expression:

Solution:

In 8y + 3 x^{2}, the terms are 8y and 3 x^{2}

The term 8 y is obtained by multiplying y by the constant 8.

The term 3 x^{2} is obtained by multiplying 3, x, and x.

Its tree diagram is as under :

In 7mn – 4, the terms are 7mn and (- 4).

To obtain 7mn, we multiply the variable m with another variable n to get mn and then multiply mn by 7.

To obtain (- 4), we take the integer – 4.

It tree diagram is as under :

In 2x 2y, the only term is 2x 2y.

For term 2x 2y, we first obtain x 2 and multiply it by another variable y to get x^{2}y and finally multiply x2y by 2 to get 2x 2y.

Its tree diagram is as under :

Question 2.

Write three expressions each having 4 terms.

Solution:

Three expressions each having 4 terms may be taken as :

x + 2y + z – 4,2x^{2} – 4y^{2} + z + 5, xy + yz + zx + 5.

Try These (Page 231)

Question 1.

Identify the coefficients of the terms of the following expressions :

4x – 3y, a + b + 5, 2y + 5, 2 XY.

Solution:

Try These (Page 233)

Question 1.

Group the like terms together from the following:

12x, 12, – 25x, – 25, – 25y, 1, x, 12y, y.

Solution:

Grouping the like terms together, we have

12x, – 25x, x; 12, – 25,1; – 25y,12y, y.

Try These (Page 233)

Question 1.

Classify the following expressions as a monomial, a binomial or a trinomial: a, a + b, ab + a + b, ab + a + b – 5, xy, xy + 5, 5x^{2} – x + 2,

4pq – 3q + 5p, 7, 4m – 7n + 10, 4m + 7.

Solution:

Monomials are a, XY, and 7.

Binomials are a + b, xy + 5 and 4mn + 7.

Trinomials are ab + a + b, 5 x^{2} – x + 2, 4pq – 3q + 5p and 4m – 7n + 10.

Try These (Page 236)

Question 1.

Think of at least two situations in each of which you need to form two algebraic expressions and add or subtract them.

First Situation

A’s father’s weight is 2 times A’s weight.

A’s grandfather’s weight is 2 kg more than the sum of A’s weight and A’s father’s weight. How do you find A’s grandfather’s weight?

Second Situation

Two planes start from a city and fly in opposite directions, one averaging a speed of 30 km/hr greater than that of the other. If they are 3000 kg apart after 5 hours. How do you find their average speeds

Try These (Page 238)

Add and subtract

(i) m – n, m + n

Solution:

(m – n) + (m + n) = m – n + m + n

=m + m – n + n

= (1 + 1)m + (-1 + 1)n

= 2m + 0n

= 2m + 0 = 2m

and, (m – n) – (m + n) = m – n – m – n

=m – m – n – n

= (1 – 1)m + (-1 -1)n

= 0 m + (- 2)n

= 0 – 2n = – 2n

(ii) mn + 5-2, mn + 3

Solution:

(mn + 5 – 2) + (mn + 3) = (mn + 3) + (mn + 3) = 2 mn + 6

and, (mn + 5 – 2) – (mn + 3) = (mn + 3) – (mn + 3) = 0

Try These (Page 245)

Question 1.

Make a similar pattern with basic figures as shown

(The number of segments required to make the figure is given to the right. Also, the expression for the number of segments required to make reshapes is also given.) Go ahead and discover more such patterns.

Solution:

More such patterns are given below :

Try These (Page 246)

Question 1.

The number of diagonals we can draw from one vertex of a polygon of resides is ( it – 3). Check it for a heptagon (7 sides) and octagon (8 sides) by drawing a figure of what is the number for a triangle (3 sides)?

(i)

Solution:

Its diagonals are

AC, AD, AE, and AF

i.e., 4 in number.

If we put n = 7

in (n – 3), we get

7 – 3 = 4.

(ii)

Solution:

Its diagonals are

AC, AD, AE, AF

and AG i.e., 5 in number

If we put n = 8 in

(n – 3), we get

8 – 3 = 5.

(iii)

Solution:

Its diagonals

are none i.e., zero.

If we put n = 3 in (n – 3), we get

3 – 3 = 0

Thus, the result is verified in each case.