CBSE students can refer to NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers InText Questions and Answers are provided by experts in order to help students secure good marks in exams.

## Class 7 Maths NCERT Solutions Chapter 9 Rational Numbers InText Questions

Try These (Page 174)

Question 1.

Is the number \(\frac{2}{-3}\) ‘rational? Think about it.

Solution:

Yes, it is a rational number, because we know that a number that can be expressed in the form \(\frac{p}{q}\), where p,q are integers and q ≠ 0 is called a rational number.

Question 2.

List ten rational numbers.

Solution:

Ten rational number may be taken as

\(\frac{1}{3}\) , \(\frac{2}{3}\) , \(\frac{5}{7}\) , \(\frac{-7}{-11}\) , \(\frac{-1}{-13}\) , \(\frac{3}{-7}\) , \(\frac{-4}{5}\) , \(\frac{5}{9}\) , \(\frac{11}{15}\) , and \(\frac{110}{119}\)

Try These (Page 174)

Question 1.

Mention five rational numbers each of whose

(a) Numerator is a negative integer and denominator is a positive integer.

(b) Numerator is a positive integer and denominator is a negative integer.

(c) Numerator and denominator both are negative integers.

(d) Numerator and denominator both are positive integers.

Solution:

(a) Five rational numbers having numerator as negative integer and denominator as positive integer may

be taken as \(\frac{-3}{5}\) , \(\frac{-4}{9}\) , \(\frac{-5}{11}\) , \(\frac{-9}{15}\) , and \(\frac{-121}{-140}\)

(b) Five rational numbers having numerator as positive integer and denominator as negative integer may be taken as

\(\frac{-3}{-5}\) , \(\frac{4}{-7}\) , \(\frac{5}{-11}\) , \(\frac{7}{-13}\) ,and \(\frac{121}{-140}\)

(c) Five rational numbers having numerator and denominator both negative integers may be taken as

\(\frac{-3}{-5}\) , \(\frac{4}{-7}\) , \(\frac{5}{-11}\) , \(\frac{-9}{-13}\) , and \(\frac{-120}{-140}\)

(d) Five rational numbers having numerator and denominator both positive integers may be taken as

\(\frac{3}{5}\) , \(\frac{4}{7}\) , \(\frac{7}{11}\) , \(\frac{9}{17}\) , and \(\frac{121}{140}\)

Question 2.

Are integers also rational numbers?

Solution:

We know that, 1 = \(=\frac{1}{1}\) , 2 = \(=\frac{2}{1}\) , 3 = \(=\frac{3}{1}\) , etc.

Also, – 1 = \(=\frac{-1}{1}\) , – 2 = \(=\frac{-2}{1}\) , – 3 = \(=\frac{-3}{1}\)etc. The integer 0 can also be written as

0 = \(\frac{0}{2}\) or \(\frac{0}{5}\) , etc.

i.e., any integer p can be written as p = \(\frac{p}{1}\), which is a rational number. Thus, every integer is a rational number.

Try These (Page 175)

(i)

Solution:

In order to fill the required box, we have to express 5 as a rational number with a denominator 16. For this, we first find an integer which when multiplied with 4 gives us 16.

Clearly, such an integer is 16 ÷ 4 = 4

Multiplying the numerator and denominator of \(\frac{5}{4}\) by 4, we have

\(\frac{5}{4}\) = \(\frac{5 \times 4}{4 \times 4}\) = \(\frac{20}{16}\)

Thus,

Again, for , we should multiply the numerator and denominator of \(\frac{5}{4}\) by 25 ÷ 5, i.e., 5, we have

\(\frac{5}{4}\) = \(\frac{5 \times 5}{4 \times 5}\) = \(\frac{25}{20}\)

Thus,

Also, for , we should multiply the numerator and denominator of \(\frac{5}{4}\) by (-15) ÷ 5 = – 3, we have

Thus,

From (1), (2), and (3), we have

(ii)

Solution:

In order to fill the required box, we have to express -3 as a rational number with a denominator 14. For this, we first find an integer which when multiplied with 7 gives 14.

clearly, such as integer is 14 ÷ 7 = 2

Multiplying the numerator and denominator of \(\frac{-3}{7}\) by 2, we have

\(\frac{-3}{7}\) = \(\frac{-3 \times 2}{7 \times 2}\) =\(\frac{-6}{14}\)

Thus,

Similarly, we have

and

From (1),(2), and (3), we have

Try These (Page 175)

Question 1.

Is 5 a positive rational number?

Solution:

We know that 5 = \(\frac{5}{1}\), where 5 and 1 are both positive.

So, 5 is a positive rational number.

Question 2.

List five more positive rational numbers.

Solution:

Five more positive rational numbers may be taken as

\(\frac{3}{5}\) , \(\frac{5}{7}\) , \(\frac{11}{17}\) , \(\frac{-15}{-19}\) , and \(\frac{-110}{-117}\)

Try These (Page 176)

Question 1.

Is – 8 a negative rational number?

Solution:

We know that – 8 = \(\frac{-8}{1}\), where 8 is negative and 1 is positive .

∴ – 8 is a negative rational number.

Question 2.

List five more negative rational numbers.

Solution:

Five more negative rational numbers may be taken as \(\frac{-3}{2}\) , \(\frac{3}{-7}\) , \(\frac{-11}{19}\) , \(\frac{15}{-17}\) , \(\frac{-110}{111}\)

Try These (Page 176)

Question 1.

Which of these are negative rational numbers?

(i) \(\frac{-2}{3}\)

Solution:

In \(\frac{-2}{3}\) , numerator and denominator are of opposite signs. Therefore, \(\frac{-2}{3}\) is a negative rational number.

(ii) \(\frac{5}{7}\)

Solution:

In \(\frac{5}{7}\) , numerator and denominator are of same signs. Therefore , \(\frac{5}{7}\) is possitive rational number.

(iii) \(\frac{3}{-5}\)

Solution:

In \(\frac{3}{-5}\) , numerator and denominator are of opposite signs. Therefore, \(\frac{3}{-5}\) is a negative rational number.

(iv) 0

Solution:

The number 0 is neither a possitive nor a negative rational number.

(v) \(\frac{6}{11}\)

Solution:

In \(\frac{6}{11}\) , numerator and denominator are of same signs. Therefore , \(\frac{6}{11}\) is possitive rational number.

(vi) \(\frac{-2}{-9}\)

Solution:

In \(\frac{-2}{-9}\) , numerator and denominator are of same signs. Therefore , \(\frac{-2}{-9}\) is possitive rational number.

Try These (Page 178)

Find the standard form of

(i) \(\frac{-18}{45}\)

Solution:

The denominator of the rational number \(\frac{-18}{45}\) is positive. In order to express it in the standard form, we divide the numenator and denominator by the H.C.F of 18 and 45. is 9

Dividing the numenator and denominator of \(\frac{-18}{45}\) by 9, we have \(\frac{-18}{45}\) = \(\frac{(-18) \div 9}{45 \div 9}\) = \(\frac{-2}{5}\)

Thus , the standard form of \(\frac{-18}{45}\) is \(\frac{-2}{5}\)

(ii) \(\frac{-12}{18}\)

Solution:

The denominator of the rational number \(\frac{-12}{18}\) is positive . In order to express it in the standard form, we divide the numenator and H.C.F. of 12 and 18 is 6.

Dividing the numenator and denominator of \(\frac{12}{18}\) by 6, we have \(\frac{-12}{18}\) = \(\frac{(-12) \div 6}{18 \div 6}\) = \(\frac{-2}{3}\)

Thus, the standard form of \(\frac{-12}{18}\) is \(\frac{-2}{3}\)

Try These (Page 181)

Can you list five rational numbers between \(\frac{-\mathbf{5}}{\mathbf{3}}\) and \(\frac{-\mathbf{8}}{\mathbf{7}}\)?

Solution:

Let us convert the given rational numbers with same denominator 21. (∵ L.C.M. of 3 and 7 = 21)

\(\frac{-5}{3}\) = \(\frac{-5 \times 7}{3 \times 7}\) = \(\frac{-35}{21}\) and \(\frac{-8}{7}\) = \(\frac{-8 \times 3}{7 \times 3}\) = \(\frac{-24}{21}\)

We have, \(\frac{-35}{21}\) < \(\frac{-34}{21}\) < \(\frac{-32}{21}\) < \(\frac{-31}{21}\) < \(\frac{-29}{21}\) < \(\frac{-28}{21}\) < \(\frac{-24}{21}\)

Thus, \(\frac{-34}{21}\) , \(\frac{-32}{21}\) , \(\frac{-31}{21}\) , \(\frac{-29}{21}\) and \(\frac{-28}{21}\) are five rational numbers between \(\frac{-5}{3}\) and \(\frac{-8}{7}\)

Try These (Page 181)

Question 1.

Find five rational numbers between \(\frac{-5}{7}\) and \(\frac{-3}{8}\)

Solution:

We have, \(\frac{-5}{7}\) = \(\frac{-5 \times 8}{7 \times 8}\) = \(\frac{-40}{56}\)

and, \(\frac{-3}{8}\) = \(\frac{-3 \times 7}{8 \times 7}\) = \(\frac{-21}{56}\)

We know that,

Hence, five rational numbers between \(\frac{-5}{7}\) and \(\frac{-3}{8}\) may be taken as \(\frac{-39}{56}\) , \(\frac{-38}{56}\) , \(\frac{-37}{56}\) , \(\frac{-36}{56}\) and \(\frac{-35}{56}\)

Try These (Page 185)

Question 1.

\(\frac{-13}{7}\) + \(\frac{6}{7}\) \(\frac{19}{5}\) + \(\left(\frac{-7}{5}\right)\)

Solution:

We have, \(\frac{-13}{7}\) + \(\frac{6}{7}\) = \(\frac{-13+6}{7}\) = \(\frac{-7}{7}\) = – 1

and, \(\frac{19}{5}\) + \(\left(\frac{-7}{5}\right)\) = \(\frac{19+(-7)}{5}\) = \(\frac{12}{5}\)

Try These (Page 185)

Question 1.

\(\frac{-3}{7}\) + \(\frac{2}{3}\)

Solution:

To find \(\frac{-3}{7}\) + \(\frac{2}{3}\)

clearly, denominators of the given numbers are positive. the L.C.M. of denominators 7 and 3 is 21.

Now, we express \(\frac{-3}{7}\) and \(\frac{2}{3}\) into forms in which both of them have the same denominator 21.

We have, \(\frac{-3}{7}\) = \(\frac{-3 \times 3}{7 \times 3}\) = \(\frac{-9}{21}\)

and, \(\frac{2}{3}\) = \(\frac{2 \times 7}{3 \times 7}\) = \(\frac{14}{21}\)

∴ \(\frac{-3}{7}\) + \(\frac{2}{3}\) = \(\frac{-9}{21}\) + \(\frac{14}{21}\) = \(\frac{-9+14}{21}\) = \(\frac{5}{21}\)

Question 2.

\(\frac{-5}{6}\) + \(\frac{(-3)}{11}\)

Solution:

To find \(\frac{-5}{6}\) + \(\frac{(-3)}{11}\)

clearly denominators of the given numbers are positive. The L.C.M. of denominators 6 and 11 is 66.

We have , \(\frac{-5}{6}\) = \(\frac{-5 \times 11}{6 \times 11}\) = \(\frac{-55}{66}\)

and, \(\frac{-3}{11}\) = \(\frac{-3 \times 6}{11 \times 6}\) = \(\frac{-18}{66}\)

∴ \(\frac{-5}{9}\) + \(\frac{(-3)}{11}\) = \(\frac{-55}{66}\) + \(\frac{(-18)}{66}\) = \(\frac{-55+(-18)}{66}\) =\(\frac{-55-18}{66}\) = \(\frac{-73}{66}\)

Try These (Page 186)

Question 1.

What will be additive inverse of \(\frac{-3}{9}\) ? \(\frac{-9}{11}\) ? \(\frac{5}{7}\) ?

Solution:

The additive inverse of \(\frac{-3}{9}\) is – \(\left(\frac{-3}{9}\right)\) = \(=\frac{3}{9}\)

The additive inverse of \(\frac{-9}{11}\) is – \(\left(\frac{-9}{11}\right)\) = \(=\frac{9}{11}\)

The additive inverse of \(\frac{5}{7}\) is – \(\left(\frac{5}{7}\right)\) = \(=\frac{-5}{7}\)

Try These (Page 187)

Question 1.

Try to find \(\frac{7}{8}\) – \(\frac{5}{9}\) , \(\frac{3}{11}\) – \(\frac{8}{7}\) in both ways. Did you get the same answer?

Solution:

To find \(\frac{7}{8}\) – \(\frac{5}{9}\)

Thus, in both ways, the answer is the same.

To find \(\frac{3}{11}\) – \(\frac{8}{7}\)

Thus, in both ways, the answer is the same.

Try These (Page 187)

Question 1.

(i) \(\frac{7}{9}\) – \(\frac{2}{5}\)

Solution:

\(\frac{7}{9}\) – \(\frac{2}{5}\) = \(\frac{7}{9}\) + \(\frac{(-2)}{5}\) = \(\frac{35+(-18)}{45}\) = \(\frac{17}{45}\)

(ii) \(2 \frac{1}{5}\) – \(\frac{(-1)}{3}\)

Solution:

\(2 \frac{1}{5}\) – \(\frac{(-1)}{3}\) = \(\frac{11}{5}\) + additive inverse of \(\left(\frac{-1}{3}\right)\) = \(\frac{11}{5}\) + \(\frac{1}{3}\) = \(\frac{33+5}{15}\) = \(\frac{38}{15}\) = \(2 \frac{8}{15}\)

Try These (Page 187)

Question 1.

Find \(\frac{-4}{7}\) × 3 , \(\frac{-6}{5}\) × 4 using both ways. What do you observe?

Solution:

I method:

On the number line,\(\frac{-4}{7}\) × 3 means three jumps of \(\frac{4}{7}\) to the left

We reach at \(\frac{-12}{7}\) so, \(\frac{-4}{7}\) × 3 = \(\frac{-12}{7}\)

II method:

\(\frac{-4}{7}\) × 3 = \(\frac{(-4) \times 3}{7}\) =\(\frac{-12}{7}\)

We observe that in both ways, the product is the same.

To find \(\frac{-6}{5}\) × 4

I method:

On the number line, \(\frac{-6}{5}\) × 4 means four jumps of \(\frac{6}{5}\) to the left. We reach at \(\frac{-24}{5}\). so ,\(\frac{-6}{5}\) × 4 = \(\frac{-24}{5}\)

II method:

\(\frac{-6}{5}\) × 4 = \(\frac{(-6) \times 4}{5}\) = \(\frac{-24}{5}\)

We observe that in both ways, the product is the same.

Try These (Page 188)

Question 1.

(i) \(\frac{-3}{5}\) × 7 ?

Solution:

\(\frac{-3}{5}\) × 7 = \(\frac{(-3) \times 7}{5}\) =\(\frac{-21}{5}\)

(ii) \(\frac{-6}{5}\) × (-2) ?

Solution:

\(\frac{-6}{5}\) × (-2) = \(\frac{(-6) \times(-2)}{5}\) = \(\frac{12}{5}\)

Try These (Page 188)

Question 1.

(i) \(\frac{-3}{4}\) × \(\frac{1}{7}\)

Solution:

we have, \(\frac{-3}{4}\) × \(\frac{1}{7}\) = \(\frac{(-3) \times 7}{5}\) = \(\frac{-21}{5}\)

(ii) \(\frac{2}{3}\) × \(\frac{-5}{9}\)

Solution:

We have, \(\frac{2}{3}\) × \(\frac{-5}{9}\) = \(\frac{2 \times(-5)}{3 \times 9}\) = \(\frac{-10}{27}\)

Try These (Page 189)

Question 1.

What will be the reciprocal of \(\frac{-6}{11}\) ? and \(\frac{-8}{5}\) ?

Solution:

The reciprocal of \(\frac{-6}{11}\) is \(\frac{11}{-6}\) i.e., \(\frac{-11}{6}\)

and, the reciprocal of \(\frac{-8}{5}\) is \(\frac{5}{-8}\) i.e., \(\frac{-5}{8}\)

Try These (Page 189)

Question 1.

Try dividing \(\frac{2}{3}\) by \(\frac{-5}{7}\) both ways and see if you get the same answer.

Solution:

I method:

\(\frac{2}{3}\) ÷ \(\frac{-5}{7}\) = \(\frac{2}{3}\) × reciprocal of latex]\frac{-5}{7}[/latex]

= \(\frac{2}{3}\) × \(\frac{7}{-5}\) = \(\frac{2 \times 7}{3 \times(-5)}\) =\(\frac{14}{-15}\) =\(\frac{-14}{15}\)

II method:

Here, \(\frac{2}{3}\) ÷ \(\frac{5}{7}\) = \(\frac{2}{3}\) x \(\frac{7}{5}\) = \(\frac{14}{15}\)

∴ \(\frac{2}{3}\) ÷ \(\frac{-5}{7}\) = – \(\frac{14}{15}\)

Try These (Page 190)

Question 1.

(i) \(\frac{2}{3}\) × \(\frac{-7}{8}\)

Solution:

we have, \(\frac{2}{3}\) x \(\frac{-7}{8}\) = \(\frac{1 \times(-7)}{3 \times 4}\) = \(\frac{-7}{12}\)

(ii) \(\frac{-6}{7}\) × \(\frac{5}{7}\)

Solution:

\(\frac{-6}{7}\) × \(\frac{5}{7}\) = \(\frac{(-6) \times 5}{7 \times 7}\) = \(\frac{-30}{49}\)