Math Ncert Exemplar 2019 Solutions for Class 8 Maths Chapter 1 - Rational Numbers

Share with your friends

Math Ncert Exemplar 2019 Solutions for Class 8 Maths Chapter 1 Rational Numbers are provided here with simple step-by-step explanations. These solutions for Rational Numbers are extremely popular among class 8 students for Maths Rational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Math Ncert Exemplar 2019 Book of class 8 Maths Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Math Ncert Exemplar 2019 Solutions. All Math Ncert Exemplar 2019 Solutions for class 8 Maths are prepared by experts and are 100% accurate.

Page No 9:

Question 1:

Choose the correct answer.
A number which can be expressed as $\frac{p}{q}$ where p and q are integers and q ≠ 0 is
(a) natural number.
(b) whole number.
(c) integer.
(d) rational number.

Answer:

A number which can be expressed as $\frac{p}{q}$ where p and q are integers and q ≠ 0 is a rational number.
Hence, the correct answer is option D.

Page No 9:

Question 2:

Choose the correct answer.
A number of the form $\frac{p}{q}$ is said to be a rational number if
(a) p and q are integers.
(b) p and q are integers and q ≠ 0
(c) p and q are integers and p ≠ 0
(d) p and q are integers and p ≠ 0 also q ≠ 0.

Answer:

A number of the form $\frac{p}{q}$ is said to be a rational number, if p and q are integers and q ≠ 0.
Hence, the correct answer is option B.

Page No 9:

Question 3:

Choose the correct answer.
The numerical expression $\frac{3}{8} + \frac{(- 5)}{7} = \frac{- 19}{56}$ shows that
(a) rational numbers are closed under addition.
(b) rational numbers are not closed under addition.
(c) rational numbers are closed under multiplication.
(d) addition of rational numbers is not commutative.

Answer:

We have $\frac{3}{8} + \frac{(- 5)}{7} = \frac{- 19}{56}$
$\frac{3}{8} and \frac{- 5}{7}$ are rational numbers and their addition is $\frac{- 19}{56}$ which is also a rational number.
$∴$ Rational numbers are closed under addition.
Hence, the correct answer is option A.

Page No 9:

Question 4:

Choose the correct answer.
Which of the following is not true?
(a) rational numbers are closed under addition.
(b) rational numbers are closed under subtraction.
(c) rational numbers are closed under multiplication.
(d) rational numbers are closed under division.

Answer:

Rational numbers are not closed under division.
As, 1 and 0 are rational numbers but $\frac{1}{0}$ is not defined.
Hence, the correct answer is option D.

Page No 9:

Question 5:

Choose the correct answer.
$- \frac{3}{8} + \frac{1}{7} = \frac{1}{7} + (\frac{- 3}{8})$ is an example to show that
(a) addition of rational numbers is commutative.
(b) rational numbers are closed under addition.
(c) addition of rational number is associative.
(d) rational numbers are distributive under addition.

Answer:

Given : $- \frac{3}{8} + \frac{1}{7} = \frac{1}{7} + (\frac{- 3}{8})$
Let a = $- \frac{3}{8} and b = \frac{1}{7}$
$∴ a + b = \frac{- 3}{8} + \frac{1}{7} = \frac{- 21 + 8}{56} = \frac{- 13}{56}$ and $b + a = \frac{1}{7} + \frac{- 3}{8} = \frac{8 - 21}{56} = \frac{- 13}{56}$
Clearly, a + b = b + a
So, addition of rational number is commutative.
Hence, the correct answer is option A.

Page No 9:

Question 6:

Choose the correct answer.
Which of the following expressions shows that rational numbers are associative under multiplication.

(a) $\frac{2}{3} \times (\frac{- 6}{7} \times \frac{3}{5}) = (\frac{2}{3} \times \frac{- 6}{7}) \times \frac{3}{5}$

(b) $\frac{2}{3} \times (\frac{- 6}{7} \times \frac{3}{5}) = \frac{2}{3} \times (\frac{3}{5} \times \frac{- 6}{7})$

(c) $\frac{2}{3} \times (\frac{- 6}{7} \times \frac{3}{5}) = (\frac{3}{5} \times \frac{2}{3}) \times \frac{- 6}{7}$

(d) $(\frac{2}{3} \times \frac{- 6}{7}) \times \frac{3}{5} = (\frac{- 6}{7} \times \frac{2}{3}) \times \frac{3}{5}$

Answer:

Associative property under multiplication can be expressed as $a \times (b \times c) = (a \times b) \times c$
The expression $\frac{2}{3} \times (\frac{- 6}{7} \times \frac{3}{5}) = (\frac{2}{3} \times \frac{- 6}{7}) \times \frac{3}{5}$
Shows that rational numbers are associated under multiplication.
Hence, the correct answer is option A.

Page No 10:

Question 7:

Choose the correct answer.
Zero (0) is
(a) the identity addition of rational numbers.
(b) the identity for subtraction of rational numbers.
(c) the identity for multiplication of rational numbers.
(d) the identity for division of rational numbers.

Answer:

Zero (0) is the identity for addition of rational numbers because if 'a' is a rational number then, a + 0 = 0 + a = a.
Hence, the correct answer is option A.

Page No 10:

Question 8:

Choose the correct answer.
One (1) is
(a) the identity for addition of rational numbers.
(b) the identity for subtraction of rational numbers.
(c) the identity for multiplication of rational numbers.
(d) the identity for division of rational numbers.

Answer:

One (1) is the identity for multiplication of rational numbers because if 'a' is a rational number then, a $\times$ 1 = 1 = a = a.
Hence, the correct answer is option C.

Page No 10:

Question 9:

Choose the correct answer.
The additive inverse of $\frac{- 7}{19}$ is
(a) $\frac{- 7}{19}$

(b) $\frac{7}{19}$

(c) $\frac{19}{7}$

(d) $\frac{- 19}{7}$

Answer:

If 'a' and 'b' are the additive inverse of each other, then, a + b = 0
Let x be the additive inverse of $\frac{- 7}{19} .$
$\begin{array}{l} ∴ x + (\frac{- 7}{19}) = 0 \\ \Rightarrow x = \frac{7}{19} \end{array}$
$∴$ Additive inverse of $\frac{- 7}{19} = \frac{7}{19} .$
Hence, the correct answer is option B.

Page No 10:

Question 10:

Choose the correct answer.
Multiplicative inverse of a negative rational number is
(a) a positive rational number.
(b) a negative rational number.
(c) 0
(d) 1

Answer:

The product of any rational number and its multiplicative inverse is 1.
Thus, the multiplicative inverse of a negative rational number must be negative so that their product is a positive rational number i.e, 1.
Hence, the correct answer is option B.

Page No 10:

Question 11:

Choose the correct answer.
If x + 0 = 0 + x = x, which is rational number, then 0 is called
(a) identity for addition of rational numbers.
(b) additive inverse of x.
(c) multiplicative inverse of x.
(d) reciprocal of x.

Answer:

The sum of any rational number and zero (0) is the rational number itself.
if $x + 0 = 0 + x = x$ , which is a rational number, thus 0 is called identity for addition of rational numbers.
Hence, the correct answer is option A.

Page No 10:

Question 12:

Choose the correct answer.
To get the product 1, we should multiply $\frac{8}{21}$ by

(a) $\frac{8}{21}$

(b) $\frac{- 8}{21}$

(c) $\frac{21}{8}$

(d) $\frac{- 21}{8}$

Answer:

To get the product 1, we should multiply $\frac{8}{21}$ by $\frac{21}{8}$
$∴$ The product of any number with its reciprocal results 1.
Reciprocal of the number $\frac{8}{21} is \frac{21}{8}$
Hence, the correct answer is option C.

Page No 11:

Question 13:

Choose the correct answer.
− (−x) is same as

(a) −x

(b) x

(c) $\frac{1}{x}$

(d) $\frac{- 1}{x}$

Answer:

Negative pf negative number is equal to a positive rational number
$∴ - (- x) = x$
Hence, the correct answer is option B.

Page No 11:

Question 14:

Choose the correct answer.
The multiplicative inverse of $- 1 \frac{1}{7}$ is
(a) $\frac{8}{7}$

(b) $\frac{- 8}{7}$

(c) $\frac{7}{8}$

(d) $\frac{7}{- 8}$

Answer:

Given : $- 1 \frac{1}{7}, i . e . \frac{- 8}{7}$
let the multiplication inverse of $\frac{- 8}{7}$ be x
$\Rightarrow \frac{- 8}{7} \times x = 1$ ( $∴$ The product of a number with its multiplicative inverse is always 1).
$\Rightarrow x = \frac{- 7}{8}$
Hence, the correct answer is option D.

Page No 11:

Question 15:

Choose the correct answer.
If x be any rational number then x + 0 is equal to
(a) x
(b) 0
(c) −x
(d) Not defined

Answer:

If x is any rational number, then $x + 0 = x$ , as 0 is the additive identity.
Hence, the correct answer is option A.

Page No 11:

Question 16:

Choose the correct answer.
The reciprocal of 1 is
(a) 1
(b) −1
(c) 0
(d) Not defined

Answer:

The reciprocal of 1 is the number itself.
Hence, the correct answer is option A.

Page No 11:

Question 17:

Choose the correct answer.
The reciprocal of −1 is
(a) 1
(b) −1
(c) 0
(d) Not defined

Answer:

The reciprocal of -1 is the number itself.
Hence, the correct answer is option B.

Page No 11:

Question 18:

Choose the correct answer.
The reciprocal of 0 is
(a) 1
(b) −1
(c) 0
(d) Not defined

Answer:

Reciprocal of 0 = $\frac{1}{0}$ , which is not defined.
Hence, the correct answer is option D.

Page No 11:

Question 19:

Choose the correct answer.
The reciprocal of any rational number $\frac{p}{q}$ , where p and q are integers and q ≠ 0 is

(a) $\frac{p}{q}$

(b) 1

(c) 0

(d) $\frac{q}{p}$

Answer:

The reciprocal of any rational number $\frac{p}{q}$ , where p and q are integers and $q \neq 0$ is $\frac{q}{p}$ .
Hence, the correct answer is option D.

Page No 11:

Question 20:

Choose the correct answer.
If y be the reciprocal of rational number x, then the reciprocal of y will be

(a) x

(b) y

(c) $\frac{x}{y}$

(d) $\frac{y}{x}$

Answer:

If $y$ be the reciprocal of x, then $y = \frac{1}{x}$ or $x = \frac{1}{y}$
Therefore, the reciprocal of $y$ will be $x$
Hence, the correct answer is option A.

Page No 11:

Question 21:

Choose the correct answer.
The reciprocal of $\frac{- 3}{8} \times (\frac{- 7}{13})$ is

(a) $\frac{104}{21}$

(b) $\frac{- 104}{21}$

(c) $\frac{21}{104}$

(d) $\frac{- 21}{104}$

Answer:

$\frac{- 3}{8} \times (\frac{- 7}{13}) = \frac{21}{104}$
Reciprocal of $\frac{21}{104}$ will be $\frac{104}{21}$
Hence, the correct answer is option A.

Page No 11:

Question 22:

Choose the correct answer.
Which of the following is an example of distributive property of multiplication over addition for rational numbers.

(a) $- \frac{1}{4} \times \{\frac{2}{3} + (\frac{- 4}{7})\} = [- \frac{1}{4} \times \frac{2}{3}] + [- \frac{1}{4} \times (\frac{- 4}{7})]$

(b) $- \frac{1}{4} \times \{\frac{2}{3} + (\frac{- 4}{7})\} = [\frac{1}{4} \times \frac{2}{3}] - (\frac{- 4}{7})$

(c) $- \frac{1}{4} \times \{\frac{2}{3} + (\frac{- 4}{7})\} = \frac{2}{3} + (- \frac{1}{4}) \times \frac{- 4}{7}$

(d) $- \frac{1}{4} \times \{\frac{2}{3} + (\frac{- 4}{7})\} = \{\frac{2}{3} + (\frac{- 4}{7})\} - \frac{1}{4}$

Answer:

The distributive property of multiplication over addition for rational numbers can be expressed as:
$a \times (b + c) = a \times b + a \times c$ , where $a, b and c$ are rational numbers:
Here, $\frac{- 1}{4} \times \{\frac{2}{3} + (\frac{- 4}{7})\} = [\frac{- 4}{4} \times \frac{2}{3}] + [\frac{- 1}{4} \times (\frac{- 4}{7})]$ is an example of the distributive property of multiplication over addition for rational numbers.
Hence, the correct answer is option A.

Page No 12:

Question 23:

Choose the correct answer.
Between two given rational numbers, we can find
(a) one and only one rational number.
(b) only two rational numbers.
(c) only ten rational numbers.
(d) infinitely many rational numbers.

Answer:

We can find infinitely many rational numbers between two given rational numbers.
Hence, the correct answer is option D.

Page No 13:

Question 24:

â€‹Choose the correct answer.
$\frac{x + y}{2}$ is a rational number.

(a) Between x and y

(b) Less than x and y both.

(c) Greater than x and y both.

(d) Less than x but greater than y.

Answer:

Here, $\frac{x + y}{2}$ is a rational number, then it always lies in between $x$ and $y$ .
Hence, the correct answer is option A.

Page No 13:

Question 25:

Choose the correct answer.
Which of the following statements is always true?

(a) $\frac{x - y}{2}$ is a rational number between x and y.

(b) $\frac{x + y}{2}$ is a rational number between x and y.

(c) $\frac{x \times y}{2}$ is a rational number between x and y.

(d) $\frac{x \div y}{2}$ is a rational number between x and y.

Answer:

$\frac{x + y}{2}$ is a rational number which lies in between $x$ and $y$ .
Hence, the correct answer is option B.

Page No 13:

Question 26:

Fill in the blanks to make the statements true.
The equivalent of $\frac{5}{7},$ whose numerator is 45 is ___________.

Answer:

To get 45 in the numerator, multiply both numerator and denominator by 9.
Then, $\frac{(5 \times 9)}{7 \times 9} = \frac{45}{63}$
So, the equivalent of $\frac{5}{7}$ , whose mumerator is 45 is $\frac{45}{63}$ .
Hence, the correct answer is $\frac{45}{63}$ .

Page No 13:

Question 27:

Fill in the blanks to make the statements true.
The equivalent rational number of $\frac{7}{9},$ whose denominator is 45 is ___________.

Answer:

To get 45 in the denominator multiply both numerator and denominator by 5.
Then, $\frac{(7 \times 5)}{(9 \times 5)} = \frac{35}{45}$
So, the equivalent of $\frac{7}{9}$ , whose denominator is 45 is $\frac{35}{45}$ .
Hence, the correct answer is $\frac{35}{45}$ .

Page No 13:

Question 28:

Fill in the blanks to make the statements true.
Between the numbers $\frac{15}{20}$ and $\frac{35}{40}$ , the greater number is __________.

Answer:

Given: $\frac{15}{20}$ and $\frac{35}{40}$
The LCM of the denominators 20 and 40 is 40.
$∴ \frac{15}{20} = \frac{15 \times 2}{20 \times 2} = \frac{30}{40}$ and $\frac{35}{20} = \frac{35 \times 1}{40 \times 1} = \frac{35}{40}$
Now, 30 < 35
$\Rightarrow \frac{30}{40} < \frac{35}{40}$
Thus, $\frac{15}{20} < \frac{35}{40}$
$∴ \frac{35}{40}$ is greater
Hence, the correct answer is $\frac{35}{40} .$

Page No 13:

Question 29:

Fill in the blanks to make the statements true.
The reciprocal of a positive rational number is ___________.

Answer:

The reciprocal of a positive rational number is always a positive rational number.
Hence, the correct answer is 'positive national number'.

Page No 13:

Question 30:

Fill in the blanks to make the statements true.
The reciprocal of a negative rational number is ___________.

Answer:

The reciprocal of a negative rational number is always a negative rational number.
Hence, the correct answer is 'negative rational number'.

Page No 13:

Question 31:

Fill in the blanks to make the statements true.
Zero has ___________ reciprocal.

Answer:

Reciprocal of 0 is $\frac{1}{0}$ which is not defined.
$∴$ Zero has no reciprocal.
Hence, the correct answer is 'no'.

Page No 13:

Question 32:

Fill in the blanks to make the statements true.
The numbers ___________ and ___________ are their own reciprocal.

Answer:

Reciprocal of 1 and -1 are $\frac{1}{1} and \frac{1}{- 1}$ i.e., 1 and -1 respectively.
$∴$ The numbers 1 and -1 are their own reciprocal.
Hence, the correct answer is '1, -1'.

Page No 13:

Question 33:

Fill in the blanks to make the statements true.
If y be the reciprocal of x, then the reciprocal of y² in terms of x will be ___________.

Answer:

Given: $y$ is the reciprocal of $x$
$∴ \frac{1}{x} = y$
Now, reciprocal of $y^{2} = \frac{1}{y^{2}} = \frac{1}{{(\frac{1}{x})}^{2}}$
$= x^{2}$
Hence, the correct answer is ' $x^{2}$ '.

Page No 13:

Question 34:

Fill in the blanks to make the statements true.
The reciprocal of $\frac{2}{5} \times (\frac{- 4}{9})$ is ___________.

Answer:

$\frac{2}{5} \times (\frac{- 4}{9}) = \frac{- 8}{45}$
The reciprocal of $\frac{- 8}{45}$ is $\frac{- 45}{8}$
Hence, the correct answer is $\frac{- 45}{8} .$

Page No 13:

Question 35:

Fill in the blanks to make the statements true.
(213 × 657)⁻¹ = 213⁻¹ × ___________.

Answer:

$\begin{array}{l} {(213 \times 657)}^{- 1} = {(213)}^{- 1} \times x \\ \frac{1}{213 \times 657} = \frac{1}{213} \times x \\ \Rightarrow x = \frac{213}{213 \times 657} \\ \Rightarrow x = \frac{1}{657} o r {(657)}^{- 1} \end{array}$
Hence, the correct answer is $\frac{1}{657}$ or ${(657)}^{- 1}$ .

Page No 13:

Question 36:

Fill in the blanks to make the statements true.
The negative of 1 is ___________.

Answer:

The negative of 1 is $- 1$ .

Page No 15:

Question 37:

Fill in the blanks to make the statements true.
For rational numbers $\frac{a}{b}, \frac{c}{d}$ and $\frac{e}{f}$ we have $\frac{a}{b} \times (\frac{c}{d} + \frac{e}{f}) =________+________.$

Answer:

For rational numbers $\frac{a}{b}, \frac{c}{d} a n d \frac{e}{f}$ we have
$\frac{a}{b} \times (\frac{c}{d} + \frac{e}{f}) = \frac{a}{b} \times \frac{c}{d} + \frac{a}{b} \times \frac{e}{f} = \frac{a c}{b d} + \frac{a e}{b f}$

Page No 15:

Question 38:

Fill in the blanks to make the statements true.
$\frac{- 5}{7}$ is ________ than –3.

Answer:

As, $- 3 = - 3 \times \frac{7}{7}$ (multiplying and dividing by 7)
$= \frac{- 21}{7}$
And, $\frac{- 5}{7} > \frac{- 21}{7}$
$∴ \frac{- 5}{7} > - 3$
Hence, $\frac{- 5}{7}$ is greater then −3

Page No 15:

Question 39:

Fill in the blanks to make the statements true.
There are ________ rational numbers between any two rational numbers.

Answer:

There are infinite rational numbers between any two rational numbers.

Page No 15:

Question 40:

Fill in the blanks to make the statements true.
The rational numbers $\frac{1}{3}$ and $\frac{- 1}{3}$ are on the ________ sides of zero on the number line.

Answer:

The rational numbers $\frac{1}{3}$ is a positive rational and hence is located on the right side of '0' on the number line.
Where as, the rational number $\frac{- 1}{3}$ is a negative rational number and hence is located on the left side of '0' on the number line.

Thus, the rational number $\frac{1}{3}$ and $\frac{- 1}{3}$ are on the opposite sides of zero on the number line.

Page No 15:

Question 41:

Fill in the blanks to make the statements true.
The negative of a negative rational number is always a ________ rational number.

Answer:

Let 'a' be a positive rational number then, -a will be a negative rational number.
Now, the negative of −a is −(−a) = a, which is a positive rational number.
Hence, the negative of a negative rational number is always a positive rational number.

Page No 15:

Question 42:

Fill in the blanks to make the statements true.
Rational numbers can be added or multiplied in any __________.

Answer:

According to the commutative property of rational numbers, rational numbers can be added or multiplied in any order.

Page No 15:

Question 43:

Fill in the blanks to make the statements true.
The reciprocal of $\frac{- 5}{7}$ is ________.

Answer:

The reciprocal of $\frac{- 5}{7}$ is $\frac{1}{(\frac{- 5}{7})},$ i.e. $\frac{- 7}{5} .$

Page No 15:

Question 44:

Fill in the blanks to make the statements true.
The multiplicative inverse of $\frac{4}{3}$ is _________.

Answer:

Let $x$ be the multiplicative invers of $\frac{4}{3}$ .
Then, by the definition of multiplicative inverse $x \times \frac{4}{3} = 1$
Hence, the multiplicative inverse of $\frac{4}{3}$ is $\frac{3}{4}$ .

Page No 15:

Question 45:

Fill in the blanks to make the statements true.
The rational number 10.11 in the from $\frac{p}{q}$ is _________.

Answer:

10.11 can be written as $\frac{1011}{100}$ , which is in the form of $\frac{p}{q}$ .
Hence, the rational number 10.11 in the form $\frac{p}{q}$ is $\frac{1011}{100}$ .

Page No 15:

Question 46:

Fill in the blanks to make the statements true.
$\frac{1}{5} \times [\frac{2}{7} + \frac{3}{8}] = [\frac{1}{5} \times \frac{2}{7}] +__________.$

Answer:

$\frac{1}{5} \times [\frac{2}{7} + \frac{3}{8}] = [\frac{1}{5} \times \frac{2}{7}] + \frac{1}{5} \times \frac{3}{8}$ (By distributive law of multiplication)

Page No 15:

Question 47:

Fill in the blanks to make the statements true.
The two rational numbers lying between −2 and −5 with denominator as 1 are _________ and _________.

Answer:

From the number line, it can be clearly observed that the two rational numbers lying between $- 2$ and $- 5$ with denominator as 1 are $- 3 and - 4 .$

Page No 15:

Question 48:

State whether the statements are true (T) or false (F).
If $\frac{x}{y}$ is a rational number, then y is always a whole number.

Answer:

False,
By the definition of rational numbers, if $\frac{x}{y}$ is a rational number then $x$ and $y$ are integers and $y \neq 0 .$
Thus, $y$ is always a non-zero integer.

Page No 15:

Question 49:

State whether the statements are true (T) or false (F).
If $\frac{p}{q}$ is a rational number, then p cannot be equal to zero.

Answer:

False,
As '0' can be written as $\frac{0}{1}$ which satisfies of the conditions of a number to be a rational number.
Therefore, '0' is a rational number.
Thus, if $\frac{p}{q}$ is a rational number, $p$ can be equal to zero as well.

Page No 15:

Question 50:

State whether the statements are true (T) or false (F).
If $\frac{r}{s}$ is a rational number, then s cannot be equal to zero.

Answer:

True,
By the definition of a rational number, if $\frac{r}{s}$ is a rational number then $s \neq 0$ .

Page No 15:

Question 51:

State whether the statements are true (T) or false (F).
$\frac{5}{6}$ lies between $\frac{2}{3}$ and 1.

Answer:

$As, \frac{2}{3} = \frac{2}{3} \times \frac{2}{2} = \frac{4}{6} and, 1 = \frac{1}{6} \times \frac{6}{6} = \frac{6}{6} So, \frac{4}{6} < \frac{5}{6} < \frac{6}{6} i . e \frac{2}{3} < \frac{5}{6} < 1$
Therefore, $\frac{5}{6}$ lines between $\frac{2}{3}$ and 1.

Page No 16:

Question 52:

State whether the statements are true (T) or false (F).
$\frac{5}{10}$ lies between $\frac{1}{2}$ and 1.

Answer:

False,
$As, \frac{1}{2} = \frac{1}{2} \times \frac{5}{5} = \frac{5}{10} and, 1 = 1 \times \frac{10}{10} = \frac{10}{10}$
Since $\frac{5}{10} = \frac{1}{2}$ , therefore, $\frac{1}{2}$ does not lie between $\frac{1}{2}$ and 1.

Page No 16:

Question 53:

State whether the statements are true (T) or false (F).
$\frac{- 7}{2}$ lies between –3 and –4.

Answer:

True,
$As, - 3 = - 3 \times \frac{2}{2} = \frac{- 6}{2} and, - 4 = - 4 \times \frac{2}{2} = \frac{- 8}{2} So, \frac{- 6}{2} > \frac{- 7}{2} > \frac{- 8}{2} i . e . - 3 > \frac{- 7}{2} > - 4$
Therefore, $\frac{- 7}{2}$ lies between -3 and -4.

Page No 16:

Question 54:

State whether the statements are true (T) or false (F).
$\frac{9}{6}$ lies between 1 and 2.

Answer:

True,
$As, 1 = 1 \times \frac{6}{6} = \frac{6}{6} and, 2 = 2 \times \frac{6}{6} = \frac{12}{6} So, \frac{6}{6} < \frac{9}{6} < \frac{12}{6} i . e 1 < \frac{9}{6} < 2$
Therefore, $\frac{9}{6}$ lies between 1 and 2.

Page No 16:

Question 55:

State whether the statements are true (T) or false (F).
If a ≠ 0, the multiplicative inverse of $\frac{a}{b}$ is $\frac{b}{a} .$

Answer:

True,
If $a = 0$ , then the multiplicative inverse of $\frac{a}{b}$ is not defined.
If $a \neq 0$ , then the multiplicative inverse of $\frac{a}{b}$ is $\frac{b}{a}$ as by the definition o multiplicative inverse $\frac{a}{b} \times \frac{b}{a} = 1 .$

Page No 16:

Question 56:

State whether the statements are true (T) or false (F).
The multiplicative inverse of $\frac{- 3}{5}$ is $\frac{5}{3} .$

Answer:

False,
The multiplicative inverse of $\frac{- 3}{5}$ is $\frac{1}{(\frac{- 3}{5})} i . e . \frac{- 5}{3} .$

Page No 16:

Question 57:

State whether the statements are true (T) or false (F).
The additive inverse of $\frac{1}{2}$ is –2.

Answer:

False,
Let $x$ be the additive inverse of $\frac{1}{2}$ .
Then, $\frac{1}{2} + x = 0$ (By the definition of additive inverse)
$\Rightarrow x = \frac{- 1}{2}$
Hence, the additive inverse of $\frac{1}{2}$ is $- \frac{1}{2}$ and not $- 2$ .

Page No 16:

Question 58:

State whether the statements are true (T) or false (F).
If $\frac{x}{y}$ is the additive inverse of $\frac{c}{d},$ then $\frac{x}{y} + \frac{c}{d} = 0 .$

Answer:

Ture,
If $\frac{x}{y}$ is the additive inverse of $\frac{c}{d}$ the, by the definition of additive inverse $\frac{x}{y} + \frac{c}{d} = 0 .$

Page No 16:

Question 59:

State whether the statements are true (T) or false (F).
For every rational number x, x + 1 = x.

Answer:

False,
Let us take an example of x = 4.
Then, clearly 4 + 1 = 5 ≠ 4.
Therefore, x + 1 ≠ x for every rational number x.

Page No 16:

Question 60:

State whether the statements are true (T) or false (F).
If $\frac{x}{y}$ is the additive inverse of $\frac{c}{d},$ then $\frac{x}{y} - \frac{c}{d} = 0$

Answer:

False,
Let us take an example of $\frac{x}{y} = \frac{3}{4} .$
Since the additive inverse of $\frac{3}{4}$ is $\frac{- 3}{4}$ .
Therefore, $\frac{c}{d} = \frac{- 3}{4}$
$\begin{array}{rcl} Now, consider \frac{x}{y} - \frac{c}{d} & = & \frac{3}{4} - (\frac{- 3}{4}) \\ = & \frac{3}{4} + \frac{3}{4} \\ = & \frac{6}{4} \neq 0 \end{array}$

Page No 16:

Question 61:

State whether the statements are true (T) or false (F).
The reciprocal of a non-zero rational number $\frac{q}{p}$ is the rational number $\frac{q}{p} .$

Answer:

False,
The reciprocal of a non-zero rational number $\frac{q}{p} is \frac{p}{q}$ .

Page No 16:

Question 62:

State whether the statements are true (T) or false (F).
If x + y = 0, then −y is known as the negative of x, where x and y are rational numbers.

Answer:

False,
If x + y = 0 then x = −y.
Thus, y here is the negative of x.

Page No 16:

Question 63:

State whether the statements are true (T) or false (F).
The negative of the negative of any rational number is the number itself.

Answer:

True,
Let x be any positive rational number.
Then, $- x$ will be the negative rational number.
Now, negative of negative rational number
= −(−x) = x
Thus, the negative of a negative rational number is the number itself.

Page No 16:

Question 64:

State whether the statements are true (T) or false (F).
The negative of 0 does not exist.

Answer:

True,
As, zero is neither negative nor positive.
Therefore, the negative of zero does not exist.

Page No 16:

Question 65:

State whether the statements are true (T) or false (F).
The negative of 1 is 1 itself.

Answer:

False
The negative of 1 is $- 1$ .

Page No 16:

Question 66:

State whether the statements are true (T) or false (F).
For all rational numbers x and y, x − y = y − x.

Answer:

False,
Let us take an example of $x = 1 and y = 2$ .
Now, consider $x - y = 1 - 2 = - 1 .$
Also, consider $y - x = 2 - 1 = 1$
Here, x − y ≠ y − x
Thus, for all rational number x and y, x − y ≠ y − x. In fact, for all rational numbers x and y, x − y = −(y − x).

Page No 16:

Question 67:

State whether the statements are true (T) or false (F).
For all rational numbers x and y, x × y = y × x.

Answer:

True,
By commutative property of multiplication, for all rational numbers $x and y, x y = y x .$

Page No 17:

Question 68:

State whether the statements are true (T) or false (F).
For every rational number x, x × 0 = x.

Answer:

False,
For every rational number $x, x \times 0 = 0$ .

Page No 17:

Question 69:

State whether the statements are true (T) or false (F).
For every rational numbers x, y and z, x + (y × z) = (x + y) × (x + z).

Answer:

False,
Let us take an example of $x = 1, y = 1 and z = 3$ .
Now, consider $x + (y \times 2) = 1 + (2 \times 3) = 7$
Also, consider $(x + y) \pm (x + 3) = (1 + 2) \times (1 + 3) = 12$
So, $x + (y \times 2) \neq (x + y) \times (x + 3)$

Page No 17:

Question 70:

State whether the statements are true (T) or false (F).
For all rational numbers a, b and c, a (b + c) = ab + bc.

Answer:

False,
Let us take an example of $a = 1, b = 1 and c = 3$
Now, consider $a \times (b + c) = 1 \times (2 + 3) = 4$
Also, consider $a b + b c = (1 \times 2) + (2 \times 3) = 8$
So, $a (b + c) \neq a b + b c$

Page No 17:

Question 71:

State whether the statements are true (T) or false (F).
1 is the only number which is its own reciprocal.

Answer:

True,
1 and −1 both are reciprocal of their own.
As, the reciprocal of 1 is i.e. 1 and the reciprocal of -1 is $\frac{1}{- 1} i . e . - 1$
So, 1 is not the only number which is its ow reciprocal.

Page No 17:

Question 72:

State whether the statements are true (T) or false (F).
−1 is not the reciprocal of any rational number.

Answer:

False,
1 and −1 are the only rational numbers that are their own reciprocals.
Hence, −1 is not the reciprocal of any other rational number.

Page No 17:

Question 73:

State whether the statements are true (T) or false (F).
For any rational number x, x + (−1) = −x.

Answer:

False,
let us take an example of $x = 3$
Now, consider $x + (- 1) = 3 + (- 1) = 2 \neq - 3$
thus, $x + (- 1) \neq - x .$

Page No 17:

Question 74:

State whether the statements are true (T) or false (F).
For rational numbers x and y, if x < y then x − y is a positive rational number.

Answer:

False,
Let us take an example of $x = 1$ and $y = 2$
Here, $x < y$ but $x - y = 1 - 2 = - 1$ is not a positive rational number.
Hence, the statement is false.

Page No 17:

Question 75:

State whether the statements are true (T) or false (F).
If x and y are negative rational numbers, then so is x + y.

Answer:

true,
Let $x = - m$ , where $m$ is a positive rational and $y = - n,$ where $n$ is a positive rational number
Then, clearly $x$ and $y$ are negative rational number.
Now, consider $x + y = - m + (- n) = - m - n = - (m + n)$
Since, $m$ and $n$ is also a positive rational number.
Therefore, $m + n$ is also a positive rational number.
Thus, $- (m + n)$ is a negative rational number and so is $x + y$ .

Page No 17:

Question 76:

State whether the statements are true (T) or false (F).
Between any two rational numbers there are exactly ten rational numbers.

Answer:

False,
There are infinite rational number between any two rational numbers.

Page No 17:

Question 77:

State whether the statements are true (T) or false (F).
Rational numbers are closed under addition and multiplication but not under subtraction.

Answer:

False,
Rational numbers are closed under subtraction as well.
Because, if $x and y$ are two rational number then $x - y or y - x$ is also a rational number.

Page No 17:

Question 78:

State whether the statements are true (T) or false (F).
Subtraction of rational numbers is commutative.

Answer:

False,
Subtraction of rational numbers is not commutative.
For example: $x = 1 and y = 2$
then, $x - y = 1 - 2 = - 1 and y - x = 2 - 1 = 1$
Thus, $x - y \neq y - x .$

Page No 17:

Question 79:

State whether the statements are true (T) or false (F).
$- \frac{3}{4}$ is smaller than −2.

Answer:

False,
$- 2 = - 2 \times \frac{4}{4} = \frac{- 8}{4} As, - 8 < - 3 ∴ \frac{- 8}{4} < \frac{- 3}{4} \Rightarrow - 2 < \frac{- 3}{4}$
Hence, $\frac{- 3}{4}$ is greater than $- 2$ .

Page No 17:

Question 80:

State whether the statements are true (T) or false (F).
0 is a rational number.

Answer:

True,
0 can be written as $\frac{0}{1}$ which is in the form of $\frac{p}{q}$ where $q \neq 0 .$
Therefore, 0 is a rational number.

Page No 17:

Question 81:

State whether the statements are true (T) or false (F).
All positive rational numbers lie between 0 and 1000.

Answer:

False,
There are infinite positive rational numbers on the right side of 0 on the number line that are not necessarily lying between 0 and 1000.
For example: 1001, 2000, $\frac{12350}{5}, \frac{17981}{10}$ etc.

Page No 17:

Question 82:

State whether the statements are true (T) or false (F).
The population of India in 2004 - 05 is a rational number.

Answer:

True,
The population of India can always be a whole number and all the whole numbers are rational numbers.

Page No 17:

Question 83:

State whether the statements are true (T) or false (F).
There are countless rational numbers between $\frac{5}{6}$ and $\frac{8}{9} .$

Answer:

True,
There are countless rational numbers between any two rational numbers.

Page No 17:

Question 84:

State whether the statements are true (T) or false (F).
The reciprocal of x⁻¹ is $\frac{1}{x}$

Answer:

False,
The reciprocal of x⁻¹ is $\frac{1}{x^{- 1}} i . e . x .$

Page No 17:

Question 85:

State whether the statements are true (T) or false (F).
The rational number $\frac{57}{23}$ lies to the left of zero on the number line.

Answer:

False,
$\frac{57}{23}$ is a positive rational number and all the positive rational numbers lie to the right of zero on the number line.

Page No 17:

Question 86:

State whether the statements are true (T) or false (F).
The rational number $\frac{7}{- 4}$ lies to the right of zero on the number line.

Answer:

False,
$\frac{7}{- 4} = \frac{- 7}{4}$ is a negative rational number and all the negative rational numbers lie to the left of zero on the number line.

Page No 17:

Question 87:

State whether the statements are true (T) or false (F).
The rational number $\frac{- 8}{- 3}$ lies neither to the right nor to the left of zero on the number line.

Answer:

False, $\frac{- 8}{- 3} = \frac{8}{3}$ is a positive rational number and all the positive rational numbers lie to the right of zero on the number line.

Page No 18:

Question 88:

State whether the statements are true (T) or false (F).
The rational numbers $\frac{1}{2}$ and −1 are on the opposite sides of zero on the number line.

Answer:

True,
$\frac{1}{2}$ is a positive rational number. So, it lies to the right side of a 0 on the number line.
$- 1$ is a negative rational number. So, it lies to the left side of 0 on the number line.
Hence, $\frac{1}{2}$ and $- 1$ are on the opposite sides of zero on the number line.

Page No 18:

Question 89:

State whether the statements are true (T) or false (F).
Every fraction is a rational number.

Answer:

True,
A fraction is a part of the whole which can be expressed in the form of $\frac{p}{q}$ where $p and q$ are whole number and $q \neq 0 .$
On the other hand, a rational number is a number which can be expressed in the form of $\frac{p}{q}$ where $p and q$ are integers and $q \neq 0 .$
Since all the whole numbers are integers.
Therefore, all the fractions are rational numbers.

Page No 18:

Question 90:

State whether the statements are true (T) or false (F).
Every integer is a rational number.

Answer:

True,
Let $a$ be any integer then $a$ can be written as $\frac{a}{1}$ which satisfies all the conditions to be a rational number.
Hence, all the integers are rational numbers.

Page No 18:

Question 91:

State whether the statements are true (T) or false (F).
The rational numbers can be represented on the number line.

Answer:

True,
All the rational numbers can be represented on the number line.

Page No 18:

Question 92:

State whether the statements are true (T) or false (F).
The negative of a negative rational number is a positive rational number.

Answer:

True,
Let $\frac{p}{q}$ be a positive rational number then $- \frac{p}{q}$ is a negative rational number.
Now, the negative of a negative rational number is $- (- \frac{p}{q}) = \frac{p}{q}$ which is a positive rational number.

Page No 18:

Question 93:

State whether the statements are true (T) or false (F).
If x and y are two rational numbers such that x > y, then x − y is always a positive rational number.

Answer:

True,
If $x > y$ , then $x - y$ is always a positive rational number.
For example: If $x = 2$ and $y = 1$ then $x - y = 1$ which is a positive rational number.

Page No 18:

Question 94:

State whether the statements are true (T) or false (F).
0 is the smallest rational number.

Answer:

False,
There are infinite rational numbers on the left of 0 on the number line. Thus, the smallest rational number does not exist.

Page No 18:

Question 95:

State whether the statements are true (T) or false (F).
Every whole number is an integer.

Answer:

True,
Whole numbers are 0, 1, 2, 3.........
and integers are ........-3, -2, -1, 0, 1, 2........
Thus, integers consists of negative numbers and whole numbers.
Hence, every whole number is an integer but the converse is not true.

Page No 18:

Question 96:

State whether the statements are true (T) or false (F).
Every whole number is a rational number.

Answer:

True,
Let ' $a$ ' be a whole number then, it can be written as $\frac{a}{1}$ which satisfies all the conditions to be a rational number.
Hence, every whole number is a rational numbers.

Page No 18:

Question 97:

State whether the statements are true (T) or false (F).
0 is whole number but it is not a rational number.

Answer:

False,
0 is a whole number and all the whole number are rational numbers as they can be written in the form of $\frac{p}{q}$ where $q = 1 \neq 0 .$

Page No 18:

Question 98:

State whether the statements are true (T) or false (F).
The rational numbers $\frac{1}{2}$ and $- \frac{5}{2}$ are on the opposite sides of 0 on the number line.

Answer:

True,
$\frac{1}{2}$ is a positive rational number. So, it lies on the right of 0 on the number line.
whereas, $\frac{- 5}{2}$ is a negative rational number.
So, it lies on the left of 0 on the number line.
Hence, $\frac{1}{2}$ and $\frac{- 5}{2}$ lies on the positive sides of the 0 on the number line.

Page No 18:

Question 99:

State whether the statements are true (T) or false (F).
Rational numbers can be added (or multiplied) in any order $\frac{- 4}{5} \times \frac{- 6}{5} = \frac{- 6}{5} \times \frac{- 4}{5}$

Answer:

True,
The commutative property of addition (or multiplication) for rational number says that the rational number can be added (or multiplied) in any order.
For example: $(\frac{- 4}{5}) \times (\frac{- 6}{5}) = (\frac{- 6}{5}) \times (\frac{- 4}{5})$

Page No 18:

Question 100:

Solve the following: Select the rational numbers from the list which are also the integers.

$\frac{9}{4}, \frac{8}{4}, \frac{7}{4}, \frac{6}{4}, \frac{9}{3}, \frac{8}{3}, \frac{7}{3}, \frac{6}{3}, \frac{5}{2}, \frac{4}{2}, \frac{3}{1}, \frac{3}{2}, \frac{1}{1}, \frac{0}{1}, \frac{- 1}{1}, \frac{- 2}{1}, \frac{- 3}{2}, \frac{- 4}{2}, \frac{- 5}{2}, \frac{- 6}{2},$

Answer:

A rational number whose denominator is either 1 or -1 or the factor of numerator, is an integer.
Hence, form the given rational numbers $\frac{8}{4}, \frac{9}{3}, \frac{6}{3}, \frac{4}{2}, \frac{3}{1}, \frac{1}{1}, \frac{- 1}{1}, \frac{- 2}{1}, \frac{- 4}{2} and \frac{- 6}{2}$ are integers.

Page No 18:

Question 101:

Select those which can be written as a rational number with denominator 4 in their lowest form:

$\frac{7}{8}, \frac{64}{16}, \frac{36}{- 12}, \frac{- 16}{17}, \frac{5}{- 4}, \frac{140}{28}$

Answer:

$\begin{array}{rcl} As, \frac{64}{16} & = & \frac{64 \div 4}{16 \div 4} = \frac{16}{4}, \\ \frac{36}{- 12} & = & \frac{36 \div (- 3)}{- 12 \div (- 3)} = \frac{- 12}{4}, \\ \frac{5}{- 4} & = & \frac{5 \div (- 1)}{- 4 \div (- 1)} = \frac{- 5}{4}, \\ \frac{140}{28} & = & \frac{140 \div 7}{28 \div 7} = \frac{20}{4} \end{array}$
$∴ \frac{64}{16}, \frac{36}{- 12}, \frac{5}{- 4} and \frac{140}{28}$ can be written as a rational number with denominator 4 in their lowest form.

Page No 19:

Question 102:

Using suitable rearrangement and find the sum:

(a) $\frac{4}{7} + (\frac{- 4}{9}) + \frac{3}{7} + (\frac{- 13}{9})$

(b) $- 5 + \frac{7}{10} + \frac{3}{7} + (- 3) + \frac{5}{14} + \frac{- 4}{5}$

Answer:

(a) $\frac{4}{7} + (\frac{- 4}{9}) + \frac{3}{7} + (\frac{- 13}{9})$
   $= \frac{4}{7} + \frac{3}{7} + (\frac{- 4}{9}) + (\frac{- 13}{9})$ (As rational numbers are commutative under addition)
   $= \frac{7}{7} - (\frac{17}{9}) = 1 - \frac{17}{9} = \frac{- 8}{9}$

(b) $- 5 + \frac{7}{10} + \frac{3}{7} + (- 3) + \frac{5}{14} + (\frac{- 4}{5})$
   $= - 5 + (- 3) + \frac{7}{10} + (\frac{- 4}{5}) + \frac{3}{7} + \frac{5}{14}$ (As rational numbers are commutative under addition)
   $= - 8 + \frac{7 - 8}{10} + \frac{6 + 5}{14} = - 8 - \frac{1}{10} + \frac{11}{14} = \frac{- 560 - 7 + 55}{70} = \frac{- 512}{70} = \frac{- 256}{35}$

Page No 19:

Question 103:

Verify − (−x) = x for

(i) $x = \frac{3}{5}$

(ii) $x = \frac{- 7}{9}$

(iii) $x = \frac{13}{- 15}$

Answer:

(a) Given: $x = \frac{3}{5}$
       $\Rightarrow - x = \frac{- 3}{5} \Rightarrow - (- x) = - (\frac{- 3}{5}) = \frac{3}{5} = x$
    Thus, $- (- x) = x .$
    Hence verified.

(b) Given: $x = \frac{- 7}{9}$
       $\Rightarrow - x = - (\frac{- 7}{9}) = \frac{7}{9} \Rightarrow - (- x) = - (\frac{7}{9}) = x \Rightarrow - (- x) = x$
    Hence verified.

(c) Given: $x = \frac{13}{- 15}$
     $\Rightarrow - x = - (\frac{13}{- 15}) = \frac{13}{15} \Rightarrow - (- x) = - \frac{13}{15} = \frac{13}{- 15} = x \Rightarrow - (- x) = x$
Hence verified.

Page No 19:

Question 104:

Give one example each to show that the rational numbers are closed under addition, subtraction and multiplication. Are rational numbers closed under division? Give two examples in support of your answer.

Answer:

Rational numbers are closed under addition.
Example: $\frac{2}{3} and \frac{- 1}{3}$ both are rationals and their sum i.e. $\frac{2}{3} + (\frac{- 1}{3}) = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$ is also a rational number.

Rational numbers are closed under subtraction.
Example: $\frac{1}{2} and \frac{- 2}{3}$ both are rationals and their subtraction i.e. $\frac{1}{2} - (\frac{- 2}{3}) = \frac{1}{2} + \frac{2}{3} = \frac{3 + 4}{6} = \frac{7}{6}$ is also a rational number.

Rational number are closed under multiplication.
Example: $\frac{1}{3} and \frac{1}{5}$ both are rational numbers and their multiplication i.e. $\frac{1}{3} \times \frac{1}{5} = \frac{1}{15}$ is also a rational number.

Rational number are not closed under division.
Example: (i) $\frac{1}{2}$ and 0 are rational numbers but $\frac{1}{2} \div 0 = \frac{1}{2} \times \frac{1}{0} = \frac{1}{0}$ is not defined.
(ii) $\frac{- 2}{3}$ and 0 are rational numbers but $\frac{- 2}{3} \div 0 = \frac{- 2}{3} \times \frac{1}{0}$ is also not defined.

Page No 19:

Question 105:

Verify the property x + y = y + x of rational numbers by taking

(a) $x = \frac{1}{2}, y = \frac{1}{2}$

(b) $x = \frac{- 2}{3}, y = \frac{- 5}{6}$

(c) $x = \frac{- 3}{7}, y = \frac{20}{21}$

(d) $x = \frac{- 2}{5}, y = \frac{- 9}{10}$

Answer:

(a) Given: $x = \frac{1}{2} and y = \frac{1}{2}$
      Consider, $x + y = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1$
      Also, consider $y + x = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1$
      Clearly, $x + y = y + x$
      Hence, verified.

(b) Given: $x = \frac{- 2}{3} and y = \frac{- 5}{6}$
      Consider, $x + y = \frac{- 2}{3} + (\frac{- 5}{6}) = \frac{- 2}{3} - \frac{5}{6} = \frac{- 4 - 5}{6} = \frac{- 9}{6}$
      Also, consider $y + x = \frac{- 5}{6} + (\frac{- 2}{3}) = \frac{- 5}{6} - \frac{2}{3} = \frac{- 5 - 4}{6} = \frac{- 9}{6}$
      Clearly, $x + y = y + x$
      Hence verified.

(c) Given: $x = \frac{- 3}{7} and y = \frac{20}{21}$
      Consider, $x + y = \frac{- 3}{7} = \frac{- 3}{7} + \frac{20}{21} = \frac{- 9 + 20}{21} = \frac{11}{21}$
      Also, consider $y + x = \frac{20}{21} + (\frac{- 3}{7}) = \frac{20}{21} - \frac{3}{7} = \frac{20 - 9}{21} = \frac{11}{21}$
      Clearly, $x + y = y + x$
Hence verified.
(d) Given: $x = \frac{- 2}{5} and y = \frac{- 9}{10}$
Consider, $x + y = \frac{- 2}{5} + (\frac{- 9}{10}) = \frac{- 2}{5} - \frac{9}{10} = \frac{- 4 - 9}{10} = \frac{- 13}{10}$
Also, consider $y + x = \frac{- 9}{10} + (\frac{- 2}{5}) = \frac{- 9}{10} - \frac{2}{5} = \frac{- 9 - 4}{10} = \frac{- 13}{10}$
Clearly, $x + y = y + x$
      Hence verified.

Page No 19:

Question 106:

Simplify each of the following by using suitable property. Also name the property.

(a) $[\frac{1}{2} \times \frac{1}{4}] + [\frac{1}{2} \times 6]$

(b) $[\frac{1}{5} \times \frac{2}{15}] - [\frac{1}{5} \times \frac{2}{5}]$

(c) $\frac{- 3}{5} \times \{\frac{3}{7} + (\frac{- 5}{6})\}$

Answer:

(a) $(\frac{1}{2} \times \frac{1}{4}) + (\frac{1}{2} \times 6) = \frac{1}{2} \times (\frac{1}{4} + 6)$ (using distributive property of multiplication over addition)
$= \frac{1}{2} \times (\frac{1 + 24}{4}) = \frac{1}{2} \times \frac{25}{4} = \frac{25}{8}$
(b) $(\frac{1}{5} \times \frac{2}{15}) - (\frac{1}{5} \times \frac{2}{5}) = \frac{1}{5} \times (\frac{2}{15} - \frac{2}{5})$ (using distributive property of multiplication over subtraction)
$= \frac{1}{5} \times (\frac{2 - 6}{15}) = \frac{1}{5} \times (\frac{- 4}{15}) = \frac{- 4}{75}$
(c) $\frac{- 3}{5} \times [\frac{3}{7} + (\frac{- 5}{6})] = \frac{- 3}{5} \times [\frac{18 + (- 35)}{42}]$
$= \frac{- 3}{5} \times (\frac{- 17}{42}) = \frac{51}{210} = \frac{17}{30}$

Page No 19:

Question 107:

Tell which property allows you to compute:

$\frac{1}{5} \times [\frac{5}{6} \times \frac{7}{9}]$ as $[\frac{1}{5} \times \frac{5}{6}] \times \frac{7}{9}$

Answer:

Associative property for multiplication states that for any rational numbers a, b and c.
$a \times (b \times c) = (a \times b) \times c$
So, $\frac{1}{5} \times (\frac{5}{6} \times \frac{7}{9})$ can be written as
$(\frac{1}{5} \times \frac{5}{6}) \times \frac{7}{9}$ by associative property for multiplication

Page No 19:

Question 108:

Verify the property x × y = y × x of rational numbers by using

(a) $x = 7 and y = \frac{1}{2}$

(b) $x = \frac{2}{3} and y = \frac{9}{4}$

(c) $x = \frac{- 5}{7} and y = \frac{14}{15}$

(d) $x = \frac{- 3}{8} and y = \frac{- 4}{9}$

Answer:

(a) Given: $x = 7 and y = \frac{1}{2}$
      Then, consider $x \times y = 7 \times \frac{1}{2} = \frac{7}{2}$
      Also, consider $y \times x = \frac{1}{2} \times 7 = \frac{7}{2}$
      Thus, $x \times y = y \times x$
      Hence, verified.
(b) Given: $x = \frac{2}{3} and y = \frac{9}{4}$
      Then, consider $x \times y = \frac{2}{3} \times \frac{9}{4} = \frac{18}{12} = \frac{3}{2}$
      Also, consider $y \times x = \frac{9}{4} \times \frac{2}{3} = \frac{18}{12} = \frac{3}{2}$
      Thus, $x \times y = y \times x$
      Hence verified.
(c) Given: $x = \frac{- 5}{7} and y = \frac{14}{15}$
      Then, consider $x \times y = \frac{- 5}{7} \times \frac{14}{15} = \frac{- 2}{3}$
      Also consider $y \times x = \frac{14}{15} \times \frac{- 5}{7} = \frac{- 2}{3}$
      Thus, $x \times y = y \times x$
      Hence verified.
(d) Given: $x = \frac{- 3}{8} and y = \frac{- 4}{9}$
      Then, consider $x \times y = \frac{- 3}{8} \times \frac{- 4}{9} = \frac{1}{6}$
      Also, consider $y \times x = \frac{- 4}{9} \times \frac{- 3}{8} = \frac{1}{6}$
      Thus, $x \times y = y \times x$
      Hence verified.

Page No 19:

Question 109:

Verify the property x × (y × z) = (x × y) × z of rational numbers by using

(a) $x = 1, y = \frac{- 1}{2} and z = \frac{1}{4}$

(b) $x = \frac{2}{3}, y = \frac{- 3}{7} and z = \frac{1}{2}$

(c) $x = \frac{- 2}{7}, y = \frac{- 5}{6} and z = \frac{1}{4}$

(d) $x = 0, y = \frac{1}{2}$

and What is the name of this property?

Answer:

(a) Given: $x = 1, y = \frac{- 1}{2} and z = \frac{1}{4}$
      Now, consider $x \times (y \times z) = 1 \times (\frac{- 1}{2} \times \frac{1}{4}) = 1 \times \frac{- 1}{8} = \frac{- 1}{8}$
      Also, consider $(x \times y) \times z = (1 \times \frac{- 1}{2}) \times \frac{1}{4} = \frac{- 1}{2} \times \frac{1}{4} = \frac{- 1}{8}$
      Thus, $x \times (y \times z) = (x \times y) \times z$
      Hence verified.
(b) Given: $x = \frac{2}{3}, y = \frac{- 3}{7} and z = \frac{1}{2}$
      Now, consider $x \times (y \times z) = \frac{2}{3} \times (\frac{- 3}{7} \times \frac{1}{2}) = \frac{2}{3} \times \frac{- 3}{14} = \frac{- 6}{42} = \frac{- 1}{7}$
      Also , consider $(x \times y) \times z = (\frac{2}{3} \times \frac{- 3}{7}) \times \frac{1}{2} = \frac{- 6}{21} \times \frac{1}{2} = \frac{- 6}{42} = \frac{- 1}{7}$
      Thus, $x \times (y \times z) = (x \times y) \times z$
      Hence verified.
(c) Given: $x = \frac{- 2}{7}, y = \frac{- 5}{6} and z = \frac{1}{4}$
      Now, consider $x \times (y \times z) = \frac{- 2}{7} \times (\frac{- 5}{6} \times \frac{1}{4}) = \frac{- 2}{7} \times \frac{- 5}{24} = \frac{10}{168} = \frac{5}{84}$
      Also, consider $(x \times y) \times z = (\frac{- 2}{7} \times \frac{- 5}{6}) \times \frac{1}{4} = \frac{10}{42} \times \frac{1}{4} = \frac{10}{168} = \frac{5}{84}$
      Thus, $x \times (y \times z) = (x \times y) \times z$
      Hence verified.
(d) Disclaimer:
      The question is incomplete.

The name of the verified property is associative property for multiplication.

Page No 20:

Question 110:

Verify the property x × (y + z) = x × y + x × z of rational numbers by taking.

(a) $x = \frac{- 1}{2}, y = \frac{3}{4}, z = \frac{1}{4}$

(b) $x = \frac{- 1}{2}, y = \frac{2}{3}, z = \frac{3}{4}$

(c) $x = \frac{- 2}{3}, y = \frac{- 4}{6}, z = \frac{- 7}{9}$

(d) $x = \frac{- 1}{5}, y = \frac{2}{15}, z = \frac{- 3}{10}$

Answer:

(a) Given: $x = \frac{- 1}{2}, y = \frac{3}{4}, z = \frac{1}{4}$
       Now, consider, $x \times (y + z) = \frac{- 1}{2} \times (\frac{3}{2} + \frac{1}{4}) = \frac{- 1}{2} \times 1 = \frac{- 1}{2}$
      Also, consider $x \times y + x \times z = \frac{- 1}{2} \times \frac{3}{4} + \frac{- 1}{2} \times \frac{1}{4} = \frac{- 3}{8} + \frac{- 1}{8} = \frac{- 4}{8} = \frac{- 1}{2}$
      Thus, $x \times (y + z) = x \times y + x \times z$
      Hence verified.
(b) Given: $x = \frac{- 1}{2}, y = \frac{2}{3}, z = \frac{3}{4}$
      Now, consider, $x \times (y + z) = \frac{1}{2} \times (\frac{2}{3} + \frac{3}{4}) = \frac{- 1}{2} \times (\frac{8 + 9}{12}) = \frac{- 1}{2} \times \frac{17}{12} = \frac{- 17}{24}$
      Also, consider $x \times y + x \times z = \frac{- 1}{2} \times \frac{2}{3} + \frac{- 1}{2} \times \frac{3}{4} = \frac{- 1}{3} - \frac{- 3}{8} = \frac{- 8 - 9}{24} = \frac{- 17}{24}$
      Thus, $x \times (y + x) = x \times y + x \times z$
      Hence verified.
(c) Given: $x = \frac{- 2}{3}, y = \frac{- 4}{6}, z = \frac{- 7}{9}$
      Now, consider,   $x \times (y + z) = \frac{- 2}{3} \times [\frac{- 4}{6} + (\frac{- 7}{9})] = \frac{- 2}{3} \times (\frac{- 12 - 14}{18}) = \frac{- 2}{3} \times \frac{26}{18} = \frac{26}{27}$
      Also, consider $x \times y + x \times z = \frac{- 2}{3} \times \frac{- 4}{6} + \frac{- 2}{3} \times \frac{- 7}{9} = \frac{8}{18} + (\frac{14}{27}) = \frac{12 + 14}{27} = \frac{26}{27}$
      Thus, $x \times (y + z) = x \times y + x \times z$
      Hence verified.
(c) Given: $x = \frac{- 1}{5}, y = \frac{2}{15}, z = \frac{- 3}{10}$
      Now, consider, $x \times (y + z) = \frac{- 1}{5} \times [\frac{2}{15} + (\frac{- 3}{10})] = \frac{- 1}{5} \times (\frac{4 - 9}{30}) = \frac{- 1}{5} \times \frac{- 5}{30} = \frac{1}{30}$
      Also, consider   $x \times y + x \times z = (\frac{- 1}{5}) \times \frac{2}{15} + (\frac{- 1}{5}) \times (\frac{- 3}{10}) = \frac{- 2}{75} + \frac{3}{50} = \frac{5}{150} = \frac{1}{30}$
      Thus, $x \times (y + z) = x \times y + x \times z$
      Hence verified.

Page No 20:

Question 111:

Use the distributivity of multiplication of rational numbers over addition to simplify

(a) $\frac{3}{5} \times [\frac{35}{24} + \frac{10}{1}]$

(b) $\frac{- 5}{4} \times [\frac{8}{5} + \frac{16}{15}]$

(c) $\frac{2}{7} \times [\frac{7}{16} - \frac{21}{4}]$

(d) $\frac{3}{4} \times [\frac{8}{9} - 40]$

Answer:

$\begin{array}{rcl} (a) \frac{3}{5} \times (\frac{35}{24} + \frac{10}{1}) & = & \frac{3}{5} \times \frac{35}{24} + \frac{3}{5} \times \frac{10}{1} (using distributive property of multiplication over addition) \\ = & \frac{7}{8} + \frac{6}{1} \\ = & \frac{7 + 48}{8} \\ = & \frac{55}{8} \end{array}$
$\begin{array}{rcl} (b) \frac{- 5}{4} \times (\frac{8}{5} + \frac{16}{15}) & = & (\frac{- 5}{4}) \times \frac{8}{5} + (\frac{- 5}{4}) \times \frac{16}{15} (using distributive property over addition of multiplication) \\ = & - 2 - \frac{4}{3} \\ = & \frac{- 6 - 4}{3} \\ = & \frac{- 10}{3} \end{array}$
$\begin{array}{rcl} (c) \frac{2}{7} \times (\frac{7}{16} - \frac{21}{4}) & = & \frac{2}{7} \times \frac{7}{16} - \frac{2}{7} \times \frac{21}{4} (using distributive property of multiplication over subtraction) \\ = & \frac{1}{8} - \frac{3}{2} \\ = & \frac{1 - 12}{8} \\ = & \frac{- 11}{8} \end{array}$
$\begin{array}{rcl} (d) \frac{3}{4} \times (\frac{8}{9} - 40) & = & \frac{3}{4} \times \frac{8}{9} - \frac{3}{4} \times 40 (using distributive property of multiplication over subtraction) \\ = & \frac{2}{3} - 30 \\ = & \frac{2 - 90}{3} \\ = & \frac{- 88}{3} \end{array}$

Page No 20:

Question 112:

Simplify:

(a) $\frac{32}{5} + \frac{23}{11} \times \frac{22}{15}$

(b) $\frac{3}{7} \times \frac{28}{15} \div \frac{14}{5}$

(c) $\frac{3}{7} + \frac{- 2}{21} \times \frac{- 5}{6}$

(d) $\frac{7}{8} + \frac{1}{16} - \frac{1}{12}$

Answer:

(a) $\frac{32}{5} + \frac{23}{11} \times \frac{22}{15} = \frac{32}{5} + \frac{46}{15} = \frac{96 + 46}{15} = \frac{142}{15}$

(b) $\frac{3}{7} \times \frac{28}{15} \div \frac{14}{5} = \frac{4}{5} \div \frac{14}{5} = \frac{4}{5} \times \frac{5}{14} = \frac{2}{7}$

(c) $\frac{3}{7} + (\frac{- 2}{21}) \times (\frac{- 5}{6}) = \frac{3}{7} + \frac{5}{63} = \frac{27 + 5}{63} = \frac{32}{63}$

(d) $\frac{7}{8} + \frac{1}{16} - \frac{1}{12} = \frac{14 + 1}{16} - \frac{1}{12} = \frac{15}{16} - \frac{1}{12} = \frac{45 - 4}{48} = \frac{41}{48}$

Page No 20:

Question 113:

Identify the rational number that does not belong with the other three. Explain your reasoning

$\frac{- 5}{11}, \frac{- 1}{2}, \frac{- 4}{9}, \frac{- 7}{3}$

Answer:

The rational number that does not belong with the other three is $\frac{- 7}{3}$ as it is smaller than -1 whereas rest of the numbers are greater then -1.

Page No 20:

Question 114:

The cost of $\frac{19}{4}$ metres of wire is Rs. $\frac{171}{2} .$ Find the cost of one metre of the wire.

Answer:

Given: The cost of $\frac{19}{4} m$ of wire = â‚¹ $\frac{171}{2}$
Therefore, the cost of 1 m of wire = $\frac{171}{2} \div \frac{19}{4}$
$= \frac{171}{2} \times \frac{4}{19} = \frac{342}{19}$
= â‚¹18
Thus, the cost of 1 m of wire is â‚¹18.

Page No 20:

Question 115:

A train travels $\frac{1445}{2}$ km in $\frac{17}{2}$ hours. Find the speed of the train in km/h.

Answer:

Given: Distance travelled by train = $\frac{1445}{2}$ km
Time taken by train to cover $\frac{1445}{2}$ km = $\frac{17}{2}$ h
$\begin{array}{rcl} ∴ The speed of the train & = & \frac{Distance Travelled}{Time taken} \\ = & \frac{1445}{2} \div \frac{17}{2} \\ = & \frac{1445}{2} \times \frac{2}{17} \\ = & \frac{1445}{17} \\ = & 85 km / h \end{array}$
Thus, the speed of the train is 85 km/h.

Page No 21:

Question 116:

If 16 shirts of equal size can be made out of 24m of cloth, how much cloth is needed for making one shirt?

Answer:

Given: The cloth required to make 16 shirts = 24 m
$∴$ The cloth required to make 1 shirt = 24 $\div$ 16
$= \frac{24}{16} = \frac{3}{2} = 1.5 m$
Thus, the cloth required to make 1 shirt is 1.5 m.

Page No 21:

Question 117:

$\frac{7}{11}$ of all the money in Hamid’s bank account is Rs. 77,000. How much money does Hamid have in his bank account?

Answer:

Let the money in Hamid's bank account be $x$ .
Given : $\frac{7}{11}$ of $x$ = â‚¹77,000
$\Rightarrow \frac{7}{11} \times x = 77, 000 \Rightarrow x = 77, 000 \times \frac{11}{7}$
= â‚¹1,21,000
Thus, the total money in Hamid's back account is â‚¹121000.

Page No 21:

Question 118:

A $117 \frac{1}{3}$ m long rope is cut into equal pieces measuring $7 \frac{1}{3}$ m each. How many such small pieces are these?

Answer:

Given: The length of the rope = $117 \frac{1}{3} m$
One piece of rope measures = $7 \frac{1}{3} m$
So, the number of pieces of the rope

= $\frac{Total length of the rope}{Length of each piece}$
$= 117 \frac{1}{3} \div 7 \frac{1}{3} = \frac{352}{3} \div \frac{22}{3} = \frac{352}{3} \times \frac{3}{22} = 16$
Hence, the number of small pieces cut from the $117 \frac{1}{3} m$ long rope is 16.

Page No 21:

Question 119:

$\frac{1}{6}$ of the class students are above average, $\frac{1}{4}$ are average and rest are below average. If there are 48 students in all, how many students are below average in the class?

Answer:

Given: Fraction of students in the class that are above average = $\frac{1}{6}$
Fraction of students in the class that are average = $\frac{1}{4}$
So, the fraction of students in the class that are below average = $1 - (\frac{1}{6} + \frac{1}{4})$
$= 1 - (\frac{2 + 3}{12}) = 1 - \frac{5}{12} = \frac{7}{12}$
Since, it is given that the total number of students in the class are 48.
Therefore, the number of students in the class that are below average $\frac{7}{12} \times 48 = 28$
Hence, the number of students in the class that are below average is 28.

Page No 21:

Question 120:

$\frac{2}{5}$ of total number of students of a school come by car while $\frac{1}{4}$ of students come by bus to school. All the other students walk to school of which $\frac{1}{3}$ walk on their own and the rest are escorted by their parents. If 224 students come to school walking on their own, how many students study in that school?

Answer:

Let the total number of students study in the school be $x$ .
Then, the number of students come by car = $\frac{2}{5} \times x$
and, the number of students come by bus = $\frac{1}{4} \times x$
Now, the number of students that walk to school
$= x - (\frac{2 x}{5} + \frac{x}{4}) = x - (\frac{8 x + 5 x}{20}) = x - \frac{13 x}{20} = \frac{7 x}{20}$
Thus, the number of students that walk to school on their own = $\frac{1}{3} of \frac{7 x}{20} = \frac{1}{3} \times \frac{7 x}{20} = \frac{7 x}{60}$
Since, 224 students come to school by walking on their own. Therefore, $\frac{7 x}{60} = 224$
$\Rightarrow x = \frac{224 \times 60}{7} = 1920$
Hence, the total number of students in that school is 1920.

Page No 21:

Question 121:

Huma, Hubna and Seema received a total of Rs. 2,016 as monthly allowance from their mother such that Seema gets $\frac{1}{2}$ of what Huma gets and Hubna gets $1 \frac{2}{3}$ times Seema’s share. How much money do the three sisters get individually?

Answer:

Let the share of Huma be $x$ .
Then, given that, Seema gets $\frac{1}{2}$ of $x$ . i.e. $\frac{x}{2}$
And, Hubna gets $1 \frac{2}{3} of \frac{x}{2} i . e . \frac{5}{3} of \frac{x}{2} i . e . \frac{5 x}{6}$
Now, total monthly allowance = $x + \frac{x}{2} + \frac{5 x}{6}$
$= \frac{6 x + 3 x + 5 x}{6} = \frac{14 x}{6} = \frac{7 x}{3}$
Also, the total monthly allowance given from their mother is â‚¹2,016.
So, $\frac{7 x}{3} = 2016$
$\Rightarrow x = \frac{2016 \times 3}{7} =$ â‚¹864
Thus, Huma's share is â‚¹864.
Seema's share is $\frac{1}{2} \times 864 =$ â‚¹432
and Hubna's share is $\frac{5}{6} \times 864 =$ â‚¹720
Hence, Huma, Seema and Hubna gets â‚¹864, â‚¹432 and â‚¹720, respectively.

Page No 21:

Question 122:

A mother and her two daughters got a room constructed for Rs. 62,000. The elder daughter contributes $\frac{3}{8}$ of her mother’s contribution while the younger daughter contributes $\frac{1}{2}$ of her mother’s share. How much do the three contribute individually?

Answer:

Let the mother's share be x.
Then, the contribution of her elder daughter = $\frac{3}{8} of x$
$= \frac{3}{8} \times x = \frac{3 x}{8}$
And, the contribution of her younger daughter = $\frac{1}{2} of x$
$= \frac{1}{2} \times x = \frac{x}{2}$
According to the question,
$x + \frac{3 x}{8} + \frac{x}{2} = 62000 \Rightarrow \frac{8 x + 3 x + 4 x}{8} = 62000 \Rightarrow \frac{15 x}{8} = 62000$
$\Rightarrow x = \frac{62000 \times 8}{15} = \frac{99200}{3} =$ â‚¹33066.6
So, mother's share is â‚¹33066.6
Thus, elder daughter's share $\frac{3}{8} \times \frac{99200}{3} =$ â‚¹12400
And, younger daughter's share = $\frac{1}{2} \times \frac{99200}{3} =$ â‚¹16533.3
Hence, mother, her elder daughter and her younger daughter contributed â‚¹33066.6, â‚¹12400 and â‚¹16533.3 respectively.

Page No 21:

Question 123:

Tell which property allows you to compare

$\frac{2}{3} \times [\frac{3}{4} \times \frac{5}{7}] and [\frac{2}{3} \times \frac{5}{7}] \times \frac{3}{4}$

Answer:

$\frac{2}{3} \times [\frac{3}{4} \times \frac{5}{7}] and [\frac{2}{3} \times \frac{5}{7}] \times \frac{3}{4}$ (By commutative property of multiplication)
= $(\frac{2}{3} \times \frac{5}{7}) \times \frac{3}{4}$ (By associative property of multiplication)
Thus, with the help of associative and commutative property of multiplication $\frac{2}{3} \times (\frac{3}{4} \times \frac{5}{7})$ can be compared with $(\frac{2}{3} \times \frac{5}{7}) \times \frac{3}{4}$ .

Page No 22:

Question 124:

Name the property used in each of the following.

(i) $- \frac{7}{11} \times \frac{- 3}{5} = \frac{- 3}{5} \times \frac{- 7}{11}$

(ii) $- \frac{2}{3} \times [\frac{3}{4} + \frac{- 1}{2}] = [\frac{- 2}{3} \times \frac{3}{4}] + [\frac{- 2}{3} \times \frac{- 1}{2}]$

(iii) $\frac{1}{3} + [\frac{4}{9} + (\frac{- 4}{3})] = [\frac{1}{3} + \frac{4}{9}] + [\frac{- 4}{3}]$

(iv) $\frac{- 2}{7} + 0 = 0 + \frac{- 2}{7} = - \frac{2}{7}$

(v) $\frac{3}{8} \times 1 = 1 \times \frac{3}{8} = \frac{3}{8}$

Answer:

(i) $\frac{- 7}{11} \times \frac{- 3}{5} = \frac{- 3}{5} \times \frac{- 7}{11}$
Here, commutative property of multiplication is used.

(ii) $\frac{- 2}{3} \times (\frac{3}{4} + \frac{- 1}{2}) = (\frac{- 2}{3} \times \frac{3}{4}) + (\frac{- 2}{3} \times \frac{- 1}{2})$
Here, distributive property of multiplication are addition is used.

(iii) $\frac{1}{3} + [\frac{4}{9} + (\frac{- 4}{3})] = [\frac{1}{3} + \frac{4}{9}] + [\frac{- 4}{3}]$
Here, associative property of addition is used.

(iv) $\frac{- 2}{7} + 0 = 0 + \frac{- 2}{7} = \frac{- 2}{7}$
Here, existence of additive identity is used.

(v) $\frac{3}{8} \times 1 = 1 \times \frac{3}{8} = \frac{3}{8}$
Here, existence of multiplication identity is used.

Page No 22:

Question 125:

Find the multiplicative inverse of

(i) $- 1 \frac{1}{8}$

(ii) $3 \frac{1}{3}$

Answer:

(i) Let the multiplication inverse of $- 1 \frac{1}{8}$ be $x$ .
Then, $- 1 \frac{1}{8} \times x = 1$ (By the definition of multiplication inverse)
$\Rightarrow \frac{- 9}{8} \times x = 1 \Rightarrow x = \frac{- 8}{9}$
Hence, the multiplication inverse of $- 1 \frac{1}{8} is \frac{- 8}{9}$ .

(ii) Let the Multiplication inverse of $3 \frac{1}{3}$ be $x$ .
Then, $3 \frac{1}{3} \times x = 1$ (By the definition of multiplication inverse)
$\Rightarrow \frac{10}{3} \times x = 1 \Rightarrow x = \frac{3}{10}$
Hence, the multiplication inverse of $3 \frac{1}{3} is \frac{3}{10}$ .

Page No 22:

Question 126:

Arrange the numbers $\frac{1}{4}, \frac{13}{16}, \frac{5}{8}$ in the descending order.

Answer:

Firstly, convert $\frac{1}{4}, \frac{13}{16} and \frac{5}{8}$ into like fractions.
Finding LCM of 4, 16 and 8.

So, LCM of 4, 16 and 8 is 16.
Now, $\frac{1}{4} \times \frac{4}{4} = \frac{4}{16},$
$\frac{13}{16} \times \frac{1}{1} = \frac{13}{16} and \frac{5}{8} \times \frac{2}{2} = \frac{10}{16}$
As, $\frac{13}{16} > \frac{10}{16} > \frac{4}{16}$
$∴ \frac{13}{16} > \frac{5}{8} > \frac{1}{4}$
Hence $\frac{13}{16}, \frac{5}{8}, \frac{1}{4}$ is the descending order of the numbers.

Page No 22:

Question 127:

The product of two rational numbers is $\frac{- 14}{27}$ . If one of the numbers be $\frac{7}{9}$ , find the other.

Answer:

Let the other number be $x$ .
Then, $\frac{7}{9} \times x = \frac{- 14}{27}$
$\Rightarrow x = \frac{- 14}{27} \times \frac{9}{7} \Rightarrow x = \frac{- 2}{3}$
Hence, the other number is $\frac{- 2}{3}$ .

Page No 22:

Question 128:

By what numbers should we multiply $\frac{- 15}{20}$ so that the product may be $\frac{- 5}{7}$ ?

Answer:

Let the number by which $\frac{- 15}{20}$ should be multiplied so that the product is $\frac{- 5}{7}$ be $x$ .
So, $\frac{- 15}{20} \times x = \frac{- 5}{7}$
$\Rightarrow x = \frac{- 5}{7} \times \frac{- 20}{15} \Rightarrow x = \frac{20}{21}$
Hence $\frac{- 15}{20}$ should be multiplied by $\frac{20}{21}$ to set the product $\frac{- 5}{7}$ .

Page No 22:

Question 129:

By what number should we multiply $\frac{- 8}{13}$ so that the product may be 24?

Answer:

Let the number by which $\frac{- 8}{13}$ should be multiplied to set the product 24 be $x$ .
$\Rightarrow \frac{- 8}{13} \times x = 24 \Rightarrow x = 24 \times \frac{13}{- 8} \Rightarrow x = - 39$
Hence, $\frac{- 8}{13}$ should be multiplied by -39 to set the product 24.

Page No 22:

Question 130:

The product of two rational numbers is −7. If one of the number is −5, find the other?

Answer:

Let the other number be $x$ .
Then, $- 5 \times x = - 7$
$\Rightarrow x = \frac{- 7}{- 5} \Rightarrow x = \frac{7}{5}$
Hence, the other number is $\frac{7}{5}$ .

Page No 22:

Question 131:

Can you find a rational number whose multiplicative inverse is −1?

Answer:

As, -1 is the multiplicative inverse of itself.
So, we can not find any other rational number. Whose multiplicative inverse is -1.

Page No 22:

Question 132:

Find five rational numbers between 0 and 1.

Answer:

0 can be written as $\frac{0}{6}$ and 1 as $\frac{6}{6}$ .
Thus we have $\frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}$ between 0 and 1.
Hence, $\frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}$ are five rational numbers between 0 and 1.

Page No 22:

Question 133:

Find two rational numbers whose absolute value is $\frac{1}{5} .$

Answer:

Absolute value describes the distance from zero that a number is on the number line, without considering direction.
Since, the distance of $\frac{- 1}{5} and \frac{1}{5}$ from 0 on the number line is $\frac{1}{5}$ .
Therefore, $\frac{- 1}{5} and \frac{1}{5}$ are the rational numbers whose absolute value is $\frac{1}{5}$ .

Page No 23:

Question 134:

From a rope 40 metres long, pieces of equal size are cut. If the length of one piece is $\frac{10}{3}$ metre, find the number of such pieces.

Answer:

Given: Total length of the rope = 40 m
Length of one piece of the rope = $\frac{10}{3}$ m
$∴$ The number of pieces = $\frac{Total length of the rope}{Length of one piece of rope}$
$= 40 \div \frac{10}{3} = 40 \times \frac{3}{10} = 12$
Hence, the number of pieces cut from the rope are 12.

Page No 23:

Question 135:

$5 \frac{1}{2}$ metres long rope is cut into 12 equal pieces. What is the length of each piece?

Answer:

Given: Total length of the rope = $5 \frac{1}{2}$ m
Number of equal pieces cut from the rope = 12
$∴$ The length of each piece = $\frac{Total length of the rope}{Total number of equal piece}$
$= 5 \frac{1}{2} \div 12 = \frac{11}{2} \div 12 = \frac{11}{2} \times \frac{1}{12} = \frac{11}{24} m$
Hence, the length of each piece is $\frac{11}{24}$ m.

Page No 23:

Question 136:

Write the following rational numbers in the descending order.

$\frac{8}{7}, \frac{- 9}{8}, \frac{- 3}{2}, 0, \frac{2}{5}$

Answer:

Finding LCM of 7, 8, 2, 1 and 5.

So, the LCM of 7, 8, 2, 1 and 5 = $2 \times 2 \times 2 \times 5 \times 7 = 280$
Now, $\frac{8}{7} = \frac{8}{7} \times \frac{40}{40} = \frac{320}{280},$
$\frac{- 9}{8} = \frac{- 9}{8} \times \frac{35}{35} = \frac{315}{280}$ ,
$\frac{- 3}{2} = \frac{- 3}{2} \times \frac{140}{140} = \frac{- 420}{280},$
$0 = \frac{0}{1} \times \frac{280}{280} = \frac{0}{280},$
and $\frac{2}{5} = \frac{2}{5} \times \frac{56}{56} = \frac{112}{280}$
Since, $\frac{320}{280} > \frac{112}{280} > \frac{0}{280} > \frac{- 315}{280} > \frac{- 420}{280}$
$∴ \frac{8}{7} > \frac{2}{5} > 0 > \frac{- 9}{8} > \frac{- 3}{2}$
Hence, $\frac{8}{7}, \frac{2}{5}, 0, \frac{- 9}{8}, \frac{- 3}{2}$ is the descending order of the given rational number.

Page No 23:

Question 137:

Find:

(i) $0 \div \frac{2}{3}$

(ii) $\frac{1}{3} \times \frac{- 5}{7} \times \frac{- 21}{10}$

Answer:

$(i) 0 \div \frac{2}{3} = 0 \times \frac{3}{2} = 0$

$\begin{array}{rcl} (i i) \frac{1}{3} \times \frac{- 5}{7} \times \frac{- 21}{10} & = & \frac{1}{3} \times (\frac{- 5}{7} \times \frac{- 21}{10}) \\ = & \frac{1}{3} \times \frac{3}{2} \\ = & \frac{1}{2} \end{array}$

Page No 23:

Question 138:

On a winter day the temperature at a place in Himachal Pradesh was −16°C. Convert it in degree Fahrenheit (°F) by using the formula.

$\frac{C}{5} = \frac{F - 32}{9}$

Answer:

Given: On a winter day the temperature at a place in Himachal Pradesh = $- 16 °$ C
By using the formula $\frac{C}{5} = \frac{F - 32}{9},$ we set $\frac{- 16}{5} = \frac{F - 32}{9}$
$\Rightarrow \frac{F - 32}{9} = \frac{- 16}{5} \Rightarrow F - 32 = \frac{- 16}{5} \times 9 \Rightarrow F = (\frac{- 16}{5} \times 9) + 32 \Rightarrow F = \frac{- 144}{5} + 32 \Rightarrow F = \frac{- 144 + (32 \times 5)}{5} \Rightarrow F = \frac{- 144 + 160}{5} \Rightarrow F = \frac{16}{5} = 3.2 ° F$
Hence, the temperature at a place in Himachal Pradesh on that winter day was 3.2 °F.

Page No 23:

Question 139:

Find the sum of additive inverse and multiplicative inverse of 7.

Answer:

The additive inverse of 7 is −7 ( $∵ 7 + (- 7) = 0$ )
The multiplicative inverse of 7 is $\frac{1}{7}$ $(∵ 7 \times \frac{1}{7} = 1)$
$∴$ The required sum = $- 7 + \frac{1}{7}$
$= \frac{- 49 + 1}{7} = \frac{- 48}{7}$
Hence, the sum of additive inverse and multiplicative inverse of 7 is $\frac{- 48}{7}$ .

Page No 23:

Question 140:

Find the product of additive inverse and multiplicative inverse of $- \frac{1}{3}$ .

Answer:

The additive inverse of $\frac{- 1}{3} is \frac{1}{3}$ $(∵ \frac{- 1}{3} + \frac{1}{3} = 0)$
The multiplicative invers of $- \frac{1}{3} is - 3$ $(∵ - \frac{1}{3} \times - 3 = 1)$
$∴$ The required product = $\frac{1}{3} \times - 3$ = $- 1$
Hence, the product of additive inverse and multiplicative inverse of $\frac{- 1}{3}$ is $- 1$ .

Page No 23:

Question 141:

The diagram shows the wingspans of different species of birds. Use the diagram to answer the question given below:

(a) How much longer is the wingspan of an Albatross than the wingspan of a Sea gull?
(b) How much longer is the wingspan of a Golden eagle than the wingspan of a Blue jay?

Answer:

(a) Given: The length of wings of an Albatross $= 3 \frac{3}{5}$ m
                  The length of wings of a sea gull $= 1 \frac{7}{10}$ m
                   $\begin{array}{rcl} ∴ The required difference & = & 3 \frac{3}{5} - 1 \frac{7}{10} \\ = & \frac{18}{5} - \frac{17}{10} \\ = & \frac{36 - 17}{10} \\ = & \frac{19}{10} m \end{array}$
Hence, the wingspan of an Albatross is $\frac{19}{10} m$ longer than the wingspan of a sea gull.

(b) Given: The length of wings of a golden eagle $= 2 \frac{1}{2} m$
                  The length of wings of a blue jay $= \frac{41}{100} m$
                   $∴ The require difference = 2 \frac{1}{2} - \frac{41}{100} = \frac{5}{2} - \frac{41}{100} = \frac{250 - 41}{100} = \frac{209}{100}$
Hence, the wingspan of golden eagle is $\frac{209}{100} m$ longer than the wingspan of a blue jay.

Page No 24:

Question 142:

Shalini has to cut out circles of diameter $1 \frac{1}{4}$ cm from an aluminium strip of dimensions $8 \frac{3}{4}$ cm by $1 \frac{1}{4}$ cm. How many full circles can Shalini cut? Also, calculate the wastage of the aluminium strip.

Answer:

Given: The diameter of the circle that has to be cut out from an aluminium strip $= 1 \frac{1}{4} cm$
The length of the aluminium strip $= 8 \frac{3}{4} cm$
and the breath of the aluminium strip $= 1 \frac{1}{4} cm$
Clearly, the breath of the aluminium strip is equal to the diameter of the circle.
$∴$ The number of full circles cut from the aluminium strip $= 8 \frac{3}{4} \div 1 \frac{1}{4}$
   $= \frac{35}{4} \div \frac{5}{4} = \frac{35}{4} \times \frac{4}{5} = 7$
Since the diameter of a full circle $= 1 \frac{1}{4} cm$
$∴$ The radius of a full circle $= 1 \frac{1}{4} \div 2$
   $= \frac{5}{4} \div 2 = \frac{5}{4} \times \frac{1}{2} = \frac{5}{8} cm$
Area of a full circle = ${πr}^{2}$
$= \frac{22}{7} \times \frac{5}{8} \times \frac{5}{8} = \frac{550}{448} {cm}^{2}$
Thus, the area of 7 such full circles $= 7 \times \frac{550}{448}$
   $= \frac{550}{64} {cm}^{2}$
Now, area of the aluminium strip = length $\times$ Breadth
$=1 \frac{1}{4} \times 8 \frac{3}{4} = \frac{5}{4} \times \frac{35}{4} = \frac{175}{16} {cm}^{2}$
$∴$ The wastage of the aluminium strip $= \frac{175}{16} - \frac{550}{64}$
   $= \frac{700 - 550}{64} = \frac{150}{64} = \frac{75}{32} = 2 \frac{11}{32} {cm}^{2}$
Hence, Shalini can cut 7 full circles and the wastage of the aluminium strip is $2 \frac{11}{32} {cm}^{2}$

Page No 24:

Question 143:

One fruit salad recipe requires $\frac{1}{2}$ cup of sugar. Another recipe for the same fruit salad requires 2 tablespoons of sugar. If 1 tablespoon is equivalent to $\frac{1}{16}$ cup, how much more sugar does the first recipe require?

Answer:

Given: Sugar required for one fruit salad recipe $= \frac{1}{2}$ cup of sugar
Sugar required for another recipe = 2 table spoons of sugar
As, 1 table spoon of sugar is equivalent to $\frac{1}{16}$ cup.
$∴$ Sugar required for another recipe = $2 \times \frac{1}{16} = \frac{1}{8}$ cup
Thus, $\frac{1}{2} - \frac{1}{8} = \frac{4 - 1}{8} = \frac{3}{8}$ cups
Hence, $\frac{3}{8}$ cups more sugar is required for first recipe.

Page No 24:

Question 144:

Four friends had a competition to see how far could they hop on one foot. The table given shows the distance covered by each.

Name	Distance covered (km)
Seema	$\frac{1}{25}$
Nancy	$\frac{1}{32}$
Megha	$\frac{1}{40}$
Soni	$\frac{1}{20}$

(a) How farther did Soni hop than Nancy?

(b) What is the total distance covered by Seema and Megha?

Answer:

(a) The distance covered by Soni = $\frac{1}{20}$ km
      The distance covered by Nancy = $\frac{1}{32}$ km
       $∴$ Soni hop more than Nancy by = $\frac{1}{20} - \frac{1}{32}$
                                                         $= \frac{8 - 5}{160} = \frac{3}{160} km$

(b) The distance covered by Seema = $\frac{1}{25} km$
The distance covered by Megha = $\frac{1}{40} km$
$∴$ The total distance covered by Seema and Megha = $\frac{1}{25} + \frac{1}{40}$
                                                                                 $= \frac{8 + 5}{200} = \frac{13}{200} km$

(c) As, 40 > 32
$∴ \frac{1}{40} < \frac{1}{32}$
Thus, Nancy walked farther than Megha.

Page No 24:

Question 145:

The table given below shows the distances, in kilometres, between four villages of a state. To find the distance between two villages, locate the square where the row for one village and the column for the other village intersect.

(a) Compare the distance between Himgaon and Rawalpur to Sonapur and Ramgarh?
(b) If you drove from Himgaon to Sonapur and then from Sonapur to Rawalpur, how far would you drive?

Answer:

(a) From the table:
      The distance between Himgaon and Rawalpur $= 98 \frac{3}{4} km$
      And, the distance between Sonapur and Ramgarh $= 40 \frac{2}{3} km$
      Difference between distance from Himgaon to Rawalpur and Sonapur to Ramgarh
       $= 98 \frac{3}{4} - 40 \frac{2}{3} = \frac{395}{4} - \frac{122}{3} = \frac{1185 - 488}{12} = \frac{697}{12} = 58 \frac{1}{12} km$

(b) From the table:
The distance between Himgaon to Sonapur = $100 \frac{5}{6} km$
The distance between Sonapur to Rawalpur $= 16 \frac{1}{2} km$
$∴$ Total distance = $100 \frac{5}{6} + 16 \frac{1}{2}$
                         $= \frac{605}{6} + \frac{33}{2} = \frac{605 + 99}{6} = \frac{704}{6} = 117 \frac{1}{3} km$
Hence, the total distance we would drive is 117 $\frac{1}{3} km$ .

Page No 25:

Question 146:

The table shows the portion of some common materials that are recycled.

Material	Recycled
Paper	$\frac{5}{11}$
Aluminium cans	$\frac{5}{8}$
Glass	$\frac{2}{5}$
Scrap	$\frac{3}{4}$

(a) Is the rational number expressing the amount of paper recycled more than $\frac{1}{2}$ or less than $\frac{1}{2}$ ?

(b) Which items have a recycled amount less than $\frac{1}{2}$ ?

(c) Is the quantity of aluminium cans recycled more (or less) than half of the quantity of aluminium cans?

(d) Arrange the rate of recycling the materials from the greatest to the smallest.

Answer:

(a) Converting $\frac{5}{11} and \frac{1}{2}$ into like fractions.
      So, $\frac{5}{11} \times \frac{2}{2} = \frac{10}{22} and \frac{1}{2} \times \frac{11}{11} = \frac{11}{22}$
      As, $\frac{10}{22} < \frac{11}{22}$
       $∴ \frac{5}{11} < \frac{1}{2}$
      Thus, the rational number expressing the amount of paper recycled is less than $\frac{1}{2}$ .

(b) Converting $\frac{5}{11}, \frac{5}{8}, \frac{2}{5}, \frac{3}{4} and \frac{1}{2}$ into like fractions
      Finding LCM of 11, 8, 5, 4 and 2.

      So, the LCM of 11, 8, 5, 4 and 2 is 440.
      Now, $\frac{5}{11} = \frac{5}{11} \times \frac{40}{40} = \frac{200}{440}, \frac{5}{8} \times \frac{55}{55} = \frac{275}{440},$
     $\frac{2}{5} = \frac{2}{5} \times \frac{88}{88} = \frac{176}{440}, \frac{3}{4} \times \frac{110}{110} = \frac{330}{440}$ and $\frac{1}{2} = \frac{1}{2} \times \frac{220}{220} = \frac{220}{440}$
    Clearly, $\frac{5}{11} < \frac{1}{2}; \frac{5}{8} > \frac{1}{2}; \frac{2}{5} < \frac{1}{2}; \frac{3}{4} > \frac{1}{2} .$
    Thus, $\frac{5}{11} and \frac{2}{5}$ are less than $\frac{1}{2}$ .
    Hence, paper and glass have a recycled amount less than $\frac{1}{2}$ .

(c) The portion of aluminium cans that are recycled = $\frac{5}{8}$
      Now, half portion of the quantity of aluminium can $= \frac{1}{2}$
      As, $\frac{1}{2} = \frac{1}{2} \times \frac{4}{4} = \frac{4}{8} and \frac{5}{8} > \frac{4}{8}$ .
       $∴$ The portion of aluminium cans that are recycled is more than half of the quantity of aluminium cans.

(d) Disclaimer: Incomplete question.
      Converting $\frac{5}{11}, \frac{5}{8}, \frac{2}{5} and \frac{3}{4}$ into like fractions.
      Finding LCM of 11, 8, 5 and 4.
      Now, taking LCM of 100, 9, 25.

So, the LCM of 11, 8, 5, 4 is 440.
Now,
$\frac{5}{11} = \frac{5}{11} \times \frac{40}{40} = \frac{200}{440}, \frac{5}{8} \times \frac{55}{55} = \frac{275}{440}, \frac{2}{5} = \frac{2}{5} \times \frac{88}{88} = \frac{176}{440}, \frac{3}{4} \times \frac{110}{110} = \frac{330}{440} ∴ \frac{3}{4} > \frac{5}{8} > \frac{5}{11} > \frac{2}{5}$
Thus, the materials having rates of recycling form the greatest to smallest is scrap, aluminium cans, paper and glass.

Page No 26:

Question 147:

State whether the statements are true (T) or false (F).
The overall width in cm of several wide-screen televisions are 97.28 cm, $98 \frac{4}{9}$ cm, $98 \frac{1}{25}$ cm and 97.94 cm. Express these numbers as rational numbers in the form $\frac{p}{q}$ and arrange the widths in ascending order.

Answer:

As 97.28 cm = $\frac{9728}{100} cm$ ,
$98 \frac{4}{9} cm = \frac{886}{9} cm, 98 \frac{1}{25} cm = \frac{2451}{25} cm, and 97.94 cm = \frac{9794}{100} cm$
So, the width of the wide-screen televisions 97.28 cm, 98 $\frac{1}{25}$ cm, 97.94 cm can be expressed in $\frac{p}{q}$ from as $\frac{9728}{100}$ cm, $\frac{886}{9}$ cm, $\frac{2451}{25}$ cm and $\frac{9794}{100}$ cm respectively.
Now, taking LCM of 100, 9, 25.

Thus, the LCM of 100, 9 and 25 is 900.
So, $\frac{9728}{100} = \frac{9728}{100} \times \frac{9}{9} = \frac{87552}{900}, \frac{886}{9} = \frac{886}{9} \times \frac{100}{100} = \frac{88600}{900}, \frac{2451}{25} = \frac{2451}{25} \times \frac{36}{36} = \frac{88236}{900}, and \frac{9794}{100} = \frac{9794}{100} \times \frac{9}{9} = \frac{88146}{900} Clearly, \frac{88600}{900} > \frac{88236}{900} > \frac{88146}{900} > \frac{87552}{900} ∴ 98 \frac{4}{9} > 98 \frac{1}{25} cm, > 97.94 > 97.28$
Thus, the width in the descending order is 98 $\frac{4}{9}$ cm, 98 $\frac{1}{25}$ cm, 97.94 cm and 97.28 cm.

Page No 26:

Question 148:

Roller Coaster at an amusement park is $\frac{2}{3}$ m high. If a new roller coaster is built that is $\frac{3}{5}$ times the height of the existing coaster, what will be the height of the new roller coaster?

Answer:

Given: The height of the roller coaster at amusement park = $\frac{2}{3} m$
The height of the new roller coaster at the amusement park = $\frac{3}{5} \times \frac{2}{3} = \frac{2}{5} m$

Page No 26:

Question 149:

Here is a table which gives the information about the total rainfall for several months compared to the average monthly rains of a town. Write each decimal in the form of rational number $\frac{p}{q} .$

Month	Above/Below normal (in cm)
May	2.6924
June	0.6096
July	−6.9088
August	−8.636

Answer:

2.6924 = $\frac{26924}{10000} = \frac{6731}{2500}$

0.6096 = $\frac{6096}{10000} = \frac{381}{625}$

−6.9088 = $\frac{- 69088}{10000} = \frac{- 4318}{625}$

−8.636 = $\frac{- 8636}{1000} = \frac{- 2159}{250}$
Thus, 2.6924 cm, 0.6096 cm, −6.9088 cm and −8.636 cm in the $\frac{p}{q}$ is $\frac{6731}{2500} cm, \frac{381}{625} cm, \frac{- 4318}{625} cm and \frac{- 2159}{250} cm$ respectively.

Page No 26:

Question 150:

The average life expectancies of males for several states are shown in the table. Express each decimal in the form $\frac{p}{q}$ and arrange the q states from the least to the greatest male life expectancy. State-wise data are included below; more indicators can be found in the “FACTFILE” section on the homepage for each state.

State	Male	$\frac{p}{q}$ form	Lowest terms
Andhra Pradesh	61.6
Assam	57.1
Bihar	60.7
Gujarat	61.9
Haryana	64.1
Himachal Pradesh	65.1
Karnataka	62.4
Kerala	70.6
Madhya Pradesh	56.5
Maharashtra	64.5
Orissa	57.6
Punjab	66.9
Rajasthan	59.8
Tamil Nadu	63.7
Uttar Pradesh	58.9
West Bengal	62.8
India	60.8

Source: Registrar General of India (2003) SRS Based Abridged Lefe Tables. SRS Analytical Studies, Report No. 3 of 2003, New Delhi: Registrar General of India. The data are for the 1995-99 period; states subsequently divided are therefore included in their pre-partition states (Chhatisgarh in MP, Uttaranchal in UP and Jharkhand in Bihar)

Answer:


State	Male	$\frac{p}{q}$ form	Lowest terms
Andhra Pradesh	61.6	$\frac{616}{10}$	$\frac{308}{5}$
Assam	57.1	$\frac{571}{10}$	$\frac{571}{10}$
Bihar	60.7	$\frac{607}{10}$	$\frac{607}{10}$
Gujarat	61.9	$\frac{619}{10}$	$\frac{619}{10}$
Haryana	64.1	$\frac{641}{10}$	$\frac{641}{10}$
Himachal Pradesh	65.1	$\frac{651}{10}$	$\frac{651}{10}$
Karnataka	62.4	$\frac{624}{10}$	$\frac{312}{5}$
Kerala	70.6	$\frac{706}{10}$	$\frac{353}{5}$
Madhya Pradesh	56.5	$\frac{565}{10}$	$\frac{113}{2}$
Maharashtra	64.5	$\frac{645}{10}$	$\frac{129}{2}$
Orissa	57.6	$\frac{576}{10}$	$\frac{288}{5}$
Punjab	66.9	$\frac{669}{10}$	$\frac{669}{10}$
Rajasthan	59.8	$\frac{598}{10}$	$\frac{299}{5}$
Tamil Nadu	63.7	$\frac{637}{10}$	$\frac{637}{10}$
Uttar Pradesh	58.9	$\frac{589}{10}$	$\frac{589}{10}$
West Bengal	62.8	$\frac{628}{10}$	$\frac{314}{5}$
India	60.8	$\frac{608}{10}$	$\frac{304}{5}$

The arrangement of the states from the least to the greatest male life expectancy is Haryana, Tamil Nadu, West Bengal, Karnataka, Gujrat, Andhra Pradesh, Bihar, Rajasthan, Uttar Pradesh, Orissa, Assam, Madhya Pradesh.

Page No 27:

Question 151:

A skirt that is $35 \frac{7}{8}$ cm long has a hem of $3 \frac{1}{8}$ cm. How long will the skirt be if the hem is let down?

Answer:

Length of the skirt = $35 \frac{7}{8} cm$
Length of the hem = $3 \frac{1}{8} cm$
Length of skirt, if hem is let down
$= 35 \frac{7}{8} + 3 \frac{1}{8} = \frac{287}{8} + \frac{25}{8} = \frac{312}{8} = 39 cm$
Hence, the length of the skirt, if the hem is let down, is 39 cm.

Page No 27:

Question 152:

Manavi and Kuber each receives an equal allowance. The table shows the fraction of their allowance each deposits into his/her saving account and the fraction each spends at the mall. If allowance of each is Rs. 1260 find the amount left with each.

Where money goes	Fraction of allowance
Where money goes	Manavi	Kuber
Saving Account	$\frac{1}{2}$	$\frac{1}{3}$
Spend at mall	$\frac{1}{4}$	$\frac{3}{5}$
Left over	?	?

Answer:

The fraction of amount left with Manavi
$= 1 - (\frac{1}{2} + \frac{1}{4}) = 1 - (\frac{2 + 1}{4}) = 1 - \frac{3}{4} = \frac{1}{4}$
The fraction of amount left with Kuber
$= 1 - (\frac{1}{3} + \frac{3}{5}) = 1 - (\frac{5 + 9}{15}) = 1 - \frac{14}{15} = \frac{1}{15}$
Given that the allowance of each of them is â‚¹1260.
Thus, the amount left with Manavi
= $\frac{1}{4} of 1260 = \frac{1}{4} \times 1260 =$ â‚¹315
and the amount left with Kuber
$= \frac{1}{15} of 1260 = \frac{1}{15} \times 1260 =$ â‚¹84

View NCERT Solutions for all chapters of Class 8