Rd Sharma 2020 2021 for Class 7 Maths Chapter 4 - Rational Numbers

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Rd Sharma 2020 2021 Solutions for Class 7 Maths Chapter 4 Rational Numbers are provided here with simple step-by-step explanations. These solutions for Rational Numbers are extremely popular among Class 7 students for Maths Rational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2020 2021 Book of Class 7 Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2020 2021 Solutions. All Rd Sharma 2020 2021 Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

Page No 4.12:

Question 1:

Determine whether the following rational numbers are in the lowest form or not:

(i) $\frac{65}{84}$
(ii) $\frac{- 15}{32}$
(iii) $\frac{24}{128}$
(iv) $\frac{- 56}{- 32}$

Answer:

(i) We observe that 65 and 84 have no common factor i..e., their HCF is 1.

Thus, $\frac{65}{84}$ is in its lowest form.

(ii) We observe that $-$ 15 and 32 have no common factor i..e., their HCF is 1.

Thus, $\frac{- 15}{32}$ is in its lowest form.

(iii) HCF of 24 and 128 is not 1.

Thus, given rational number is not in its simplest form.

(iv) HCF of 56 and 32 is 8.

Thus, given rational number is not in its simplest form.

Page No 4.12:

Question 2:

Express each of the following rational numbers to the lowest form:
(i) $\frac{4}{22}$
(ii) $\frac{- 36}{180}$
(iii) $\frac{132}{- 428}$
(iv) $\frac{- 32}{- 56}$

Answer:

Lowest form of:
$(i) \frac{4}{22} is : 4 = 2 \times 2 22 = 2 \times 11 HCF of 4 and 22 is 2 . D ividing the fraction by 2, we get \frac{2}{11} . (ii) \frac{- 36}{180} is : 36 = 3 \times 3 \times 2 \times 2 180 = 5 \times 3 \times 3 \times 2 \times 2 HCF of 36 and 180 is 36 . D ividing the fraction by 36, we get \frac{- 1}{5}$
$(iii) \frac{132}{- 428} is : 132 = 2 \times 3 \times 2 \times 11 428 = 2 \times 2 \times 107 HCF of 132 and 428 is 4 . D ividing the fraction by 4, we get \frac{33}{- 107} (iv) \frac{- 32}{- 56} is : 32 = 2 \times 2 \times 2 \times 2 \times 2 56 = 2 \times 2 \times 2 \times 7 HCF of 32 and 56 is 8 . D ividing the fraction by 8, we get \frac{4}{7}$

Page No 4.12:

Question 3:

Fill in the blanks:
(i) $\frac{- 5}{7} = \frac{\dots}{35} = \frac{\dots}{49}$
(ii) $\frac{- 4}{- 9} = \frac{\dots}{18} = \frac{12}{\dots}$
(iii) $\frac{6}{- 13} = \frac{- 12}{\dots} = \frac{24}{\dots}$
(iv) $\frac{- 6}{\dots} = \frac{3}{11} = \frac{\dots}{- 55}$

Answer:

$(i) Here, \frac{- 5 \times 5}{7 \times 5} = \frac{- 25}{35} A lso, \frac{- 5 \times 7}{7 \times 7} = \frac{- 35}{49} . Therefore, \frac{- 5}{7} = \frac{- 25}{35} = \frac{- 35}{49} (ii) Here, \frac{- 4 \times - 2}{- 9 \times - 2} = \frac{8}{18} A lso, \frac{- 4 \times - 3}{- 9 \times - 3} = \frac{12}{27} Therefore, \frac{- 4}{- 9} = \frac{8}{18} = \frac{12}{27} (iii) Here, \frac{6 \times - 2}{- 13 \times - 2} = \frac{- 12}{26} Also, \frac{6 \times 4}{- 13 \times 4} = \frac{24}{- 52} Therefore, \frac{6}{- 13} = \frac{- 12}{26} = \frac{24}{- 52} (iv) Here, \frac{- 6 / - 2}{- 22 / - 2} = \frac{3}{11} Also, \frac{- 6}{- 22} = \frac{3 \times - 5}{11 \times - 5} = \frac{- 15}{- 55} Therefore, \frac{- 6}{- 22} = \frac{3}{11} = \frac{- 15}{- 55}$

Page No 4.15:

Question 1:

Write each of the following rational numbers in the standard form:
(i) $\frac{2}{10}$
(ii) $\frac{- 8}{36}$
(iii) $\frac{4}{- 16}$
(iv) $\frac{- 15}{- 35}$
(v) $\frac{299}{- 161}$
(vi) $\frac{- 63}{- 210}$
(vii) $\frac{68}{- 119}$
(viii) $\frac{- 195}{275}$

Answer:

(i) The denominator is positive and HCF of 2 and 10 is 2.

$∴$ Dividing the numerator and denominator by 2, we get:

$\frac{2}{10} = \frac{2 / 2}{10 / 2} = \frac{1}{5}$

(ii) The denominator is positive and HCF of 8 and 36 is 4.

$∴$ Dividing the numerator and denominator by 4, we get:

$\frac{- 8}{36} = \frac{- 8 / 4}{36 / 4} = \frac{- 2}{9}$

(iii)

The denominator is negative.

$\frac{4 \times - 1}{- 16 \times - 1} = \frac{- 4}{16}$

HCF of 4 and 16 is 4.

$∴$ Dividing the numerator and denominator by 4, we get:

$\frac{- 4 / 4}{16 / 4} = \frac{- 1}{4}$

(iv) The denominator is negative.

$\frac{- 15 \times - 1}{- 35 \times - 1} = \frac{15}{35}$

HCF of 15 and 35 is 5.

$∴$ Dividing the numerator and denominator by 5, we get:

$\frac{15 / 5}{35 / 5} = \frac{3}{7}$

(v) The denominator is negative.

$\frac{299 \times - 1}{- 161 \times - 1} = \frac{- 299}{161}$

HCF of 299 and 161 is 23.

$∴$ Dividing the numerator and denominator by 23, we get:

$\frac{- 299 / 23}{161 / 23} = \frac{- 13}{7}$

(vi) The denominator is negative.

$\frac{- 63 \times - 1}{- 210 \times - 1} = \frac{63}{210}$

HCF of 63 and 210 is 21.

$∴$ Dividing the numerator and denominator by 21, we get:

$\frac{63 / 21}{210 / 21} = \frac{3}{10}$

(vii) The denominator is negative.

$\frac{68 \times - 1}{- 119 \times - 1} = \frac{- 68}{119}$

HCF of 68 and 119 is 17.

$∴$ Dividing the numerator and denominator by 17, we get:

$\frac{- 68 / 17}{119 / 17} = \frac{- 4}{7}$

(viii) The denominator is positive and HCF of 195 and 275 is 5.

$∴$ Dividing divide the numerator and denominator by 5, we get:

$\frac{- 195 / 5}{275 / 5} = \frac{- 39}{55}$

Page No 4.20:

Question 1:

Which of the following rational numbers are equal?
(i) $\frac{- 9}{12} and \frac{8}{- 12}$
(ii) $\frac{- 16}{20} and \frac{20}{- 25}$
(iii) $\frac{- 7}{21} and \frac{3}{- 9}$
(iv) $\frac{- 8}{14} and \frac{13}{21}$

Answer:

(i)
$The standard form of \frac{- 9}{12} is \frac{- 9 / 3}{12 / 3} = \frac{- 3}{4} The standard form of \frac{8}{- 12} is \frac{8 / - 4}{- 12 / - 4} = \frac{- 2}{3} Since, the standard forms of two rational numbers are not same . Hence, they are not equal .$

(ii)
$Since, LCM of 20 and 25 is 100 . Therefore making the denominators equal, \frac{- 16}{20} = \frac{- 16 \times 5}{20 \times 5} = \frac{- 80}{100} and \frac{20}{- 25} = \frac{- 20 \times 4}{25 \times 4} = \frac{- 80}{100} . Therefore, \frac{- 16}{20} = \frac{20}{- 25} .$

(iii)
$Since, LCM of 21 and 9 is 63 . Therefore making the denominators equal, \frac{- 7}{21} = \frac{- 7 \times 3}{21 \times 3} = \frac{- 21}{63} and \frac{3}{- 9} = \frac{- 3 \times 7}{9 \times 7} = \frac{- 21}{63} . Therefore, \frac{- 7}{21} = \frac{3}{- 9} .$

(iv)
$Since, LCM of 14 and 21 is 42 . Therefore making the denominators equal, \frac{- 8}{14} = \frac{- 8 \times 3}{14 \times 3} = \frac{- 24}{42} and \frac{13}{21} = \frac{13 \times 2}{21 \times 2} = \frac{26}{42} . Therefore, \frac{- 8}{14} is not equal to \frac{13}{21} .$

Page No 4.20:

Question 2:

If each of the following pairs represents a pair of equivalent rational numbers, find the values of x:
(i) $\frac{2}{3} and \frac{5}{x}$
(ii) $\frac{- 3}{7} and \frac{x}{4}$
(iii) $\frac{3}{5} and \frac{x}{- 25}$
(iv) $\frac{13}{6} and \frac{- 65}{x}$

Answer:

(i) $\frac{2}{3} = \frac{5}{x}, then x = 5 \times \frac{3}{2} = \frac{15}{2}$
(ii) $\frac{- 3}{7} = \frac{x}{4}, then x = \frac{- 3}{7} \times 4 = \frac{- 12}{7}$
(iii) $\frac{3}{5} = \frac{x}{- 25}, then x = \frac{3}{5} \times (- 25) = \frac{- 75}{5} = - 15$
(iv) $\frac{13}{6} = \frac{- 65}{x}, then x = \frac{6}{13} \times (- 65) = 6 \times (- 5) = - 30$

Page No 4.20:

Question 3:

In each of the following, fill in the blanks so as to make the statement true:
(i) A number which can be expressed in the form $\frac{p}{q}$ , where p and q are integers and q is not equal to zero, is called a .....

(ii) If the integers p and q have no common divisor other than 1 and q is positive, then the rational number $\frac{p}{q}$ is said to be in the ....
(iii) Two rational numbers are said to be equal, if they have the same .... form.

\frac{a}{b} = \frac{a \div m}{. . . .}

(v) If p and q are positive integers, then

\frac{p}{q}

is a ..... rational number and

\frac{p}{- q}

is a ..... rational number.

(vi) The standard form of −1 is ...

(vii) If

\frac{p}{q}

is a rational number, then q cannot be ....
(viii) Two rational numbers with different numerators are equal, if their numerators are in the same .... as their denominators.

Answer:

(i) rational number
(ii) standard rational number
(iii) standard form
(iv) $\frac{a}{b} = \frac{a \div m}{b \div m}$
(v) positive rational number, negative rational number
(vi) $\frac{- 1}{1}$
(vii) zero
(viii) ratio

Page No 4.20:

Question 4:

In each of the following state if the statement is true (T) or false (F):
(i) The quotient of two integers is always an integer.
(ii) Every integer is a rational number.
(iii) Every rational number is an integer.
(iv) Every fraction is a rational number.
(v) Every rational number is a fraction
(vi) If $\frac{a}{b}$ is a rational number and m any integer, then $\frac{a}{b} = \frac{a \times m}{b \times m}$
(vii) Two rational numbers with different numerators cannot be equal.
(viii) 8 can be written as a rational number with any integer as denominator.
(ix) 8 can be written as a rational number with any integer as numerator.
(x) $\frac{2}{3} is equal to \frac{4}{6} .$

Answer:

(i) False; not necessary
(ii) True; every integer can be expressed in the form of p/q, where q is not zero.
(iii) False; not necessary
(iv) True; every fraction can be expressed in the form of p/q, where q is not zero.
(v) False; not necessary
(vi) True
(vii) False; they can be equal, when simplified further.
(viii) False
(ix) False
(x) True; in the standard form, they are equal.

Page No 4.26:

Question 1:

Draw the number line and represent the following rational numbers on it:
(i) $\frac{2}{3}$
(ii) $\frac{3}{4}$
(iii) $\frac{3}{8}$
(vi) $\frac{- 5}{8}$
(v) $\frac{- 3}{16}$
(vi) $\frac{- 7}{3}$
(vii) $\frac{22}{- 7}$
(viii) $\frac{- 31}{3}$

Answer:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

Page No 4.26:

Question 2:

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
(i) $\frac{- 3}{8}, 0$
(ii) $\frac{5}{2}, 0$
(iii) $\frac{- 4}{11}, \frac{3}{11}$
(iv) $\frac{- 7}{12}, \frac{5}{- 8}$
(v) $\frac{4}{- 9}, \frac{- 3}{- 7}$
(vi) $\frac{- 5}{8}, \frac{3}{- 4}$
(vii) $\frac{5}{9}, \frac{- 3}{- 8}$
(viii) $\frac{5}{- 8}, \frac{- 7}{12}$

Answer:

(i) We know that every positive rational number is greater than zero and every negative rational number is smaller than zero. Thus,

$\frac{- 3}{8} > 0$
(ii) $\frac{5}{2} > 0 .$ Because every positive rational number is greater than zero and every negative rational number is smaller than zero.
(iii) $\frac{- 4}{8} < \frac{3}{11} .$ Because every positive rational number is greater than zero and every negative rational number is smaller than zero.
(iv)

$\frac{- 7}{12} = \frac{- 7 \times 2}{12 \times 2} = \frac{- 14}{24} and \frac{5}{- 8} = \frac{- 5 \times 3}{8 \times 3} = \frac{- 15}{24} Therefore, \frac{- 7}{12} > \frac{5}{- 8}$

(v)

$\frac{4}{- 9} = \frac{- 4 \times 7}{9 \times 7} = \frac{- 28}{63} and \frac{- 3}{- 7} = \frac{3 \times 7}{7 \times 9} = \frac{21}{63} Therefore, \frac{4}{- 9} < \frac{- 3}{- 7}$
(vi)

$\frac{- 5}{8} and \frac{3}{- 4} = \frac{- 3 \times 2}{4 \times 2} = \frac{- 6}{8} Therefore, \frac{- 5}{8} > \frac{3}{- 4}$

(vii)

$\frac{5}{9} = \frac{5 \times 8}{9 \times 8} = \frac{40}{72} and \frac{- 3}{- 8} = \frac{3 \times 9}{8 \times 9} = \frac{27}{72} Therefore, \frac{5}{9} > \frac{- 3}{- 8}$

(viii)
$\frac{- 7}{12} = \frac{- 7 \times 2}{12 \times 2} = \frac{- 14}{24} and \frac{5}{- 8} = \frac{- 5 \times 3}{8 \times 3} = \frac{- 15}{24} Therefore, \frac{- 7}{12} > \frac{5}{- 8}$

Page No 4.26:

Question 3:

Which of the two rational numbers in each of the following pairs of rational numbers is smaller?
(i) $\frac{- 6}{- 13}, \frac{7}{13}$
(ii) $\frac{16}{- 5}, 3$
(iii) $\frac{- 4}{3}, \frac{8}{- 7}$
(iv) $\frac{- 12}{5}, - 3$

Answer:

(i) $\frac{- 6}{- 13} = \frac{6}{13} < \frac{7}{13}$
(ii) $\frac{16}{- 5} < 3$
(iii)
$\frac{- 4}{3} = \frac{- 4 \times 7}{3 \times 7} = \frac{- 28}{21} and \frac{8}{- 7} = \frac{- 8 \times 3}{7 \times 3} = \frac{- 24}{21} Therefore, \frac{- 4}{3} < \frac{8}{- 7}$
(iv)
$\frac{- 12}{5} and - 3 = \frac{- 3 \times 5}{1 \times 5} = \frac{- 15}{5} Therefore \frac{- 12}{5} > - 3$

Page No 4.27:

Question 4:

Fill in the blanks by the correct symbol out of >, =, or <:
(i) $\frac{- 6}{7} . . . . . \frac{7}{13}$
(ii) $\frac{- 3}{5} . . . . . \frac{- 5}{6}$
(iii) $\frac{- 2}{3} . . . . . \frac{5}{- 8}$
(iv) $0 . . . . . \frac{- 2}{5}$

Answer:

$(i) Because every positive number is greater than a negative number, \frac{- 6}{7} < \frac{7}{13} . (ii) On multiplying \frac{- 3}{5} by \frac{6}{6}, we get \frac{- 18}{30} . On multiplying \frac{- 5}{6} by \frac{5}{5}, we get \frac{- 25}{30} . Because - 18 > - 25, \frac{- 3}{5} > \frac{- 5}{6} . (iii) On multiplying \frac{- 2}{3} by \frac{8}{8}, we get \frac{- 16}{24} . On multiplying \frac{5}{- 8} by \frac{3}{3}, we get \frac{15}{- 24} = \frac{- 15}{24} . Because - 15 > - 16, \frac{- 2}{3} < \frac{5}{- 8} . (iv) Because every positive number is greater than a negative number, 0 > \frac{- 2}{5} .$

Page No 4.27:

Question 5:

Arrange the following rational numbers in ascending order:
(i) $\frac{3}{5}, \frac{- 17}{- 30}, \frac{8}{- 15}, \frac{- 7}{10}$
(ii) $\frac{- 4}{9}, \frac{5}{- 12}, \frac{7}{- 18}, \frac{2}{- 3}$

Answer:

(i) Ascending order:

$Since, LCM of 5, - 30, - 15, 10 is 30 . Multiplying the numerators and denominators to get the denominator equal to the LCM, \frac{3}{5} = \frac{3 \times 6}{5 \times 6} = \frac{18}{30}, \frac{17}{30} = \frac{17 \times 1}{30 \times 1} = \frac{17}{30}, \frac{8}{- 15} = \frac{- 8 \times 2}{15 \times 2} = \frac{- 16}{30}, \frac{- 7}{10} = \frac{- 7 \times 3}{10 \times 3} = \frac{- 21}{30} . Order is - 21 < - 16 < 17 < 8 . Order is \frac{- 7}{10} < \frac{8}{- 15} < \frac{17}{30} < \frac{3}{5} .$

(ii)

$Since, LCM of 9, - 12, - 18, 3 is 36 . Multiplying the numerators and denominators to get the denominator equal to the LCM, \frac{- 4}{9} = \frac{- 4 \times 4}{9 \times 4} = \frac{- 16}{36} \frac{5}{- 12} = \frac{- 5 \times 3}{12 \times 3} = \frac{- 15}{36} \frac{7}{- 18} = \frac{- 7 \times 2}{18 \times 2} = \frac{- 14}{36} \frac{2}{- 3} = \frac{- 2 \times 12}{3 \times 12} = \frac{- 24}{36} . Order is - 24 < - 16 < - 15 < - 14 . Order is \frac{2}{- 3} < \frac{- 4}{9} < \frac{5}{- 12} < \frac{7}{- 18} .$

Page No 4.27:

Question 6:

Arrange the following rational numbers in descending order:
(i) $\frac{7}{8}, \frac{64}{16}, \frac{36}{- 12}, \frac{5}{- 4}, \frac{140}{28}$
(ii) $\frac{- 3}{10}, \frac{17}{- 30}, \frac{7}{- 15}, \frac{- 11}{20}$

Answer:

We have to arrange them in descending order.

(i)
$Since, LCM of 8, 16, - 12, - 4, 28 is 336 . Multiplying the numerators and denominators, to get the denominator equal to the LCM, \frac{7}{8} = \frac{7 \times 42}{8 \times 42} = \frac{294}{336}, \frac{64}{16} = \frac{64 \times 21}{16 \times 21} = \frac{1344}{336}, \frac{36}{- 12} = \frac{- 36 \times 28}{12 \times 28} = \frac{- 1008}{336}, \frac{5}{- 4} = \frac{- 5 \times 84}{4 \times 84} = \frac{- 420}{336}, \frac{140}{28} = \frac{140 \times 12}{28 \times 12} = \frac{1680}{336} . Order is 1680 > 1344 > 294 > - 420 > - 1008 . Order is \frac{140}{28} > \frac{64}{16} > \frac{7}{8} > \frac{5}{- 4} > \frac{36}{- 12} .$
(ii)

$Since, LCM of 10, - 30, - 15, 20 is 60 . Multiplying the numerators and denominators, to get the denominator equal to the LCM, \frac{- 3}{10} = \frac{- 3 \times 6}{10 \times 6} = \frac{- 18}{60}, \frac{17}{- 30} = \frac{- 17 \times 2}{30 \times 2} = \frac{- 34}{60}, \frac{7}{- 15} = \frac{- 7 \times 4}{15 \times 4} = \frac{- 28}{60}, \frac{- 11}{20} = \frac{- 11 \times 3}{20 \times 3} = \frac{- 33}{60}, Order is, - 18 > - 28 > - 33 > - 34 . Order is \frac{- 3}{10} > \frac{7}{- 15} > \frac{- 11}{20} > \frac{17}{- 30} .$

Page No 4.27:

Question 7:

Which of the following statements are true:
(i) The rational number $\frac{29}{23}$ lies to the left of zero on the number line.
(ii) The rational number $\frac{- 12}{- 17}$ lies to the left of zero on the number line.
(iii) The rational number $\frac{3}{4}$ lies to the right of zero on the number line.
(iv) The rational numbers $\frac{- 12}{- 5} and \frac{- 7}{17}$ are on the opposite side of zero on the number line.
(v) The rational numbers $\frac{- 21}{- 5} and \frac{7}{- 31}$ are on the opposite side of zero on the number line.
(vi) The rational number $\frac{- 3}{- 5}$ is one the right of $\frac{- 4}{7}$ on the number line.

Answer:

(i) False; it lies to the right of zero because it is a positive number.
(ii) False; it lies to the right of zero because it is a positive number.
(iii) True
(iv) True; they are of opposite signs.
(v) False; they both are of same signs.
(v) True; they both are of opposite signs and positive number is greater than the negative number. Thus, it is on the right of the negative number.

Page No 4.27:

Question 1:

Mark the correct alternative in each of the following:

$\frac{44}{- 77}$ in standard form is

(a) $\frac{4}{- 7}$ (b) $- \frac{4}{7}$ (c) $- \frac{44}{77}$ (d) None of these

Answer:

The denominator of $\frac{44}{- 77}$ is negative.

Firstly, multiply the numerator and denominator by −1 to make it positive.

$\frac{44}{- 77} = \frac{44 \times (- 1)}{- 77 \times (- 1)} = \frac{- 44}{77}$

Now,

HCF of 44 and 77 = 11

Dividing the numerator and denominator of $\frac{- 44}{77}$ by 11, we have

$\frac{- 44}{77} = \frac{- 44 \div 11}{77 \div 11} = \frac{- 4}{7} = - \frac{4}{7}$

Thus, the standard form of $\frac{44}{- 77}$ is $- \frac{4}{7}$ .

Hence, the correct answer is option (b).

Page No 4.27:

Question 2:

Mark the correct alternative in each of the following:

$- \frac{102}{119}$ in standard form is

(a) $- \frac{6}{7}$ (b) $\frac{6}{7}$ (c) $- \frac{6}{17}$ (d) None of these

Answer:

The denominator of the rational number $- \frac{102}{119}$ is positive.

In order to write the rational number in standard form, divide its numerator and denominator by the HCF of 102 and 119.

HCF of 102 and 119 = 17

Dividing the numerator and denominator of $- \frac{102}{119}$ by 17, we have

$- \frac{102}{119} = - \frac{102 \div 17}{119 \div 17} = - \frac{6}{7}$

Thus, the standard form of $- \frac{102}{119}$ is $- \frac{6}{7}$ .

Hence, the correct answer is option (a).

Page No 4.28:

Question 3:

Mark the correct alternative in each of the following:

A rational number equal to $\frac{- 2}{3}$ is

(a) $\frac{- 10}{25}$ (b) $\frac{10}{- 15}$ (c) $\frac{- 9}{6}$ (d) None of these

Answer:

We know that two rational numbers are equal if they have the same standard form.

The rational number $\frac{- 2}{3}$ is in its standard form.

Consider the rational number $\frac{10}{- 15}$ .

This rational number can be expressed in standard form as follows:

$\frac{10}{- 15} = \frac{10 \times (- 1)}{- 15 \times (- 1)} = \frac{- 10}{15}$                   (Multiplying numerator and denominator by −1 to make denominator positive)

HCF of 10 and 15 = 5

Dividing the numerator and denominator of $\frac{- 10}{15}$ by 5, we have

$\frac{- 10}{15} = \frac{- 10 \div 5}{15 \div 5} = \frac{- 2}{3}$

Thus, the standard form of $\frac{- 10}{15}$ is $\frac{- 2}{3}$ , which is same as the given rational number.

So, the rational number equal to $\frac{- 2}{3}$ is $\frac{10}{- 15}$ .

Let us check why options (a) and (c) are not correct.

The standard form of $\frac{- 10}{25}$ is $\frac{- 2}{5}$ .

HCF of 10 and 25 = 5

Dividing the numerator and denominator of $\frac{- 10}{25}$ by 5, we have

$\frac{- 10}{25} = \frac{- 10 \div 5}{25 \div 5} = \frac{- 2}{5}$

The standard form of $\frac{- 9}{6}$ is $\frac{- 3}{2}$ .

HCF of 6 and 9 = 3

Dividing the numerator and denominator of $\frac{- 9}{6}$ by 3, we have

$\frac{- 9}{6} = \frac{- 9 \div 3}{6 \div 2} = \frac{- 3}{2}$

Hence, the correct answer is option (b).

Page No 4.28:

Question 4:

Mark the correct alternative in each of the following:

If $\frac{- 3}{7} = \frac{x}{35}$ , then x =

(a) 15 (b) 21 (c) −15 (d) −21

Answer:

Firstly, write $\frac{- 3}{7}$ as a rational number with denominator 35.

Multiplying the numerator and denominator of $\frac{- 3}{7}$ by 5, we have

$\frac{- 3}{7} = \frac{- 3 \times 5}{7 \times 5} = \frac{- 15}{35}$

$∴ \frac{- 3}{7} = \frac{x}{35} \Rightarrow \frac{- 15}{35} = \frac{x}{35} \Rightarrow x = - 15$

Hence, the correct answer is option (c).

Page No 4.28:

Question 5:

Mark the correct alternative in each of the following:

Which of the following is correct?

(a) $\frac{5}{9} > \frac{- 3}{- 8}$ (b) $\frac{5}{9} < \frac{- 3}{- 8}$ (c) $\frac{2}{- 3} < \frac{- 8}{7}$ (d) $\frac{4}{- 3} > \frac{- 8}{7}$

Answer:

Consider the rational numbers $\frac{5}{9}$ and $\frac{- 3}{- 8}$ .

We write the rational number $\frac{- 3}{- 8}$ with positive denominator.
$\frac{- 3}{- 8} = \frac{- 3 \times (- 1)}{- 8 \times (- 1)} = \frac{3}{8}$

Now, we write the rational numbers so that they have a common denominator.

LCM of 8 and 9 = 72

So, $\frac{5}{9} = \frac{5 \times 8}{9 \times 8} = \frac{40}{72}$ and $\frac{3}{8} = \frac{3 \times 9}{8 \times 9} = \frac{27}{72}$

Now,

$40 > 27 \Rightarrow \frac{40}{72} > \frac{27}{72} \Rightarrow \frac{5}{9} > \frac{3}{8} \Rightarrow \frac{5}{9} > \frac{- 3}{- 8}$

It can also be checked that $\frac{2}{- 3} > \frac{- 8}{7}$ and $\frac{4}{- 3} < \frac{- 8}{7}$ .

Hence, the correct answer is option (a).

Page No 4.28:

Question 6:

Mark the correct alternative in each of the following:

If the rational numbers $\frac{- 2}{3}$ and $\frac{4}{x}$ represent a pair of equivalent rational numbers, then x =

(a) 6 (b) −6 (c) 3 (d) −3

Answer:

It is given that the rational numbers $\frac{- 2}{3}$ and $\frac{4}{x}$ represent a pair of equivalent rational numbers.

We know that the values of two equivalent rational numbers is equal.

$∴ \frac{- 2}{3} = \frac{4}{x} \Rightarrow - 2 \times x = 4 \times 3 (\frac{a}{b} = \frac{c}{d} \Rightarrow a d = b c) \Rightarrow - 2 x = 12 \Rightarrow \frac{- 2 x}{- 2} = \frac{12}{- 2} (Dividing both sides by - 2) \Rightarrow x = - 6$

Hence, the correct answer is option (b).

Page No 4.28:

Question 7:

Mark the correct alternative in each of the following:

What is the additive identity element in the set of whole numbers?

(a) 0 (b) 1 (c) −1 (d) None of these

Answer:

If a is a whole number then a + 0 = a = 0 + a.

Therefore, 0 is the additive identity element for addition of whole number because it does not change the identity or value of the whole number during the operation of addition.

Hence, the correct answer is option (a).

Page No 4.28:

Question 8:

Mark the correct alternative in each of the following:

What is the multiplicative identity element in the set of whole numbers?

(a) 0 (b) 1 (c) −1 (d) None of these

Answer:

We know that if a is a whole number, then a × 1 = a = 1 × a.

Therefore, 1 is the multiplicative identity element for multiplication of whole numbers because it does not change the identity or value of the whole number during the operation of multiplication.

Hence, the correct answer is option (b).

Page No 4.28:

Question 9:

Mark the correct alternative in each of the following:

Which of the following is not zero?

(a) 0 × 0 (b) $\frac{0}{3}$ (c) $\frac{7 - 7}{3}$ (d) 9 ÷ 0

Answer:

If any number is multiplied by 0, the product is 0.

∴ 0 × 0 = 0

If 0 is divided by any number (≠ 0), the quotient is always 0.

∴ $\frac{0}{3} = 0$ and $\frac{7 - 7}{3} = \frac{0}{3} = 0$

Division of any number by 0 is meaningless and is not defined.

∴ 9 ÷ 0 is not defined.

Hence, the correct answer is option (d).

Page No 4.28:

Question 10:

Mark the correct alternative in each of the following:

The whole number nearest to 457 and divisible by 11 is

(a) 450 (b) 451 (c) 460 (d) 462

Answer:

The numbers 450 and 460 are not divisible by 11.

Now, both the numbers 451 and 462 are divisible by 11.

Distance between 457 and 451 on the number line = 457 − 451 = 6

Distance between 457 and 462 on the number line = 462 − 457 = 5

Thus, the whole number nearest to 457 and divisible by 11 is 462.

Hence, the correct answer is option (d).

Page No 4.28:

Question 11:

Mark the correct alternative in each of the following:

If $- \frac{3}{8}$ and $\frac{x}{- 24}$ are equivalent rational numbers, then x =

(a) 3 (b) 6 (c) 9 (d) 12

Answer:

It is given that the rational numbers $- \frac{3}{8}$ and $\frac{x}{- 24}$ are equivalent rational numbers.

We know that the values of two equivalent rational numbers is equal.

$∴ \frac{x}{- 24} = - \frac{3}{8} \Rightarrow x \times 8 = - 3 \times (- 24) (\frac{a}{b} = \frac{c}{d} \Rightarrow a d = b c) \Rightarrow 8 x = 72 \Rightarrow \frac{8 x}{8} = \frac{72}{8} (Dividing both sides by 8) \Rightarrow x = 9$

Hence, the correct answer is option (c).

Page No 4.28:

Question 12:

Mark the correct alternative in each of the following:

If $\frac{27}{- 45}$ is expressed as a rational number with denominator 5, then the numerator is

(a) 3 (b) −3 (c) 6 (d) −6

Answer:

In order to express $\frac{27}{- 45}$ as a rational number with denominator 5, firstly find a number which gives 5 when −45 is divided by it.

This number is −45 ÷ 5 = −9.

Dividing the numerator and denominator of $\frac{27}{- 45}$ by −9, we have

$\frac{27}{- 45} = \frac{27 \div (- 9)}{- 45 \div (- 9)} = \frac{- 3}{5}$

Thus, the numerator is −3.

Hence, the correct answer is option (b).

Page No 4.28:

Question 13:

Mark the correct alternative in each of the following:

Which of the following pairs of rational numbers are on the opposite side of the zero on the number line?

(a) $\frac{3}{7}$ and $\frac{5}{12}$ (b) $- \frac{3}{7}$ and $\frac{- 5}{12}$ (c) $\frac{3}{7}$ and $\frac{- 5}{12}$ (d) None of these

Answer:

The rational numbers $\frac{3}{7}$ and $\frac{5}{12}$ are positive rational numbers. We know that every positive rational number is greater than 0, so both the rational numbers $\frac{3}{7}$ and $\frac{5}{12}$ are represented by points on the right of the zero on the number line.

The rational numbers $- \frac{3}{7}$ and $\frac{- 5}{12}$ are negative rational numbers. We know that every negative rational number is less than 0, so both the rational numbers $\frac{3}{7}$ and $\frac{5}{12}$ are represented by points on the left of the zero on the number line.

The rational numbers $\frac{3}{7}$ is a positive rational number whereas the rational number $\frac{- 5}{12}$ is a negative rational numbers. We know that every negative rational number is less than 0 and every positive rational number is greater than 0, so the rational number $\frac{3}{7}$ is represented by point on the right of the zero and $\frac{- 5}{12}$ is represented by point on the left of the zero on the number line.
Thus, the rational numbers $\frac{3}{7}$ and $\frac{- 5}{12}$ are on the opposite side of the zero on the number line.

Hence, the correct answer is option (c).

Page No 4.28:

Question 14:

Mark the correct alternative in each of the following:

The rational number equal to $\frac{2}{- 3}$ is

(a) $\frac{14}{- 18}$ (b) $\frac{- 6}{9}$ (c) $\frac{- 8}{- 12}$ (d) $\frac{3}{- 2}$

Answer:

We know that two rational numbers are equal if they have the same standard form.

$\frac{2}{- 3} = \frac{2 \times (- 1)}{- 3 \times (- 1)} = \frac{- 2}{3}$

The standard form of $\frac{2}{- 3}$ is $\frac{- 2}{3}$ .

Consider the rational number $\frac{- 6}{9}$ .

HCF of 6 and 9 = 3

Dividing the numerator and denominator of $\frac{- 6}{9}$ by 3, we have

$\frac{- 6}{9} = \frac{- 6 \div 3}{9 \div 3} = \frac{- 2}{3}$

Thus, the standard form of $\frac{- 6}{9}$ is $\frac{- 2}{3}$ .

So, the rational number $\frac{- 6}{9}$ is equal to $\frac{2}{- 3}$ .

It can be checked that

Standard form of $\frac{14}{- 18}$ = $\frac{- 7}{9}$

Standard form of $\frac{- 8}{- 12}$ = $\frac{2}{3}$

Standard form of $\frac{3}{- 2}$ = $\frac{- 3}{2}$

Hence, the correct answer is option (b).

Page No 4.29:

Question 15:

Mark the correct alternative in each of the following:

If $- \frac{3}{4} = \frac{6}{x}$ , then x =

(a) −8 (b) 4 (c) −4 (d) 8

Answer:

$- \frac{3}{4} = \frac{6}{x} \Rightarrow - 3 \times x = 6 \times 4 (\frac{a}{b} = \frac{c}{d} \Rightarrow a d = b c) \Rightarrow - 3 x = 24 \Rightarrow \frac{- 3 x}{- 3} = \frac{24}{- 3} (Dividing both sides by - 3) \Rightarrow x = - 8$

Hence, the correct answer is option (a).

Page No 4.3:

Question 1:

Write down the numerator of each of the following rational numbers:
(i) $\frac{- 7}{5}$
(ii) $\frac{15}{- 4}$
(iii) $\frac{- 17}{- 21}$
(iv) $\frac{8}{9}$
(v) 5

Answer:

Numerators are:
(i) $-$ 7
(ii) 15
(iii) $-$ 17
(iv) 8
(v) 5

Page No 4.3:

Question 2:

Write down the denominator of each of the following rational numbers:
(i) $\frac{- 4}{5}$
(ii) $\frac{11}{- 34}$
(iii) $\frac{- 15}{- 82}$
(iv) 15
(v) 0

Answer:

Denominators are:
(i) 5
(ii) $-$ 34
(iii) $-$ 82
(iv) 1
(v) 1

Page No 4.3:

Question 3:

Write down the rational number whose numerator is (−3) × 4, and whose denominator is (34 − 23) × (7 − 4).

Answer:

According to the question:

Numerator = ( $-$ 3) × 4 = $-$ 12

Denominator = (34 $-$ 23) × (7 $-$ 4) = 11 × 3 = 33

∴ Rational number = $\frac{- 12}{33}$

Page No 4.3:

Question 4:

Write the following rational numbers as integers:
$\frac{7}{1}, \frac{- 12}{1}, \frac{34}{1}, \frac{- 73}{1}, \frac{95}{1}$

Answer:

Integers are 7, $-$ 12, 34, $-$ 73 and 95.

Page No 4.3:

Question 5:

Write the following integers as rational numbers with denominator 1:
$- 15, 17, 85, - 100$

Answer:

Rational numbers of given integers with denominator 1 are:

$\frac{- 15}{1}, \frac{17}{1}, \frac{85}{1}, \frac{- 100}{1}$

Page No 4.3:

Question 6:

Write down the rational number whose numerator is the smallest three digit number and denominator is the largest four digit number.

Answer:

Smallest three-digit number = 100

Largest four-digit number = 9999

∴ Required rational number = $\frac{100}{9999}$

Page No 4.3:

Question 7:

Separate positive and negative rational numbers from the following rational numbers:

$\frac{- 5}{- 7}, \frac{12}{- 5}, \frac{7}{4}, \frac{13}{- 9}, 0, \frac{- 18}{- 7}, \frac{- 95}{116}, \frac{- 1}{- 9}$

Answer:

Given rational numbers can be rewritten as:

$\frac{5}{7} . - \frac{12}{5}, \frac{7}{4}, - \frac{13}{9}, 0, \frac{18}{7}, - \frac{95}{116}, \frac{1}{9}$

Thus, positive rational numbers are:

$\frac{5}{7}, \frac{7}{4}, \frac{18}{7}, \frac{1}{9}$
or, $\frac{- 5}{- 7}, \frac{7}{4}, 0, \frac{- 18}{- 7} . \frac{- 6}{- 9}$

Negative rational numbers are:

$- \frac{12}{5}, - \frac{13}{9}, - \frac{95}{116}$
or, $\frac{12}{- 5}, \frac{13}{- 9}, \frac{- 95}{116}$

Page No 4.4:

Question 8:

Which of the following rational numbers are positive:
(i) $\frac{- 8}{7}$
(ii) $\frac{9}{8}$
(iii) $\frac{- 19}{- 13}$
(iv) $\frac{- 21}{13}$

Answer:

The numbers can be rewritten as:

$(i) - \frac{8}{7} (i i) \frac{9}{8} (i i i) \frac{19}{13} (i v) - \frac{21}{13}$

Positive rational numbers are (ii) and (iii), i.e., $\frac{9}{8} and \frac{- 19}{- 13}$ .

Page No 4.4:

Question 9:

Which of the following rational numbers are negative?
(i) $\frac{- 3}{7}$
(ii) $\frac{- 5}{- 8}$
(iii) $\frac{9}{- 83}$
(iv) $\frac{- 115}{- 197}$

Answer:

The numbers can be rewritten as:
$(i) - \frac{3}{7} (i i) \frac{5}{8} (i i i) - \frac{9}{83} (i v) \frac{115}{197}$

Negative rational numbers are (i) and (iii).

Page No 4.8:

Question 1:

Express each of the following as a rational number with positive denominator:
(i) $\frac{- 15}{- 28}$
(ii) $\frac{6}{- 9}$
(iii) $\frac{- 28}{- 11}$
(iv) $\frac{19}{- 7}$

Answer:

Rational number with positive denominators:

(i) Multiplying the number by $-$ 1, we get:

$\frac{- 15}{- 28} = \frac{- 15 \times - 1}{- 28 \times - 1} = \frac{15}{28}$

(ii) Multiplying the number by $-$ 1, we get:

$\frac{6}{- 9} = \frac{6 \times - 1}{- 9 \times - 1} = \frac{- 6}{9}$

(iii) Multiplying the number by $-$ 1, we get:

$\frac{- 28}{- 11} = \frac{- 28 \times - 1}{- 11 \times - 1} = \frac{28}{11}$

(iv) Multiplying the number by $-$ 1, we get:

$\frac{19}{- 7} = \frac{19 \times - 1}{- 7 \times - 1} = \frac{- 19}{7}$

Page No 4.8:

Question 2:

Express $\frac{3}{5}$ as a rational number with numerator:
(i) 6
(ii) −15
(iii) 21
(iv) −27

Answer:

Rational number with numerator:

(i) 6 is:

$\frac{3 \times 2}{5 \times 2} = \frac{6}{10} (multiplying numerator and denominator by 2)$

(ii)

$- 15 is : \frac{3 \times - 5}{5 \times - 5} = \frac{- 15}{- 25} (multiplying numerator and denominator by - 5)$

(iii)

$21 is : \frac{3 \times 7}{5 \times 7} = \frac{21}{35} (multiplying numerator and denominator by 7)$

(iv)

$- 27 is : \frac{3 \times - 9}{5 \times - 9} = \frac{- 27}{- 45} (multiplying numerator and denominator by - 9)$

Page No 4.8:

Question 3:

Express $\frac{5}{7}$ as a rational number with denominator:
(i) −14
(ii) 70
(iii) −28
(iv) −84

Answer:

$\frac{5}{7}$ as a rational number with denominator:
(i) −14 is:

$\frac{5 \times - 2}{7 \times - 2} = \frac{- 10}{- 14} (Multiplyin g numerator and denominator by - 2)$
(ii) 70 is:

$\frac{5 \times 10}{7 \times 10} = \frac{50}{70} (Multiplying numerator and denominator by 10)$

(iii) −28 is:

$\frac{5 \times - 4}{7 \times - 4} = \frac{- 20}{- 28} (Multiplying numerator and denominator by - 4)$

(iv) −84 is:

$\frac{5 \times - 12}{7 \times - 12} = \frac{- 60}{- 84} (Multiplying numerator and denominator by - 12)$

Page No 4.9:

Question 4:

Express $\frac{3}{4}$ as a rational number with denominator:
(i) 20
(ii) 36
(iii) 44
(iv) −80

Answer:

3/4 as rational number with denominator:

(i)
$20 is : \frac{3 \times 5}{4 \times 5} = \frac{15}{20} (multiplying numerator and denominator by 5)$

(ii)
$36 is : \frac{3 \times 9}{4 \times 9} = \frac{27}{36} (multiplying numerator and denominator by 9)$

(iii)
$44 is : \frac{3 \times 11}{4 \times 11} = \frac{33}{44} (multiplying numerator and denominator by 11)$

(iv)
$- 80 is : \frac{3 \times - 20}{4 \times - 20} = \frac{- 60}{- 80} (multiplying numerator and denominator by - 20)$

Page No 4.9:

Question 5:

Express $\frac{2}{5}$ as a rational number with numerator:
(i) −56
(ii) 154
(iii) −750
(iv) 500

Answer:

2/5 as a rational number with numerator:

(i)
$- 56 is : \frac{2 \times 28}{5 \times - 28} = \frac{- 56}{- 140} (multiplying numerator and denominator by - 28)$

(ii)
$154 is : \frac{2 \times 77}{5 \times 77} = \frac{154}{385} (multiplying numerator and denominator by 77)$

(iii)
$- 750 is : \frac{2 \times - 375}{5 \times - 375} = \frac{- 750}{- 1875} (multiplying numerator and denominator by - 375)$

(iv)
$500 is : \frac{2 \times 250}{5 \times 250} = \frac{500}{1250} (multiplying numerator and denominator by 250)$

Page No 4.9:

Question 6:

Express $\frac{- 192}{108}$ as a rational number with numerator:
(i) 64
(ii) −16
(iii) 32
(iv) −48

Answer:

Rational number with numerator:

$(i) 64 as numerator : \frac{- 192 / - 3}{108 / - 3} = \frac{64}{- 36} (Dividing the numerator and denomintor by - 3) (ii) - 16 as numerator : \frac{- 192 / 12}{108 / 12} = \frac{- 16}{9} (Dividing the numerator and denomintor by 12) (iii) 32 as numerator : \frac{- 192 / - 6}{108 / - 6} = \frac{32}{- 18} (Dividing the numerator and denomintor by - 6) (iv) - 48 as numerator : \frac{- 192 / 4}{108 / 4} = \frac{- 48}{27} (Dividing the numerator and denomintor by 4)$

Page No 4.9:

Question 7:

Express $\frac{168}{- 294}$ as a rational number with denominator:
(i) 14
(ii) −7
(iii) −49
(iv) 1470

Answer:

Rational number with denominator:

$(i) 14 as denominator : \frac{168 / - 21}{- 294 / - 21} = \frac{- 8}{14} (Dividing the numerator and denomintor by - 21) (ii) - 7 as denominator : \frac{168 / 42}{- 294 / 42} = \frac{4}{- 7} (Dividing the numerator and denomintor by 42) (iii) - 49 as denominator : \frac{168 / 6}{- 294 / 6} = \frac{28}{- 49} (Dividing the numerator and denomintor by 6) (iv) 1470 as denominator : \frac{168 \times - 5}{- 294 \times - 5} = \frac{- 840}{1470} (Multiplying the numerator and denomintor by - 5)$

Page No 4.9:

Question 8:

Write $\frac{- 14}{42}$ in a form so that the numerator is equal to:
(i) −2
(ii) 7
(iii) 42
(iv) −70

Answer:

Rational number with numerator:

$(i) - 2 is : \frac{- 14 / 7}{42 / 7} = \frac{- 2}{6} (Dividing numerator and denominator by 7)$
$(i i) 7 is : \frac{- 14 / - 2}{42 / - 2} = \frac{7}{- 21} (Dividing numerator and denominator by - 2)$
$(i i i) 42 is : \frac{- 14 \times - 3}{42 \times - 3} = \frac{42}{- 126} (M ultiplying numerator and denominator by - 3)$
$(i v) - 70 is : \frac{- 14 \times 5}{42 \times 5} = \frac{- 70}{210} (Multiplying numerator and denominator by 5)$

Page No 4.9:

Question 9:

Select those rational numbers which can be written as a rational number with numerator 6:

\frac{1}{22}, \frac{2}{3}, \frac{3}{4}, \frac{4}{- 5}, \frac{5}{6}, \frac{- 6}{7}, \frac{- 7}{8}

Answer:

Given rational numbers that can be written as a rational number with numerator 6 are:

$\frac{1}{22} (On multiplying by 6) = \frac{6}{132} \frac{2}{3} (On multiplying by 3) = \frac{6}{9} \frac{3}{4} (On multiplying by 2) = \frac{6}{8} \frac{- 6}{7} (On multiplying by - 1) = \frac{6}{- 7}$

Page No 4.9:

Question 10:

Select those rational numbers which can be written as a rational number with denominator 4:

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

Answer:

Given rational numbers that can be written as a rational number with denominator 4 are:

$\frac{7}{8} (On dividing by 2) = \frac{3.5}{4} \frac{64}{16} (On dividing by 4) = \frac{16}{4} \frac{36}{- 12} (On dividing by 3) = \frac{12}{- 4} = \frac{- 12}{4} \frac{- 16}{17} can' t be expressed with a denominator 4 . \frac{5}{- 4} (On multiplying by - 1) = \frac{- 5}{4} \frac{140}{28} (On dividing by 7) = \frac{20}{4}$
122 (On multiplying by 6) = 613223 (On multiplying by 3) = 6934 (On multiplying by 2) = 68−67 (On multiplying by −1) = 6−7

Page No 4.9:

Question 11:

In each of the following, find an equivalent form of the rational number having a common denominator:
(i) $\frac{3}{4} and \frac{5}{12}$
(ii) $\frac{2}{3}, \frac{7}{6} and \frac{11}{12}$
(iii) $\frac{5}{7}, \frac{3}{8}, \frac{9}{14} and \frac{20}{21}$

Answer:

Equivalent forms of the rational number having common denominator are:

(i) $\frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12} and \frac{5}{12}$ .

(ii)
$\frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12} and \frac{7}{6} = \frac{7 \times 2}{6 \times 2} = \frac{14}{12} a n d \frac{11}{12} Forms are \frac{8}{12}, \frac{14}{12} and \frac{11}{12}$

(iii)
$\frac{5}{7} = \frac{5 \times 24}{7 \times 24} = \frac{120}{168}, \frac{3}{8} = \frac{3 \times 21}{8 \times 21} = \frac{63}{168}, \frac{9}{14} = \frac{9 \times 12}{14 \times 12} = \frac{108}{168} a n d \frac{20}{21} = \frac{20 \times 8}{21 \times 8} = \frac{160}{168} Forms are \frac{120}{168}, \frac{63}{168}, \frac{108}{168} and \frac{160}{168}$

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