Rd Sharma 2020 2021 for Class 7 Maths Chapter 2 - Fractions

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Rd Sharma 2020 2021 Solutions for Class 7 Maths Chapter 2 Fractions are provided here with simple step-by-step explanations. These solutions for Fractions are extremely popular among Class 7 students for Maths Fractions 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 2 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 2.12:

Question 1:

Compare the following fractions by using the symbol > or < or =:
(i) $\frac{7}{9} and \frac{8}{13}$
(ii) $\frac{11}{9} and \frac{5}{9}$
(iii) $\frac{37}{41} and \frac{19}{30}$
(iv) $\frac{17}{15} and \frac{119}{105}$

Answer:

First, we need to find the LCM of denominators in each case. After that, we will equate the denominators in order to compare the two fractions.

(i)
LCM of 9 and 13 is 117.

$Now make both fraction equivalent with denominator as 117 \frac{7}{9} = \frac{7}{9} \times \frac{13}{13} \Rightarrow \frac{7}{9} = \frac{91}{117} \frac{8}{13} = \frac{8}{13} \times \frac{9}{9} \Rightarrow \frac{8}{13} = \frac{72}{117} we know 91 > 72 \Rightarrow \frac{91}{117} > \frac{72}{117} \Rightarrow \frac{7}{9} > \frac{8}{13}$

(ii)

$both fraction have same denominator as 9 we know 11 > 5 \Rightarrow \frac{11}{9} > \frac{5}{9}$
(iii)

$LCM of 41 and 30 is 1230 Now convert both fraction to their equivalent fractions with denominator as 1230 \frac{37}{41} = \frac{37}{41} \times \frac{30}{30} \Rightarrow \frac{37}{41} = \frac{1110}{1230} \frac{19}{30} = \frac{19}{30} \times \frac{41}{41} \Rightarrow \frac{19}{30} = \frac{779}{1230} we know 1110 > 779 \Rightarrow \frac{1110}{1230} > \frac{779}{1230} \Rightarrow \frac{37}{41} > \frac{19}{30}$

(iv)

LCM of 15 and 105 is 105.

$Now convert fraction to its equivalent fractions with denominator as 105 \frac{17}{15} = \frac{17}{15} \times \frac{7}{7} \Rightarrow \frac{17}{15} = \frac{119}{105}$

Page No 2.12:

Question 2:

Arrange the following fractions in ascending order:
(i) $\frac{3}{8}, \frac{5}{6}, \frac{6}{8} \frac{2}{4}, \frac{1}{3}$
(ii) $\frac{4}{3}, \frac{3}{8}, \frac{6}{12}, \frac{5}{16}$

Answer:

(i)
$LCM of the denominators 8, 6, 4 and 3 is 24 . Now, convert all fractions into their equivalent fractions with denominator 24 . \frac{3}{8} = \frac{3}{8} \times \frac{3}{3} \Rightarrow \frac{3}{8} = \frac{9}{24} \frac{5}{6} = \frac{5}{6} \times \frac{4}{4} \Rightarrow \frac{5}{6} = \frac{20}{24} \frac{6}{8} = \frac{6}{8} \times \frac{3}{3} \Rightarrow \frac{6}{8} = \frac{18}{24} \frac{2}{4} = \frac{2}{4} \times \frac{6}{6} \Rightarrow \frac{2}{4} = \frac{12}{24} \frac{1}{3} = \frac{1}{3} \times \frac{8}{8} \Rightarrow \frac{1}{3} = \frac{8}{24} We know : 8 < 9 < 12 < 18 < 20 \Rightarrow \frac{8}{24} < \frac{9}{24} < \frac{12}{24} < \frac{18}{24} < \frac{20}{24} \Rightarrow \frac{1}{3} < \frac{3}{8} < \frac{2}{4} < \frac{6}{8} < \frac{5}{6}$
(ii)
$LCM of the denominators 8, 12, 16 and 3 is 48 . Now, convert all fractions into their equivalent fractions with denominator 48 . \frac{4}{3} = \frac{4}{3} \times \frac{16}{16} \Rightarrow \frac{4}{3} = \frac{64}{48} \frac{3}{8} = \frac{3}{8} \times \frac{6}{6} \Rightarrow \frac{3}{8} = \frac{18}{48} \frac{6}{12} = \frac{6}{12} \times \frac{4}{4} \Rightarrow \frac{6}{12} = \frac{24}{48} \frac{5}{16} = \frac{5}{16} \times \frac{3}{3} \Rightarrow \frac{5}{16} = \frac{15}{48} we know 15 < 18 < 24 < 64 \Rightarrow \frac{15}{48} < \frac{18}{48} < \frac{24}{48} < \frac{64}{48} \Rightarrow \frac{5}{16} < \frac{3}{8} < \frac{6}{12} < \frac{4}{3}$

Page No 2.12:

Question 3:

Arrange the following fractions in descending order:
(i) $\frac{4}{5}, \frac{7}{10}, \frac{11}{15}, \frac{17}{20}$
(ii) $\frac{2}{7}, \frac{11}{35}, \frac{9}{14}, \frac{13}{28}$

Answer:

(i)
$LCM of the denominators 5, 10, 15 and 20 is 60 . Now, convert all fractions to their equivalent fractions with denominator 60 . \frac{4}{5} = \frac{4}{5} \times \frac{12}{12} = \frac{48}{60} \frac{7}{10} = \frac{7}{10} \times \frac{6}{6} = \frac{42}{60} \frac{11}{15} = \frac{11}{15} \times \frac{4}{4} = \frac{44}{60} \frac{17}{20} = \frac{17}{20} \times \frac{3}{3} = \frac{51}{60} We know : 51 > 48 > 44 > 42 \Rightarrow \frac{51}{60} > \frac{48}{60} > \frac{44}{60} > \frac{42}{60} \Rightarrow \frac{17}{20} > \frac{4}{5} > \frac{11}{15} > \frac{7}{10}$

(ii)
$LCM of the denominators 7, 35, 14 and 28 is 140 . Now, convert all fractions to their equivalent fractions with denominator 140 . \frac{2}{7} = \frac{2}{7} \times \frac{20}{20} = \frac{40}{140} \frac{11}{35} = \frac{11}{35} \times \frac{4}{4} = \frac{44}{140} \frac{9}{14} = \frac{9}{14} \times \frac{10}{10} = \frac{90}{140} \frac{13}{28} = \frac{13}{28} \times \frac{5}{5} = \frac{65}{140} We know : 90 > 65 > 44 > 40 \Rightarrow \frac{90}{140} > \frac{65}{140} > \frac{44}{140} > \frac{40}{140} \Rightarrow \frac{9}{14} > \frac{13}{28} > \frac{11}{35} > \frac{2}{7}$

Page No 2.12:

Question 4:

Write five equivalent fractions of $\frac{3}{5}$ .

Answer:

Five equivalent fractions of $\frac{3}{5}$ are:

$(i) \frac{3}{5} = \frac{3}{5} \times \frac{2}{2} \Rightarrow \frac{3}{5} = \frac{6}{10} (ii) \frac{3}{5} = \frac{3}{5} \times \frac{3}{3} \Rightarrow \frac{3}{5} = \frac{9}{15} (iii) \frac{3}{5} = \frac{3}{5} \times \frac{4}{4} \Rightarrow \frac{3}{5} = \frac{12}{20} (iv) \frac{3}{5} = \frac{3}{5} \times \frac{5}{5} \Rightarrow \frac{3}{5} = \frac{15}{25} (v) \frac{3}{5} = \frac{3}{5} \times \frac{6}{6} \Rightarrow \frac{3}{5} = \frac{18}{30}$

Page No 2.12:

Question 5:

Find the sum:
(i) $\frac{5}{8} + \frac{3}{10}$
(ii) $4 \frac{3}{4} + 9 \frac{2}{5}$
(iii) $\frac{5}{6} + 3 + \frac{3}{4}$
(iv) $2 \frac{3}{5} + 4 \frac{7}{10} + 2 \frac{4}{15}$

Answer:

(i)

$\frac{5}{8} + \frac{3}{10} L C M o f 8, 10 i s 40 . \Rightarrow \frac{(5 \times 5) + (3 \times 4)}{40} = \frac{37}{40}$
(ii)

$4 \frac{3}{4} + 9 \frac{2}{5} o r \frac{(4 \times 4) + 3}{4} + \frac{(9 \times 5) + 2}{5} \Rightarrow \frac{19}{4} + \frac{47}{5} L C M o f 5, 4 i s 20 . \frac{(19 \times 5) + (47 \times 4)}{20} = \frac{283}{20}$
(iii)

$\frac{5}{6} + 3 + \frac{3}{4} L C M o f 6, 4 i s 24 . \frac{(5 \times 4) + (3 \times 24) + (3 \times 6)}{24} = \frac{110}{24}$
(iv)

$2 \frac{3}{5} + 4 \frac{7}{10} + 2 \frac{4}{15} o r \frac{(2 \times 5) + 3}{5} + \frac{(4 \times 10) + 7}{10} + \frac{(2 \times 15) + 4}{15} \Rightarrow \frac{13}{5} + \frac{47}{10} + \frac{34}{15} L C M o f 15, 10 a n d 5 i s 30 . \frac{(13 \times 6) + (47 \times 3) + (34 \times 2)}{30} = \frac{287}{30}$

Page No 2.12:

Question 6:

Find the difference of
(i) $\frac{13}{24} and \frac{7}{16}$
(ii) $6 and \frac{23}{3}$
(iii) $\frac{21}{25} and \frac{18}{20}$
(iv) $3 \frac{3}{10} and 2 \frac{7}{15}$

Answer:

(i)

$\frac{13}{24} - \frac{7}{16} L C M o f 24 a n d 16 i s 48 . \frac{(13 \times 2) - (7 \times 3)}{48} = \frac{5}{48}$

(ii)

$\frac{23}{3} - 6 L C M o f 3 a n d 1 i s 3 . \frac{(23 \times 1) - (6 \times 3)}{3} = \frac{5}{3}$

(iii)

$\frac{18}{20} - \frac{21}{25} L C M o f 20, 25 i s 100 . \Rightarrow \frac{(18 \times 5) - (21 \times 4)}{100} = \frac{6}{100} = \frac{3}{50}$

(iv)

$3 \frac{3}{10} - 2 \frac{7}{15} \Leftrightarrow \frac{(3 \times 10) + 3}{10} - \frac{(2 \times 15) + 7}{15} \Leftrightarrow \frac{33}{10} - \frac{37}{15} L C M o f 10 a n d 15 i s 30 . \frac{(33 \times 3) - (37 \times 2)}{30} = \frac{25}{30} \Leftrightarrow \frac{5}{6}$

Page No 2.13:

Question 7:

Find the difference:
(i) $\frac{6}{7} - \frac{9}{11}$
(ii) $8 - \frac{5}{9}$
(iii) $9 - 5 \frac{2}{3}$
(iv) $4 \frac{3}{10} - 1 \frac{2}{15}$

Answer:

(i)
$\frac{6}{7} - \frac{9}{11} L C M o f 7 a n d 11 i s 77 . \Rightarrow \frac{6}{7} - \frac{9}{11} \Leftrightarrow \frac{(6 \times 11) - (9 \times 7)}{77} \Leftrightarrow \frac{3}{77}$

(ii)
$8 - \frac{5}{9} L C M o f 1 a n d 9 i s 9 . \Rightarrow \frac{8}{1} - \frac{5}{9} \Leftrightarrow \frac{(8 \times 9) - (5 \times 1)}{9} \Leftrightarrow \frac{67}{9}$

(iii)
$9 - 5 \frac{2}{3} \Leftrightarrow \frac{9}{1} - \frac{(5 \times 3) + 2}{3} L C M o f 1 a n d 3 i s 3 . \Rightarrow \frac{9}{1} - \frac{17}{3} \Leftrightarrow \frac{(9 \times 3) - (17 \times 1)}{3} \Leftrightarrow \frac{10}{3}$

(iv)
$4 \frac{3}{10} - 1 \frac{2}{15} \Leftrightarrow \frac{(4 \times 10) + 3}{10} - \frac{(15 \times 1) + 2}{15} L C M o f 10 a n d 15 i s 30 . \Rightarrow \frac{43}{10} - \frac{17}{15} \Leftrightarrow \frac{(43 \times 3) - (17 \times 2)}{30} \Leftrightarrow \frac{95}{30} \Leftrightarrow \frac{19}{6}$

Page No 2.13:

Question 8:

Simplify:
(i) $\frac{2}{3} + \frac{1}{6} - \frac{2}{9}$
(ii) $12 - 3 \frac{1}{2}$
(iii) $7 \frac{5}{6} - 4 \frac{3}{8} + 2 \frac{7}{12}$

Answer:

(i)

$\frac{2}{3} + \frac{1}{6} - \frac{2}{9} LCM of 3.6 and 9 is 18 . = \frac{2}{3} + \frac{1}{6} - \frac{2}{9} = \frac{(2 \times 6) + (1 \times 3) - (2 \times 2)}{18} = \frac{11}{18}$
(ii)

$12 - 3 \frac{1}{2} = \frac{12}{1} - \frac{(3 \times 2) + 1}{2} LCM of 2 and 1 is 2 . = \frac{12}{1} - \frac{7}{2} = \frac{(12 \times 2) - (7 \times 1)}{2} = \frac{17}{2}$

(iii)

$7 \frac{5}{6} - 4 \frac{3}{8} + 2 \frac{7}{12} = \frac{(7 \times 6) + 5}{6} - \frac{(4 \times 8) + 3}{8} + \frac{(2 \times 12) + 7}{12} LCM of 6, 8 and 12 is 24 . = \frac{47}{6} - \frac{35}{8} + \frac{31}{12} = \frac{(47 \times 4) - (35 \times 3) + (31 \times 2)}{24} = \frac{145}{24}$

Page No 2.13:

Question 9:

What should be added to $5 \frac{3}{7}$ to get 12?

Answer:

Let x be the required fraction.

According to the question:

$x + 5 \frac{3}{7} = 12 \Rightarrow x + \frac{(5 \times 7) + 3}{7} = 12 \Rightarrow x = 12 - \frac{38}{7} \Rightarrow x = \frac{(12 \times 7) - (38 \times 1)}{7} \Leftrightarrow \frac{46}{7}$

Page No 2.13:

Question 10:

What should be added to $5 \frac{4}{15}$ to get $12 \frac{3}{5} ?$

Answer:

Let x be the required fraction.

According to the question:

$x + 5 \frac{4}{15} = 12 \frac{3}{5} \Rightarrow x + \frac{(15 \times 5) + 4}{15} = \frac{(12 \times 5) + 3}{5} \Rightarrow x = \frac{63}{5} - \frac{79}{15} L C M o f 5 a n d 15 i s 15 . \Rightarrow x = \frac{(63 \times 3) - (79 \times 1)}{15} \Leftrightarrow \frac{110}{15} \Rightarrow x = \frac{110}{15} \Leftrightarrow \frac{22}{3}$

Page No 2.13:

Question 11:

Suman studies for $5 \frac{2}{3}$ hours daily. She devotes $2 \frac{4}{5}$ hours of her time for Science and Mathematics. How much time does she devote for other subjects?

Answer:

Suman studies for $5 \frac{2}{3} hours$ daily. Therefore, we have
$5 \frac{2}{3} hours = \frac{(5 \times 3) + 2}{3} = \frac{17}{3} hours$
She studies science and mathematics for $2 \frac{4}{5} hours$ . Therefore, we have
$2 \frac{4}{5} hours = \frac{(2 \times 5) + 4}{5} = \frac{14}{5} hours$
Time devoted to other subjects = Total study time $-$ Time devoted to science and mathematics

= $\frac{17}{3} - \frac{14}{5} = \frac{(17 \times 5) - (14 \times 3)}{15} = \frac{43}{15} hours$

Page No 2.13:

Question 12:

A piece of wire is of length $12 \frac{3}{4} m$ . If it is cut into two pieces in such a way that the length of one piece is $5 \frac{1}{4} m,$ what is the length of the other piece?

Answer:

Let the length of second piece be x.

Total length of wire = Length of one piece + Length of second piece

$12 \frac{3}{4} = 5 \frac{1}{4} + x \Rightarrow \frac{(12 \times 4) + 3}{4} = \frac{(5 \times 4) + 1}{4} + x \Rightarrow x = \frac{(12 \times 4) + 3}{4} - \frac{(5 \times 4) + 1}{4} \Rightarrow x = \frac{51}{4} - \frac{21}{4} \Leftrightarrow \frac{30}{4} \Rightarrow x = \frac{30}{4} \Leftrightarrow \frac{15}{2}$

Page No 2.13:

Question 13:

A rectangular sheet of paper is $12 \frac{1}{2} cm$ long and $10 \frac{2}{3} cm$ wide. Find its perimeter.

Answer:

Perimeter of rectangle = 2(length + width)

$2 \times [12 \frac{1}{2} + 10 \frac{2}{3}] = 2 \times [\frac{(12 \times 2) + 1}{2} + \frac{(10 \times 3) + 2}{3}] = 2 \times [\frac{(25 \times 3) + (32 \times 2)}{6}] = 2 \times [\frac{139}{6}] = \frac{139}{3} cm$

Page No 2.13:

Question 14:

In a "magic square", the sum of the numbers in each row, in each column and along the diagonal is the same. Is this a magic square?

$\frac{4}{11}$	$\frac{9}{11}$	$\frac{2}{11}$
$\frac{3}{11}$	$\frac{5}{11}$	$\frac{7}{11}$
$\frac{8}{11}$	$\frac{1}{11}$	$\frac{6}{11}$

Answer:

$Sum along col umns and rows : \frac{4}{11} + \frac{9}{11} + \frac{2}{11} = \frac{15}{11} \frac{3}{11} + \frac{5}{11} + \frac{7}{11} = \frac{15}{11} \frac{8}{11} + \frac{1}{11} + \frac{6}{11} = \frac{15}{11} \frac{4}{11} + \frac{3}{11} + \frac{8}{11} = \frac{15}{11} \frac{9}{11} + \frac{5}{11} + \frac{1}{11} = \frac{15}{11} \frac{2}{11} + \frac{7}{11} + \frac{6}{11} = \frac{15}{11} S um along diagonals : \frac{4}{11} + \frac{5}{11} + \frac{6}{11} = \frac{15}{11} \frac{2}{11} + \frac{5}{11} + \frac{8}{11} = \frac{15}{11} Since, all the sums in the square are equal along rows, columns and diagonals, it is a magic square .$

Page No 2.13:

Question 15:

The cost of Mathematics book is Rs $25 \frac{3}{4}$ and that of Science book is Rs $20 \frac{1}{2}$ . Which costs more and by how much?

Answer:

$Cost of mathematics book = Rs 25 \frac{3}{4} = \frac{(25 \times 4) + 3}{4} = Rs \frac{103}{4} Cost of Science book = Rs 20 \frac{1}{2} = \frac{(20 \times 2) + 1}{2} = \frac{41}{2} = \frac{41}{2} \times \frac{2}{2} = Rs \frac{82}{4} We know 82 < 103 = > \frac{82}{4} < \frac{103}{4} Thus, Mathematics book costs more . Difference in the cost of Mathematics and Science book = cost of Mathematics book - Cost of Science book = \frac{103}{4} - \frac{82}{4} = \frac{21}{4} = Rs 5 \frac{1}{4} So, Mathematics book costs more by Rs 5 \frac{1}{4}$

Page No 2.13:

Question 16:

(i) Provide the number in box and also give its simplest from in each of the following:
(i) $\frac{2}{3} \times = \frac{10}{30}$
(ii) $\frac{3}{5} \times = \frac{24}{75}$ .

Answer:

(i) $\frac{2}{3} \times x = \frac{10}{30} x = \frac{10}{30} \times \frac{3}{2} = \frac{1}{2}$

(ii) $\frac{3}{5} \times x = \frac{24}{75} x = \frac{24}{75} \times \frac{5}{3} = \frac{8}{15}$

Page No 2.19:

Question 1:

Multiply:
(i) $\frac{7}{11} by \frac{3}{5}$
(ii) $\frac{3}{5} by 25$
(iii) $3 \frac{4}{15} by 24$
(iv) $3 \frac{1}{8} by \times 4 \frac{10}{14}$

Answer:

(i)
$\frac{7}{11} \times \frac{3}{5} = \frac{7 \times 3}{11 \times 5} \Rightarrow \frac{21}{55}$

(ii)
$\frac{3}{5} \times 25 = \frac{3 \times 25}{5 \times 1} \Rightarrow \frac{75}{5} \Leftrightarrow 15$

(iii)
$[3 \frac{4}{15}] \times 24 [\frac{(3 \times 15) + 4}{15}] \times 24 = \frac{49 \times 24}{15 \times 1} \Rightarrow \frac{392}{5} = 78 \frac{2}{5}$

(iv)
$[3 \frac{1}{8}] \times [4 \frac{10}{11}] [\frac{(3 \times 8) + 1}{8}] \times [\frac{(4 \times 11) + 10}{11}] = \frac{25 \times 54}{8 \times 11} \Rightarrow \frac{1350}{88} \Leftrightarrow \frac{675}{44} = 15 \frac{15}{44}$

Page No 2.19:

Question 2:

Find the product:
(i) $\frac{4}{7} \times \frac{14}{25}$
(ii) $7 \frac{1}{2} \times 2 \frac{4}{15}$
(iii) $3 \frac{6}{7} \times 4 \frac{2}{3}$
(iv) $6 \frac{11}{14} \times 3 \frac{1}{2}$

Answer:

(i)
$\frac{4}{7} \times \frac{14}{25} = \frac{4 \times 14}{7 \times 25} \Rightarrow \frac{8}{25}$

(ii)
$7 \frac{1}{2} \times 2 \frac{4}{15} \Leftrightarrow \frac{(7 \times 2) + 1}{2} \times \frac{(2 \times 15) + 4}{15} \Rightarrow \frac{15}{2} \times \frac{34}{15} \Leftrightarrow \frac{17}{1}$
(iii)
$3 \frac{6}{7} \times 4 \frac{2}{3} \Leftrightarrow \frac{(3 \times 7) + 6}{7} \times \frac{(4 \times 3) + 2}{3} \Rightarrow \frac{27^{9}}{7} \times \frac{14^{2}}{3} \Leftrightarrow \frac{18}{1}$
(iv)
$6 \frac{11}{14} \times 3 \frac{1}{2} \Leftrightarrow \frac{(14 \times 6) + 11}{14} \times \frac{(3 \times 2) + 1}{2} \Rightarrow \frac{95}{14^{2}} \times \frac{7}{2} \Leftrightarrow \frac{95}{4}$

Page No 2.20:

Question 3:

Simplify:
(i) $\frac{12}{25} \times \frac{15}{28} \times \frac{35}{36}$
(ii) $\frac{10}{27} \times \frac{39}{56} \times \frac{28}{65}$
(iii) $2 \frac{2}{17} \times 7 \frac{2}{9} \times 1 \frac{33}{52}$

Answer:

(i)
$\frac{12}{25} \times \frac{15}{28} \times \frac{35}{36} \Leftrightarrow \frac{12 \times 15^{3} \times 35^{5}}{25^{5} \times 28^{4} \times 36^{3}} \Rightarrow \frac{3 \times 5}{5 \times 4 \times 3} \Leftrightarrow \frac{1}{4}$

(ii)
$\frac{10}{27} \times \frac{39}{56} \times \frac{28}{65} \Leftrightarrow \frac{10 \times 39^{3^{1}} \times 28}{27^{9} \times 56^{2} \times 65^{5}} \Rightarrow \frac{1}{9}$
(iii)
$2 \frac{2}{17} \times 7 \frac{2}{9} \times 1 \frac{33}{52} \Leftrightarrow \frac{(2 \times 17) + 2}{17} \times \frac{(7 \times 9) + 2}{9} \times \frac{(52 \times 1) + 33}{52} \Leftrightarrow \frac{36^{4} \times 65^{5} \times 85^{5}}{17 \times 9 \times 52^{4}} \Rightarrow \frac{25}{1}$

Page No 2.20:

Question 4:

Find:
(i) $\frac{1}{2} of 4 \frac{2}{9}$
(ii) $\frac{5}{8} of 9 \frac{2}{3}$
(iii) $\frac{2}{3} of \frac{9}{16}$

Answer:

$(i) \frac{1}{2} of 4 \frac{2}{9} = \frac{1}{2} \times (4 \frac{2}{9}) = \frac{1}{2} \times \frac{(4 \times 9) + 2}{9} = \frac{1 \times 38}{2 \times 9} = \frac{19}{9} = 2 \frac{1}{9} (ii) \frac{5}{8} of 9 \frac{2}{3} = \frac{5}{8} \times (9 \frac{2}{3}) = \frac{5}{8} \times \frac{(9 \times 3) + 2}{3} = \frac{5 \times 29}{8 \times 3} = \frac{145}{24} = 6 \frac{1}{24} (iii) \frac{2}{3} of \frac{9}{16} = \frac{2 \times 9}{3 \times 16} = \frac{3}{8}$

Page No 2.20:

Question 5:

Which is greater? $\frac{1}{2} of \frac{6}{7} or \frac{2}{3} of \frac{3}{7}$

Answer:

$\frac{1}{2} of \frac{6}{7} = \frac{1}{2} \times \frac{6}{7} = \frac{6}{14} \frac{2}{3} of \frac{3}{7} = \frac{2}{3} \times \frac{3}{7} = \frac{2}{7}$
$c onvert \frac{2}{7} to its equivalent fraction with denominator as 14$
$\frac{2}{7} = \frac{2}{7} \times \frac{2}{2} = \frac{4}{14}$
$w e know 6 > 4$
$\Rightarrow \frac{6}{14} > \frac{2}{7}$
$\Rightarrow \frac{1}{2} of \frac{6}{7} > \frac{2}{3} of \frac{3}{7}$

Page No 2.20:

Question 6:

Find:
(i) $\frac{7}{11}$ of Rs 330
(ii) $\frac{5}{9}$ of 108 metres
(iii) $\frac{3}{7}$ of 42 litres
(iv) $\frac{1}{12}$ of an hour
(v) $\frac{5}{6}$ of an year
(vi) $\frac{3}{20}$ of a kg
(vii) $\frac{7}{20}$ of a litre
(viii) $\frac{5}{6}$ of a day
(ix) $\frac{2}{7}$ of a week

Answer:

$(i) \frac{7}{11} o f R s 330 = \frac{7 \times 330}{11} \Rightarrow R s 210 (i i) \frac{5}{9} o f 108 m e t r e s = \frac{5}{9} \times 108 \Rightarrow 60 m e t r e s (i i i) \frac{3}{7} o f 42 l i t r e s = \frac{3}{7} \times 42 \Rightarrow 18 l i t r e s (i v) \frac{1}{12} o f 1 h o u r = \frac{1}{12} \times 1 \Rightarrow \frac{1}{12} h o u r 1 h o u r = 60 m i n u t e s \Rightarrow \frac{1}{12} h o u r = \frac{1}{12} \times 60 = 5 m i n u t e s (v) \frac{5}{6} o f 1 y e a r = \frac{5}{6} \times 1 \Rightarrow \frac{5}{6} y e a r 1 y e a r = 12 m o n t h s \Rightarrow \frac{5}{6} y e a r = \frac{5}{6} \times 12 = 10 m o n t h s (v i) \frac{3}{20} o f 1 k g = \frac{3}{20} \times 1 \Rightarrow \frac{3}{20} k g 1 k g = 1000 g \Rightarrow \frac{3}{20} k g = \frac{3}{20} \times 1000 = 150 g (v i i) \frac{7}{20} o f 1 l i t r e = \frac{7}{20} \times 1 \Rightarrow \frac{7}{20} l i t r e 1 l = 1000 m l \Rightarrow \frac{7}{20} l i t r e = \frac{7}{20} \times 1000 = 350 m l (v i i i) \frac{5}{6} o f 1 d a y = \frac{5}{6} \times 1 \Rightarrow \frac{5}{6} d a y 1 d a y = 24 h o u r s \Rightarrow \frac{5}{6} d a y = \frac{5}{6} \times 24 = 20 h o u r s (i x) \frac{2}{7} o f 1 w e e k = \frac{2}{7} \times 1 \Rightarrow \frac{2}{7} w e e k 1 w e e k = 7 d a y s \Rightarrow \frac{2}{7} w e e k = \frac{2}{7} \times 7 = 2 d a y s$

Page No 2.20:

Question 7:

Shikha plants 5 saplings in a row in her garden. The distance between two adjacent saplings is $\frac{3}{4} m$ . Find the distance between the first and the last sapling.

Answer:

Distance between the first and second saplings = $\frac{3}{4} m$
Distance between the first and third saplings = $2 \times \frac{3}{4} m = \frac{3}{2} m$
Distance between the first and fourth saplings = $3 \times \frac{3}{4} m = \frac{9}{4} m$

Distance between the first and fifth saplings = 4 × $\frac{3}{4} m = 3 m$

Page No 2.20:

Question 8:

Ravish reads $\frac{1}{3}$ part of a book in 1 hour. How much part of the book will he read in $2 \frac{1}{5}$ houurs?

Answer:

$2 \frac{1}{5} hours = \frac{(2 \times 5) + 1}{5} = \frac{11}{5} hours In 1 hour Ravish reads \frac{1}{3} of the book Part of book Ravish will read in \frac{11}{5} hours = P art read in 1 hour \times \frac{11}{5} Part of book Ravish will read in \frac{11}{5} hours = \frac{1}{3} \times \frac{11}{5} = \frac{11}{15}$

Page No 2.20:

Question 9:

Lipika reads a book for $1 \frac{3}{4}$ hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book?

Answer:

In one day, Lipika reads for $1 \frac{3}{4} hours$ .
$1 \frac{3}{4} hours = \frac{(1 \times 4) + 3}{4} = \frac{7}{4} hours$
Total hours required = $\frac{7}{4} \times 6 = \frac{42}{4} hours = \frac{21}{2} hours = 10 \frac{1}{2} hours$

Page No 2.20:

Question 10:

Find the area of a rectangular park which is $41 \frac{2}{3} m$ long and $18 \frac{3}{5} m$ broad.

Answer:

$Length of rectangular park = 41 \frac{2}{3} m = \frac{(41 \times 3) + 2}{3} m \Rightarrow Length of rectangular park = \frac{125}{3} m Width of rectangular park = 18 \frac{3}{5} m = \frac{(18 \times 5) + 3}{5} m \Rightarrow Width of rectangular park = \frac{93}{5} m Area of rectangular park = Length of rectangular park \times Width of rectangular park Area of rectangular park = \frac{125}{3} \times \frac{93}{5} = 775 m^{2}$

Page No 2.20:

Question 11:

If milk is available at Rs $17 \frac{3}{4}$ per litre, find the cost of $7 \frac{2}{5}$ litres of milk.

Answer:

Cost of 1 litre milk = $Rs . 17 \frac{3}{4}$
$17 \frac{3}{4} = \frac{(17 \times 4) + 3}{4} = \frac{71}{4}$
$\Rightarrow Cost of 1 litre milk = Rs . \frac{71}{4}$
$7 \frac{2}{5} litres = \frac{(7 \times 5) + 2}{5} = \frac{37}{5} litres Cost of \frac{37}{5} litres milk = C ost of 1 litre milk \times \frac{37}{5} Cost of \frac{37}{5} litres milk = \frac{71}{4} \times \frac{37}{5} = Rs . \frac{2627}{20} = Rs . 131 \frac{7}{20}$

Page No 2.20:

Question 12:

Sharda can walk $8 \frac{1}{3} km$ in one hour. How much distance will she cover in $2 \frac{2}{5}$ hours?

Answer:

$8 \frac{1}{3} km = \frac{(8 \times 3) + 1}{3} \Rightarrow 8 \frac{1}{3} km = \frac{25}{3} km$

Distance covered by Sharda in 1 hour = $\frac{25}{3}$ km
$2 \frac{2}{5} hours = \frac{(2 \times 5) + 2}{5} = \frac{12}{5} hours$
Distance covered by Sharda in $\frac{12}{5} hours$ = Distance covered in 1 hour $\times \frac{12}{5}$
$\Rightarrow$ Distance covered by Sharda in $\frac{12}{5} hours$ = $\frac{25}{3} \times \frac{12}{5} = 20 km$

Page No 2.20:

Question 13:

A sugar bag contains 30 kg of sugar. After consuming $\frac{2}{3}$ of it, how much sugar is left in the bag?

Answer:

$Total amount of sugar in the bag = 30 kg \frac{2}{3} of 30 kg = \frac{2}{3} \times 30 = 20 kg \Rightarrow Amount of sugar consumed = 20 kg Sugar left in the bag = Total amount of sugar in the bag - Amount of sugar consumed Sugar left in the bag = 30 - 20 = 10 kg$

Page No 2.20:

Question 14:

Each side of a square is $6 \frac{2}{3} m$ long. Find its area.

Answer:

$6 \frac{2}{3} m = \frac{(6 \times 3) + 2}{3} m \Rightarrow 6 \frac{2}{3} m = \frac{20}{3} m Area of s quare = Side \times Side Area of s quare = \frac{20}{3} \times \frac{20}{3} \Rightarrow Area of s quare = \frac{400}{9} m^{2} = 44 \frac{4}{9} m^{2}$

Page No 2.20:

Question 15:

There are 45 students in a class and $\frac{3}{5}$ of them are boys. How many girls are there in the class?

Answer:

$\frac{3}{5} of 45 = \frac{3}{5} \times 45 = 27 Therefore, 27 students are boys . Number of girls = Total number of students in the class - Number of boys Number of girls = 45 - 27 = 18 Therefore, 18 students are girls .$

Page No 2.24:

Question 1:

Find the reciprocal of each of the following fractions and classify them as proper, improper and whole numbers:
(i) $\frac{3}{7}$
(ii) $\frac{5}{8}$
(iii) $\frac{9}{7}$
(iv) $\frac{6}{5}$
(v) $\frac{12}{7}$
(vi) $\frac{1}{8}$

Answer:

Reciprocal of a non-zero fraction $\frac{a}{b} is \frac{b}{a}$
(i)
$Reciprocal of \frac{3}{7} is \frac{7}{3} It' s improper fraction because n umerator is greater than denominator$

(ii)
$Reciprocal of \frac{5}{8} is \frac{8}{5} It' s improper fraction because n umerator is greater than denominator$

(iii)
$Reciprocal of \frac{9}{7} is \frac{7}{9} It' s proper fraction because n umerator is less than denominator$

(iv)
$Reciprocal of \frac{6}{5} is \frac{5}{6} It' s proper fraction because n umerator is less than denominator$

(v)
$Reciprocal of \frac{12}{7} is \frac{7}{12} It' s proper fraction because n umerator is less than denominator$

(vi)
$Reciprocal of \frac{1}{8} is \frac{8}{1} = 8 It i s a w hole number$

Page No 2.25:

Question 2:

Divide:
(i) $\frac{3}{8} by \frac{5}{9}$
(ii) $3 \frac{1}{4} by \frac{2}{3}$
(iii) $\frac{7}{8} by 4 \frac{1}{2}$
(iv) $6 \frac{1}{4} by 2 \frac{3}{5}$

Answer:

(i)
$\frac{3}{8} \div \frac{5}{9} = \frac{3}{8} \times \frac{9}{5} \Rightarrow \frac{3 \times 9}{8 \times 5} \Leftrightarrow \frac{27}{40}$

(ii)
$3 \frac{1}{4} \div \frac{2}{3} = \frac{(3 \times 4) + 1}{4} \times \frac{3}{2} \Rightarrow \frac{13}{4} \times \frac{3}{2} \Leftrightarrow \frac{39}{8} = 4 \frac{7}{8}$

(iii)
$\frac{7}{8} \div 4 \frac{1}{2} = \frac{7}{8} \div \frac{(4 \times 2) + 1}{2} \Rightarrow \frac{7}{8^{4}} \times \frac{2}{9} \Rightarrow \frac{7}{4} \times \frac{1}{9} \Leftrightarrow \frac{7}{36}$
(iv)
$6 \frac{1}{4} \div 2 \frac{3}{5} = \frac{(6 \times 4) + 1}{4} \div \frac{(2 \times 5) + 3}{5} \Rightarrow \frac{25}{4} \div \frac{13}{5} = \frac{25}{4} \times \frac{5}{13} \Rightarrow \frac{125}{52} = 2 \frac{21}{52}$

Page No 2.25:

Question 3:

Divide:
(i) $\frac{3}{8} by 4$
(ii) $\frac{9}{16} by 6$
(iii) $9 by \frac{3}{16}$
(iv) $10 by \frac{100}{3}$

Answer:

(i)
$\frac{3}{8} \div 4 = \frac{3}{8} \times \frac{1}{4} = \frac{3}{32}$

(ii)
$\frac{9}{16} \div 6 = \frac{9}{16} \times \frac{1}{6} = \frac{9^{3}}{16 \times 6^{2}} = \frac{3}{32}$

(iii)
$9 \div \frac{3}{16} = \frac{9}{1} \times \frac{16}{3} \Leftrightarrow \frac{9^{3} \times 16}{3} = 48$

(iv)
$10 \div \frac{100}{3} = \frac{10}{1} \times \frac{3}{100} \Leftrightarrow \frac{10 \times 3}{100^{10}} = \frac{3}{10}$

Page No 2.25:

Question 4:

Simplify:
(i) $\frac{3}{10} \div \frac{10}{3}$
(ii) $4 \frac{3}{5} \div \frac{4}{5}$
(iii) $5 \frac{4}{7} \div 1 \frac{3}{10}$
(iv) $4 \div 2 \frac{2}{5}$

Answer:

(i)
$\frac{3}{10} \div \frac{10}{3} = \frac{3}{10} \times \frac{3}{10} \Leftrightarrow \frac{3 \times 3}{10 \times 10} \Rightarrow \frac{9}{100}$
(ii)
$4 \frac{3}{5} \div \frac{4}{5} = \frac{(4 \times 5) + 3}{5} \div \frac{4}{5} \Rightarrow 4 \frac{3}{5} \div \frac{4}{5} = \frac{23}{5} \div \frac{4}{5} \Leftrightarrow \frac{23}{5} \times \frac{5}{4} \Rightarrow \frac{23}{4} = 5 \frac{3}{4}$

(iii)
$5 \frac{4}{7} \div 1 \frac{3}{10} = \frac{(5 \times 7) + 4}{7} \div \frac{(1 \times 10) + 3}{10} \Rightarrow 5 \frac{4}{7} \div 1 \frac{3}{10} = \frac{39}{7} \div \frac{13}{10} \Leftrightarrow \frac{39^{3}}{7} \times \frac{10}{13} \Rightarrow \frac{30}{7} = 4 \frac{2}{7}$

(iv)
$4 \div 2 \frac{2}{5} = \frac{4}{1} \div \frac{(2 \times 5) + 2}{5} \Leftrightarrow \frac{4}{1} \times \frac{5}{12^{3}} \Rightarrow \frac{5}{3} = 1 \frac{2}{3}$

Page No 2.25:

Question 5:

A wire of length $12 \frac{1}{2} m$ is cut into 10 pieces of equal length. Find the length of each piece.

Answer:

$12 \frac{1}{2} m = \frac{(12 \times 2) + 1}{2} m = \frac{25}{2} m$

Length of one piece = $\frac{Length of wire}{10} = \frac{25}{2} \times \frac{1}{10} = \frac{25}{20}$
Length of one piece = $\frac{25^{5}}{20^{4}} = \frac{5}{4} m$

Page No 2.25:

Question 6:

The length of a rectangular plot of area $65 \frac{1}{3} m^{2} is 12 \frac{1}{4} m$ . What is the width of the plot?

Answer:

Area of rectangle = Length of rectangle $\times$ Width of rectangle
$65 \frac{1}{3} = 12 \frac{1}{4} \times Width of the rectangle \frac{(65 \times 3) + 1}{3} = [\frac{(12 \times 4) + 1}{4}] \times Width of the rectangle \frac{196}{3} = \frac{49}{4} \times Width of the rectangle Width of the rectangle = \frac{196^{4}}{3} \times \frac{4}{49} = \frac{16}{3} m$

Page No 2.25:

Question 7:

By what number should $6 \frac{2}{9}$ be multiplied to get $4 \frac{4}{9}$ ?

Answer:

Let the required number be x.

According to the question:

$6 \frac{2}{9} \times x = 4 \frac{4}{9} \frac{(6 \times 9) + 2}{9} \times x = \frac{(4 \times 9) + 4}{9} \frac{56}{9} \times x = \frac{40}{9} x = \frac{40}{9} \times \frac{9}{56} x = \frac{40}{56} = \frac{5}{7}$

Page No 2.25:

Question 8:

The product of two numbers is $25 \frac{5}{6}$ . If one of the numbers is $6 \frac{2}{3}$ , find the other.

Answer:

Let the required number be x.

According to the question:

$6 \frac{2}{3} \times x = 25 \frac{5}{6} \frac{(6 \times 3) + 2}{3} \times x = \frac{(25 \times 6) + 5}{6} \frac{20}{3} \times x = \frac{155}{6} x = \frac{3}{20} \times \frac{155}{6} = \frac{3 \times 155^{31}}{20^{4} \times 6^{2}} = \frac{31}{8} = 3 \frac{7}{8}$

Page No 2.25:

Question 9:

The cost of $6 \frac{1}{4} kg$ of apples is Rs 400. At what rate per kg are the apples being sold?

Answer:

$6 \frac{1}{4} kg = \frac{(6 \times 4) + 1}{4} kg \Rightarrow \frac{25}{4} kg$
Cost of $\frac{25}{4} kg$ of apples = Rs. 400
Cost of 1 kg of apples = $400 \div \frac{25}{4} = 400 \times \frac{25}{4} = Rs . 64$

Page No 2.25:

Question 10:

By selling oranges at the rate of Rs $5 \frac{1}{4}$ per orange, a fruit-seller gets Rs 630. How many dozens of oranges does he sell?

Answer:

Cost of 1 orange = $Rs 5 \frac{1}{4} = \frac{(5 \times 4) + 1}{4} = Rs \frac{21}{4}$
Number of oranges sold = $630 \div \frac{21}{4}$
$\Rightarrow 630^{30} \times \frac{4}{21} = 120$
$∵$ 12 oranges = 1 dozen

$∴$ 120 oranges = $\frac{120}{12} = 10 dozen$

Page No 2.25:

Question 11:

In mid-day meal scheme $\frac{3}{10}$ litre of milk is given to each student of a primary school. If 30 litres of milk is distributed every day in the school, how many students are there in the school?

Answer:

Number of students in the school = $\frac{Total amount of milk distributed per day}{Amount of milk given to one student}$
$= 30 \div \frac{3}{10} = 30^{10} \times \frac{10}{3} = 100$

Page No 2.25:

Question 12:

In a charity show Rs 6496 were collected by selling some tickets. If the price of each ticket was Rs $50 \frac{3}{4}$ , how many tikets were sold?

Answer:

Number of tickets sold = $\frac{Total amount of money collected}{Price of one ticket}$
Price of one ticket:

$50 \frac{3}{4} \Rightarrow \frac{(50 \times 4) + 3}{4} \Rightarrow Rs \frac{203}{4}$

Number of tickets sold = $6496 \div \frac{203}{4}$
$\Rightarrow 6496^{32} \times \frac{4}{203} \Rightarrow 128$

Page No 2.26:

Question 1:

Mark the correct alternative in each of the following:

If a fraction $\frac{a}{b}$ is a lowest terms, then HCF of a and b is

(a) a (b) b (c) 1 (d) ab

Answer:

We know that a fraction is in its lowest terms if its numerator and denominator have no common factor other than 1.

Thus, if the fraction $\frac{a}{b}$ is in its lowest terms, then the HCF of a and b is 1.

Hence, the correct answer is option (c).

Page No 2.26:

Question 2:

Mark the correct alternative in each of the following:

The fraction $\frac{84}{98}$ in its lowest terms is

(a) $\frac{42}{49}$ (b) $\frac{12}{14}$ (c) $\frac{6}{7}$ (d) $\frac{3}{7}$

Answer:

Factors of 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84

Factors of 98: 1, 2, 7, 14, 49, 98

Common factors of 84 and 98: 1, 2, 14

∴ HCF of 84 and 98 = 14

Now,

$\frac{84}{98} = \frac{84 \div 14}{98 \div 14} = \frac{6}{7}$ (Dividing numerator and denominator by the HCF of 84 and 98 i.e. 14)

Hence, the correct answer is option (c).

Page No 2.26:

Question 3:

Mark the correct alternative in each of the following:

Which of the following is a vulgar fraction?

(a) $\frac{7}{10}$ (b) $\frac{13}{1000}$ (c) $2 \frac{9}{10}$ (d) $\frac{7}{9}$

Answer:

The fractions with denominator not equal to 10, 100, 1000 etc are called vulgar fractions.

Thus, the fraction $\frac{7}{9}$ is a vulgar fraction.

Hence, the correct answer is option (d).

Page No 2.26:

Question 4:

Mark the correct alternative in each of the following:

Which of the following fraction is an irreducible (or in its lowest terms)?

(a) $\frac{91}{104}$ (b) $\frac{105}{112}$ (c) $\frac{51}{85}$ (d) $\frac{43}{83}$

Answer:

We know that a fraction is irreducible (or is in its lowest terms) if the HCF of its numerator and denominator is 1.

Consider the fraction $\frac{91}{104}$ .

HCF of 91 and 104 = 13 ≠ 1

So, the fraction $\frac{91}{104}$ is reducible.

Consider the fraction $\frac{105}{112}$ .

HCF of 105 and 112 = 7 ≠ 1

So, the fraction $\frac{105}{112}$ is reducible.

Consider the fraction $\frac{51}{85}$ .

HCF of 51 and 85 = 17 ≠ 1

So, the fraction $\frac{51}{85}$ is reducible.

Now,

Consider the fraction $\frac{43}{83}$ .

HCF of 43 and 83 = 1

So, the fraction $\frac{43}{83}$ is irreducible (or is in its lowest terms).

Hence, the correct answer is option (d).

Page No 2.26:

Question 5:

Mark the correct alternative in each of the following:

Which of the following is a proper fraction?

(a) $\frac{13}{17}$ (b) $\frac{17}{13}$ (c) $\frac{12}{5}$ (d) $1 \frac{3}{4}$

Answer:

A fraction whose numerator is less than the denominator is called a proper fraction.

The numerator in each of the fractions $\frac{17}{13}$ , $\frac{12}{5}$ , $1 \frac{3}{4} = \frac{7}{4}$ is more than the denominator, so these fractions are improper fractions.

The numerator of the fraction $\frac{13}{17}$ is less than the denominator, so this fraction is a proper fraction.

Hence, the correct answer is option (a).

Page No 2.26:

Question 6:

Mark the correct alternative in each of the following:

The reciprocal of the fraction $2 \frac{3}{5}$ is

(a) $2 \frac{5}{3}$ (b) $\frac{13}{5}$ (c) $\frac{5}{13}$ (d) $2 \frac{2}{5}$

Answer:

The reciprocal of a non-zero fraction $\frac{a}{b}$ is the fraction $\frac{b}{a}$ .

$2 \frac{3}{5} = \frac{2 \times 5 + 3}{5} = \frac{13}{5}$

Now,

Reciprocal of the fraction $\frac{13}{5}$ = $\frac{5}{13}$

∴ Reciprocal of the fraction $2 \frac{3}{5}$ = $\frac{5}{13}$

Hence, the correct answer is option (c).

Page No 2.26:

Question 7:

Mark the correct alternative in each of the following:

$4 \frac{1}{3} - 2 \frac{1}{3} =$

(a) $2 \frac{1}{3}$ (b) 2 (c) $3 \frac{1}{3}$ (d) $\frac{1}{2}$

Answer:

$4 \frac{1}{3} - 2 \frac{1}{3} = \frac{13}{3} - \frac{7}{3} = \frac{13 - 7}{3} = \frac{6}{3} = 2$

Hence, the correct answer is option (b).

Page No 2.26:

Question 8:

Mark the correct alternative in each of the following:

$2 \frac{3}{5} \div \frac{5}{7} =$

(a) $\frac{13}{7}$ (b) $\frac{13}{25}$ (c) $\frac{91}{25}$ (d) $\frac{25}{91}$

Answer:

$2 \frac{3}{5} \div \frac{5}{7} = \frac{13}{5} \div \frac{5}{7} = \frac{13}{5} \times \frac{7}{5} = \frac{13 \times 7}{5 \times 5} = \frac{91}{25}$

Hence, the correct answer is option (c).

Page No 2.26:

Question 9:

Mark the correct alternative in each of the following:

By what number should $1 \frac{3}{4}$ be divided to get $2 \frac{1}{2}$ ?

(a) $\frac{3}{7}$ (b) $1 \frac{2}{5}$ (c) $\frac{7}{10}$ (d) $1 \frac{3}{7}$

Answer:

Let the required number be x.

Now,

$1 \frac{3}{4} \div x = 2 \frac{1}{2} \Rightarrow \frac{7}{4} \times \frac{1}{x} = \frac{5}{2} \Rightarrow x = \frac{7}{4} \times \frac{2}{5} \Rightarrow x = \frac{7 \times 1}{2 \times 5} \Rightarrow x = \frac{7}{10}$

Thus, the required number is $\frac{7}{10}$ .

Hence, the correct answer is option (c).

Page No 2.26:

Question 10:

Mark the correct alternative in each of the following:

By what number $4 \frac{3}{5}$ be multiplied to get $2 \frac{3}{7}$ ?

(a) $\frac{391}{35}$ (b) $\frac{85}{91}$ (c) $\frac{91}{85}$ (d) None of these

Answer:

Product of two numbers = $2 \frac{3}{7} = \frac{17}{7}$

One of the numbers = $4 \frac{3}{5} = \frac{23}{5}$

∴ Other number = Product of two numbers ÷ One of the numbers

$= \frac{17}{7} \div \frac{23}{5} = \frac{17}{7} \times \frac{5}{23} = \frac{17 \times 5}{7 \times 23} = \frac{85}{161}$

Hence, the correct answer is option (d).

Page No 2.27:

Question 11:

Mark the correct alternative in each of the following:

$(5 \frac{1}{4} - 3 \frac{1}{3}) =$

(a) $\frac{12}{23}$ (b) 2 (c) $1 \frac{11}{12}$ (d) $\frac{11}{12}$

Answer:

$5 \frac{1}{4} - 3 \frac{1}{3} = \frac{21}{4} - \frac{10}{3} = \frac{21 \times 3}{4 \times 3} - \frac{10 \times 4}{3 \times 4} (LCM of 3 and 4 is 12) = \frac{63}{12} - \frac{40}{12}$
$= \frac{63 - 40}{12} = \frac{23}{12} = 1 \frac{11}{12}$

Hence, the correct answer is option (c).

Page No 2.27:

Question 12:

Mark the correct alternative in each of the following:

The fraction equivalent to $1 \frac{2}{3}$ is

(a) $\frac{10}{3}$ (b) $\frac{3}{5}$ (c) $\frac{10}{6}$ (d) $\frac{6}{10}$

Answer:

The given fraction is $1 \frac{2}{3} = \frac{5}{3}$ .

We know that if $\frac{a}{b}$ and $\frac{c}{d}$ are two equivalent fractions, then

$a \times d = b \times c$

Now,

$5 \times 6 = 3 \times 10$

So, the fractions $\frac{5}{3}$ and $\frac{10}{6}$ are equivalent fractions.

Thus, the fraction equivalent to $1 \frac{2}{3}$ is $\frac{10}{6}$ .

Hence, the correct answer is option (c).

Page No 2.27:

Question 13:

Mark the correct alternative in each of the following:

By what number $9 \frac{4}{5}$ be multiplied to get 42?

(a) $\frac{30}{7}$ (b) $\frac{7}{30}$ (c) $4 \frac{1}{7}$ (d) $4 \frac{3}{7}$

Answer:

Product of two numbers = 42

One of the numbers = $9 \frac{4}{5} = \frac{49}{5}$

∴ Other number = Product of two numbers ÷ One of the numbers

$= 42 \div \frac{49}{5} = \frac{42}{1} \times \frac{5}{49} = \frac{6 \times 5}{1 \times 7} = \frac{30}{7}$

Hence, the correct answer is option (a).

Page No 2.27:

Question 14:

Mark the correct alternative in each of the following:

Which of the following statements is true?

(a) $\frac{7}{12} < \frac{4}{21}$ (b) $\frac{7}{12} = \frac{4}{21}$ (c) $\frac{7}{12} > \frac{4}{21}$ (d) None of these

Answer:

Consider the fractions $\frac{7}{12}$ and $\frac{4}{21}$ .

Prime factorisation of 12 = 2 × 2 × 3

Prime factorisation of 21 = 3 × 7

∴ LCM of 12 and 21 = 2 × 2 × 3 × 7 = 84

Firstly, convert the fractions to equivalent fractions with denominator 84.

$\frac{7}{12} = \frac{7 \times 7}{12 \times 7} = \frac{49}{84} \frac{4}{21} = \frac{4 \times 4}{21 \times 4} = \frac{16}{84}$

Now,

49 > 16

$∴ \frac{49}{84} > \frac{16}{84} \Rightarrow \frac{7}{12} > \frac{4}{21}$

Hence, the correct answer is option (c).

Page No 2.27:

Question 15:

Mark the correct alternative in each of the following:

Which one of the following is the correct statement?

(a) $\frac{3}{4} < \frac{2}{3} < \frac{12}{15}$ (b) $\frac{2}{3} < \frac{3}{4} < \frac{12}{15}$ (c) $\frac{2}{3} < \frac{12}{15} < \frac{3}{4}$ (d) $\frac{12}{15} < \frac{2}{3} < \frac{3}{4}$

Answer:

Consider the fractions $\frac{3}{4}$ , $\frac{2}{3}$ and $\frac{12}{15}$ .

LCM of 4, 3 and 15 = 60

Firstly, convert the fractions into equivalent fractions with denominator 60.

$\frac{3}{4} = \frac{3 \times 15}{4 \times 15} = \frac{45}{60} \frac{2}{3} = \frac{2 \times 20}{3 \times 20} = \frac{40}{60} \frac{12}{15} = \frac{12 \times 4}{15 \times 4} = \frac{48}{60}$

Now,

40 < 45 < 48

$∴ \frac{40}{60} < \frac{45}{60} < \frac{48}{60} \Rightarrow \frac{2}{3} < \frac{3}{4} < \frac{12}{15}$

Hence, the correct answer is option (b).

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Question 16:

Mark the correct alternative in each of the following:

Which of the following fractions lies between $\frac{2}{3}$ and $\frac{5}{7}$ ?

(a) $\frac{3}{4}$ (b) $\frac{4}{5}$ (c) $\frac{5}{6}$ (d) None of these

Answer:

Consider the fractions $\frac{2}{3}$ , $\frac{5}{7}$ , $\frac{3}{4}$ , $\frac{4}{5}$ and $\frac{5}{6}$ .

LCM of 3, 4, 5, 6 and 7 = 420

Firstly, convert the fractions into equivalent fractions with denominator 420.

$\frac{2}{3} = \frac{2 \times 140}{3 \times 140} = \frac{280}{420} \frac{5}{7} = \frac{5 \times 60}{7 \times 60} = \frac{300}{420} \frac{3}{4} = \frac{3 \times 105}{4 \times 105} = \frac{315}{420} \frac{4}{5} = \frac{4 \times 84}{5 \times 84} = \frac{336}{420} \frac{5}{6} = \frac{5 \times 70}{6 \times 70} = \frac{350}{420}$

Now,

280 < 300 < 315 < 336 < 350

$∴ \frac{280}{420} < \frac{300}{420} < \frac{315}{420} < \frac{336}{420} < \frac{350}{420} \Rightarrow \frac{2}{3} < \frac{5}{7} < \frac{3}{4} < \frac{4}{5} < \frac{5}{6}$

Thus, none of the fractions $\frac{3}{4}$ , $\frac{4}{5}$ , $\frac{5}{6}$ lies between the fractions $\frac{2}{3}$ and $\frac{5}{7}$ .

Hence, the correct answer is option (d).

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Question 17:

Mark the correct alternative in each of the following:

Which one of the following is true?

(a) $\frac{1}{2} < \frac{9}{13} < \frac{3}{4} < \frac{12}{17}$ (b) $\frac{3}{4} < \frac{9}{13} < \frac{1}{2} < \frac{12}{17}$

(c) $\frac{1}{2} < \frac{3}{4} < \frac{9}{13} < \frac{12}{17}$ (d) $\frac{1}{2} < \frac{9}{13} < \frac{12}{17} < \frac{3}{4}$

Answer:

Consider the fractions $\frac{1}{2}$ , $\frac{9}{13}$ , $\frac{3}{4}$ and $\frac{12}{17}$ .

LCM of 2, 4, 13 and 17 = 884

Firstly, convert the fractions into equivalent fractions with denominator 884.

$\frac{1}{2} = \frac{1 \times 442}{2 \times 442} = \frac{442}{884} \frac{9}{13} = \frac{9 \times 68}{13 \times 68} = \frac{612}{884} \frac{3}{4} = \frac{3 \times 221}{4 \times 221} = \frac{663}{884} \frac{12}{17} = \frac{12 \times 52}{17 \times 52} = \frac{624}{884}$

Now,

442 < 612 < 624 < 663

$∴ \frac{442}{884} < \frac{612}{884} < \frac{624}{884} < \frac{663}{884} \Rightarrow \frac{1}{2} < \frac{9}{13} < \frac{12}{17} < \frac{3}{4}$

Hence, the correct answer is option (d).

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Question 18:

Mark the correct alternative in each of the following:

The smallest of the fractions $\frac{2}{3}, \frac{4}{7}, \frac{8}{11}$ and $\frac{5}{9}$ is

(a) $\frac{2}{3}$ (b) $\frac{4}{7}$ (c) $\frac{8}{11}$ (d) $\frac{5}{9}$

Answer:

The given fractions are $\frac{2}{3}, \frac{4}{7}, \frac{8}{11}$ and $\frac{5}{9}$ .

LCM of 3, 7, 9 and 11 = 693

Firstly, convert the fractions into equivalent fractions with denominator 693.

$\frac{2}{3} = \frac{2 \times 231}{3 \times 231} = \frac{462}{693} \frac{4}{7} = \frac{4 \times 99}{7 \times 99} = \frac{396}{693} \frac{8}{11} = \frac{8 \times 63}{11 \times 63} = \frac{504}{693} \frac{5}{9} = \frac{5 \times 77}{9 \times 77} = \frac{385}{693}$

Now,

385 < 396 < 462 < 504

$∴ \frac{385}{693} < \frac{396}{693} < \frac{462}{693} < \frac{504}{693} \Rightarrow \frac{5}{9} < \frac{4}{7} < \frac{2}{3} < \frac{8}{11}$

Thus, the smallest of the given fractions is $\frac{5}{9}$ .

Hence, the correct answer is option (d).

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Question 19:

Mark the correct alternative in each of the following:

$9 \times (- \frac{1}{3}) \times (- 3) \times (- \frac{1}{9}) =$

(a) 1 (b) −1 (c) −3 (d) 3

Answer:

Since the number of negative terms in the product is odd. Therefore, their product is negative.

$9 \times (- \frac{1}{3}) \times (- 3) \times (- \frac{1}{9}) = 9 \times (- \frac{1}{9}) \times (- \frac{1}{3}) \times (- 3) = - (9 \times \frac{1}{9} \times \frac{1}{3} \times 3) = - (1 \times 1) = - 1$

Hence, the correct answer is option (b).

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Question 20:

Mark the correct alternative in each of the following:

Which of the following is correct?

(a) $\frac{2}{3} < \frac{3}{5} < \frac{11}{15}$ (b) $\frac{3}{5} < \frac{2}{3} < \frac{11}{15}$ (c) $\frac{11}{15} < \frac{3}{5} < \frac{2}{3}$ (d) $\frac{3}{5} < \frac{11}{15} < \frac{2}{3}$

Answer:

Consider the fractions $\frac{2}{3}$ , $\frac{3}{5}$ and $\frac{11}{15}$ .

LCM of 3, 5 and 15 = 15

Firstly, convert the fractions into equivalent fractions with denominator 15.

$\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15} \frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15}$

Now,

9 < 10 < 11

$∴ \frac{9}{15} < \frac{10}{15} < \frac{11}{15} \Rightarrow \frac{3}{5} < \frac{2}{3} < \frac{11}{15}$

Hence, the correct answer is option (b).

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Question 21:

Mark the correct alternative in each of the following:

Which is the smallest of the following fractions?

(a) $\frac{4}{9}$ (b) $\frac{2}{5}$ (c) $\frac{3}{7}$ (d) $\frac{1}{4}$

Answer:

Consider the fractions $\frac{4}{9}$ , $\frac{2}{5}$ , $\frac{3}{7}$ and $\frac{1}{4}$ .

LCM of 4, 5, 7 and 9 = 1260

Firstly, convert the fractions into equivalent fractions with denominator 1260.

$\frac{4}{9} = \frac{4 \times 140}{9 \times 140} = \frac{560}{1260} \frac{2}{5} = \frac{2 \times 252}{5 \times 252} = \frac{504}{1260} \frac{3}{7} = \frac{3 \times 180}{7 \times 180} = \frac{540}{1260} \frac{1}{4} = \frac{1 \times 315}{4 \times 315} = \frac{315}{1260}$

Now,

315 < 504 < 540 < 560

$∴ \frac{315}{1260} < \frac{504}{1260} < \frac{540}{1260} < \frac{560}{1260} \Rightarrow \frac{1}{4} < \frac{2}{5} < \frac{3}{7} < \frac{4}{9}$

Thus, the smallest fraction is $\frac{1}{4}$ .

Hence, the correct answer is option (d).

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Question 22:

Mark the correct alternative in each of the following:

The difference between the greatest and the least fractions out of $\frac{6}{7}, \frac{7}{8}, \frac{8}{9}$ and $\frac{9}{10}$ is

(a) $\frac{3}{10}$ (b) $\frac{1}{56}$ (c) $\frac{1}{40}$ (d) $\frac{1}{72}$

Answer:

Consider the fractions $\frac{6}{7}, \frac{7}{8}, \frac{8}{9}$ and $\frac{9}{10}$ .

LCM of 7, 8, 9 and 10 = 2520

Firstly, convert the fractions into equivalent fractions with denominator 2520.

$\frac{6}{7} = \frac{6 \times 360}{7 \times 360} = \frac{2160}{2520} \frac{7}{8} = \frac{7 \times 315}{8 \times 315} = \frac{2205}{2520} \frac{8}{9} = \frac{8 \times 280}{9 \times 280} = \frac{2240}{2520} \frac{9}{10} = \frac{9 \times 252}{10 \times 252} = \frac{2268}{2520}$

Now,

2160 < 2205 < 2240 < 2268

$∴ \frac{2160}{2520} < \frac{2205}{2520} < \frac{2240}{2520} < \frac{2268}{2520} \Rightarrow \frac{6}{7} < \frac{7}{8} < \frac{8}{9} < \frac{9}{10}$

So,

Greatest fraction = $\frac{9}{10}$

Least fraction = $\frac{6}{7}$

∴ Required difference

$= \frac{9}{10} - \frac{6}{7} = \frac{2268}{2520} - \frac{2160}{2520} = \frac{2268 - 2160}{2520} = \frac{108}{2520}$

$= \frac{108 \div 36}{2520 \div 36} (HCF of 108 and 2520 = 36) = \frac{3}{70}$

Disclaimer: None of the options given in the question matches with the answer.

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Question 23:

Mark the correct alternative in each of the following:

Which of the following fractions is greater than $\frac{3}{4}$ and less than $\frac{5}{6}$ ?

(a) $\frac{2}{3}$ (b) $\frac{1}{2}$ (c) $\frac{4}{5}$ (d) $\frac{9}{10}$

Answer:

Consider the fractions $\frac{3}{4}, \frac{5}{6}, \frac{2}{3}, \frac{1}{2}, \frac{4}{5}$ and $\frac{9}{10}$ .

LCM of 2, 3, 4, 5, 6 and 10 = 60

Firstly, convert the fractions into equivalent fractions with denominator 60.

$\frac{3}{4} = \frac{3 \times 15}{4 \times 15} = \frac{45}{60} \frac{5}{6} = \frac{5 \times 10}{6 \times 10} = \frac{50}{60} \frac{2}{3} = \frac{2 \times 20}{3 \times 20} = \frac{40}{60} \frac{1}{2} = \frac{1 \times 30}{2 \times 30} = \frac{30}{60} \frac{4}{5} = \frac{4 \times 12}{5 \times 12} = \frac{48}{60} \frac{9}{10} = \frac{9 \times 6}{10 \times 6} = \frac{54}{60}$

Now,

30 < 40 < 45 < 48 < 50 < 54

$∴ \frac{30}{60} < \frac{40}{60} < \frac{45}{60} < \frac{48}{60} < \frac{50}{60} < \frac{54}{60} \Rightarrow \frac{1}{2} < \frac{2}{3} < \frac{3}{4} < \frac{4}{5} < \frac{5}{6} < \frac{9}{10}$

Thus, the fraction $\frac{4}{5}$ is greater than $\frac{3}{4}$ and less than $\frac{5}{6}$ .

Hence, the correct answer is option (c).

Page No 2.28:

Question 24:

Mark the correct alternative in each of the following:

Which of the following fractions is more than one-third?

(a) $\frac{23}{70}$ (b) $\frac{205}{819}$ (c) $\frac{26}{75}$ (d) $\frac{118}{335}$

Answer:

Let $\frac{a}{b}$ and $\frac{c}{d}$ be two fractions. Then, $\frac{a}{b} > \frac{c}{d}$ if $a \times d > b \times c$ .

Consider the fractions $\frac{23}{70}$ and $\frac{1}{3}$ .
$23 \times 3 = 69 1 \times 70 = 70 ∴ 23 \times 3 < 1 \times 70 \Rightarrow \frac{23}{70} < \frac{1}{3}$

Consider the fractions $\frac{205}{819}$ and $\frac{1}{3}$ .
$205 \times 3 = 615 1 \times 819 = 819 ∴ 205 \times 3 < 1 \times 819 \Rightarrow \frac{205}{819} < \frac{1}{3}$

Consider the fractions $\frac{26}{75}$ and $\frac{1}{3}$ .
$26 \times 3 = 78 1 \times 75 = 75 ∴ 26 \times 3 > 1 \times 75 \Rightarrow \frac{26}{75} > \frac{1}{3}$

Consider the fractions $\frac{118}{335}$ and $\frac{1}{3}$ .
$118 \times 3 = 354 1 \times 335 = 335 ∴ 118 \times 3 > 1 \times 335 \Rightarrow \frac{118}{335} > \frac{1}{3}$

Thus, the fractions $\frac{26}{75}$ and $\frac{118}{335}$ are more than the fraction $\frac{1}{3}$ .

Hence, the correct answers are options (c) and (d).

Disclaimer: There are two correct options in the question. One of the two options among (c) and (d) must be changed accordingly to get only one correct answer.

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