Rd Sharma 2018 Solutions for Class 7 Math Chapter 2 Fractions are provided here with simple step-by-step explanations. These solutions for Fractions are extremely popular among Class 7 students for Math Fractions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2018 Book of Class 7 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rd Sharma 2018 Solutions. All Rd Sharma 2018 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

Question 1:

Compare the following fractions by using the symbol > or < or =:
(i)
(ii)
(iii)
(iv)

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.

(ii)

(iii)

(iv)

LCM of 15 and 105 is 105.

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) $\frac{4}{3},\frac{3}{8},\frac{6}{12},\frac{5}{16}$

(i)

(ii)

Question 3:

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

(i)

(ii)

Question 4:

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

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

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) $2\frac{3}{5}+4\frac{7}{10}+2\frac{4}{15}$

(i)

(ii)

(iii)

(iv)

Question 6:

Find the difference of
(i)
(ii)
(iii)
(iv)

(i)

(ii)

(iii)

(iv)

Question 7:

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

(i)

(ii)

(iii)

(iv)

Question 8:

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

(i)

(ii)

(iii)

Question 9:

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

Let x be the required fraction.

According to the question:

$x+5\frac{3}{7}=12\phantom{\rule{0ex}{0ex}}⇒x+\frac{\left(5×7\right)+3}{7}=12\phantom{\rule{0ex}{0ex}}⇒x=12-\frac{38}{7}\phantom{\rule{0ex}{0ex}}⇒x=\frac{\left(12×7\right)-\left(38×1\right)}{7}⇔\frac{46}{7}$

Question 10:

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

Let x be the required fraction.

According to the question:

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?

Suman studies for  $5\frac{2}{3}\mathrm{hours}\phantom{\rule{0ex}{0ex}}$ daily. Therefore, we have

She studies science and mathematics for  $2\frac{4}{5}\mathrm{hours}$. Therefore, we have
$2\frac{4}{5}\mathrm{hours}=\frac{\left(2×5\right)+4}{5}=\frac{14}{5}\mathrm{hours}$
Time devoted to other subjects = Total study time $-$ Time devoted to science and mathematics

$\frac{17}{3}-\frac{14}{5}=\frac{\left(17×5\right)-\left(14×3\right)}{15}\phantom{\rule{0ex}{0ex}}=\frac{43}{15}\mathrm{hours}$

Question 12:

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

Let the length of second piece be x.

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

$12\frac{3}{4}=5\frac{1}{4}+x\phantom{\rule{0ex}{0ex}}⇒\frac{\left(12×4\right)+3}{4}=\frac{\left(5×4\right)+1}{4}+x\phantom{\rule{0ex}{0ex}}⇒x=\frac{\left(12×4\right)+3}{4}-\frac{\left(5×4\right)+1}{4}\phantom{\rule{0ex}{0ex}}⇒x=\frac{51}{4}-\frac{21}{4}⇔\frac{30}{4}\phantom{\rule{0ex}{0ex}}⇒x=\frac{30}{4}⇔\frac{15}{2}$

Question 13:

A rectangular sheet of paper is long and wide. Find its perimeter.

Perimeter of rectangle = 2(length + width)

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?

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

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?

Question 16:

(i) Provide the number in box and also give its simplest from in each of the following:
(i)
(ii) .

(i) $\frac{2}{3}×x=\frac{10}{30}\phantom{\rule{0ex}{0ex}}x=\frac{10}{30}×\frac{3}{2}=\frac{1}{2}$

(ii) $\frac{3}{5}×x=\frac{24}{75}\phantom{\rule{0ex}{0ex}}x=\frac{24}{75}×\frac{5}{3}=\frac{8}{15}$

Question 1:

Multiply:
(i)
(ii)
(iii)
(iv)

(i)
$\frac{7}{11}×\frac{3}{5}=\frac{7×3}{11×5}\phantom{\rule{0ex}{0ex}}⇒\frac{21}{55}$

(ii)
$\frac{3}{5}×25=\frac{3×25}{5×1}\phantom{\rule{0ex}{0ex}}⇒\frac{75}{5}⇔15$

(iii)

(iv)

Question 2:

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

(i)
$\frac{4}{7}×\frac{14}{25}=\frac{4×14}{7×25}\phantom{\rule{0ex}{0ex}}⇒\frac{8}{25}$

(ii)
$7\frac{1}{2}×2\frac{4}{15}⇔\frac{\left(7×2\right)+1}{2}×\frac{\left(2×15\right)+4}{15}\phantom{\rule{0ex}{0ex}}⇒\frac{15}{2}×\frac{34}{15}⇔\frac{17}{1}$
(iii)
$3\frac{6}{7}×4\frac{2}{3}⇔\frac{\left(3×7\right)+6}{7}×\frac{\left(4×3\right)+2}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{{\overline{)27}}^{9}}{\overline{)7}}×\frac{{\overline{)14}}^{2}}{\overline{)3}}⇔\frac{18}{1}$
(iv)
$6\frac{11}{14}×3\frac{1}{2}⇔\frac{\left(14×6\right)+11}{14}×\frac{\left(3×2\right)+1}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{95}{\overline{){14}^{2}}}×\frac{\overline{)7}}{2}⇔\frac{95}{4}$

Question 3:

Simplify:
(i)
(ii)
(iii)

(i)
$\frac{12}{25}×\frac{15}{28}×\frac{35}{36}⇔\frac{\overline{)12}×{\overline{)15}}^{3}×{\overline{)35}}^{5}}{{\overline{)25}}^{5}×{\overline{)28}}^{4}×{\overline{)36}}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{\overline{)3}×\overline{)5}}{\overline{)5}×4×\overline{)3}}⇔\frac{1}{4}$

(ii)
$\frac{10}{27}×\frac{39}{56}×\frac{28}{65}⇔\frac{\overline{)10}×{\overline{)39}}^{{\overline{)3}}^{1}}×\overline{)28}}{{\overline{)27}}^{9}×{\overline{)56}}^{\overline{)2}}×{\overline{)65}}^{\overline{)5}}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{9}$
(iii)
$2\frac{2}{17}×7\frac{2}{9}×1\frac{33}{52}⇔\frac{\left(2×17\right)+2}{17}×\frac{\left(7×9\right)+2}{9}×\frac{\left(52×1\right)+33}{52}\phantom{\rule{0ex}{0ex}}⇔\frac{\overline{){36}^{\overline{)4}}}×{\overline{)65}}^{5}×{\overline{)85}}^{5}}{\overline{)17}×\overline{)9}×{\overline{)52}}^{\overline{)4}}}\phantom{\rule{0ex}{0ex}}⇒\frac{25}{1}$

Find:
(i)
(ii)
(iii)

Question 5:

Which is greater?

$\frac{2}{7}=\frac{2}{7}×\frac{2}{2}=\frac{4}{14}$

$⇒\frac{6}{14}>\frac{2}{7}$

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

Question 7:

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

Distance between the first and second saplings = $\frac{3}{4}\mathrm{m}$
Distance between the first and third saplings = $2×\frac{3}{4}\mathrm{m}=\frac{3}{2}\mathrm{m}\phantom{\rule{0ex}{0ex}}$
Distance between the first and fourth saplings $3×\frac{3}{4}\mathrm{m}=\frac{9}{4}\mathrm{m}$

Distance between the first and fifth saplings  = 4 ×

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?

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?

In one day, Lipika reads for $1\frac{3}{4}\mathrm{hours}\phantom{\rule{0ex}{0ex}}$.
$1\frac{3}{4}\mathrm{hours}=\frac{\left(1×4\right)+3}{4}=\frac{7}{4}\mathrm{hours}$
Total hours required =

Question 10:

Find the area of a rectangular park which is long and broad.

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.

Cost of 1 litre milk =
$17\frac{3}{4}=\frac{\left(17×4\right)+3}{4}=\frac{71}{4}\phantom{\rule{0ex}{0ex}}$

Question 12:

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

$8\frac{1}{3}\mathrm{km}=\frac{\left(8×3\right)+1}{3}\phantom{\rule{0ex}{0ex}}⇒8\frac{1}{3}\mathrm{km}=\frac{25}{3}\mathrm{km}$

Distance covered by â€‹Sharda in 1 hour =  $\frac{25}{3}$km

Distance covered by Sharda in $\frac{12}{5}\mathrm{hours}$ = Distance covered in 1 hour $×\frac{12}{5}$
$⇒$Distance covered by Sharda in $\frac{12}{5}\mathrm{hours}$ =

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?

Question 14:

Each side of a square is long. Find its area.

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?

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) $\frac{1}{8}$

Reciprocal of a non-zero fraction
(i)

(ii)

(iii)

(iv)

(v)

(vi)

Question 2:

Divide:
(i)
(ii)
(iii)
(iv)

(i)
$\frac{3}{8}÷\frac{5}{9}=\frac{3}{8}×\frac{9}{5}\phantom{\rule{0ex}{0ex}}⇒\frac{3×9}{8×5}⇔\frac{27}{40}$

(ii)
$3\frac{1}{4}÷\frac{2}{3}=\frac{\left(3×4\right)+1}{4}×\frac{3}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{13}{4}×\frac{3}{2}⇔\frac{39}{8}=4\frac{7}{8}$

(iii)
$\frac{7}{8}÷4\frac{1}{2}=\frac{7}{8}÷\frac{\left(4×2\right)+1}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{7}{\overline{){8}^{4}}}×\frac{\overline{)2}}{9}\phantom{\rule{0ex}{0ex}}⇒\frac{7}{4}×\frac{1}{9}⇔\frac{7}{36}$
(iv)
$6\frac{1}{4}÷2\frac{3}{5}=\frac{\left(6×4\right)+1}{4}÷\frac{\left(2×5\right)+3}{5}\phantom{\rule{0ex}{0ex}}⇒\frac{25}{4}÷\frac{13}{5}=\frac{25}{4}×\frac{5}{13}\phantom{\rule{0ex}{0ex}}⇒\frac{125}{52}=2\frac{21}{52}$

Question 3:

Divide:
(i)
(ii)
(iii)
(iv)

(i)
$\frac{3}{8}÷4=\frac{3}{8}×\frac{1}{4}\phantom{\rule{0ex}{0ex}}=\frac{3}{32}$

(ii)
$\frac{9}{16}÷6=\frac{9}{16}×\frac{1}{6}=\frac{\overline{){9}^{3}}}{16×\overline{){6}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{3}{32}$

(iii)
$9÷\frac{3}{16}=\frac{9}{1}×\frac{16}{3}⇔\frac{\overline{){9}^{3}}×16}{\overline{)3}}\phantom{\rule{0ex}{0ex}}=48$

(iv)
$10÷\frac{100}{3}=\frac{10}{1}×\frac{3}{100}⇔\frac{\overline{)10}×3}{\overline{){100}^{10}}}\phantom{\rule{0ex}{0ex}}=\frac{3}{10}$

Question 4:

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

(i)
$\frac{3}{10}÷\frac{10}{3}=\frac{3}{10}×\frac{3}{10}⇔\frac{3×3}{10×10}\phantom{\rule{0ex}{0ex}}⇒\frac{9}{100}\phantom{\rule{0ex}{0ex}}$
(ii)
$4\frac{3}{5}÷\frac{4}{5}=\frac{\left(4×5\right)+3}{5}÷\frac{4}{5}\phantom{\rule{0ex}{0ex}}⇒4\frac{3}{5}÷\frac{4}{5}=\frac{23}{5}÷\frac{4}{5}⇔\frac{23}{5}×\frac{5}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{23}{4}=5\frac{3}{4}$

(iii)
$5\frac{4}{7}÷1\frac{3}{10}=\frac{\left(5×7\right)+4}{7}÷\frac{\left(1×10\right)+3}{10}\phantom{\rule{0ex}{0ex}}⇒5\frac{4}{7}÷1\frac{3}{10}=\frac{39}{7}÷\frac{13}{10}⇔\frac{\overline{){39}^{3}}}{7}×\frac{10}{\overline{)13}}\phantom{\rule{0ex}{0ex}}⇒\frac{30}{7}=4\frac{2}{7}$

(iv)
$4÷2\frac{2}{5}=\frac{4}{1}÷\frac{\left(2×5\right)+2}{5}⇔\frac{\overline{)4}}{1}×\frac{5}{\overline{){12}^{3}}}\phantom{\rule{0ex}{0ex}}⇒\frac{5}{3}=1\frac{2}{3}$

Question 5:

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

$12\frac{1}{2}\mathrm{m}=\frac{\left(12×2\right)+1}{2}\mathrm{m}\phantom{\rule{0ex}{0ex}}=\frac{25}{2}\mathrm{m}$

Length of one piece =
Length of one piece = $\frac{\overline{){25}^{5}}}{{\overline{)20}}^{4}}=\frac{5}{4}\mathrm{m}$

Question 6:

The length of a rectangular plot of area . What is the width of the plot?

Area of rectangle = Length of rectangle $×$ Width of rectangle

Question 7:

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

Let the required number be x.

According to the question:

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.

Let the required number be x.

According to the question:

$6\frac{2}{3}×x=25\frac{5}{6}\phantom{\rule{0ex}{0ex}}\frac{\left(6×3\right)+2}{3}×x=\frac{\left(25×6\right)+5}{6}\phantom{\rule{0ex}{0ex}}\frac{20}{3}×x=\frac{155}{6}\phantom{\rule{0ex}{0ex}}x=\frac{3}{20}×\frac{155}{6}=\frac{\overline{)3}×{\overline{)155}}^{31}}{\overline{){20}^{4}}×\overline{){6}^{2}}}=\frac{31}{8}\phantom{\rule{0ex}{0ex}}=3\frac{7}{8}$

Question 9:

The cost of of apples is Rs 400. At what rate per kg are the apples being sold?

$6\frac{1}{4}\mathrm{kg}=\frac{\left(6×4\right)+1}{4}\mathrm{kg}\phantom{\rule{0ex}{0ex}}⇒\frac{25}{4}\mathrm{kg}$
Cost of  $\frac{25}{4}\mathrm{kg}$ of apples = Rs. 400
Cost of 1 kg of apples =

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?

Cost of 1 orange =
Number of oranges sold = $630÷\frac{21}{4}$

$\because$ 12 oranges = 1 dozen

$\therefore$ 120 oranges =

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?

Number of students in the school =

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?

Number of tickets sold =
Price of one ticket:

Number of tickets sold = $6496÷\frac{203}{4}$

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

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).

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) $\frac{3}{7}$

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,

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

Hence, the correct answer is option (c).

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) $\frac{7}{9}$

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).

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) $\frac{43}{83}$

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).

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) $1\frac{3}{4}$

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).

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) $2\frac{2}{5}$

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

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

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).

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) $\frac{1}{2}$

$4\frac{1}{3}-2\frac{1}{3}\phantom{\rule{0ex}{0ex}}=\frac{13}{3}-\frac{7}{3}\phantom{\rule{0ex}{0ex}}=\frac{13-7}{3}\phantom{\rule{0ex}{0ex}}=\frac{6}{3}\phantom{\rule{0ex}{0ex}}=2$

Hence, the correct answer is option (b).

Question 8:

Mark the correct alternative in each of the following:

$2\frac{3}{5}÷\frac{5}{7}=$

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

$2\frac{3}{5}÷\frac{5}{7}\phantom{\rule{0ex}{0ex}}=\frac{13}{5}÷\frac{5}{7}\phantom{\rule{0ex}{0ex}}=\frac{13}{5}×\frac{7}{5}\phantom{\rule{0ex}{0ex}}=\frac{13×7}{5×5}\phantom{\rule{0ex}{0ex}}=\frac{91}{25}$

Hence, the correct answer is option (c).

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) $1\frac{3}{7}$

Let the required number be x.

Now,

$1\frac{3}{4}÷x=2\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{7}{4}×\frac{1}{x}=\frac{5}{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{7}{4}×\frac{2}{5}\phantom{\rule{0ex}{0ex}}⇒x=\frac{7×1}{2×5}\phantom{\rule{0ex}{0ex}}⇒x=\frac{7}{10}$

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

Hence, the correct answer is option (c).

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

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

$=\frac{17}{7}÷\frac{23}{5}\phantom{\rule{0ex}{0ex}}=\frac{17}{7}×\frac{5}{23}\phantom{\rule{0ex}{0ex}}=\frac{17×5}{7×23}\phantom{\rule{0ex}{0ex}}=\frac{85}{161}$

Hence, the correct answer is option (d).

Question 11:

Mark the correct alternative in each of the following:

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

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

$=\frac{63-40}{12}\phantom{\rule{0ex}{0ex}}=\frac{23}{12}\phantom{\rule{0ex}{0ex}}=1\frac{11}{12}$

Hence, the correct answer is option (c).

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) $\frac{6}{10}$

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

$a×d=b×c$

Now,

$5×6=3×10$

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).

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) $4\frac{3}{7}$

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

$=42÷\frac{49}{5}\phantom{\rule{0ex}{0ex}}=\frac{42}{1}×\frac{5}{49}\phantom{\rule{0ex}{0ex}}=\frac{6×5}{1×7}\phantom{\rule{0ex}{0ex}}=\frac{30}{7}$

Hence, the correct answer is option (a).

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

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.

$\frac{7}{12}=\frac{7×7}{12×7}=\frac{49}{84}\phantom{\rule{0ex}{0ex}}\frac{4}{21}=\frac{4×4}{21×4}=\frac{16}{84}$

Now,

49 > 16

$\therefore \frac{49}{84}>\frac{16}{84}\phantom{\rule{0ex}{0ex}}⇒\frac{7}{12}>\frac{4}{21}$

Hence, the correct answer is option (c).

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) $\frac{12}{15}<\frac{2}{3}<\frac{3}{4}$

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.

$\frac{3}{4}=\frac{3×15}{4×15}=\frac{45}{60}\phantom{\rule{0ex}{0ex}}\frac{2}{3}=\frac{2×20}{3×20}=\frac{40}{60}\phantom{\rule{0ex}{0ex}}\frac{12}{15}=\frac{12×4}{15×4}=\frac{48}{60}$

Now,

40 < 45 <  48

$\therefore \frac{40}{60}<\frac{45}{60}<\frac{48}{60}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{3}<\frac{3}{4}<\frac{12}{15}$

Hence, the correct answer is option (b).

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

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.

$\frac{2}{3}=\frac{2×140}{3×140}=\frac{280}{420}\phantom{\rule{0ex}{0ex}}\frac{5}{7}=\frac{5×60}{7×60}=\frac{300}{420}\phantom{\rule{0ex}{0ex}}\frac{3}{4}=\frac{3×105}{4×105}=\frac{315}{420}\phantom{\rule{0ex}{0ex}}\frac{4}{5}=\frac{4×84}{5×84}=\frac{336}{420}\phantom{\rule{0ex}{0ex}}\frac{5}{6}=\frac{5×70}{6×70}=\frac{350}{420}$

Now,

280 < 300 < 315 < 336 < 350

$\therefore \frac{280}{420}<\frac{300}{420}<\frac{315}{420}<\frac{336}{420}<\frac{350}{420}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{3}<\frac{5}{7}<\frac{3}{4}<\frac{4}{5}<\frac{5}{6}$

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).

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

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.

$\frac{1}{2}=\frac{1×442}{2×442}=\frac{442}{884}\phantom{\rule{0ex}{0ex}}\frac{9}{13}=\frac{9×68}{13×68}=\frac{612}{884}\phantom{\rule{0ex}{0ex}}\frac{3}{4}=\frac{3×221}{4×221}=\frac{663}{884}\phantom{\rule{0ex}{0ex}}\frac{12}{17}=\frac{12×52}{17×52}=\frac{624}{884}$

Now,

442 < 612 < 624 < 663

$\therefore \frac{442}{884}<\frac{612}{884}<\frac{624}{884}<\frac{663}{884}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}<\frac{9}{13}<\frac{12}{17}<\frac{3}{4}$

Hence, the correct answer is option (d).

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) $\frac{5}{9}$

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.

$\frac{2}{3}=\frac{2×231}{3×231}=\frac{462}{693}\phantom{\rule{0ex}{0ex}}\frac{4}{7}=\frac{4×99}{7×99}=\frac{396}{693}\phantom{\rule{0ex}{0ex}}\frac{8}{11}=\frac{8×63}{11×63}=\frac{504}{693}\phantom{\rule{0ex}{0ex}}\frac{5}{9}=\frac{5×77}{9×77}=\frac{385}{693}$

Now,

385 < 396 < 462 < 504

$\therefore \frac{385}{693}<\frac{396}{693}<\frac{462}{693}<\frac{504}{693}\phantom{\rule{0ex}{0ex}}⇒\frac{5}{9}<\frac{4}{7}<\frac{2}{3}<\frac{8}{11}$

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

Hence, the correct answer is option (d).

Question 19:

Mark the correct alternative in each of the following:

$9×\left(-\frac{1}{3}\right)×\left(-3\right)×\left(-\frac{1}{9}\right)=$

(a) 1                                      (b) −1                                       (c) −3                                      (d) 3

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

$9×\left(-\frac{1}{3}\right)×\left(-3\right)×\left(-\frac{1}{9}\right)\phantom{\rule{0ex}{0ex}}=9×\left(-\frac{1}{9}\right)×\left(-\frac{1}{3}\right)×\left(-3\right)\phantom{\rule{0ex}{0ex}}=-\left(9×\frac{1}{9}×\frac{1}{3}×3\right)\phantom{\rule{0ex}{0ex}}=-\left(1×1\right)\phantom{\rule{0ex}{0ex}}=-1$

Hence, the correct answer is option (b).

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) $\frac{3}{5}<\frac{11}{15}<\frac{2}{3}$

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.

$\frac{2}{3}=\frac{2×5}{3×5}=\frac{10}{15}\phantom{\rule{0ex}{0ex}}\frac{3}{5}=\frac{3×3}{5×3}=\frac{9}{15}$

Now,

9 < 10 < 11

$\therefore \frac{9}{15}<\frac{10}{15}<\frac{11}{15}\phantom{\rule{0ex}{0ex}}⇒\frac{3}{5}<\frac{2}{3}<\frac{11}{15}$

Hence, the correct answer is option (b).

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) $\frac{1}{4}$

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.

$\frac{4}{9}=\frac{4×140}{9×140}=\frac{560}{1260}\phantom{\rule{0ex}{0ex}}\frac{2}{5}=\frac{2×252}{5×252}=\frac{504}{1260}\phantom{\rule{0ex}{0ex}}\frac{3}{7}=\frac{3×180}{7×180}=\frac{540}{1260}\phantom{\rule{0ex}{0ex}}\frac{1}{4}=\frac{1×315}{4×315}=\frac{315}{1260}$

Now,

315 < 504 < 540 < 560

$\therefore \frac{315}{1260}<\frac{504}{1260}<\frac{540}{1260}<\frac{560}{1260}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{4}<\frac{2}{5}<\frac{3}{7}<\frac{4}{9}$

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

Hence, the correct answer is option (d).

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) $\frac{1}{72}$

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.

$\frac{6}{7}=\frac{6×360}{7×360}=\frac{2160}{2520}\phantom{\rule{0ex}{0ex}}\frac{7}{8}=\frac{7×315}{8×315}=\frac{2205}{2520}\phantom{\rule{0ex}{0ex}}\frac{8}{9}=\frac{8×280}{9×280}=\frac{2240}{2520}\phantom{\rule{0ex}{0ex}}\frac{9}{10}=\frac{9×252}{10×252}=\frac{2268}{2520}$

Now,

2160 < 2205 < 2240 < 2268

$\therefore \frac{2160}{2520}<\frac{2205}{2520}<\frac{2240}{2520}<\frac{2268}{2520}\phantom{\rule{0ex}{0ex}}⇒\frac{6}{7}<\frac{7}{8}<\frac{8}{9}<\frac{9}{10}$

So,

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

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

∴ Required difference

$=\frac{9}{10}-\frac{6}{7}\phantom{\rule{0ex}{0ex}}=\frac{2268}{2520}-\frac{2160}{2520}\phantom{\rule{0ex}{0ex}}=\frac{2268-2160}{2520}\phantom{\rule{0ex}{0ex}}=\frac{108}{2520}$

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

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) $\frac{9}{10}$

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.

$\frac{3}{4}=\frac{3×15}{4×15}=\frac{45}{60}\phantom{\rule{0ex}{0ex}}\frac{5}{6}=\frac{5×10}{6×10}=\frac{50}{60}\phantom{\rule{0ex}{0ex}}\frac{2}{3}=\frac{2×20}{3×20}=\frac{40}{60}\phantom{\rule{0ex}{0ex}}\frac{1}{2}=\frac{1×30}{2×30}=\frac{30}{60}\phantom{\rule{0ex}{0ex}}\frac{4}{5}=\frac{4×12}{5×12}=\frac{48}{60}\phantom{\rule{0ex}{0ex}}\frac{9}{10}=\frac{9×6}{10×6}=\frac{54}{60}$

Now,

30 < 40 < 45 < 48 < 50 < 54

$\therefore \frac{30}{60}<\frac{40}{60}<\frac{45}{60}<\frac{48}{60}<\frac{50}{60}<\frac{54}{60}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<\frac{4}{5}<\frac{5}{6}<\frac{9}{10}$

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).

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) $\frac{118}{335}$

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

Consider the fractions $\frac{23}{70}$ and $\frac{1}{3}$.
$23×3=69\phantom{\rule{0ex}{0ex}}1×70=70\phantom{\rule{0ex}{0ex}}\therefore 23×3<1×70\phantom{\rule{0ex}{0ex}}⇒\frac{23}{70}<\frac{1}{3}$

Consider the fractions $\frac{205}{819}$ and $\frac{1}{3}$.
$205×3=615\phantom{\rule{0ex}{0ex}}1×819=819\phantom{\rule{0ex}{0ex}}\therefore 205×3<1×819\phantom{\rule{0ex}{0ex}}⇒\frac{205}{819}<\frac{1}{3}$

Consider the fractions $\frac{26}{75}$ and $\frac{1}{3}$.
$26×3=78\phantom{\rule{0ex}{0ex}}1×75=75\phantom{\rule{0ex}{0ex}}\therefore 26×3>1×75\phantom{\rule{0ex}{0ex}}⇒\frac{26}{75}>\frac{1}{3}$

Consider the fractions $\frac{118}{335}$ and $\frac{1}{3}$.
$118×3=354\phantom{\rule{0ex}{0ex}}1×335=335\phantom{\rule{0ex}{0ex}}\therefore 118×3>1×335\phantom{\rule{0ex}{0ex}}⇒\frac{118}{335}>\frac{1}{3}$

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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