RS Aggarwal 2019,2020 solution class 7 chapter 2 Fractions Exercise 2C

Exercise 2C

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Question 1:

Write down the reciprocal of:

(i) $\frac{5}{8}$
(ii) 7
(iii) $\frac{1}{12}$
(iv) $12\frac{3}{5}$

Answer 1:

(i) Reciprocal of $\frac{5}{8}$ = $\frac{8}{5}$            [ ∵ $\frac{5}{8}\times \frac{8}{5}=1$]

(ii) Reciprocal of  7 =$\frac{1}{7}$             [ ∵ $7\times \frac{1}{7}=1$]

(iii) Reciprocal of  $\frac{1}{12}$ = 12       [ ∵ $\frac{1}{12}\times 12=1$]

(iv) Reciprocal of $12\frac{3}{5}$ = Reciprocal of $\frac{63}{5}$ =$\frac{5}{63}$            [∵ $\frac{63}{5}\times \frac{5}{63}=1$]

Question 2:

Simplify:

(i)  $\frac{4}{7}÷\frac{9}{14}$
(ii) $\frac{7}{10}÷\frac{3}{5}$
(iii) $\frac{8}{9}÷16$
(iv) $9÷\frac{1}{3}$
(v) $24÷\frac{6}{7}$
(vi) $3\frac{3}{5}÷\frac{4}{5}$
(vii) $3\frac{3}{7}÷\frac{8}{21}$
(viii) $5\frac{4}{7}÷1\frac{3}{10}$
(ix) $15\frac{3}{7}÷1\frac{23}{49}$

Answer 2:

(i) $\frac{4}{7}÷\frac{9}{14}=\frac{4}{7}\times \frac{14}{9}$              [∵ Reciprocal of $\frac{9}{14}$ = $\frac{14}{9}$]

    = $\frac{8}{9}$

(ii) $\frac{7}{10}÷\frac{3}{5}=\frac{7}{10}\times \frac{5}{3}$            [∵ Reciprocal of $\frac{3}{5}$ = $\frac{5}{3}$]

     = $\frac{7}{6}=1\frac{1}{6}$

(iii) $\frac{8}{9}÷16=\frac{8}{9}\times \frac{1}{16}$              [∵ Reciprocal of 16 = $\frac{1}{16}$]

       = $\frac{1}{18}$

(iv) $9÷\frac{1}{3}=9\times 3$                      [∵ Reciprocal of $\frac{1}{3}$ = 3]

     = 27


(v) $24÷\frac{6}{7}=24\times \frac{7}{6}$                [∵ Reciprocal of $\frac{6}{7}$ = $\frac{7}{6}$]

    = 4 $\times$ 7 = 28


(vi) $3\frac{3}{5}÷\frac{4}{5}=\frac{18}{5}÷\frac{4}{5}$

     = $\frac{18}{5}\times \frac{5}{4}$            [∵ Reciprocal of $\frac{4}{5}$ = $\frac{5}{4}$]

      = $\frac{18}{4}=\frac{9}{2}=4\frac{1}{2}$

(vii) $3\frac{3}{7}÷\frac{8}{21}=\frac{24}{7}÷\frac{8}{21}$

       = $\frac{24}{7}\times \frac{21}{8}$          [∵ Reciprocal of $\frac{8}{21}$ = $\frac{21}{8}$]

       = 3  3 = 9


(viii) $5\frac{4}{7}÷1\frac{3}{10}$ =$\frac{39}{7}÷\frac{13}{10}$

       = $\frac{39}{7}\times \frac{10}{13}$             [∵ Reciprocal of $\frac{13}{10}$ = $\frac{10}{13}$]

       = $\frac{30}{7}=4\frac{2}{7}$


(ix) $15\frac{3}{7}÷1\frac{23}{49}$ = $\frac{108}{7}÷\frac{72}{49}$

      = $\frac{108}{7}\times \frac{49}{72}$          [∵ Reciprocal of $\frac{72}{49}$ = $\frac{49}{72}$]

      = $\frac{9\times 7}{1\times 6}=\frac{3\times 7}{1\times 2}=\frac{21}{2}=10\frac{1}{2}$

Question 3:

Divide:

(i) $\frac{11}{24} \mathrm{by} \frac{7}{8}$
(ii) $6\frac{7}{8} \mathrm{by} \frac{11}{16}$
(iii) $5\frac{5}{9} \mathrm{by} 3\frac{1}{3}$
(iv) $32 \mathrm{by} 1\frac{3}{5}$
(v) $45 \mathrm{by} 1\frac{4}{5}$
(vi) $63 \mathrm{by} 2\frac{1}{4}$

Answer 3:

(i)  $\frac{11}{24}÷\frac{7}{8}$

      = $\frac{11}{24}\times \frac{8}{7}$                           [∵ Reciprocal of $\frac{7}{8}$ = $\frac{8}{7}$]

      = $\frac{11}{21}$

(ii) $6\frac{7}{8}÷\frac{11}{16}$ = $\frac{55}{8}÷\frac{11}{16}$

       =$\frac{55}{8}\times \frac{16}{11}$         [∵ Reciprocal of $\frac{11}{16}$ = $\frac{16}{11}$]

       = 5 $\times$ 2 = 10

(iii) $5\frac{5}{9}÷3\frac{1}{3}$ = $\frac{50}{9}÷\frac{10}{3}$

        = $\frac{50}{9}\times \frac{3}{10}$          [∵ Reciprocal of $\frac{10}{3}$ = $\frac{3}{10}$]

        =  $\frac{5}{3}$ = $1\frac{2}{3}$

(iv) $32÷1\frac{3}{5}$ = $32÷\frac{8}{5}$

      = $32\times \frac{5}{8}$                [∵ Reciprocal of $\frac{8}{5}$ = $\frac{5}{8}$]

      = 4 $\times$ 5 = 20

(v) $45÷1\frac{4}{5}$ = $45÷\frac{9}{5}$

      = $45\times \frac{5}{9}$               [∵ Reciprocal of $\frac{9}{5}$ = $\frac{5}{9}$]

      = 5 $\times$ 5 = 25

(vi) $63÷2\frac{1}{4}$ = $63÷\frac{9}{4}$

      = $63\times \frac{4}{9}$            [∵ Reciprocal of $\frac{9}{4}$ = $\frac{4}{9}$]

      = 7 $\times$ 4 = 28

Question 4:

A rope of length $13\frac{1}{2}$ m has been divided into 9 pieces of the same length. What is the length of each piece?

Answer 4:

Length of the rope = $13\frac{1}{2}$ m =$\frac{27}{2}$ m
Number of equal pieces = 9

∴ Length of each piece = $\left(\frac{27}{2}÷9\right)$ m
                                      = $\left(\frac{27}{2}\times \frac{1}{9}\right)$ m      [∵ Reciprocal of 9 = $\frac{1}{9}$]
                                      = $\frac{3}{2}$ m =$1\frac{1}{2}$ m
Hence, the length of each piece of rope is $1\frac{1}{2}$ m.

Question 5:

18 boxes of nails weigh equally and their total weight is $49\frac{1}{2}$ kg. How much does each box weigh?

Answer 5:

Weight of 18 boxes of nails = $49\frac{1}{2}$ kg = $\frac{99}{2}$ kg
∴ Weight of 1 box = $\left(\frac{99}{2}÷18\right)$ kg
                            = $\left(\frac{99}{2}\times \frac{1}{18}\right)$ kg         [∵ Reciprocal of 18 = $\frac{1}{18}$]
                            = $\left(\frac{99\times 1}{2\times 18}\right)$ kg = $\left(\frac{11\times 1}{2\times 2}\right)$ kg =$\frac{11}{4}$ kg = $2\frac{3}{4}$ kg

Hence, the weight of each box is $2\frac{3}{4}$ kg.

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Question 6:

By selling oranges at the rate of Rs $6\frac{3}{4}$ per orange, a man gets Rs 378. How many oranges does he sell?

Answer 6:

Selling price of an orange = Rs $6\frac{3}{4}$ = Rs$\frac{27}{4}$
Total money received after selling oranges = Rs 378
Total no. of oranges = $\frac{378}{27}\times 4=56$
Hence, total no. of oranges = 56

Question 7:

Mangoes are sold at Rs $43\frac{1}{2}$ per kg. What is the weight of mangoes available for Rs $326\frac{1}{4}?$

Answer 7:

Selling price of 1kg mango = ₹ $43\frac{1}{2}$ = ₹ $\frac{87}{2}$

Weight of mangoes at ₹ $\frac{1305}{4}=\frac{1305}{4}\times \frac{2}{87}=\frac{435}{58}=\frac{15}{2}=7\frac{1}{2}$
Hence, the weight of mangoes= $7\frac{1}{2}$ kg

Question 8:

Vikas can cover a distance of $20\frac{2}{3}$ km in $7\frac{3}{4}$ hours on foot. How many km per hour does he walk?

Answer 8:

Distance covered by Vikas in $7\frac{3}{4}$ h = $20\frac{2}{3}$ km
∴ Distance covered by him in 1 h = $\left(20\frac{2}{3}÷7\frac{3}{4}\right)$ km
                                                   = $\left(\frac{62}{3}÷\frac{31}{4}\right)$ km
                                                   = $\left(\frac{62}{3}\times \frac{4}{31}\right)$ km
                                                    = $\left(\frac{2\times 4}{3}\right)$ km =$\left(\frac{8}{3}\right)$ km = $2\frac{2}{3}$ km

Hence, the distance covered by Vikas in 1 h is $2\frac{2}{3}$ km.

Question 9:

Preeti bought $8\frac{1}{2}$ kg of sugar for Rs $242\frac{1}{4}$. Find the price of sugar per kg.

Answer 9:

Cost of $\frac{17}{2}$ kg of sugar = ₹ $\frac{969}{4}$
Cost of 1 kg of sugar = $\frac{{\displaystyle \frac{969}{4}}}{{\displaystyle \frac{17}{2}}}=\frac{969}{4}\times \frac{2}{17}=\frac{57}{2}=28\frac{1}{2}$

Hence, the cost of sugar is Rs $28\frac{1}{2}$ per kg

Question 10:

If the cost of a notebook is Rs $27\frac{3}{4}$, how many notebooks can be purchased for Rs $249\frac{3}{4}?$

Answer 10:

Cost of 1 notebook = ₹ $27\frac{3}{4}$ = ₹ $\frac{111}{4}$
Number of notebooks purchased for ₹ $249\frac{3}{4}$ = $=\frac{\frac{999}{4}}{{\displaystyle \frac{111}{4}}}=\frac{999}{4}\times \frac{4}{111}=9$

Hence, the number of notebooks purchased are 9.

Question 11:

At a charity show the price of each ticket was Rs $32\frac{1}{2}.$ The total amount collected by a boy was Rs $877\frac{1}{2}.$ How many tickets were sold by him?

Answer 11:

Total amount collected = $₹877\frac{1}{2}=₹\frac{1755}{2}$
Price of 1 ticket = ₹ $32\frac{1}{2}$ = ₹ $\frac{65}{2}$
Number of tickets sold = $\frac{{\displaystyle \frac{1755}{2}}}{{\displaystyle \frac{65}{2}}}=\frac{1755}{2}\times \frac{2}{65}=27$
Hence, the number of tickets sold were 27.

Question 12:

A group of students arranged a picnic. Each student contributed Rs $261\frac{1}{2}$. The total contribution was Rs $2876\frac{1}{2}$. How many students are there in the group?

Answer 12:

Total contribution = ₹ $2876\frac{1}{2}=\frac{5753}{2}$      
Contribution of each student = ₹ $261\frac{1}{2}$ = ₹ $\frac{523}{2}$
Number of students = $\frac{5753}{2}÷\frac{523}{2}$ = $\frac{5753}{2}\times \frac{2}{523}=11$
Hence, number of students in the group are 11.

Question 13:

24 litres of milk was distributed equally among all the students of a hostel. If each student got $\frac{2}{5}$ litre of milk, how many students are there in the hostel?

Answer 13:

Quantity of milk given to each student  = $\frac{2}{5}$ L
Total quantity of milk distributed among all the students = 24 L

∴ Number of students = $\left(24÷\frac{2}{5}\right)$

                                    = $\left(24\times \frac{5}{2}\right)$       [∵ Reciprocal of $\frac{2}{5}$ = $\frac{5}{2}$]

                                    = (12 $\times$ 5) = 60

Hence, there are 60 students in the hostel.

Question 14:

A bucket contains $20\frac{1}{4}$ litres of water. A small jug has a capacity of $\frac{3}{4}$ litre. How many times the jug has to be filled with water from the bucket to get it emptied?

Answer 14:

Capacity of the small jug = $\frac{3}{4}$ L
Capacity of the bucket = $20\frac{1}{4}$ L = $\frac{81}{4}$ L
∴ Required number of small jugs = $\left(\frac{81}{4} ÷ \frac{3}{4}\right)$
                                                   = $\left(\frac{81}{4}\times \frac{4}{3}\right)$      [∵ Reciprocal of $\frac{3}{4}$ = $\frac{4}{3}$]
                                                   = $\left(\frac{81}{3}\right)$ = 27

Hence, the small jug has to be filled 27 times to empty the water from the bucket.

Question 15:

The product of two numbers is $15\frac{5}{6}$. If one of the numbers is $6\frac{1}{3}$, find the other.

Answer 15:

Product of the two numbers = $15\frac{5}{6}$ =$\frac{95}{6}$

One of the numbers = $6\frac{1}{3}$ =$\frac{19}{3}$

∴ The other number = $\left(\frac{95}{6} ÷ \frac{19}{3}\right)$

                                 = $\left(\frac{95}{6} \times \frac{3}{19}\right)$     [∵ Reciprocal of $\frac{19}{3}$ = $\frac{3}{19}$]

                               = $\left(\frac{5}{2}\right) = 2\frac{1}{2}$

Hence, the other number is $2\frac{1}{2}$.

Question 16:

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

Answer 16:

Product of the two numbers = 42
One of the numbers = $9\frac{4}{5}$ = $\frac{49}{5}$
∴ The other number = $\left(42 ÷ \frac{49}{5}\right)$
                                 =$\left(42 \times \frac{5}{49}\right)$           [∵ Reciprocal of $\frac{49}{5}$ = $\frac{5}{49}$]
                                 =$\left(\frac{6 \times 5}{7}\right) = \frac{30}{7} = 4\frac{2}{7}$

Hence, the required number is $4\frac{2}{7}$.

Question 17:

By what number should $6\frac{2}{9}$ be divided to obtain $4\frac{2}{3}$?

Answer 17:

Required number = $\left(6\frac{2}{9} ÷ 4\frac{2}{3}\right)$
                            = $\left(\frac{56}{9} ÷ \frac{14}{3}\right)$
                            = $\left(\frac{56}{9} \times \frac{3}{14}\right)$     [ ∵ Reciprocal of $\frac{14}{3}$ = $\frac{3}{14}$]
                            = $\left(\frac{4}{3}\right)=1\frac{1}{3}$

Hence, we have to divide $6\frac{2}{9}$ by $1\frac{1}{3}$ to get $4\frac{2}{3}$.

Question 1:

Mark (✓) against the correct answer
Which of the following is a vulgar fraction?

(a) $\frac{3}{10}$
(b) $\frac{13}{10}$
(c) $\frac{10}{3}$
(d) none of these

Answer 1:

(c) $\frac{10}{3}$

$\frac{10}{3}$ is a vulgar fraction, because its denominator is other than 10, 100, 1000, etc.

Question 2:

Mark (✓) against the correct answer
Which of the following is an improper fraction?

(a) $\frac{7}{10}$
(b) $\frac{7}{9}$
(c) $\frac{9}{7}$
(d) none of these

Answer 2:

(c) $\frac{9}{7}$
$\frac{9}{7}$ is an improper fraction, because its numerator is greater than its denominator.

Question 3:

Mark (✓) against the correct answer
Which of the following is a reducible fraction?

(a) $\frac{105}{112}$
(b) $\frac{104}{121}$
(c) $\frac{77}{72}$
(d) $\frac{46}{63}$

Answer 3:

(a) $\frac{105}{112}$

A fraction that is reducible can be reduced by dividing both the numerator and denominator by a common factor.

$\frac{105÷7}{112÷7}=\frac{15}{16}$

Thus, $\frac{105}{112}$ is a reducible fraction.