RS Aggarwal 2019,2020 solution class 7 chapter 2 Fractions Test Paper 2

Test Paper 2

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Question 1:

Define:

(i) Fractions
(ii) Vulgar fractions
(iii) Improper fractions

Give two examples of each.

Answer 1:

(i) A number of the form $\frac{a}{b}$, where a and b are natural numbers, is called a natural number.
Here, a is the numerator and b is the denominator.

$\frac{2}{3}$ is a fraction with 2 as the numerator and 3 as the denominator.

$\frac{12}{5}$ is a fraction with 12 as the numerator and 5 as the denominator.

(ii) A fraction whose denominator is a whole number other than 10, 100, 1000, etc., is called a vulgar faction.
Examples: $\frac{2}{5}$ and $\frac{4}{15}$

(iii) A fraction whose numerator is greater than or equal to its denominator is called an improper fraction.
Examples: $\frac{11}{3}$ and $\frac{41}{35}$

Question 2:

What should be added to $6\frac{3}{5}$ to get 15?

Answer 2:

Required number to be added = $15-6\frac{3}{5}$

                                             = $\frac{15}{1}-\frac{33}{5}$

                                              = $\frac{75-33}{5}$  [∵ LCM of 1 and 5 = 5]

                                              = $\frac{42}{5}=8\frac{2}{5}$

Hence, the required number is $8\frac{2}{5}$.

Question 3:

Simplify: $9\frac{5}{6}-4\frac{3}{8}+2\frac{7}{12}.$

Answer 3:

We have,

$9\frac{5}{6}-4\frac{3}{8}+2\frac{7}{12}$

= $\frac{59}{6}-\frac{35}{8}+\frac{31}{12}$

= $\frac{236-105+62}{24}$   [∵ LCM of 6, 8 and 12 = 24]

= $\frac{298-105}{24}$= $\frac{193}{24}=8\frac{1}{24}$

Question 4:

Find:

(i) $\frac{12}{25}$ of a litre
(ii) $\frac{5}{8}$ of a kilogram
(iii) $\frac{3}{5}$ of an hour

Answer 4:

We have:

(i) $\frac{12}{25}$ of 1 L = $\frac{12}{25}$ of 1000 ml = $\left(1000\times \frac{12}{25}\right)$ ml = (40 $\times$ 12) ml = 480 ml

(ii)$\frac{5}{8}$ of 1 kg = $\frac{5}{8}$ of 1000 g = $\left(1000\times \frac{5}{8}\right)$ g = (125 $\times$5) g = 625 g

(iii) $\frac{3}{5}$ of 1 h = $\frac{3}{5}$ of 60 min = $\left(60\times \frac{3}{5}\right)$ min = (12 $\times$ 3) min = 36 min

Question 5:

Milk is sold at Rs $37\frac{3}{4}$ per litre. Find the cost of $6\frac{2}{5}$ litres milk.

Answer 5:

Cost of 1 L of milk = Rs $37\frac{3}{4}$ =  Rs $\frac{151}{4}$
Cost of $6\frac{2}{5}$ L of milk = Rs $\left(\frac{151}{4}\times 6\frac{2}{5}\right)$
                                 = Rs $\left(\frac{151}{4}\times \frac{32}{5}\right)$
                                 = Rs $\left(\frac{151\times 8}{1\times 5}\right)$ =  Rs $\frac{1208}{5}$ = Rs $241\frac{3}{5}$
Hence, the cost of $6\frac{2}{5}$ L of milk is Rs $241\frac{3}{5}$.

Question 6:

The cost of $5\frac{1}{4}$ kg of mangoes is Rs 189. At what rate per kg are the mangoes being sold?

Answer 6:

Cost of  $5\frac{1}{4}$ kg of mangoes = Rs 189
Cost of 1 kg of mango = Rs $\left(189÷5\frac{1}{4}\right)$
                                   = Rs $\left(189÷\frac{21}{4}\right)$ 
                                   = Rs $\left(189\times \frac{4}{21}\right)$      [∵ Reciprocal of $\frac{21}{4}$ = $\frac{4}{21}$]
                                   = Rs (9 $\times$ 4) = Rs 36

Hence, the mangoes are being sold at Rs 36 per kg.

Question 7:

Simplify:

(i) $1\frac{3}{4}\times 2\frac{2}{5}\times 3\frac{4}{7}$
(ii) $5\frac{5}{9}÷3\frac{1}{3}$

Answer 7:

We have:

(i)$1\frac{3}{4}\times 2\frac{2}{5}\times 3\frac{4}{7}$

     = $\frac{7}{4}\times \frac{12}{5}\times \frac{25}{7}$

     = $\frac{7\times 12\times 25}{4\times 5\times 7}=\frac{1\times 3\times 5}{1\times 1\times 1}=15$


(ii) $5\frac{5}{9}÷3\frac{1}{3}$

     = $\frac{50}{9}÷\frac{10}{3}$

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

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

Question 8:

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

Answer 8:

Required number = $6\frac{2}{9}÷4\frac{2}{3}$

                            = $\frac{56}{9}÷\frac{14}{3}$

                            = $\frac{56}{9}\times \frac{3}{14}$    [∵ Reciprocal of $\frac{14}{3}$ = $\frac{3}{14}$]

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

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

Question 9:

Each side of a square is $5\frac{2}{3}$ m long. Find its area.

Answer 9:

Side of the square = $5\frac{2}{3}$ m = $\frac{17}{3}$ m

Its area = (side)² = ${\left(\frac{17}{3}\mathrm{m}\right)}^2$ = $\left(\frac{17}{3}\mathrm{m}\times \frac{17}{3}\mathrm{m}\right)=\left(\frac{289}{9}\right){\mathrm{m}}^2=32\frac{1}{9} {\mathrm{m}}^2$

Hence, the area of the square is $32\frac{1}{9}{\mathrm{m}}^2$.

Question 10:

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

(a) $\frac{7}{10}$
(b) $\frac{19}{100}$
(c) $3\frac{3}{10}$
(d) $\frac{5}{8}$

Answer 10:

(d) $\frac{5}{8}$

$\frac{5}{8}$ is a vulgar fraction, because its denominator is other than 10, 100, 1000, etc.

Question 11:

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

(a) $\frac{105}{112}$
(b) $\frac{66}{77}$
(c) $\frac{46}{63}$
(d) $\frac{51}{85}$

Answer 11:

(c) $\frac{46}{63}$

A fraction $\frac{a}{b}$ is said to be irreducible or in its lowest terms if the HCF of a and b is 1.
46 = 2 $\times$ 23 $\times$1
63 = 3 $\times$ 3$\times$ 21 $\times$1

Clearly, the HCF of 46 and 63 is 1.

Hence, $\frac{46}{63}$ is an irreducible fraction.

Question 12:

Mark (✓) against the correct answer
Reciprocal of $1\frac{3}{5}$ is
(a) $1\frac{5}{3}$
(b) $5\frac{1}{3}$
(c) $3\frac{1}{5}$
(d) none of these

Answer 12:

(d) none of these

Reciprocal of $1\frac{3}{5}$ = Reciprocal of $\frac{8}{5}$ = $\frac{5}{8}$

Question 13:

Mark (✓) against the correct answer
$1\frac{3}{5}÷\frac{2}{3}=?$

(a) $1\frac{9}{10}$
(b) $1\frac{1}{15}$
(c) $2\frac{2}{5}$
(d) none of these

Answer 13:

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

$1\frac{3}{5}÷\frac{2}{3}$= $\frac{8}{5}÷\frac{2}{3}$

             = $\frac{8}{5}\times \frac{3}{2}$        [∵ Reciprocal of $\frac{2}{3}$ = $\frac{3}{2}$]

             = $\left(\frac{4\times 3}{5}\right)=\frac{12}{5}=2\frac{2}{5}$

Question 14:

Mark (✓) against the correct answer
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 14:

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

The given fractions are $\frac{2}{3}, \frac{3}{5}$ and $\frac{11}{15}$.

LCM of 5, 3 and 15 = 15

Now, we have:

$\frac{2}{3}\times \frac{5}{5}=\frac{10}{15}$, $\frac{3}{5}\times \frac{3}{3}=\frac{9}{15}$ and $\frac{11}{15}\times \frac{1}{1}=\frac{11}{15}$

Clearly, $\frac{9}{15}<\frac{10}{15}<\frac{11}{15}$

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

Question 15:

Mark (✓) against the correct answer
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 15:

(c) $\frac{7}{10}$

Required number = $1\frac{3}{4}÷2\frac{1}{2}$

                             = $\frac{7}{4}÷\frac{5}{2}$

                             = $\frac{7}{4}\times \frac{2}{5}$   [∵ Reciprocal of $\frac{5}{2}$ = $\frac{2}{5}$]

                             = $\frac{7\times 1}{2\times 5}=\frac{7}{10}$

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Question 16:

Mark (✓) against the correct answer
A car runs 9 km using 1 litre of petrol. How much distance will it cover in $3\frac{2}{3}$ litres o petrol?
(a) 36 km
(b) 33 km
(c) $2\frac{5}{11}$ km
(d) 22 km

Answer 16:

(b) 33 km
Distance covered by the car on $3\frac{2}{3}$ L of petrol = $\left(9\times 3\frac{2}{3}\right)$ km
                                                                      = $\left(9\times \frac{11}{3}\right)$ km
                                                                       = (3 $\times$ 11) km = 33 km

Question 17:

Fill in the blanks.

(i) Reciprocal of $8\frac{2}{5}$ is ...... .
(ii) $13\frac{1}{2}÷8=......$
(iii) $69\frac{3}{4}÷7\frac{3}{4}=......$
(iv) $41\frac{2}{3}\times 18\frac{3}{5}=......$
(v) $\frac{84}{98}$(in irreducible form)= ......

Answer 17:

(i) The reciprocal of $8\frac{2}{5}$ is $\frac{5}{42}$.

Reciprocal of $8\frac{2}{5}$ = Reciprocal of $\frac{42}{5}$ = $\frac{5}{42}$

(ii) $13\frac{1}{2}÷8=1\frac{11}{16}$

     $13\frac{1}{2}÷8=\frac{27}{2}\times \frac{1}{8}=\frac{27}{16}=1\frac{11}{16}$

(iii) $69\frac{3}{4}÷7\frac{3}{4}=9$

$69\frac{3}{4}÷7\frac{3}{4}=\frac{279}{4}÷\frac{31}{4}$

 =$\frac{279}{4}\times \frac{4}{31}=\frac{279}{31}$ = 9

(iv) $41\frac{2}{3}÷18\frac{3}{5}=775$

        $41\frac{2}{3}\times 18\frac{3}{5}=\frac{125}{3}\times \frac{93}{5}$
      
       = $\frac{125}{3}\times \frac{93}{5}=\frac{25\times 31}{1\times 1}=775$

(v) $\frac{84}{98}$(irreducible form) = $\frac{6}{7}$
  
The HCF of 84 and 98 is 14.
      
∴ $\frac{84÷14}{98÷14}=\frac{6}{7}$

Question 18:

Write 'T' for true and 'F' for false

(i) $\frac{9}{16}<\frac{13}{24}.$
(ii) Among $\frac{2}{5},\frac{16}{35}$ and $\frac{9}{14}$, the largest is $\frac{16}{35}.$
(iii) $\frac{11}{15}-\frac{9}{20}=\frac{17}{60}.$
(iv) $\frac{11}{25}$ of a litre = 440 mL.
(v) $16\frac{3}{4}\times 6\frac{2}{5}=107\frac{3}{10}.$

Answer 18:

(i) F

     By cross multiplication, we have:

     9 $\times$ 24 = 216 and 13 $\times$ 16 = 208

     However, 216 > 208

     ∴ $\frac{9}{16}>\frac{13}{24}$

(ii) F

      The LCM of 5, 35 and 14 is 70.

      Now, $\frac{2}{5}=\frac{2}{5}\times \frac{14}{14}=\frac{28}{70}; \frac{16}{35}=\frac{16}{35}\times \frac{2}{2}=\frac{32}{70} \mathrm{and} \frac{9}{14}=\frac{9}{14}\times \frac{5}{5}=\frac{45}{70}$

      Clearly, $\frac{28}{70}<\frac{32}{70}<\frac{45}{70}$

      ∴ $\frac{2}{5}<\frac{16}{35}<\frac{9}{14}$

(iii) T

       The LCM of 15 and 20 = (5 $\times$ 3 $\times$ 4) = 60

        ∴ $\frac{11}{15}-\frac{9}{20}=\frac{44-27}{60}=\frac{17}{60}$
(iv) T

       $\frac{11}{25}$ of 1 L = $\frac{11}{25}$ of 1000 ml = $\left(1000\times \frac{11}{25}\right)$ ml = (40 $\times$ 11) ml = 440 ml

(v) F

      $16\frac{3}{4}\times 6\frac{2}{5}$= $\left(\frac{67}{4}\times \frac{32}{5}\right)=\left(\frac{67\times 32}{4\times 5}\right)=\left(\frac{67\times 8}{5}\right)=\frac{536}{5}=107\frac{1}{5}$