Question 8:

Aneeta bought $3 \frac{3}{4}$ kg apples and $4 \frac{1}{2}$ kg guava. What is the total weight of fruits purchased by her?

Answer:

Total weight of fruits bought by Aneeta = $(3 \frac{3}{4} + 4 \frac{1}{2}) kg$
Now, we have:

$3 \frac{3}{4} + 4 \frac{1}{2} = \frac{15}{4} + \frac{9}{2}$

$= \frac{15 + 18}{4}$ [∵ LCM of 2 and 4 = 4]

$= \frac{15 + 18}{4} = \frac{33}{4} = 8 \frac{1}{4}$

Hence, the total weight of the fruits purchased by Aneeta is $8 \frac{1}{4} kg$ .

Page No 22:

Question 9:

A rectangular sheet of paper is $15 \frac{3}{4}$ cm long and $12 \frac{1}{2}$ cm wide. Find its perimeter.

Answer:

We have:

Perimeter of the rectangle ABCD = AB + BC + CD +DA
= $(15 \frac{3}{4} + 12 \frac{1}{2} + 15 \frac{3}{4} + 12 \frac{1}{2}) cm$
= $(\frac{63}{4} + \frac{25}{2} + \frac{63}{4} + \frac{25}{2}) cm$
= $(\frac{63 + 50 + 63 + 50}{4}) cm$ [∵ LCM of 2 and 4 = 4]
= $(\frac{226}{4}) cm = (\frac{113}{2}) cm = 56 \frac{1}{2} cm$

Hence, the perimeter of ABCD is $56 \frac{1}{2} cm$ .

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

(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) $13 \frac{1}{2} \div 8 = 1 \frac{11}{16}$

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

(iii) $69 \frac{3}{4} \div 7 \frac{3}{4} = 9$

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

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

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

        $41 \frac{2}{3} \times 18 \frac{3}{5} = \frac{125}{3} \times \frac{93}{5}$

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

(v) $\frac{84}{98}$ (irreducible form) = $\frac{6}{7}$

The HCF of 84 and 98 is 14.

∴ $\frac{84 \div 14}{98 \div 14} = \frac{6}{7}$

Page No 35:

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:

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

      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 = $(1000 \times \frac{11}{25})$ ml = (40 $\times$ 11) ml = 440 ml

(v) F

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

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