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

Exercise 2B

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Question 1:

Find the product:

(i) $\frac{3}{5}\times \frac{7}{11}$
(ii) $\frac{5}{8}\times \frac{4}{7}$
(iii) $\frac{4}{9}\times \frac{15}{16}$
(iv) $\frac{2}{5}\times 15$
(v) $\frac{8}{15}\times 20$
(vi) $\frac{5}{8}\times 1000$
(vii) $3\frac{1}{8}\times 16$
(viii) $2\frac{4}{15}\times 12$
(ix) $3\frac{6}{7}\times 4\frac{2}{3}$
(x) $9\frac{1}{2}\times 1\frac{9}{19}$
(xi) $4\frac{1}{8}\times 2\frac{10}{11}$
(xii) $5\frac{5}{6}\times 1\frac{5}{7}$

Answer 1:

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

(ii) $\frac{5}{8}\times \frac{4}{7}=\frac{5\times 4}{8\times 7}=\frac{5\times 1}{2\times 7}=\frac{5}{14}$

(iii) $\frac{4}{9}\times \frac{15}{16}=\frac{4\times 15}{9\times 16}=\frac{1\times 5}{3\times 4}=\frac{5}{12}$

(iv) $\frac{2}{5}\times 15=\frac{2}{5}\times \frac{15}{1}=\frac{2\times 15}{5\times 1}=\frac{2\times 3}{1\times 1}=6$

(v) $\frac{8}{15}\times 20=\frac{8}{15}\times \frac{20}{1}=\frac{8\times 20}{15\times 1}=\frac{8\times 4}{3\times 1}=\frac{32}{3}=10\frac{2}{3}$

(vi) $\frac{5}{8}\times 1000=\frac{5}{8}\times \frac{1000}{1}=\frac{5\times 1000}{8\times 1}=\frac{5\times 125}{1\times 1}=625$

(vii) $3\frac{1}{8}\times 16=\frac{25}{8}\times \frac{16}{1}=\frac{25\times 16}{8\times 1}=\frac{25\times 2}{1\times 1}=50$

(viii) $2\frac{4}{15}\times 12=\frac{34}{15}\times \frac{12}{1}=\frac{34\times 12}{15\times 1}=\frac{34\times 4}{5\times 1}=\frac{136}{5}=27\frac{1}{5}$

(ix) $3\frac{6}{7}\times 4\frac{2}{3}=\frac{27}{7}\times \frac{14}{3}=\frac{27\times 14}{7\times 3}=\frac{9\times 2}{1\times 1}=18$

(x) $9\frac{1}{2}\times 1\frac{9}{19}=\frac{19}{2}\times \frac{28}{19}=\frac{19\times 28}{2\times 19}=\frac{1\times 14}{1\times 1}=14$

(xi) $4\frac{1}{8}\times 2\frac{10}{11}=\frac{33}{8}\times \frac{32}{11}=\frac{33\times 32}{8\times 11}=\frac{3\times 4}{1\times 1}=12$

(xii) $5\frac{5}{6}\times 1\frac{5}{7}=\frac{35}{6}\times \frac{12}{7}=\frac{35\times 12}{6\times 7}=\frac{5\times 2}{1\times 1}=10$

Question 2:

Simplify:

(i) $\frac{2}{3}\times \frac{5}{44}\times \frac{33}{35}$
(ii) $\frac{12}{25}\times \frac{15}{28}\times \frac{35}{36}$
(iii) $\frac{10}{27}\times \frac{28}{65}\times \frac{39}{56}$
(iv) $1\frac{4}{7}\times 1\frac{13}{22}\times 1\frac{1}{15}$
(v) $2\frac{2}{17}\times 7\frac{2}{9}\times 1\frac{33}{52}$
(vi) $3\frac{1}{16}\times 7\frac{3}{7}\times 1\frac{25}{39}$

Answer 2:

We have the following:

(i) $\frac{2}{3}\times \frac{5}{44}\times \frac{33}{35}=\frac{2\times 5\times 33}{3\times 44\times 35}=\frac{1\times 1\times 11}{1\times 22\times 7}=\frac{1\times 1\times 1}{1\times 2\times 7}=\frac{1}{14}$

(ii)$\frac{12}{25}\times \frac{15}{28}\times \frac{35}{36}=\frac{1\times 3\times 5}{5\times 4\times 3}=\frac{1\times 1\times 1}{1\times 4\times 1}=\frac{1}{4}$

(iii) $\frac{10}{27}\times \frac{28}{65}\times \frac{39}{56}=\frac{10\times 1\times 3}{27\times 5\times 2}=\frac{1\times 1\times 3}{27\times 1\times 1}=\frac{3}{27}=\frac{1}{9}$

(iv) $1\frac{4}{7}\times 1\frac{13}{22}\times 1\frac{1}{15}$

      =$\frac{11}{7}\times \frac{35}{22}\times \frac{16}{15}=\frac{11\times 35\times 16}{7\times 22\times 15}=\frac{1\times 5\times 16}{1\times 2\times 15}=\frac{1\times 1\times 8}{1\times 1\times 3}=\frac{8}{3}=2\frac{2}{3}$

(v) $2\frac{2}{17}\times 7\frac{2}{9}\times 1\frac{33}{52}$

     =$\frac{36}{17}\times \frac{65}{9}\times \frac{85}{52}=\frac{36\times 65\times 85}{17\times 9\times 52}=\frac{4\times 5\times 5}{1\times 1\times 4}=\frac{1\times 5\times 5}{1\times 1\times 1}=25$

(vi) $3\frac{1}{16}\times 7\frac{3}{7}\times 1\frac{25}{39}$

      =$\frac{49}{16}\times \frac{52}{7}\times \frac{64}{39}=\frac{7\times 4\times 4}{1\times 1\times 3}=\frac{112}{3}=37\frac{1}{3}$

Question 3:

Find:

(i) $\frac{1}{3}$ of 24
(ii) $\frac{3}{4}$ of 32
(iii) $\frac{5}{9}$ of 45
(iv) $\frac{7}{50}$ of 1000
(v) $\frac{3}{20}$ of 1020
(vi) $\frac{5}{11}$ of Rs 220
(vii) $\frac{4}{9}$ of 54 metres
(viii) $\frac{6}{7}$ of 35 litres
(ix) $\frac{1}{6}$ of an hour
(x) $\frac{5}{6 }$ of an year
(xi) $\frac{7}{20}$  of a kg
(xii) $\frac{9}{20}$ of a metre
(xiii) $\frac{7}{8}$ of a day
(xiv) $\frac{3}{7}$ of a week
(xv) $\frac{7}{50}$ of a litre

Answer 3:

We have the following:

(i) $\frac{1}{3}$ of 24 = $24\times \frac{1}{3}=\frac{24}{1}\times \frac{1}{3}=\frac{24\times 1}{1\times 3}=8$

(ii) $\frac{3}{4}$ of 32 = $32\times \frac{3}{4}=\frac{32}{1}\times \frac{3}{4}=\frac{32\times 3}{1\times 4}=\frac{8\times 3}{1\times 1}=24$

(iii) $\frac{5}{9}$ of 45 = $45\times \frac{5}{9}=\frac{45}{1}\times \frac{5}{9}=\frac{45\times 5}{1\times 9}=\frac{5\times 5}{1\times 1}=25$

(iv) $\frac{7}{50}$ of 1000 = $1000\times \frac{7}{50}=\frac{1000}{1}\times \frac{7}{50}=\frac{20\times 7}{1\times 1}=140$

(v) $\frac{3}{20}$ of 1020 = $1020\times \frac{3}{20}=\frac{1020}{1}\times \frac{3}{20}=\frac{51\times 3}{1\times 1}=153$

(vi) $\frac{5}{11}$ of Rs 220 = Rs $\left(220\times \frac{5}{11}\right)$ = Rs (20 $\times$ 5 ) = Rs 100

(vii) $\frac{4}{9}$of 54 m = $\left(\frac{4}{9}\times 54\right)\mathrm{m}$ = (4 $\times$ 6) m = 24 m

(viii) $\frac{6}{7}$ of 35 L = $\left(\frac{6}{7}\times 35\right)\mathrm{L}$ = (6 $\times$ 5) L = 30 L

(ix) $\frac{1}{6}$ of 1 h = $\frac{1}{6}$ of 60 min = $\left(60\times \frac{1}{6}\right)$ min = 10 min

(x) $\frac{5}{6}$ of an year = $\frac{5}{6}$ of 12 months = $\left(12\times \frac{5}{6}\right)$ months = (2 $\times$ 5) months = 10 months

(xi) $\frac{7}{20}$ of a kg = $\frac{7}{20}$ of 1000 g = $\left(1000\times \frac{7}{20}\right)$ g = (50 $\times$ 7) gm = 350 g

(xii) $\frac{9}{20}$ of 1 m = $\frac{9}{20}$ of 100 cm = $\left(100\times \frac{9}{20}\right)$ cm = (5 $\times$ 9) cm = 45 cm

(xiii) $\frac{7}{8}$ of a day = $\frac{7}{8}$ of 24 h = $\left(24\times \frac{7}{8}\right)$ h = (3 $\times$ 7) = 21 h

(xiv) $\frac{3}{7}$ of a week = $\frac{3}{7}$ of 7 days = $\left(7\times \frac{3}{7}\right)$ days = 3 days

(xv) $\frac{7}{50}$ of 1 L = $\frac{7}{50}$ of 1000 ml = $\left(1000\times \frac{7}{50}\right)$ ml = (20 $\times$ 7) ml = 140 ml

Question 4:

Apples are sold at Rs $48\frac{4}{5}$ per kg. What is the cost of $3\frac{3}{4}$ kg of apples?

Answer 4:

Cost of 1 kg apple = ₹ ^{$48\frac{4}{5}=\frac{244}{5}$}

Cost of 3$\frac{3}{4}$kg apples = 3$\frac{3}{4}$ $\times$$\frac{244}{5}=\frac{15}{4}\times \frac{244}{5}=₹ 183$

Hence, the cost of $3\frac{3}{4}$ apples is ₹ 183.

Question 5:

Cloth is being sold at Rs $42\frac{1}{2}$ per metre. What is the cost of $5\frac{3}{5}$ metres of this cloth?

Answer 5:

Cost of 1 m of cloth = $\mathrm{Rs} 42\frac{1}{2} = \mathrm{Rs} \frac{85}{2}$
∴ Cost of $5\frac{3}{5} \mathrm{m}$of cloth = Rs $\left(\frac{85}{2}\times 5\frac{3}{5}\right)$
                                        = Rs $\left(\frac{85}{2}\times \frac{28}{5}\right)=\mathrm{Rs} \left(\frac{85\times 28}{2\times 5}\right)=\mathrm{Rs} \left(17\times 14\right)=\mathrm{Rs} 238$
Hence, the cost of $5\frac{3}{5}\mathrm{m}$of cloth is Rs 238.

Question 6:

A car covers a certain distance at a uniform speed of $66\frac{2}{3}$ km per hour. How much distance will it cover in 9 hours?

Answer 6:

Distance covered by the car in 1 h = $66\frac{2}{3} \mathrm{km}$
Distance covered by the car in 9 h  =$\left(66\frac{2}{3}\times 9\right) \mathrm{km}$
                                                            =$\left(\frac{200}{3}\times 9\right) \mathrm{km}=\left(\frac{200\times 9}{3\times 1}\right) \mathrm{km}=\left(200\times 3\right) \mathrm{km}=600 \mathrm{km}$

Hence, the distance covered by the car in 9 h will be 600 km.

Question 7:

One tin holds $12\frac{3}{4}$ litres of oil. How many litres of oil can 26 such tins hold?

Answer 7:

Capacity of 1 tin = $12\frac{3}{4} \mathrm{L}=\frac{51}{4} \mathrm{L}$
∴ Capacity of 26 such tins =$\left(26\times \frac{51}{4}\right) \mathrm{L}$
                                           =$\left(\frac{26}{1}\times \frac{51}{4}\right) \mathrm{L}=\left(\frac{26\times 51}{1\times 4}\right) \mathrm{L}=\left(\frac{13\times 51}{1\times 2}\right) \mathrm{L}=\left(\frac{663}{2}\right) \mathrm{L}=331\frac{1}{2} \mathrm{L}$

Hence, 26 such tins can hold $331\frac{1}{2}$ L of oil.

Question 8:

For a particular show in a circus, each ticket costs Rs $35\frac{1}{2}$. If 308 tickets are sold for the  show, how much amount has been collected?

Answer 8:

Cost of 1 ticket = Rs $35\frac{1}{2}$= Rs $\frac{71}{2}$
∴ Cost of 308 tickets = Rs $\left(\frac{71}{2}\times 308\right)=\mathrm{Rs} \left(\frac{71}{2}\times \frac{308}{1}\right)=\mathrm{Rs} (71\times 154)=\mathrm{Rs} 10934$

Hence, 308 tickets were sold for Rs 10,934.

Question 9:

Nine boards are stacked on the top of each other. The thickness of each board is $3\frac{2}{3}$ cm. How high is the stack?

Answer 9:

Thickness of 1 board = $3\frac{2}{3}$ cm
∴ Thickness of 9 boards = $\left(9\times 3\frac{2}{3}\right) \mathrm{cm}$
                                        =$\left(\frac{9}{1}\times \frac{11}{3}\right) \mathrm{cm}$ = (3 $\times$ 11) cm = 33 cm

Hence, the height of the stack is 33 cm.

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Question 10:

Rohit takes $4\frac{4}{5}$ minutes to make complete round of a circular park. How much time will he take to make 15 rounds?

Answer 10:

Time taken by Rohit to complete one round of the circular park = $4\frac{4}{5}$ min = $\frac{24}{5}$min

∴ Time taken to complete 15 rounds =$\left(15\times \frac{24}{5}\right)$ min
                                                                      = (3 $\times$ 24) min
                                                                      = 72 min
                                                                      = 1 h 12 min   [∵ 1 hr = 60 min]

Hence, Rohit will take 1 h 12 min to make 15 complete rounds of the circular park.

Question 11:

Amit weighs 35 kg. His sister Kavita's weight is $\frac{3}{5}$ of Amit's weight. How much does Kavita weigh?

Answer 11:

Weight of Amit = 35 kg
Weight of Kavita = $\frac{3}{5}$ of Amit's weight
                             = 35 kg x $\frac{3}{5}$ = $\left(35\times \frac{3}{5}\right)\mathrm{kg}=(7\times 3) \mathrm{kg}=21 \mathrm{kg}$
Hence, Kavita's weight is 21 kg.

Question 12:

There are 42 students in a class and $\frac{5}{7}$ of the students are boys. How many girls are there in the class?

Answer 12:

Number of boys in the class = $\frac{5}{7}$ of the total no. of students
                                           =$\frac{5}{7}$ $\times$ 42 = $\left(\frac{5\times 42}{7}\right)=5\times 6=30$

∴ Number of girls in the class = 42 − 30 = 12

Hence, there are 12 girls in the class.

Question 13:

Sapna earns Rs 24000 per month. She spends $\frac{7}{8}$ of her income and deposits rest of the money in a bank. How much money does she deposit in the bank each month?

Answer 13:

Total monthly income =  ₹ 24000
Monthly expenditure = $\frac{7}{8}$of  ₹ 24000
                                 =24000$\times \frac{7}{8}$ = ₹ 21000
Monthly savings = 24000−21000 = ₹ 3000

Question 14:

Each side of a square field is $4\frac{2}{3}$ m. Find its area.

Answer 14:

Side of the square field = $4\frac{2}{3} \mathrm{m}$
∴ Area of the square = (side)²
                                  = ${\left(4\frac{2}{3} \mathrm{m}\right)}^2$
                                  = ${\left(\frac{14}{3} \mathrm{m}\right)}^2=\frac{14}{3} \mathrm{m} \times \frac{14}{3} \mathrm{m}=\left(\frac{14\times 14}{3\times 3}\right) {\mathrm{m}}^2=\frac{196}{9} {\mathrm{m}}^2=21\frac{7}{9} {\mathrm{m}}^2$

Hence, the area of the square field is $21\frac{7}{9} {\mathrm{m}}^2$.

Question 15:

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

Answer 15:

Length of the rectangular park = $41\frac{2}{3} \mathrm{m}=\frac{125}{3} \mathrm{m}$

Its breadth = $18\frac{3}{5} \mathrm{m}=\frac{93}{5} \mathrm{m}$

∴ Its area = length $\times$ breadth

               = $\left(\frac{125}{3}\times \frac{93}{5}\right) \mathrm{m}$²
                = (25 $\times$ 31) m = 775 m²

Hence, the area of the rectangular park is 775 m².