RS Aggarwal 2021-2022 for Class 6 Maths Solutions Chapter 10- Ratio, Proportion And Unitary Method
RS Aggarwal Class 6 Math Solution Chapter 10- Ratio, Proportion And Unitary Method Exercise 10 A is available here. RS Aggarwal Class 6 Math Solutions are solved by expert teachers in step by step, which help the students to understand easily. RS Aggarwal textbooks are responsible for a strong foundation in Maths. These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students’ understanding.
Rs Aggarwal Class 6 Math Solution Chapter 10- Ratio, Proportion And Unitary Method
Exercise 10A
1. Find each of the following ratios in the simplest form:
(i) 24 to 56
(ii) 84 paise to Rs.3
(iii) 4 kg to 750 g
(iv) 1.8 kg to 6 kg
(v) 48 minutes to 1 hour
(vi) 2.4 km to 900 m
Solution:
(i) 24 to 56
First, we find H.C.F of 24 and 56.
So, H.C.F of 24 and 56 = 8
Now, we divide 24 and 56 by 8 to get simplest form.
= `frac\{24 ÷ 8}{56 ÷ 8}` = `frac\{3}{7}`
or, 3 : 7 is in simplest form.
(ii) 84 paise to Rs.3 = 84 p : 300 p
First, we find H.C.F of 84 and 300.
So, H.C.F of 84 and 300 = 12
Now, we divide 84 and 300 by 12 to get simplest form.
= `frac\{84 ÷ 12}{300 ÷ 12}` = `frac\{7}{25}`
or, 7 : 25 is in simplest form.
(iii) 4 kg to 750 g = 4000 g : 750g
First, we find H.C.F of 4000 and 750.
So, H.C.F of 84 and 300 = 250
Now, we divide 84 and 300 by 12 to get simplest form.
= `frac\{4000 ÷ 250}{750 ÷ 250}` = `frac\{16}{3}`
or, 16 : 3 is in simplest form.
(iv) 1.8 kg to 6 kg = 1800 g : 6000 g
First, we find H.C.F of 1800 and 6000.
So, H.C.F of 1800 and 6000 = 600
Now, we divide 1800 and 6000 by 600 to get simplest form.
= `frac\{1800 ÷ 600}{6000 ÷ 600}` = `frac\{3}{10}`
or, 3 : 10 is in simplest form.
(v) 48 minutes to 1 hour = 48 min : 60 min
First, we find H.C.F of 48 and 60.
So, H.C.F of 48 and 60 = 12
Now, we divide 48 and 60 by 12 to get simplest form.
= `frac\{48 ÷ 12}{60 ÷ 12}` = `frac\{4}{5}`
or, 4 : 5 is in simplest form.
(vi) 2.4 km to 900 m = 2400 m : 900 m
First, we find H.C.F of 2400 and 900.
So, H.C.F of 2400 and 900 = 300
Now, we divide 2400 and 900 by 300 to get simplest form.
= `frac\{2400 ÷ 300}{900 ÷ 300}` = `frac\{8}{3}`
or, 8 : 3 is in simplest form.
2. Express each of the following ratios in the simplest form:
(i) 36 : 90
(ii) 324 : 144
(iii) 85 : 561
(iv) 480 : 384
(v) 186 : 403
(vi) 777 : 1147
Solution:
(i) 36 : 90
= `frac\{36}{90}` (H.C.F of 36 and 90 = 18)
= `frac\{36 ÷ 18}{90 ÷ 18}`
= `frac\{2}{5}` = 2 : 5 is in simplest form.
(ii) 324 : 144
= `frac\{324}{144}` (H.C.F of 324 and 144 = 36)
= `frac\{324 ÷ 36}{144 ÷ 36}`
= `frac\{9}{4}` = 9 : 4 is in simplest form.
(iii) 85 : 561
= `frac\{85}{561}` (H.C.F of 85 and 561 = 17)
= `frac\{85 ÷ 17}{561 ÷ 17}`
= `frac\{5}{33}` = 5 : 33 is in simplest form.
(iv) 480 : 384
= `frac\{480}{384}` (H.C.F of 480 and 384 = 96)
= `frac\{480 ÷ 96}{384 ÷ 96}`
= `frac\{5}{4}` = 5 : 4 is in simplest form.
(v) 186 : 403
= `frac\{186}{403}` (H.C.F of 186 and 403 = 31)
= `frac\{186 ÷ 31}{403 ÷ 31}`
= `frac\{6}{13}` = 6 : 13 is in simplest form.
(vi) 777 : 1147
= `frac\{777}{1147}` (H.C.F of 777 and 1147= 37)
= `frac\{777 ÷ 37}{1147 ÷ 37}`
= `frac\{21}{31}` = 21 : 31 is in simplest form.
3. Write each of the following ratios in the simplest form:
(i) Rs. 6.30 : Rs. 16.80
(ii) 3 weeks: 30 days
(iii) 3 m 5 cm : 35 cm
(iv) 48 min : 2 hours 40 min
(v) 1 L 35 ml : 270 mL.
(vi) 4 kg : 2 kg 500 g
Solution:
(i) Rs. 6.30 : Rs. 16.80 = 630 p : 1680 p
First, we find H.C.F of 630 and 1680.
So, H.C.F of 630 and 1680 = 210
Now, we divide 630 and 1680 by 210 to get simplest form.
= `frac\{630 ÷ 210}{1680 ÷ 210}` = `frac\{3}{8}`
or, 3 : 8 is in simplest form.
(ii) 3 weeks: 30 days = 21 days : 30 days
First, we find H.C.F of 21 : 30.
So, H.C.F of 21 and 30 = 3
Now, we divide 21 and 30 by 3 to get simplest form.
= `frac\{21 ÷ 3}{30 ÷ 3}` = `frac\{7}{10}`
or, 7 : 10 is in simplest form.
(iii) 3 m 5 cm : 35 cm = 305 cm : 35 cm
First, we find H.C.F of 305 : 35.
So, H.C.F of 305 and 35 = 5
Now, we divide 305 and 35 by 5 to get simplest form.
= `frac\{305 ÷ 5}{35 ÷ 5}` = `frac\{61}{7}`
or, 61 : 7 is in simplest form.
(iv) 48 min : 2 hours 40 min = 48 min : 160 min
First, we find H.C.F of 48 : 160.
So, H.C.F of 48 and 160 = 16
Now, we divide 48 and 160 by 16 to get simplest form.
= `frac\{48 ÷ 16}{160 ÷ 16}` = `frac\{3}{10}`
or, 3 : 10 is in simplest form.
(v) 1 L 35 ml : 270 mL. = 1035 ml : 270 ml
First, we find H.C.F of 1035 : 270.
So, H.C.F of 1035 and 270 = 45
Now, we divide 1035 and 270 by 45 to get simplest form.
= `frac\{1035 ÷ 45}{270 ÷ 45}` = `frac\{23}{6}`
or, 23 : 6 is in simplest form.
(vi) 4 kg : 2 kg 500 g = 4000 g : 2500 g
First, we find H.C.F of 4000 : 2500.
So, H.C.F of 4000 and 2500 = 500
Now, we divide 4000 and 2500 by 500 to get simplest form.
= `frac\{4000 ÷ 500}{2500 ÷ 500}` = `frac\{8}{5}`
or, 8 : 5 is in simplest form.
4. Mr Sahai and his wife are both school teachers and earn Rs. 16800 and Rs. 10500 per month respectively. Find the ratio of
(i) Mr Sahai's income to his wife's income:
(ii) Mrs Sahal's income to her husband's income:
(iii) Mr Sahai's income to the total income of the two.
Solution:
Mr Sahai's income = Rs. 16800
His wife's income = Rs. 10500
Total income of both = Rs. (16800 + 10500)
= Rs. 27300
(i) Ratio of Mr Sahai's income to his wife's income
= 16800 : 10500
= `frac\{16800}{10500}`
= `frac\{16800 ÷ 2100}{10500 ÷ 2100}`
(H.C.F 16800 and 10500 = 2100)
= `frac\{8}{5}` = 8 : 5
(ii) Ratio of Mrs Sahal's income to her husband's income:
= = 10500 : 16800
= `frac\{10500}{16800}`
= `frac\{10500 ÷ 2100}{16800 ÷ 2100}`
(H.C.F 16800 and 10500 = 2100)
= `frac\{5}{8}` = 5 : 8
(iii) Ratio of Mr Sahai's income to the total income of the two
= 16800 : 27300
= `frac\{16800}{27300}`
= `frac\{16800 ÷ 2100}{27300 ÷ 2100}`
(H.C.F 16800 and 27300 = 2100)
= `frac\{8}{13}` = 8 : 13
5. Rohit earns Rs. 15300 and saves Rs. 1224 per month. Find the ratio of
(i) his income and savings;
(ii) his income and expenditure;
(iii) his expenditure and savings.
Solution:
Rohit monthly earnings = Rs. 15300
His savings = Rs. 1224
So, his expenditure = Rs. 15300 – 1224
= Rs. 14076
Now,
= `frac\{15300 ÷ 612}{1224 ÷ 612}`
(H.C.F 15300 and 1224 = 612)
= `frac\{25}{2}` = 25 : 2
(ii) Ratio of his income and expenditure;
= 15300 : 14076
= `frac\{15300}{14076}`
= `frac\{15300 ÷ 612}{14076 ÷ 612}`
(H.C.F 15300 and 14076 = 612)
= `frac\{25}{23}` = 25 : 23
(iii) Ratio of his expenditure and savings.
= 14076 : 1224
= `frac\{14076}{1224}`
= `frac\{14076 ÷ 612}{1224 ÷ 612}`
(H.C.F 14076 and 1224 = 612)
= `frac\{23}{2}` = 23 : 2
6. The ratio of the number of male and female workers in a textile mill is 5 :3. If there are 115 male workers, what is the number of female workers in the mill?
Solution:
Let the number of male be 5x
and number of female be 3x.
Then,
5x = 115
⇒ x = `frac\{cancel 115^23}{cancel 5}`
⇒ x = 23
Number of female workers = 3x
= 3 × 23 = 69.
7. The boys and the girls in a school are in the ratio 9 : 5. If the total strength of the school is 448, find the number of girls.
Solution:
Let the number of boys = 9x
Let the number of girls = 5x
Total strength of the school = 448
According to given question, we have:
9x + 5x = 448
⇒ 14x = 448
⇒ x = `frac\{cancel 448^32}{cancel 14}`
∴ x = 32
Number of boys = 9x = 9 × 32 = 288
Number of girls = 5x = 5 × 32 = 160
8. Divide Rs. 1575 between Kamal and Madhu in the ratio 7 : 2.
Solution:
Let the share of Kamal be = 7x
and the share of Madhu be = 2x
Total amount = Rs. 1575
According to question,
7x + 2x = 1575
⇒ 9x = 1575
⇒ x = `frac\{cancel 1575^175}{cancel 9}`
∴ x = 175
Share of Kamal = 7x = 7 × 175 = Rs. 1225
Share of Madhu = 2x = 2 × 175 = Rs. 350
9. Divide Rs. 3450 among A, B and C in the ratio 3 : 5 : 7.
Solution:
A : B : C = 3 : 5 : 7
Sum of the ratio terms = 3 + 5 + 7 = 15
A's share = `frac\{3}{cancel 15}` × `cancel 3450^230` = Rs 690
B's share = `frac\{5}{cancel 15}` × `cancel 3450^230` = Rs 1150
C's share = `frac\{7}{cancel 15}` × `cancel 3450^230` = Rs 1610
10. Two numbers are in the ratio 11 : 12 and their sum is 460. Find the numbers.
Solution:
Let the numbers be 11x and 12x.
According to question,
⇒ 11x + 12x = 460
⇒ 23x = 460
⇒ x = `frac\{cancel 460^20}{cancel 23}`
∴ x = 20
First number = 11x = 11 × 20 = 220
Second number = 12x = 12 × 20 = 240
Hence, the numbers are 220 and 240.
11. A 35-cm line segment is divided into two parts in the ratio 4 : 3. Find the length of each part.
Solution:
Ratio of the two parts of line segment = 4 : 3
Sum of the ratio terms = 4 + 3 = 7
First part = `frac\{4}{cancel 7}` × `cancel 35^5` cm
= 4 × 5 cm = 20 cm
Second part = `frac\{3}{cancel 7}` × `cancel 35^5` cm
= 3 × 5 cm = 15 cm
12. A factory produces electric bulbs. If 1 out of every 10 bulbs is defective and the factory produces 630 bulbs per day, find the number of defective bulbs produced each day.
Solution:
Number of bulbs produced each day = 630
Number of defective bulb out of 10 bulbs = 1
Number of defective bulbs = `frac\{cancel 630^63}{cancel 10}` = 63
∴ Number of defective bulbs produced each day = 63
13. Find the ratio of the price of a pencil to that of a ball pen if pencils cost Rs. 96 per score and ball pens cost Rs. 50.40 per dozen.
Solution:
1 score = 20 unit.
1 dozen = 12 unit.
Cost of 20 pencils = Rs 96
Cost of 1 pencil = `frac\{96}{20}` = Rs. 4.8
Cost of 12 ball pens = Rs 50.40
Cost of 1 ball pen = `frac\{50.40}{12}` = Rs. 4.2
Ratio of a pencil and a ball pen
= `frac\{4.8}{4.2}`
= `frac\{cancel 48^8}{cancel 42^7}`
= `frac\{8}{7}` = 8 : 7
14. The ratio of the length of a field to its width is 5 : 3. Find its length if the width is 42 metres.
Solution:
Length : Width = 5 : 3
Let the length be 5x and width be 3x.
Width = 42 m (given)
3x = 42
⇒ x = `frac\{cancel 42^14}{cancel 3}` = 14
∴ Length = 5x = 5 × 14 = 70 metres
15. The ratio of income to savings of a family is 11 : 2. Find the expenditure if the savings is Rs. 1520.
Solution:
Income : Savings = 11 : 2
Let the income and the saving be Rs 11x and Rs 2x, respectively.
Saving = Rs 1520
2x = 1520
⇒ x = `frac\{cancel 1520^760}{cancel 2}` = 760
∴ Income = Rs 11x = Rs (11 × 760) = Rs 8360
Expenditure = Income − Saving
= Rs (8360 − 1520 )
= Rs 6840
16. The ratio of income to expenditure of a family is 7 : 6. Find the savings if the income is Rs. 14000.
Solution:
Income : Expenditure = 7 : 6
Let the income and the expenditure be Rs 7x and Rs 6x, respectively.
Income = Rs 14000
7x = 14000
⇒ x = `frac\{cancel 14000^2000}{cancel 7}` = 2000
Expenditure = Rs 6x = Rs 6 × 2000 = Rs 12000
∴ Saving = Income − Expenditure
= Rs (14000 − 12000)
= Rs 2000
17. The ratio of zinc and copper in an alloy is 7 : 9. If the weight of copper in the alloy is 11.7 kg find the weight of zinc in it.
Solution:
Let the weight of zinc be x kg.
Ratio of zinc and copper = 7 : 9
Weight of copper in the alloy = 11.7 kg
`frac{7}{9}` = `frac\{x}{11.7}`
⇒ x = `frac\{11.7 × 7}{9}` = `frac\{81.9}{9}` = 9.1
Weight of zinc = 9.1 kg
18. A bus covers 128 km in 2 hours and a train covers 240 km in 3 hours. Find the ratio of their speeds.
Solution:
Cover distance by bus = 128 km.
Time taken = 2 hours
∴ Speed of bus = `frac\{distance}{time}`
= `frac\{cancel 128^64}{cancel 2}`
= 64 km/h
Cover distance by train = 240 km.
Time taken = 3 hours
∴ Speed of train = `frac\{distance}{time}`
= `frac\{cancel 240^80}{cancel 3}`
= 80 km/h
Ratio of their speeds = 64:80
= `frac\{cancel 64^4}{cancel 80^5}`
= `frac\{4}{5}`
∴ Ratio of the speeds of the bus and the train = 4:5
19. From each of the given pairs, find which ratio is larger:
(i) (3 : 4) or (9 : 16)
(ii) (5 : 12) or (17 : 30)
(iii) (3 : 7) or (4 : 9)
(iv) (1 : 2) or (13 : 27)
Solution:
(i) (3 : 4) or (9 : 16)
`frac\{3}{4}` or `frac\{9}{16}`
Making the denominator equal
`frac\{3}{4}` = `frac\{3 × 4}{4 × 4}` = `frac\{12}{16}`
`frac\{9}{16}` = `frac\{9 × 1}{16 × 1}` = `frac\{9}{16}`
∵ 12 > 9
∴ `frac\{12}{16}` > `frac\{9}{16}`
∴ (3 : 4) > (9 : 16)
(ii) (5 : 12) or (17 : 30)
`frac\{5}{12}` or `frac\{17}{30}`
Making the denominator equal
`frac\{5}{12}` = `frac\{5 × 5}{12 × 5}` = `frac\{25}{60}`
`frac\{17}{30}` = `frac\{17 × 2}{30 × 2}` = `frac\{34}{60}`
∵ 25 < 34
∴ `frac\{25}{60}` < `frac\{34}{60}`
∴ (5 : 12) < (17 : 30)
(iii) (3 : 7) or (4 : 9)
`frac\{3}{7}` or `frac\{4}{9}`
Making the denominator equal
`frac\{3}{7}` = `frac\{3 × 9}{7 × 9}` = `frac\{27}{63}`
`frac\{4}{9}` = `frac\{4 × 7}{9 × 7}` = `frac\{28}{63}`
∵ 27 < 28
∴ `frac\{27}{63}` < `frac\{28}{63}`
∴ (3 : 7) < (4 : 9)
20. Fill in the place holders:
(i) `frac\{24}{40}` = `frac\{⬜}{5}` = `frac\{12}{⬜}`
(ii) `frac\{36}{63}` = `frac\{4}{⬜}` = `frac\{⬜}{21}`
(iii) `frac\{5}{7}` = `frac\{⬜}{528` = `frac\{35}{⬜}`
Solution:
(i) `frac\{24}{40}` = `frac\{⬜}{5}` = `frac\{12}{⬜}`
5 = 40 ÷ 8
So, `frac\{24}{40}` = `frac\{24 ÷ 8}{40 ÷ 8}` = `frac\{3}{5}`
12 = 24 ÷ 2
So, `frac\{24}{40}` = `frac\{24 ÷ 2}{40 ÷ 2}` = `frac\{12}{20}`
Therefore, `frac\{24}{40}` = `frac\{3}{5}` = `frac\{12}{20}`
(ii) `frac\{36}{63}` = `frac\{4}{⬜}` = `frac\{⬜}{21}`
4 = 36 ÷ 9
`frac\{36}{63}` = `frac\{36 ÷ 9}{63 ÷ 9}` = `frac\{4}{7}`
21 = 63 ÷ 3
`frac\{36}{63}` = `frac\{36 ÷ 3}{63 ÷ 3}` = `frac\{12}{21}`
Therefore, `frac\{36}{63}` = `frac\{4}{7}` = `frac\{12}{21}`
(iii) `frac\{5}{7}` = `frac\{⬜}{28}` = `frac\{35}{⬜}`
28 = 7 × 4
`frac\{5}{7}` = `frac\{5 × 4}{7 × 4}` = `frac\{20}{28}`
35 = 5 × 7
So, `frac\{5}{7}` = `frac\{5 × 7}{7 × 7}` = `frac\{35}{49}`
Therefore, `frac\{5}{7}` = `frac\{20}{28}` = `frac\{35}{49}`
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