 RS Aggarwal Class 6 Solutions Chapter 10 - Ratio Proportion And Unitary Method Ex 10A(10.1)

RS Aggarwal Class 6 Chapter 10 - Ratio Proportion And Unitary Method Ex 10A(10.1) Solutions Free PDF

Q1: Determine if the following numbers are in proportion:

(i) 4, 6, 8, 12

(ii) 7, 42, 13, 78

(iii) 33, 121, 9, 96

(iv) 22, 33, 42, 63

(v) 32, 48, 70, 210

(vi) 150, 200, 250, 300

Sol:

(i) 4, 6, 8, 12

$\frac{4}{6} = \frac{4 \div 2}{6\div 2} = \frac{2}{3}\; ; \;\; \frac{8}{12} = \frac{8 \div 4}{12 \div 4} = \frac{2}{3}$

Hence 4 : 6 :: 8 : 12 are in proportion.

(ii) 7, 42, 13, 78

$\frac{7}{42} = \frac{7 \div 7}{42 \div 7} = \frac{1}{6}\; ; \;\; \frac{13}{78} = \frac{13 \div 13}{78 \div 13} = \frac{1}{6}$

Hence 7 : 42 :: 13 : 78 are in proportion.

(iii) 33, 121, 9, 96

$\frac{33}{121} = \frac{33 \div 11}{121 \div 11} = \frac{3}{11}\; ; \;\; \frac{9}{96} = \frac{9 \div 3}{96 \div 3} = \frac{3}{32}$

Hence 33 : 121 :: 9 : 96 are not in proportion.

(iv) 22, 33, 42, 63

$\frac{22}{33} = \frac{22 \div 11}{33 \div 11} = \frac{2}{3}\; ; \;\; \frac{42}{63} = \frac{42 \div 21}{63 \div 21} = \frac{2}{3}$

Hence 4 : 6 :: 8 : 12 are in proportion.

(v) 32, 48, 70, 210

$\frac{32}{48} = \frac{32 \div 16}{48 \div 16} = \frac{2}{3}\; ; \;\; \frac{70}{210} = \frac{70 \div 70}{210 \div 70} = \frac{1}{3}$

Hence 4 : 6 :: 8 : 12 are not in proportion.

(vi) 150, 200, 250, 300

$\frac{150}{200} = \frac{150 \div 50}{200 \div 50} = \frac{3}{4}\; ; \;\; \frac{250}{300} = \frac{250 \div 50}{300 \div 50} = \frac{5}{6}$

Hence 150 : 200 :: 250 : 300 are not in proportion.

Q2: Verify the following:

(i) 60 : 105 :: 84 : 147

(ii) 91 : 104 :: 119 : 136

(iii) 108 : 72 :: 129 : 86

(iv) 39 : 65 :: 141 : 235

Sol:

(i) 60 : 105 :: 84 : 147

$\frac{60}{105} = \frac{60 \div 15}{105 \div 15} = \frac{4}{7}$  (H.C.F. of 60 and 105 is 15.)

$\frac{84}{147} = \frac{84 \div 21}{147 \div 21} = \frac{4}{7}$        (H.C.F. of 84 and 147 is 21.)

Hence 60 : 105 :: 84 : 147 are in proportion.

(ii) 91 : 104 :: 119 : 136

$\frac{91}{104} = \frac{91 \div 13}{104 \div 13} = \frac{7}{8}$  (H.C.F. of 91 and 104 is 13.)

$\frac{119}{136} = \frac{119 \div 17}{136 \div 17} = \frac{7}{8}$        (H.C.F. of 119 and 136 is 17.)

Hence 91 : 104 :: 119 : 136 are in proportion.

(iii) 108 : 72 :: 129 : 86

$\frac{108}{72} = \frac{108 \div 36}{72 \div 36} = \frac{3}{2}$  (H.C.F. of 108 and 72 is 36.)

$\frac{129}{86} = \frac{129 \div 43}{86 \div 43} = \frac{3}{2}$        (H.C.F. of 129 and 86 is 43.)

Hence 108 : 72 :: 129 : 86 are in proportion.

(iv) 39 : 65 :: 141 : 235

$\frac{39}{65} = \frac{39 \div 13}{65 \div 13} = \frac{3}{5}$  (H.C.F. of 39 and 65 is 13.)

$\frac{141}{235} = \frac{141 \div 47}{235 \div 47} = \frac{3}{5}$        (H.C.F. of 141 and 235 is 47.)

Hence 39 : 65 :: 141 : 235 are in proportion.

Q3: Find the value of x in each of the following proportions:

(i) 55 : 11 :: x : 6

(ii) 27 : x :: 63 : 84

(iii) 51 : 85 :: 57 : x

(iv) x : 92 :: 87 : 116

Sol:

(i) 55 : 11 :: x : 6

Product of extremes = Product of means

66 $\times$ 6 = 11 $\times$ x

$\Rightarrow$ 11x = 330

x = $\frac{330}{11} = 30$

(ii) 27 : x :: 63 : 84

Product of extremes = Product of means

27 $\times$ 84 = x $\times$ 63

$\Rightarrow$ 63x = 2268

x = $\frac{2268}{63} = 36$

(iii) 51 : 85 :: 57 : x

Product of extremes = Product of means

51 $\times$ x = 85 $\times$ 57

$\Rightarrow$ 51x = 4845

x = $\frac{4845}{51} = 95$

(iv) x : 92 :: 87 : 116

Product of extremes = Product of means

x $\times$ 116 = 92 $\times$ 87

$\Rightarrow$ 116x = 8004

x = $\frac{8004}{116} = 69$

Q4: Write (T) for true and False (F) in case of each of the following:

(i) 51 : 68 :: 85 : 102

(ii) 36 : 45 :: 80 : 100

(iii) 30 bags : 18 bags :: Rs. 450 : Rs. 270

(iv) 81 kg : 45 kg :: 18 men : 10 men

(v) 45 km : 60 km :: 12 h : 15 h

(vi) 32 kg : Rs. 36 :: 8 kg : Rs. 9

Sol:

(i) 51 : 68 :: 85 : 102

Product of means = 68 $\times$ 85 = 5780

Product of extremes = 51 $\times$ 102 = 5202

Product of means $\neq$ Product of extremes

Hence (F).

(ii) 36 : 45 :: 80 : 100

Product of means = 45 $\times$ 80 = 3600

Product of extremes = 36 $\times$ 100 = 3600

Product of means = Product of extremes

Hence (T).

(iii) 30 bags : 18 bags :: Rs. 450 : Rs. 270

Product of means = 18 $\times$ 450 = 8100

Product of extremes = 30 $\times$ 270 = 8100

Product of means = Product of extremes

Hence (T).

(iv) 81 kg : 45 kg :: 18 men : 10 men

Product of means = 45 $\times$ 18 = 8100

Product of extremes = 30 $\times$ 270 = 8100

Product of means = Product of extremes

Hence (T).

(v) 45 km : 60 km :: 12 h : 15 h

Product of means = 60 $\times$ 12 = 720

Product of extremes = 45 $\times$ 15 = 675

Product of means $\neq$ Product of extremes

Hence (F).

(vi) 32 kg : Rs. 36 :: 8 kg : Rs. 9

Product of means = 36 $\times$ 8 = 288

Product of extremes = 32 $\times$ 9 = 299

Product of means = Product of extremes

Hence (T).

Q5: Determine if the following ratios form a proportion:

(i) 25 cm : 1 m and Rs. 40 and Rs. 160

(ii) 39 litres : 65 litres and 6 bottles : 10 bottles

(iii) 200 mL : 2.5 L and Rs. 4 : Rs. 50

(iv) 2 kg : 80 kg and 25 g : 625 kg

Sol:

(i) 25 cm : 1 m and Rs. 40 and Rs. 160

25 cm : 100 cm :: Rs. 40 : Rs. 160

$\frac{25}{100} = \frac{25 \div 25}{100 \div 25} = \frac{1}{4} \;\; and \;\; \frac{40}{160} = \frac{40 \div 40 }{160 \div 40} = \frac{1}{4}$

Hence, they are in proportion.

(ii) 39 litres : 65 litres and 6 bottles : 10 bottles

$\frac{39}{65} = \frac{39 \div 13}{65 \div 13} = \frac{3}{5} \;\; and \;\; \frac{6}{10} = \frac{6 \div 2}{10 \div 2} = \frac{3}{5}$

Hence, they are in proportion.

(iii) 200 mL : 2.5 L and Rs. 4 : Rs. 50

200 mL : 2500 mL :: Rs. 4 : Rs. 50

$\frac{200}{2500} = \frac{200 \div 100}{2500 \div 100} = \frac{2}{25} \;\; and \;\; \frac{4}{50} = \frac{4 \div 2}{50 \div 2} = \frac{2}{25}$

Hence, they are in proportion.

(iv) 2 kg : 80 kg and 25 g : 625 kg

2 kg : 80 kg :: 25 g : 625000 g

$\frac{2}{80} = \frac{1} {40} \;\; and \;\; \frac{25}{625000} = \frac{25 \div 25 }{625000 \div 25} = \frac{1}{25000}$

Hence, they are not in proportion.

Q6: In a proportion the 1st, 2nd and 4th term are 51, 68 and 108 respectively. Find the 3rd term.

Sol:

Let the 3rd term be x.

Thus, 51 : 68 :: x : 108

We know,

Product of extremes = Product of means

51 $\times$ 108 = 68 $\times$ x

$\Rightarrow$ 5508 = 68 x

$\Rightarrow$ x = 81

Hence, the third term is 81.

Q7: The 1st, 3rd and 4h terms of a proportion are 12, 8 and 14 respectively. Find the 2nd term.

Sol:

Let the second term be x.

Then, 12 : x :: 8 : 14

We know:

Product of extremes = Product of means

12 $\times$ 14 = 8x

$\Rightarrow$ 168 = 8x

$\Rightarrow$ x = $\frac{168}{8} = 21$

Hence the second term must be 21.

Q8: Show that the following numbers are in continued proportion:

(i) 48, 60, 75

(ii) 36, 90, 225

(iii) 16, 84, 441

Sol:

(i) 48 : 60 : 75

Product of means = 60 $\times$ 60 = 3600

Product of extremes = 48 $\times$ 75 = 3600

Product of means = Product of extremes

Hence, 48 : 60 :: 60 : 75 are in continued proportion.

(ii) 36 : 90, 90 : 225

Product of means = 90 $\times$ 90 = 8100

Product of extremes = 36 $\times$ 225 = 8100

Product of means = Product of extremes

Hence, 36 : 90, 90 : 225 are in continued proportion.

(iii) 16 : 84, 84 : 441

Product of means = 84 $\times$ 84 = 7056

Product of extremes = 16 $\times$ 441 = 7056

Product of means = Product of extremes

Hence, 16 : 84, 84 : 441 are in continued proportion.

Q9: If 9, x, x, 49 are in proportion, find the value of x.

Sol:

Given: 9 : x :: x : 49

We know,

Product of means = Product of extremes

x $\times$ x = 9 $\times$ 49

$\Rightarrow x^{2} = 441$

$\Rightarrow x^{2} = (21)^{2}$

$\Rightarrow x = 21$

Q10: An electric pole casts a shadow 20m at a time when a tree 6 m high casts a shadow of length 8m. Find the height of the pole.

Sol:

Let the height of the pole = x m

Then, we have:

x : 20 :: 6 : 8

Now we have:

Product of extremes = Product of means

8x = 20 $\times$ 6

x = $\frac{120}{8} = 15$

Hence, the height of the pole is 15m.

Q11: Find the value of x if 5 : 3 :: x : 6

Sol:

5: 3 :: x : 6

We know,

Product of means = Product of extremes

3x = 5 $\times$ 6

$\Rightarrow \frac{30}{3} = 10$

Therefore x = 10

Exercise 10C

Q1: If the cost of 14 m of cloth is Rs. 1890, find the cost of 6 m of cloth.

Sol:

Cost of 14 m of cloth = Rs. 1890

Cost of 1 m of cloth = $\frac{1890}{14}$ = Rs. 135

Cost of 6 m of cloth = $6 \times 135 =$ Rs. 810

Q2: If the cost of a dozen soaps is Rs. 285.60, what will be the cost of 15 such soaps?

Sol:

Cost of dozen soaps = Rs. 285.60

Cost of 1 soap = $\frac{285.60}{12}$

Cost of 15 soaps = 15 $\times \frac{285.60}{12} = \frac{4824}{12} =$ Rs. 357

Q3: If 9 kg rice costs Rs. 327.60, what will be the cost of 50 kg of rice?

Sol:

Cost of 9 gram of rice = Rs. 327.60

Cost of 1 kg of rice = $\frac{327.60}{9}$

Cost of 50 kg of rice = $50 \times \frac{327.60}{9} = \frac{16380}{9} =$ Rs. 1820

Hence, the cost of 50 kg of rice is Rs. 1820.

Q4: If 22.5 m of a uniform iron weights 85.5 kg, what will be the weight of 5 m of the same rod?

Sol:

Weight of 22.5 m of uniform rod = 85.5 kg

Weight of 1 m f uniform rod = $\frac{85.5}{22.5}$ kg

Weight of 5 m of uniform rod = $5 \times \frac{85.5}{22.5} = \frac{427.5}{22.5} = 19$ kg

Thus the weight of 5 m of iron rod is 19 kg.

Q5: If 15 tins of the same size contains 234 kg oil, how much oil will be there in 10 such tins?

Sol:

Oil contained by 15 tins = 234 kg

Oil contained by 1 tin = $\frac{234}{15}$ kg

Oil contained by 10 tins = $10 \times \frac{234}{15} = \frac{2340}{15} = 156$ kg

Q6: If 2 L of diesel is consumed by a car in covering a distance of 222 km, how many kilometers will it go in 22 L of diesel?

Sol:

Distance covered by a car in 12 L of diesel = 222 km

Distance covered by it in 1 L = $\frac{222}{12}$ km

Distance covered by it in 22 L of diesel = $22 \times \frac{222}{12} = \frac{4884}{12} = 407$ km

Q7: A transport company charges Rs. 540 to carry 25 tonnes of weight. What will it charge to carry 35 tonnes?

Sol:

Cost of transporting 25 tonnes of weight = Rs. 540

Cost of transporting 1 tonne of weight = $\frac{540}{25}$

Cost of transporting 35 tonnes of weight = $35 \times \frac{540}{25} = \frac{18900}{25} =$ = Rs. 756

Q8: 4.5 g of an alloy of copper and zinc contains 3.5 g of copper. What weight of copper will be there in 18.9 g of the alloy?

Sol:

Let the weight of copper be x g.

Then 4.5 : 3.5 :: 18.9 : x

Product of extremes = Product of means

4.5 $\times$ x = 3.5 $\times$ 18.9

$\Rightarrow x = \frac{66.15}{4.5} = 14.7$

So, the weight of copper is 14.7 g.

Q9: 35 inland letter cost Rs. 87.50. How many such letters can we buy for Rs. 315?

Sol:

Number of inland letters whose total cost is Rs. 87.50 = 35

Number of inland letters whose cost is Re 1 = $\frac{35} {87.50}$

Number of inland letters whose cost is Rs. 315 = $315 \times \frac{35}{87.50} = \frac{11025}{87.50} = 126$

Hence, we can buy 126 inland letters for Rs. 315

Q10: Cost of 4 dozen of bananas is Rs. 104. How many bananas can be purchased for Rs. 6.50?

Sol:

Number of bananas that can be purchased for Rs. 104 = 48 (4 dozen)

Number of bananas that can be purchased for Re 1 = $\frac{48}{104}$

Number of bananas that can be purchased for Rs. 6.50 = $6.50 \times \frac{48}{104} = \frac{312}{104} = 3$

Hence, 3 bananas can be purchased for Rs. 6.50

Q11: The cost of 18 chairs is Rs. 22770. How many such chairs can be bought for Rs. 10120?

Sol:

Number of chairs that can be bought for Rs. 22770 = 18

Number of chairs that can be bought for Re 1 = $\frac{18}{22770}$

Number of chairs that can be bought for Rs. 10120 = $10120 \times \frac{18}{22770} = \frac{182160}{22770} = 8$

Q12: A car travels 195 km in 3 hours.

(i) How long will it take to travel 520 km?

(ii) How far will it travel in 7 hours with the same speed?

Sol:

(i) Time taken by the car to travel 195 km = 3 hours

Time taken by the car to travel 1 km = $\frac{3}{195}$ hours

Time taken by the car to travel 520 km = $520 \times \frac{3}{195} = \frac{1560}{195}$ = 8 hours

(ii) Distance covered by the car in 3 hours = 195 km

Distance covered by it in 1 hour = $\frac{195}{3}$ km

Distance covered by it in 7 hours = $7 \times 65 = 455$ km

Q13: A labourer earns Rs. 1980 in 12 days.

(i) How much does he earns in 7 days?

(ii) In how many days will he earn Rs. 2640?

Sol:

(i) Earning of a labourer in 12 days = Rs. 1980

Earning of the labourer in 1 day = $\frac{1980}{12} =$ Rs. 165

Earning of labourer in 7 days = 7 $\times$ 165 = Rs. 1155

(ii) Number of days taken by the labourer to earn Rs. 1980 = 12 days

Number of days taken by him to earn Re 1 = $\frac{12}{1980}$ days

Number of days taken by him to earn Rs 2640 = $2640 \times \frac{12}{1980} = \frac{31680}{1980} = 16$ days

Q14: The weight of 65 books is 13 kg.

(i) What is the weight of 80 such books?

(ii) How many such books weight 6.4 kg.

Sol:

(i) Weight of 65 books = 13 kg

Weight of 1 book = $\frac{13}{65}$ kg

Weight of 80 books = $80 \times \frac{13}{65} = \frac{1040}{65} = 16$ kg

(ii) Number of books weighing 13 kg = 65

Number of books weighing 1 kg = $\frac{65}{13} = 5$

Number of books weighing 6.4 kg = $6.4 \times 5 = 32$

Q15: If 48 boxes contains 6000 pens. How many such boxes will be needed for 1875 pens?

Sol:

Number of boxes containing 6000 pens = 48

Number of boxes containing 1 pen = $\frac{48}{6000}$

Number of boxes containing 1875 pens = $1875 \times \frac{48}{6000} = \frac{90000}{6000} = 15$

15 boxes are needed for 1875 pens.

Q16: 24 workers can build a wall in 15 days. How many men will be needed to finish it in 16 days?

Sol:

Number of days taken by 24 workers to build a wall = 15 days

Number of days taken by 1 worker to build the wall = 15 $\times$ 24 = 360 days      (Less worker means more days i.e. inverse relation)

Number of days taken by 9 workers to build the wall = $\frac{360}{9} = 40$ days

Q17: 40 men can finish a piece of work in 26 days. How many men will be needed to finish it in 16 days?

Sol:

Number of men required to complete the work in 26 days = 40

Number of men required to complete the work in 1 day = 40 $\times$ 26 = 1040 men  ( less men more days, inverse relation)

Number of men required to complete the work in 16 days = $\frac{1040}{16} = 65$

Q18: In an army camp, there were provision for 550 men for 28 days. But 700 men attended the camp. How long did the provisions last?

Sol:

Number of days the provision will last for 550 men = 28 days

Number of days the provision will last for 1 man = 28 $\times$ 550 = 15400 days   ( less man more days)

Number of days the provision will last for 700 men = $\frac{15400}{700} = 22$ days

Thus the provision will last for 22 days.

Q19: A given quality of rice is sufficient for 60 persons for 3 days. How many days would the rice last for 18 persons?

Sol:

Number of days for which the given quantity of rice is sufficient for 60 days = 3 days

Number of days for which it is sufficient for 1 person = 3 $\times$ 60 = 180 days    (Less men means more days)

Number of days for which it is sufficient for 18 persons = $\frac{180}{18} = 10$ days’

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The cockroach has a pair of sensitive antennae.