RS Aggarwal 2020 2021 Solutions for Class 8 Maths Chapter 12 Direct And Inverse Proportions are provided here with simple step-by-step explanations. These solutions for Direct And Inverse Proportions are extremely popular among class 8 students for Maths Direct And Inverse Proportions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2020 2021 Book of class 8 Maths Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal 2020 2021 Solutions. All RS Aggarwal 2020 2021 Solutions for class 8 Maths are prepared by experts and are 100% accurate.

(i)

(ii)

(iii)

(i)

(ii)

(iii)

#### Question 3:

Let the required distance be x km. Then, we have:

 Quantity of diesel (in litres) 34 20 Distance (in km) 510 x

Clearly, the less the quantity of diesel consumed, the less is the distance covered.
So, this is a case of direct proportion.

Therefore, the required distance is 300 km.

#### Question 4:

Let the required distance be x km. Then, we have:

 Quantity of diesel (in litres) 34 20 Distance (in km) 510 x

Clearly, the less the quantity of diesel consumed, the less is the distance covered.
So, this is a case of direct proportion.

Therefore, the required distance is 300 km.

Let the charge for a journey of 124 km be â‚¹x.

 Price(in â‚¹) 2550 x Distance(in km) 150 124
More is the distance travelled, more will be the price.
So, it is a case of direct proportion.
$\therefore \frac{2550}{150}=\frac{x}{124}\phantom{\rule{0ex}{0ex}}⇒x=\frac{2550×124}{150}=2108$
Thus, the taxi charges â‚¹2,108 for the distance of 124 km.

#### Question 5:

Let the charge for a journey of 124 km be â‚¹x.

 Price(in â‚¹) 2550 x Distance(in km) 150 124
More is the distance travelled, more will be the price.
So, it is a case of direct proportion.
$\therefore \frac{2550}{150}=\frac{x}{124}\phantom{\rule{0ex}{0ex}}⇒x=\frac{2550×124}{150}=2108$
Thus, the taxi charges â‚¹2,108 for the distance of 124 km.

Let the required distance be x km. Then, we have:
.

 Distance (in km) 16 x Time (in min) 25 300

Clearly, the more the time taken, the more will be the distance covered.

So, this is a case of direct proportion.

Therefore, the required distance is 192 km.

#### Question 6:

Let the required distance be x km. Then, we have:
.

 Distance (in km) 16 x Time (in min) 25 300

Clearly, the more the time taken, the more will be the distance covered.

So, this is a case of direct proportion.

Therefore, the required distance is 192 km.

Let the required number of dolls be x. Then, we have:

 No of dolls 18 x Cost of dolls (in rupees) 630 455

Clearly, the less the amount of money, the less will be the number of dolls bought.
So, this is a case of direct proportion.

Therefore, 13 dolls can be bought for Rs 455.

#### Question 7:

Let the required number of dolls be x. Then, we have:

 No of dolls 18 x Cost of dolls (in rupees) 630 455

Clearly, the less the amount of money, the less will be the number of dolls bought.
So, this is a case of direct proportion.

Therefore, 13 dolls can be bought for Rs 455.

Let the quantity of sugar bought for â‚¹371 be x kg.

 Quantity(in kg) 9 x Price(in â‚¹) 238.5 371
The price increases as the quantity increases. Thus, this is a case of direct proportion.
$\therefore \frac{9}{238.50}=\frac{x}{371}\phantom{\rule{0ex}{0ex}}⇒x=\frac{9×371}{238.50}=14$
Thus, the quantity of sugar bought for â‚¹371 is 14 kg.

#### Question 8:

Let the quantity of sugar bought for â‚¹371 be x kg.

 Quantity(in kg) 9 x Price(in â‚¹) 238.5 371
The price increases as the quantity increases. Thus, this is a case of direct proportion.
$\therefore \frac{9}{238.50}=\frac{x}{371}\phantom{\rule{0ex}{0ex}}⇒x=\frac{9×371}{238.50}=14$
Thus, the quantity of sugar bought for â‚¹371 is 14 kg.

Let the length of cloth be x m. Then, we have:
â€‹

 Length of cloth (in metres) 15 x Cost of cloth (in rupees) 981 1308

Clearly, more length of cloth can be bought by more amount of money.
So, this is a case of direct proportion.

Therefore, 20 m of cloth can be bought for Rs 1,308.

#### Question 9:

Let the length of cloth be x m. Then, we have:
â€‹

 Length of cloth (in metres) 15 x Cost of cloth (in rupees) 981 1308

Clearly, more length of cloth can be bought by more amount of money.
So, this is a case of direct proportion.

Therefore, 20 m of cloth can be bought for Rs 1,308.

Let x m be the length of the model of the ship. Then, we have:

 Length of the mast (in cm) Length of the  ship (in cm) Actual ship 1500 3500 Model of the ship 9 x

Clearly, if the length of the actual ship is more, then the length of the model ship will also be more.
So, this is a case of direct proportion.

Therefore, the length of the model of the ship is 21 cm.

#### Question 10:

Let x m be the length of the model of the ship. Then, we have:

 Length of the mast (in cm) Length of the  ship (in cm) Actual ship 1500 3500 Model of the ship 9 x

Clearly, if the length of the actual ship is more, then the length of the model ship will also be more.
So, this is a case of direct proportion.

Therefore, the length of the model of the ship is 21 cm.

Let x kg be the required amount of dust. Then, we have:

 No. of days 8 15 Dust (in kg) $6.4×{10}^{7}$ x

Clearly, more amount of dust will be collected in more number of days.
So, this is a case of direct proportion.

Therefore, 12,00,00,000 kg of dust will be picked up in 15 days.

#### Question 11:

Let x kg be the required amount of dust. Then, we have:

 No. of days 8 15 Dust (in kg) $6.4×{10}^{7}$ x

Clearly, more amount of dust will be collected in more number of days.
So, this is a case of direct proportion.

Therefore, 12,00,00,000 kg of dust will be picked up in 15 days.

Let x km be the required distance. Then, we have:

 Distance covered (in km) 50 x Time (in min) 60 72

Clearly, more distance will be covered in more time.
So, this is a case of direct proportion.

Therefore, the distance travelled by the car in 1 h 12 min is 60 km.

#### Question 12:

Let x km be the required distance. Then, we have:

 Distance covered (in km) 50 x Time (in min) 60 72

Clearly, more distance will be covered in more time.
So, this is a case of direct proportion.

Therefore, the distance travelled by the car in 1 h 12 min is 60 km.

Let x km be the required distance covered by Ravi in 2 h 24 min.
Then, we have:

 Distance covered (in km) 5 x Time (in min) 60 144

Clearly, more distance will be covered in more time.
So, this is a case of direct proportion.
$\mathrm{Now},\frac{5}{60}=\frac{x}{144}\phantom{\rule{0ex}{0ex}}⇒x=\frac{5×144}{60}\phantom{\rule{0ex}{0ex}}⇒x=12\phantom{\rule{0ex}{0ex}}$

Therefore, the distance covered by Ravi in 2 h 24 min is 12 km.

#### Question 13:

Let x km be the required distance covered by Ravi in 2 h 24 min.
Then, we have:

 Distance covered (in km) 5 x Time (in min) 60 144

Clearly, more distance will be covered in more time.
So, this is a case of direct proportion.
$\mathrm{Now},\frac{5}{60}=\frac{x}{144}\phantom{\rule{0ex}{0ex}}⇒x=\frac{5×144}{60}\phantom{\rule{0ex}{0ex}}⇒x=12\phantom{\rule{0ex}{0ex}}$

Therefore, the distance covered by Ravi in 2 h 24 min is 12 km.

Let x mm be the required thickness. Then, we have:

 Thickness of cardboard (in mm) 65 x No. of cardboards 12 312

Clearly, when the number of cardboard is more, the thickness will also be more.
So, it is a case of direct proportion.

Therefore, the thickness of the pile of 312 cardboards is 1690 mm.

#### Question 14:

Let x mm be the required thickness. Then, we have:

 Thickness of cardboard (in mm) 65 x No. of cardboards 12 312

Clearly, when the number of cardboard is more, the thickness will also be more.
So, it is a case of direct proportion.

Therefore, the thickness of the pile of 312 cardboards is 1690 mm.

Let x be the required number of men.

Then, we have:

 Number of men 11 x Length of trench (in metres) $\frac{27}{4}$ 27

Clearly, the longer the trench, the greater will be the number of men required.
So, it is a case of direct proportion.

Therefore, 44 men should be employed to dig a trench of length 27 m.

#### Question 15:

Let x be the required number of men.

Then, we have:

 Number of men 11 x Length of trench (in metres) $\frac{27}{4}$ 27

Clearly, the longer the trench, the greater will be the number of men required.
So, it is a case of direct proportion.

Therefore, 44 men should be employed to dig a trench of length 27 m.

Let Reenu type x words in 8 minutes.

 No. of words 540 x Time taken (in min) 30 8

Clearly, less number of words will be typed in less time.
So, it is a case of direct proportion.
$\mathrm{Now},\frac{540}{30}=\frac{x}{8}\phantom{\rule{0ex}{0ex}}⇒x=\frac{540×8}{30}\phantom{\rule{0ex}{0ex}}⇒x=144\phantom{\rule{0ex}{0ex}}$

Therefore, Reenu will type 144 words in 8 minutes.

#### Question 1:

Let Reenu type x words in 8 minutes.

 No. of words 540 x Time taken (in min) 30 8

Clearly, less number of words will be typed in less time.
So, it is a case of direct proportion.
$\mathrm{Now},\frac{540}{30}=\frac{x}{8}\phantom{\rule{0ex}{0ex}}⇒x=\frac{540×8}{30}\phantom{\rule{0ex}{0ex}}⇒x=144\phantom{\rule{0ex}{0ex}}$

Therefore, Reenu will type 144 words in 8 minutes.

(i)

(ii)

(iii)

(i)

(ii)

(iii)

#### Question 3:

Let x be the required number of days. Then, we have:

 No. of days 8 x No. of men 35 20

Clearly, less men will take more days to reap the field.
So, it is a case of inverse proportion.

Therefore, 20 men can reap the same field in 14 days.

#### Question 4:

Let x be the required number of days. Then, we have:

 No. of days 8 x No. of men 35 20

Clearly, less men will take more days to reap the field.
So, it is a case of inverse proportion.

Therefore, 20 men can reap the same field in 14 days.

Let x be the required number of men. Then, we have:

 No. of days 8 6 No. of men 12 x

Clearly, more men will require less number of days to dig the pond.
So, it is a case of inverse proportion.

Therefore, 16 men can dig the pond in 6 days.

#### Question 5:

Let x be the required number of men. Then, we have:

 No. of days 8 6 No. of men 12 x

Clearly, more men will require less number of days to dig the pond.
So, it is a case of inverse proportion.

Therefore, 16 men can dig the pond in 6 days.

Let x be the number of days. Then, we have:

 No. of days 28 x No. of cows 6 14

Clearly, more number of cows will take less number of days to graze the field.
So, it is a case of inverse proportion.

Therefore, 14 cows will take 12 days to graze the field.

#### Question 6:

Let x be the number of days. Then, we have:

 No. of days 28 x No. of cows 6 14

Clearly, more number of cows will take less number of days to graze the field.
So, it is a case of inverse proportion.

Therefore, 14 cows will take 12 days to graze the field.

Let x h be the required time taken. Then, we have:

 Speed (in km/h) 60 75 Time (in h) 5 x

Clearly, the higher the speed, the lesser will be the the time taken.
So, it is a case of inverse proportion.

Therefore, the car will reach its destination in 4 h if it travels at a speed of 75 km/h.

#### Question 7:

Let x h be the required time taken. Then, we have:

 Speed (in km/h) 60 75 Time (in h) 5 x

Clearly, the higher the speed, the lesser will be the the time taken.
So, it is a case of inverse proportion.

Therefore, the car will reach its destination in 4 h if it travels at a speed of 75 km/h.

Let x be the number of machines required to produce same number of articles in 48.
Then, we have:

 No. of machines 42 x No. of days 56 48

Clearly, less number of days will require more number of machines.
So, it is a case of inverse proportion.

Therefore, 49 machines would be required to produce the same number of articles in 48 days.

#### Question 8:

Let x be the number of machines required to produce same number of articles in 48.
Then, we have:

 No. of machines 42 x No. of days 56 48

Clearly, less number of days will require more number of machines.
So, it is a case of inverse proportion.

Therefore, 49 machines would be required to produce the same number of articles in 48 days.

Let x be the required number of taps. Then, we have:
1 h = 60 min
i.e., 1 h 36 min = (60+36) min = 96 min

 No. of taps 7 8 Time (in min) 96 x

Clearly, more number of taps will require less time to fill the tank.
So, it is a case of inverse proportion.

Therefore, 8 taps of the same size will take 84 min or 1 h 24 min to fill the tank.

#### Question 9:

Let x be the required number of taps. Then, we have:
1 h = 60 min
i.e., 1 h 36 min = (60+36) min = 96 min

 No. of taps 7 8 Time (in min) 96 x

Clearly, more number of taps will require less time to fill the tank.
So, it is a case of inverse proportion.

Therefore, 8 taps of the same size will take 84 min or 1 h 24 min to fill the tank.

Let x min be the required number of time. Then, we have:

 No. of taps 8 6 Time (in min) 27 $x$

Clearly, less number of taps will take more time to fill the tank .
So, it is a case of inverse proportion.

Therefore, it will take 36 min to fill the tank.

#### Question 10:

Let x min be the required number of time. Then, we have:

 No. of taps 8 6 Time (in min) 27 $x$

Clearly, less number of taps will take more time to fill the tank .
So, it is a case of inverse proportion.

Therefore, it will take 36 min to fill the tank.

Let x be the required number of days. Then, we have:

 No. of days 9 x No. of animals 28 36

Clearly, more number of animals will take less number of days to finish the food.
So, it is a case of inverse proportion.

Therefore, the food will last for 7 days.

#### Question 11:

Let x be the required number of days. Then, we have:

 No. of days 9 x No. of animals 28 36

Clearly, more number of animals will take less number of days to finish the food.
So, it is a case of inverse proportion.

Therefore, the food will last for 7 days.

Let x be the required number of days. Then, we have:

 No. of men 900 1400 No. of days 42 x

Clearly, more men will take less number of days to finish the food.
So, it is a case of inverse proportion.

Therefore, the food will now last for 27 days.

#### Question 12:

Let x be the required number of days. Then, we have:

 No. of men 900 1400 No. of days 42 x

Clearly, more men will take less number of days to finish the food.
So, it is a case of inverse proportion.

Therefore, the food will now last for 27 days.

Let x be the required number of days. Then, we have:

 No. of students 75 60 No. of days 24 x

Clearly, less number of students will take more days to finish the food.
So, it is a case of inverse proportion.

Therefore, the food will now last for 30 days.

#### Question 13:

Let x be the required number of days. Then, we have:

 No. of students 75 60 No. of days 24 x

Clearly, less number of students will take more days to finish the food.
So, it is a case of inverse proportion.

Therefore, the food will now last for 30 days.

Let x min be the duration of each period when the school has 8 periods a day.

 No. of periods 9 8 Time (in min) 40 x

Clearly, if the number of periods reduces, the duration of each period will increase.
So, it is a case of inverse proportion.

Therefore, the duration of each period will be 45 min if there were eight periods a day.

#### Question 14:

Let x min be the duration of each period when the school has 8 periods a day.

 No. of periods 9 8 Time (in min) 40 x

Clearly, if the number of periods reduces, the duration of each period will increase.
So, it is a case of inverse proportion.

Therefore, the duration of each period will be 45 min if there were eight periods a day.

 $x$ 15 9 $y$ 6 ${y}_{1}$

∴ Value of $y=10$, when x =9

#### Question 15:

 $x$ 15 9 $y$ 6 ${y}_{1}$

∴ Value of $y=10$, when x =9

 $x$ 18 ${x}_{1}$ $y$ 8 16

∴ Value of $x=9$

#### Question 1:

 $x$ 18 ${x}_{1}$ $y$ 8 16

∴ Value of $x=9$

Let 22 kg of pulses cost â‚¹x.

 Quantity(in kg) 14 22 Price(in â‚¹) 882 x
As the quantity increases, the price also increases. So, it is a case of direct proportion.
$\therefore \frac{14}{882}=\frac{22}{x}\phantom{\rule{0ex}{0ex}}⇒x=\frac{22×882}{14}=1386$
Thus, the cost of 22 kg of pulses is â‚¹1,386.

Hence, the correct answer is option (d).

#### Question 2:

Let 22 kg of pulses cost â‚¹x.

 Quantity(in kg) 14 22 Price(in â‚¹) 882 x
As the quantity increases, the price also increases. So, it is a case of direct proportion.
$\therefore \frac{14}{882}=\frac{22}{x}\phantom{\rule{0ex}{0ex}}⇒x=\frac{22×882}{14}=1386$
Thus, the cost of 22 kg of pulses is â‚¹1,386.

Hence, the correct answer is option (d).

Let the number of oranges that can be bought for â‚¹169 be x.

 Quantity 8 x Price(in â‚¹) 52 169
As the quantity increases the price also increases. So, this is a case of direct proportion.
$\therefore \frac{8}{52}=\frac{x}{169}\phantom{\rule{0ex}{0ex}}⇒x=\frac{8×169}{52}=26$
Thus, 26 oranges can be bought for â‚¹169.

Hence, the correct answer is option (c).

#### Question 3:

Let the number of oranges that can be bought for â‚¹169 be x.

 Quantity 8 x Price(in â‚¹) 52 169
As the quantity increases the price also increases. So, this is a case of direct proportion.
$\therefore \frac{8}{52}=\frac{x}{169}\phantom{\rule{0ex}{0ex}}⇒x=\frac{8×169}{52}=26$
Thus, 26 oranges can be bought for â‚¹169.

Hence, the correct answer is option (c).

(b) 700

Let x be the number of bottles filled in 5 hours.

 No. of bottles 420 $x$ Time (h) 3 5

More number of bottles will be filled in more time.

Therefore, 700 bottles would be filled in 5 h.

#### Question 4:

(b) 700

Let x be the number of bottles filled in 5 hours.

 No. of bottles 420 $x$ Time (h) 3 5

More number of bottles will be filled in more time.

Therefore, 700 bottles would be filled in 5 h.

(a) 25 km

Let x km be the required distance.
Now, 1 h = 60 min

 Distance (in km) 75 $x$ Time (in min) 60 20

Less distance will be covered in less time.

#### Question 5:

(a) 25 km

Let x km be the required distance.
Now, 1 h = 60 min

 Distance (in km) 75 $x$ Time (in min) 60 20

Less distance will be covered in less time.

(c) 300
Let x sheets weigh 1 kg.
Now, 1 kg = 1000 g

 No. of sheets 12 $x$ Weight (in  g) 40 1000

#### Question 6:

(c) 300
Let x sheets weigh 1 kg.
Now, 1 kg = 1000 g

 No. of sheets 12 $x$ Weight (in  g) 40 1000

(b) 9.8 m
Let x m be the height of the tree.

 Height of object 14 $x$ Length of shadow 10 7

The more the length of the shadow, the more will be the height of the tree.

Therefore, a 9.8 m tall tree will cast a shadow of length 7 m.

#### Question 7:

(b) 9.8 m
Let x m be the height of the tree.

 Height of object 14 $x$ Length of shadow 10 7

The more the length of the shadow, the more will be the height of the tree.

Therefore, a 9.8 m tall tree will cast a shadow of length 7 m.

(c)
Let x cm be the actual length of the bacteria.
The larger the object, the larger its image will be.

Hence, the actual length of the bacteria is â€‹.

#### Question 8:

(c)
Let x cm be the actual length of the bacteria.
The larger the object, the larger its image will be.

Hence, the actual length of the bacteria is â€‹.

(b) 144 min
Let x min be the time taken by 5 pipes to fill the tank.

 No. of pipes 6 5 Time (in min) 120 $x$

Therefore, 5 pipes will take 144 min to fill the tank.

#### Question 9:

(b) 144 min
Let x min be the time taken by 5 pipes to fill the tank.

 No. of pipes 6 5 Time (in min) 120 $x$

Therefore, 5 pipes will take 144 min to fill the tank.

(b) 3 days
Let x be number of days taken by 4 persons to build the wall.

 No. of persons 3 4 No. of days 4 $x$

More number of persons will take less time to build the wall.
So, it is a case of inverse proportion.

Therefore, 4 persons can build the wall in 3 days.

#### Question 10:

(b) 3 days
Let x be number of days taken by 4 persons to build the wall.

 No. of persons 3 4 No. of days 4 $x$

More number of persons will take less time to build the wall.
So, it is a case of inverse proportion.

Therefore, 4 persons can build the wall in 3 days.

(a) 1 h 30 min
Let x h be the time taken by the car travelling at 80 km/hr.

 Speed (km/h) 60 80 Time (in h) 2 $x$

#### Question 1:

(a) 1 h 30 min
Let x h be the time taken by the car travelling at 80 km/hr.

 Speed (km/h) 60 80 Time (in h) 2 $x$

Let x be the required number of boxes.

 No. of boxes 350 $x$ No. of cartons 25 16

Less number of boxes will require less number of cartons.
So, it is a case of direct proportion.

∴ 224 boxes can be placed in 16 cartoons.

#### Question 2:

Let x be the required number of boxes.

 No. of boxes 350 $x$ No. of cartons 25 16

Less number of boxes will require less number of cartons.
So, it is a case of direct proportion.

∴ 224 boxes can be placed in 16 cartoons.

Let Rs x be the cost of 24 tennis balls.

 No. of balls 140 24 Cost of balls 4900 $x$

More tennis balls will cost more.

∴ The cost of 2 dozen tennis balls is Rs 840.

#### Question 3:

Let Rs x be the cost of 24 tennis balls.

 No. of balls 140 24 Cost of balls 4900 $x$

More tennis balls will cost more.

∴ The cost of 2 dozen tennis balls is Rs 840.

Let Rs x be the railway fare for a journey of distance 53 km.

 Distance (in km) 61 53 Railway fare (in rupees) 183 $x$

The lesser the distance, the lesser will be the fare.
So, it is a case of direct proportion .

The railway fare for a journey of distance 53 km is Rs 159.

#### Question 4:

Let Rs x be the railway fare for a journey of distance 53 km.

 Distance (in km) 61 53 Railway fare (in rupees) 183 $x$

The lesser the distance, the lesser will be the fare.
So, it is a case of direct proportion .

The railway fare for a journey of distance 53 km is Rs 159.

Let x people dig the trench in 4 days.

 No. of people 10 $x$ No. of days 6 4

More people will take less number of days to dig the trench. Hence, this is a case of inverse proportion.

∴ 15 people can dig the trench in 4 days.

#### Question 5:

Let x people dig the trench in 4 days.

 No. of people 10 $x$ No. of days 6 4

More people will take less number of days to dig the trench. Hence, this is a case of inverse proportion.

∴ 15 people can dig the trench in 4 days.

Let x be the number of days taken by 21 men to finish the piece of work.

 No. of men 30 21 No. of days 28 $x$

More men will take less time to complete the work.
So, this is a case of inverse proportion.

∴ 21 men will take 40 days to finish the piece of work.

#### Question 6:

Let x be the number of days taken by 21 men to finish the piece of work.

 No. of men 30 21 No. of days 28 $x$

More men will take less time to complete the work.
So, this is a case of inverse proportion.

∴ 21 men will take 40 days to finish the piece of work.

Clearly, the remaining food is sufficient for 200 men for (45 − 15), i.e., 30 days.
Total number of men = 200 + 40 = 240
Let the remaining food last for x days.

 No. of men 200 240 No. of days 30 $x$

Clearly, more men will take less number of days to finish the food.
So, it is a case of inverse proportion.

∴ The remaining food will last for 25 days.

#### Question 7:

Clearly, the remaining food is sufficient for 200 men for (45 − 15), i.e., 30 days.
Total number of men = 200 + 40 = 240
Let the remaining food last for x days.

 No. of men 200 240 No. of days 30 $x$

Clearly, more men will take less number of days to finish the food.
So, it is a case of inverse proportion.

∴ The remaining food will last for 25 days.

(d) 144 minutes

Let one pipe take x min to fill the tank.

 No. of pipe 6 1 Time(in min) 24 $x$

Clearly, one pipe will take more time to fill the tank.
So, it is a case of inverse proportion.

∴ One pipe can fill the tank in 144 minutes.

#### Question 8:

(d) 144 minutes

Let one pipe take x min to fill the tank.

 No. of pipe 6 1 Time(in min) 24 $x$

Clearly, one pipe will take more time to fill the tank.
So, it is a case of inverse proportion.

∴ One pipe can fill the tank in 144 minutes.

(d) 588 days
Let one worker take x days to build the wall.

 No. of workers 14 1 No. of days 42 $x$

Clearly, one worker will take more days to finish the work.
So, it is a case of inverse proportion.

∴ One worker can build the wall in 588 days.

#### Question 9:

(d) 588 days
Let one worker take x days to build the wall.

 No. of workers 14 1 No. of days 42 $x$

Clearly, one worker will take more days to finish the work.
So, it is a case of inverse proportion.

∴ One worker can build the wall in 588 days.

(a) 14 days
Let 20 men take x days to reap the field.

 No. of days 8 $x$ No. of men 35 20

Clearly, less number of men will take more days.
So, it is a case of inverse proportion.

∴ 20 men can reap the field in 14 days.

#### Question 10:

(a) 14 days
Let 20 men take x days to reap the field.

 No. of days 8 $x$ No. of men 35 20

Clearly, less number of men will take more days.
So, it is a case of inverse proportion.

∴ 20 men can reap the field in 14 days.

(b) 72 km
Let x km be the distance covered in 1 h 12 min.
Now, 1 h 12 min = (60+12) min = 72 min

 Distance(in km) 60 $x$ Time(in min) 60 72

More distance will be covered in more time.
So, it is a cas of direct proportion.

∴ The car will cover a distance of 72 $\mathrm{km}$ in 1 h 12 min.â€‹

#### Question 11:

(b) 72 km
Let x km be the distance covered in 1 h 12 min.
Now, 1 h 12 min = (60+12) min = 72 min

 Distance(in km) 60 $x$ Time(in min) 60 72

More distance will be covered in more time.
So, it is a cas of direct proportion.

∴ The car will cover a distance of 72 $\mathrm{km}$ in 1 h 12 min.â€‹

(c) 170 words

Let x be the number of words typed by Rashmi in 10 minutes.

 No. of words 510 $x$ Time(in min) 30 10

Less time will be taken to type less number of words.
So, it is a case of direct variation.

∴ Rashmi will type 170 words in 10 minutes.

#### Question 12:

(c) 170 words

Let x be the number of words typed by Rashmi in 10 minutes.

 No. of words 510 $x$ Time(in min) 30 10

Less time will be taken to type less number of words.
So, it is a case of direct variation.

∴ Rashmi will type 170 words in 10 minutes.

(c) 8

 $x$ 3 ${x}_{1}$ $y$ 36 96

∴ Value of $x=8$

#### Question 13:

(c) 8

 $x$ 3 ${x}_{1}$ $y$ 36 96

∴ Value of $x=8$

(a) 10

 $x$ 15 9 $y$ 6 ${y}_{1}$

∴ Value of y = 10, when x = 9.

#### Question 14:

(a) 10

 $x$ 15 9 $y$ 6 ${y}_{1}$

∴ Value of y = 10, when x = 9.

(i)
Let x be the number of days taken by 4 persons to complete the work.

 No. of days 4 $x$ No. of persons 3 4

Clearly, more workers will take less number of days.
So, it is a case of inverse proportion.

Therefore, 4 persons can do the piece of work in 3 days.

(ii)
Let x min be the time taken by 6 pipes to fill the tank.
 No. of pipes 5 6 Time (in min) 144 $x$

Clearly, more number of pipes will take less time to fill the tank.
So, it is a case of inverse proportion.

∴ 6 pipes can fill the tank in 120 min.

(iii)
Let x min be the time taken by the car travelling at 45 km/h.
Now, 1 h 30 min = (60+30) min
 Speed(in km/hr) 60 45 Time(in min) 90 $x$

Clearly, a car travelling at a less speed will take more time.
So, it is a case of inverse proportion.

∴ The car will take 2 h if it travels at a speed of 45 km/h.

(iv)
Let Rs x be the cost of 5 oranges.
 No. of oranges 8 5 Cost of oranges 20.8 $x$

Clearly, less number of oranges will cost less.
So, it is a case of direct variation.

∴ The cost of 5 oranges is Rs 13.

(v)
Let x be the number of sheets that weigh 500 g.
 No. of sheets 12 $x$ Weight(in grams) 50 500

More number of sheets will weigh more.
So, it is a case of direct variation.

∴ 120 sheets will weigh 500 g.

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