NCERT Solutions for Class 8 Maths Chapter 12 - Direct And Inverse Proportions
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NCERT 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 NCERT 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 NCERT Solutions. All NCERT Solutions for class 8 Maths are prepared by experts and are 100% accurate.

Page No 162:

Question 1:

Answer:

(i)
$Clearly, \frac{x}{y} = \frac{3}{9} = \frac{5}{15} = \frac{8}{24} = \frac{11}{33} = \frac{26}{78} = \frac{1}{3} (constant) Therefore, x and y are proportional .$

(ii)
$Clearly, \frac{x}{y} = \frac{2.5}{10} = \frac{4}{16} = \frac{7.5}{30} = \frac{10}{40} = \frac{1}{4}, while \frac{14}{42} = \frac{1}{3} i . e ., \frac{2.5}{10} = \frac{4}{16} = \frac{7.5}{30} = \frac{10}{40} is not equal to \frac{14}{42} . Therefore, x and y are not proportional .$

(iii)
$Clearly, \frac{x}{y} = \frac{5}{15} = \frac{7}{21} = \frac{9}{27} = \frac{25}{75} = \frac{1}{3}, while \frac{15}{60} = \frac{18}{72} = \frac{1}{4} i . e ., \frac{5}{15} = \frac{7}{21} = \frac{9}{27} = \frac{25}{75} is not equal to \frac{15}{60} and \frac{18}{72} . Therefore, x and y are not proportional .$

Page No 162:

Question 2:

Answer:

$Since x and y are directly propotional, we have : \frac{3}{72} = \frac{x_{1}}{120} = \frac{x_{2}}{192} = \frac{10}{y_{1}} Now, \frac{3}{72} = \frac{x_{1}}{120} \Rightarrow x_{1} = \frac{120 \times 3}{72} = 5$

$And, \frac{3}{72} = \frac{x_{2}}{192} \Rightarrow x_{2} = \frac{3 \times 192}{72} = 8$
$And, \frac{3}{72} = \frac{10}{y_{1}} \Rightarrow y_{1} = \frac{72 \times 10}{3} = 240$
$Therefore, x_{1} = 5, x_{2} = 8 and y_{1} = 240$

Page No 162:

Question 3:

Answer:

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.

Now, \frac{34}{510} = \frac{20}{x} \Rightarrow \frac{1}{15} = \frac{20}{x} \Rightarrow x \times 1 = 20 \times 15 = 300

Therefore, the required distance is 300 km.

Page No 162:

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.

Now, \frac{34}{510} = \frac{20}{x} \Rightarrow \frac{1}{15} = \frac{20}{x} \Rightarrow x \times 1 = 20 \times 15 = 300

Therefore, the required distance is 300 km.

Answer:

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.

∴ \frac{2550}{150} = \frac{x}{124} \Rightarrow x = \frac{2550 \times 124}{150} = 2108

Thus, the taxi charges â‚¹2,108 for the distance of 124 km.

Page No 162:

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.

∴ \frac{2550}{150} = \frac{x}{124} \Rightarrow x = \frac{2550 \times 124}{150} = 2108

Thus, the taxi charges â‚¹2,108 for the distance of 124 km.

Answer:

Let the required distance be x km. Then, we have:
$1 h = 60 \min i . e ., 5 h = 5 \times 60 = 300 \min$ .

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.

Now, \frac{16}{25} = \frac{x}{300} \Rightarrow x = (\frac{16 \times 300}{25}) \Rightarrow x = 192

Therefore, the required distance is 192 km.

Page No 162:

Question 6:

Let the required distance be x km. Then, we have:
$1 h = 60 \min i . e ., 5 h = 5 \times 60 = 300 \min$ .

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.

Now, \frac{16}{25} = \frac{x}{300} \Rightarrow x = (\frac{16 \times 300}{25}) \Rightarrow x = 192

Therefore, the required distance is 192 km.

Answer:

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.

Now, \frac{18}{630} = \frac{x}{455} \Rightarrow \frac{1}{35} = \frac{x}{455} \Rightarrow x = \frac{455}{35} \Rightarrow x = 13

Therefore, 13 dolls can be bought for Rs 455.

Page No 162:

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.

Now, \frac{18}{630} = \frac{x}{455} \Rightarrow \frac{1}{35} = \frac{x}{455} \Rightarrow x = \frac{455}{35} \Rightarrow x = 13

Therefore, 13 dolls can be bought for Rs 455.

Answer:

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

Quantity(in kg)	9	x
Price(in â‚¹)	238.50	371

The price increases as the quantity increases. Thus, this is a case of direct proportion.

∴ \frac{9}{238.50} = \frac{x}{371} \Rightarrow x = \frac{9 \times 371}{238.50} = 14

Thus, the quantity of sugar bought for â‚¹371 is 14 kg.

Page No 162:

Question 8:

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

Quantity(in kg)	9	x
Price(in â‚¹)	238.50	371

The price increases as the quantity increases. Thus, this is a case of direct proportion.

∴ \frac{9}{238.50} = \frac{x}{371} \Rightarrow x = \frac{9 \times 371}{238.50} = 14

Thus, the quantity of sugar bought for â‚¹371 is 14 kg.

Answer:

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.

Now, \frac{15}{981} = \frac{x}{1308} \Rightarrow x = \frac{15 \times 1308}{981} \Rightarrow x = 20

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

Page No 163:

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.

Now, \frac{15}{981} = \frac{x}{1308} \Rightarrow x = \frac{15 \times 1308}{981} \Rightarrow x = 20

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

Answer:

Let x m be the length of the model of the ship. Then, we have:
$1 m = 100 cm Therefore, 15 m = 1500 cm 35 m = 3500 cm$

	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.

Now, \frac{1500}{9} = \frac{3500}{x} \Rightarrow x = \frac{3500 \times 9}{1500} \Rightarrow x = 21 cm

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

Page No 163:

Question 10:

Let x m be the length of the model of the ship. Then, we have:
$1 m = 100 cm Therefore, 15 m = 1500 cm 35 m = 3500 cm$

	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.

Now, \frac{1500}{9} = \frac{3500}{x} \Rightarrow x = \frac{3500 \times 9}{1500} \Rightarrow x = 21 cm

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

Answer:

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

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

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

Now, \frac{8}{6.4 \times 10^{7}} = \frac{15}{x} \Rightarrow x = \frac{15 \times 6.4 \times 10^{7}}{8} \Rightarrow x = 12 \times 10^{7}

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

Page No 163:

Question 11:

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

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

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

Now, \frac{8}{6.4 \times 10^{7}} = \frac{15}{x} \Rightarrow x = \frac{15 \times 6.4 \times 10^{7}}{8} \Rightarrow x = 12 \times 10^{7}

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

Answer:

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

$1 h = 60 \min i . e ., 1 h 12 \min = (60 + 12) \min = 72 \min$

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.

Now, \frac{50}{60} = \frac{x}{72} \Rightarrow x = \frac{50 \times 72}{60} \Rightarrow x = 60

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

Page No 163:

Question 12:

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

$1 h = 60 \min i . e ., 1 h 12 \min = (60 + 12) \min = 72 \min$

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.

Now, \frac{50}{60} = \frac{x}{72} \Rightarrow x = \frac{50 \times 72}{60} \Rightarrow x = 60

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

Answer:

Let x km be the required distance covered by Ravi in 2 h 24 min.
Then, we have:
$1 h = 60 \min i . e ., 2 h 24 \min = (120 + 24) \min = 144 \min$

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.

Now, \frac{5}{60} = \frac{x}{144} \Rightarrow x = \frac{5 \times 144}{60} \Rightarrow x = 12

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

Page No 163:

Question 13:

Let x km be the required distance covered by Ravi in 2 h 24 min.
Then, we have:
$1 h = 60 \min i . e ., 2 h 24 \min = (120 + 24) \min = 144 \min$

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.

Now, \frac{5}{60} = \frac{x}{144} \Rightarrow x = \frac{5 \times 144}{60} \Rightarrow x = 12

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

Answer:

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.

Now, \frac{65}{12} = \frac{x}{312} \Rightarrow x = \frac{65 \times 312}{12} \Rightarrow x = 1690

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

Page No 163:

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.

Now, \frac{65}{12} = \frac{x}{312} \Rightarrow x = \frac{65 \times 312}{12} \Rightarrow x = 1690

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

Answer:

Let x be the required number of men.

$Now, 6 \frac{3}{4} m = \frac{27}{4} m$

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.

Now, \frac{11}{\frac{27}{4}} = \frac{x}{27} \Rightarrow \frac{11 \times 4}{27} = \frac{x}{27} \Rightarrow x = 44

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

Page No 163:

Question 15:

Let x be the required number of men.

$Now, 6 \frac{3}{4} m = \frac{27}{4} m$

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.

Now, \frac{11}{\frac{27}{4}} = \frac{x}{27} \Rightarrow \frac{11 \times 4}{27} = \frac{x}{27} \Rightarrow x = 44

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

Answer:

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.

Now, \frac{540}{30} = \frac{x}{8} \Rightarrow x = \frac{540 \times 8}{30} \Rightarrow x = 144

Therefore, Reenu will type 144 words in 8 minutes.

Page No 165:

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.

Now, \frac{540}{30} = \frac{x}{8} \Rightarrow x = \frac{540 \times 8}{30} \Rightarrow x = 144

Therefore, Reenu will type 144 words in 8 minutes.

Answer:

(i)
$Clearly, 6 \times 9 \neq 10 \times 15 \neq 14 \times 21 \neq 16 \times 24 T herefore, x and y are not inversely proportional .$

(ii)
$Clearly, 5 \times 18 = 9 \times 10 = 15 \times 6 = 3 \times 30 = 45 \times 2 = 90 = (consant) Therefore, x and y are inversely proportional .$

(iii)
$Clearly, 9 \times 4 = 3 \times 12 = 36 \times 1 = 36, while 6 \times 9 = 54 i . e ., 9 \times 4 = 3 \times 12 = 36 \times 1 \neq 6 \times 9 Therefore, x and y are not inversely proportional .$

Page No 165:

Question 2:

Answer:

$Since x and y are inversely proportional, x y must be a constant .$
$Therefore, 8 \times y_{1} = x_{1} \times 4 = 16 \times 5 = x_{2} \times 2 = 80 \times y_{2} Now, 16 \times 5 = 8 \times y_{1} \Rightarrow \frac{80}{8} = y_{1} ∴ y_{1} = 10 16 \times 5 = x_{1} \times 4 \Rightarrow \frac{80}{4} = x_{1} ∴ x_{1} = 20 16 \times 5 = x_{2} \times 2 \Rightarrow \frac{80}{2} = x_{2} ∴ x_{2} = 40 16 \times 5 = 80 \times y_{2} \Rightarrow \frac{80}{80} = y_{2} ∴ y_{2} = 1 Hence, y_{1} = 10, x_{1} = 20, x_{2} = 40 and y_{2} = 1$

Page No 165:

Question 3:

Answer:

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.

Now, 8 \times 35 = x \times 20 \Rightarrow \frac{8 \times 35}{20} = x \Rightarrow 14 = x

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

Page No 165:

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.

Now, 8 \times 35 = x \times 20 \Rightarrow \frac{8 \times 35}{20} = x \Rightarrow 14 = x

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

Answer:

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.

Now, 8 \times 12 = 6 \times x \Rightarrow x = \frac{8 \times 12}{6} \Rightarrow x = 16

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

Page No 166:

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.

Now, 8 \times 12 = 6 \times x \Rightarrow x = \frac{8 \times 12}{6} \Rightarrow x = 16

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

Answer:

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.

Now, 28 \times 6 = x \times 14 \Rightarrow x = \frac{28 \times 6}{14} \Rightarrow x = 12

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

Page No 166:

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.

Now, 28 \times 6 = x \times 14 \Rightarrow x = \frac{28 \times 6}{14} \Rightarrow x = 12

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

Answer:

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.

Now, 60 \times 5 = 75 \times x \Rightarrow x = \frac{60 \times 5}{75} \Rightarrow x = 4

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

Page No 166:

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.

Now, 60 \times 5 = 75 \times x \Rightarrow x = \frac{60 \times 5}{75} \Rightarrow x = 4

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

Answer:

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.

Now, 42 \times 56 = x \times 48 \Rightarrow x = \frac{42 \times 56}{48} \Rightarrow x = 49

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

Page No 166:

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.

Now, 42 \times 56 = x \times 48 \Rightarrow x = \frac{42 \times 56}{48} \Rightarrow x = 49

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

Answer:

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.

Now, 7 \times 96 = 8 \times x \Rightarrow x = \frac{7 \times 96}{8} \Rightarrow x = 84

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

Page No 166:

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.

Now, 7 \times 96 = 8 \times x \Rightarrow x = \frac{7 \times 96}{8} \Rightarrow x = 84

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

Answer:

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.

Now, 8 \times 27 = 6 \times x \Rightarrow x = \frac{8 \times 27}{6} \Rightarrow x = 36

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

Page No 166:

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.

Now, 8 \times 27 = 6 \times x \Rightarrow x = \frac{8 \times 27}{6} \Rightarrow x = 36

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

Answer:

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.

Now, 9 \times 28 = x \times 36 \Rightarrow x = \frac{9 \times 28}{36} \Rightarrow x = 7

Therefore, the food will last for 7 days.

Page No 166:

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.

Now, 9 \times 28 = x \times 36 \Rightarrow x = \frac{9 \times 28}{36} \Rightarrow x = 7

Therefore, the food will last for 7 days.

Answer:

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.

Now, 900 \times 42 = 1400 \times x \Rightarrow x = \frac{900 \times 42}{1400} \Rightarrow x = 27

Therefore, the food will now last for 27 days.

Page No 166:

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.

Now, 900 \times 42 = 1400 \times x \Rightarrow x = \frac{900 \times 42}{1400} \Rightarrow x = 27

Therefore, the food will now last for 27 days.

Answer:

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.

Now, 75 \times 24 = 60 \times x \Rightarrow x = \frac{75 \times 24}{60} \Rightarrow x = 30

Therefore, the food will now last for 30 days.

Page No 166:

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.

Now, 75 \times 24 = 60 \times x \Rightarrow x = \frac{75 \times 24}{60} \Rightarrow x = 30

Therefore, the food will now last for 30 days.

Answer:

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.

Now, 9 \times 40 = 8 \times x \Rightarrow x = \frac{9 \times 40}{8} \Rightarrow x = 45

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

Page No 166:

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.

Now, 9 \times 40 = 8 \times x \Rightarrow x = \frac{9 \times 40}{8} \Rightarrow x = 45

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

Answer:

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

x and y var y inversely . i . e . x y = constant Now, 15 \times 6 = 9 \times y_{1} \Rightarrow y_{1} = \frac{15 \times 6}{9} \Rightarrow y_{1} = 10

∴ Value of

y = 10

, when x =9

Page No 166:

Question 15:

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

x and y var y inversely . i . e . x y = constant Now, 15 \times 6 = 9 \times y_{1} \Rightarrow y_{1} = \frac{15 \times 6}{9} \Rightarrow y_{1} = 10

∴ Value of

y = 10

, when x =9

Answer:

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

x and y var y inversely . i . e . x y = constant Now, 18 \times 8 = x_{1} \times 16 \Rightarrow \frac{18 \times 8}{16} = x_{1} \Rightarrow 9 = x_{1}

∴ Value of

x = 9

Page No 166:

Question 1:

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

x and y var y inversely . i . e . x y = constant Now, 18 \times 8 = x_{1} \times 16 \Rightarrow \frac{18 \times 8}{16} = x_{1} \Rightarrow 9 = x_{1}

∴ Value of

x = 9

Answer:

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.

∴ \frac{14}{882} = \frac{22}{x} \Rightarrow x = \frac{22 \times 882}{14} = 1386

Thus, the cost of 22 kg of pulses is â‚¹1,386.

Hence, the correct answer is option (d).

Page No 166:

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.

∴ \frac{14}{882} = \frac{22}{x} \Rightarrow x = \frac{22 \times 882}{14} = 1386

Thus, the cost of 22 kg of pulses is â‚¹1,386.

Hence, the correct answer is option (d).

Answer:

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.

∴ \frac{8}{52} = \frac{x}{169} \Rightarrow x = \frac{8 \times 169}{52} = 26

Thus, 26 oranges can be bought for â‚¹169.

Hence, the correct answer is option (c).

Page No 166:

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.

∴ \frac{8}{52} = \frac{x}{169} \Rightarrow x = \frac{8 \times 169}{52} = 26

Thus, 26 oranges can be bought for â‚¹169.

Hence, the correct answer is option (c).

Answer:

(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.

Now, \frac{420}{3} = \frac{x}{5} \Rightarrow x = \frac{420 \times 5}{3} \Rightarrow x = 700

Therefore, 700 bottles would be filled in 5 h.

Page No 166:

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.

Now, \frac{420}{3} = \frac{x}{5} \Rightarrow x = \frac{420 \times 5}{3} \Rightarrow x = 700

Therefore, 700 bottles would be filled in 5 h.

Answer:

(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.

Now, \frac{75}{60} = \frac{x}{20} \Rightarrow x = \frac{75 \times 20}{60} \Rightarrow x = 25 km

Page No 166:

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.

Now, \frac{75}{60} = \frac{x}{20} \Rightarrow x = \frac{75 \times 20}{60} \Rightarrow x = 25 km

Answer:

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

Now, \frac{12}{40} = \frac{x}{1000} \Rightarrow x = \frac{12 \times 1000}{40} \Rightarrow x = 300

Page No 166:

Question 6:

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

Now, \frac{12}{40} = \frac{x}{1000} \Rightarrow x = \frac{12 \times 1000}{40} \Rightarrow x = 300

Answer:

(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.

Now, \frac{14}{10} = \frac{x}{7} \Rightarrow x = \frac{14 \times 7}{10} \Rightarrow x = 9.8

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

Page No 167:

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.

Now, \frac{14}{10} = \frac{x}{7} \Rightarrow x = \frac{14 \times 7}{10} \Rightarrow x = 9.8

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

Answer:

(c) $10^{- 4} cm$
Let x cm be the actual length of the bacteria.
The larger the object, the larger its image will be.
$Now, \frac{x}{1} = \frac{5}{50000} = 10^{- 4} cm$

Hence, the actual length of the bacteria is â€‹ $10^{- 4} cm$ .

Page No 167:

Question 8:

Answer:

(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	$x$

Now, 6 \times 120 = 5 \times x \Rightarrow x = 144

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

Page No 167:

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	$x$

Now, 6 \times 120 = 5 \times x \Rightarrow x = 144

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

Answer:

(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.

Now, 3 \times 4 = 4 \times x \Rightarrow x = 3

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

Page No 167:

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.

Now, 3 \times 4 = 4 \times x \Rightarrow x = 3

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

Answer:

(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	$x$

The greater the speed, the lesser will be the time taken . So, it is a case of inverse proportion . Now, 60 \times 2 = 80 \times x \Rightarrow x = \frac{120}{80} \Rightarrow x = 1.5 Therefore, the car will take 1 h 30 \min to reach its destination if it travels at a speed of 80 km / h .

Page No 168:

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	$x$

The greater the speed, the lesser will be the time taken . So, it is a case of inverse proportion . Now, 60 \times 2 = 80 \times x \Rightarrow x = \frac{120}{80} \Rightarrow x = 1.5 Therefore, the car will take 1 h 30 \min to reach its destination if it travels at a speed of 80 km / h .

Answer:

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.

Now, \frac{350}{25} = \frac{x}{16} \Rightarrow x = \frac{350 \times 16}{25} \Rightarrow x = 224

∴ 224 boxes can be placed in 16 cartoons.

Page No 168:

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.

Now, \frac{350}{25} = \frac{x}{16} \Rightarrow x = \frac{350 \times 16}{25} \Rightarrow x = 224

∴ 224 boxes can be placed in 16 cartoons.

Answer:

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.

Now, \frac{140}{4900} = \frac{24}{x} \Rightarrow x = \frac{24 \times 4900}{140} \Rightarrow x = 840

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

Page No 168:

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.

Now, \frac{140}{4900} = \frac{24}{x} \Rightarrow x = \frac{24 \times 4900}{140} \Rightarrow x = 840

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

Answer:

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 .

Now, \frac{61}{183} = \frac{53}{x} \Rightarrow x = \frac{53 \times 183}{61} \Rightarrow x = 159

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

Page No 168:

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 .

Now, \frac{61}{183} = \frac{53}{x} \Rightarrow x = \frac{53 \times 183}{61} \Rightarrow x = 159

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

Answer:

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.

Now, 10 \times 6 = x \times 4 \Rightarrow x = \frac{60}{4} \Rightarrow x = 15

∴ 15 people can dig the trench in 4 days.

Page No 168:

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.

Now, 10 \times 6 = x \times 4 \Rightarrow x = \frac{60}{4} \Rightarrow x = 15

∴ 15 people can dig the trench in 4 days.

Answer:

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.

Now, 30 \times 28 = 21 \times x \Rightarrow x = \frac{30 \times 28}{21} \Rightarrow x = 40

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

Page No 168:

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.

Now, 30 \times 28 = 21 \times x \Rightarrow x = \frac{30 \times 28}{21} \Rightarrow x = 40

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

Answer:

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.

Now, 200 \times 30 = 240 \times x \Rightarrow x = \frac{200 \times 30}{240} \Rightarrow x = 25

∴ The remaining food will last for 25 days.

Page No 168:

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.

Now, 200 \times 30 = 240 \times x \Rightarrow x = \frac{200 \times 30}{240} \Rightarrow x = 25

∴ The remaining food will last for 25 days.

Answer:

(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.

Now, 6 \times 24 = 1 \times x \Rightarrow x = 6 \times 24 \Rightarrow x = 144

∴ One pipe can fill the tank in 144 minutes.

Page No 168:

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.

Now, 6 \times 24 = 1 \times x \Rightarrow x = 6 \times 24 \Rightarrow x = 144

∴ One pipe can fill the tank in 144 minutes.

Answer:

(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.

Now, 14 \times 42 = 1 \times x \Rightarrow x = 14 \times 42 \Rightarrow x = 588

∴ One worker can build the wall in 588 days.

Page No 168:

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.

Now, 14 \times 42 = 1 \times x \Rightarrow x = 14 \times 42 \Rightarrow x = 588

∴ One worker can build the wall in 588 days.

Answer:

(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.

Now, 8 \times 35 = x \times 20 \Rightarrow x = \frac{8 \times 35}{20} \Rightarrow x = 14

∴ 20 men can reap the field in 14 days.

Page No 168:

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.

Now, 8 \times 35 = x \times 20 \Rightarrow x = \frac{8 \times 35}{20} \Rightarrow x = 14

∴ 20 men can reap the field in 14 days.

Answer:

(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.

Now, \frac{60}{60} = \frac{x}{72} \Rightarrow x = 72 km

∴ The car will cover a distance of 72

km

in 1 h 12 min.â€‹

Page No 168:

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.

Now, \frac{60}{60} = \frac{x}{72} \Rightarrow x = 72 km

∴ The car will cover a distance of 72

km

in 1 h 12 min.â€‹

Answer:

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.

Now, \frac{510}{30} = \frac{x}{10} \Rightarrow x = 170

∴ Rashmi will type 170 words in 10 minutes.

Page No 168:

Question 12:

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.

Now, \frac{510}{30} = \frac{x}{10} \Rightarrow x = 170

∴ Rashmi will type 170 words in 10 minutes.

Answer:

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

x and y var y directly . Then x = k y, where k is the constant of proportionality . \Rightarrow k = \frac{x}{y} Now, \frac{3}{36} = \frac{x_{1}}{96} \Rightarrow \frac{96 \times 3}{36} = x_{1} \Rightarrow 8 = x_{1}

∴ Value of

x = 8

Page No 168:

Question 13:

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

x and y var y directly . Then x = k y, where k is the constant of proportionality . \Rightarrow k = \frac{x}{y} Now, \frac{3}{36} = \frac{x_{1}}{96} \Rightarrow \frac{96 \times 3}{36} = x_{1} \Rightarrow 8 = x_{1}

∴ Value of

x = 8

Answer:

(a) 10

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

Since x and y var y inversely, x y = constant . Now, 15 \times 6 = 9 \times y_{1} \Rightarrow \frac{90}{9} = y_{1} \Rightarrow 10 = y_{1}

∴ Value of y = 10, when x = 9.

Page No 168:

Question 14:

(a) 10

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

Since x and y var y inversely, x y = constant . Now, 15 \times 6 = 9 \times y_{1} \Rightarrow \frac{90}{9} = y_{1} \Rightarrow 10 = y_{1}

∴ Value of y = 10, when x = 9.

Answer:

(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.

Now, 4 \times 3 = x \times 4 \Rightarrow x = 3

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.

Now, 5 \times 144 = 6 \times x \Rightarrow x = \frac{5 \times 144}{6} \Rightarrow x = 120 \min

∴ 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.

Now, 60 \times 90 = 45 \times x \Rightarrow x = \frac{60 \times 90}{45} \Rightarrow x = 120 \min = 2 h

∴ 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	20.80	$x$

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

Now, \frac{8}{20.80} = \frac{5}{x} \Rightarrow x = \frac{5 \times 20.80}{8} \Rightarrow x = 13

∴ 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.

Now, \frac{12}{50} = \frac{x}{500} \Rightarrow x = \frac{12 \times 500}{50} \Rightarrow x = 120

∴ 120 sheets will weigh 500 g.

View NCERT Solutions for all chapters of Class 8