RS Aggarwal 2018 Solutions for Class 8 Math 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 Math Direct And Inverse Proportions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2018 Book of class 8 Math Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal 2018 Solutions. All RS Aggarwal 2018 Solutions for class 8 Math are prepared by experts and are 100% accurate.

#### Page No 162:

#### Question 1:

Observe the tables given below and in each one find whether *x* and *y* are proportional:

(i)

x |
3 | 5 | 8 | 11 | 26 |

y |
9 | 15 | 24 | 33 | 78 |

(ii)

x |
2.5 | 4 | 7.5 | 10 | 14 |

y |
10 | 16 | 30 | 40 | 42 |

(iii)

x |
5 | 7 | 9 | 15 | 18 | 25 |

y |
15 | 21 | 27 | 60 | 72 | 75 |

#### Answer:

(i)

$\mathrm{Clearly},\frac{x}{y}=\frac{3}{9}=\frac{5}{15}=\frac{8}{24}=\frac{11}{33}=\frac{26}{78}=\frac{1}{3}\left(\mathrm{constant}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{are}\mathrm{proportional}.$

(ii)

$\mathrm{Clearly},\frac{x}{y}=\frac{2.5}{10}=\frac{4}{16}=\frac{7.5}{30}=\frac{10}{40}=\frac{1}{4},\mathrm{while}\frac{14}{42}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.,\frac{2.5}{10}=\frac{4}{16}=\frac{7.5}{30}=\frac{10}{40}\mathrm{is}\mathrm{not}\mathrm{equal}\mathrm{to}\frac{14}{42}.\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},x\mathrm{and}y\mathrm{are}\mathrm{not}\mathrm{proportional}.$

(iii)

$\mathrm{Clearly},\frac{x}{y}=\frac{5}{15}=\frac{7}{21}=\frac{9}{27}=\frac{25}{75}=\frac{1}{3},\mathrm{while}\frac{15}{60}=\frac{18}{72}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.,\frac{5}{15}=\frac{7}{21}=\frac{9}{27}=\frac{25}{75}\mathrm{is}\mathrm{not}\mathrm{equal}\mathrm{to}\frac{15}{60}\mathrm{and}\frac{18}{72}.\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},x\mathrm{and}y\mathrm{are}\mathrm{not}\mathrm{proportional}.$

#### Page No 162:

#### Question 2:

If *x *and *y* are directly proportional, find the values of *x*_{1} , *x*_{2} and *y*_{1} in the table given below:

x |
3 | x_{1} |
x_{2} |
10 |

y |
72 | 120 | 192 | y_{1} |

#### Answer:

_{}

$\mathrm{And},\frac{3}{72}=\frac{{x}_{2}}{192}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{2}=\frac{3\times 192}{72}=8$

$\mathrm{And},\frac{3}{72}{=}\frac{10}{{y}_{1}}\phantom{\rule{0ex}{0ex}}\Rightarrow {y}_{1}=\frac{72\times 10}{3}=240$

$\mathrm{Therefore},{x}_{1}=5,{x}_{2}=8\mathrm{and}{y}_{1}=240$

#### Page No 162:

#### Question 3:

A truck covers a distance of 510 km in 34 litres of diesel. How much distance would it cover in 20 litres of diesel?

#### 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.

$\mathrm{Now},\frac{34}{510}=\frac{20}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{15}=\frac{20}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x\times 1=20\times 15=300\phantom{\rule{0ex}{0ex}}$

Therefore, the required distance is 300 km.

#### Page No 162:

#### Question 4:

A taxi charges a fare of Rs 2550 for a journey of 150 km. How much would it charge for a journey of 124 km?

#### Answer:

Let the charge for a journey of 124 km be ₹*x*.

Price(in ₹) | 2550 | x |

Distance(in km) | 150 | 124 |

So, it is a case of direct proportion.

$\therefore \frac{2550}{150}=\frac{x}{124}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{2550\times 124}{150}=2108$

Thus, the taxi charges ₹2,108 for the distance of 124 km.

#### Page No 162:

#### Question 5:

A loaded truck covers 16 km in 25 minutes. At the same speed, how far can it travel in 5 hours?

#### Answer:

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

$1\mathrm{h}=60\mathrm{min}\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.,5\mathrm{h}=5\times 60=300\mathrm{min}$.

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.

$\mathrm{Now},\frac{16}{25}=\frac{x}{300}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\left(\frac{16\times 300}{25}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow x=192\phantom{\rule{0ex}{0ex}}$

Therefore, the required distance is 192 km.

#### Page No 162:

#### Question 6:

If 18 dolls cost Rs 630, how many dolls can be bought for Rs 455?

#### 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.

$\mathrm{Now},\frac{18}{630}=\frac{x}{455}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{35}=\frac{x}{455}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{455}{35}\phantom{\rule{0ex}{0ex}}\Rightarrow x=13\phantom{\rule{0ex}{0ex}}$

Therefore, 13 dolls can be bought for Rs 455.

#### Page No 162:

#### Question 7:

If 9 kg of sugar costs ₹ 238.50, how much sugar can be bought for ₹ 371?

#### Answer:

Let the quantity of sugar bought for ₹371 be *x *kg.

Quantity(in kg) | 9 | x |

Price(in ₹) | 238.50 | 371 |

$\therefore \frac{9}{238.50}=\frac{x}{371}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{9\times 371}{238.50}=14$

Thus, the quantity of sugar bought for ₹371 is 14 kg.

#### Page No 162:

#### Question 8:

The cost of 15 metres of a cloth is Rs 981. What length of this cloth can be purchased for Rs 1308?

#### Answer:

Let the length of cloth be *x* m. Then, we have:

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.

$\mathrm{Now},\frac{15}{981}=\frac{x}{1308}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{15\times 1308}{981}\phantom{\rule{0ex}{0ex}}\Rightarrow x=20\phantom{\rule{0ex}{0ex}}$

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

#### Page No 163:

#### Question 9:

In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 15m high. If the length of the ship is 35 metres, how long is the model ship?

#### Answer:

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

$1\mathrm{m}=100\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},15\mathrm{m}=1500\mathrm{cm}\phantom{\rule{0ex}{0ex}}35\mathrm{m}=3500\mathrm{cm}$

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.

$\mathrm{Now},\frac{1500}{9}=\frac{3500}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{3500\times 9}{1500}\phantom{\rule{0ex}{0ex}}\Rightarrow x=21\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

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

#### Page No 163:

#### Question 10:

In 8 days, the earth picks up (6.4 × 10^{7}) kg of dust from the atmosphere. How much dust will it pick up in 15 days?

#### 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.

$\mathrm{Now},\frac{8}{6.4\times {10}^{7}}=\frac{15}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{15\times 6.4\times {10}^{7}}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow x=12\times {10}^{7}\phantom{\rule{0ex}{0ex}}$

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

#### Page No 163:

#### Question 11:

A car is travelling at the average speed of 50 km/hr. How much distance would it travel in 1 hour 12 minutes?

#### Answer:

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

$1\mathrm{h}=60\mathrm{min}\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.,1\mathrm{h}12\mathrm{min}=(60+12)\mathrm{min}=72\mathrm{min}$

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.

$\mathrm{Now},\frac{50}{60}=\frac{x}{72}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{50\times 72}{60}\phantom{\rule{0ex}{0ex}}\Rightarrow x=60\phantom{\rule{0ex}{0ex}}$

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

#### Page No 163:

#### Question 12:

Ravi walks at the uniform rate of 5 km/hr. What distance would he cover in 2 hours 24 minutes?

#### Answer:

Let *x* km be the required distance covered by Ravi in 2 h 24 min.

Then, we have:

$1\mathrm{h}=60\mathrm{min}\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.,2\mathrm{h}24\mathrm{min}=(120+24)\mathrm{min}=144\mathrm{min}$

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.

$\mathrm{Now},\frac{5}{60}=\frac{x}{144}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{5\times 144}{60}\phantom{\rule{0ex}{0ex}}\Rightarrow x=12\phantom{\rule{0ex}{0ex}}$

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

#### Page No 163:

#### Question 13:

If the thickness of a pile of 12 cardboards is 65 mm, find the thickness of a pile of 312 such cardboards.

#### 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.

$\mathrm{Now},\frac{65}{12}=\frac{x}{312}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{65\times 312}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow x=1690\phantom{\rule{0ex}{0ex}}$

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

#### Page No 163:

#### Question 14:

11 men can dig $6\frac{3}{4}$-metre-long trench in one day. How many men should be employed for digging 27-metre-long trench of the same type in one day?

#### Answer:

Let *x *be the required number of men.

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

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.

$\mathrm{Now},\frac{11}{{\displaystyle \frac{27}{4}}}=\frac{x}{27}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{11\times 4}{27}=\frac{x}{27}\phantom{\rule{0ex}{0ex}}\Rightarrow x=44\phantom{\rule{0ex}{0ex}}$

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

#### Page No 163:

#### Question 15:

Reenu types 540 words during half an hour. How many words would she type in 8 minutes?

#### 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.

$\mathrm{Now},\frac{540}{30}=\frac{x}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{540\times 8}{30}\phantom{\rule{0ex}{0ex}}\Rightarrow x=144\phantom{\rule{0ex}{0ex}}$

Therefore, Reenu will type 144 words in 8 minutes.

#### Page No 165:

#### Question 1:

Observe the tables given below and in each case find whether *x* and *y* are inversely proportional:

(i)

x |
6 | 10 | 14 | 16 |

y |
9 | 15 | 21 | 24 |

(ii)

x |
5 | 9 | 15 | 3 | 45 |

y |
18 | 10 | 6 | 30 | 2 |

(iii)

x |
9 | 3 | 6 | 36 |

y |
4 | 12 | 9 | 1 |

#### Answer:

(i)

$\mathrm{Clearly},6\times 9\ne 10\times 15\ne 14\times 21\ne 16\times 24\phantom{\rule{0ex}{0ex}}\mathrm{T}\mathrm{herefore},x\mathrm{and}y\mathrm{are}\mathrm{not}\mathrm{inversely}\mathrm{proportional}.$

(ii)

$\mathrm{Clearly},5\times 18=9\times 10=15\times 6=3\times 30=45\times 2=90=\left(\mathrm{consant}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},x\mathrm{and}y\mathrm{are}\mathrm{inversely}\mathrm{proportional}.$

(iii)

$\mathrm{Clearly},9\times 4=3\times 12=36\times 1=36,\mathrm{while}6\times 9=54\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.,9\times 4=3\times 12=36\times 1\ne 6\times 9\phantom{\rule{0ex}{0ex}}{\mathrm{Therefore}}{,}{}{x}{}{\mathrm{and}}{}{y}{}{\mathrm{are}}{}{\mathrm{not}}{}{\mathrm{inversely}}{}{\mathrm{proportional}}{.}$

#### Page No 165:

#### Question 2:

If *x* and *y* are inversely proportional, find the values of *x*_{1}, *x*_{2}, *y*_{1} and *y*_{2} in the table given below:

x |
8 | x_{1} |
16 | x_{2} |
80 |

y |
y_{1} |
4 | 5 | 2 | y_{2} |

#### Answer:

$\mathrm{Since}x\mathrm{and}y\mathrm{are}\mathrm{inversely}\mathrm{proportional},xy\mathrm{must}\mathrm{be}\mathrm{a}\mathrm{constant}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$\mathrm{Therefore},8\times {y}_{1}={x}_{1}\times 4=16\times 5={x}_{2}\times 2=80\times {y}_{2}\phantom{\rule{0ex}{0ex}}\mathrm{Now},16\times 5=8\times {y}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{80}{8}={y}_{1}\phantom{\rule{0ex}{0ex}}\therefore {y}_{1}=10\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}16\times 5={x}_{1}\times 4\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{80}{4}={x}_{1}\phantom{\rule{0ex}{0ex}}\therefore {x}_{1}=20\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}16\times 5={x}_{2}\times 2\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{80}{2}={x}_{2}\phantom{\rule{0ex}{0ex}}\therefore {x}_{2}=40\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}16\times 5=80\times {y}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{80}{80}={y}_{2}\phantom{\rule{0ex}{0ex}}\therefore {y}_{2}=1\phantom{\rule{0ex}{0ex}}\mathrm{Hence},{y}_{1}=10,{x}_{1}=20,{x}_{2}=40\mathrm{and}{y}_{2}=1$

#### Page No 165:

#### Question 3:

If 35 men can reap a field in 8 days, in how many days can 20 men reap the same field?

#### 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.

$\mathrm{Now},8\times 35=x\times 20\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{8\times 35}{20}=x\phantom{\rule{0ex}{0ex}}\Rightarrow 14=x$

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

#### Page No 165:

#### Question 4:

12 men can dig a pond in 8 days. How many men can dig it in 6 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.

$\mathrm{Now},8\times 12=6\times x\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{8\times 12}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow x=16$

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

#### Page No 166:

#### Question 5:

6 cows can graze a field in 28 days. How long would 14 cows take to graze the same field?

#### 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.

$\mathrm{Now},28\times 6=x\times 14\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{28\times 6}{14}\phantom{\rule{0ex}{0ex}}\Rightarrow x=12$

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

#### Page No 166:

#### Question 6:

A car takes 5 hours to reach a destination by travelling at the speed of 60 km/hr. How long will it take when the car travels at the speed of 75 km/hr?

#### 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.

$\mathrm{Now},60\times 5=75\times x\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{60\times 5}{75}\phantom{\rule{0ex}{0ex}}\Rightarrow x=4$

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:

A factory requires 42 machines to produce a given number of articles in 56 days. How many machines would be required to produce the same number of articles in 48 days?

#### 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.

$\mathrm{Now},42\times 56=x\times 48\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{42\times 56}{48}\phantom{\rule{0ex}{0ex}}\Rightarrow x=49$

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

#### Page No 166:

#### Question 8:

7 teps of the same size fill a tank in 1 hour 36 minutes. How long will 8 taps of the same size take to fill the tank?

#### 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.

$\mathrm{Now},7\times 96=8\times x\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{7\times 96}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow x=84$

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:

8 taps of the same size fill a tank in 27 minutes. If two taps go out of order, how long would the remaining taps take 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.

$\mathrm{Now},8\times 27=6\times x\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{8\times 27}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow x=36$

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

#### Page No 166:

#### Question 10:

A farmer has enough food to feed 28 animals in his cattle for 9 days. How long would the food last, if there were 8 more animals in his cattle?

#### 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.

$\mathrm{Now},9\times 28=x\times 36\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{9\times 28}{36}\phantom{\rule{0ex}{0ex}}\Rightarrow x=7$

Therefore, the food will last for 7 days.

#### Page No 166:

#### Question 11:

A garrison of 900 men had provisions for 42 days. However, a reinforcement of 500 men arrived. For how many days will the food last now?

#### 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.

$\mathrm{Now},900\times 42=1400\times x\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{900\times 42}{1400}\phantom{\rule{0ex}{0ex}}\Rightarrow x=27$

Therefore, the food will now last for 27 days.

#### Page No 166:

#### Question 12:

In a hostel, 75 students had food provision for 24 days. If 15 students leave the hostel, for how many days would the food provision last?

#### 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.

$\mathrm{Now},75\times 24=60\times x\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{75\times 24}{60}\phantom{\rule{0ex}{0ex}}\Rightarrow x=30$

Therefore, the food will now last for 30 days.

#### Page No 166:

#### Question 13:

A school has 9 periods a day each of 40 minutes duration. How long would each period be, if the school has 8 periods a day, assuming the number of school hours to be the same?

#### 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.

$\mathrm{Now},9\times 40=8\times x\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{9\times 40}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow x=45$

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

#### Page No 166:

#### Question 14:

If *x* and *y* vary inversely and *x* = 15 when *y* = 6, find *y* when *x* = 9.

#### Answer:

$x$ | 15 | 9 |

$y$ | 6 | ${y}_{1}$ |

$x\mathrm{and}y\mathrm{var}y\mathrm{inversely}.\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.xy=\mathrm{constant}\phantom{\rule{0ex}{0ex}}\mathrm{Now},15\times 6=9\times {y}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow {y}_{1}=\frac{15\times 6}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow {y}_{1}=10\phantom{\rule{0ex}{0ex}}$

∴ Value of $y=10$, when

*x*=9

#### Page No 166:

#### Question 15:

If *x *and *y* vary inversely and *x* = 18 when *y* = 8, find *x* when *y* = 16.

#### Answer:

$x$ | 18 | ${x}_{1}$ |

$y$ | 8 | 16 |

$x\mathrm{and}y\mathrm{var}y\mathrm{inversely}.\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.xy=\mathrm{constant}\phantom{\rule{0ex}{0ex}}\mathrm{Now},18\times 8={x}_{1}\times 16\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{18\times 8}{16}={x}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow 9={x}_{1}\phantom{\rule{0ex}{0ex}}$

∴ Value of $x=9$

#### Page No 166:

#### Question 1:

If 14 kg of pulses cost ₹ 882, what is the cost of 22 kg of pulses?

(a) ₹ 1254

(b) ₹ 1298

(c) ₹ 1342

(d) ₹ 1386

#### Answer:

Let 22 kg of pulses cost ₹*x*.

Quantity(in kg) | 14 | 22 |

Price(in ₹) | 882 | x |

$\therefore \frac{14}{882}=\frac{22}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{22\times 882}{14}=1386$

Thus, the cost of 22 kg of pulses is ₹1,386.

Hence, the correct answer is option (d).

#### Page No 166:

#### Question 2:

If 8 oranges cost ₹ 52, how many oranges can be bought for ₹ 169?

(a) 13

(b) 18

(c) 26

(d) 24

#### Answer:

Let the number of oranges that can be bought for ₹169 be *x*.

Quantity | 8 | x |

Price(in ₹) | 52 | 169 |

$\therefore \frac{8}{52}=\frac{x}{169}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{8\times 169}{52}=26$

Thus, 26 oranges can be bought for ₹169.

Hence, the correct answer is option (c).

#### Page No 166:

#### Question 3:

**Tick (✓) the correct answer:**

A machine fills 420 bottles in 3 hours. How many bottles will it fill in 5 hours?

(a) 252

(b) 700

(c) 504

(d) 300

#### 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.

$\mathrm{Now},\frac{420}{3}=\frac{x}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{420\times 5}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow x=700$

Therefore, 700 bottles would be filled in 5 h.

#### Page No 166:

#### Question 4:

**Tick (✓) the correct answer:**

A car is travelling at a uniform speed of 75 km/hr. How much distance will it cover in 20 minutes?

(a) 25 km

(b) 15 km

(c) 30 km

(d) 20 km

#### 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.

$\mathrm{Now},\frac{75}{60}=\frac{x}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{75\times 20}{60}\phantom{\rule{0ex}{0ex}}\Rightarrow x=25\mathrm{km}$

#### Page No 166:

#### Question 5:

**Tick (✓) the correct answer:**

The weight of 12 sheets of a thick paper is 40 grams. How many sheets would weight 1 kg?

(a) 480

(b) 360

(c) 300

(d) none of these

#### Answer:

(c) 300

Let x sheets weigh 1 kg.

Now, 1 kg = 1000 g

No. of sheets | 12 | $x$ |

Weight (in g) | 40 | 1000 |

$\mathrm{Now},\frac{12}{40}=\frac{x}{1000}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{12\times 1000}{40}\phantom{\rule{0ex}{0ex}}\Rightarrow x=300$

#### Page No 166:

#### Question 6:

**Tick (✓) the correct answer:**

A pole 14 m high casts a shadow of 10 m. At the same time, what will be the height of a tree, the length of whose shadow is 7 metres?

(a) 20 m

(b) 9.8 m

(c) 5 m

(d) none of these

#### 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.

$\mathrm{Now},\frac{14}{10}=\frac{x}{7}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{14\times 7}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow x=9.8$

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

#### Page No 167:

#### Question 7:

**Tick (✓) the correct answer:**

A photograph of a bacteria enlarged 50000 times attains a length of 5 cm. The actual length of bacteria is

(a) 1000 cm

(b) 10^{−3} cm

(c) 10^{−4} cm

(d) 10^{−2} cm

#### Answer:

(c) ${10}^{-4}\mathrm{cm}$

Let *x* cm be the actual length of the bacteria.

The larger the object, the larger its image will be.

$\mathrm{Now},\frac{x}{1}=\frac{5}{50000}={10}^{-4}\mathrm{cm}$

Hence, the actual length of the bacteria is ${10}^{-4}\mathrm{cm}$.

#### Page No 167:

#### Question 8:

**Tick (✓) the correct answer:**

6 pipes fill a tank in 120 minutes, then 5 pipes will fill it in

(a) 100 min

(b) 144 min

(c) 140 min

(d) 108 min

#### 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 | $x$ |

$\mathrm{Now},6\times 120=5\times x\phantom{\rule{0ex}{0ex}}\Rightarrow x=144$

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

#### Page No 167:

#### Question 9:

**Tick (✓) the correct answer:**

3 persons can build a wall in 4 days, then 4 persons can build it in

(a) $5\frac{1}{3}$ days

(b) 3 days

(c) $4\frac{1}{3}$ days

(d) none of these

#### 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.

$\mathrm{Now},3\times 4=4\times x\phantom{\rule{0ex}{0ex}}\Rightarrow x=3$

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

#### Page No 167:

#### Question 10:

**Tick (✓) the correct answer:**

A car takes 2 hours to reach a destination by travelling at 60 km/hr. How long will it take while travelling at 80 km/hr?

(a) 1 hr 30 min

(b) 1 hr 40 min

(c) 2 hrs 40 min

(d) none of these

#### 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 | $x$ |

$\mathrm{The}\mathrm{greater}\mathrm{the}\mathrm{speed},\mathrm{the}\mathrm{lesser}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{time}\mathrm{taken}.\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{case}\mathrm{of}\mathrm{inverse}\mathrm{proportion}.\phantom{\rule{0ex}{0ex}}\mathrm{Now},60\times 2=80\times x\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{120}{80}\phantom{\rule{0ex}{0ex}}\Rightarrow x=1.5\mathrm{Therefore},\mathrm{the}\mathrm{car}\mathrm{will}\mathrm{take}1\mathrm{h}30\mathrm{min}\mathrm{to}\mathrm{reach}\mathrm{its}\mathrm{destination}\mathrm{if}\mathrm{it}\mathrm{travels}\mathrm{at}\mathrm{a}\mathrm{speed}\mathrm{of}80\mathrm{km}/\mathrm{h}.$

#### Page No 168:

#### Question 1:

350 boxes can be placed in 25 cartons. How many boxes can be placed in 16 cartons?

#### 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.

$\mathrm{Now},\frac{350}{25}=\frac{x}{16}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{350\times 16}{25}\phantom{\rule{0ex}{0ex}}\Rightarrow x=224$

∴ 224 boxes can be placed in 16 cartoons.

#### Page No 168:

#### Question 2:

The cost of 140 tennis balls is Rs 4900. Find the cost of 2 dozen such balls.

#### 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.

$\mathrm{Now},\frac{140}{4900}=\frac{24}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{24\times 4900}{140}\phantom{\rule{0ex}{0ex}}\Rightarrow x=840$

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

#### Page No 168:

#### Question 3:

The railway fare for 61 km is Rs 183. Find the fare for 53 km.

#### 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 .

$\mathrm{Now},\frac{61}{183}=\frac{53}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{53\times 183}{61}\phantom{\rule{0ex}{0ex}}\Rightarrow x=159$

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

#### Page No 168:

#### Question 4:

10 people can dig a trench in 6 days. How many people can dig it in 4 days?

#### 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.

$\mathrm{Now},10\times 6=x\times 4\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{60}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow x=15$

∴ 15 people can dig the trench in 4 days.

#### Page No 168:

#### Question 5:

30 men can finish a piece of work in 28 days. How many days will be taken by 21 men to finish it?

#### 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.

$\mathrm{Now},30\times 28=21\times x\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{30\times 28}{21}\phantom{\rule{0ex}{0ex}}\Rightarrow x=40$

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

#### Page No 168:

#### Question 6:

A garrison of 200 men had provisions for 45 days. After 15 days, 40 more men join the garrison. Find the number of days for which the remaining food will last.

#### 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.

$\mathrm{Now},200\times 30=240\times x\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{200\times 30}{240}\phantom{\rule{0ex}{0ex}}\Rightarrow x=25$

∴ The remaining food will last for 25 days.

#### Page No 168:

#### Question 7:

**Mark (✓) against the correct answer:**

6 pipes can fill a tank in 24 minutes. One pipe can fill it in

(a) 4 minutes

(b) 30 minutes

(c) 72 minutes

(d) 144 minutes

#### 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.

$\mathrm{Now},6\times 24=1\times x\phantom{\rule{0ex}{0ex}}\Rightarrow x=6\times 24\phantom{\rule{0ex}{0ex}}\Rightarrow x=144$

∴ One pipe can fill the tank in 144 minutes.

#### Page No 168:

#### Question 8:

**Mark (✓) against the correct answer:**

14 workers can build a wall in 42 days. One worker can build it in

(a) 3 days

(b) 147 days

(c) 294 days

(d) 588 days

#### 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.

$\mathrm{Now},14\times 42=1\times x\phantom{\rule{0ex}{0ex}}\Rightarrow x=14\times 42\phantom{\rule{0ex}{0ex}}\Rightarrow x=588$

∴ One worker can build the wall in 588 days.

#### Page No 168:

#### Question 9:

**Mark (✓) against the correct answer:**

35 men can reap a field in 8 days. In how many days can 20 men reap it?

(a) 14 days

(b) 28 days

(c) $87\frac{1}{2}$ days

(d) none of these

#### 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.

$\mathrm{Now},8\times 35=x\times 20\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{8\times 35}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow x=14$

∴ 20 men can reap the field in 14 days.

#### Page No 168:

#### Question 10:

**Mark (✓) against the correct answer:**

A car is travelling at an average speed of 60 km per hour. How much distance will it cover in 1 hour 12 minutes?

(a) 50 km

(b) 72 km

(c) 63 km

(d) 67.2 km

#### 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.

$\mathrm{Now},\frac{60}{60}=\frac{x}{72}\phantom{\rule{0ex}{0ex}}\Rightarrow x=72\mathrm{km}\phantom{\rule{0ex}{0ex}}$

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

#### Page No 168:

#### Question 11:

**Mark (✓) against the correct answer:**

Rashmi types 510 words in half an hour. How many words would she type in 10 minutes?

(a) 85

(b) 150

(c) 170

(d) 153

#### Answer:

(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.

$\mathrm{Now},\frac{510}{30}=\frac{x}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow x=170\phantom{\rule{0ex}{0ex}}$

∴ Rashmi will type 170 words in 10 minutes.

#### Page No 168:

#### Question 12:

**Mark (✓) against the correct answer:**

*x* and *y* vary directly. When *x* = 3, then *y* = 36. What will be the value of *x* when *y* = 96?

(a) 18

(b) 12

(c) 8

(d) 4

#### Answer:

(c) 8

$x$ | 3 | ${x}_{1}$ |

$y$ | 36 | 96 |

$x\mathrm{and}y\mathrm{var}y\mathrm{directly}.\phantom{\rule{0ex}{0ex}}\mathrm{Then}x=ky,\mathrm{where}k\mathrm{is}\mathrm{the}\mathrm{constant}\mathrm{of}\mathrm{proportionality}.\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{x}{y}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\frac{3}{36}=\frac{{x}_{1}}{96}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{96\times 3}{36}={x}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow 8={x}_{1}\phantom{\rule{0ex}{0ex}}$

∴ Value of $x=8$

#### Page No 168:

#### Question 13:

**Mark (✓) against the correct answer:**

*x* and *y* vary inversely. When *x* = 15, then *y* = 6. What will be the value of *y* when *x* = 9?

(a) 10

(b) 15

(c) 54

(d) 135

#### Answer:

(a) 10

$x$ | 15 | 9 |

$y$ | 6 | ${y}_{1}$ |

$\mathrm{Since}x\mathrm{and}y\mathrm{var}y\mathrm{inversely},xy=\mathrm{constant}.\phantom{\rule{0ex}{0ex}}\mathrm{Now},15\times 6=9\times {y}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{90}{9}={y}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow 10={y}_{1}\phantom{\rule{0ex}{0ex}}$

∴ Value of

*y*= 10, when

*x*= 9.

#### Page No 168:

#### Question 14:

**Fill in the blanks.**

(i) If 3 persons can do a piece of work in 4 days, then 4 persons can do it in ......... days.

(ii) If 5 pipes can fill tank in 144 minutes, then 6 pipes can fill it in ......... minutes.

(iii) A car covers a certain distance in 1 hr 30 minutes at 60 km per hour. If it moves at 45 km per hour, it will take ......... hours.

(iv) If 8 oranges cost Rs 20.80, the cost of 5 oranges is Rs .........

(v) The weight of 12 sheets of a paper is 50 grams. How many sheets will weigh 500 grams?

#### 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.

$\mathrm{Now},4\times 3=x\times 4\phantom{\rule{0ex}{0ex}}\Rightarrow x=3\phantom{\rule{0ex}{0ex}}$

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.

$\mathrm{Now},5\times 144=6\times x\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{5\times 144}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow x=120\mathrm{min}\phantom{\rule{0ex}{0ex}}$

∴ 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.

$\mathrm{Now},60\times 90=45\times x\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{60\times 90}{45}\phantom{\rule{0ex}{0ex}}\Rightarrow x=120\mathrm{min}=2\mathrm{h}\phantom{\rule{0ex}{0ex}}$

∴ 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.

$\mathrm{Now},\frac{8}{20.80}=\frac{5}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{5\times 20.80}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow x=13\phantom{\rule{0ex}{0ex}}$

∴ 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.

$\mathrm{Now},\frac{12}{50}=\frac{x}{500}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{12\times 500}{50}\phantom{\rule{0ex}{0ex}}\Rightarrow x=120\phantom{\rule{0ex}{0ex}}$

∴ 120 sheets will weigh 500 g.

View NCERT Solutions for all chapters of Class 8