Rs Aggarwal 2017 Solutions for Class 8 Math Chapter 1 Rational Numbers are provided here with simple step-by-step explanations. These solutions for Rational Numbers are extremely popular among Class 8 students for Math Rational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 8 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class 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