# NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions | class 8th maths Chapter 13

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# NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions

class 8th maths

**NCERT Solutions for Class 8 Maths Chapter 13**

class 8th maths

Ex 13.1

class 8th maths

Ex 13.1

EXERCISE 13.1

**1. Following are the car parking charges near a railway station upto**

4 hours | ₹ 60 |

8 hours | ₹ 100 |

12 hours | ₹ 140 |

24 घंटो तक | ₹ 180 |

Check if the parking charges are in direct proportion to the parking time.

**Solutions :- **

As the parking hours increase, the charges will increase proportionately.

\(\displaystyle \frac{{{{x}_{1}}}}{{{{y}_{1}}}}=\frac{{{{x}_{2}}}}{{{{y}_{2}}}}\)

*x*_{1} = 4, *y _{1}* = 60 और

*x*= 8,

_{2}*y*= 100

_{2}∴ \(\displaystyle \begin{array}{l}\frac{4}{{60}}=\frac{8}{{100}}\\\frac{1}{{15}}=\frac{2}{{25}}\end{array}\)

Therefore \(\displaystyle \frac{{{{x}_{1}}}}{{{{y}_{1}}}}\ne \frac{{{{x}_{2}}}}{{{{y}_{2}}}}\)

Hourly charges are not the same in both the parking charges, so we can conclude that the parking charges are not in direct proportion.

**2. A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.**

Parts of Red Pigments | 1 | 4 | 7 | 12 | 20 |

Parts of Base | 8 | …… | …… | …… | …… |

**Solutions :-**

We know that –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{y}_{1}}}}=\frac{{{{x}_{2}}}}{{{{y}_{2}}}}\)

*x*_{1} = 1, *y _{1}* = 8 and

*x*= 4,

_{2}*y*=

_{2}*y*

_{2}Therefore \(\displaystyle \frac{1}{8}=\frac{4}{{{{y}_{2}}}}\)

*y*= 4 × 8

_{2}*y*= 32

_{2}same as –

\(\displaystyle \frac{{{{x}_{2}}}}{{{{y}_{2}}}}=\frac{{{{x}_{3}}}}{{{{y}_{3}}}}\)

*x*= 4,

_{2}*y*= 32 and

_{2}*x*= 7,

_{3}*y*=

_{3}*y*

_{3}therefore \(\displaystyle \frac{4}{{32}}=\frac{7}{{{{y}_{3}}}}\)

4

*y*= 32 × 7

_{3}\(\displaystyle {{y}_{3}}=\frac{{32\times 7}}{4}\)

*y*= 8 × 7

_{3}*y*= 56

_{3}same as –

\(\displaystyle \frac{{{{x}_{3}}}}{{{{y}_{3}}}}=\frac{{{{x}_{4}}}}{{{{y}_{4}}}}\)

*x*= 7,

_{3}*y*= 56 and

_{3}*x*= 12,

_{4}*y*=

_{4}*y*

_{4}Therefore \(\displaystyle \frac{7}{{56}}=\frac{12}{{{{y}_{4}}}}\)

7

*y*= 12 × 56

_{4}So, \(\displaystyle {{y}_{4}}=\frac{{12\times 56}}{7}\)

7

*y*= 12 × 8

_{4}*y*= 96

_{4}same as –

\(\displaystyle \frac{{{{x}_{4}}}}{{{{y}_{4}}}}=\frac{{{{x}_{5}}}}{{{{y}_{5}}}}\)

*x*= 12,

_{4}*y*= 96 and

_{4}*x*= 20,

_{5}*y*=

_{5}*y*

_{5}Therefore \(\displaystyle \frac{12}{{96}}=\frac{20}{{{{y}_{5}}}}\)

12

*y*= 20 × 96

_{4}So, \(\displaystyle {{y}_{5}}=\frac{{20\times 96}}{{12}}\)

12

*y*= 20 × 8

_{5}*y*= 160

_{5}Parts of Red Pigments | 1 | 4 | 7 | 12 | 20 |

Parts of Base | 8 | 32 |
56 |
96 |
160 |

**3. In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?**

**Solutions :-**

We know that –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{y}_{1}}}}=\frac{{{{x}_{2}}}}{{{{y}_{2}}}}\)

*x _{1}* = 1,

*y*= 75 and

_{1}*x*= x

_{2}_{2},

*y*= 1800

_{2}75

*x*= 1800

_{2}\(\displaystyle {{x}_{2}}=\frac{{1800}}{{75}}\)

*x*= 24

_{2}Therefore, in 1800 mL of the Base mixture, we should add 24 mL of red colored pigments.

**4. A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?**

**Solutions :-**

We know that –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{y}_{1}}}}=\frac{{{{x}_{2}}}}{{{{y}_{2}}}}\)

*x*_{1} = 840, *y _{1}* = 6 and

*x*=

_{2}*x*,

_{2}*y*= 5

_{2}\(\displaystyle \frac{{840}}{6}=\frac{{{{x}_{2}}}}{5}\)

6

*x*= 840 × 5

_{2}\(\displaystyle {{x}_{2}}=\frac{{840\times 5}}{6}\)

*x*= 140 × 5

_{2}*x*= 700

_{2}So that machine will fill 700 bottles in five hours.

**5. A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?
**

**Solutions :-**

Actual length of the bacteria is

\(\displaystyle =\frac{5}{{50000}}=\frac{1}{{10000}}\)

= 10

^{-4}cm

Enlarged | 50,000 | 20,000 |

Length | 5 cm | y_{2} |

We know that –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{y}_{1}}}}=\frac{{{{x}_{2}}}}{{{{y}_{2}}}}\)

*x*_{1} = 50,000, *y _{1}* = 5 and

*x*= 20,000,

_{2}*y*=

_{2}*y*

_{2}\(\displaystyle \frac{{50000}}{5}=\frac{{20000}}{{{{y}_{2}}}}\)

10000

*y*= 20000

_{2}\(\displaystyle {{y}_{2}}=\frac{{20000}}{{10000}}\)

*y*= 2 cm

_{2}So if the photograph is enlarged only 20,000 times, then its actual length will be 2 cm.

**6. In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?**

**Solutions :-**

Let the length of the mast of the model be = *x*

ship model mast length (in cm) | 9 | x |

length of mast (in cm) | 12 | 28 |

We know that –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{y}_{1}}}}=\frac{{{{x}_{2}}}}{{{{y}_{2}}}}\)

where *x*_{1} = 9, *y _{1}* = 12 and

*x*= x,

_{2}*y*= 28

_{2}\(\displaystyle \frac{9}{{12}}=\frac{x}{{28}}\)

12

*x*= 9 × 28

\(\displaystyle x=\frac{{9\times 28}}{{12}}\)

\(\displaystyle x=\frac{{3\times 28}}{4}\)

*x*= 3 × 7

*x*= 21

Hence, the length of the model of the ship will be 21 cm.

**7. Suppose 2 kg of sugar contains 9 × 10 ^{6} crystals.
How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?
**

**Solutions :-**

Let the number of crystals in 5 kg of sugar be = y

amount of sugar (in kg) | 2 | 5 |

Number of crystals | 9 × 10^{6} |
y |

We know that –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{y}_{1}}}}=\frac{{{{x}_{2}}}}{{{{y}_{2}}}}\)

*x*_{1} = 2, *y _{1}* = 9 × 10

^{6}and

*x*= 5,

_{2}*y*= y

_{2}\(\displaystyle \frac{2}{{9\times {{{10}}^{6}}}}=\frac{5}{y}\)

2

*y*= 5 × 9 × 10

^{6}

\(\displaystyle y=\frac{{5\times 9\times {{{10}}^{6}}}}{2}\)

*y*= 2.5 × 9 × 10

^{6}

*y*= 22.5 × 10

^{6}

*y*= 2.25 × 10

^{7}

Hence, the number of crystals in 5 kg of sugar will be 2.25 × 10

^{7}.

Let the number of crystals in 1.2 kg of sugar be = *y*

amount of sugar (in kg) | 2 | 1.2 |

Number of crystals | 9 × 10^{6} |
y |

We know that –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{y}_{1}}}}=\frac{{{{x}_{2}}}}{{{{y}_{2}}}}\)

*x*_{1} = 2, *y _{1}* = 9 × 10

^{6}and

*x*= 1.2,

_{2}*y*= y

_{2}\(\displaystyle \frac{2}{{9\times {{{10}}^{6}}}}=\frac{1.2}{y}\)

2

*y*= 1.2 × 9 × 10

^{6}

\(\displaystyle y=\frac{{1.2\times 9\times {{{10}}^{6}}}}{2}\)

*y*= 0.6 × 9 × 10

^{6}

*y*= 5.4 × 10

^{6}

Hence, the number of crystals in 1.2 kg of sugar will be 5.4 × 10

^{6}.

**8. Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?**

**Solutions :-**

Let the distance covered by them be shown in the map = x

scale in map (in cm) | 1 | x |

length of road (in km) | 18 | 72 |

We know that –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{y}_{1}}}}=\frac{{{{x}_{2}}}}{{{{y}_{2}}}}\)

*x*_{1} = 1, *y _{1}* = 18 and

*x*= x,

_{2}*y*= 72

_{2}\(\displaystyle \frac{1}{{18}}=\frac{x}{{72}}\)

18x = 72

\(\displaystyle x=\frac{{72}}{{18}}\)

*x*= 4 cm

Hence, the distance covered by him will be 4 cm on the map.

**9. A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time (i) the length of the shadow cast by another pole 10 m 50 cm high (ii) the height of a pole which casts a shadow 5m long.
**

**Solutions :-**

**(i) the length of the shadow cast by another pole 10 m 50 cm high**

Let the length of the shadow of the pole be = y

height of pole (in m) | 5.60 | 10.50 |

height of shadow (in m) | 3.20 | y |

We know that –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{y}_{1}}}}=\frac{{{{x}_{2}}}}{{{{y}_{2}}}}\)

*x*_{1} = 5.60, *y _{1}* = 3.20 and

*x*= 10.50,

_{2}*y*= y

_{2}\(\displaystyle \frac{{5.60}}{{3.20}}=\frac{{10.50}}{y}\)

5.60 ×

*y*= 10.50 × 3.20

\(\displaystyle y=\frac{{10.50\times 3.20}}{{5.60}}\)

\(\displaystyle y=\frac{{10.50\times 0.4}}{{0.7}}\)

*y*= 15 × 0.4

*y*= 6.0 m

So, height of shadow of a 10 m 50 cm height pole is 6.0 m.

**(ii) the height of a pole which casts a shadow 5m long.**

let’s height of pole = x

height of pole (in m) | 5.60 | x |

height of shadow (in m) | 3.20 | 5 |

We know that –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{y}_{1}}}}=\frac{{{{x}_{2}}}}{{{{y}_{2}}}}\)

*x*_{1} = 5.60, *y _{1}* = 3.20 and

*x*=

_{2}*x*,

*y*= 5

_{2}\(\displaystyle \frac{{5.60}}{{3.20}}=\frac{x}{5}\)

3.20 ×

*x*= 5.60 × 5

\(\displaystyle x=\frac{{5.60\times 5}}{{3.20}}\)

\(\displaystyle x=\frac{{7\times 5}}{4}\)

\(\displaystyle x=\frac{{35}}{4}\)

*x*= 8.75 m

So, the height of pole is 8 m 75 cm.

**10. A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?**

**Solutions :-**

Let the distance covered be = y

time (in min) | 25 | 300 (5 hours) |

distance (in km) | 14 | y |

We know that –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{y}_{1}}}}=\frac{{{{x}_{2}}}}{{{{y}_{2}}}}\)

where* x*_{1} = 25, *y _{1}* = 14 and

*x*= 300,

_{2}*y*= y

_{2}\(\displaystyle \frac{{25}}{{14}}=\frac{{300}}{y}\)

25 ×

*y*= 300 × 14

\(\displaystyle y=\frac{{300\times 14}}{{25}}\)

*y*= 12 × 14

*y*= 168 km

Hence he will be able to cover a distance of 168 km in 5 hours.

**NCERT Solutions for Class 8 Maths Chapter 13**

class 8th maths

Ex 13.2

class 8th maths

Ex 13.2

**1. Which of the following are in inverse proportion?
**(i) The number of workers on a job and the time to complete the job.

(ii) The time taken for a journey and the distance travelled in a uniform speed.

(iii) Area of cultivated land and the crop harvested.

(iv) The time taken for a fixed journey and the speed of the vehicle.

(v) The population of a country and the area of land per person.

**Solutions :-**

**(i) The number of workers on a job and the time to complete the job.**

**Answer –** This is a inverse proportion because If there is an increase in the number of workers, the time taken to work will decrease.

**(ii) The time taken for a journey and the distance travelled in a uniform speed.**

**Answer –** This is a direct proportion because when time is increased then the distance travelled is also increased.

**(iii) Area of cultivated land and the crop harvested.**

**Answer –** This is a direct proportion because the greater the area of the field, the greater will be the crop harvested in it.

**(iv) The time taken for a fixed journey and the speed of the vehicle.**

**Answer –** This is a inverse proportion because the higher the speed of the vehicle, the lesser the time taken.

**(v) The population of a country and the area of land per person.**

**Answer –** This is a inverse proportion because The more the population increases, the less will be the area of land per person.

**2. In a Television game show, the prize money of ₹ 1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners? **

No. of Winner | 1 | 2 | 4 | 5 | 8 | 10 | 20 |

Prize of each winner (in Rs.) | 1,00,000 | 50,000 | ….. | ….. | ….. | ….. | ….. |

**Solutions :-**

The above table shows that the number of winners and prize money are in inverse proportion. If the number of winners increases, the amount of prize awarded will decrease.

Hence, if the number of winners is 4, then the prize money to be given is = 100000/4

= ₹ 25,000

Hence, if the number of winners is 5, then the prize money to be given is = 100000/5

= ₹ 20,000

Hence, if the number of winners is 8, then the prize money to be given is = 100000/8

= ₹ 12,500

Hence, if the number of winners is 10, then the prize money to be given is = 100000/10

= ₹ 10,000

Hence, if the number of winners is 20, then the prize money to be given is = 100000/20

= ₹ 5,000

No. of Winner | 1 | 2 | 4 | 5 | 8 | 10 | 20 |

Prize of each winner (in Rs.) | 1,00,000 | 50,000 | 25,000 |
20,000 |
12,500 |
10,000 |
5,000 |

**3. Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table.**

Number of spokes | 4 | 6 | 8 | 10 | 12 |

Angle between a pair of consecutive spokes | 90° | 60° | ….. | ….. | ….. |

(i) Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?

(ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.

(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?

Solutions :-

The appropriate table will have the inverse ratio because if the number of spokes increases, the angle between the consecutive spokes will decrease.

So, if the number of spokes were 8, then the angle between the consecutive spokes would be = 360/8

= 45°

If the number of matchsticks is 10, then the angle between the consecutive spokes is = 360/10

= 36°

If the number of matchsticks is 12, then the angle between the consecutive spokes is = 360/12

= 30°

Number of spokes | 4 | 6 | 8 | 10 | 12 |

Angle between a pair of consecutive spokes | 90° | 60° | 45° | 36° | 30° |

**(i) Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?**

**Ans –** Yes, there is an inverse ratio between the number of spokes and the angle formed between the consecutive spokes.

**(ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.**

**Ans –** When the number of matchsticks is 15, then the angle between each pair of spokes is = 360/15

= 24°

**(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?**

**Ans –** If the angle between each pair of consecutive spokes is 40°, then the number of spokes required is = 360/40

= 9°

**4. If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?**

**Solutions :-**

number of children | 24 | (24 – 4) = 20 |

number of sweets | 5 | x |

It is clear that if the number of children increases, each child will get less sweets.

Hence it is an inverse ratio.

in inverse ratio –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{x}_{2}}}}=\frac{{{{y}_{2}}}}{{{{y}_{1}}}}\)

where *x _{1}* = 24,

*y*= 5 and

_{1}*x*= 20,

_{2}*y*=

_{2}*x*

\(\displaystyle \frac{{24}}{{20}}=\frac{x}{5}\)

20 ×

*x*= 5 × 24

\(\displaystyle \begin{array}{l}x=\frac{{5\times 24}}{{20}}\\x=\frac{{24}}{4}\end{array}\)

*x*= 6

So each child will get 6 sweets.

**5. A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?**

**Solutions :-**

number of animals | 20 | (20 + 10) = 30 |

Food days | 6 | x |

It is clear that if there is an increase in the number of animals in the cattle shed, the food will last for a shorter period of time.

Hence it is an inverse ratio.

in inverse ratio –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{x}_{2}}}}=\frac{{{{y}_{2}}}}{{{{y}_{1}}}}\)

where *x _{1}* =20,

*y*= 6 and

_{1}*x*= 30,

_{2}*y*=

_{2}*x*

\(\displaystyle \frac{{20}}{{30}}=\frac{x}{6}\)

30 ×

*x*= 20 × 6

\(\displaystyle x=\frac{{20\times 6}}{{30}}\)

\(\displaystyle x=\frac{{2\times 6}}{3}\)

*x*= 2 × 2

*x*= 4

Hence, food for 30 animals will be sufficient for 6 days.

**6. A contractor estimates that 3 persons could rewire Jasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?**

**Solutions :- **

No. of person | 3 | 4 |

No. of days | 4 | x |

It is clear that if the contractor increases the number of persons, the work will be completed in less days.

Hence it is an inverse ratio.

in inverse ratio –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{x}_{2}}}}=\frac{{{{y}_{2}}}}{{{{y}_{1}}}}\)

where *x _{1}* = 3,

*y*= 4 and

_{1}*x*= 4,

_{2}*y*=

_{2}*x*

\(\displaystyle \frac{3}{4}=\frac{x}{4}\)

4 ×

*x*= 4 × 3

\(\displaystyle x=\frac{{4\times 3}}{4}\)

*x*= 3

Four men will do that work in 3 days.

**7. A batch of bottles were packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?**

**Solutions :-**

No. of boxes | 25 | x |

No. of bottles | 12 | 20 |

It is clear that if the number of bottles is less, then the number of boxes will increase.

Hence it is an inverse ratio.

in inverse ratio –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{x}_{2}}}}=\frac{{{{y}_{2}}}}{{{{y}_{1}}}}\)

where *x _{1}* =25,

*y*= 12 and

_{1}*x*=

_{2}*x*,

*y*= 20

_{2}\(\displaystyle \frac{{25}}{x}=\frac{{20}}{{12}}\) 20 × x = 25 × 12

\(\displaystyle \begin{array}{l}x=\frac{{25\times 12}}{{20}}\\x=\frac{{5\times 12}}{4}\end{array}\)

*x*= 5 × 3

*x*= 15

So if each box contains 20 bottles, then 15 boxes will be filled.

**8. A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?
**

**Solutions :-**

No. of days | 63 | 54 |

No. of machines | 42 | x |

It is clear that if the goods are to be manufactured in less days, more machines will be required.

Hence it is an inverse ratio.

in inverse ratio –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{x}_{2}}}}=\frac{{{{y}_{2}}}}{{{{y}_{1}}}}\)

where *x _{1}* =63,

*y*= 42 and

_{1}*x*= 54,

_{2}*y*=

_{2}*x*

\(\displaystyle \frac{{63}}{{54}}=\frac{x}{{42}}\) 54 × x = 63 × 42

\(\displaystyle \begin{array}{l}x=\frac{{63\times 42}}{{54}}\\x=\frac{{7\times 42}}{6}\end{array}\)

*x*= 7 × 7

*x*= 49

Hence, to make the same number of articles in 54 days, 49 machines will be required.

**9. A car takes 2 hours to reach a destination by travelling at the speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?**

**Solutions : –**

Speed (in km/h) | 60 | 80 |

Time (in hours) | 2 | x |

It is clear that if the speed of the car increases, then the time taken to decide will be less.

Hence it is an inverse ratio.

in inverse ratio –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{x}_{2}}}}=\frac{{{{y}_{2}}}}{{{{y}_{1}}}}\)

where *x _{1}* = 60,

*y*= 2 and

_{1}*x*= 80,

_{2}*y*=

_{2}*x*

\(\displaystyle \frac{{60}}{{80}}=\frac{x}{2}\) 80 ×

*x*= 60 × 2

\(\displaystyle \begin{array}{l}x=\frac{{60\times 2}}{{80}}\\x=\frac{{6\times 2}}{8}\\x=\frac{{3\times 2}}{4}\\x=\frac{3}{2}\\x=1\frac{1}{2}\end{array}\) So, car take \(\displaystyle 1\frac{1}{2}\) hours with the speed of 80 km/h.

**10. Two persons could fit new windows in a house in 3 days.
**(i) One of the persons fell ill before the work started. How long would the job take now?

(ii) How many persons would be needed to fit the windows in one day?

**Solutions :-**

**(i) One of the persons fell ill before the work started. How long would the job take now?
**

**Solution :-**

No. of person | 2 | 1 |

Time | 3 | x |

It is clear that if the number of persons decreases, the number of days will increase.

Hence it is an inverse ratio.

in inverse ratio –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{x}_{2}}}}=\frac{{{{y}_{2}}}}{{{{y}_{1}}}}\)

where *x _{1}* = 2,

*y*= 3 and

_{1}*x*= 1,

_{2}*y*=

_{2}*x*

\(\displaystyle \frac{2}{1}=\frac{x}{3}\)

*x*= 3 × 2

*x*= 6

So now this work will be completed in 6 days.

**(ii) How many persons would be needed to fit the windows in one day?**

No. of person | 2 | x |

Time | 3 | 1 |

It is clear that if the number of persons decreases, the number of days will increase.

Hence it is an inverse ratio.

in inverse ratio –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{x}_{2}}}}=\frac{{{{y}_{2}}}}{{{{y}_{1}}}}\)

where *x _{1}* = 2,

*y*= 3 and

_{1}*x*=

_{2}*x*,

*y*= 1

_{2}\(\displaystyle \frac{2}{x}=\frac{1}{3}\)

*x*= 3 × 2

*x*= 6

Hence, to get the windows installed in a single day, 6 persons would be required.

**11. A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?**

**Solutions :-**

Time (in min) | 45 | x |

Periods | 8 | 9 |

It is clear that if the number of periods increases, then the duration of the period will decrease.

Hence it is an inverse ratio.

in inverse ratio –

\(\displaystyle \frac{{{{x}_{1}}}}{{{{x}_{2}}}}=\frac{{{{y}_{2}}}}{{{{y}_{1}}}}\)

where *x*_{1} = 45, *y*_{1} = 8 and *x*_{2} = *x*, *y*_{2} = 9

\(\displaystyle \frac{{45}}{x}=\frac{9}{8}\)
9 × *x* = 8 × 45

\(\displaystyle x=\frac{{8\times 45}}{9}\)
*x* = 8 × 5

*x* = 40 min

So if the school has 9 periods of equal duration, then each period will be of 40 minutes.