# RS Aggarwal Class 8 Math Twelfth Chapter Direct Inverse Proportions Exercise 12B Solution

## EXERCISE 12B

**(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 |

Solution: Clearly, we have:

xy = 6 × 9 ≠ 10 × 15 ≠ 14 × 21 ≠ 16 × 24.

∴ x and y are not inversely proportional.

(ii)

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

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

Solution: Clearly, we have:

xy = 5 × 18 = 9 × 10 = 15 × 6 = 3 × 30 = 45 × 2 = 90 = constant.

∴ x and y are inversely proportional.

(iii)

x | 9 | 3 | 6 | 36 |

y | 4 | 12 | 9 | 1 |

Solution: xy = 9 × 4 = 3 × 12 = 36 × 1 ≠ 6 × 9.

∴ x and y are not inversely proportional.

**(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} |

Solution: Since x and y are inversely proportional, we must have xy = constant.

∴ 8 × y_{1} = x_{1} × 4 = 16 × 5 = x_{2} × 2 = 80 × y_{2}.

Now, 16 × 5 = 8 × y_{1} ⇒ 8y_{1} = 80 ⇒y_{1} = 10.

16 × 5 = x_{1} × 4 ⇒ 4x_{1} = 80 ⇒ x_{1} = 20.

16 × 5 = x_{2} × 2 ⇒ 2x_{2} = 80 ⇒ x_{2} = 40.

16 × 5 = 80 × y_{2} ⇒ 80y_{2} = 80 ⇒ y_{2} = 1.

∴ x_{1 }= 20, x_{2} = 40, y_{1} = 10 and y_{2} = 1.

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

Solution: Let the required time be x days.

Number of men | 35 | 20 |

Time (In day) | 8 | x |

Less men will be finish reap the field more days. So, it is case of inverse proportion.

∴ 35 × 8 = 20 × x

⇒ 20x = 280

⇒ x = 14

∴ Required number of days =14.

**(4) 12 men can dig a pond in 8 days. How many men can dig it in 6 days.**

Solution: Let the required number of men be x.

Number of men | 12 | x |

Time (in days) | 8 | 6 |

For finishing the work in less time, more workers will be needed. So it is case of inverse proportion.

∴ 12 × 8 = x × 6

⇒ 6x = 96

⇒ x = 16

∴ Required number of workers = 16.

**(5) 6 cows can graze a field in 28 days. How long would 14 cows take to graze the same field?**

Solution: Let the required number of time be x days.

Number of cows | 6 | 14 |

Time (In days) | 28 | x |

For finish grazing with more cows take less time. So, it is case of inverse proportion.

∴ 6 × 28 = 14 × x

⇒ 14x = 168

⇒ x = 12

∴ Required number of days = 12 .

**(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?**

Solution: The car travel 5 hours at the speed 60 km/hr = (5 × 60) = 300 km.

Now, For 75 km traveled the car take 1 hour

** (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?**

Solution: Let the required number of machine be x.

Number of machine | 42 | x |

Time (in days) | 56 | 48 |

∴ 42 ×56 = x × 48

⇒ 48x = 2352

⇒ x = 49

∴ Required number of machine is 49.

**(8) 7 taps 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?**

Solution: Let the required time be x minutes.

Here, 1 hour 36 minutes = 96 minutes

Number of taps | 7 | 8 |

Time (in minutes) | 96 | x |

For fill the tank more taps take less time. So, it is case of inverse proportion.

∴ 7 × 96 = 8 × x

⇒ 8x = 672

⇒ x = 84

∴ Required time to fill the tank be 8 taps = 84 minutes = 1 hour 24 minutes.

**(9) 8 taps of the same size fill a tank in 27 minutes. If two taps go out order, how long would the remaining taps take to fill the tank?**

Solution: Let the required time be x minutes.

Number of taps | 8 | 8 – 2 = 6 |

Time (in minutes) | 27 | x |

Less tap take more time to fill the tank. So, it is case of inverse proportion.

∴ 8 × 27 = 6 × x

⇒ 6x = 216

⇒ x = 36

∴ Required time to fill the tank = 36 minutes.

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

Solution: Let the required number of days be x.

Time (in days) | 9 | x |

Number of animals | 28 | (28+8) = 36 |

More animals will have food for less time. So, it is case of inverse proportion.

∴ 9 × 28 = 36 × x

⇒ 36x = 252

⇒ x = 7

∴ Required number of days = 7.

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

Solution: Let the required time be x days.

Number of men | 900 | (900+500) = 1400 |

Time (in days) | 42 | x |

More men will have food for less time. So, it is case of inverse proportion.

∴ 900 × 42 = 1400 × x

⇒ 1400x = 37800

⇒ 14x = 378

⇒ x = 27

∴ Required number of days = 27.

**(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?**

Solution: Let the required number of days be x.

Number of students | 75 | (75 – 15) = 60 |

Time (in days) | 24 | x |

Less number of students will have food for more days. So, it is case of inverse proportion.

∴ 75 × 24 = 60 × x

⇒ 60x = 1800

⇒ 6x = 180

⇒ x = 30

∴ The required number of days = 30.

**(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 the school hours to be the same?**

Solution: If duration of each period 40 minutes. Therefore, durations of 9 periods = (9 × 40) = 360 minutes

∴ Therefore required time of each periods if there are 8 periods in the school= (360 ÷ 8) = 45 minutes.

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

Solution: We can write,

xy = 15 × 6

⇒ 9y = 90

⇒ y = 10

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

Solution: We can write,

xy = 18 × 8

⇒ 16x = 144

⇒ x = 9

Thank you so much rs aggarwal