myhelper: RS Aggarwal solution class 8 chapter 12 Direct and Inverse Proportions Exercise 12B

Exercise 12B

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Q1 | Ex-12B | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

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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 1:

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

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

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

Q2 | Ex-12B | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

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Question 2:

If x and y are inversely proportional, find the values of x₁, x₂, y₁ and y₂ in the table given below:

x	8	x₁	16	x₂	80
y	y₁	4	5	2	y₂

Answer 2:

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

Q3 | Ex-12B | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

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Question 3:

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

Answer 3:

Let x be the required number of days. Then, we have:

No. of days	8	x
No. of men	35	20

Clearly, less men will take more days to reap the field.
So, it is a case of inverse proportion.

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

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

Q4 | Ex-12B | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

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Question 4:

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

Answer 4:

Let x be the required number of men. Then, we have:

No. of days	8	6
No. of men	12	x

Clearly, more men will require less number of days to dig the pond.
So, it is a case of inverse proportion.

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

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

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Q5 | Ex-12B | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

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Question 5:

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

Answer 5:

Let x be the number of days. Then, we have:

No. of days	28	x
No. of cows	6	14

Clearly, more number of cows will take less number of days to graze the field.
So, it is a case of inverse proportion.

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

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

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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 6:

Let x h be the required time taken. Then, we have:

Speed (in km/h)	60	75
Time (in h)	5	x

Clearly, the higher the speed, the lesser will be the the time taken.
So, it is a case of inverse proportion.

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

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

Q7 | Ex-12B | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

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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 7:

Let x be the number of machines required to produce same number of articles in 48.
Then, we have:

No. of machines	42	x
No. of days	56	48

Clearly, less number of days will require more number of machines.
So, it is a case of inverse proportion.

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

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

Q8 | Ex-12B | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

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Question 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?

Answer 8:

Let x be the required number of taps. Then, we have:
1 h = 60 min
i.e., 1 h 36 min = (60+36) min = 96 min

No. of taps	7	8
Time (in min)	96	x

Clearly, more number of taps will require less time to fill the tank.
So, it is a case of inverse proportion.

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

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

Q9 | Ex-12B | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

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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 9:

Let x min be the required number of time. Then, we have:

No. of taps	8	6
Time (in min)	27	$x$

Clearly, less number of taps will take more time to fill the tank .
So, it is a case of inverse proportion.

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

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

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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 10:

Let x be the required number of days. Then, we have:

No. of days	9	x
No. of animals	28	36

Clearly, more number of animals will take less number of days to finish the food.
So, it is a case of inverse proportion.

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

Therefore, the food will last for 7 days.

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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 11:

Let x be the required number of days. Then, we have:

No. of men	900	1400
No. of days	42	x

Clearly, more men will take less number of days to finish the food.
So, it is a case of inverse proportion.

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

Therefore, the food will now last for 27 days.

Q12 | Ex-12B | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

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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 12:

Let x be the required number of days. Then, we have:

No. of students	75	60
No. of days	24	x

Clearly, less number of students will take more days to finish the food.
So, it is a case of inverse proportion.

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

Therefore, the food will now last for 30 days.

Q13 | Ex-12B | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

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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 13:

Let x min be the duration of each period when the school has 8 periods a day.

No. of periods	9	8
Time (in min)	40	x

Clearly, if the number of periods reduces, the duration of each period will increase.
So, it is a case of inverse proportion.

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

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

Q14 | Ex-12B | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

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Question 14:

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

Answer 14:

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

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

∴ Value of

y = 10

, when x =9

Q15 | Ex-12B | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

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Question 15:

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

Answer 15:

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

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

∴ Value of

x = 9

RS Aggarwal solution class 8 chapter 12 Direct and Inverse Proportions Exercise 12B

Exercise 12B

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Q1 | Ex-12B | Class 8 | RS AGGARWAL | chapter 12 | Direct and Inverse Proportions | myhelper

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