Direct and Indirect Proportions Class 8 Extra Questions with Answers

These Class 8 Maths Extra Questions Chapter 13 Direct and Indirect Proportions will help students prepare well for the exams.

Class 8 Maths Chapter 13 Extra Questions Direct and Indirect Proportions

Direct and Indirect Proportions Extra Questions Class 8

Question 1.
A classroom of 24 students can finish a project within 5 days.
a) How many more students are required to complete the same project within 3 days?
b) How many days it will take to complete the same project if 6 more students join in the classroom ?
Solution:

Number of students (x)	x1 = 24	x2 = ?	x3 = 30
Number of days (y)	y1 = 5	y2 = 3	y3 = ?

Number of students (x) is inversely proportional to number of days (y)
a) So, x₁y₁ = x₂y₂
24 × 5 = x₂ × 3
x₂ = \(\frac{24 \times 5}{3}\)
x₂ = 40
x₂ – x₁ = 40 – 24 = 16
∴ 16 more students are required to complete the same project with in 3 days.

b) x₁y₁ = x₃y₃
24 × 5 = 30 × y₃
y₃ = \(\frac{24 \times 5}{30}\)
y₃ = 4
∴ 4 days will be taken to complete the same project if 6 more students join in the classroom.

Extra Questions of Direct and Indirect Proportions Class 8

Question 1.
Imagine you invest an unknown amount (P) at 10% interest for 2 years. You earn Rs. 200 in simple interest. How much MORE would you earn using compound interest instead of simple interest under the same conditions ?
Solution:
I = ₹ 200 ; R = 10% ; T = 2 years

= ₹ 1210 – 1000 = ₹ 210
Difference = CI – SI
= ₹ 210 – 200 = ₹ 10.
∴ Instead of simple interest if we use compound interest then we get ₹ 10

Direct and Indirect Proportions Class 8 Extra Questions

Question 1.
The cost of 5 metres of a particular quality of cloth is ₹ 210. Tabulate the cost of 2, 4, 10 and 13 metres of cloth of the same type.

Solution:
Suppose the length of cloth is x metres and its cost, in ₹, is y.

x	2	4	5	10	13
y	y2	y3	210	y4	y5

As the length of cloth increases, cost of the cloth also increases in the same ratio. It is a case of direct proportion.
We make use of the relation of type
\(\frac{\mathrm{x}_1}{\mathrm{y}_1}\) = \(\frac{\mathrm{x}_2}{\mathrm{y}_2}\)

i) Here x₁ = 5, y₁ = 210 and x₂ = 2
Therefore, \(\frac{\mathrm{x}_1}{\mathrm{y}_1}\) = \(\frac{\mathrm{x}_2}{\mathrm{y}_2}\) gives \(\frac{5}{210}\) =
\(\frac{2}{y_2}\) or 5y₂ = 2 × 210 or y₂ = \(\frac{2 \times 210}{5}\) = 84

ii) If x₃ = 4, then = \(\frac{5}{210}\) = \(\frac{4}{y_3}\) or 5y₃ = 4 × 210 or y₃ = \(\frac{4 \times 210}{5}\) = 168

iii) If x₄ = 10, then = \(\frac{5}{210}\) = \(\frac{10}{y_4}\) or y₄ = \(\frac{10 \times 210}{5}\) = 420

iv) If x₅ = 13, then \(\frac{5}{210}\) = \(\frac{13}{y_5}\) or
y₅ = \(\frac{13 \times 210}{5}\) = 546
Note that here we can also use \(\frac{2}{84}\) or \(\frac{4}{168}\) or \(\frac{10}{420}\) in the place of \(\frac{5}{210}\) ]

Question 2.
An electric pole, 14 metres high, casts a shadow of 10 metres. Find the height of a tree that casts a shadow of 15 metres under similar conditions.
Solution:
Let the height of the tree be x metres.
We form a table as shown below :

height of the object in (in metres)	14	x
length of the shadow (in meters)	10	15

Note the cloth also increases in the same ratio. It is a case of direct proportion. That is, \(\frac{\mathrm{x}_1}{\mathrm{y}_1}\) = \(\frac{\mathrm{x}_2}{\mathrm{y}_2}\)
we have \(\frac{14}{10}\) = \(\frac{x}{15}\)
or \(\frac{14}{10}\) × 15 = x
or \(\frac{14 \times 3}{2}\) = x
so 21 = x

Thus, height of the tree is 21 metres. Alternately, we can write
\(\frac{\mathrm{x}_1}{\mathrm{y}_1}\) = \(\frac{\mathrm{x}_2}{\mathrm{y}_2}\) as \(\frac{\mathrm{x}_1}{\mathrm{x}_2}\) = \(\frac{\mathrm{y}_1}{\mathrm{y}_2}\)
so x₁ : x₂ = y₁ : y₂
or 14 : x = 10 : 15
Therefore, 10 × x = 15 × 14
or x = \(\frac{15 \times 14}{10}\) = 21

Question 3.
If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the same paper would weigh 2\(\frac{1}{2}\) kilograms ?
Solution:
Let the number of sheets which weigh 2\(\frac{1}{2}\) kg be x. We put the above information in the form of a table as shown below:

Number of sheets	12	x
Weight of sheets (in grams)	40	2500

1 kilogram =1000 grams
2\(\frac{1}{2}\) kilograms =2500 grams
More the number of sheets, the more would their weight be. So, the number of sheets and their weights are directly proportional to each other.

So, \(\frac{12}{40}\) = \(\frac{x}{2500}\)
or \(\frac{12 \times 2500}{40}\) = x or 750 = x
Thus, the required number of sheets of paper = 750.

Alternate method:
Two quantities x and y which vary in direct proportion have the relation
x = ky or \(\frac{x}{y}\) = k
Here, k = \(\frac{\text { number of sheets }}{\text { weight of sheets in grams }}\) = \(\frac{12}{40}\) = \(\frac{3}{10}\)
Now x is the number of sheets of the paper which weigh 2\(\frac{1}{2}\)kg [2500 g]. Using the relation x = ky,
x = \(\frac{3}{10}\) × 2500 = 750
Thus, 750 sheets of paper would weigh 2\(\frac{1}{2}\) kg.

Question 4.
A train is moving at a uniform speed of 75 km/hour.
i) How far will it travel in 20 minutes ?
ii) Find the time required to cover a distance of 250 km.
Solution:
Let the distance travelled (in km) in 20 minutes be x and time taken (in minutes) to cover 250 km be y.

Distance travelled (in km)	75	x	250
Time taken (in minutes)	60	20	y

1 hour = 60 minutes
Since the speed is uniform, therefore, the distance covered would be directly proportional to time.
i) we have \(\frac{75}{60}\) = \(\frac{x}{20}\)
or \(\frac{75}{60}\) × 20 = x
or x = 25
So, the train will cover a distance of 25 km in 20 minutes.

ii) Also, \(\frac{75}{60}\) = \(\frac{250}{y}\)
or y = \(\frac{250 \times 60}{75}\) = 200 minutes
or 3 hours 20 minutes.
Therefore, 3 hours 20 minutes will be required to cover a distance of 250 kilometres.
Alternatively, when x is known, then one can determine y from the relation
\(\frac{x}{20}\) = \(\frac{250}{y}\).

Question 5.
The scale of a map is given as 1 : 30000000. Two cities are 4 cm apart on the map. Find the actual distance between them.

Solution:
Let the map distance be x cm and actual distance be y cm, then
1 : 30000000 = x : y
or \(\frac{1}{3 \times 10^7}\) = \(\frac{x}{y}\)
Since x = 4 so, \(\frac{1}{3 \times 10^7}\) = \(\frac{4}{y}\)
or y = 4 × 3 × 10⁷ = 12 × 10⁷ cm = 1200km.
Thus, two cities’, which are 4 cm apart on the map, are actually 1200 km away from each other.

Question 6.
6 pipes are required to fill a tank in 1 hour 20 minutes. How long will it take if only 5 pipes of the same type are used?

Solution:
Let the desired time to fill the tank be x minutes. Thus, we have the following table.

Number of pipes	6	5
Time (m minutes)	80	x

Lesser the number of pipes, more will be the time required by it to fill the tank. So, this is a case of inverse proportion.
Hence, 80 × 6 = x × 5 [x₁y₁ = x₂y₂]
or \(\frac{80 \times 6}{5}\) = x
or x = 96
Thus, time taken to fill the tank by 5 pipes is 96 minutes or 1 hour 36 minutes.

Question 7.
There are 100 students in a hostel. Food provision for them is for 20 days. How long will these provisions last, if 25 more students join the group ?

Suppose the provisions last for y days when the num ber of students is 125. We have the following table.

Number of students	100	125
Number of days	20	y

Note that more the number of students, the sooner would the provisions exhaust. Therefore, this is a case of inverse proportion.
So, 100 × 20 = 125 × y
or \(\frac{100 \times 20}{125}\) y or 16 = y
Thus, the provisions will last for 16 days, if 25 more students join the hostel.
Alternately, we can write x₁y₁ = x₂y₂
as \(\frac{\mathrm{x}_1}{\mathrm{x}_2}\) = \(\frac{\mathrm{y}_1}{\mathrm{y}_2}\).
That is, x₁ : x₂ = y₂ : y₁
or 100 : 125 y : 20
or y = \(\frac{100 \times 20}{125}\) = 16

Question 8.
If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours ?

Solution:
Let the number of workers employed to build the wall in 30 hours be y.
We have the following table.

Number of hours	48	30
Number of workers	15	y

Obviously more the number of workers, faster will they build the wall. So, the number of hours and number of workers vary in inverse proportion.
So 48 × 15 = 30 × y
There fore, \(\frac{48 \times 15}{30}\) = y or y = 24
i.e., to finish the work in 30 hours, 24 workers are required.