# Free Direct and Inverse Proportions 02 Practice Test - 8th Grade

### Question 1

In a college there are 10 periods per day each of 40 minutes. The students were protesting to reduce the number of periods but the Principal did not want to reduce the working hours. What should be the working hours of each period if Principal reduces the number of periods to 8?

30 minutes

55 minutes

50 minutes

45 minutes

#### SOLUTION

Solution :C

When number of periods is 10, time for each period is 40 mins.

Let the time for each period be x mins when number of periods is changed to 8.

AB = constant

A1×B1 = A2×B2

⇒x=10×408=50 minutes.The principal should make each period of 50 mins.

### Question 2

Sophia was counting the number of crystals in 5kg of sugar. Can you help her to find the number of crystals in 5kg without actual counting, if 12kg of sugar contains 24 x 104 crystals?

2 x 104

1 × 105

2.4 x 106

24 x 102

#### SOLUTION

Solution :B

Weight of sugar12 5 Number of crystals24×104y

This situation follows direct proportionality.1224×104=5y

y=105

### Question 3

10 men can reap a field in 20 days. 10 men reaped the field for 10 days and after 10 days 10 more men join them. When will the work be finished?

15 days

25 days

30 days

12 days

#### SOLUTION

Solution :A

Total number of work days required =10 x 20 = 200

Since 10 men worked for 10 days, 10 x 10 = 100 days work is already done.

So, 200 - 100 = 100 days of work is left.

Once, 10 more men joined, total number of workers=20

So, number of days required more =10020=5

Thus, total number of days = 10 + 5 = 15.

### Question 4

If an SUV can transport 7 people, car can transport 5, jeep can transport 8 and auto can transport 2 people, which of the following combination can be used to exactly transport 25 people?

Two rounds of jeep, one round of SUV and one round of auto.

2 rounds of SUV, 2 rounds of car

One round of SUV and jeep each, two rounds of car

Three rounds of car, two rounds of SUV and on round of jeep

#### SOLUTION

Solution :A and C

Let the number of rounds done by SUV be x, car be y, jeep be z and auto be p.

Thus the number of passengers transported will be 7x+5y+8z+2p .... (1)

Checking options respectively:

1→x=1,z=2,p=1

Substituting these in (1)

7+8 × 2+2=25.

So option A is correct.

2→x=2,y=2

Substituting these in (1)

7×2+85×2=24

So option B is incorrect.

3→x=1,y=2,z=1

Substituting these in (1)

7+5×2+8=25.

So option C is correct.

4→ Similarly substituting the values of option D. We will find number of people transported=26

### Question 5

Ipsa bought a car which covers 363 km in 24.2 litres of petrol. What will be the distance covered using 7 litres of petrol?

213 km

105 km

67 km

86 km

#### SOLUTION

Solution :B

Let the required distance be y.

Distance (in km)363yAmount of petrol(in l)24.27

The distance travelled and the amount of petrol consumed are directly proportional

AB= Constant

A1B1 = A2B2

24.2363 = 7y

24.2×y=363×7y=7×36324.2=105 km

### Question 6

An army camp of 250 men has provision for 350 days. After 25 days, 50 men died due to epidemic. The food will last for

#### SOLUTION

Solution :Remaining number of men = 250-50 =200 men

Remaining number of days = (350-25) days = 325 days.

250 men had provisions for 325 days

1 man had provisions for (250 × 325) days

200 men had provisions for (250×325)200 days

= 406.25days =406 days

hence, the remaining food will last for 406 days.

### Question 7

I decided to purchase an anniversary gift for my parents. I selected gifts, 3 of which costs Rs.550. If I want to spend Rs.1100, I could purchase 6 gifts.

True

False

#### SOLUTION

Solution :A

Number of gifts3yPrice (In Rupees)5501100

3550=y1100⇒y=3×1100550=6 gifts

### Question 8

4 oxen or 12 cows can graze a field in 32 days. How long would 7 oxen and 3 cows take to graze the same field?

12 days

17 days

32 days

16 days

#### SOLUTION

Solution :D

4 oxen = 12 cows

⇒ 1 ox = 124 cows

⇒ 7 oxen≡124×7 cows=21 cows

⇒ (7 oxen + 3 cows) ≡ (21 cows + 3 cows) = 24 cows

Now, 12 cows can graze the field in 32 days.

So, 1 cow can graze the field in (32×12) days.

[less cows, more days]

24 cows can graze the field in 32×1224=16 days

[more cows, less days]

Hence, 9 oxen and 3 cows can graze the field in 16 days.

### Question 9

Shane has to fill a water tank in 3 hours 40 minutes and he used 7 pipes for that. How many pipes should he use to fill the tank in 1 hour 21 minutes?

21 pipes

19 pipes

7 pipes

30 pipes

#### SOLUTION

Solution :B

Let the number of pipes be y.

Number of pipes7yTime (in minutes)22081

The number of pipes and the time are inversely proportional.

AB = constant

A1×B1 = A2×B2

7y = 81220⇒y=7×22081=19.01≈19

Shane has to use 19 pipes to fill the tank in 1 hour 21 minutes.

### Question 10

A bottle filling machine can fill 238 bottles in 2hrs. In 7hrs, it can fill

#### SOLUTION

Solution :Number of bottles238yTime (In hours)27

The time taken and the number of bottles filled are directly proportional.

AB= Constant

A1B1 = A2B2

2238=7y

⇒y=7×2382

⇒y=833 bottles