# Free Comparing Quantities 01 Practice Test - 8th Grade

### Question 1

A bacterial population doubles every hour. If there are 1000 bacteria now, when were 125 bacteria present?

1 hour ago

2 hours ago

3 hours ago

5 hours ago

#### SOLUTION

Solution :C

If there are 1000 bacteria now, 1 hour ago there would have been 500 since the number of bacteria doubles every hour. 2 hours ago there would have been half of 500 bacteria, that is 250 bacteria. 3 hours ago there would have been 125 bacteria.

### Question 2

A man invests ₹ 5000 at a certain rate of interest, compounded annually. At the end of one year, it amounts to ₹ 5600. The rate of interest per annum is

#### SOLUTION

Solution :Since the tenure is 1 year, the compound interest will be same as simple interest.

Amount = Principal + Interest

Interest = 5600 - 5000 = 600

Interest for one year =PRT100=600R=100×6005000×1=12%

### Question 3

A table marked at ₹15000 is available for ₹14400. What is the discount %?

4%

10%

40%

60%

#### SOLUTION

Solution :A

Discount %=Marked price−Selling priceMarked price×100

=15000−1440015000×100=4%

### Question 4

Ravi has purchased a vehicle at ₹ 60,000. He sold it at a price 30% less than its value to Amit. Amit spent 5% of the amount he purchased in repairs. Find the percentage decrease of the price of the vehicle, including the repair price also.

#### SOLUTION

Solution :B

Amount at which Ravi sold the vehicle =60,000–(60,000×30100)=42,000

Amount spent by Amit on repairs = 42,000×5100= 2,100

Total amount spent by Amit =42,000+2,100=44,100

Decrease in price of vehicle =60,000−44,100=15,900

Percentage decrease =1590060000×100=26.5%

### Question 5

There are 6 apples and 6 oranges. Three more apples should be added so that the ratio of apples to oranges becomes 3 : 2.

True

False

#### SOLUTION

Solution :A

There are 6 apples and 6 oranges.

The ratio is 66, that is 1:1.

Let the number of apples added be x.

Then, 6+x6=32

⇒6+x=32×6

⇒x=9−6

⇒x=3

Hence, 3 apples should be added.

Thus, the given statement is true.

### Question 6

What is the value of 200% of 12?

#### SOLUTION

Solution :A

200% of 12

=200100×12=2×12=24

### Question 7

I bought a second-hand car for ₹50,000, had it transported for ₹5,000 and sold it for ₹59,400. What is my profit %?

1%

0%

3%

8%

#### SOLUTION

Solution :D

Profit is the difference between the selling price and total cost price

Total cost price = Cost price + Transportation costs = ₹50,000 + ₹5,000 = ₹55,000

Selling price = ₹59,400

Profit = ₹59400 - ₹55000 = ₹4400

Profit%=ProfitNet expenses×100

=440055000×100=8%.

### Question 8

The compound interest on a certain sum of money at 5% per annum for two years is ₹246. Calculate the simple interest on the same sum for three years at 6% per annum.

₹430

₹432

₹442

₹452

#### SOLUTION

Solution :B

We know that, C.I=P[(1+r100)n−1]where P is the principal, r is the rateof interest and t is the time period.

Hence, 246=P[(1+5100)2−1]246=P[(2120)×(2120)−1]246=P×(41400)

⇒P=246×40041=₹ 2400

Now, P= ₹ 2400,r=6% per annum and t = 3 years

S.I=P×r×t100=2400×6×3100=₹432

### Question 9

On what is Goods and Service Tax (GST) levied?

Cost Price

Selling Price

Marked Price

Discount

#### SOLUTION

Solution :B

Goods and Service Tax (GST) is levied on selling price.

### Question 10

The cost of a car depreciates at the rate of 20% every year. If its present worth is ₹ 315600, find the purchase value of the car, if it was purchased two years back.

₹ 4,93,125

₹ 2,01,984

₹ 4,10884

₹ 2,22,884

#### SOLUTION

Solution :A

Given that A = ₹ 3,15,600 , r = 20 % ; n = 2 years.

Let P be the value of the car two years ago.

A=P(1−20100)2

315600=P(1−15)2

315600=P(4×45×5)

P=315600×5×54×4

P=₹ 4,93,125

∴Purchase value of the car=₹ 4,93,125