# Free Comparing Quantities 01 Practice Test - 8th Grade

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

A.

1 hour ago

B.

2 hours ago

C.

3 hours ago

D.

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.

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=600

R=100×6005000×1=12%

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

A.

4%

B.

10%

C.

40%

D.

60%

#### SOLUTION

Solution : A

Discount %=Marked priceSelling priceMarked price×100

=150001440015000×100=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.

A. 30%
B. 26.5%
C. 30.5%
D. 32%

#### 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 =60,00044,100=15,900
Percentage decrease =1590060000×100=26.5%

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

True

B.

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=96
x=3

Hence, 3 apples should be added.
Thus, the given statement is true.

What is the value of 200% of 12?

A. 24
B. 20
C. 6
D. 16

#### SOLUTION

Solution : A

200% of 12

=200100×12=2×12=24

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 %?

A.

1%

B.

0%

C.

3%

D.

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%.

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.

A.

430

B.

432

C.

442

D.

452

#### SOLUTION

Solution : B

We know that, C.I=P[(1+r100)n1]where P is the principal, r is the rateof interest and t is the time period.

Hence, 246=P[(1+5100)21]

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

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

A.

Cost Price

B.

Selling Price

C.

Marked Price

D.

Discount

#### SOLUTION

Solution : B

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

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.

A.

₹ 4,93,125

B.

₹ 2,01,984

C.

₹ 4,10884

D.

₹ 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(120100)2
315600=P(115)2
315600=P(4×45×5)

P=315600×5×54×4

P= 4,93,125

Purchase value of the car= 4,93,125