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