Basic Numeracy - Online Test

Q1. 3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day must 4 pumps work to empty the tank in 1 day?
Answer : Option D
Explaination / Solution:

Let the required number of working hours per day be x.

More pumps, Less working hours per day (Indirect Proportion)

Less days, More working hours per day (Indirect Proportion)

Pumps4:3:: 8 : x
Days1:2

 4 x 1 x x = 3 x 2 x 8

 x =(3 x 2 x 8)
(4)

 x = 12.


Q2. Running at the same constant rate, 6 identical machines can produce a total of 270 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes?
Answer : Option B
Explaination / Solution:

Let the required number of bottles be x.

More machines, More bottles (Direct Proportion)

More minutes, More bottles (Direct Proportion)

Machines6:10:: 270 : x
Time (in minutes)1:4

 6 x 1 x x = 10 x 4 x 270

 x =(10 x 4 x 270)
(6)

 x = 1800.


Q3. A boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 68 km downstream.
Answer : Option C
Explaination / Solution:

Speed downstream = (13 + 4) km/hr = 17 km/hr.

Time taken to travel 68 km downstream =68hrs = 4 hrs.
17


Q4. A man's speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man's speed against the current is:
Answer : Option C
Explaination / Solution:

Man's rate in still water = (15 - 2.5) km/hr = 12.5 km/hr.

Man's rate against the current = (12.5 - 2.5) km/hr = 10 km/hr.


Q5. Which of the following statements is not correct?
Answer : Option B
Explaination / Solution:

(a) Since loga a = 1, so log10 10 = 1.

(b) log (2 + 3) = log 5 and log (2 x 3) = log 6 = log 2 + log 3

      log (2 + 3)  log (2 x 3)

(c) Since loga 1 = 0, so log10 1 = 0.

(d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3.

So, (b) is incorrect.


Q6. If log 2 = 0.3010 and log 3 = 0.4771, the value of log5 512 is:
Answer : Option C
Explaination / Solution:

log5 512
=log 512
log 5
=log 29
log (10/2)
=9 log 2
log 10 - log 2
=(9 x 0.3010)
1 - 0.3010
=2.709
0.699
=2709
699
= 3.876

Q7. In order to obtain an income of Rs. 650 from 10% stock at Rs. 96, one must make an investment of:
Answer : Option B
Explaination / Solution:

To obtain Rs. 10, investment = Rs. 96.

To obtain Rs. 650, investment = Rs.96x 650= Rs. 6240.
10


Q8. A man buys Rs. 20 shares paying 9% dividend. The man wants to have an interest of 12% on his money. The market value of each share is:
Answer : Option B
Explaination / Solution:

Dividend on Rs. 20 = Rs.9x 20= Rs.9.
1005

Rs. 12 is an income on Rs. 100.

 Rs.9is an income on Rs.100x9= Rs. 15.
5125


Q9. A man purchased a cow for Rs. 3000 and sold it the same day for Rs. 3600, allowing the buyer a credit of 2 years. If the rate of interest be 10% per annum, then the man has a gain of:
Answer : Option A
Explaination / Solution:

C.P. = Rs. 3000.

S.P. = Rs.3600 x 100= Rs. 3000.
100 + (10 x 2)

Gain = 0%.


Q10. A trader owes a merchant Rs. 10,028 due 1 year hence. The trader wants to settle the account after 3 months. If the rate of interest 12% per annum, how much cash should he pay?
Answer : Option B
Explaination / Solution:

Required money= P.W. of Rs. 10028 due 9 months hence
= Rs.10028 x 100
100 +12 x9
12
= Rs. 9200.