# 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)

 Pumps 4 : 3 :: 8 : x Days 1 : 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)

 Machines 6 : 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 = 68 hrs = 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. 96 x 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. 9 x 20 = Rs. 9 . 100 5

Rs. 12 is an income on Rs. 100. Rs. 9 is an income on Rs. 100 x 9 = Rs. 15. 5 12 5

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 x 9 12
= Rs. 9200.