Dear Reader, Below are 3 sample generic aptitude questions with solutions for your practice.
Question 1
Chandrasekaran worked very sincerely in the software company which selected him in the campus recruitment. At the end of the first year he was given a salary increase of 5%. In the second year another friend from his college also joined the company and they spent more time together. As a result of this performance of Chandrasekaran came down in terms of the team leader. The team leader recommended a cut of 2.5% from the salary of Chandrasekaran. In the third year at the end of the first month he received Rs.34398 as salary and felt aggrieved. He started thinking about resigning from this company and try for other companies which pay a higher salary. What was the salary of Chandrasekaran at the beginning of first year?
a) Rs.32400 b) Rs.36800 c) Rs.22000 d) Rs.33600
Answer : d) Rs. 33600
Solution :
Let the salary at the beginning of first year be Rs.X
At the beginning of second year it will then be 5% more than his beginning salary. Therefore his second year salary = 105% of X = 105X/100
On this 2.5% reduction , his salary now will be 2.5% lesser than his second year salary = 97.5% x Second Year Salary = (97.5/100)(105X/100) – But this value is given as Rs.34398
Solving (97.5/100)(105X/100) = 34398, we get,
X = 34398 x 100 x 100 divided by 105 x 97.5
= Rs. 33600
At the beginning of second year it will then be 5% more than his beginning salary. Therefore his second year salary = 105% of X = 105X/100
On this 2.5% reduction , his salary now will be 2.5% lesser than his second year salary = 97.5% x Second Year Salary = (97.5/100)(105X/100) – But this value is given as Rs.34398
Solving (97.5/100)(105X/100) = 34398, we get,
X = 34398 x 100 x 100 divided by 105 x 97.5
= Rs. 33600
Question 2
Miss. Ranjitha bought a new wall clock which rings at every one hour. At 3 O’ clock the clock ticks 3 times and at 4 O’ clock the clock ticks 4 times. At 4 O’ clock the time between the first tick and the last tick is 21 seconds. How long does it tick at 12 O’ clock?
a) 66 seconds b) 55 seconds c) 77 seconds d)44 seconds
Answer : c) 77 seconds
Solution :
This is a simple problem. The gap between the first tick and the last tick at 4 O' clock is 3 gaps. Time taken = 21 seconds. That means between two gaps it takes 7 seconds.
At 12 O clock there will be 11 gaps. Total time for 11 gaps = 77 seconds.
Though the question could look confusing at first, this is actually a very simple questions. With practice, you should be ready to identify these kinds of questions which could potentially save a lot of your time during tests.
At 12 O clock there will be 11 gaps. Total time for 11 gaps = 77 seconds.
Though the question could look confusing at first, this is actually a very simple questions. With practice, you should be ready to identify these kinds of questions which could potentially save a lot of your time during tests.
Question 3
Anushka has a jewel chest containing Rings, Pins and Ear-rings. The chest contains 26 pieces. Anushka has 2 1/2 times as many rings as pins, and the number of pairs of ear rings is 4 less than the number of rings. How many earrings does Anushka have?
a) 12 b)8 c)6 d)10
Answer : a) 12
Solution :
Assume that there are R rings, P pins and E pair of ear-rings.
It is given that, she has 2 1/2 times as many rings as pins.
R = (5/2) * P or P = (2*R)/5
R = (5/2) * P or P = (2*R)/5
And, the number of pairs of earrings is 4 less than the number of rings.
E = R - 4 or R = E + 4
E = R - 4 or R = E + 4
Also, there are in total 26 pieces.
R + P + 2*E = 26
R + (2*R)/5 + 2*E = 26
5*R + 2*R + 10*E = 130
7*R + 10*E = 130
7*(E + 4) + 10*E = 130
7*E + 28 + 10*E = 130
17*E = 102
E = 6
R + P + 2*E = 26
R + (2*R)/5 + 2*E = 26
5*R + 2*R + 10*E = 130
7*R + 10*E = 130
7*(E + 4) + 10*E = 130
7*E + 28 + 10*E = 130
17*E = 102
E = 6
Hence, there are 6 pairs of Ear-rings i.e. total 12 Ear-rings
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