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Aptitude and Reasoning Questions

What are Different sections in Aptitude and Reasoning Question and Answers?

In this highly competitive world, Aptitude and reasoning tests are prominently important. Typically, there are multiple sections in this type of tests. Broadly they are: Verbal Reasoning(Mental Ability, Logical Deduction), Non-Verbal Reasoning, Quantitative Aptitude(Data interpretation, Arithmetic Ability).


Why Aptitude and Reasoning Questions?

Aptitude and Reasoning tests simply mean to measure or determine a person's ability in a particular skill or field of knowledge. These days most of Aptitude and Reasoning tests are in online format. With proper practice of these aptitude tests, They are easy to crack. Preparing for Aptitude and Reasoning tests will often avoid disappointments in Entrance Exams for various competitive exams and job interviews.


What type of questions are there in Aptitude and Reasoning tests?

Aptitude and Reasoning tests consists of various Arithmetic, Data interpretation, Diagrammatic and Psychometric question and answers. Though these tests seem confusing at first, with proper practice and applied logic, they are very easy to crack. Later, it becomes interesting to solve such puzzles. Most of these are based on a particular sequence, hence it is important to understand the sequence to solve the problem. Practicing these more and more will yield better results.

Popular Latest Rated
Q:

Ajay and his wife Reshmi appear in an interview for two vaccancies in the same post. The Probability of Ajay's selection is 1/7 and that of his wife Reshmi's selection is 1/5. What is the probability that only one of them will be selected?

A) 5/7 B) 1/5
C) 2/7 D) 2/35
 
Answer & Explanation Answer: C) 2/7

Explanation:

P( only one of them will be selected) = p[(E and not F) or (F and not E)] 

 = PE∩F∪F∩E 

 

= PEPF+PFPE

 

 =17×45+15×67=27

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Filed Under: Probability
Exam Prep: AIEEE , Bank Exams , CAT , GATE
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17 6315
Q:

If log330 = 1a  and  log530=1b then the value of 3log302 is:

A) 3(1+a+b) B) 2(1-a-b)
C) 3(1-a-b) D) 3(1+a-b)
 
Answer & Explanation Answer: C) 3(1-a-b)

Explanation:
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Filed Under: Logarithms
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25 6301
Q:

In the following question, various terms of an alphanumerical series are given with one or more terms missing as shown by (?). Choose the missing terms out of the given alternatives. Q1F, S2E, U6D, W21C, ?

A) Z88B B) Y88B
C) Y44B D) K25P
 
Answer & Explanation Answer: B) Y88B

Explanation:

The first letter moves two steps forward. The last letter moves one steps backward. The number series runs as follows :
1 x 1 + 1 = 2
2 x 2 + 2 = 6
6 x 3 + 3 = 21
21 x 4 + 4 = 88.

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Filed Under: Number Series

9 6298
Q:

Prasanna invested certain amount in three different schemes X, Y and Z with the rate of interest 10% p.a, 12% p.a and 15% p.a respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in Scheme Z was 150% of the amount invested in Scheme X and 240% of the amount invested in Scheme Y, what was the amount invested in Scheme Y by Prasanna ?

A) Rs.6000 B) Rs.4500
C) Rs.7500 D) Rs.5000
 
Answer & Explanation Answer: D) Rs.5000

Explanation:

Let a, b and c be the amounts invested in schemes X, Y and Z respectively. Then,

As we know:

Simple interest (S.I.) = PTR/100

(a × 10 × 1/100) + (b × 12 × 1/100) + (c × 15 × 1/100) = 3200

= 10a + 12b + 15c = 320000 .........(1)

Now, c = 240% of b = 12b/5 .........(2)

And, c = 150% of a = 3a/2 => a = 2/3 c = (2 × 12)b/(3 × 5) = 8b/5 .......(3)

From (1), (2) and (3), we have

16b + 12b + 36b = 320000 => 64b = 320000 => b = 5000

∴ Sum invested in Scheme Y = Rs.5000.

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Filed Under: Percentage
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9 6295
Q:

In the following question, various terms of an alphanumerical series are given with one or more terms missing as shown by (?). Choose the missing terms out of the given alternatives. EJO, TYD, INS, XCH ?

A) NRW B) NSX
C) MRW D) MSX
 
Answer & Explanation Answer: C) MRW

Explanation:

There is a gap of four letters between the first and second, the second and third letters.
There is a gap of four letters between the last letter and the first letter of the next term.

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Filed Under: Number Series

17 6294
Q:

In an election between two candidates 70% of the voters cast their votes, out of which 2% of the votes were declared invalid. A candidate got 7203 votes which was 60% of the total valid votes. Find the total number of voters enrolled in that election.

A) 18050 B) 17500
C) 17000 D) 7203
 
Answer & Explanation Answer: B) 17500

Explanation:

Let total number of votes enrolled = x
No. of voted cast = 70% of x = 0.7x
Valid votes = 98% of 0.7x
⇒ 60% of total valid votes = 7203 = 60% of 98% of 0.7x
⇒ 0.6 X 0.98 X 0.7x = 7203
⇒ x = 7203 / (0.6 X 0.98 X 0.7)
⇒ x = 17500

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Filed Under: Percentage
Exam Prep: Bank Exams

1 6285
Q:

What is the missing number in 5, 14, 42, 107, 233, ?

A) 316 B) 384
C) 450 D) 479
 
Answer & Explanation Answer: C) 450

Explanation:

Given Series is 5, 14, 42, 107, 233, _?

5, 14, 42, 107, 233, _?

Differences is 9, 28, 65, 126

9 = 2^3 + 1

28 = 3^3 + 1

65 = 4^3 + 1

126 = 5^3 + 1

Now, next difference must be 6^3 + 1 = 217

Hence, next number in the given series = 233 + 217 = 450.

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Filed Under: Odd Man Out
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4 6281
Q:

A jogger running at 9 kmph alongside a railway track is 240 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction. In how much time will the train pass the jogger  ?

A) 10 sec B) 48 sec
C) 36 sec D) 56 sec
 
Answer & Explanation Answer: C) 36 sec

Explanation:

Given speed of jogger = 9 kmph
Speed of train = 45 kmph
As they are in same direction,
Relative speed = 45 - 9 = 36 kmph = 36x5/18 = 10 m/s
Distance = 240 + 120 = 360 mts
Time = distance/speed = 360/10 = 36 sec.

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Filed Under: Problems on Trains
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2 6281