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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:

Jeopardy : Peril :: Jealousy : ?

A) Lust B) Sin
C) Envy D) Insecurity
 
Answer & Explanation Answer: C) Envy

Explanation:

First is a more intense form of the second. Here Jeopardy is the more intense form of Peril and we know that Jealousy is the more intense form of Envy.

 

Hence, Jeopardy : Peril :: Jealousy : Envy.

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Filed Under: Analogy
Exam Prep: AIEEE , Bank Exams , CAT
Job Role: Analyst , Bank Clerk , Bank PO

5 7166
Q:

Which value is equal to 5% of 1,500?

A) 35 B) 55
C) 75 D) 45
 
Answer & Explanation Answer: C) 75

Explanation:

In the question it is asked for 5% of 1500

5 x 1500/100 = 15 x 5 = 75.

 

Hence, 5% of 1,500 is equal to 75.

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Filed Under: Percentage
Exam Prep: AIEEE , Bank Exams , CAT , GATE
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10 7155
Q:

If each of the 8 teams in a league must play each other three times,how many games will be played?

A) 72 B) 84
C) 36 D) 79
 
Answer & Explanation Answer: B) 84

Explanation:

Since,each member of the league must meet every other member of the league.If they only played each other once,there would be 8C2 games.Since,each pairing of teams will occur three times,the answer will be triple.

 

Therfore, 8C2*3=84

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Filed Under: Permutations and Combinations

1 7152
Q:

Jay wants to buy a total of 100 plants using exactly a sum of Rs 1000. He can buy Rose plants at Rs 20 per plant or marigold or Sun flower plants at Rs 5 and Re 1 per plant respectively. If he has to buy at least one of each plant and cannot buy any other type of plants, then in how many distinct ways can Jay make his purchase?

A) 3 B) 6
C) 4 D) 2
 
Answer & Explanation Answer: A) 3

Explanation:

Let the number of Rose plants be ‘a’.
Let number of marigold plants be ‘b’.
Let the number of Sunflower plants be ‘c’.
20a+5b+1c=1000; a+b+c=100

 

Solving the above two equations by eliminating c,
19a+4b=900

b = (900-19a)/4 

b = 225 - 19a/4----------(1)


b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e .:0 < b < 99--------(2)

Substituting (1) in (2),

 0 < 225 - 19a/4 < 99

225 <  -19a/4 < (99 -225)

=> 4 x 225 > 19a > 126 x 4

=> 900/19 > a > 505

 

a is the integer between 47 and 27 ----------(3)
From (1), it is clear, a should be multiple of 4.


Hence possible values of a are (28,32,36,40,44)


For a=28 and 32, a+b>100
For all other values of a, we get the desired solution:
a=36,b=54,c=10
a=40,b=35,c=25
a=44,b=16,c=40


Three solutions are possible.

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Filed Under: Permutations and Combinations

2 7151
Q:

If all the letters in the word  MERCIFUL are rearranged in alphabetical order and substituted by the alphabet preceding them in the English alphabet. What will be the new arrangement of letters?

A) BDHEKLQT B) BDEHKLQT
C) BDEHLKQT D) BDEJMLQT
 
Answer & Explanation Answer: B) BDEHKLQT

Explanation:

MERCIFUL   =>   CEFILMRU   => BDEHKLQT

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Filed Under: Alphabet Test

8 7150
Q:

One card is drawn from a pack of 52 cards , each of the 52 cards being equally likely to be drawn. Find the probability that the card  drawn is neither a spade nor a king.

A) 0 B) 9/13
C) 1/2 D) 4/13
 
Answer & Explanation Answer: B) 9/13

Explanation:

There are 13 spades ( including one king). Besides there are 3 more kings in remaining 3 suits

 

Thus   n(E) = 13 + 3 = 16

 

Hence nE¯=52-16=36 

  

Therefore, PE=3652=913

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Filed Under: Probability

6 7147
Q:

A take twice as much time as B or thrice as much time to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone in  ?

A) 8 hrs B) 12 hrs
C) 6 hrs D) 4 hrs
 
Answer & Explanation Answer: C) 6 hrs

Explanation:

Suppose A, B and C take x, x/2 and x/3 respectively to finish the work. 

Then, (1/x + 2/x + 3/x) = 1/2 

=> 6/x = 1/2      => x = 12 

So, B takes 6 hours to finish the work.

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Filed Under: Time and Work
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4 7145
Q:

A can do a work in 9 days, B can do a work in 7 days, C can do a work in 5 days. A works on the first day, B works on the second day and C on the third day respectively that is they work on alternate days. When will they finish the work ?

A) [7 + (215/345)] days B) [6 + (11/215)] days
C) [6 + (261/315)] days D) [5 + (112/351)] days
 
Answer & Explanation Answer: C) [6 + (261/315)] days

Explanation:

After day 1, A finishes 1/9 of the work.

After day 2, B finishes 1/7 more of the total work. (1/9) + (1/7) is finished.

After day 3, C finishes 1/5 more of total work. Total finished is 143/315.

So, after day 6, total work finished is 286/315.

 

Now remaining work =  29 /315

 

On day 7, A will work again 

 

Work will be completed on day 7 when A is working. He must finish 29/315 of total remaining work.

 

Since he takes 9 days to finish the total task, he will need 261/315 of the day. 

 

Total days required is 6 + (261/315) days.

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Filed Under: Time and Work
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6 7137