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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:

Find the regular polygon whose interior angel is 150 more than exterior angle. note the total sum of interior and exterior angle should be 180 at one point ?

A) 12 B) 24
C) 36 D) 6
 
Answer & Explanation Answer: B) 24

Explanation:

Sum of Interior angle is 180 .
and it is given that interior angle is 150 more than exterior,
so let exterior angle is x then interior will be x+150

now , interior + exterior =180
x+150+x=180
x=15

now , number of sides of polygon = 360 / exterior angle
n = 360 / 15
n = 24

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Filed Under: Numbers
Exam Prep: AIEEE , Bank Exams , CAT
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3 5112
Q:

There are 15 lines in plane. How many intersections (Maximum) can be made ?

A) 55 B) 105
C) 215 D) 148
 
Answer & Explanation Answer: B) 105

Explanation:

First line will cut all other 14, similarly second will cut 13, and so on
Total = 14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 105.

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Filed Under: Problems on Numbers
Exam Prep: AIEEE , Bank Exams , CAT , GATE , GRE
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6 5112
Q:

If '/' denotes 'subtraction', '-' denotes 'addition', '+' denotes 'multiplication' and 'x' denotes 'division', which of the following is the value of the expression

 

182/91x7+3-1

 

A) 16 B) 24
C) 144 D) 150
 
Answer & Explanation Answer: C) 144

Explanation:
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Filed Under: Coding and Decoding
Exam Prep: Bank Exams

0 5109
Q:

A sum is equally invested in two different schemes on CI at the rate of 15% and 20% for two years. If interest gained from the sum invested at 20% is Rs. 528.75 more than the sum invested at 15%, find the total sum?

A) Rs. 7000 B) Rs. 4500
C) Rs. 9000 D) Rs. 8200
 
Answer & Explanation Answer: C) Rs. 9000

Explanation:

Let Rs. K invested in each scheme

Two years C.I on 20% = 20 + 20 + 20x20/100 = 44%

Two years C.I on 15% = 15 + 15 + 15x15/100 = 32.25%

Now,

(P x 44/100) - (P x 32.25/100) = 528.75

=> 11.75 P = 52875

=> P = Rs. 4500

 

Hence, total invested money = P + P = 4500 + 4500 = Rs. 9000.

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Filed Under: Compound Interest
Exam Prep: AIEEE , Bank Exams , CAT , GATE
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19 5104
Q:

Rohan spends 24% of an amount of money on an insurance policy, 34% on food, 19% on children’s education and 17% on recreation. He deposits the remaining amount of Rs. 540 in bank. How much total amount does he spend on food and insurance policy together?

A) Rs. 6350 B) Rs. 5220
C) Rs. 5890 D) Rs. 6458
 
Answer & Explanation Answer: B) Rs. 5220

Explanation:

Remaining % of amount with Rohan

= 100 - (24 + 34 + 17 + 19)

= 100 - 94

= 6%

=> 6% = 540

=> 100% = ?

540 x 100/6 = Rs. 9000

Now, Money spent on insurance and food is

= 34 + 24 = 58% of 9000

= 58 x 9000/100

= 58 x 90

= Rs. 5220

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Filed Under: Percentage
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5 5103
Q:

If A1, A2, A3, A4, ..... A10 are speakers for a meeting and A1 always speaks after, A2 then the number of ways they can speak in the meeting is 

A) 9! B) 9!/2
C) 10! D) 10!/2
 
Answer & Explanation Answer: D) 10!/2

Explanation:

As A1 speaks always after A2, they can speak only in  1st  to 9th places and 

 

A2 can speak in 2nd to 10 the places only when A1 speaks in 1st place 

 

A2 can speak in 9 places the remaining 

 

 A3, A4, A5,...A10  has no restriction. So, they can speak in 9.8! ways. i.e

 

when A2 speaks in the first place, the number of ways they can speak is 9.8!.

 

When A2 speaks in second place, the number of ways they can speak is  8.8!.

 

When A2 speaks in third place, the number of ways they can speak is  7.8!. When A2 speaks in the ninth place, the number of ways they can speak is 1.8!

 

 

 

Therefore,Total Number of ways they can  speak = (9+8+7+6+5+4+3+2+1) 8! = 92(9+1)8! = 10!/2

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Filed Under: Permutations and Combinations
Exam Prep: CAT , Bank Exams , AIEEE
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5 5097
Q:

The school has always been the most important means of transferring the wealth of tradition form one generation to the next. This applies today in an even higher degree than in former times for, through the modern development of economy, the family as bearer of tradition and education has become weakened.

 

This passage best supports the statement that for transferring the wealth of tradition from one generation to the next  _______

A) there are means other than the school. B) several different sources must be tried.
C) economic development plays a crucial role D) modern technology must be put to use.
 
Answer & Explanation Answer: C) economic development plays a crucial role

Explanation:
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Filed Under: Theme Detection

18 5097
Q:

The length of a rectangle is 10 cm more than the breadth. Find the diagonal length of rectangle if perimeter is 44 cm ?

A) 12.50 cm B) 14.21 cm
C) 15.98 cm D) 17.08 cm
 
Answer & Explanation Answer: D) 17.08 cm

Explanation:

Let the length of the rectangle = L cm

 

Then the breadth of the rectangle = (L - 10) cm

 

Perimeter of a rectangle = 2(L + B) cm

 

=> 44 = 2(L + L - 10)

 

=> 44 = 4L - 20

 

=> 4L = 64

 

=> L = 16 cm

 

=> B = L - 10 = 6 cm

 

Diagonal = L2+B2 = 162+62 = 17.08 cm

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Filed Under: Area
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11 5090