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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:

One year ago the ratio between Mohan and Ashok's age was 4:3 after one yerar the rato of their ages will be 5:4 . what is the sunm of their present ages in years.

Answer

Let their ages one year  ago be 4x and 3x years


  


   


 Sum of their present ages =(4x +1+3x+1) = 16 years.

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Subject: Problems on Ages

6 6913
Q:

How many six digit odd numbers can be formed from the digits 0, 2, 3, 5, 6, 7, 8, and 9 (repetition not allowed)?

A) 8640 B) 720
C) 3620 D) 4512
 
Answer & Explanation Answer: A) 8640

Explanation:

Let the 6 digits of the required 6 digit number be abcdef

Then, the number to be odd number the last digit must be odd digit i.e 3, 5, 7 or 9

The first digit cannot be ‘0’ => possible digits = 3, 5, 7, 2, 6, 8

Remaining 4 places can be of 6 x 5 x 4 x 3 ways

This can be easily understood by

Therefore, required number of ways = 6 x 6 x 5 x 4 x 3 x 4 = 36 x 20 x 12 = 720 x 12

= 8640 ways.

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Filed Under: Permutations and Combinations
Exam Prep: AIEEE , Bank Exams , CAT
Job Role: Bank Clerk , Bank PO

10 6909
Q:

Den Bosch and Eastbourne are two famous cities 300 km apart. Maradona starts from Den Bosch at 8:24 am . An hour later pele starts from Den Bosch. After travelling for 1 hour , Pele reaches Nottingham that Maradona had passed 40 minutes earlier. Nottingham falls on the way from Den Bosch to Eastbourne. If pele and Maradona just reaches  Eastbourne at the same time, what are thespeeds of the Maradona and Pele Respectively? 

A) 100 km/h , 125 km/h B) 60 km/h , 80 km/h
C) 60 km/h , 75 km/h D) 75 km/h , 100 km/h
 
Answer & Explanation Answer: D) 75 km/h , 100 km/h

Explanation:

Let Pele x km in 1 hour. so Maradona takes (2h-40min)=1 h 20 min to cover x km. Let speed of Maradona and Pele be M and P respectively than 

                            x =M x (4/3)  and x= P x 1

  => M/P = 3/4 

 Again      300M=300P=1

⇒3003k-3004k=1 

 => k= 25

  M= 3k = 75 km/h   and   P= 4k = 100 km/h 

( Through options it is very easy to solve)

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Filed Under: Time and Distance

9 6905
Q:

A class has 8 football players. A 5-member team and a captain will be selected out of these 8 players. How many different selections can be made ? 

A) 210 B) 168
C) 1260 D) 10!/6!
 
Answer & Explanation Answer: B) 168

Explanation:

we can select the 5 member team out of the 8 in 8C5 ways = 56 ways.

The captain can be selected from amongst the remaining 3 players in 3 ways.

Therefore, total ways the selection of 5 players and a captain can be made = 56x3 = 168 ways.

 

(or)

 

Alternatively, A team of 6 members has to be selected from the 8 players. This can be done in 8C6 or 28 ways.

Now, the captain can be selected from these 6 players in 6 ways.

Therefore, total ways the selection can be made is 28x6 = 168 ways.

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Filed Under: Permutations and Combinations
Exam Prep: GATE , CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

8 6898
Q:

A sum of Rs.1890 has to be used to give 9 prizes to the customers of a super market for their overall academic purchases. If each prize is Rs.30 less than its preceding price, what is the least value of the price ?

A) 90 B) 95
C) 85 D) 80
 
Answer & Explanation Answer: A) 90

Explanation:

Let the least value of the prize = Rs. x

Then the next value of the prize is x+30 , x+60, x+90, ....x+240.

Given total amount of cash prizes = Rs.1890

--> x + (x+30) + (x+60) + (x+90) + ....+ (x+240) = 1890

--> 9x + (30 + 60 + 90 + 120 + 150 + 180 + 210 + 240) = 1890

--> 9x + 30(1 + 2 + 3 + 4....+ 8) = 1890

--> 9x + 30(36) = 1890

--> 9x = 810 --> x=90

Hence the least value of the prize x=90

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Filed Under: Mathematical Operations
Exam Prep: CAT , Bank Exams
Job Role: Bank PO , Bank Clerk

29 6897
Q:

When 5% is lost in grinding Maize, a country has to import 20 million bags; but when only 2% is lost, it has to import only 15 million bags. Find the quantity of Maize, which grows in the country ?

A) 188 million B) 166 2/3 million
C) 192 3/4 million D) None
 
Answer & Explanation Answer: B) 166 2/3 million

Explanation:

5% - 2% = 3%
=> 3% ------ 5 million bags
100% ---- ?
=> 166 2/3 million bags.

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Filed Under: Arithmetical Reasoning
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Bank Clerk , Bank PO

8 6896
Q:

When six fairs coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads ?

A) 15 B) 42
C) 16 D) 40
 
Answer & Explanation Answer: B) 42

Explanation:

The question requires you to find number of the outcomes in which at most 3 coins turn up as heads.

i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads.

 

The number of outcomes in which 0 coins turn heads is 6C0=1 

The number of outcomes in which 1 coin turns head is =6C1=6 

The number of outcomes in which 2 coins turn heads is6C2=15 

The number of outcomes in which 3 coins turn heads is6C3=20

 

Therefore, total number of outcomes =1+6+15+20= 42 outcomes

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Filed Under: Permutations and Combinations

3 6892
Q:

If the following series is written in the reverse order, which number will be fourth to the right of the seventh number from the left?

7, 3, 9, 7, 0, 3, 8, 4, 6, 2, 1, 0, 5, 11, 13

 

A) 5 B) 11
C) 9 D) 0
 
Answer & Explanation Answer: D) 0

Explanation:

The given series when written in reverse order becomes:

13, 11, 5, 0, 1, 2, 6, 4, 8, 3, 0, 7, 9, 3, 7

The seventh number from the left is 6, the fourth number to the right of 6 is 0

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Filed Under: Number, Ranking and Time Sequence Test

14 6883