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Aptitude - Simplification
Why Aptitude Simplification?
In this section you can learn and practice Aptitude Questions based on "Simplification" and improve your skills in order to face the interview, competitive examination and various entrance test (CAT, GATE, GRE, MAT, Bank Exam, Railway Exam etc.) with full confidence.
Where can I get Aptitude Simplification questions and answers with explanation?
IndiaBIX provides you lots of fully solved Aptitude (Simplification) questions and answers with Explanation. Solved examples with detailed answer description, explanation are given and it would be easy to understand. View the solution for the problems with feel and good user interface, easily go through all questions and answers.
How to solve Aptitude Simplification problems?
You can easily solve all kind of Aptitude questions based on Simplification by practicing the objective type exercises given below, also get shortcut methods to solve Aptitude Simplification problems.
A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?
There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:
The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Rs. 4000. The total price of 12 chairs and 3 tables is:
The price of 2 sarees and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ?
Let the price of a saree and a shirt be Rs. x and Rs. y respectively.
Then, 2x + 4y = 1600 .... (i)
and x + 6y = 1600 .... (ii)
Divide equation (i) by 2, we get the below equation.
=> x + 2y = 800. --- (iii)
Now subtract (iii) from (ii)
x + 6y = 1600 (-)
x + 2y = 800
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4y = 800
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Therefore, y = 200.
Now apply value of y in (iii)
=> x + 2 x 200 = 800
=> x + 400 = 800
Therefore x = 400
16x = 480
x = 30.
