# 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. All students, freshers can download Aptitude Simplification quiz questions with answers as PDF files and eBooks.

## Where can I get Aptitude Simplification Interview Questions and Answers (objective type, multiple choice)?

Here you can find objective type Aptitude Simplification questions and answers for interview and entrance examination. Multiple choice and true or false type questions are also provided.

## 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.

### Exercise :: Simplification - General Questions

1.

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 ?

 A. 45 B. 60 C. 75 D. 90

Explanation:

Let number of notes of each denomination be x.

Then x + 5x + 10x = 480

16x = 480

x = 30.

Hence, total number of notes = 3x = 90.

2.

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:

 A. 20 B. 80 C. 100 D. 200

Explanation:

Let the number of students in rooms A and B be x and y respectively.

Then, x - 10 = y + 10      x - y = 20 .... (i)

and x + 20 = 2(y - 20)      x - 2y = -60 .... (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

The required answer A = 100.

3.

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:

 A. Rs. 3500 B. Rs. 3750 C. Rs. 3840 D. Rs. 3900

Explanation:

Let the cost of a chair and that of a table be Rs. x and Rs. y respectively.

 Then, 10x = 4y   or   y = 5 x. 2

15x + 2y = 4000

 15x + 2 x 5 x = 4000 2

20x = 4000

x = 200.

 So, y = 5 x 200 = 500. 2

Hence, the cost of 12 chairs and 3 tables = 12x + 3y

= Rs. (2400 + 1500)

= Rs. 3900.

4.

If a - b = 3 and a2 + b2 = 29, find the value of ab.

 A. 10 B. 12 C. 15 D. 18

Explanation:

2ab = (a2 + b2) - (a - b)2

= 29 - 9 = 20

ab = 10.

5.

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 ?

 A. Rs. 1200 B. Rs. 2400 C. Rs. 4800 D. Cannot be determined E. None of these

Explanation:

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
----------------
4y =  800
----------------

Therefore, y = 200.

Now apply value of y in (iii)

=>  x + 2 x 200 = 800

=>  x + 400 = 800

Therefore x = 400

```

Solving (i) and (ii) we get x = 400, y = 200.

Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.

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