# Solve Simplification solutions Quantitative Aptitude Problems and Solutions ## Simplification Solutions Quantitative Aptitude Problems and Solutions

Banking Classes Online : Takshila Learning institute regularly offers you different blogs and articles on important topics, news, and information related to banking exams. In this article, the following topic ’Simplification’ from Quantitative section is explained.

## SIMPLIFICATION SOLUTIONS

Q: 1 Aman has Rs. 480 in the denomination of 1-rupee notes, 5-rupee notes and 10-rupee notes. The number of notes of each denomination are equal. What is the total number of notesthat Aman has?

(A) 40                          (B) 65      (C) 85                               (D) 90

Sol: Let the number of notes of each denomination be x.

Then x + 5x + 10x = 480

16x = 480

So, x = 30.

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

Q: 2 Amit has some hens and horses. If the number of heads is counted to be 48 and the number of feet equals140, then the number of hens will be:

(A) 22                                     (B) 21                          (C) 25                              (D) 26

Sol: Let the number of hens be x and the number of horses be y.

Then, x + y = 48 …. (i)

and 2x + 4y = 140            So, x + 2y = 70 …. (ii)

Solving (i) and (ii) we get: x = 26, y = 22.

So, the number of hens = 26.

Q: 3 In a regular week, there are 5 working days and for each day, the working hours are fixed at 8. Vinodgets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then for how many hours does he work for?

(A) 165                                   (B) 175                                (C) 185                              (D) 195

Sol: Suppose Vinod works overtime for x hours.

Now, working hours in 4 weeks = (5 * 8 * 4) = 160.

160 * 2.40 + x * 3.20 = 432

3.20x = 432 – 384 = 48

x = 15.

Hence, total number of hours he worked = (160 + 15) = 175.

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Q: 4 If x – y = 3 and x2 + y2 = 29, then find the value of xy.

(A) 10                                (B) 13                                    (C) 16                                  (D) 20

2xy = (x2 + y2) – (x – y)2

= 29 – 9 = 20

xy = 10.

Q: 5 There are two examination rooms X and Y. If 10 students are sent from X to Y, then the number of students in each examination room is the same. If 20 candidates are sent from Y to X, then the number of students in X is double the number of students in Y. The number of students in room X is:

(A)120 (B)180                                 (C) 100                               (D) 200

Let the number of students in rooms X and Y be x and y respectively.

Then, x – 10 = y + 10           So, x – y = 20 …. (i)

and x + 20 = 2(y – 20)         So, x – 2y = -60 …. (ii)

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

Hence, the required answer is X = 100.

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