XtraStudy Header

⁞ Problems On Numbers

Aptitude ➤ Problems On Numbers ➤ 1

In this section, questions involving a set of numbers are put in the form of a puzzle. You have to analyze the given conditions, assume the unknown numbers and form equations accordingly, which on solving yield the unknown numbers.

Question 1
Q1.  A number is as much greater than 36 as is less than 86. Find the number.
 

Answer :  61


Let the number be x.
Then, x - 36 = 86 - x
⇒ 2x = 86 + 36 = 122
⇒ x = 61.

Hence, the required number is 61.
⚔ DISCUSS

Question 2
Q2.  If one-third of one-fourth of a number is 15, then three-tenth of that number is :
 

Answer :  54


Let the number be x.
Then, 1/3 of 1/4 of x = 15
x = 15 x 12 = 180.

So, required number = (3/10) x 180 = 54.
⚔ DISCUSS

Question 3
Q3.  Find the number, when 15 is subtracted from 7 times the number, the result is 10 more than twice of the number ?
 

Answer :  5


Let the number be x.
Then, According to question,
7x - 15 = 2x + 10
⇒ 5x = 25
⇒ x = 5
⚔ DISCUSS

Question 4
Q4.  The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number ?
 

Answer :  4


Let the ten's digit be x and unit's digit be y.
Then,
(10x + y) - (10y + x) = 36
⇒ 9(x - y) = 36
⇒ x - y = 4.
⚔ DISCUSS

Question 5
Q5.  The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is :
 

Answer :  20


Let the numbers be a, b and c.
Then,
a2 + b2 + c2 = 138 and (ab + bc + ca) = 131
Now,
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
= 138 + 2 x 131 = 400
⇒ (a + b + c) = √400 = 20.
⚔ DISCUSS

Comments
Comments (0)


XtraStudy
Explore
Explore
About
App Link
xtrastudy android app
Others

Earn While You Learn
FAQ

Register Now

© 2022 Company, Inc. All rights reserved.