# Problems on Numbers - Aptitude test, questions, shortcuts, solved example videos

Video on Problems on Numbers - shortcuts, tips and tricks

## Problems on Numbers

Basic Formulae: (Must Remember)

1) (a - b)2 = (a2 + b2 - 2ab)
2) (a + b)2 = (a2 + b2 + 2ab)
3) (a + b) (a – b) = (a2 – b2 )
4) (a3 + b3) = (a + b) (a2 – ab + b2)
5) (a3 - b3) = (a - b) (a2 – ab + b2)
6) (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)
7) (a3 + b3 + c3 – 3abc) = (a + b + c) (a2 + b2 + c2 – ab – bc – ac)

Question Variety

Type 1: Find the numbers, if we are given
a) Sum of squares/sum of products
b) Sum/average of consecutive numbers

Q 1. If the sum two numbers is 31 and their product is 240, then find the absolute difference between the numbers.

a. 1
b. 3
c. 4
d. 5
View solution

Correct Option: (a)

Let two numbers be x and y
We are given that, sum of two numbers x + y = 31 and product = xy = 240
Therefore,
x – y = (x + y)2 – 4xy
Substituting the values, we get
x – y = (31)2 – 4 × y
= 961 – 960
= 1
= 1
The required difference between the numbers is 1.

Q 2. Find the three consecutive odd numbers whose sum of the squares is 2531.

a. 19, 21, 23
b. 23, 25, 27
c. 27, 29, 31
d. 31, 33, 35
View solution

Correct Option: (c)

Let three consecutive odd numbers be x, x+2, x+4.
x2+(x+2)2+(x+4)2=2531
Simplifying we get,
x2+4x-837=0
27 x 31 = 837 and also the difference between 27 and 31 is 4
Therefore,
x2+31x-27x-837=0
(x + 31) (x – 27)
X = 27 or x = –31
Hence, the value of
x = 27
(x+2) = 27 + 2 = 29
(x+4) = 27 + 4 = 31

Q 3. The sum of squares of three numbers is 138 and the sum of their products taken two at a time is 131. Find their sum.

a. 35
b. 42
c. 20
d.18
View solution

Correct Option: (c)

Let the three numbers be x, y and z.
Given: Sum of squares of three numbers is 138 and sum of their products taken two at a time is 131
Therefore,
x2+y2+z2=138
xy+yz+zx=131
Formula: (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)
This formula can be used to easily find the sum of three numbers. Substituting the values, we get
(x + y + z) 2 = x2 + y2 + z2 + 2 (xy + yz + zx)
(x + y + z ) 2 = 138 + 2(131)
(x + y + z ) 2 = 400
Hence, (x + y + z) = 20

Q 4. The average of 3 consecutive even numbers is 18, find the largest of these numbers.

a. 15
b. 16
c. 20
d. 26
View solution

Correct Option: (c)

Let the numbers be x, x +2, and x + 4
Hence, sum of these numbers = (average x no. of numbers) = (18 x 3) = 54
x + (x +2) + (x + 4) = 54
3x = 48
x = 16
Largest number = (x + 4) = (16 + 4) = 20

Type 2: n Times the numbers

Q 5. Find a positive number which when increased by 11 is equal to 60 times the reciprocal of the number
a. 3
b. 4
c. 6
d. 9
View solution

Correct Option: (b)

Let the number be x.
A positive number (x) increased by 11 is equal to 60 times the reciprocal of the number (1/x)

 x + 11 = 60 x
x2+11x-60=0
x2+15x-4x-60=0
x(x+15)-4(x+15)=0
(x+15)(x-4)=0
x = 4

The positive number (x) = 4

Q 6. 5 times a positive number is less than its square by 24. What is the integer?
a. 5
b. 8
c. 7.5
d. 9
View solution

Correct Option: (b)
Let the unknown number be x.
5 times a positive number = 5x
5 times a positive number is less than its square by 24
x2 – 5x = 24
x2 + 3x – 8x – 24
x (x + 3) – 8(x + 3)
(x - 8) (x + 3)
x = 8
8 is the required integer.

Type 3: Fraction

Q 7. The product of fraction A and B is (3/49). Find the value of fraction A, if fraction A is thrice fraction B.

a. 7/9
b. 5/7
c. 3/7
d. 5/9
View solution

Correct Option: (c)

Product of fraction A and B is (3/49).

 AB = 3 49

But fraction A is thrice fraction B i.e A = 3B.
 Value of B = 1 A 3

Therefore, substitute the value of B in AB
 1 AA = 3 3 49
 1 A2 = 3 3 49
 A2 = 9 49
 Therefore, A = 3 7

Q 8. The sum of numerator and denominator of a fraction is 30. If 2 is added to numerator and 2 is subtracted from denominator, then it becomes 2/3. Find the fraction.

a. 1/2
b. 1/3
c. 2/3
d. 1/4
View solution

Correct Option: (a)

Read the question breaking stepwise and understand given parameters.

 1) Let the fraction be a b
2) Sum of numerator and denominator=a+b=30-----(1)
3) 2 is added to numerator and 2 is subtracted from denominator. Hence,
 a + 2 = 2 b – 2 3
3(a+2)=2(b-2)
Solving, we get
3a+6=2b-4
3a-2b=-10-----(2)
Solve equations (1) and (2), we get
a = 10 and b = 20
 Therefore, the fraction = 10 = 1 20 2

Q 9. The denominator of a fraction is 2 more than numerator. If the numerator as well as denominator is increased by 4, the fraction becomes 8/10. Find the original fraction.

a. 2/3
b. 1/3
c. 4 /7
d. 2/5
View solution

Correct Option: (a)

Let the numerator be x.
Denominator of a fraction is 2 more than numerator. Therefore, denominator = x + 2

 Now, Numerator = 8 Denominator 10
 x = 8 x + 2 10
The numerator and denominator are increased by 4. Therefore,
 x + 4 = 8 (x + 2) + 4 10
10 (x + 4) = 8 ( x + 6)
10 x + 40 = 8x + 48
2x = 8
x = 4
 Hence, the fraction is 4 = 2 6 3

Practice questions on Numbers
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