Aptitude question and answer on Problems on numbers

In this section we are going to discuss about problems on numbers problems. Not just the overview of the topic, but also you are going to learn the important formulas on problems on numbers along with explanation which is easy to understand.

1
The sum of two numbers is 184. If one-third of the one exceeds one-seventh of the other by 8, find the smaller number.
A.    69
B.    70
C.    71
D.    72

Answer : D.  72

Explanation:
Let the numbers be x and (184 - x).

Then,
\({X \over3 }- {184-x \over7} = 8\)

=> 7x – 3(184 – x) = 168

=> 10x = 720 => x = 72.

So, the numbers are 72 and 112. Hence, smaller number = 72.
 

2
The ratio between a two-digit number and the sum of the digits of that number is 4 : 1.If the digit in the unit's place is 3 more than the digit in the ten’s place, what is the number?
 
A.    34
B.    35
C.    36
D.    37

Answer : C.  36

Explanation:
Let the ten's digit be x. Then, unit's digit = (x + 3).

Sum of the digits = x + (x + 3) = 2x + 3.

Number = l0x + (x + 3) = llx + 3.

\({{11x+3} \over {2x + 3 } }= {4 \over 1 }\)

=> \(11x + 3 = 4 (2x + 3)\) 

=>  3x = 9     

=> x = 3.

Hence, required number = 11x + 3 = 36.
 

3
The sum of two numbers is 15 and the sum of their squares is 113. Find the numbers.
 
A.    6 and 9
B.    7 and 8
C.    4 and 11
D.    10 and 5

Answer : B.   7 and 8

Explanation:
Let the numbers be x and (15 - x).
          Then,\( x^2 + (15 - x)^2 = 113?\)     
    
=>         \(x^2 + 225 + x^2 - 30x = 113\)   

=>       \(2x^2 - 30x + 112 = 0 \)        

=>        \(x^2 - 15x + 56 = 0\)

=>       (x - 7) (x - 8) = 0         
    
=>      x = 7  or  x = 8.

           So, the numbers are 7 and 8.
 

4
If three numbers are added in pairs, the sums equal 10, 19 and 21. Find the numbers 
A.    6, 4 and 15
B.    7, 4 and 14
C.    6, 2 and 17
D.    6, 5 and 13

Answer : A.  6, 4 and 15

Explanation:
Let the numbers be x, y and z. Then,

x+ y = 10      ...(i)     

y
+ z = 19               ...(ii)      

x + z = 21       …(iii)

Adding (i) ,(ii) and (iii), we get: 

2 (x + y + z ) = 50  or   (x + y + z) = 25.

Thus, x= (25 - 19) = 6;  y = (25 - 21) = 4;  z = (25 - 10) = 15.

Hence, the required numbers are 6, 4 and 15.
 

5
50 is divided into two parts such that the sum of their reciprocals is \(1\over 12\).Find the two parts.
A.    10 and 40
B.    15 and 35
C.    45 and 5
D.    30 and 20

Answer : D.  30 and 20

Explanation:
Let the two parts be x and  (50 - x).

Then, 1 / x + 1 / (50 – x) =  \(1 \over 12\)

=> (50 – x + x) / x ( 50 – x) = \(1 \over 12\)

=> x2 – 50x + 600 = 0

=> (x – 30) ( x – 20) = 0

=> x = 30 or x = 20.

So, the parts are 30 and 20.
 

6
The average  of  four consecutive even numbers is 27. Find the largest of these numbers.
 
A.    29
B.    30
C.    31
D.    32

Answer : B.   30

Explanation:
Let the four consecutive even numbers be x, x + 2, x + 4 and x + 6.

Then, sum of these numbers = (27 x 4) = 108.

So, x + (x + 2) + (x + 4) + (x + 6) = 108 

or  4x = 96  or  x = 24.

\(\dot{..}\)Largest number = (x + 6) = 30.
 



7
A number is as much greater than 36 as is less than 86. Find the number.
A.    71
B.    61
C.    51
D.    41

Answer : B.   61

Explanation:
Let the number be x.

Then, x - 36 = 86 - x 

=> 2x = 86 + 36 = 122

=> x = 61.

Hence, the required number is 61.
 

8
Find a number such that when 15 is subtracted from 7 times the number, the Result is 10 more than twice the number
A.    5
B.    6
C.    7
D.    8

Answer : A.  5

Explanation:
Let the number be x.

Then, 7x - 15 = 2x + 10

=> 5x = 25

=>x = 5.

Hence, the required number is 5.