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Aptitude question and answer on Problems on numbers



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.