## IBM Sample Problems with digits of a number

Below are four sample problems on two digit numbers involving calculations on the digits of the number.

Question 1
If the digit in the ten's place of a two digit number is 3 less than the digit unit’s place and if the ratio of the sum of the digits of that number to that of the number is 1:4 then the number is:
a)32 b)34 c)36 d)38

Solution :

Let the ten’s digit of the number be x.
Then the unit’s digit is x+3
Sum of the digits = x + x + 3 = 2x+3.
Then the number can be expressed as 10x +(x+3) = 11x + 3
The given ratio between the sum of digits to the sum = 1 / 4 = 2x+3 / 11x+3
4(2x + 3) = 11x + 3
3x = 9
x = 3
Then the required number is 11x + 3 = 11(3) + 3 = 36.

Question 2
The sum of the digits of a two digit number is equal to 9 and the number obtained by interchanging the digits of that number is 63 less than the number. Then the ten’s place of the two digit number is :
a)1 b)8 c)9 d)3

Solution :

Let the ten’s place of the two digit number be x.
Since the sum of the digit is 9, the unit's place will be 9-x.
Then the number is represented by = 10x +(9-x) = 9x + 9

For the number with digits reversed, tens's place will be 9 - x and unit's place will be 9.
Therefore, the reversed digits number can be represented as 10(9-x)+x = 90 - 9x

Given that the new number is 63 less than the original number.
Therefore, New Number = Original Number - 63
i.e 90-9x = 9x + 9 - 63
18x = 144
x = 144/18
x = 8.
Then the number is 9(8) + 9 = 81.
Hence the required digit is 8.

Question 3
If unit’s place of a two digit number is 2 more than the ten’s place and 144 is the product of the number and the sum of the digits of the number then the unit place of the number is:
a)4 b)1 c)5 d)3

Solution :

Let the ten’s digit of the two digit number be x.
Then the unit’s digit is x + 2.
And the number = 10x + x + 2 = 11x + 2
The sum of the digits of the number = x + x + 2 = 2x + 2
Product of the number and sum of the digits = 144
i.e (11x + 2)(2x + 2) = 144
22x2 + 22x + 4x + 4 = 144
22x2 + 26x - 140 = 0
11x2 + 13x - 70 = 0
(x - 2)(11x + 35) = 0
x = 2 or x = -35/11
Since digit of a number cannot be a fraction, x has to be 2.
i.e., ten’s place of the number is 2.
And unit’s place = x + 2 = 4.

Question 4
If x and y are the numbers of ten’s and unit’s place of a two digit number(xy). If y is halved and x is doubled, the number obtained is equal to the number yx (i.e., interchange the digits of xy). Then the relation between x and y is:
a) x+y=2 b) x=2y c) y=2x d) x-y=2

Solution :

Given that the ten’s and unit’s digit is x and y respectively.
Then the number = 10x + y.
The number obtained by interchanging the digits = 10y + x.
If y is halved (y/2) and x is doubled then the new number is 10(2x) + y/2
But the new number is equal to the number obtained by interchanging the digits.
Therefore, 10(2x) + y/2 = 10y+x
20x - x = 19y / 2
19x = 19y / 2
x = y/2
y = 2x
Hence the answer is option c.

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