## IBM Sample Problems with digits of a number

Dear Reader,

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

**Answer : **c)36

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.

Hence the answer is 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

**Answer :** b)8

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

**Answer : **a)4

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

22x^{2} + 22x + 4x + 4 = 144

22x^{2} + 26x - 140 = 0

11x^{2} + 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.

Hence the answer is 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

**Answer :** c) y=2x.

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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