WORD PROBLEMS ON DIGITS AND NUMBERS

Problem 1 :

The sum of the digits of a two digit number is 12. The number obtained by interchanging the two digits exceeds the given number by 18. Find the number.

Solution:

Let the tens digit of the required number be x and the units digit be y.

x + y = 12 --- > (1)

Required Number = (10x + y)

Number obtained on reversing the digits = (10y + x)

(10y + x) - (10x + y) = 18

9y - 9x = 18

y - x = 2 --- > (2)

Adding (1) and (2), we get

2y = 14

y = 14/2

y = 7

By applying y = 7 in (1),

x + 7 = 12

x = 12 - 7

x = 5

Hence, the required number is 57.

Problem 2 :

Seven times a two-digit number is equal to four times the number obtained by reversing the order of its digit. If the difference between the digits is 3, then find the number.

Solution:

Let numbers be x at onces place and y at tens place.

Reversed digit = 10x + y

7(10y + x) = 4(10x + y)

70y + 7x = 40x + 4y

70y - 4y = 40x - 7x

66y = 33x

x = 2y --- > (1)

Now, 

Given, x - y = 3

Substitute x = 2y in equation (1)

2y - y = 3

y = 3

So, x = 2y

x = 2(3) 

x = 6

So, required original number = 10y + x

= 10(3) + 6

= 36

Problem 3 :

The sum of the digits of a two digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution:

Let, the tens digit number be x and ones digit number be y.

So, the number will be 10x + y

x + y = 9 --- > (1)

9(10x + y)= 2(10y + x)

90x + 9y = 20y + 2x

90x - 2x = 20y - 9y

88x = 11y

y = 8x

Substitute y = 8x in equation (1)

x + 8x = 9

9x = 9

x = 1

By applying x = 1 in equation (1)

1 + y = 9

y = 9 - 1

y = 8

Hence, the number is10x + y

= 10(1) + 8

= 18

Problem 4 :

The sum of the digits of a two digit number is 9. If 27 is added to it, the digits of the numbers get reversed. Find the number.

Solution:

Let the digit at the tens place be x and that at the ones place be y.

x + y = 9 --- > (1)

If 27 is added to it, the digits get reversed

10x + y + 27 = 10y + x

10x - x - 10y + y = -27

9x - 9y = -27 --- > (2)

(1) × 9 ==> 9x + 9y = 81

Adding (1) and (2),

18x = 54

x = 54/18

x = 3

By applying x = 3 in equation (1),

3 + y = 9

y = 9 - 3

y = 6

So, the original number is 36.

Problem 5 :

The sum of a two-digit number and the number by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?

Solution:

Let the two digit number be x and y.

Number = 10x + y

Sum of two digit and reverse of it

10x + y + 10y + x = 66

11x + 11y = 66

x + y = 6 --- > (1)

Digits differ by 2,

x - y = 2 --- > (2)

Adding (1) and (2),

2x = 8

x = 8/2

x = 4

By applying x = 4 in equation (1)

4 + y = 6

y = 6 - 4

y = 2

So, two digit numbers are 42 and 24.

Problem 6 :

A two-digit number is 4 more than 6 times the sum of its digit. If 18 is subtracted from the number, the digits are reversed. Find the number.

Solution:

Let, the ones digit number be x and tens digit number be y.

x + 10y = 6(x + y) + 4

x + 10y = 6x + 6y + 4

10y - 6y = 6x - x + 4

4y = 5x + 4 ---> (1)

And, from if 18 is subtracted from the number, the digits are reversed.

x + 10y - 18 = y + 10x

10y - 18 = y + 9x

9y - 18 = 9x

x = y - 2 --- > (2)

Substitute x = y - 2 in (1)

4y = 5(y - 2) + 4

4y = 5y - 10 + 4

4y - 5y = -6

-y = -6

y = 6

By applying y = 6 in (2)

x = 6 - 2

x = 4

So, the number is 64.