# SOLVING WORD PROBLEMS WITH DIGITS

Problem 1:

The sum of digits of a two digit numbers is 15 and if 9 is added to the number the digits are interchanged. Find the required number.

(A) 36        (B) 57      (C) 27      (D) 78

Solution :

Let xy be the required two digit number.

The sum of digits of a two digit number  =  15

x + y  =  15  ------(1)

If 9 is added to the number the digits are interchanged

x y + 9  =  y x

Now let us write this number in expanded form

10 x + y + 9  =  10y + x

10 x - x + y - 10 y  =  -9

9x - 9y  =  -9 ------(2)

By solving these two equations we will get the value of x and y.

(1)  9 => 9 x + 9 y  =  135

(1) + (2)   9 x - 9 y  =  -9

______________

18 x  =  126 ==> x  =  7

By applying x  =  7 in (1), we get

7 + y  =  15

y  =  15 - 7

y  =  8

Therefore the required number is 78.

Problem 2 :

The sum of the digits of a two digit number is 10. If the number formed by reversing the digits is less than the original number by 36,find the required number.

(A) 73      (B) 55      (C) 64           (D) 28

Solution :

Let "x y" be the required two digit number. Here "x" is in ten's place and "y" is in unit place.

The sum of the digits of a two digit number  =  10

x + y  =  10  --------(1)

If the number formed by reversing the digits is less than the original number by 36.

y x  =  x y - 36

Let us write these as expanded form

10 y + x = 10 x + y - 36

x - 10 x + 10 y - y = - 36

- 9 x + 9 y = - 36

By divide this equation by 9 we will get

- x + y = -4  --------(2)

(1) + (2)    x + y = 10

-x + y = -4

_________

2 y = 6  ==>  y = 3

By applying y = 3 in (1), we get

x + 3  =  10

x  =  7

Therefore the required number is 73.

Problem 3 :

The unit's digit of a two digit number is twice its ten's digit. If 18 is added to the number, the digits interchange their places. Find the number.

(A) 48        (B) 24     (C) 36       (D) 84

Solution :

Let xy  be the required two digit number.

The unit's digit of a two digit number is twice its ten's digit

y  =  2 x

2x - y  =  0-------(1)

xy + 18  =  yx

10 x + y + 18  =  10 y + x

10 x - x + y - 10 y  =  -18

9 x - 9 y  =  -18

Dividing this equation by 9

x - y = - 2  --------(2)

(1) - (2) 2x - y = 0    ------(1)

x - y = - 2  ------(2)

(-)  (+)   (+)

_____________

x = 2

By applying x = 2 in the first equation

2 (2) - y  =  0

y  =  4

Therefore the required number is 24.

Problem 4 :

The sum of the digits of two digit number is 12. If the  new number formed by reversing the digits is greater than the original number by 54, find the original number.

(A) 48        (B) 75      (C) 39        (D) 84

Solution :

Let xy  be the required two digit number.

The sum of the digits of two digit number  =  12

x + y  =  12  ------(1)

If the  new number formed by reversing the digits is greater than the original number by 54

y x = x y + 54

Let us write this as expanded form

10 y + x = 10 x + y + 54

x - 10 x + 10 y - y = 54

- 9 x + 9 y = 54

Dividing this equation by 9. We will get

-x + y = 6 ------(2)

(1) + (2)      x + y = 12

- x + y = 6

___________

2 y = 18  ==> y  =  9

By applying y  =  9 in (1), we get

x + 9  =  12

x  =  12 - 9

x  =  3

Therefore the required number is 39.

Problem 5 :

The number consists of two digits whose sum is 9. If 45 is added to the number, the digits are reversed. Find the number.

(A) 63    (B) 27     (C) 36    (D) 81

Solution :

Let xy  be the required two digit number.

The sum of its digits = 9

x + y  =  9   --------- (1)

If 45 is added to the number, the digits are reversed.The given number is x y and its reversed digits is y x.

xy + 45  =  y x

10x + y + 45  =  10 y + x

10x - x + y - 10 y  =  - 45

9x - 9y  =  - 45

Divide this equation by 9.

x - y = - 5 --------- (2)

By solving these two equations we will get the value of x and y.

x + y  =  9   --------- (1)

x - y  =  -5   -------- (2)

(1) + (2)    x + y  =  9

x - y  =  -5

__________

2 x  =  4  ==> x  =  2

By applying x = 2 in (1), we get

2 + y  =  9

y  =  9 - 2

y  =  7

Therefore the required number is 27.

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