Question

a number consists of two digits whose sum is 9. if 45 is subtracted from the number its digits are interchanged. find the number.

Answer 1

let the digits be x and y.
so the number becomes 10x+y
ATQ,
X+Y=9-----------(1)

CASE 2 AFTER INTERCHANGING THE DIGITS , THE NUMBER BECOMES 10Y+X
10X + Y - 45 = 10Y+X
10X+Y- 10Y - X = 45
9X - 9Y = 45
TAKING 9 COMMON
9(X - Y) = 45
X - Y= 45/9
X -Y = 5 ------------(2)

ELIMINATING THE TWO EQUATIONS
X+Y = 9
X-Y = 5
----------------
- + -
2Y = 4
Y = 4/2
Y = 2


PUTTING VALUE OF Y IN EQ 1
X+ 2 = 9
X= 9-2= 7

THEREFORE THE NUMBER BECOMES
10 (7)+ 2= 70+2= 72
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Answer 2

The required number is 72

Given :

• A number consists of two digits whose sum is 9.

• If 45 is subtracted from the number its digits are interchanged.

To find :

The number

Solution :

Step 1 of 2 :

Form the equations to find the number

Let the digit in unit place is b and tenth place is a

Then the number is 10a + b

It is given that sum of the digits is 9

By the given condition

a + b = 9 - - - - - - (1)

Again it is given that when 45 is subtracted from the number its digits are interchanged

So by the above condition

(10a + b) - 45 = (10b + a) - - - - - (2)

Step 2 of 2 :

Find the number

a + b = 9 - - - - - - (1)

(10a + b) - 45 = (10b + a) - - - - - (2)

⇒ 10a + b - 10b - a = 45

⇒ 9a - 9b = 45

⇒ 9(a - b) = 45

⇒ a - b = 45/9

⇒ a - b = 5 - - - - - - - (3)

Adding Equation 1 and Equation 2 we get

2a = 14

⇒ a = 14/2

⇒ a = 7

From Equation 1 we get

7 + b = 9

⇒ b = 9 - 7

⇒ b = 2

Hence the required number

= 10a + b

= (10 × 7) + 2

= 70 + 2

= 72

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