Question

There\ is\ a\ number\ of\ two\ digits,\ which\ is\ four\ times\ the\ sum\ of\ its\ digits.if\ 45\ is\ added\ to\ the\ number,the\ digits\ change\ places.find\ the\ number.

Answer 1

Hi,

Let ten's place digit = x

units place digit = y

number = 10x + y ---( 1 )

according to the problem given ,

10x+ y = 4 ( x + y )

10x + y = 4x + 4y

10x - 4x = 4y - y

6x = 3y

6x/3 = y

2x = y ----( 2 )

if 45 is add to the number , the digits change

places

10x + y + 45 = 10y + x

10x - x + 45 = 10y - y

9x +45 = 9y

divide each term with 9 , we get

x + 5 = y ---( 3 )

( 3 ) = ( 2 )

x + 5 = 2x

5 = x

Therefore ,

x = 5 ,

y = 2x

y = 2 × 5 = 10

Required number ,

10x + y

= 10 × 5 + 10

= 50 + 10

= 60

I hope this helps you.

:)