Question

1/3 rd of a two-digits number is greater than 1/4 th of a number by 8. Find the sum of digits of that number​

Answer 1

Answer:

15

Step-by-step explanation:

Given :-

1/3 rd of a two-digits number is greater than 1/4 th of a number by 8.

To find :-

Find the sum of digits of that number ?

Solution :-

Let the digit at 10's place be X

Then the place value of X = 10× X = 10X

Let the digit at 1's place be Y

The place value of Y = Y×1 = Y

Then the two digit number = 10X+Y

1/3rd of the number

= (1/3) of (10X+Y)

= (1/3)×(10X+Y)

=> (10X+Y)/3 -----------(1)

1/4th of the number

= (1/4) of (10X+Y)

= (1/4) of (10X+Y)

= (1/4)×(10X+Y)

= (10X+Y)/4 ------------(2)

Given that

1/3 rd of a two-digits number is greater than 1/4 th of a number by 8.

=> 1/3rd of the number = 1/4th of the number +8

=> (10X+Y)/3 = [ (10X+Y)/4 ]+ 8

=> [ ( 10X+Y)/3 ] - [ (10X+Y)/4 ] = 8

LCM of 3 and 4 = 12

=>[ {(10X+Y)×4} - {(10X+Y)×3} ] / 12 = 8

=> [ (40X+4Y) - (30X+3Y) ] / 12 = 8

=> (40X+4Y-30X-3Y)/12 = 8

=> (10X+Y) /12 = 8

=> 10X + Y = 8×12

=> 10X + Y = 96

=> 10X+Y = 90+6

=> 10X + Y = 10×9 + 6

On Comparing both sides then

=> X = 9 and Y = 6

Now, X+Y = 9+6 = 15

Therefore, X+Y = 15

Answer:-

The sum of the digits in the two digits number for the given problem is 15