Question

(e) If a two-digit number is divided by the sum of the digits, the quotient is 9. But the number
obtained by reducing the digit at the tens place by four times the digit at the units place, leaves
the remainder 5 when divided by the sum of the digits of the original number. Find the
original number.​

Answer 1

Answer:

72

Step-by-step explanation:

Let the tens digit be x

Let the units digit be y

The required number is 10x+y

The sum of the digits is x+y

Given, (10x+y)/(x+y) = 8

=> 10x+y = 8(x+y)

=> 10x+y = 8x+8y

=> 10x-8x = 8y-y

=> 2x = 7y

=> 2x-7y = 0 ------------(1)

Given, x-3y = 1

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

Solving (1) and (2), we get x=7 and y=2

The answer is 72

Answer 2

Answer:

81

Step-by-step explanation:

THE NUMBER IS

10x + y

THEN,

$\frac{10x + y}{ x + y} = 9$

(1)

AND

$\frac{10(x - 4y) + y}{x + y} = k + \frac{5}{x + y}$

(2)

,k Being the quotient

FROM (1) we get 10x + y = 9x + 9y

=> 8y.

Putting x=8y in (2),we get :

$\frac{41y}{9y} = k + \frac{5}{9y}$

$k + \frac{5}{9y} = \frac{41}{9} = 4 + \frac{5}{9}$

So y=1 and x=8y=8

● xy = 81