Question

67. A number consists of two digits, where the number is divided by the sum of
its digits, the quotient is 7. If 27 is subtracted from the number, the digits
interchange their places, find the number. (CBSE 2010] ​

Answer 1

Answer :

Number = 10x+y=63

where x = 6 & y =3

Step by step Explanation :

Let the tens and the units digits of the required number be x and y. Then ,t

• Required number=10x+y
• Reversed number = 10y+x

According to the question :

If number is divided by the sum of

its digits, the quotient is 7.then ,

$\sf\implies\dfrac{10x+y}{x+y}=7$

$\sf\implies\:10x+y=7x+7y$

$\sf\implies\:3x-6y=0$

$\sf\implies\:3(x-2y)=0$

$\sf\implies\:x-2y=0...(1)$

And , If 27 is subtracted from the number, the digits interchange their places.

$\sf\implies\:10x+y-27=10y+x$

$\sf\implies\:9(x-y)=27$

$\sf\implies\:x-y=3....(2)$

On Add equation (1) & (2)

$x-2y=0 \\ x-y = 3 \\ - + = - \\ \rule{50}1 \\ -y = -3$

$\sf\implies\:y=3$

Now put y =3 in equation (2)

$\sf\implies\:x-y=3$

$\sf\implies\:x=3+3=6$

Thus , x = 6 & y=3

Therefore the number is 63

Answer 2

Answer:

Answer is in the attachment...

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