Question

a number consists of two digits. when 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 thee number.

Answer 1

Here is the solution and the two digit number is 10x+y = 10*6+3= 63 :-

Answer 2

let the digit at ten's place be a and one's place be b

so number = ( 10a + b)

=) according to the question
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=) (10a+b)/(a+b) =7

=) (10a + b) = 7(a + b)

=) 10a + b = 7a + 7b

=) 10a - 7a = 7b -b

=) 3a -6b = 0

dividing this equation by 3 we get,

=) a - 2b = 0 ----------------------(1)

again,

=) (10a + b ) -27 = (10b + a)

=) 10a + b -27 =10b + a

=) 10a- a - 27 = 10b - b

=) 9a - 27 = 9b

=) 9a - 9b = 27

dividing this eqn by 9 we get,

=) a - b = 3 -----------------------(2)

now subtracting eqn 2 - eqn 1

a - b = 3
a -2b = 0
- +
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=). 0 + b = 3

so, b = 3

putting value of b in equation (2) we get

=) a - 3 = 3

=) a = 6

so the number = (10a + b) = (10×6 + 3)

= 63

◆ The required number is 63

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