Question

a number consist of two digit when the number is divided by the sum of its digit the quotient is 7 if 27 is subtracted from the number digit interchange the places find the number

Answer 1

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

so number = ( 10a + b)

=) according to the question
______________________

=) (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
- +
__________________

=). 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

Hope it helps

Answer 2

Hey

Here is your answer,

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

so number = ( 10a + b)

=) according to the question
______________________

=) (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
- +
__________________

=). 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

Hope it helps you!