Question

When the sum of a two digit number ab and the number obtained by reversing the digits is divided by the (a+b), then the quotient is?

Answer 1

Answer:

The quotient does obtain is 11

Step-by-step explanation:

Given as :

The two digits number = 10 a + b

The number obtain by reversing the digit = 10 b + a

Now, The sum of two digits number and number obtain by reversing the digit = ( 10 a + b ) + ( 10 b + a )

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

= 11 a + 11 b

= 11 (a + b)

Now, The sum is divided by (a + b)

i.e Dividend = 11 (a + b)

Let The quotient = Q

∵ Dividend = divisor × quotient + remainder

i.e  11 (a + b) =   (a + b) × Q + 0

or, Q = $\dfrac{11 (a + b)}{a+b}$

∴   Q = 11

So, The quotient = Q = 11

Hence, The quotient does obtain is 11 . Answer

Answer 2

Answer:

11 ans is true

Step-by-step explanation:

Given as :

The two digits number = 10 a + b

The number obtain by reversing the digit = 10 b + a

Now, The sum of two digits number and number obtain by reversing the digit = ( 10 a + b ) + ( 10 b + a )

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

= 11 a + 11 b

= 11 (a + b)

Now, The sum is divided by (a + b)

i.e Dividend = 11 (a + b)

Let The quotient = Q

∵ Dividend = divisor × quotient + remainder

i.e 11 (a + b) = (a + b) × Q + 0

or, Q = \dfrac{11 (a + b)}{a+b}

a+b

11(a+b)

∴ Q = 11

So, The quotient = Q = 11

Hence, The quotient does obtain is 11 . Answer