Question

A number consists of two digits . The digit is the tens place exceeds the digit in the unit place by 4. The sum if the digit is1/7 . Find numbers

Answer 1

Answer:

Step-by-step explanation:

Let the two digit number be defined as ‘xy’ where x is the 10’s digit and y is the 1’s digit. The number is then 10*x + y (For instance if the number is 45, it is 10*4 + 5)

Based on the facts given, x = y + 4

We also know that x^2 + y^2 = (10*x + y) - 15 . Lets replace x by y+4 here:

(y+4)^2 + y^2 = (10*(y+4) + y) - 15.

or

y^2 + 8y + 16 + y^2 = 10*y + 40 + y - 15

or

2*y^2 + 8y + 16 = 10*y + 25 +y.

or, by moving all terms to the left:

2*y^2 - 3*y - 9 = 0

or

(y - 3) (2y + 3) = 0

or y = 3 or y = -3/2 .

y cannot be a negative number. So y has to be 3. And x = y+4 = 7.

Therefore the number is 73.

Let's validate our answer against the facts provided:

The sum of the square of digits is 7^2 + 3^2 = 58 and this is 15 lesser than the number (58 = 73 - 15).