When a two-digit number is multiplied by sum of its digits, the product is 715 but when the reverse of a two-digit number is multiplied by sum of its digits the product obtained is 99 less, The sum of the digits of a number is
Let the two-digit number be (10x+y), then (10x+y)*(x+y)=715 ⇒ 10x² + y² +11xy = 715 ---(i)
And (10y+x)*(x+y)=(715-99) ⇒ x² + 10y² + 11xy = 616 --- (ii)
Subtracting eqn (ii) from (i), we get x² - y² = 11
⇒ (x+y) (x-y) = 11
As 11 is prime, only product 11*1 = 11 is possible
.'. (6+5) (6-5) = 11, so x=6, y=5
Hence, the number is 65 and sum of the digits=11
Alternate Method:
If x and y are the digits of a number, then (10x+y)*(x+y)=715 and (10y+x)*(x+y)=616
As we know (x+y) is common factor in both the equations.
Only 11 is the common factor.
Therefore sum (x+y)=11
Correct Option 7)