When a two-digit number is multiplied by sum of its digits, the product is 715 but when the reverse

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12 Sep 2022 Number System

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

OPtion

1) 7
2) 8
3) 6
4) 9
5) 12
6) 15
7) 11
8) 10
9) 13
10) None of these

Solution

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)

Correct Option: 7 Best Solution (2)

G.S.Kantharao   4 Months AGO

H.C.F of 715,616=11
Hence my option is 7👈

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RAMADURAI NATESAN   4 Months AGO

Let A and B be the digits . It is given that:
(A + B ) ( 10A + B) =715 and (A + B ) ( 10B + A ) =616.Thus,
(10A+B)/(10B+A)=717/616 = 65/56. Solving, A /B = 6/5.
Being digits of a 2 digit No.only possibility is A=6,B=5.
A+B= 11

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