21. While adding ten two-digit numbers, the digits of one
of the numbers (X) were interchanged. As a result,
the sum of the ten numbers increased by a value
which was four less than X. Three times the sum of
the digits of X is ten less than X. What is the product
of the digits of X?
(A) 56 (B) 18
(C) 36
(D) 42
Answers
Answer:
36
Step-by-step explanation:
let the 10th digit be interchanged and in form 10x+y
therefore original sum
=> (a1+a2+...+a9) + 10x + y = S
but the sum obtained was::: (a1+a2+...+a9) + 10y + x = S' bcoz the digits were interchanged
"the sum of all the ten numbers increased by a value which was four less than that number"
therefore!!
S' - S = (10x+y) - 4
the a1, a2...a9 gets cancelled out & what is left is
=> 9(y-x)=10x+y-4 => 8y-19x=(-4)-----------------------------...
"three times the sum of digits of that original number is ten less than the number"
therefore
=> 3(x+y) = 10x+y - 10 => 2y-7x = -10-------------------------------------...
by (1) & (2), the y can be cancelled out by multiplying (2) with -4 & adding (1) & (2)
the result we get is
=> 9x = 36 => x=4
therefore by (2)
2y -7.4= -10 => y=9
therefore x.y = 36