Question

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

Answer 1

Answer:

c) 36

Step-by-step explanation:

as we know

X=10a+b (32=3*10+2)

so all numbers sum is

c+d+(10a+b)+........=Z

then interchanged digits X=10b+a

so sum

c+d+(10b+a)+........=Z+10a+b-4

so by solving we get the equation

20a-10b=4.........2

next we know that

3(a+b)=X-10

=3a+3b=10a+b-10

=2b=7a-10

10=7a-2b.........1

by elimination method,

20a-10b=4

-35a+10b=-50(as by multiplying 10=7a-2b *-5)

so we get

15a=46

so a =15/46

value of b=7a-10/2

multiply a and b

we get 172*46/225

by solving we get

35.7634.....

so 36 is approx answer