a sum on 2 digit number with t digits interchanged-pleas solve-- i need the full explanation with the answers?

Question

A salesman enters the quantity sold and the price into the computer. Both the numbers are two-digit numbers. Once, by mistake, both the numbers were entered with their digits interchanged. The total sales value remained the same, i.e. Rs. 1148, but the inventory reduced by 54.

1.What is the actual price per piece?
[1] 82 [2] 41 [3] 56 [4] 28
2. What is the actual quantity sold?
[1] 28 [2] 14 [3] 82 [4] 41

Answer 1

(10x + y)(10a + b) = 1148 = (10y+x)(10b+a)
=> 99(ax -- by) = 0
=> ax = by where a, b, x, y all are < 10.
from given answers we get ax = by in the pair 14, 82 we have
14x82 = 1148 = 41x28 so
price is 14, number sold is 82
when reversed digits
price is 41 and number sold is 28 which is
82 -- 28 = 54 less.

Answer 2

correct ab x cd
entered ba x dc

and ba - ab = 54 so 17 / 71 , 28 / 82 or 39 / 93

ony the pair 28 / 82 are factors of 1148 => the prices are 41 / 14

The wrong quantity is higher - 82; the real 28.

Actual price 41; quantity 28

Answer 3

I think there is some mistake in your question. If u factorize 1154,
1154=2x2x7x41
as both the qty & price are 2 digit, only factor possible are,
1) 1154=28x41 ; 41-28=13
2) 1154=14x82 ; 82-14= 68
In neither case, the difference is 54
Pl. check again & revert back.

Answer 4

44