If m and n are 2 digit numbers, so that their digits are interchanged when they are increased by 75 percent. Then the maximum of difference between m and n is
Answers
Answer to the question is 36
Solution :-
Let us assume that, the 2 digits numbers whose when digits are interchanged, they are increased by 75% are (10x + y).
After digits are interchanged = (10y + x) .
So,
→ number increased by = (10y + x) - (10x + y) = 10y - y + x - 10x = (9y - 9x) = 9(y - x) .
now, this increased number is 75% of the original number.
Therefore,
→ 75 * (10x + y)/100 = 9(y - x)
→ 75(10x + y) = 100 * 9(y - x)
→ 3(10x + y) = 4 * 9(y - x)
→ 10x + y = 12(y - x)
→ 10x + y = 12y - 12x
→ 10x + 12x = 12y - y
→ 22x = 11y
→ 2x = y .
So, Possible values of x and y are :-
- x = 1 , y = 2.
- x = 2 , y = 4 .
- x = 3 , y = 6 .
- x = 4 , y = 8 .
So, maximum of difference between m and n is when one is greatest number and one is smallest number.
Therefore,
- m = 12 .
- n = 48 .
Hence,
→ m - n = 48 - 12 = 36 .(Ans .)
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