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Bunuel
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In order to maximize the value of checks that were lost, we need to minimize the value that was cashed.

Now the denominations of 50 and 100 are related by the equation that f+h =14 and f= h+/-2

Again to maximize the value lost, we need to minimize the value cashed.
Hence, we assume there have been more 50 denomination (n+2) checks cashed and less 100 denomination cheques (n).

Hence n+n+2 =14
This gives us n= 6.

Hence 6 X 100 denomination cashed
and 8 X 50 denomination cashed.

Hence remaining value, which can be lost = 5000- (6*100 + 50*8) = 4000

Checking if 100 and 50 denominations can make up 4000.
(100*30 + 50*20). Hence possible

IMO Ans C
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Let x be no. of $ 100 checks cashed
Let y be no. of $ 50 checks cashed

given x+y = 14
and y=x+2 or x-2
hence x & y can be 8,6 or 6,8 respectively
hence maximum value of checks lost = 5000 - x(100) - y(50)
maximum value we get = $4000 (Option C)
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