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sandalphon
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To make the value of lost to minimum we should have maximum number of $50 checks encashed

4*$50 = 200 +
3*$`0 = 30 = 230 so subtract this from 1500 we get 1270

if you use 4 $10 checks then the values are 190 and we get 1310.
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stretchad
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I did it as follows:

x = number of $10 checks
y = number of $50 checks

According to the statements in the problem, x=y+1 or x=y-1, and x+y = 7
substitute x+y=7 into x=y+1 to get:
x=7-x+1 ---> 2x=8, x=4

If we had 4 $10 checks that means we had 3 50 checks. For a total amount cashed of $190. 1500-190 cashed = $310

BUT, since we need to MINIMIZE the amount LOST (not the amount cashed), then we need to maximize the amount CASHED, which would mean we need to substitute x+y=7 into x=y-1...

doing that you get
x=7-x-1
2x=6
x=3

which means the traveler cashed $230, and lost at the most, 1500-230=$1270.

I second answer D
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Let X represent $10 and Y represent $50
Cashed: X + y = 7
We are given X = Y + 1 or X=y - 1
If X = y + 1 then 2y = 6, y = 3, and X= 4 which means 50*3+10*4 was cashed or 190$ was cashed, and 1500 - 190 = 1310$ was lost.
If X = y - 1 then 2y = 8 , y = 4 and X = 3. wjocj 50*4 + 10 *3 was cahsed or 230$ was cashed, and 1500 - 230 = 1270$ was lost.
The minimum of this is 1270 hence D.
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automan
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7 checks

1. 4 $10 and 3 $50 ---- $190----- he lost $1310

2. 3 $10 and 4 $50 ---- $230----- he lost $1270


D is the solution.
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Agreed. D sounds right
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sandalphon
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You guys are right D is the OA.



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