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Difficult one: Tourist ourchased ticket

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Difficult one: Tourist ourchased ticket [#permalink] New post 03 Dec 2005, 08:43
Please help, I have tried over half an hour to solve that problem, but I just can*t get it right. Can someone please explain it?

A tourist purchased a total of $1500 worth of traveler's
check's in $10 and $50 denominations. During the trip
the tourist cashed 7 checks and then lost all
of the rest. If the number of $10 checks cashed was one
more or one less than the number of
$50 checks cashed, what is the minimum possible value of
the checks that were lost?

A 1430
B 1310
C 1290
D 1270
E 1150
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Re: Difficult one: Tourist ourchased ticket [#permalink] New post 03 Dec 2005, 10:06
sandalphon wrote:
Please help, I have tried over half an hour to solve that problem, but I just can*t get it right. Can someone please explain it?

A tourist purchased a total of $1500 worth of traveler's
check's in $10 and $50 denominations. During the trip
the tourist cashed 7 checks and then lost all
of the rest. If the number of $10 checks cashed was one
more or one less than the number of
$50 checks cashed, what is the minimum possible value of
the checks that were lost?

A 1430
B 1310
C 1290
D 1270
E 1150


I pick D, since if there is a total of 7 cashed checks, with the relationship of x+1 or x-1, its 4:3. If we want the minimum lost, then we want the maximum cashed in, so take 4:3 as the ratio of 4 $50 and 3:$10, totalling a maximum cashed in amount of $230, minimising our lost to 1500-230=1270

I hope there is no flaw in the logic.
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Re: Difficult one: Tourist ourchased ticket [#permalink] New post 04 Dec 2005, 09:13
wlee76 wrote:
sandalphon wrote:
Please help, I have tried over half an hour to solve that problem, but I just can*t get it right. Can someone please explain it?

A tourist purchased a total of $1500 worth of traveler's
check's in $10 and $50 denominations. During the trip
the tourist cashed 7 checks and then lost all
of the rest. If the number of $10 checks cashed was one
more or one less than the number of
$50 checks cashed, what is the minimum possible value of
the checks that were lost?

A 1430
B 1310
C 1290
D 1270
E 1150


I pick D, since if there is a total of 7 cashed checks, with the relationship of x+1 or x-1, its 4:3. If we want the minimum lost, then we want the maximum cashed in, so take 4:3 as the ratio of 4 $50 and 3:$10, totalling a maximum cashed in amount of $230, minimising our lost to 1500-230=1270

I hope there is no flaw in the logic.


I think I saw this Q recently in a princeton review online test. If I remember right the OA and OE are as you have given.
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 [#permalink] New post 04 Dec 2005, 18:37
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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 [#permalink] New post 05 Dec 2005, 12:20
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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Re: Difficult one: Tourist ourchased ticket [#permalink] New post 06 Dec 2005, 00:38
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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 [#permalink] New post 06 Dec 2005, 03:54
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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 [#permalink] New post 07 Dec 2005, 12:34
Agreed. D sounds right
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 [#permalink] New post 08 Dec 2005, 00:45
You guys are right D is the OA.
  [#permalink] 08 Dec 2005, 00:45
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