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# Difficult one: Tourist ourchased ticket

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Manager
Joined: 05 Nov 2005
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Location: Germany
Difficult one: Tourist ourchased ticket [#permalink]

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03 Dec 2005, 08:43
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Kudos [?]: 98 [0], given: 0

Intern
Joined: 05 Nov 2005
Posts: 33

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Location: London
Re: Difficult one: Tourist ourchased ticket [#permalink]

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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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Current Student
Joined: 23 Oct 2005
Posts: 243

Kudos [?]: 14 [0], given: 2

Schools: Cranfield SOM
Re: Difficult one: Tourist ourchased ticket [#permalink]

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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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SVP
Joined: 16 Oct 2003
Posts: 1798

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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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Manager
Joined: 20 Nov 2005
Posts: 53

Kudos [?]: 18 [0], given: 0

Location: Indianapolis, IN

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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.

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Senior Manager
Joined: 14 Apr 2005
Posts: 414

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Location: India, Chennai
Re: Difficult one: Tourist ourchased ticket [#permalink]

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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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Director
Joined: 09 Jul 2005
Posts: 589

Kudos [?]: 69 [0], given: 0

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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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Senior Manager
Joined: 03 Nov 2005
Posts: 380

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Location: Chicago, IL

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07 Dec 2005, 12:34
Agreed. D sounds right
_________________

Hard work is the main determinant of success

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Manager
Joined: 05 Nov 2005
Posts: 223

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Location: Germany

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08 Dec 2005, 00:45
You guys are right D is the OA.

Kudos [?]: 98 [0], given: 0

08 Dec 2005, 00:45
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# Difficult one: Tourist ourchased ticket

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