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# Difficult one: Tourist ourchased ticket

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Manager
Joined: 05 Nov 2005
Posts: 232
Location: Germany
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Difficult one: Tourist ourchased ticket [#permalink]  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 Intern Joined: 05 Nov 2005 Posts: 33 Location: London Followers: 0 Kudos [?]: 1 [0], given: 0 Re: Difficult one: Tourist ourchased ticket [#permalink] 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. Current Student Joined: 23 Oct 2005 Posts: 243 Schools: Cranfield SOM Followers: 2 Kudos [?]: 13 [0], given: 2 Re: Difficult one: Tourist ourchased ticket [#permalink] 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. SVP Joined: 16 Oct 2003 Posts: 1813 Followers: 4 Kudos [?]: 56 [0], given: 0 [#permalink] 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. Manager Joined: 20 Nov 2005 Posts: 55 Location: Indianapolis, IN Followers: 1 Kudos [?]: 3 [0], given: 0 [#permalink] 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.

Senior Manager
Joined: 14 Apr 2005
Posts: 418
Location: India, Chennai
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Re: Difficult one: Tourist ourchased ticket [#permalink]  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.
Director
Joined: 09 Jul 2005
Posts: 595
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Kudos [?]: 28 [0], given: 0

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.
Senior Manager
Joined: 03 Nov 2005
Posts: 397
Location: Chicago, IL
Followers: 3

Kudos [?]: 29 [0], given: 17

Agreed. D sounds right
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Manager
Joined: 05 Nov 2005
Posts: 232
Location: Germany
Followers: 2

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

You guys are right D is the OA.
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