Last visit was: 23 May 2024, 14:02 It is currently 23 May 2024, 14:02
Decision Tracker
My Rewards
Unanswered
Active Topics
New posts
New comers' posts
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

Difficult one: Tourist ourchased ticket

User avatar
sandalphon
Manager
Manager
Joined: 05 Nov 2005
Posts: 114
Own Kudos [?]: 939 [0]
Given Kudos: 0
Location: Germany
Send PM
Difficult one: Tourist ourchased ticket [#permalink] Difficult one: Tourist ourchased ticket   03 Dec 2005, 09:43
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



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
avatar
wlee76
Intern
Intern
Joined: 05 Nov 2005
Posts: 29
Own Kudos [?]: 1 [0]
Given Kudos: 0
Location: London
Send PM
Re: Difficult one: Tourist ourchased ticket [#permalink] Difficult one: Tourist ourchased ticket   03 Dec 2005, 11:06
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.
User avatar
scorpioguy
Current Student
Joined: 23 Oct 2005
Posts: 205
Own Kudos [?]: 17 [0]
Given Kudos: 2
Schools: Cranfield SOM
 Q46  V40
Send PM
Re: Difficult one: Tourist ourchased ticket [#permalink] Difficult one: Tourist ourchased ticket   04 Dec 2005, 10:13
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.
User avatar
Bhai
VP
VP
Joined: 16 Oct 2003
Posts: 1018
Own Kudos [?]: 735 [0]
Given Kudos: 0
Send PM
Re: Difficult one: Tourist ourchased ticket [#permalink] Difficult one: Tourist ourchased ticket   04 Dec 2005, 19:37
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.
User avatar
stretchad
Intern
Intern
Joined: 20 Nov 2005
Posts: 32
Own Kudos [?]: 66 [0]
Given Kudos: 0
Location: Indianapolis, IN
Send PM
Re: Difficult one: Tourist ourchased ticket [#permalink] Difficult one: Tourist ourchased ticket   05 Dec 2005, 13:20
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
User avatar
krisrini
Manager
Manager
Joined: 15 Apr 2005
Posts: 159
Own Kudos [?]: 32 [0]
Given Kudos: 0
Location: India, Chennai
Send PM
Re: Difficult one: Tourist ourchased ticket [#permalink] Difficult one: Tourist ourchased ticket   06 Dec 2005, 01:38
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.
User avatar
automan
Senior Manager
Senior Manager
Joined: 09 Jul 2005
Posts: 320
Own Kudos [?]: 145 [0]
Given Kudos: 0
Send PM
Re: Difficult one: Tourist ourchased ticket [#permalink] Difficult one: Tourist ourchased ticket   06 Dec 2005, 04:54
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.
User avatar
rlevochkin
Manager
Manager
Joined: 03 Nov 2005
Posts: 149
Own Kudos [?]: 109 [0]
Given Kudos: 17
Location: Chicago, IL
Send PM
Re: Difficult one: Tourist ourchased ticket [#permalink] Difficult one: Tourist ourchased ticket   07 Dec 2005, 13:34
Agreed. D sounds right
User avatar
sandalphon
Manager
Manager
Joined: 05 Nov 2005
Posts: 114
Own Kudos [?]: 939 [0]
Given Kudos: 0
Location: Germany
Send PM
Re: Difficult one: Tourist ourchased ticket [#permalink] Difficult one: Tourist ourchased ticket   08 Dec 2005, 01:45
You guys are right D is the OA.



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
GMAT Club Bot
gmatclubot
Re: Difficult one: Tourist ourchased ticket [#permalink]
08 Dec 2005, 01:45
Moderator:
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts