Tom purchased a total of $5000 worth of travelers checks in $50

Question

Tom purchased a total of $5000 worth of traveler’s checks in $50 and $100 denominations for his vacation. During his vacation Tom cashed 14 checks and then lost all the rest. If the number of $50 checks cashed was two more or two less than the number of $100 checks cashed, what is the maximum possible value of checks that were lost?

A. $3800
B. $3900
C. $4000
D. $4100
E. None of the above

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bhargavhhhhhhhh · Answer

Let x be the number of $50 checks cashed, and y be the number of $100 checks cashed. 
We are told x = y ± 2 and x + y = 14.
Case 1: x = y + 2
Substitute x = y + 2 into x + y = 14: 
(y + 2) + y = 14 
2y + 2 = 14 
2y = 12 
y = 6, so x = 8.
The value of checks cashed = 50 * 8 + 100 * 6 = 1000. 
The lost value = 5000 - 1000 = 4000.

Case 2: x = y - 2
Substitute x = y - 2 into x + y = 14: 
(y - 2) + y = 14 
2y - 2 = 14 
2y = 16 
y = 8, so x = 6.
The value of checks cashed = 50 * 6 + 100 * 8 = 1100. 
The lost value = 5000 - 1100 = 3900.

The maximum value of the checks that were lost is $4000.

tgsankar10 · Answer

Lets say the no of    $50    &    $100    checks are    a    &    b

a=b+2  or  a=b-2

cashed checks,  a+b=14

when  a=b+2 ,  2b+2=14 ,  b=6 ,  a=8  & the cashed amount is  600+400=$1000

when  a=b-2 ,  2b-2=14 ,  b=8 ,  a=6  & the cashed amount is  800+300=$1100

The max amount of lost check will be  $5000-$1000=$4000

Answer: C

Whoisdmx15 · Answer

In order to maximize the value of checks that were lost, we need to minimize the value that was cashed.

Now the denominations of 50 and 100 are related by the equation that f+h =14 and f= h+/-2

Again to maximize the value lost, we need to minimize the value cashed. 
Hence, we assume there have been more 50 denomination (n+2) checks cashed and less 100 denomination cheques (n).

Hence n+n+2 =14
This gives us n= 6.

Hence 6 X 100 denomination cashed
and 8 X 50 denomination cashed.

Hence remaining value, which can be lost = 5000- (6*100 + 50*8) = 4000

Checking if 100 and 50 denominations can make up 4000.
(100*30 + 50*20). Hence possible

IMO Ans C

eega · Answer

Let x be no. of $ 100 checks cashed 
Let y be no. of $ 50 checks cashed

given x+y = 14
and y=x+2 or x-2
hence x & y can be 8,6 or 6,8 respectively
hence maximum value of checks lost = 5000 - x(100) - y(50)
maximum value we get = $4000  (Option C)

Tom purchased a total of $5000 worth of travelers checks in $50

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Tom purchased a total of $5000 worth of travelers checks in $50

Prep Toolkit

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