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Fri Apr 09, 2021 3:45 am
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D
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Difficulty:
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Question Stats:
81% (02:28) correct
19%
(03:23)
wrong
based on 47
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An overseas business purchased a total of $2,000 worth of traveler’s checks in $20 and $50 denominations. During the trip, he cashed only ten checks and lost all the remaining checks. If the number of $20 checks cashed was two more or two less than the number of $50 checks cashed, what is the minimum possible value of the checks that were lost?
(A) $1200
(B) $1440
(C) $1500
(D) $1620
(E) $1880
(A) $1200
(B) $1440
(C) $1500
(D) $1620
(E) $1880
Fri Apr 09, 2021 4:07 am
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An overseas business purchased a total of $2,000 worth of traveler’s checks in $20 and $50 denominations.
Let x be number of 20$ checks and y be number of 50$ checks
So per statement
2000= 20x+50y
case 1: If the number of $20 checks cashed was two more than the number of $50 checks cashed
x=y+2,
Substituting in the above equation,
2000=20(y+2)+50y,
solving this will yield y=28, and so as x=y+2, we have x=30
Case 2: If the number of $20 checks cashed was two less than the number of $50 checks cashed
x=y-2,
Substituting in the above equation,
2000=20(y-2)+50y,
Solving this will yield a non integer value of y, which is physically impossible.
Hence we shall proceed with values obtained in case 1:
As we need minimum possible value of the checks that were lost, we shall try to assign maximum possible value of the 10 checks that were cashed. That is possible when all 10 checks be 50$ checks.
Now number of remaining checks:
y=28-10=18, x=30
Their total value = 18*50 + 30*20 = $1500
Let x be number of 20$ checks and y be number of 50$ checks
So per statement
2000= 20x+50y
case 1: If the number of $20 checks cashed was two more than the number of $50 checks cashed
x=y+2,
Substituting in the above equation,
2000=20(y+2)+50y,
solving this will yield y=28, and so as x=y+2, we have x=30
Case 2: If the number of $20 checks cashed was two less than the number of $50 checks cashed
x=y-2,
Substituting in the above equation,
2000=20(y-2)+50y,
Solving this will yield a non integer value of y, which is physically impossible.
Hence we shall proceed with values obtained in case 1:
As we need minimum possible value of the checks that were lost, we shall try to assign maximum possible value of the 10 checks that were cashed. That is possible when all 10 checks be 50$ checks.
Now number of remaining checks:
y=28-10=18, x=30
Their total value = 18*50 + 30*20 = $1500
Sat Oct 02, 2021 5:56 am
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Bunuel
Let the number of 20$ be x and 50$ be y respectively.
x + y = 10
Also, 20x + 50y = 2000
2x + 5y = 200(Remember this!)
2 possible cases: x = 2 + y or x = y - 2
As per the question, we need to minimize the number of cheques that were lost. To minimize the number, we need to maximize the below equation:
2x + 5y = 200
To maximize the above equation, the number of 50$ should be more than number of 20$
So, for y = 6, x = 4
We have: 200 - 38 = 162
As, we had divided the whole eq by 10, the minimum value is 162*10 = 1620
IMO, (D)!
