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Tourist purchased a total of 30 travelers checks in $50 and

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Joined: 25 Mar 2016
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Location: India
Concentration: Finance, General Management
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Re: Tourist purchased a total of 30 travelers checks in $50 and [#permalink]

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New post 13 Jun 2016, 19:29
GMATBaumgartner wrote:
Tourist purchased a total of 30 travelers checks in $50 and $100 denominations. The total worth of the travelers checks is $1800. How many checks of $50 denominations can he spend so that average amount (arithmetic mean) of the remaining travelers checks is $80?

A. 4
B. 12
C. 15
D. 20
E. 24



Let the Number of $50 that we can spend be X.

Then, 1800- 50X/30-X = 80

We can solve for X and the answer will be X= 20

Option D

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Re: Tourist purchased a total of 30 travelers checks in $50 and [#permalink]

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New post 13 Jun 2016, 22:22
GMATBaumgartner wrote:
Tourist purchased a total of 30 travelers checks in $50 and $100 denominations. The total worth of the travelers checks is $1800. How many checks of $50 denominations can he spend so that average amount (arithmetic mean) of the remaining travelers checks is $80?

A. 4
B. 12
C. 15
D. 20
E. 24


I present an alternate and fast method to do such problems. These are essentially "mixture" problems and can be done by using alligation approach.

Refer following figure:

Attachment:
Moneybill-1.JPG
Moneybill-1.JPG [ 30.34 KiB | Viewed 307 times ]


Total checks are 30. Total value is 1800. This means average value is (1800/30) = 60.

Once we have individual values, we can get the ratios.

We find that 100$ checks and 50$ checks are in the ratio of 1:4.

say 100$ checks are x in no. then 50$ checks are 4x in number

x+4x = 30, so 5x = 30; x = 6.

100$ --> 6 in Number
50$--> 24 in Number

Now let us come to second part. Refer following figure


Attachment:
Moneybill-2.JPG
Moneybill-2.JPG [ 28.84 KiB | Viewed 307 times ]


Revised average = 80

We find that 100$ checks and 50$ checks are in the ratio of 3:2

Now we know that 100$ checks are 6 in number.

3: 2 is same is 6:4.

hence 50$ checks will have to be 4 in Number.

Originally 50$ checks were 24 in number. So 20 checks will be given away.

D is the answer.

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Re: Tourist purchased a total of 30 travelers checks in $50 and [#permalink]

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New post 30 Aug 2017, 10:10
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Re: Tourist purchased a total of 30 travelers checks in $50 and [#permalink]

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New post 05 Sep 2017, 18:04
GMATBaumgartner wrote:
Tourist purchased a total of 30 travelers checks in $50 and $100 denominations. The total worth of the travelers checks is $1800. How many checks of $50 denominations can he spend so that average amount (arithmetic mean) of the remaining travelers checks is $80?

A. 4
B. 12
C. 15
D. 20
E. 24


We can let x = the number of $50 traveler’s checks and y = the number of $100 traveler’s checks. Thus, we can create a “number of traveler’s checks” equation:

x + y = 30

x = 30 - y

We can also create a “monetary value” equation:

50x + 100y = 1800

We substitute (30 - y) for x:

50(30 - y) + 100y = 1800

1500 - 50y + 100y = 1800

50y = 300

y = 6

Since y = 6 and x = 30 - y, x = 24.

We can let n = the number of $50 traveler’s checks spent and create the following equation:

[50(24 - n) + 100(6)]/(30 - n) = 80

1200 - 50n + 600 = 2400 - 80n

30n = 600

n = 20

Answer: D
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Re: Tourist purchased a total of 30 travelers checks in $50 and [#permalink]

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New post 24 Sep 2017, 15:54
Solving Algebraically:

X = the number of checks in the $50 denominations
Y = the number of checks in the $100 denominations
Total number of checks = X + Y = 30

Total value of the checks = $50X + $100Y = $1800

let n = the number of $50 denomination checks the tourist spends

Calculate the average amount of the remaining checks:

=($1800 - $50n)/(X+Y-n) = $80
=($1800 - $50n)/(30- n) = $80
=$1800 - $50n = $80(30- n)
=$1800 - $50n = $2400 - $80n
=$30n = $600
n = 20
Answer: D

Kudos [?]: [0], given: 75

Re: Tourist purchased a total of 30 travelers checks in $50 and   [#permalink] 24 Sep 2017, 15:54

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