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

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

by vinni.k » Sun Oct 22, 2017 8:31 am
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

OA is D

How can i solve this question quickly. I am trying scale method or allegation method but i have a doubt in that too
(100 - 80)/(80 - 50) = $50 denomination/$100 denomination = 2/3 ( i am stuck here)

Now the doubt : In some forums i have seen that ratio of 2/3 is multiplied by 2 (both numerator and denominator) = 4/6
This means if we have six 100 and four 50 we will get average 80 considering all notes.
So, extra 50 notes are 24-4= 20

Now, how would i know that 6 is $100 denomination with scale method or multiplying it by 2.

I was able to get through with average concept and understood how 6 came. Here's below but my issue is with scale method. I am not able to proceed after 2/3

Here's with average concept
[(24 - m)50 + (6)100 ] / (30 - m) = 80

Please let me know how to proceed in scale method or any other quick approach.

Thanks
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by [email protected] » Sun Oct 22, 2017 10:17 am
Hi vinni.k,

There are a couple of different ways to approach this question - depending on how you 'see' the math involved. This question can be solved rather quickly by TESTing THE ANSWERS.

From the prompt, we know that there are 30 checks in total (some are 50s and some are 100s) and the the SUM of their value is $1800. That means the 'starting average value' is 1800/30 = $60. As we remove 50s from that calculation, the average will INCREASE. The question asks how many of the 50s we have to remove to get the average up to $80.

Since $80 is a nice 'round number', it's likely that we'll end up removing a number of checks that will create a 'nice' number to divide by (such as a multiple of 5 or a multiple of 10). Looking at the answers, I would start with answer D, since removing 20 checks would leave us with 10 checks (and 10 will be easy to divide into whatever the new numerator is...

Let's TEST Answer D: 20

If we remove twenty 50s from the total value, then the new total value will be 1800 - (20)(50) = $800.
After removing those twenty checks, there will be 30 - 20 = 10 checks remaining.
The average of these remaining checks is $800/10 = $80
This matches what we were told, so this MUST be the answer.

Final Answer: D

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by vinni.k » Mon Oct 23, 2017 10:42 am
Rich your answer always amazes me :D both in beatthegmat and in GMATclub

Thanks
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by GMATGuruNY » Mon Oct 23, 2017 1:28 pm
vinni.k 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
Average value of the 30 checks = 1800/30 = 60.
This is a MIXTURE problem.
$50 checks are combined with $100 checks to form an original mixture with an average value of $60 and a new mixture with an average value of $80.
One approach is to use ALLIGATION to determine the ratio of $50 checks to $100 checks in each mixture.
Let F = the number of $50 checks and H = the number of $100 checks.

Case 1: Average for the mixture = $60

Step 1: Plot the 3 dollar values on a number line, with F and H on the ends and the average for the mixture in the middle.
F 50------------60------------100 H

Step 2: Calculate the distances between the averages.
F 50----10----60----40----100 H

Step 3: Determine the ratio in the mixture.
The required ratio of F to H is equal to the RECIPROCAL of the distances in red.
F:H = 40:10 = 4:1= 24:6.

The ratio in blue indicates that the 30 original checks are composed of 24 $50 checks and 6 $100 checks.

Case 2: Average for the mixture = $80

Step 1: Plot the 3 dollar values on a number line, with F and H on the ends and the average for the mixture in the middle.
F 50------------80------------100 H

Step 2: Calculate the distances between the averages.
F 50----30----80----20----100 H

Step 3: Determine the ratio in the mixture.
The required ratio of F to H is equal to the RECIPROCAL of the distances in red.
F:H = 20:30 = 2:3 = 4:6.

The ratio in green indicates that -- to yield an average value of $80 -- 4 $50 checks must be combined with the 6 original $100 checks.
Since the number of $50 checks decreases from 24 to 4, the number of $50 checks that must be removed = 24-4 = 20.

The correct answer is D.
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by vinni.k » Tue Oct 24, 2017 11:11 am
Thanks Mitch. Really appreciate it. Another solution with an awesome explanation.
These all alternative explanations will really help in other similar type of questions and save valuable time.

Thanks again
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by Scott@TargetTestPrep » Mon Dec 04, 2017 10:38 am
vinni.k 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 a = the number of checks in 50-dollar denomination and b = the number of checks in 100-dollar denomination and create the equation:

50a + 100b = 1800

a + 2b = 36

We also know that a + b = 30.

Subtracting our second equation from our first, we have:

b = 6, so a = 24

We can let n = the number of 50-dollar checks to be spent and create the equation:

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

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

5(24 - n) + 10(6) = 8(30 - n)

120 - 5n + 60 = 240 - 8n

3n = 60

n = 20

Answer: D

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