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

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26 Aug 2012, 18:57

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30% (02:31) wrong based on 369 sessions

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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?

Re: Tourist purchased a total of 30 travellers checks [#permalink]

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26 Aug 2012, 19:22

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vinay911 wrote:

Tourist purchased a total of 30 travellers 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(arithemetic mean) of the remaining travelers checks is $80?

a)4 b)12 c)15 d)20 e)24

This one took me 2+ min.

x+y = 30 50x+100y=1800 Solving both the equation will give you - x = 24 and y = 6. Now one can start plug-n-play and can figure out that D fits the answer. x=4, y=6 makes the average amount of the remaining travelers checks to $80.

D wins.
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Re: Tourist purchased a total of 30 travellers checks [#permalink]

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27 Aug 2012, 08:12

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This can be solved easily using algorithm, x + y = 30 ---(i) 50x +100y=1800 => x+2y=36 ---- (ii)

Therefore from equation (i) and (ii) we get, y=6 Let, we can use the algorithm to find how 50 and 100 make the average to 80

................50......................100 ............................80 ...........(100-80)...............(80-50) ..............=20......................=30 >> ............2 .........................3 (this is the ratio by which 50 & 100 are to be combined to make average of 80) >> .............4.........................6 (just multiplying both sides with 2, it does not change the ratio)

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
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Re: Tourist purchased a total of 30 travellers checks [#permalink]

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27 Sep 2013, 05:21

Hello from the GMAT Club BumpBot!

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Tourist purchased a total of 30 travellers 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(arithemetic mean) of the remaining travelers checks is $80?

Re: Tourist purchased a total of 30 travellers checks [#permalink]

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27 Sep 2013, 07:59

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GMATBaumgartner wrote:

Tourist purchased a total of 30 travellers 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(arithemetic mean) of the remaining travelers checks is $80?

a)4 b)12 c)15 d)20 e)24

actually there is no need to calculate the numbers of two checks. Here is how I got the correct answer:

say the number of checks need to be spent to get the final average amount of $80 is x Average amount= amount of $100 checks / numbers of remaining checks

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

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24 Oct 2014, 12:03

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 took more than 2min to solve this question. Is there any quick way? My approach was this.

Initially, Tourist had total 30 travelers checks. Let say say, he initially had x number of $50 travelers checks. So he will have 30-x number of $100 travelers checks. The total worth of the travelers checks is $1800.

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

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06 May 2015, 14:04

ammuseeru wrote:

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 took more than 2min to solve this question. Is there any quick way? My approach was this.

Initially, Tourist had total 30 travelers checks. Let say say, he initially had x number of $50 travelers checks. So he will have 30-x number of $100 travelers checks. The total worth of the travelers checks is $1800.

So he had 24 fifty dollar travelers checks and (30-24=6) hundred dollar travelers checks.

Suppose he spends few checks, remaining number of $50 is y

(5*y+6*100)/(y+6) = 80 ----comments:- Weight Average Mean Formula. He did not spend any $100 checks so 6 is as it is. y=4

he was left with only 4 checks of $50. So he spent 20 checks of $ 50.

Answer is D

I Also had 2+ minutes until i figured out, that it might be the best way to start plugging in. Start in the middle with plugging in and check if the answer fits the average ov $80. Start Plugging in 15 and you get 15*50 = 750 spent leaving 1050 USD and 15 tickets averaging $70 > eliminate this answer choice. If youre not sure which way to go, choose one direction, plug in 20 and you get 20*50 = 1000 spent leaving 800 USD and 20 tickets averaging $80.

Very easy, but it took me more than 5min to figure out which method to take ... poor me
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Re: Tourist purchased a total of 30 travelers checks in $50 and [#permalink]

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07 May 2015, 03:36

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 x be the number of $50 checks he had to start with. 50x + 100(30-x) = 1800 Solving, we get x=24

therefore he had 6 $100 checks (value $600)

After he spends some $50 checks, let y be the total number of checks he has remaining. Therefore, 80y = 6*100 + 50(y-6) ....... eqauting the value of the checks we get y=10 Hence we had 10 total checks remaining, out of which 6 were $100 and 4 $50. Therefore, he spend 20 $50 checks.

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

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08 May 2016, 05:14

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Suppose the person can spend 'n' $50 checks, then the equation becomes

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

Solving the equation we get 'n'= 20

chetan2u, please advise if there is any shortcut method for it.

Hi, after you have got x= 24 and y=6, you can use weighted average method to find the rest of the answer.. for average of 80, the ratio of # of 50 and # of 100 = \(\frac{100-80}{80-50} = \frac{2}{3}\).. therefore you should have 2 * 50$ for every 3 *100$... but we have 6 *100$, so # of 50$ = \(2 *\frac{6}{3}= 4\).. and he can spend the REST = \(24-4 = 20\)
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Suppose the person can spend 'n' $50 checks, then the equation becomes

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

Solving the equation we get 'n'= 20

chetan2u, please advise if there is any shortcut method for it.

Two more methods would be - 1)substitute - you can substitute one by one and find... 2) Logic.. # of 50 is 24 and # of 100 is 6.... so if both are 6, the average is 75$, but we are looking for higher average, 80$, so # of 50 $ has to be less than 6... ONLY 24 - 20 =4 fits in..
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Re: Tourist purchased a total of 30 travelers checks in $50 and [#permalink]

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09 May 2016, 07:06

chetan2u wrote:

Divyadisha wrote:

It took me more than 2 mins.

x+y= 30 50x +100y= 1800

Solving the above equation we get x= 24, y=6

Suppose the person can spend 'n' $50 checks, then the equation becomes

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

Solving the equation we get 'n'= 20

chetan2u, please advise if there is any shortcut method for it.

Two more methods would be - 1)substitute - you can substitute one by one and find... 2) Logic.. # of 50 is 24 and # of 100 is 6.... so if both are 6, the average is 75$, but we are looking for higher average, 80$, so # of 50 $ has to be less than 6... ONLY 24 - 20 =4 fits in..

Thanks for suggesting all approaches
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