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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  [#permalink]

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New post 26 Aug 2012, 18:57
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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?

A. 4
B. 12
C. 15
D. 20
E. 24
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Re: Tourist purchased a total of 30 travellers checks  [#permalink]

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New post 26 Aug 2012, 19:37
2
6
X + y = 30
50x + 100y = 1800
=> X +2y = 36
=> Y = 6 x = 24
Let n be the number
So ((24-n) * 50 + 6*100)/(30-n) = 80

=> 120 – 5n + 60 = 240 -8n
=> 3n = 60
=> n = 20

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

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New post 26 Aug 2012, 19:22
2
1
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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New post 27 Aug 2012, 08:12
4
1
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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New post 27 Sep 2013, 05:28
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


Similar question to practice: a-tourist-purchased-a-total-of-1-500-worth-of-traveler-s-111369.html
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Re: Tourist purchased a total of 30 travellers checks  [#permalink]

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New post 27 Sep 2013, 07:37
subhashghosh wrote:
X + y = 30
50x + 100y = 1800
=> X +2y = 36
=> Y = 6 x = 24
Let n be the number
So ((24-n) * 50 + 6*100)/(30-n) = 80

=> 120 – 5n + 60 = 240 -8n
=> 3n = 60
=> n = 20

Answer - D



I am confused as to why cant we do the opposite for this

(24*50 + (6-n)*100)/(30-n)) = 80 ?
The answer doesnt hold then ....
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Re: Tourist purchased a total of 30 travellers checks  [#permalink]

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New post 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

then 1800-50*x / 30-x =80
x= 20
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Re: Tourist purchased a total of 30 travelers checks in $50 and  [#permalink]

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New post 01 Dec 2013, 23:08
you could set-up a quick table and brute force the answer.

A 4 * 50 200 1800 -200 1600 26 61.54
B 12 * 50 600 1800 -600 1200 18 66.67
C 15 * 50 750 1800 -750 1050 15 70.00
D 20 * 50 1000 1800 -1000 800 10 80.00
E 24 * 50 1200 1800 -1200 600 6 100.00
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Re: Tourist purchased a total of 30 travelers checks in $50 and  [#permalink]

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New post 10 Mar 2014, 08:18
average is 80 $, only pssible total amount for this is 800 $. Hence he can spend 1000 $ which equals to 20 50$ checks.

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 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.


x*50 + (30-x)*100 = 1800
x*5+(30-x)*10=180
300-x*5=180
x=24

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

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New post 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.


x*50 + (30-x)*100 = 1800
x*5+(30-x)*10=180
300-x*5=180
x=24

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

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New post 08 May 2016, 09:17
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.
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Re: Tourist purchased a total of 30 travelers checks in $50 and  [#permalink]

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New post 08 May 2016, 09:27
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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.


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

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New post 08 May 2016, 09:32
1
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..
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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

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New post 08 May 2016, 20:42
equation 1: 50x+100y=1800
equation 2: 50(x-z)+100y=80(30-z)➡
50x+100y=2400-30z
therefore,
1800=2400-30z
z=20 $50 checks he needs to spend
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Re: Tourist purchased a total of 30 travelers checks in $50 and  [#permalink]

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

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New post 09 Jun 2016, 06:18
Took me almost 3 min. but only because I figured out after 2 min. that the best method to solve this question is by plugging in the numbers.
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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 1569 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 1568 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.
Re: Tourist purchased a total of 30 travelers checks in $50 and &nbs [#permalink] 13 Jun 2016, 22:22

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