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Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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10 Jan 2016, 15:24
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Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3, respectively. Narcisse gives Aristide some of his tokens, and the new ratio is 6 : 5. What is the least number of tokens that Narcisse could have given to Aristide? a. 9 b. 17 c. 21 d. 27 e. 53 Do you guys have a suggestion on how to approach the problem?
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Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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10 Jan 2016, 20:49
dflorez wrote: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3, respectively. Narcisse gives Aristide some of his tokens, and the new ratio is 6 : 5. What is the least number of tokens that Narcisse could have given to Aristide?
a. 9 b. 17 c. 21 d. 27 e. 53
Do you guys have a suggestion on how to approach the problem? Hi, whenever you have ratios, change them into numeric values by multiplying with a common term as a 'variable'.. let it be x here and y is the number that N gives to A..so N will have 7xy and A will have 3x + y...the new ratio is 6/5.. so \(\frac{{7x5}}{{3x+5}}=\frac{6}{5}\).. 35x15=18x+6y... 17x=11y... since 17 and 11 are prime, x will be a multiple of 11 and y will be a multiple of 17....here we are to find value of y, which will be a multiple of 17.... only B is a multiple of 17.. ans B
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Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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10 Jan 2016, 15:50
dflorez wrote: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3, respectively. Narcisse gives Aristide some of his tokens, and the new ratio is 6 : 5. What is the least number of tokens that Narcisse could have given to Aristide?
a. 9 b. 17 c. 21 d. 27 e. 53
Do you guys have a suggestion on how to approach the problem? Good question to learn factors, multiples and the related concepts. Let N and A stand for tokens with Narcisse and Aristide. Thus per the original ratio: N/A = 7/3 > N = (7/3)A ...(1) Let x be the number of tokens given to Aristide by Narcisse. Thus after redistribution of the tokens you get, \(\frac{Nx}{A+x} = \frac{6}{5}\) > rearranging you get, 5N=6A+11x ... (2) When you subsititue N in terms of A from (1) in (2), you get, 11x = (17/3)A > x = (17/33)A....(3) Now as tokens can only take INTEGER values and so does 'x', A must be a multiple of 33 in order to get an integer value for 'x' in (3). Once you realize this, you will see that x will then become a multiple of 17 and out of all the options provided, only option B is a multiple of 17 and is hence the correct answer. x = (17/33)A, as A= 33p, where p is a positive integer, you get, x = (17/33)*33p = 17p Hope this helps.



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Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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10 Jan 2016, 19:45
dflorez wrote: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3, respectively. Narcisse gives Aristide some of his tokens, and the new ratio is 6 : 5. What is the least number of tokens that Narcisse could have given to Aristide?
a. 9 b. 17 c. 21 d. 27 e. 53
Do you guys have a suggestion on how to approach the problem? At first Ratio N:A = 7:3 i.e. N=7x and K =3x Now let constant x=1 that gives N=7 and K=3 Now if N gives K arcades to A we have new ratio 7K:3+K=6:5 =>K=17/11 So K is a multiple of 17/11 and the least multiple that makes K an integer is 17 [i.e. (17/11).1 , (17/11).2 ...... first integer multiple is (17/11).11=17 ] Thats answer K=17 Ans B
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Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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12 Jan 2016, 21:48
x  unknown multiplier y  token changed
7xy/3x+y=6/5
x/y=11/17 is maximally reduced fraction, so minimal y=17
B



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Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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13 Jan 2016, 06:52
I figured this out, but I used a different way than all in the thread. I'm not sure if I got lucky or not.
I chose to use a real number for the ratio, which made it easier to see for me. I used 110 total tokens.
Original ratio of 7:3 meant there were 77 and 33 tokens. Changing the ratio to 6:5 meant that there were now 60:50 tokens.
7760 = 17
I'm curious now though that if I used a different number that I'd have gotten the wrong answer.



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Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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13 Jan 2016, 10:17
Suppose that initially Narcisse has 7x tokens and Aristide has 3x tokens. Assuming that "n" tokens were given. => (7x  n)/(3x + n) = 6/5 Solving, 11n = 17x => n = 17x/11
So, the minimum value of x, so that above expression is an integer, is x = 11. So, with x = 11, n = 17



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Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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14 Jan 2016, 09:06
I have a question
I was discussing this problem with a friend. She suggested the following method. Can someone confirm that this is foolproof:
Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3, respectively. Narcisse gives Aristide some of his tokens, and the new ratio is 6 : 5. What is the least number of tokens that Narcisse could have given to Aristide?
Initially, the number of tokens = 7x + 3x = 10x Finally, the number of tokens = 6y + 5y = 11y
But, as the number of tokens remains the same, 10x = 11y So, x/y = 11/10
The minimum value of for the above fraction will be when x = 11 and y = 10
=> Initially, tokens = (77,33) (corresponding to 7x and 3x) Finally, tokens = (60,50) (corresponding to 6y and 5y)
So, tokens that need to be transferred = 77  60 = 17



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Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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14 Jan 2016, 10:23
PrijitDebnath wrote: I have a question
I was discussing this problem with a friend. She suggested the following method. Can someone confirm that this is foolproof:
Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3, respectively. Narcisse gives Aristide some of his tokens, and the new ratio is 6 : 5. What is the least number of tokens that Narcisse could have given to Aristide?
Initially, the number of tokens = 7x + 3x = 10x Finally, the number of tokens = 6y + 5y = 11y
But, as the number of tokens remains the same, 10x = 11y So, x/y = 11/10
The minimum value of for the above fraction will be when x = 11 and y = 10
=> Initially, tokens = (77,33) (corresponding to 7x and 3x) Finally, tokens = (60,50) (corresponding to 6y and 5y)
So, tokens that need to be transferred = 77  60 = 17 That's what I did and my thinking is it will always work. The total of the first ratio has to be a multiple of 10. The total of the second ration has to be a multiple of 11. The lowest common multiple of 10 and 11 is 110. So that is the smallest number the two totals can be equal to.



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Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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14 Jan 2016, 10:45
PrijitDebnath wrote: I have a question
I was discussing this problem with a friend. She suggested the following method. Can someone confirm that this is foolproof:
Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3, respectively. Narcisse gives Aristide some of his tokens, and the new ratio is 6 : 5. What is the least number of tokens that Narcisse could have given to Aristide?
Initially, the number of tokens = 7x + 3x = 10x Finally, the number of tokens = 6y + 5y = 11y
But, as the number of tokens remains the same, 10x = 11y So, x/y = 11/10
The minimum value of for the above fraction will be when x = 11 and y = 10
=> Initially, tokens = (77,33) (corresponding to 7x and 3x) Finally, tokens = (60,50) (corresponding to 6y and 5y)
So, tokens that need to be transferred = 77  60 = 17 This is perfectly fine method to tackle such questions. A bit too on the lengthier side though! You should always learn multiple methods to tackle the same problem but make sure to employ the method that will also help you in saving time on GMAT. Time management is a very critical component in GMAT. An alternate method is discussed at narcisseandaristidehavenumbersofarcadetokensintheratio211705.html#p1628739



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Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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14 Jan 2016, 17:34
let y=tokens N gives to A (7xy)/(3x+y)=6/5 17x=11y x/y=11/17 y=17



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Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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04 Apr 2016, 19:50
chetan2u wrote: dflorez wrote: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3, respectively. Narcisse gives Aristide some of his tokens, and the new ratio is 6 : 5. What is the least number of tokens that Narcisse could have given to Aristide?
a. 9 b. 17 c. 21 d. 27 e. 53
Do you guys have a suggestion on how to approach the problem? Hi, whenever you have ratios, change them into numeric values by multiplying with a common term as a 'variable'.. let it be x here and y is the number that N gives to A..so N will have 7xy and A will have 3x + y...the new ratio is 6/5.. so \(\frac{{7x5}}{{3x+5}}=\frac{6}{5}\).. 35x15=18x+6y... 17x=11y... since 17 and 11 are prime, x will be a multiple of 11 and y will be a multiple of 17....here we are to find value of y, which will be a multiple of 17.... only B is a multiple of 17.. ans B Hi chetan2u, How would you rate this question? Nice solution. Do such kind of questions appear on GMAT? I mean I reached up to the two equations but did not deduce it further because for some reason I thought I was going the wrong way . I ended up guessing because I could not conclude on an answer.



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Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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13 May 2016, 02:04
I have a question regarding my calculation, I answered the question correctly but I guess i got lucky. Could someone explain what is wrong with my calculation?
I converted both ratio's to a common denominator: 7:3 > 35/15 6:5 > 18/15 to get 17.
Thanks in advance,
Regards,



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Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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13 May 2016, 03:24
rachitshah wrote: chetan2u wrote: dflorez wrote: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3, respectively. Narcisse gives Aristide some of his tokens, and the new ratio is 6 : 5. What is the least number of tokens that Narcisse could have given to Aristide?
a. 9 b. 17 c. 21 d. 27 e. 53
Do you guys have a suggestion on how to approach the problem? Hi, whenever you have ratios, change them into numeric values by multiplying with a common term as a 'variable'.. let it be x here and y is the number that N gives to A..so N will have 7xy and A will have 3x + y...the new ratio is 6/5.. so \(\frac{{7x5}}{{3x+5}}=\frac{6}{5}\).. 35x15=18x+6y... 17x=11y... since 17 and 11 are prime, x will be a multiple of 11 and y will be a multiple of 17....here we are to find value of y, which will be a multiple of 17.... only B is a multiple of 17.. ans B Hi chetan2u, How would you rate this question? Nice solution. Do such kind of questions appear on GMAT? I mean I reached up to the two equations but did not deduce it further because for some reason I thought I was going the wrong way . I ended up guessing because I could not conclude on an answer. Hi, sorry, missed out on this Q earlier, I do not think this is sub600 should be rated higher.. And the logic can be tested on GMAT
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Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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22 Jul 2018, 16:49
chetan2u wrote: dflorez wrote: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3, respectively. Narcisse gives Aristide some of his tokens, and the new ratio is 6 : 5. What is the least number of tokens that Narcisse could have given to Aristide?
a. 9 b. 17 c. 21 d. 27 e. 53
Do you guys have a suggestion on how to approach the problem? Hi, whenever you have ratios, change them into numeric values by multiplying with a common term as a 'variable'.. let it be x here and y is the number that N gives to A..so N will have 7xy and A will have 3x + y...the new ratio is 6/5.. so \(\frac{{7x5}}{{3x+5}}=\frac{6}{5}\).. 35x15=18x+6y... 17x=11y... since 17 and 11 are prime, x will be a multiple of 11 and y will be a multiple of 17....here we are to find value of y, which will be a multiple of 17.... only B is a multiple of 17.. ans B Hi, sorry don't mean to bump an old post, but I got to 17x = 11y, but not sure how you determined that Y will be a multiple of 17? if we are asked to find Y, when I get to 17x = 11y I get that x must be a multiple of 17, but why does y have to be?



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Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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22 Jul 2018, 17:38
bp2013 wrote: chetan2u wrote: dflorez wrote: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3, respectively. Narcisse gives Aristide some of his tokens, and the new ratio is 6 : 5. What is the least number of tokens that Narcisse could have given to Aristide?
a. 9 b. 17 c. 21 d. 27 e. 53
Do you guys have a suggestion on how to approach the problem? Hi, whenever you have ratios, change them into numeric values by multiplying with a common term as a 'variable'.. let it be x here and y is the number that N gives to A..so N will have 7xy and A will have 3x + y...the new ratio is 6/5.. so \(\frac{{7x5}}{{3x+5}}=\frac{6}{5}\).. 35x15=18x+6y... 17x=11y... since 17 and 11 are prime, x will be a multiple of 11 and y will be a multiple of 17....here we are to find value of y, which will be a multiple of 17.... only B is a multiple of 17.. ans B Hi, sorry don't mean to bump an old post, but I got to 17x = 11y, but not sure how you determined that Y will be a multiple of 17? if we are asked to find Y, when I get to 17x = 11y I get that x must be a multiple of 17, but why does y have to be? Hi 17x=11y. Here x need not be multiple of 17 but 17x is a multiple of 17. Now 17x=11y. 11 of course does not contain 17 so y should contain 17.. Or you can see this way.. 17x=11y.......y=17x/11=17*(X/11) X/11 is an integer so X is multiple of 11 and y is multiple of 17 as y = 17*some integer (X/11)
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Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3
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05 Aug 2018, 08:46
dflorez wrote: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3, respectively. Narcisse gives Aristide some of his tokens, and the new ratio is 6 : 5. What is the least number of tokens that Narcisse could have given to Aristide?
a. 9 b. 17 c. 21 d. 27 e. 53
Do you guys have a suggestion on how to approach the problem? OA:B Total number of Arcade token would be multiple of \(10 (7+3)\) and a multiple of \(11(6+5)\). Lowest integer that would be multiple of \(10\) and \(11\) simultaneously is \(110\). Intial token with Narcisse : \(\frac{7}{10}*110 = 77\) Final token with Narcisse : \(\frac{6}{11}*110 = 60\) least number of token Narcisse >Aristide \(=7760=17\)




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