Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3

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,

\(\frac{N-x}{A+x} = \frac{6}{5}\) ---> 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.

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
7-K: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

x - unknown multiplier
y - token changed

7x-y/3x+y=6/5

x/y=11/17 is maximally reduced fraction, so minimal y=17

B

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.

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 narcisse-and-aristide-have-numbers-of-arcade-tokens-in-the-ratio-211705.html#p1628739

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 \(=77-60=17\)

7:3 = 77:33
6:5 = 60:50

77-60 = 17. Therefore, Narcisse has given 17 tokens.

Solution:

Let 7n and 3n be the number of tokens that Narcisse and Aristide, respectively, originally had, and let x be the number of tokens Narcisse gave to Aristide. We can create the equation:

(7n - x) / (3n + x) = 6/5

6(3n + x) = 5(7n - x)

18n + 6x = 35n - 5x

11x = 17n

x = 17n/11

Since 17 is not divisible by 11, we see that the least positive value of n is 11. Therefore, the least positive value of x is 17(11)/11 = 17.

Answer: B

I was wondering if my method was alright because I'm not too sure about that even though I got the right answer in about a minute.

We have 7/3 and 6/5
Then I did (7/3)*6 = 42/18 and (6/5)*7 = 42/35
35-18 = 17 (answer B)

OA from ManhattenPrep

Because Narcisse and Aristide are exchanging tokens (rather than acquiring new ones or spending the ones they have), the total number of tokens must remain the same. Thus, this total must be a multiple of both 10 (7 + 3) and 11 (6 + 5). The lowest common multiple of 10 and 11 is 110, so the least number of tokens they could have is 110.

If the two of them have a total of 110 tokens, then Narcisse has 77 to start with, and Aristide has 33. After the exchange, Narcisse will have 60 tokens and Aristide will have 50. 17 tokens will have changed hands.

The correct answer is (B).

Since tokens are TRANSFERRED from Narcisse to Aristide but not removed, the total number of tokens for each ratio must be THE SAME.

Narcisse and Aristide have arcade tokens in the ratio 7 : 3.
Here, the sum of the parts of the ratio = 7+3 = 10, implying that the total number of tokens must be a MULTIPLE OF 10.

Narcisse gives Aristide some of his tokens, and the new ratio is 6 : 5.
Here, the sum of the parts of the ratio = 6+5 = 11, implying that the total number of tokens must be a MULTIPLE OF 11.

The smallest possible total for the two ratios = the LCM of 10 and 11 = 110.

Old ratio of 7 to 3:
Since the parts here sum to 10 -- and the LCM above is 11 times as great -- each part of this ratio must be increased by a factor of 11:
(old N) : (old A) = (7*11) : (3*11) = 77:33 --> sum = 77+33 = 110

New ratio of 6 to 5:
Since the parts here sum to 11 -- and the LCM above is 10 times as great -- each part of this ratio must be increased by a factor of 10:
(new N) : (new A) = (6*10) : (5*10) = 60:50 --> sum = 60+50 = 110

What is the least number of tokens that Narcisse could have given to Aristide?
(old N) - (new N) = 77-60 = 17