GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 05 Apr 2020, 18:00

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 22 Dec 2015
Posts: 5
Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

10 Jan 2016, 14:24
4
22
00:00

Difficulty:

25% (medium)

Question Stats:

80% (02:12) correct 20% (02:27) wrong based on 211 sessions

### HideShow timer Statistics

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?
Math Expert
Joined: 02 Aug 2009
Posts: 8308
Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

10 Jan 2016, 19:49
3
5
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 7x-y and A will have 3x + y...
the new ratio is 6/5..
so $$\frac{{7x-5}}{{3x+5}}=\frac{6}{5}$$..
35x-15=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
_________________
CEO
Joined: 20 Mar 2014
Posts: 2543
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

10 Jan 2016, 14:50
5
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,

$$\frac{N-x}{A+x} = \frac{6}{5}$$ ---> 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.
##### General Discussion
Manager
Joined: 23 Jul 2014
Posts: 86
Location: India
WE: Information Technology (Computer Software)
Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

10 Jan 2016, 18:45
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
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 ]

Ans B
_________________
The Mind ~ Muscle Connection
My GMAT Journey is Complete.Going to start the MBA in Information Management from 2016
Good Luck everyone.
Director
Joined: 23 Jan 2013
Posts: 511
Schools: Cambridge'16
Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

12 Jan 2016, 20:48
1
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
Intern
Joined: 29 Dec 2015
Posts: 10
GPA: 2.8
Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

13 Jan 2016, 05:52
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.

77-60 = 17

I'm curious now though that if I used a different number that I'd have gotten the wrong answer.
Manager
Joined: 22 Feb 2015
Posts: 74
Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

13 Jan 2016, 09:17
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
Manager
Joined: 22 Feb 2015
Posts: 74
Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

14 Jan 2016, 08:06
5
I have a question

I was discussing this problem with a friend. She suggested the following method. Can someone confirm that this is fool-proof:

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
Manager
Joined: 22 Dec 2015
Posts: 87
Concentration: General Management
GMAT 1: 760 Q48 V47
GPA: 3.89
WE: Accounting (Energy and Utilities)
Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

14 Jan 2016, 09:23
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 fool-proof:

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.
CEO
Joined: 20 Mar 2014
Posts: 2543
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

14 Jan 2016, 09:45
1
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 fool-proof:

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 narcisse-and-aristide-have-numbers-of-arcade-tokens-in-the-ratio-211705.html#p1628739
VP
Joined: 07 Dec 2014
Posts: 1241
Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

14 Jan 2016, 16:34
let y=tokens N gives to A
(7x-y)/(3x+y)=6/5
17x=11y
x/y=11/17
y=17
Manager
Joined: 25 Sep 2015
Posts: 99
Location: United States
GMAT 1: 700 Q48 V37
GPA: 3.26
Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

04 Apr 2016, 18:50
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 7x-y and A will have 3x + y...
the new ratio is 6/5..
so $$\frac{{7x-5}}{{3x+5}}=\frac{6}{5}$$..
35x-15=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.
Intern
Joined: 14 Mar 2016
Posts: 1
Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

13 May 2016, 01:04
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.

Regards,
Math Expert
Joined: 02 Aug 2009
Posts: 8308
Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

13 May 2016, 02:24
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 7x-y and A will have 3x + y...
the new ratio is 6/5..
so $$\frac{{7x-5}}{{3x+5}}=\frac{6}{5}$$..
35x-15=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 sub-600 should be rated higher..
And the logic can be tested on GMAT
_________________
Current Student
Joined: 07 May 2015
Posts: 35
Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

22 Jul 2018, 15:49
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 7x-y and A will have 3x + y...
the new ratio is 6/5..
so $$\frac{{7x-5}}{{3x+5}}=\frac{6}{5}$$..
35x-15=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?
Math Expert
Joined: 02 Aug 2009
Posts: 8308
Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

22 Jul 2018, 16:38
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 7x-y and A will have 3x + y...
the new ratio is 6/5..
so $$\frac{{7x-5}}{{3x+5}}=\frac{6}{5}$$..
35x-15=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)
_________________
Manager
Joined: 18 Jun 2018
Posts: 248
Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

05 Aug 2018, 07:46
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 $$=77-60=17$$
Manager
Joined: 25 May 2016
Posts: 84
Location: Singapore
Concentration: Finance, General Management
GMAT 1: 620 Q46 V30
Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

15 Dec 2018, 22:57
1
7:3 = 77:33
6:5 = 60:50

77-60 = 17. Therefore, Narcisse has given 17 tokens.
Director
Joined: 25 Dec 2018
Posts: 612
Location: India
Concentration: General Management, Finance
GMAT Date: 02-18-2019
GPA: 3.4
WE: Engineering (Consulting)
Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3  [#permalink]

### Show Tags

10 Feb 2020, 00:10
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?

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.

Re: Narcisse and Aristide have numbers of arcade tokens in the ratio 7 : 3   [#permalink] 10 Feb 2020, 00:10
Display posts from previous: Sort by