This year, x people won an Olympic medal for water competitions. One

Everyone won at least one medal.
Also, the number of people who won medal for both diving and swimming is (x/3)*1/4 = x/12
Now,
(AuB) - (AnB) = complement of (AnB)

here
AuB = x
AnB = x/12

The number of people who won single medal(or compliment of AnB) is
x-x/12
=11x/12

Answer:-A

VERITAS PREP OFFICIAL SOLUTION:

Let’s work on a Venn diagram and fill in what we do know – an important first step as many of these problems come down in large part to “getting organized”. We know there will be some medal winners who swam but did not dive, some who dove but did not swim, and some who swam AND dove.



“x” here will be at the top of our Venn because it is the total for ALL PARTS of the Venn. That is, all three categories will sum to x. x/3 represents the 1/3 of the total (“x”) who swam, including those who swam only AND those who swam and dove.

We can make up a variable, let’s say “y,” to represent the total number of divers. The key to understanding this question lies in the last sentence and the phrase “not both.”

We need to know the people who ONLY swam but did NOT dive, and the people who ONLY dove but did NOT swim. I made up variables for these two groups: “a” and “z.”

Let’s use the answer choices to our advantage! Since they have the denominators of 12 and 7, let’s use one of those and work backwards! 12 appears more often, so we can start there.

If x = 12, there are 12/3 = 4 swimmers total, (12/3)/4 = 1 of whom swam and dove. That means a = 3. If 4 people swam, then 12-4 = 8 dove, so z = 8.

The two categories we’re looking for (a + z) are 3 + 8 = 11. We are looking for an answer choice that gives us 11 when x = 12.

The answer is (A).

Hi All,

This question can be solved by TESTing VALUES (and taking a few notes).

We're told that X people won a medal for water competitions. Of those X people, 1/3 won a medal for swimming; of those who won a medal for SWIMMING, 1/4 also won a medal for diving.

The common denominator between 1/3 and 1/4 is 12, so let's TEST X = 12....

X = 12 medal winners

(1/3)(12) = 4 won a medal for swimming
(1/4)(4) = 1 of the swimming winners ALSO won a medal for diving

We're asked how many of the X people did NOT win a medal for BOTH swimming and diving.

Since there were 12 people, and only 1 won BOTH medals, the other 11 won JUST ONE medal - thus, the answer to the question is 11 (when X = 12). The answers are written in such a way that you don't have to do much math to find the correct answer.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

In the last part,

When only ppl eho dove has to be found out, why havent we done 8-1 to get ppl who dove only.Because for ppl who Swam only we did Swam-Both+Swam only, so why not Dove only=Total -Swam-Both.
Bunuel kindly clear my doubt here.

Total = x = 12
Swam only = a = 3
Both swam and dove = x/12 = 1

Dove only = z = Total - Swam only - Both = 12 - 3 - 1 = 8.

Total= x
Swimming= x/3
Both = x/12
Diving= y

x= x/3+y- x/12
x= 3x/4

Answer would be: x/4+2x/3= 11x/12

I got the answer but is this the correct way?

where's the neither part in the formula? can any1 tell me? the part " people who neither won medal for swimming nor won medal for diving" . I saw many solutions omit this but I don't know why.

Here the sample is those who won a medal: x people won an Olympic medal for water competitions.

Thank you a lot!! my mistake

We can just substitute numbers in this problem. Since we have x/3 and x/4, take 12 as it divides evenly both 3 and 4. then, we have 4 people winning for swimming out of which (1/4) won for diving or 1 person. So 12-1=11. Only A gives 11 when substituted for x=12

hey pushpitkc is my understanding correct ? have a great weekend

Let SWIMMERS be S
Let DIVERS be D
let total # of winners be x

One-third of the winners earned a medal for swimming \(\frac{1}{3}x\) (S) ( remaining part \(\frac{2}{3}x\); we shall use it later)

One-fourth of those who earned a medal for swimming also earned a medal for diving \(\frac{1}{3}x\) * \(\frac{1}{4}\)= \(\frac{1}{12}x\) (S+D)

Now we know that both D and S = \(\frac{1}{12}x\)

To find only DIVERS subtract both from remaining part i.e \(\frac{2}{3}x\) \(-\) \(\frac{1}{12}x\)= \(\frac{7}{12}x\) (D)

(NOTE: we subtract only once because there are two group overlaps, in other words no need to make similar subtraction like this 1/3x -1/12x to find SWIMMERS )


To find only swimmers and only divers \(\frac{1}{3}x\)+ \(\frac{7}{12}x\) = \(\frac{11}{12}x\)

Have dave13

Your method finally lands you at the answer, but IMO, you should try and use numbers

Assume P(Total) = x = 120 (who win medals in all water games)

Given: \(\frac{1}{3}\)rd win medals in Swimming -> P(Only Swimming) = 40
\(\frac{1}{4}\)th of the medal winners in Swimming also win medals in Diving -> P(Both) = 10

Total number of people who won an Olympic medal for water competitions but did not both
receive a medal for swimming and a medal for diving is: P(Total) - P(Both) = 120 - 10 = 110

Therefore, we can write the number 110 as \(\frac{11}{12}*x\)(Option A) as x = 120

Hope this helps you!

\(x\,\,\left( {{\rm{water}}} \right)\,\,\left\{ \matrix{\\
\,{x \over 3}\,\,\left( {{\rm{swim}}} \right)\,\,\, \to \,\,\,\left\{ \matrix{\\
\,{1 \over 4}\left( {{x \over 3}} \right)\,\,\left( {{\rm{swim}}\,\,{\rm{\& }}\,\,{\rm{dive}}} \right) \hfill \cr \\
\,{3 \over 4}\left( {{x \over 3}} \right)\,\,\left( {{\rm{swim}}\,\,{\rm{\& }}\,\,{\rm{not}}\,\,{\rm{dive}}} \right) \hfill \cr} \right. \hfill \cr \\
\,{{2x} \over 3}\,\,\left( {{\rm{not}}\,\,{\rm{swim}}} \right) \hfill \cr} \right.\)

\(? = \left( {{\rm{swim}}} \right) - \left( {{\rm{swim}}\,\,{\rm{\& }}\,\,{\rm{dive}}} \right) = x - {1 \over 4}\left( {{x \over 3}} \right) = {{11} \over {12}}x\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

We see that ⅓(¼x) = (1/12)x = x/12 people won a medal in both swimming and dividing.

Thus x - x/12 = 12x/12 - x/12 = 11x/12 people did not win a medal in both swimming and dividing.

Answer: A

Total number of people= x
Both= x/12
Required= x - x/12 = 11x/12

Hence, A. Cheers.

kindly explain how did you get for only swimming ... i got confused

Here it goes.
x = no. of medals
'[1x][/3]' = swimming
'[1][/4]' of '[1x][/3]' = swimming+diving

=> X-'[1x][/12]'='[11x][/12]'

ANSWER A