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In its annual fest, a college, which has 510 students, is organizing t 28 Apr 2020, 03:16
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In its annual fest, a college, which has 510 students, is organizing three events. The students were allowed to participate in any number of events they liked. While viewing the statistics of the performance, the general secretary noticed:

1. The number of students who participated in at least two events were 52% more than those who participated in exactly one event.
2. The number of students participating in 1, 2 or 3 events respectively was at least equal to 1.
3. The number of students who did not participate in any of the three events was the minimum possible integral value under these conditions.

What can be the maximum number of students who participated in exactly 3 games?

A. 200
B. 300
C. 303
D. 304
E. 305


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C
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In its annual fest, a college, which has 510 students, is organizing t Updated on: 18 Jun 2024, 08:41
Originally posted by lacktutor on 28 Apr 2020, 04:24.
Last edited by Bunuel on 18 Jun 2024, 08:41, edited 2 times in total.
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Bunuel
In its annual fest, a college, which has 510 students, is organizing three events. The students were allowed to participate in any number of events they liked. While viewing the statistics of the performance, the general secretary noticed:

1. The number of students who participated in at least two events were 52% more than those who participated in exactly one event.
2. The number of students participating in 1, 2 or 3 events respectively was at least equal to 1.
3. The number of students who did not participate in any of the three events was the minimum possible integral value under these conditions.

What can be the maximum number of students who participated in exactly 3 games?

A. 200
B. 300
C. 303
D. 304
E. 305

Are You Up For the Challenge: 700 Level Questions
Show more

Total = 510
The number of students who participated:
--> Only one event =a+b+c
--> Only two events = x+y+z
--> All three events = P

\((x+y+z) + p = 1.52 (a+b+c)\)
Neither =minimum possible integral value under these conditions

--> \((a+b+c)+ (x+y+z)+ P+ Neither = 510\)

\(2.52 (a+b+c)= 510 - Neither \)

In order \((a+b+c )\) to be integer, 'Neither' should be \(6\).
--> \((a+b+c)= \frac{510-6}{2.52 }= 200\)

\((x+y+z)+ P= 510-6-200= 304\)
In order P to be the maximum value,

--> \(x+y+z =1\) --> \(P =304-1= 303\)

Answer (C).­
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In its annual fest, a college, which has 510 students, is organizing t 28 Apr 2020, 15:10
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\(N(2 or 3) = (\frac{152}{100})N(1)\)

\(N(2 or 3) = (\frac{38}{25})N(1)\)

N(2 or 3) must be a multiple of 38 and N(1) must be a multiple of 25.

Total students who participate in at least 1 event = 38k+25k = 63k

Now we need to minimize the number of students who did not participate in any of the three event, so we take the maximum value of k that satisfies the following equation (multiple of 63 that is closest to 510).

510-63k>0

k<8.xyz

k= 8

Number of participants who participate in 2 or 3 events = 38*8= 304

the maximum number of students who participated in exactly 3 games = 304-1 =303
Bunuel
In its annual fest, a college, which has 510 students, is organizing three events. The students were allowed to participate in any number of events they liked. While viewing the statistics of the performance, the general secretary noticed:

1. The number of students who participated in at least two events were 52% more than those who participated in exactly one event.
2. The number of students participating in 1, 2 or 3 events respectively was at least equal to 1.
3. The number of students who did not participate in any of the three events was the minimum possible integral value under these conditions.

What can be the maximum number of students who participated in exactly 3 games?

A. 200
B. 300
C. 303
D. 304
E. 305

Are You Up For the Challenge: 700 Level Questions
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In its annual fest, a college, which has 510 students, is organizing t 28 Apr 2020, 23:07
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Alternatively
Let x represent the number who participated in exactly 1 event, y represent the number who participated in exactly 2 events, and z represent the number who participated in all three events.
Then y+z=1.52x -------(1)
Since the total number of students in the school is 510, then the last two digits of x is either 00 or 50. In other words, x must be a multiple of 50.
Possible values of x are: 50,100,150, and 200. Note that when x=250, the total number of students exceeds 510. i.e. (1.52+1)*250=630 and 2.52x≤510. Hence x cannot exceed 200.
From the information given, we need the number of students who did not participate in any of the 3 events to be minimum. Hence, x must be 200.
y+z=1.52*200 = 304
But from the second information provided above, y ≥ 1.
Hence z is maximum when y=1.
z=304-1=303

The answer is C.
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In its annual fest, a college, which has 510 students, is organizing t 28 Apr 2020, 23:28
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You got lucky brother. Try to solve the question if there were 570 total number of students.


eakabuah
Alternatively
Let x represent the number who participated in exactly 1 event, y represent the number who participated in exactly 2 events, and z represent the number who participated in all three events.
Then y+z=1.52x -------(1)
Since the total number of students in the school is 510, then the last two digits of x is either 00 or 50. In other words, x must be a multiple of 50.
Possible values of x are: 50,100,150, and 200. Note that when x=250, the total number of students exceeds 510. i.e. (1.52+1)*250=630 and 2.52x≤510. Hence x cannot exceed 200.
From the information given, we need the number of students who did not participate in any of the 3 events to be minimum. Hence, x must be 200.
y+z=1.52*200 = 304
But from the second information provided above, y ≥ 1.
Hence z is maximum when y=1.
z=304-1=303

The answer is C.
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In its annual fest, a college, which has 510 students, is organizing t 29 Apr 2020, 00:05
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nick1816 thanks for the feedback. However, as long as the total number of students is less than 630, the answer to this question per the reasoning provided in my solution will still be 303. Why? The test criterion is 2.52x≤the total number of students.
So 2.52x≤570 Hence x≤226. Possible values of x are 50,100,150,200,250,300,350, etc. Clearly, for the minimum possible value of students who do not participate in any of the 3 events, x must be 200. Still, any value of x greater than 200 will mean that the total number of students will still be more than 570, hence 250 and above are still not applicable. Unless I don't understand the point you're making the solution provided above should still work.

I only relied on numbers theory to come up with the possible values of x to have their last two digits to be either 00 or 50(implying x is a multiple of 50) since 1.52*x must be an integer.

eakabuah
Alternatively
Let x represent the number who participated in exactly 1 event, y represent the number who participated in exactly 2 events, and z represent the number who participated in all three events.
Then y+z=1.52x -------(1)
Since the total number of students in the school is 510, then the last two digits of x is either 00 or 50. In other words, x must be a multiple of 50.
Possible values of x are: 50,100,150, and 200. Note that when x=250, the total number of students exceeds 510. i.e. (1.52+1)*250=630 and 2.52x≤510. Hence x cannot exceed 200.
From the information given, we need the number of students who did not participate in any of the 3 events to be minimum. Hence, x must be 200.
y+z=1.52*200 = 304
But from the second information provided above, y ≥ 1.
Hence z is maximum when y=1.
z=304-1=303

The answer is C.
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In its annual fest, a college, which has 510 students, is organizing t 29 Apr 2020, 00:09
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Let E(1), E(2), E(3) be the students who participate in 1,2,3 of events respectively. Given that E(2)+E(3) =1.52E(1)
So 2.52 E(1)=510---assuming e(0) is 0
But 510 is not divisible by 2.52 so take a multiple of 2.52 nearest to 510..504
E(1) =504/2.52=200 ; E(0) =6 ;
E(2)+E(3) =304 we have to find max value of E(3)-but we have condition that min value of each is 1 so let E(2) =1 so E(3) =303-----Answer(C)

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In its annual fest, a college, which has 510 students, is organizing t 29 Apr 2020, 00:32
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If y = (152/100)*x

Now if you say that last 2 digits of x gonna be only 50 or 00, then you'll miss a lot of cases. That's why your answer would be wrong if there were 570 students. X will be equal to 225.

We can clearly see that 152 and 100 not co-prime. So deduce the ratio in the lowest form.

Y/x = 38/25

x can be any multiple of 25. So last 2 digits of x can be 25,50,75 or 00.





eakabuah
nick1816 thanks for the feedback. However, as long as the total number of students is less than 630, the answer to this question per the reasoning provided in my solution will still be 303. Why? The test criterion is 2.52x≤the total number of students.
So 2.52x≤570 Hence x≤226. Possible values of x are 50,100,150,200,250,300,350, etc. Clearly, for the minimum possible value of students who do not participate in any of the 3 events, x must be 200. Still, any value of x greater than 200 will mean that the total number of students will still be more than 570, hence 250 and above are still not applicable. Unless I don't understand the point you're making the solution provided above should still work.

I only relied on numbers theory to come up with the possible values of x to have their last two digits to be either 00 or 50(implying x is a multiple of 50) since 1.52*x must be an integer.

eakabuah
Alternatively
Let x represent the number who participated in exactly 1 event, y represent the number who participated in exactly 2 events, and z represent the number who participated in all three events.
Then y+z=1.52x -------(1)
Since the total number of students in the school is 510, then the last two digits of x is either 00 or 50. In other words, x must be a multiple of 50.
Possible values of x are: 50,100,150, and 200. Note that when x=250, the total number of students exceeds 510. i.e. (1.52+1)*250=630 and 2.52x≤510. Hence x cannot exceed 200.
From the information given, we need the number of students who did not participate in any of the 3 events to be minimum. Hence, x must be 200.
y+z=1.52*200 = 304
But from the second information provided above, y ≥ 1.
Hence z is maximum when y=1.
z=304-1=303

The answer is C.
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In its annual fest, a college, which has 510 students, is organizing t 29 Apr 2020, 00:37
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nick1816 I now understand your point clearly. You're perfectly right. Thanks once again for the feedback.

nick1816
If y = (152/100)*x

Now if you say that last 2 digits of x gonna be only 50 or 00, then you'll miss a lot of cases. That's why your answer would be wrong if there were 570 students. X will be equal to 225.

We can clearly see that 152 and 100 not co-prime. So deduce the ratio in the lowest form.

Y/x = 38/25

x can be any multiple of 25. So last 2 digits of x can be 25,50,75 or 00.





eakabuah
nick1816 thanks for the feedback. However, as long as the total number of students is less than 630, the answer to this question per the reasoning provided in my solution will still be 303. Why? The test criterion is 2.52x≤the total number of students.
So 2.52x≤570 Hence x≤226. Possible values of x are 50,100,150,200,250,300,350, etc. Clearly, for the minimum possible value of students who do not participate in any of the 3 events, x must be 200. Still, any value of x greater than 200 will mean that the total number of students will still be more than 570, hence 250 and above are still not applicable. Unless I don't understand the point you're making the solution provided above should still work.

I only relied on numbers theory to come up with the possible values of x to have their last two digits to be either 00 or 50(implying x is a multiple of 50) since 1.52*x must be an integer.

eakabuah
Alternatively
Let x represent the number who participated in exactly 1 event, y represent the number who participated in exactly 2 events, and z represent the number who participated in all three events.
Then y+z=1.52x -------(1)
Since the total number of students in the school is 510, then the last two digits of x is either 00 or 50. In other words, x must be a multiple of 50.
Possible values of x are: 50,100,150, and 200. Note that when x=250, the total number of students exceeds 510. i.e. (1.52+1)*250=630 and 2.52x≤510. Hence x cannot exceed 200.
From the information given, we need the number of students who did not participate in any of the 3 events to be minimum. Hence, x must be 200.
y+z=1.52*200 = 304
But from the second information provided above, y ≥ 1.
Hence z is maximum when y=1.
z=304-1=303

The answer is C.

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In its annual fest, a college, which has 510 students, is organizing t 29 Apr 2020, 04:55
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hi, nick1816
I got the question incorrect, first.
Actually, I didn't get the 2nd statement well.
Anyway, what do you think my approach solving the question?
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In its annual fest, a college, which has 510 students, is organizing t 29 Apr 2020, 08:54
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Based on the question data, a Venn diagram can be drawn to include all the constraints. The Venn diagram looks like the one below:

Attachment:
29th April 2020 - Reply 5.jpg
29th April 2020 - Reply 5.jpg [ 73.71 KiB | Viewed 9787 times ]

The areas under the circle comprises of people who participated in at least one event. This should be a multiple of 63 since this is the sum of 38 and 25.

The number of students who did not participate in any of the three events is to be kept to a minimum integer. This can be done by calculating the biggest multiple of 63 less than 510. This is 504. Therefore, 6 students did not participate in any event.

504 = 63 X 8. Therefore,
a+b+c+x = 38 X 8 = 304 and Exactly One = 25 X 8 = 200.

We need to have at least one student participating in 2 events. Therefore, minimum value of (a+b+c) = 1 (as per condition 2 in the question).
Hence, maximum value of x = 304 – 1 = 303.

The correct answer option is C.

Hope that helps!
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In its annual fest, a college, which has 510 students, is organizing t 30 Sep 2024, 07:31
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lacktutor
Bunuel
In its annual fest, a college, which has 510 students, is organizing three events. The students were allowed to participate in any number of events they liked. While viewing the statistics of the performance, the general secretary noticed:

1. The number of students who participated in at least two events were 52% more than those who participated in exactly one event.
2. The number of students participating in 1, 2 or 3 events respectively was at least equal to 1.
3. The number of students who did not participate in any of the three events was the minimum possible integral value under these conditions.

What can be the maximum number of students who participated in exactly 3 games?

A. 200
B. 300
C. 303
D. 304
E. 305

Are You Up For the Challenge: 700 Level Questions

Total = 510
The number of students who participated:
--> Only one event =a+b+c
--> Only two events = x+y+z
--> All three events = P

\((x+y+z) + p = 1.52 (a+b+c)\)
Neither =minimum possible integral value under these conditions

--> \((a+b+c)+ (x+y+z)+ P+ Neither = 510\)

\(2.52 (a+b+c)= 510 - Neither \)

In order \((a+b+c )\) to be integer, 'Neither' should be \(6\).
--> \((a+b+c)= \frac{510-6}{2.52 }= 200\)

\((x+y+z)+ P= 510-6-200= 304\)
In order P to be the maximum value,

--> \(x+y+z =1\) --> \(P =304-1= 303\)

Answer (C).­
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Hi Can you please explain how you got 2.52(a+b+c)?
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In its annual fest, a college, which has 510 students, is organizing t 31 Oct 2024, 02:35
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Hi lacktutor,

How did we figure out that Neither should be 6. I'm not able to figure out. Please help.

Regards,
Amulya
lacktutor
Bunuel
In its annual fest, a college, which has 510 students, is organizing three events. The students were allowed to participate in any number of events they liked. While viewing the statistics of the performance, the general secretary noticed:

1. The number of students who participated in at least two events were 52% more than those who participated in exactly one event.
2. The number of students participating in 1, 2 or 3 events respectively was at least equal to 1.
3. The number of students who did not participate in any of the three events was the minimum possible integral value under these conditions.

What can be the maximum number of students who participated in exactly 3 games?

A. 200
B. 300
C. 303
D. 304
E. 305

Are You Up For the Challenge: 700 Level Questions

Total = 510
The number of students who participated:
--> Only one event =a+b+c
--> Only two events = x+y+z
--> All three events = P

\((x+y+z) + p = 1.52 (a+b+c)\)
Neither =minimum possible integral value under these conditions

--> \((a+b+c)+ (x+y+z)+ P+ Neither = 510\)

\(2.52 (a+b+c)= 510 - Neither \)

In order \((a+b+c )\) to be integer, 'Neither' should be \(6\).
--> \((a+b+c)= \frac{510-6}{2.52 }= 200\)

\((x+y+z)+ P= 510-6-200= 304\)
In order P to be the maximum value,

--> \(x+y+z =1\) --> \(P =304-1= 303\)

Answer (C).­
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In its annual fest, a college, which has 510 students, is organizing t 26 Nov 2024, 20:45
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Bunuel How did you get Neither should be 6, and what's the approach to arrive at it in 2 mins that we have to answer a question?
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In its annual fest, a college, which has 510 students, is organizing t 07 Feb 2025, 11:23
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Let x be students participating in only 1 event, y in 2 events, z in all 3 events.

From 1, y+z= 1.52*x

We know that,
x+y+z+n= 510
Substitute 1 in above eqn,
1.52x+x+n= 510
2.52x + n= 510
Both x and n have to be integers, and from 3rd condition n has to be minimum possible integer value starting from 0.
If we apply both the above information, we should take the value of n in such a way that 510-n should be multiple of 2.52. Therefore n= 6
2.52x= 510-6=504
x=200
Substitute in eqn 1
x+y+z+n=510
200+y+z+6= 510
y+z=304

From condition 2, x, y and z all are >=1

To maximize z, y=1
z=304-1
z=303
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