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A certain high school with a total enrollment of 900 student 21 Oct 2022, 07:41
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A certain high school with a total enrollment of 900 students held a science fair for three days last week. How many of the students enrolled in the high school attended the science fair on all three days?

(1) Of the students enrolled in the school, 30 percent attended the science fair on two or more days.
(2) Of the students enrolled in the school, 10 percent of those that attended the science fair on at least one day attended on all three days.­

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A certain high school with a total enrollment of 900 student 21 Oct 2022, 07:53
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Bunuel
A certain high school with a total enrollment of 900 students held a science fair for three days last week. How many of the students enrolled in the high school attended the science fair on all three days?

(1) Of the students enrolled in the school, 30 percent attended the science fair on two or more days.
(2) Of the students enrolled in the school, 10 percent of those that attended the science fair on at least one day attended on all three days.
Show more

The trick here is that we don't know whether all of 900 students attended on at least one day. Meaning that there can be some students who didn't take part in the fair.

So we have the following groups:

    A. Attended on only one day;
    B. Attended on two days exactly;
    C. Attended on all three days;
    D. Did not attend at all.

And the sum of these groups is A+B+C+D=900. We want to determine the value of C.

(1) Of the students enrolled in the school, 30 percent attended the science fair on two or more days.

So, 900*0.3=270 attended 2 or more days:

    B + C = 270.

Not sufficient.

(2) Of the students enrolled in the school, 10 percent of those that attended the science fair on at least one day attended on all three days.

    10%*(A + B + C) = C;
    A + B = 9C.


Not sufficient.

(1)+(2) B+C=270, A+B=9C and A+B+C+D=900. Three linear equations four unknowns. Not sufficient.

Answer: E.
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A certain high school with a total enrollment of 900 student 21 Oct 2022, 09:28
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Statement 1: 270 students attended for 2 or more days.
270 can be for 2 days only or for 3 days. Not Sufficient.

Statement 2: Here we don't know how many students attended atleast one day or didn't attended on any day. Not sufficient.

Combine: Even after combining we still don't know how many attended on atleast one day or not attended single day.
Not sufficient.

Ans E

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A certain high school with a total enrollment of 900 student 01 Apr 2024, 04:31
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Bunuel
Bunuel
A certain high school with a total enrollment of 900 students held a science fair for three days last week. How many of the students enrolled in the high school attended the science fair on all three days?

(1) Of the students enrolled in the school, 30 percent attended the science fair on two or more days.
(2) Of the students enrolled in the school, 10 percent of those that attended the science fair on at least one day attended on all three days.

The trick here is that we don't know whether all of 900 students attended on at least one day. Meaning that there can be some students who didn't take part in the fair.

So we have the following groups:

    A. Attended on only one day;
    B. Attended on two days exactly;
    C. Attended on all three days;
    D. Did not attend at all.

And the sum of these groups is A+B+C+D=900. We want to determine the value of C.

(1) Of the students enrolled in the school, 30 percent attended the science fair on two or more days.

So, 900*0.3=270 attended 2 or more days:

    B + C = 270.

Not sufficient.

(2) Of the students enrolled in the school, 10 percent of those that attended the science fair on at least one day attended on all three days.

    10%*(A + B + C) = C;
    A + B = 9C.

Not sufficient.

(1)+(2) B+C=270, A+B=9C and A+B+C+D=900. Three linear equations four unknowns. Not sufficient.

Answer: E.
Show more
­
But does "at least on one day" not imply that all 3 cases are possible: on one day, two days, and/or three days? Meaning, there is even higher uncertainty to answer this question. Equally, in S2, how can we consider A+B+C for at least one day? If we use a Venn Diagram, it could be all 3 cases, correct? Or did you just not go too much into detail as even if we consider one case, we are left with no sufficient answer?
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A certain high school with a total enrollment of 900 student 10 Sep 2024, 11:04
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SergejK
Bunuel
Bunuel
A certain high school with a total enrollment of 900 students held a science fair for three days last week. How many of the students enrolled in the high school attended the science fair on all three days?

(1) Of the students enrolled in the school, 30 percent attended the science fair on two or more days.
(2) Of the students enrolled in the school, 10 percent of those that attended the science fair on at least one day attended on all three days.

The trick here is that we don't know whether all of 900 students attended on at least one day. Meaning that there can be some students who didn't take part in the fair.

So we have the following groups:

    A. Attended on only one day;
    B. Attended on two days exactly;
    C. Attended on all three days;
    D. Did not attend at all.

And the sum of these groups is A+B+C+D=900. We want to determine the value of C.

(1) Of the students enrolled in the school, 30 percent attended the science fair on two or more days.

So, 900*0.3=270 attended 2 or more days:

    B + C = 270.

Not sufficient.

(2) Of the students enrolled in the school, 10 percent of those that attended the science fair on at least one day attended on all three days.

    10%*(A + B + C) = C;
    A + B = 9C.

Not sufficient.

(1)+(2) B+C=270, A+B=9C and A+B+C+D=900. Three linear equations four unknowns. Not sufficient.

Answer: E.
­
But does "at least on one day" not imply that all 3 cases are possible: on one day, two days, and/or three days? Meaning, there is even higher uncertainty to answer this question. Equally, in S2, how can we consider A+B+C for at least one day? If we use a Venn Diagram, it could be all 3 cases, correct? Or did you just not go too much into detail as even if we consider one case, we are left with no sufficient answer?
Show more
­

Sorry, I now realize that my question is completely wrong on so many levels. S1 talks about 2 or more days, hence it is correct and the calculation is accounting for that. As A+B+C+D=900, and 3/10 of that total is at least at 2 days, we are not looking at D (neither) or A (exactly one day) as we are provided information regarding those that took part at the fair on at least 2 (>=2) days, but focus only on B and C, as those represent exaclty 2 and exactly 3 days. Hence: 900*0.3=270 leading to B + C = 270.

For S2 it is equally correct to use the formula A+B+C for at least one day. Like in S1, we are provided with the information who takes part at the fair, without focussing on who doesn't, hence D is irrelevant. So 10 percent of those that attended the science fair on at least one day would be 1/10(A+B+C) and equate that to all 3 days, which is C: 1/10(A+B+C)=C ->A+B=9C

S1 and S2 also can't work: We are not able to solve 3 equations with 4 variables: B + C = 270 and A+B=9C and and A+B+C+D=900

Sorry for the confusion.
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A certain high school with a total enrollment of 900 student 16 Sep 2025, 23:36
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