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Updated on: 02 Nov 2022, 10:30
Originally posted by udaymathapati on 23 Sep 2010, 12:15.
Last edited by Bunuel on 02 Nov 2022, 10:30, edited 1 time in total.
Last edited by Bunuel on 02 Nov 2022, 10:30, edited 1 time in total.
Edited the second statement: "less" changed to "more".
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C
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Each person attending a fund-raising party for a certain club was charged the same admission fee. How many people attended the party?
(1) If the admission fee had been 0.75$ less and 100 more people had attended, the club would have received the same amount in admission fee
(2) If the admission fee had been 1.50$ more and 100 fewer people had attended, the club would have received the same amount in admission fee
Image7.JPG [ 13.33 KiB | Viewed 40738 times ]
(1) If the admission fee had been 0.75$ less and 100 more people had attended, the club would have received the same amount in admission fee
(2) If the admission fee had been 1.50$ more and 100 fewer people had attended, the club would have received the same amount in admission fee
Attachment:
Image7.JPG [ 13.33 KiB | Viewed 40738 times ]
23 Sep 2010, 13:02
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Each person attending a fund raising party for a certain club was charged the same admission fee. How many people attended the party?
Let the original admission fee be \(x\) and the original # of people be \(n\). Question: \(n=?\)
(1) If the admission fee had been 0.75$ less and 100 more people had attended, the club would have received the same amount in the admission fee
\(xn=(x-0.75)(n+100)\) --> \(100x-0.75n=75\). Two unknowns, one equation, not sufficient to calculate \(n\).
(2 ) If the admission fee had been 1.50$ less and 100 fewer people had attended, the club would have received the same amount in the admission fee
\(xn=(x-1.5)(n-100)\) --> \(100x-1.5n=-150\). Two unknowns, one equation, not sufficient to calculate \(n\).
(1)+(2) \(100x-0.75n=75\) and \(100x-1.5n=-150\) --> we have two distinct linear equation with two unknowns, hence we can solve for \(n\). Sufficient.
Answer: C.
Let the original admission fee be \(x\) and the original # of people be \(n\). Question: \(n=?\)
(1) If the admission fee had been 0.75$ less and 100 more people had attended, the club would have received the same amount in the admission fee
\(xn=(x-0.75)(n+100)\) --> \(100x-0.75n=75\). Two unknowns, one equation, not sufficient to calculate \(n\).
(2 ) If the admission fee had been 1.50$ less and 100 fewer people had attended, the club would have received the same amount in the admission fee
\(xn=(x-1.5)(n-100)\) --> \(100x-1.5n=-150\). Two unknowns, one equation, not sufficient to calculate \(n\).
(1)+(2) \(100x-0.75n=75\) and \(100x-1.5n=-150\) --> we have two distinct linear equation with two unknowns, hence we can solve for \(n\). Sufficient.
Answer: C.
24 Sep 2010, 15:17
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TomB, how can you assign a random value (100) to xy and claim that x can be solved??
Let a = admission fee and n = number of people attended
(1) If the admission fee had been $0.75 less and 100 more people attended, the club would have received the same amount in admission fees.
an = (a - 0.75)(n + 100)
an = an - 0.75n + 100a -75
0.75n - 100a = -75
Insufficient
(2) If the admission fee had been $1.50 more and 100 fewer people had atteneded, the club would have received the same amount in admission fees.
an = (a + 1.50)(n - 100)
an = an + 1.5n - 100a - 150
1.5n - 100a = 150
Insufficent
Subtrace (1) from (2)
1.5n - 100a = 150
0.75n - 100a = -75
0.75n = 225 >> n = 300 & a can be solved
Let a = admission fee and n = number of people attended
(1) If the admission fee had been $0.75 less and 100 more people attended, the club would have received the same amount in admission fees.
an = (a - 0.75)(n + 100)
an = an - 0.75n + 100a -75
0.75n - 100a = -75
Insufficient
(2) If the admission fee had been $1.50 more and 100 fewer people had atteneded, the club would have received the same amount in admission fees.
an = (a + 1.50)(n - 100)
an = an + 1.5n - 100a - 150
1.5n - 100a = 150
Insufficent
Subtrace (1) from (2)
1.5n - 100a = 150
0.75n - 100a = -75
0.75n = 225 >> n = 300 & a can be solved
