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Frank is organizing a charity event in an effort to r

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Frank is organizing a charity event in an effort to r [#permalink] New post   19 Aug 2017, 08:18
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Frank is organizing a charity event in an effort to raise money to build a neigh­borhood park. The park will cost $1,200 to build. Each person attending the event will donate $300, less $20 for each person who attends the event. Thus, for example, if 3 people attend the event, each person will donate $300 - $20(3) = $240. What is the minimum number of people who must attend the event in order to raise enough money to build the park?

A. 4

B. 5

C. 6

D. 8

E. Frank cannot raise enough money at this charity event to build the park.
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IlCreatore
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Re: Frank is organizing a charity event in an effort to r [#permalink] New post   19 Sep 2017, 06:03
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Is it right to set up the equation as (300-20(x))x=1200, where x is the number of participants? Then, it is a second degree equation 20x^2-300x+1200=0 that has no solution. Thus, the answer is E!
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Re: Frank is organizing a charity event in an effort to r [#permalink] New post   21 Feb 2018, 12:12
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draw two graphs. one for 20x^2 and one for 300x.
They will met at 15. After 15, 20x^2 will always larger than 300x.
The largest gap lies between 7 and 8. since there can only be integer number of guest, put 7 and 8 into the equation. we get 1120, which is below 1200.
Answer is E.
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Re: Frank is organizing a charity event in an effort to r [#permalink] New post   22 Feb 2018, 05:23
Tricky question
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Re: Frank is organizing a charity event in an effort to r [#permalink] New post   14 Dec 2018, 13:20
hugebigmac wrote:
draw two graphs. one for 20x^2 and one for 300x.
They will met at 15. After 15, 20x^2 will always larger than 300x.
The largest gap lies between 7 and 8. since there can only be integer number of guest, put 7 and 8 into the equation. we get 1120, which is below 1200.
Answer is E.


Why 20x^2 and not 20x? I am not following.

Can anyone give an alternative solution as well?
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Re: Frank is organizing a charity event in an effort to r [#permalink] New post   16 Dec 2018, 01:50
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Forget about fancy yet powerful ways of solution.

Your goal is when you are in front of a certain question is to approach it in the fastest and efficient way. Simple and straight. the clock is running. down to the business.

In this case, pick C as a first possible answer. From this is C is too high as a number you do know that the answer must be D or E. Or the other way around.


330-20(6)=180*6=1080 not enough

300-20(8)=140*8= 1120 not enough

E is the answer.

Of course, you do not need to perform all such calculations. It is pretty easy to see they are not enough
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Re: Frank is organizing a charity event in an effort to r [#permalink] New post   28 Feb 2021, 11:42
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QuantumWonder wrote:
hugebigmac wrote:
draw two graphs. one for 20x^2 and one for 300x.
They will met at 15. After 15, 20x^2 will always larger than 300x.
The largest gap lies between 7 and 8. since there can only be integer number of guest, put 7 and 8 into the equation. we get 1120, which is below 1200.
Answer is E.


Why 20x^2 and not 20x? I am not following.

Can anyone give an alternative solution as well?


Approach 1:

Each person will donate = (300−20x)
Total Donation = (300−20x) x x = $1200


i.e. 20x2−300x+1200=0
Upon solving, you will get Imaginary roots, Hence you cannot raise the desired amount of $1200
So, option E

Approach 2:
You can simply plug in the option choices, and you will notice even when x=8, the raised amount is $1120
So we can go with option E
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Re: Frank is organizing a charity event in an effort to r [#permalink] New post   23 Aug 2022, 23:11
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Re: Frank is organizing a charity event in an effort to r [#permalink]
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