hard question!!!

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hard question!!!

by bupbebeo » Sun Apr 04, 2010 8:55 am
a club collected EXACTLY $599 from its members. IF each member contributed at least $12, what is the greatest number of members the club could have?

a. 43
b. 44
c. 49
d. 50
e. 51
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by JGaynor » Sun Apr 04, 2010 9:06 am
My guess would be 49.

599/12= 49 with a remainder..

The stem states that each member gave at least $12... to refute that the answer choice D of 50, if each member gave the minimum of $12, that would give $600.

But 49 people could give $12 and some cents to derive at $599.

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by thephoenix » Sun Apr 04, 2010 9:35 am
price * number=total
in order to maximize no. we need to minimize price becoz no.=tot/price

since minmum value price can have=$12

and 12*50=600>599 not possible
12*49=588 possible and is the greatest value

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by srinivasarajui » Sun Apr 04, 2010 11:30 pm
I also agree with thephoenix and JGaynor

So the final ans is C.49
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by bupbebeo » Mon Apr 05, 2010 2:24 am
srinivasarajui wrote:I also agree with thephoenix and JGaynor

So the final ans is C.49
If we have 49 members. So, the total amount of money we have is just 588. The problem says that the club receives EXACTLY 599. So, I don't think 49 is a right answer

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by ajith » Mon Apr 05, 2010 3:14 am
bupbebeo wrote:
srinivasarajui wrote:I also agree with thephoenix and JGaynor

So the final ans is C.49
If we have 49 members. So, the total amount of money we have is just 588. The problem says that the club receives EXACTLY 599. So, I don't think 49 is a right answer
What about 48 members contributing 12 and one contributing 23?

49 is indeed the right answer
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by thephoenix » Mon Apr 05, 2010 4:03 am
bupbebeo wrote:
srinivasarajui wrote:I also agree with thephoenix and JGaynor

So the final ans is C.49
If we have 49 members. So, the total amount of money we have is just 588. The problem says that the club receives EXACTLY 599. So, I don't think 49 is a right answer
yeah but the q says they contribute at least $12 ......which means >12

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by szy » Tue May 18, 2010 8:39 am
the question states that each member pays NO LESS THAN $12 and the club collects EXACTLY $599.

The answer 49 still doesn't make much sense to me. If there were 49 members in the club and each paid at least $12 then....
49x12 = 588 which gives us a remainder of $11 left which is less than $12


confused..

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by sk818020 » Tue May 18, 2010 12:27 pm
szy wrote:the question states that each member pays NO LESS THAN $12 and the club collects EXACTLY $599.

The answer 49 still doesn't make much sense to me. If there were 49 members in the club and each paid at least $12 then....
49x12 = 588 which gives us a remainder of $11 left which is less than $12


confused..
Think about it this way;

599/12=49+11/12

So, 48 people will pay $12 and the 49th person will have to pay 12 plus the remainder.

48*$12= $576

$599-$576=$23

Because the least anyone can pay is $12, the 49th person must pay the remaining $23.

So 49 people is the greatest number of people that pay if everyone must pay a minimum of $12.

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by frank1 » Tue May 18, 2010 6:47 pm
I think it is not as diffcult as it seems or sounds.....(atleast one doesnt get stumped...)

here i used AT WORST CASE,each member will contribute 12 each (because they can't go beyond that)

now 12X5=60

so cannot be 50 as there is one 1 dollar less
so must be 49
(extra 11 dollars were contributed by some persons among 49 users)

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by rockeyb » Tue May 18, 2010 10:48 pm
In order to have max number of members we need to minimize the amount each member pays .

Since each member have to pay at least $12 the max .

So in order to find the max number of people who contribute only $12 to the total of $599 , divide $599 by 12 .

The result will not be an integer that means there must be at least 49 people who contribute $12 to the total .

How much do these 49 people contribute 12 x 49 = $588 . This amount is 21 short of $599 so there must be a member who contributed more than 12 .

But if we are to maximize the number of people we need to minimize their individual contribution . But we can not fit 2 people who would contribute $21 as each must contribute at least $12 .

So only one person contributed $21 and 49 people contributed $12 .

Total members in club 50 .
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by sk818020 » Tue May 18, 2010 11:06 pm
rockeyb wrote: 12 x 49 = $588 . This amount is 21 short of $599
588 is 11 short of 599.

599-588=11.

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by rockeyb » Tue May 18, 2010 11:17 pm
Sorry made a mistake .


Yup agree with 49 .

Here is how I got to it . .

11 people contribute $13 = $143

and 38 people contribute $12 each = $456

Total amount = $599 .

Number of people 38+ 11 = 49 .
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