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# 20 students took the test. The arithmetic mean of the scores

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Senior Manager
Joined: 10 Apr 2012
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20 students took the test. The arithmetic mean of the scores [#permalink]

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04 Aug 2013, 03:00
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Question Stats:

68% (01:46) correct 32% (02:13) wrong based on 280 sessions

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20 students took the test. The arithmetic mean of the scores was 4.75 and the range was 2. If the only possible scores were {2,3,4,5}, what is the greatest possible number of excellent scores (5)?
(A)19
(B)17
(C)14
(D)13
(E)12
[Reveal] Spoiler: OA
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Re: 20 students took the test. The arithmetic mean of the scores [#permalink]

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04 Aug 2013, 03:13
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guerrero25 wrote:
20 students took the test. The arithmetic mean of the scores was 4.75 and the range was 2. If the only possible scores were {2,3,4,5}, what is the greatest possible number of excellent scores (5)?
(A)19
(B)17
(C)14
(D)13
(E)12

As the range is 2, and we want to maximize the no of 5's, there can not be any 2 in the given test.Also, the total score = 20*4.75 = 95.

Now, let the no of 3's,4's and 5's be x,y,z repectively.

Thus, 3x+4y+5z = 95.Now, for x=2 and y=1, we get 5z = 95-10=85 and z =17.

B.
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Re: 20 students took the test. The arithmetic mean of the scores [#permalink]

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04 Aug 2013, 03:13
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20 students took the test. The arithmetic mean of the scores was 4.75 and the range was 2. If the only possible scores were {2,3,4,5}, what is the greatest possible number of excellent scores (5)?

The sum of the scores is $$4.75*20=95$$. If the range is 2 and we want to find the max number of 5s, the lowest score possible is 3.
To maximize the 5s we have to reduce the 3s and the 4s.

5*x+4*z+3*y=95 and x+y+z=20.

So y=2 and z=1, 5*x+4+6=85 and x=17.
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Re: 20 students took the test. The arithmetic mean of the scores [#permalink]

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04 Aug 2013, 03:24
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guerrero25 wrote:
20 students took the test. The arithmetic mean of the scores was 4.75 and the range was 2. If the only possible scores were {2,3,4,5}, what is the greatest possible number of excellent scores (5)?
(A)19
(B)17
(C)14
(D)13
(E)12

Total Score = 20 x 4.75 = 95
Range is 2
Possible scores {2,3,4,5}

Rather Than solving this question using algebra, i would prefer testing options directly. My purpose would be to maximize the no of scores that value 5
Option A - 19 x 5= 95
Possible but this option fails to satisfy range criteria. As per this option range would be 0, but the range is 2.
Thus Incorrect

Option B - 17 x 5= 85
We are left with 3 scores that must sum to 10
Because the range is 2 we can not scores that value 2. Thus we can use only 3 & 4
remaining 3 scores - 3,3,4
Possible
Thus correct

Thus Incorrect
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Re: 20 students took the test. The arithmetic mean of the scores [#permalink]

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24 Nov 2014, 00:11
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x1-x20 = 2 also we have to make the number of 5s max. so we have to reject the 2 and 4 pair. Hence max of the 20 numbers is 5 and the minimum is 3. Now x1+x2+.....x20 = 4.75*20 =95,
or x2+x3+......+x19 = 95-(3+5) = 87
to find the maximum of 5s divide 87 by 5 and the whole number part will be the answer i.e 17
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Re: 20 students took the test. The arithmetic mean of the scores [#permalink]

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07 Aug 2017, 21:21
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KUDOS
OUTSTANDING question.

Took me a lot of time to figure out but here is how I will be solving it the next time -->

Average of 20 => 4.75

Sum(20)/20 = 4.75
Hence the sum of all the scores => 20*4.75 = 95

Now range is 2.
We have to maximise the number of 5's.

USE HIT AND TRIAIL.

Number of 5's = 20 => 5*20 = 100=> Overflow => REJECT.

Number of 5's = 19 => Sum = 19*5=95.
But wait that would mean that all the terms are 5 and the range would be zero => Reject.

Number of 5's = 18 => Sum = 18*5 = 90. The leftout value = 5. There Is no way for us to write that in terms of 3,4.

Number of 5's = 17 => Sum => 17*5 = 85. The leftout value = 10 => 3+3+4 =>Bingo.

Hence B.

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Re: 20 students took the test. The arithmetic mean of the scores [#permalink]

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07 Aug 2017, 22:10
Since the only possible scores are 2,3,4,5 and the range is 2 wherein the max score should be 5 thus the lowest score should be 3
Now as we have the avg =4.75 and the no.of terms =20 we can find the sum of all the 20 scores = avg * no.of students=4.75*20=95
Now to find the max no.of students who scored 5 we will consider that only 1 student scored 3 and only 1 student scored 4
Thus 95-(3+4)=88 which is not divisible by 5 thus option A is incorrect.
Likewise when we increase the no.of students who scored 3 or 4 we get that option B is correct.
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Re: 20 students took the test. The arithmetic mean of the scores [#permalink]

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12 Aug 2017, 06:58
guerrero25 wrote:
20 students took the test. The arithmetic mean of the scores was 4.75 and the range was 2. If the only possible scores were {2,3,4,5}, what is the greatest possible number of excellent scores (5)?
(A)19
(B)17
(C)14
(D)13
(E)12

4.75*20=
17*5 +3*2+4
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Re: 20 students took the test. The arithmetic mean of the scores [#permalink]

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02 Jan 2018, 22:24
guerrero25 wrote:
20 students took the test. The arithmetic mean of the scores was 4.75 and the range was 2. If the only possible scores were {2,3,4,5}, what is the greatest possible number of excellent scores (5)?
(A)19
(B)17
(C)14
(D)13
(E)12

A mere algebra approach:
As the range was 2, we are left with 3, 4 & 5 as possible scores.
Let x,y,z be the no. of students who got 3, 4 & 5 respectively.
=> x + y + z = 20
=> x = 20 - y -z (1)
Also we have: 3x + 4y + 5z = 4.75x20 = 95 (2)
(1) & (2) => 3(20 - y -z) + 4y + 5z = 95
=> 60 + y + 2z = 95
=> y + 2z = 35
Z would be maximized when Y is lowest.
Minimum of Y would be 1, so Maximum of Z equals to (35-1)/2 = 17.
Re: 20 students took the test. The arithmetic mean of the scores   [#permalink] 02 Jan 2018, 22:24
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