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Each of the 20 students has 9, or 10, or 11 books. What is the average

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Each of the 20 students has 9, or 10, or 11 books. What is the average  [#permalink]

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New post 13 Apr 2019, 03:07
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43% (01:18) correct 57% (01:46) wrong based on 28 sessions

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Each of the 20 students has 9, or 10, or 11 books. What is the average number of the books that the students have?

(1) The number of the students who have 9 books is equal to the number of the students who have 11 books.

(2) The number of the students who have 9 books is greater than the number of the students who have 10 books.

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Re: Each of the 20 students has 9, or 10, or 11 books. What is the average  [#permalink]

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New post 13 Apr 2019, 05:24
Can someone explain how is it A? I think its E

Let students with 9 books =x, 10 books=y and 11 books=z

We can x=7,y=6,z=7 or x=8,y=4,z=8. Average will be different

The above example will give different average for C as well...
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Each of the 20 students has 9, or 10, or 11 books. What is the average  [#permalink]

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New post 13 Apr 2019, 05:47
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Sanjeetgujrall wrote:
Can someone explain how is it A? I think its E

Let students with 9 books =x, 10 books=y and 11 books=z

We can x=7,y=6,z=7 or x=8,y=4,z=8. Average will be different

The above example will give different average for C as well...


Statement 1 tells you that the number of people who hold, respectively, 9 and 11 books is the same. So, \(x + z\), as a whole, have 10 books on average. Since you know that \(y\) students have 10 books, \(x + y + z\) students have 10 books each one, on average.
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Each of the 20 students has 9, or 10, or 11 books. What is the average  [#permalink]

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New post 14 Apr 2019, 03:12
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Sanjeetgujrall wrote:
Can someone explain how is it A? I think its E

Let students with 9 books =x, 10 books=y and 11 books=z

We can x=7,y=6,z=7 or x=8,y=4,z=8. Average will be different

The above example will give different average for C as well...


Having same number of students with 9 and 11 books so
average will be \(\frac {9 n + 10 k + 11 n} {(2n+k)}\)

we can represent this as

\(\frac{n ( 9 +11 ) + 10k}{2n + k }\) = \(\frac {n (20) + 10k}{ 2n+k}\) = \(\frac {2n(10) + 10k }{2n+k}\) = \(\frac {(2n + k)10 }{2n+k}\) = 10

hope this is helpful.
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Each of the 20 students has 9, or 10, or 11 books. What is the average   [#permalink] 14 Apr 2019, 03:12
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