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A student worked 20 days. For each of the amount shown in

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A student worked 20 days. For each of the amount shown in [#permalink]

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23 Oct 2005, 11:24
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A student worked 20 days. For each of the amount shown in the first row of the table, second row gives the number of days the student earned that amount. Median amount of money earned per day for 20 days is?

A. 96
B. 84
C. 80
D. 70
E. 48
[Reveal] Spoiler: OA

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Last edited by Bunuel on 27 Feb 2013, 03:37, edited 2 times in total.
Edited the question and added the OA

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Re: A student worked 20 days. For each of the amount shown (see [#permalink]

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05 Mar 2012, 08:58
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Arrange the amount in ascending order

48 - 2
70 - 3
80 - 4
84 - 7
96 - 4

Median = (Amount in 10th day+Amount in 11th Day)/2

= (84+84)/2
=84

B
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Re: A student worked 20 days. For each of the amount shown (see [#permalink]

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05 Mar 2012, 09:16
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ENAFEX wrote:
Arrange the amount in ascending order

48 - 2
70 - 3
80 - 4
84 - 7
96 - 4

Median = (Amount in 10th day+Amount in 11th Day)/2

= (84+84)/2
=84

B

A student worked 20 days. For each of the amount shown in the first row of the table, second row gives the number of days the student earned that amount. Median amount of money earned per day for 20 days is?
A. 96
B. 84
C. 80
D. 70
E. 48

A student worked 20 days.

The median of a set with even number of terms is the average of the two middle terms when arranged in ascending or descending order. Since the data (amount earned per day) is already arranged in descending order there is no need to rearrange it.

The median will be the average of the 10th and 11th values (either in ascending or descending order) --> both of those values are 84 --> the median is (84+84)/2.

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Manager
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Re: A student worked 20 days. For each of the amount shown in [#permalink]

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05 Mar 2012, 10:22
Expert.. Thanks Bunuel. I didn't notice that at all.. Got used to arranging a group in ascending order. :D
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Re: A student worked 20 days. For each of the amount shown in [#permalink]

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03 Feb 2014, 10:43
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For people who want a make this one a "Quickie"

Look at the values given:

48 - 2
70 - 3
80 - 4
84 - 7
96 - 4

Both the highest values (84 and 96) have largest no. of days associated with them(7 and 4). Hence, your immediate reaction should be that the median is going to heavily lean towards the higher values. Now look at the options:

1)$96 2)$84
3)$80 4)$70
5)\$48

It is obvious that it cannot be either of 96,48 (or) if you think about it even 70. The only decision now is between 84 and 80. Now out of these two 84 leans more towards the highest values and also has 7 days assigned to it. So........

P.S: Only meant for scenarios wherein you desperately want a " Quickie "

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A student worked 20 days. For each of the amount shown in [#permalink]

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23 Jul 2016, 07:14
vikramm wrote:
Attachment:
table.gif
A student worked 20 days. For each of the amount shown in the first row of the table, second row gives the number of days the student earned that amount. Median amount of money earned per day for 20 days is?

A. 96
B. 84
C. 80
D. 70
E. 48

20 days ... 20 is an even number so the median will be $$\frac{10^{th}+11^{th} term}{2}$$
Now the amount of money that corresponds to 10 and 11th day is 84. So it is the median
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Re: A student worked 20 days. For each of the amount shown in [#permalink]

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21 Oct 2017, 10:25
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Re: A student worked 20 days. For each of the amount shown in   [#permalink] 21 Oct 2017, 10:25
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