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# The average (arithmetic mean) of the integers from 200 to

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Math Expert
Joined: 02 Sep 2009
Posts: 51188
The average (arithmetic mean) of the integers from 200 to  [#permalink]

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14 Jun 2012, 01:27
4
23
00:00

Difficulty:

5% (low)

Question Stats:

86% (01:07) correct 14% (01:29) wrong based on 1832 sessions

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The average (arithmetic mean) of the integers from 200 to 400, inclusive, is how much greater than the average of the integers from 50 to 100, inclusive?

(A) 150
(B) 175
(C) 200
(D) 225
(E) 300

Diagnostic Test
Question: 2
Page: 20
Difficulty: 550

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Joined: 02 Sep 2009
Posts: 51188
Re: The average (arithmetic mean) of the integers from 200 to  [#permalink]

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14 Jun 2012, 01:28
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SOLUTION

The average (arithmetic mean) of the integers from 200 to 400, inclusive, is how much greater than the average of the integers from 50 to 100, inclusive?

(A) 150
(B) 175
(C) 200
(D) 225
(E) 300

The average of the integers from 200 to 400, inclusive is $$\frac{first+last}{2}=\frac{200+400}{2}=300$$;
The average of the integers from 50 to 100, inclusive is $$\frac{first+last}{2}=\frac{50+100}{2}=75$$;

The difference is $$300-75=225$$.

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Re: The average (arithmetic mean) of the integers from 200 to  [#permalink]

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14 Jun 2012, 04:49
3
1
Hi,

This is an easy one (Difficulty level: 600, depending on your approach)

My approach is:
For an AP the mean or average of series is average of first and last term.
So, average of numbers between 200 to 400, inclusive = (200+400)/2 = 300
average of numbers between 50 to 100, inclusive = (50+100)/2 = 75
difference = 300 - 75
= 225

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Joined: 02 Jun 2011
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Re: The average (arithmetic mean) of the integers from 200 to  [#permalink]

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14 Jun 2012, 06:04
200+400/2 = 300
100+50/2 = 75

300 -75 = 225
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Re: The average (arithmetic mean) of the integers from 200 to  [#permalink]

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16 Jun 2012, 02:37
Quite easy if you know that F+L/2 is average. To me its below 600.

Solution: First+ Last/2= Average

1. 200+400/2= 300
2. 50+150/2=75

D: 225
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Re: The average (arithmetic mean) of the integers from 200 to  [#permalink]

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Updated on: 17 Jun 2012, 11:11

The mean of the set is equal to the average of the FIRST and LAST.

(400+200) / 2 = 300

and

(100+50) / 2 = 75

and

300-75 = 225

Originally posted by damstamnd on 17 Jun 2012, 09:27.
Last edited by damstamnd on 17 Jun 2012, 11:11, edited 1 time in total.
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Re: The average (arithmetic mean) of the integers from 200 to  [#permalink]

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17 Jun 2012, 09:49
damstamnd wrote:

The mean of the set is equal to the average of the FIRST and LAST.

(400-200) / 2 = 300

and

(100-50) / 2 = 75

and

300-75 = 225

Correction: (400+200) / 2 = 300
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Re: The average (arithmetic mean) of the integers from 200 to  [#permalink]

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17 Dec 2016, 15:50
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Re: The average (arithmetic mean) of the integers from 200 to  [#permalink]

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21 Dec 2016, 07:14
Bunuel wrote:
The average (arithmetic mean) of the integers from 200 to 400, inclusive, is how much greater than the average of the integers from 50 to 100, inclusive?

(A) 150
(B) 175
(C) 200
(D) 225
(E) 300

Diagnostic Test
Question: 2
Page: 20
Difficulty: 550

X = Mean1 - Mean2
=>X=[(400+200)-(100+50)]/2 [Since, Mean=average of First & Last numbers in a set]
=>X=450/2
=>X=225 (D)
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Re: The average (arithmetic mean) of the integers from 200 to  [#permalink]

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13 Jan 2017, 13:21
Avg 200 to 400 inclusive = 300
Avg 50 to 100 inclusive = 75
225 more
D
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Re: The average (arithmetic mean) of the integers from 200 to  [#permalink]

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13 Feb 2018, 17:27
Bunuel wrote:
The average (arithmetic mean) of the integers from 200 to 400, inclusive, is how much greater than the average of the integers from 50 to 100, inclusive?

(A) 150
(B) 175
(C) 200
(D) 225
(E) 300

Since we have an evenly spaced set of integers, the average of the integers from 200 to 400, inclusive, is (200 + 400)/2 = 300.

Similarly, the average of the integers from 50 to 100, inclusive, is (50 + 100)/2 = 75.

Thus, the difference is 300 - 75 = 225.

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Re: The average (arithmetic mean) of the integers from 200 to  [#permalink]

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29 Jun 2018, 02:09
Mean of 1 set of numbers = (200+400)/2=300
Mean of 2nd set of numbers = (50+100)/2=75
300-75=225

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Re: The average (arithmetic mean) of the integers from 200 to &nbs [#permalink] 29 Jun 2018, 02:09
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