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

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

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New post 14 Jun 2012, 02:27
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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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Re: The average (arithmetic mean) of the integers from 200 to [#permalink]

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New post 14 Jun 2012, 02:28
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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 \(300-75=225\).

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

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New post 14 Jun 2012, 05:49
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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

Answer is (D).

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

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New post 14 Jun 2012, 07:04
200+400/2 = 300
100+50/2 = 75

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

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New post 16 Jun 2012, 03:37
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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New post Updated on: 17 Jun 2012, 12:11
Answer: D

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, 10:27.
Last edited by damstamnd on 17 Jun 2012, 12:11, 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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New post 17 Jun 2012, 10:49
damstamnd wrote:
Answer: D

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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New post 17 Dec 2016, 16:50
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Re: The average (arithmetic mean) of the integers from 200 to [#permalink]

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New post 21 Dec 2016, 08:14
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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New post 13 Jan 2017, 14:21
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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New post 13 Feb 2018, 18:27
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.

Answer: D
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Re: The average (arithmetic mean) of the integers from 200 to   [#permalink] 13 Feb 2018, 18:27
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