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200+400/2 = 300
100+50/2 = 75

300 -75 = 225
Answer - D
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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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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
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damstamnd
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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Here is my solution to this one =>
Mean(1)=200+400/2 = 300
Mean(2)=50+100/2 = 75

Hence Mean (1) is 225 Greater than Mean 2

Hence D
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Bunuel
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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Avg 200 to 400 inclusive = 300
Avg 50 to 100 inclusive = 75
225 more
D
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Bunuel
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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Mean of 1 set of numbers = (200+400)/2=300
Mean of 2nd set of numbers = (50+100)/2=75
300-75=225

Answer : D
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200+400/2 = 300
100+50/2 = 75

300 -75 = 225
Answer = D
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Hi can someone tell
why only first and the last integer are added and divided by 2 rather than finding the sum of integers from 200 to 400 and dividing them by 201 (as both are inclusive)?
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Middle number from 200 to 400 is 300 and middle number from 50 to 100 is 75
Therefore 300-75 =225 is the answer

Posted from my mobile device
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Thanks Gurmukh ,

However my query was
why don’t we use the formula of mean.
i.e -Sum of all observations /total number of obs. instead we take average of first and last term.

Also I’ve found the answer to it.
We can use average of first and last term only in equally spaced set or in a set where all observations are same else we have to use the actual formula.

Posted from my mobile device
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Hey, I found this shortcut - am I correct in thinking about the problem this way?

The average for 200 - 400, inc = (L+F)/2 = (400+200)/2 = 300.

Notice that 200 - 400 is 4x greater than 50 - 100.

Therefore, the average for 50 - 100, inc = 75. (Shortcut)

300 - 75 = 225. D.
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­Evenly spaced set makes it easy:

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Bunuel
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





Nick Slavkovich, GMAT/GRE tutor with 20+ years of experience

[email protected]
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