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The average (arithmetic mean) of a set of 6 numbers is 28. If a certai
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25 May 2023, 22:45
25 May 2023, 22:45
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The average (arithmetic mean) of a set of 6 numbers is 28. If a certain number, y, is removed from the set, the average of the remaining numbers in the set is 24.
A Quantity A is greater.
B Quantity B is greater.
C The two quantities are equal.
D The relationship cannot be determined from the information given.
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Quantity A |
Quantity B |
y |
48 |
A Quantity A is greater.
B Quantity B is greater.
C The two quantities are equal.
D The relationship cannot be determined from the information given.
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
\(\Longrightarrow\) Cracking the GRE Premium
\(\Longrightarrow\) GRE Math Essentials - A most comprehensive handout!!
Re: The average (arithmetic mean) of a set of 6 numbers is 28. If a certai
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28 May 2023, 10:29
28 May 2023, 10:29
Expert Reply
OE
The problem says that the average of a set of six numbers is 28, . If a certain number, y, is removed from the set, there are now five numbers left. We already know that the new average is 24. The difference between the totals must be equal to y: 168 − 120 = 48.
Thus, the two quantities are equal, and the answer is (C).
The problem says that the average of a set of six numbers is 28, . If a certain number, y, is removed from the set, there are now five numbers left. We already know that the new average is 24. The difference between the totals must be equal to y: 168 − 120 = 48.
Thus, the two quantities are equal, and the answer is (C).
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Joined: 11 Sep 2023
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GRE 1: Q155 V141
Re: The average (arithmetic mean) of a set of 6 numbers is 28. If a certai
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02 Oct 2023, 05:53
02 Oct 2023, 05:53
(a+b+c+d+e+y)/6=28
hence, a+b+c+d+e+y=168
Now, (a+b+c+d+e)/5=24
hence, a+b+c+d+e=120
a+b+c+d+e+y - (a+b+c+d+e)=168-120=48
Hence C
hence, a+b+c+d+e+y=168
Now, (a+b+c+d+e)/5=24
hence, a+b+c+d+e=120
a+b+c+d+e+y - (a+b+c+d+e)=168-120=48
Hence C
