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SAT:Punch Of the Week (05-Aug-09)

Author: Sureshbala

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What is "Punch Of The Week"?

Every week Learnhub's team will post one question each on Verbal Ability and Quantitative Ability of the GMAT Test Preparation
Members can post their answers along with explanations for the next one week.
After one week, Learnhub's team will post the official answer along with the explanation.Besides answering the previous week's questions, Learnhub's Team will come up with the next set of questions for that week as well.

What are the rules of "Punch Of The Week"?

	Every user must refrain from posting sarcastic comments on the answers posted by others.
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	If you think that there is something wrong with the explanation of the Learnhub's Team, which seldom happens, please send a personal message to the respective team member informing him/her the same. Posting the same here will not be considered and will be deleted.
	Please bear in mind that only Learnhub's Team members have the privilege of posting questions here. Questions posted by other members will not be entertained and hence will be deleted immediately.

"Punch Of The Week" Team Members

This Week's Punches

Sentence Completion
That so shameless, sentimental, and dishonest a film attracts a large and enthusiastic audience is _ of how low the popular taste has.

A. an indication . . . soared
B. a gauge . . . refined
C. a signatory . . . fallen
D. a refutation . . . descended
E. a barometer . . . plummeted

Problem Solving
Let A = {2, 5, 10, 12, 17, 19, 22}. Find the total number of non-empty subsets of A which contain 10 but do not contain 19.

A. 25
B. 32
C. 36
D. 48
E. None of these

Click here for last week's Punch and explanatory answers for the same

Posted on 13-Aug-09

So,folks here are the correct answers

Sentence Correction
The correct answer is E.

The word low tells us the second word must be fallen, descended, or plummeted. The words indication, gauge, or barometer fit the first blank, so the right choice must be E.

Problem Solving
The correct answer is B.

The total number of elements in the set A is 7
We have to include 10 and exclude 19. So from the remaining 5 elements we may select at least one of them or even we can discard all the 5 elements.
Hence the total number of subsets is ^5C_0 + ^5C_1 + ^5C_2 + ^5C_3 + ^5C_4 + ^5C_5=2^5=32