Area, Members of Club, Candy Manufacturer

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qduong: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Wed Jun 01, 2011 11:02 pmFollowed by: 1 member

by qduong » Thu Jun 16, 2011 8:10 pm

Please help explain. Thank you.

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Source: Beat The GMAT — Problem Solving | Thu Jun 16, 2011 8:10 pm

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Thu Jun 16, 2011 8:52 pm

Please try to do the following when you're posting questions in this forum...

1. Do not post more than one question in the same thread.
2. Please take the burden of typing the question instead of posting a screenshot. The forum is not for your benefit only. Make the post searchable.

Thanks.

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Thu Jun 16, 2011 8:58 pm

PS 1:

A certain candy manufacturer reduced the weight of candy bar M by 20 percent but left the price unchanged. what was the resulting percent increase in the price per ounce of the candy bar M?

Say, initially the weight of each candy was W pounce and price of each candy was P.
Hence, initial price per ounce of each candy = P/W

After reducing the weight, weight of each candy bar = (W - 20% of W) = 0.8*W
After reducing the weight, price per ounce of each candy bar = P/0.8*W = 5P/4W

Hence, increase in price per ounce of each candy bar = (5P/4W - P/W) = P/4W
Hence, percentage increase in price per ounce of each candy bar = 100*(P/4W)/(P/W) = 100/4 = 25%

The correct answer is E.

Last edited by Anurag@Gurome on Thu Jun 16, 2011 9:19 pm, edited 1 time in total.

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Thu Jun 16, 2011 9:05 pm

PS 2:

Two members of a club are to be selected to represent the club at a national meeting. if there are 190 different possible selections of the 2 members, how many members does the club have?

Say, there are n members.
Then number of different possible selections of the 2 members out of n = nC2

So, nC2 = 190
--> n!/[(n - 2)!*2!] = 190
--> n(n - 1)/2 = 190
--> n(n - 1) = 380

Now we can solve this quadratic for n or we can look for a two-factor factorization of 380 so that the factors are consecutive integer.

As 380 can be expressed as 19*20, n = 20

The correct answer is A.

Last edited by Anurag@Gurome on Thu Jun 16, 2011 9:20 pm, edited 1 time in total.

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GMAT/MBA Expert

Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Thu Jun 16, 2011 9:13 pm

PS 3:

In the figures above, if the area of the triangle on the right is twice the area of the triangle on the left, then in terms of s, S =

As the triangles are similar, the ratio of their area will be equal to the ratio of the square of their corresponding sides.

So, sÂ²/SÂ² = 1/2
--> SÂ² = 2sÂ²
--> S = âˆš2s

The correct answer is C.

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qduong: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Wed Jun 01, 2011 11:02 pmFollowed by: 1 member

by qduong » Thu Jun 16, 2011 10:18 pm

Anurag@Gurome wrote:Please try to do the following when you're posting questions in this forum...
1. Do not post more than one question in the same thread.
2. Please take the burden of typing the question instead of posting a screenshot. The forum is not for your benefit only. Make the post searchable.
Thanks.

Thank you for the suggestion. I didn't know we could "Search" for the problems and now I do.
As for the problems that needed help, I'll type them out so others can search, unless the problems include graphs, figures, or complicated signs.

Thanks

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