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by Baton » Mon Jan 12, 2015 1:38 pm
I am solving these sums half and seem to be missing out the final step. Please help
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by MartyMurray » Mon Jan 12, 2015 2:36 pm
Last month 15 homes were sold in Town X. The average (arithmetic mean) sale price of the homes was $150,000 and the median sale price was $130,000. Which of the following statements must be true?

I. At least one of the homes was sold for more than $165,000.
II. At least one of the homes was sold for more than $130,0000 and less than $150,000
III. At least one of the homes was sold for less than $130,000.

A. I only
B. II only
C. III only
D. I and II
E. I and III

Ok so the mean is 150,000 and the median is 130,000.

The median is the middle number in the group, so it's in the middle of of the 15 prices, and you could call it the 8th highest.

If we were to have only the median to consider, then anything goes as long as the middle price is 130,000. All of them could cost 130,000. One could cost 130,000 and the rest could be evenly split above and below 130,000. Anything that keeps 130,000 in the middle of the fifteen prices works.

We do have another piece of information though; the mean is 150,000. Now we know something else. At least one of the homes has to have been sold for more than 130,000 in order to create that mean.

So let's go to the statements.

I. Does one have to cost more than 165,000? What this is really asking is if the median is 130,000 does another have to cost over 165,000 to make the mean 150,000?

In other words this is asking how low the highest price can be and still create a mean of 150,000.

Start by maximizing the prices at or below the median. By working with this maximum, we can find the minimum prices above the median that create that 150,000 mean. There will be eight prices at or below the median, and the maximum is 130,000 each. So we have a total of 8 * 130,000 = 1,040,000.

If the mean is 150000, and there are 15 houses, the total of all of their costs is 150,000 * 15 = 2,250,000.

So by taking away from the total of the fifteen the total of the prices at or below the median, 2,250,000 - 1,040,000 = 1,210,000, we get the total value of the other 7 houses. Divide by 7 to get 1,210,000/7 = approx. 173,000 So even if all of prices above the mean were the same and as high priced as possible, the minimum price of any of them would be about 173,000. So statement I has to be true.

II. We already figured out that 8 of them can go for 130,000 and the other 7 for about 173,000. So this does not have to be true.

III. This is maybe a little bit of a trap in case you thought some had to be below the median, but really it is not true either, as was discussed above.

So only I is true.

Choose A.

As for the other problem, how about posting it in a separate post so that we can keep it to one question per post.
Last edited by MartyMurray on Fri Jan 16, 2015 5:23 pm, edited 1 time in total.
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by Brent@GMATPrepNow » Mon Jan 12, 2015 2:51 pm
Last month, 15 homes were sold in Town X. The average sale price of the homes was $150,000 and the median sale price was $130,000. Which of the following statements must be true?

I. at least one of the homes was sold for more than $165,000
II. at least one of the homes was sold for more than $13,000 but less than $150,000
III. at least one of the homes was sold for less than $130,000

A) I only
B) II only
C) III only
D) I and II only
E) I and III only

OA A
The key word in this question is MUST. As in, which of the following MUST be true. So, if it's possible that one scenario is not true, we can eliminate it.

So, let's looks at one possible scenario and see which answer choices we can eliminate.

Aside: To make things simpler, let's divide all of the prices by 1000.

First, we'll use a nice rule that says: sum of all values = (mean)(number of values)
So, the sum of all 15 prices = ($150)(15) = $2250.

If the median is $130, then the middlemost value is $130

So, one possible scenario is:
130, 130, 130, 130, 130, 130, 130, 130, 130, 130, 130, 130, 130, 130, 130, 430

Aside: To find the last value (430), I took the sum of all 15 numbers (2250) and subtracted (14)(130)

Notice that this scenario tells us that II and III need not be true (since our scenario does not conform to either one).

Since answer choices B, C, D and E all include either II or III, we can eliminate them.

This leaves us with A, which must be the correct answer.

Cheers,
Brent
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by Brent@GMATPrepNow » Mon Jan 12, 2015 2:53 pm
If n and y are positive integers and 450y = n³, which of the following must be an integer?

I. y/(3 x 2² x 5)
II. y/(3² x 2 x 5)
III. y/(3 x 2 x 5²)

a. None
b. I only
c. II only
d. III only
e. I, II, and III
It often helps to find the prime factorization in these question types where we ask whether a certain rational expression is an integer.

450y = n^3
2*3*3*5*5*y = n^3
For 2*3*3*5*5*y to be a cube, we need the number of 2's, 3's and 5's in the prime factorization to each be divisible by 3.
So, for example, 2*2*2*2*2*2*3*3*3*5*5*5 = (2*2*3*5)^3

For 2*3*3*5*5*y to be a cube, it must be the case that the prime factorization of y includes at least two additional 2's, one additional 3 and one additional 5.
So, y = 2*2*3*5*(other possible numbers)

Now check the option.

I. Must y/(3 * 2^2 * 5) be an integer?
Plug in y to get: 2*2*3*5*(other possible numbers)/(3 * 2^2 * 5)
= some integer
Since this must be an integer, we can eliminate A, C and D, which leaves us with B or E.

II. Must y/(3^2 * 2 * 5) be an integer?
Plug in y to get: 2*2*3*5*(other possible numbers)/(3^2 * 2 * 5)
= 2*(other possible numbers)/3
Not necessarily an integer
Since this need not be an integer, we can eliminate E, which leaves us with B.

NOTE: At this point we have the correct answer. But let's check III for "fun"

III. Must y/(3 * 2 * 5^2) be an integer?
Plug in y to get: 2*2*3*5*(other possible numbers)/(3 * 2 * 5^2)
= 2*(other possible numbers)/5
Not necessarily an integer

Answer: B

Cheers,
Brent
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