Problem Solving

Last month 15 homes were sold in town X. the average price was $150,000 and median was $130,000. Which of the following must be true:
1) atleast one home sold for more than $165,000
2) at least one home was sold for more than $130,000 and less than $150,000
3) at least one home was sold for less than $130,000

Can someone pl direct me towards more questions like this. I can solve regular mean, median questions just get a bit confused in these ones
Thanks very much indeed for all the help
OA is 1 only

Thanks so much.
Any links to more questions like these?

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

The most important part of the process is to read the question carefully.

Here question says Which of the following statements must be true?

Which means if you could identify even one case that proves that given statement can default then that statement and option concerned can be eliminated

Step 1) Make 15 places and place median at 8th place in ascending order and sum of all numbers against the places
__ __ __ __ __ __ __ 130 __ __ __ __ __ __ __ = (150 x 15) = 2250

I. at least one of the homes was sold for more than $165,000
Try to bring the Highest Selling price of home less than $165 we must maximize the price of other 14 houses including median 130
130 130 130 130 130 130 130 130 X X X X X X X = 2250 Where X is the maximum price for other houses beyond median
7X = 2250-(130x8) = 2250 - 1040 = 1210
X = 172 [i.e. Minimum price is exceeding 165 therefore It MUST be true]

ELIMINATE OPTION B and C that Don't include <I>

Now we have got to check other statement to get the correct answer, so choose that looks easier to check

I find III easier
III. at least one of the homes was sold for less than $130,000

Try to bring the Lowest Selling price of homes Greater than or equal to $130
But we have done this already
130 130 130 130 130 130 130 130 X X X X X X X = 2250 Where X is the maximum price for other houses beyond median
7X = 2250-(130x8) = 2250 - 1040 = 1210
X = 172 [i.e. Minimum price is exceeding 165 therefore It MUST NOT be true]

Now Check Statement II
II. at least one of the homes was sold for more than $13,000 but less than $150,000
We have proved it wrong already while solving statement I

Therefore Answer: Option A

Thanks Rich, my exam is this week so trying to tie up all loose end

Hey Nijo -

Here's one:

If I have $4.55 entirely in dimes and quarters, what are:

(1) the maximum number of coins I could have,
(2) the minimum number of coins I could have,

and

(3) is it possible for me to have the same number of each type of coin?

Matt@VeritasPrep wrote:Hey Nijo -

Here's one:

If I have $4.55 entirely in dimes and quarters, what are:

(1) the maximum number of coins I could have,
(2) the minimum number of coins I could have,

and

(3) is it possible for me to have the same number of each type of coin?

Hey Matt, no one answered this but I think its

(1) 44, where we want to maximize the dimes and minimize the quarters. So 43 dimes =4.30 and 1 quarter =0.25 total $4.55
(2) 20, where we maximize the quarters and minimize the dimes. 17 quarters= $4.25 and 3 dimes =0.30 total $4.55

(3) Wasn't sure about this one, Im guessing we can use two equations?
d=q
and
d+q=4.55
or
0.1d+0.25q=4.55?
We could also plug values for q and d but that would be time consuming.

prada wrote:

Matt@VeritasPrep wrote:Hey Nijo -

Here's one:

If I have $4.55 entirely in dimes and quarters, what are:

(1) the maximum number of coins I could have,
(2) the minimum number of coins I could have,

and

(3) is it possible for me to have the same number of each type of coin?

Hey Matt, no one answered this but I think its

(1) 44, where we want to maximize the dimes and minimize the quarters. So 43 dimes =4.30 and 1 quarter =0.25 total $4.55
(2) 20, where we maximize the quarters and minimize the dimes. 17 quarters= $4.25 and 3 dimes =0.30 total $4.55

(3) Wasn't sure about this one, Im guessing we can use two equations?
d=q
and
d+q=4.55
or
0.1d+0.25q=4.55?
We could also plug values for q and d but that would be time consuming.

Nice! You've got the right idea.

Another approach for the third one would be to say that (d + q) = .35, so if we have the same number of each, we should be able to solve

.35 * (d + q) = 4.55 and have (d + q) be an integer.

Luckily for us, that works: d + q = 13, so we'd have 13 of each coin. (The notation is a little sloppy, since d doesn't equal the number of dimes, but instead the value of a dime ... but you get the idea!)

Brent@GMATPREPNOW

You should remove one 130 from your answer as below, because its 16 values not 15
130, 130, 130, 130, 130, 130, 130, 130, 130, 130, 130, 130, 130, 130, 130, 430