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X is the average price of 10 items of which no two cost

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by guerrero » Tue Jun 11, 2013 12:35 pm
X is the average price of 10 items of which no two cost the same. If each of the cheapest 5 items loses 10% in price while each of the most expensive 5 items gains 10% in price, which of the following is possible about X?

I. X will decrease

II. X will not change

III. X will increase

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

OA A
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Source: — Problem Solving |

by Brent@GMATPrepNow » Tue Jun 11, 2013 1:28 pm
guerrero wrote:X is the average price of 10 items of which no two cost the same. If each of the cheapest 5 items loses 10% in price while each of the most expensive 5 items gains 10% in price, which of the following is possible about X?

I. X will decrease

II. X will not change

III. X will increase

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

OA A
Rather than talk about the average price of the 10 items, we can just talk about the total price of the 10 items. After all, if the total price of the 10 items increases then the average price of the 10 items increases (and vice versa)

So, let's just think of X as being the total price of all 10 items.

Let C = the total price of the 5 cheapest items.
Let E = the total price of the 5 most expensive items.

This means that C + E = X (the original/old value of X is C + E)

We can also conclude that E - C > 0
In other words, E - C = some positive value
(this part will be useful later.


Okay, on with the solution . . .
The cheapest 5 items lose 10% in price
Total price of the 5 cheapest items = 0.9C

The most expensive 5 items gain 10% in price
Total price of the 5 most expensive items = 1.1E

The new value of X = 0.9C + 1.1E
How does this compare to the old value of X (when X = C + E)?

Notice that
new value of X = 0.9C + 1.1E
= (C - 0.1C) + (E + 0.1E)
= C + E + 0.1E - 0.1C
= C + E + 0.1(E - C)
= C + E + 0.1(some positive value)

This means that the new value of X must be greater than the old value of X

Answer: A

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by guerrero » Tue Jun 11, 2013 1:40 pm
thank you Brent , You 're amazing .
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by GMATGuruNY » Wed Jun 12, 2013 3:33 am
guerrero wrote:X is the average price of 10 items of which no two cost the same. If each of the cheapest 5 items loses 10% in price while each of the most expensive 5 items gains 10% in price, which of the following is possible about X?

I. X will decrease

II. X will not change

III. X will increase

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

OA A
Brent's solution is great.
Some test-takers might find it easier to plug in values.

Average = sum/number.
Thus, if the SUM increases, so does the AVERAGE.
Let L = the total cost of the 5 least expensive items and E = the total cost of the 5 most expensive items.

Case 1:
Old sum:
Let L = 10 and E = 100.
Old sum = 10+100 = 110.

New sum:
L reduced by 10% = 10 - .1(10) = 9.
E increased by 10% = 100 + 10 = 110.
New sum = 9 + 110 = 119.

The new sum is greater, implying that the value of X -- the new average -- also will be greater.

Case 2:
Old sum:
Let L = 10 and E = 11.
Old sum = 10+11 = 21.

New sum:
L reduced by 10% = 10 - .1(10) = 9.
E increased by 10% = 11 + 1.1 = 12.1.
New sum = 9 + 12.1 = 21.1.

The new sum is greater, implying that the value of X -- the new average -- also will be greater.

When L and E are far apart, the value of X increases.
When L and E are very close together, the value of X increases.
Thus, we can be reasonably certain that the value of X will increase in every case.

The correct answer is A.
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by Atekihcan » Thu Jun 13, 2013 12:57 am
As the price of the items are different, price of 5 cheapest items will be less than the price of the 5 most expensive items. So, decrease in the total price = 10% decrease in the 5 cheapest items must be less than the increase in the total price = 10% increase in the 5 most expensive items.

So, finally total price will increase
So, average will increase.

Answer : A
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