M14-26

Official Solution:


\(X\) 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


Let \(m\) denote the sum of original prices of 5 cheapest items and \(M\) denote the sum of original prices of 5 most expensive items. The original total cost of the lot is \(m + M\); the new total cost of the lot will be \(0.9m + 1.1M\).

Is I possible? \(\frac{0.9m + 1.1M}{10} \lt \frac{m + M}{10}\) or \(0.9m + 1.1M \lt m + M\) or \(0.1M \lt 0.1m\) or \(M \lt m\), which is not true. Thus, I is not possible.

Is II possible? \(\frac{0.9m + 1.1M}{10} = \frac{m + M}{10}\) or \(0.9m + 1.1M = m + M\) or \(0.1M = 0.1m\) or \(M = m\), which is not true. Thus, II is not possible.

III is possible because \(M \gt m\).


Answer: A
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I think this is a high-quality question and I agree with explanation. What do you think of solving this question like this :

Let's say you have Price 1 = 3 (reduce by 50% -> 1.5
Price 2 =5 Increase by 50% ->7,5
(1,5+7.5)/2>(3+5)/2
Boom I go with III only

Since the number of items remain the same. the average depends on the total... If total increases average increases.

And given that the cheapest 5 items (half the total number) has price reduced by 10 percent. - ok. so the total reduced by some amount( say D - for drop).

now each of the costliest items increased by 10 percent - Firstly these 5 items have a greater price than the cheapest items obviously.So we are increasing 10 percent each of a 5 greater numbers. This raises the total by lets say (I - for increase)

Its clear that 10 percent of bigger numbers is def greater that 10 percent of smaller numbers. so I - D > 0 . that is it is a positive value. meaning the increase (due to 10 percent increase ) is greater that the drop (due to 10 percent reduction)

The reduction of 10 percent from the cheaper set is not enough to balance the 10 percent increase
of the bigger number. - So the total will increase.

and thus the average will increase.

A decrease in 10% can be compensated only by an increase in 11.11%
while a decrease in 9.09% can be compensated by an increase in 10%

Here we have lower value decreased by 10% while the higher value is increased by 10%.
Even for two consecutive numbers, the final average tends to increase(10 decreased by 10% gives 9 while 11 increased by 10% gives 12.1; the average is increased from 10.5 to 10.55), hence the same will follow for 10 numbers giving answer 3.

Do this theoretically, w/o pen and paper:
Firstly, When the number of items is same, Avg only depends upon the Total sum. Sum increases, Avg increases and vice-versa.
Next, X% of smaller value will obviously be less than X% of bigger value.

Hence, in this case, the total loss (X% of cheaper items) is less than the total gain (X% of expensive items). Therefore, in effect, the Total Sum actually increased.
Therefore the Avg too shall increase.

Option A

I got the solution with a somewhat simpler reasoning:

10% of something smaller is always smaller than 10% of something bigger, then:

The 5 (half of total) most expensive items are the ones having a 10% gain => Net gain will always be positive.

x1 ,x2 , x3, x4, x5, x6, x7, x8, x9, x10....
10,20, 30, 40 ,50, 60, 70, 80,90, 100

average is 55

first five decrease by 10%, second five increase by 10%

9, 18, 27, 36, 45, 66, 77, 88, 99 , 110

average 59.5

only number 3 or option A

How would making the numbers NOT distinct change the answer here? Bunuel do you know?