Max-Min Strategies: Establishing Base Case

Quarter Wit, Quarter WisdomContinuing our discussion on maximizing/minimizing strategies, let’s look at another question today. Today we discuss the strategy of establishing a base case, a strategy which often comes in handy in DS questions. The base case gives us a starting point and direction to our thoughts. Otherwise, with the number of possible cases in any given scenario, we may find our mind wandering from one direction to another without reaching any conclusions. That is a huge waste of time, a precious commodity.

Question: Four friends go to Macy’s for shopping and buy a top each. Three of them buy a pillow case each too. The prices of the seven items were all different integers, and every top cost more than every pillow case. What was the price, in dollars, of the most expensive pillow case if the total price of the seven items was $89?

Statement 1: The most expensive top cost $16.

Statement 2: The least expensive pillow case cost $9.

Solution: The first problem here is figuring out the starting point. There must be many ways in which you can price the seven items such that the total cost is $89. So we need to establish a base case (which conforms to all the conditions given in the question stem) first and then we will tweak it around according to the additional information obtained from our statements.

‘Seven items for $89’ means the average price for each item is approximately $12. But 12 is not the exact average. 12*7 = 84 which means another $5 were spent.

A sequence with an average of 12 and different integers is $9, $10, $11, $12, $13, $14, $15.

But actually another $5 were spent so the prices could be any one of the following variations (and many others):

$9, $10, $11, $12, $13, $14, $20 (Add $5 to the highest price)

$9, $10, $11, $12, $13, $16, $18 (Split $5 into two and add to the two highest prices)

$9, $10, $12, $13, $14, $15, $16 (Split $5 into five parts of $1 each and add to the top 5 prices)

$7, $9, $13, $14, $15, $16, $17 (Take away some dollars from the lower prices and add them to the higher prices along with the $5)

etc

Let’s focus on another piece of information given in the question stem: “every top cost more than every pillow case.”

This means that when we arrange all the prices in the increasing order (as done above), the last four are the prices of the four tops and the first three are the prices of the three pillow cases. The most expensive pillow case is the third one.

Now that we have accounted for all the information given in the question stem, let’s focus on the statements.

Statement 1: The most expensive top cost $16.

We have already seen a case above where the maximum price was $16. Is this the only case possible? Let’s look at our base case again:

$9, $10, $11, $12, $13, $14, $15

(a further $5 needs to be added to bring the total price up to $89)

Since the prices need to be all unique, if we add 1 to any one price, we also need to add at least $1 to each subsequent price. E.g. if we increase the price of the least expensive pillow case by $1 and make it $10, we will need to increase the price of every subsequent item by $1 too. But we have only $5 more to give.

If the maximum price is $16, it means the rightmost price can increase by only $1. So all prices before it can also only increase by $1 only and except the first two prices, they must increase by $1 to adjust the extra $5.

Hence the only possible case is $9, $10, $12, $13, $14, $15, $16.

So the cost of the most expensive pillow case must have been $12.

Statement 1 is sufficient alone.

Statement 2: The least expensive pillow case cost $9.

A restriction on the lowest price is much less restrictive. Starting from our base case

$9, $10, $11, $12, $13, $14, $15,

we can distribute the extra $5 in various ways. We can do what we did above in statement 1 i.e. give $1 to each of the 5 highest prices: $9, $10, $12, $13, $14, $15, $16

We can also give the entire $5 to the highest price: $9, $10, $11, $12, $13, $14, $20

So the price of the most expensive pillow case could take various values. Hence, statement 2 alone is not sufficient.

Answer (A)

Note that the answer is a little unexpected, isn’t it? If we were to read the question and guess within 20 secs, we would probably guess that the answer is (C), (D) or (E). The two statements give similar but complementary information. It would be hard to guess that one will be sufficient alone while other will not be. This is what makes this question interesting and hard too.

Our strategy here was to establish a base case and tweak it according to the information given in the statements. This strategy is often useful in DS – not just in max-min questions but others too.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog series!

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