We haven’t dealt with maximizing/minimizing strategies in our QWQW series yet (except in sets). The reason for this is that the strategy to be used varies from question to question. What works in one question may not work in another. You might have to think up on what to do in a question from scratch and you have only 2 mins to do it in. The saving grace is that once you know what you have to do, the actual work involved to arrive at the answer is very little.

Let’s look at some maximizing minimizing strategies in the next few weeks. We start with an OG question today with a convoluted question stem.

**Question**: List T consists of 30 positive decimals, none of which is an integer, and the sum of the 30 decimals is S. The estimated sum of the 30 decimals, E, is defined as follows. Each decimal in T whose tenths digit is even is rounded up to the nearest integer, and each decimal in T whose tenths digits is odd is rounded down to the nearest integer. If 1/3 of the decimals in T have a tenths digit that is even, which of the following is a possible value of E – S?

I. -16

II. 6

III. 10

A. I only

B. I and II only

C. I and III only

D. II and III only

E. I, II, and III

**Solution**:

There is a lot of information in the question stem and a lot of variables are explained. Let’s review the given data in our own words first.

T has 30 decimals. The sum of all the decimals is S.

10 decimals have even tenths digit. They will be rounded up.

20 decimals have odd tenths digit. They will be rounded down.

The sum of rounded numbers is E.

E – S can take many values so how do we figure which ones it cannot take? We need to find the minimum value E – S can take and the maximum value it can take. That will help us figure out the values that E – S cannot take. Note that E could be greater than S and it could be less than S. So E – S could be positive or negative.

Step 1: Getting Minimum Value of E – S

Let’s try to make E as small as possible. For that, we need to do two things:

1. When we round up the decimals (even tenths digit), the difference between actual and estimate should be very small. The estimate should add a very small number to round it up so that E is not much greater than S. Say the numbers are something similar to 3.8999999 (the tenths digit is the largest even digit) and they will be rounded up to 4 i.e. the estimate gains about 0.1 per number. Since there are 10 even tenths digit numbers, the estimate will be approximately .1*10 = 1 more than actual.

2. When we round down the decimals (odd tenths digit), the difference between actual and estimate should be as large as possible. Say the numbers are something similar to 3.999999 (tenths digit is the largest odd digit) and they will be rounded down to 3 i.e. the estimate loses approximately 1 per number. Since there are 20 such numbers, the estimate is 1*20 = 20 less than actual.

Overall, the estimate will be approximately 20 – 1 = 19 less than actual.

Minimum value of E – S = -19

Step 2: Getting Maximum Value of E – S

Now let’s try to make E as large as possible. For that, we need to do two things:

1. When we round up the decimals (even tenths digit), the difference between actual and estimate should be very high. Say the numbers are something similar to 3.000001 (tenths digit is the smallest even digit) and they will be rounded up to 4 i.e. the estimate gains 1 per number. Since there are 10 even tenths digit numbers, the estimate will be approximately 1*10 = 10 more than actual.

2. When we round down the decimals (odd tenths digit), the difference between actual and estimate should be very little. Say the numbers are something similar to 3.1 (tenths digit is the smallest odd digit). They will be rounded down to 3 i.e. the estimate loses approximately 0.1 per number. Since there are 20 such numbers, the estimate is approximately 0.1*20 = 2 less than actual.

Overall, the estimate will be approximately 10 – 2 = 8 more than actual.

Maximum value of E – S = 8.

The minimum value of E – S is -19 and the maximum value of E – S is 8.

So E – S can take the values -16 and 6 but cannot take the value 10.

**Answer (B)**

*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!*

Karishma,

Wonderful question.I spent lot of time deep diving into analysis.Great.