Standard shortcut to solve a variety of quants

Often times, two quantitative questions that would seem to fall into separate categories can be solved using the same approach. When we have to sift through dozens of techniques and strategies under pressure, we’re likely to become paralyzed by indecision. If, however, we have a small number of go-to approaches, we can quickly consider all available options and arrive at one that will work in any given context.
Say, for example, that we have a classroom of students from two countries, which we’ll call “A” and “B.” They all take the same exam. The average score of the students from country A is 92 and the average score of the students from country B is 86. If the overall average is 90, what is the ratio of the number of students from A to the number of students in B? We could solve this algebraically. If we call the number of students from county A, “a” and the number of students from country B “b,” we’ll have a total of a + b students, and we can set up the following chart.

Average Number of Terms Sum
Country A 92 a 92a
Country B 86 b 86b
Total 90 a + b 90a + 90b


The sum of the scores of the students from A when added to the sum of the scores of the students from B will equal the sum of all the students together. So we’ll get the following equation: 92a + 86b = 90a + 90b.

Subtract 90a from both sides: 2a + 86b = 90b

Subtract 86b from both sides: 2a = 4b

Divide both sides by b: 2a/b = 4

Divide both sides by 2: a/b =4/2 =2/1. So we have our ratio. There are twice as many students from A as there are from B.

Not terrible. But watch how much faster we can tackle this question if we use the number line approach, and use the difference between each group’s average and the overall average to get the ratio:

b              Tot       a


Gap:  4           2

Ratio a/b = 4/2 = 2/1. Much faster. (We know that the ratio is 2:1 and not 1:2 because the overall average is much closer to A than to B, so there must be more students from A than from B. Put another way, because the average is closer to A, A is exerting a stronger pull. Generally speaking, each group corresponds to the gap that’s farther away.)

The thing to see is that this approach can be used on a broad array of questions. First, take this mixture question from the Official Guide*:

Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 % fescue. If a mixture of X and Y contains 30% ryegrass, what percent of the weight of the mixture is X?

A. 10%
B. 33 1/3%
C. 40%
D. 50%
E. 66 2/3%

In a mixture question like this, we can focus exclusively on what the mixtures have in common. In this case, they both have ryegrass. Mixture X has 40% ryegrass, Mixture Y has 25% ryegrass, and the combined mixture has 30% ryegrass.

Using a number line, we’ll get the following:

Y          Tot             X


Gap: 5             10

So our ratio of X/Y = 5/10 = ½. (Because X is farther away from the overall average, there must be less X than Y in the mixture.) Be careful here. We’re asked what percent of the overall mixture is represented by X. If we have 1 part X for every 2 parts of Y, and we had a mixture of 3 parts, then only 1 of those parts would be X. So the answer is 1/3 = 33.33% or B.

So now we see that this approach works for the weighted average example we saw earlier, and it also works for this mixture question, which, as we’ve seen, is simply another variation of a weighted average question.

Let’s try another one*:

During a certain season, a team won 80 percent of its first 100 games and 50 percent of its remaining games. If the team won 70 percent of its games for the entire season, what was the total number of games that the team played?

a) 180
b) 170
c) 156
d) 150
e) 105

First, we’ll plot the win percentages on a number line.

Remaining             Total           First 100


Gap           20                     10

Remaining Games/First 100 = 10/20  = ½.

Put another way, the number of the remaining games is ½ the number of the first 100. That means there must be (½) * 100 = 50 games remaining. This gives us a total of 100 + 50 = 150 games played. The answer is D.

Note the pattern of all three questions. We’re taking two groups and then mixing them together to get a composite. We could have worded the last question, “mixture X is 80% ryegrass and weighs 100 grams, and mixture Y is 50% ryegrass. If a mixture of 100 grams of X and some amount of Y were 70% ryegrass, how much would the combined mixture weigh?” This is what I mean by making horizontal connections. One problem is about test scores, one is about ryegrass, and one is about baseball, but they’re all testing the same underlying principle, and so the same technique can be applied to any of them.

Takeaway: always try to pay attention to what various questions have in common. If you find that one technique can solve a variety of questions, this is a technique that you’ll want to make an effort to consciously consider throughout the exam. Any time we’re stuck, we can simply toggle through our most useful approaches. Can I pick numbers? Can I back-solve? Can I make a chart? Can I use the number line? The chances are, one of those approaches will not only work but will save you a fair amount of time in the process.

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