Number Sense

What is number sense and how can you recognize number sense problems on the QUANT? Before we get into the details, let's start with a few number sense practice problems. Remember, no calculator.

Warm-Up Problems

Q1. Rank those three in order from smallest to biggest.

I. \dfrac{47}{150}

II. \dfrac{111}{330}

III. \dfrac{299}{900}

(A) I, II, III

(B) I, III, II

(C) II, I, III

(D) II, III, I

(E) III, I, II

Q2. Let P = 36000. Let Q equal the sum of all the factors of 36000, not including 36000 itself. Let R be the sum of all the prime numbers less than 36000. Rank the numbers P, Q, and R in numerical order from smallest to biggest.

(A) P, Q, R

(B) P, R, Q

(C) Q, P, R

(D) R, P, Q

(E) R, Q, P

Q3. Rank those three in order from smallest to biggest.

I. 2\sqrt{5}

II. 3\sqrt{2}

III. 3\sqrt[4]{401}

(A) I, II, III

(B) I, III, II

(C) II, I, III

(D) II, III, I

(E) III, II, I

Q4. In the below addition A, B, C, D, E, F, and G represent the digits 0, 1, 2, 3, 4, 5 and 6. If each variable has a different value, and E ≠ 0, then G equals


(A) 2

(B) 3

(C) 4

(D) 5

(E) 6

Solutions for these number sense problems will come at the end of this blog article.

What is Number Sense?

Many Quantitative Problems, like the foregoing pair, test number sense. What is number sense? Number sense is a good intuition for what happens to different kinds of numbers (positive, negative, fractions, etc.) when you perform various arithmetic operations on them.

Number sense is what allows some folks to “see” shortcuts such as estimation or visual solutions. For example, in any of the problems above, there's absolutely no need to do any detailed calculations: in fact, folks with number sense can probably do all the math they need to do in their heads.

Examples of a few number sense facts

1. Making the numerator of a fraction bigger makes the whole fraction bigger.

2. Making the denominator of a fraction bigger makes the whole fraction smaller.

3. (big positive) + (small negative) = something positive.

4. (small positive) + (big negative) = something negative.

5. Multiplying by a positive decimal less than one makes something smaller.

6. Dividing by a positive decimal less than one makes something bigger.

Of course, it would be near impossible to make anything like a complete list. The left-brain reductionist dreams of something like an exhaustive list one could study, but number sense is all about right-brain pattern matching. If you're not familiar with the distinction of left/right hemisphere, see this “How to do Math Faster” which touches on similar issues.

How do you get number sense?

If you don't have it, how do you get it? That's not an easy question. There's no magical shortcut

to number sense, but here are some concrete suggestions.

1. Do only mental math. You shouldn't be using a calculator to practice for the QUANT anyway. Try to do simpler math problems without even writing anything down. Furthermore, look for opportunities every day, in every situation, to do some simple math or simple estimation (e.g. there are about 20 cartons of milk on the grocery store's shelf—about how much would it cost to buy all twenty?)

2. Look for patterns with numbers. Add & subtract & multiply & divide all kinds of numbers—positive integers, negative integers, positive fractions, negative fractions, and look for patterns. Number sense is all about pattern with numbers!!

3. This is a BIG one — in any QUANT practice problem that seemed (to you) to demand incredibly long calculations, but which had a very elegant solution of which you would have never dreamt — that problem & its solution are pure gold. In a journal, write down what insights were used to simplify the problem dramatically. Force yourself to articulate this, and return to this solution and to your notes on it often. Over time, you should develop an array of problems like this, and if you study those solutions, you probably will start to see patterns.

4. Here's a variant on a game you can play, alone or with others who also want practice. Pick four single digit numbers at random—some repeats are allowed. You could roll a die four times, and use the results. Now, once you have those four numbers, your job is to use all four of them, each of them only once, and any arithmetic, to generate each number from 1 to 20. By “any arithmetic,” I mean any combination of: (a) add, subtract, multiply, divide; (b) exponents; and (c) parentheses & fractions

For example, if the four numbers I picked were {1, 2, 3, 4}, I could get 2 from

\bold{2 = (4-3)+(2-1)} or \bold{2 = 2*1^{3+4}} or \bold{2 = 4 \times \dfrac{1^3}{2}}

For any one number, you only need to come up with it in one way (although you can consider it a bonus to come up with multiple ways for a single number!) Here, I show three ways just to demonstrate the possibilities. A few examples for some of the higher numbers:

\bold{12 = (2 \times 4) + 3 +1}

\bold{13 = (3 \times 4) + 1^2}

\bold{14 = (3 \times 4) + (1 \times 2)}

\bold{15 = (2 + 3) \times (4 - 1)}

Notice that I used a variant of the expression for 13 to create an expression for 14. Also, if I changed the plus sign in the expression for 13 to a minus sign, I would get an expression for 11. Also, notice that if 1 is one of the four numbers, then if you don't need it, you can simply multiply by it; furthermore, notice in the expression for 13 and in the second expression for 2 above, the exponent of 1 is a useful place to stash other numbers you don't need!

As you practice, you will start to develop a sense of how expressions for one number can be tweaked to give you another number. Overall, using similar combinations, you have to get every number from 1-20 with these four, or with whatever four you pick. Actually, the set {1, 2, 3, 4} is a very good warm-up set. When you want more of a challenge, use {2,3,3,5}.

One of our Remote Test Prep Experts, Jeff Derrenberger, created an awesome web game based on this mental math game. Click on the banner to check it out!

Practice Problem Solution

Q1. Notice that all three of these are close to fractions that equal \dfrac{1}{3}. The fractions that equal \dfrac{1}{3} would be, respectively, 50/150, 110/330, and 300/900. First of all, only the second one has a higher numerator, so the second one is more than \dfrac{1}{3} and the other two are less than \dfrac{1}{3}. Therefore, II is the greatest.

Now, from I and III, which is greater? Well, think about it this way. \dfrac{50}{150} = \dfrac{300}{900}, because both of those equal \dfrac{1}{3}. How much less than one third is each one of these? Well, 47/150 is \dfrac{3}{150} less than \dfrac{50}{150} = \dfrac{1}{3}, and \dfrac{299}{900} is \dfrac{1}{900} less than \dfrac{300}{900} = \dfrac{1}{3}. Well, clearly, \dfrac{3}{150} \gt \dfrac{1}{900} (the latter has a smaller numerator and a larger denominator!) Therefore, starting from \dfrac{1}{3}, \dfrac{47}{150} goes down further than does \dfrac{299}{900}. Therefore, \dfrac{47}{150}, dropping down a larger distance, must be the minimum value. Therefore, the correct order is I, III, II.

Q2. We know that some of the factors of 36,000 are 18,000, 12,000, and 9,000. Right there, those three add up to 39,000 more than 36,000. Right there, we know that P \lt Q. We can eliminate (C) and (E).

Now, R is a little trickier. We don't need to have detailed knowledge here. We know there are several prime numbers less than 100. Obviously, the density of prime numbers gets slightly less as we get bigger. Let's assume, extremely conservatively, that when we get up into the 20 and 30 thousands, there is at least one prime number every thousand: one between 20K and 21K, one between 21K and 22K, all the way up to 36K. The 6 primes in the thirty thousands are all greater than 30K, so let's estimate their sum as (30K)*6 = 180K. The 10 primes in the twenty thousands are all greater than 20K, so let's estimate their sum as (20K)*10 = 200K. Right away, that's 380K on an extremely conservative estimate–we didn't even include any of the primes less than 20,000. There is no way that the sum of the factors of 36K, not including 36K itself, is going to be more than ten times 36K! Thus, R is much larger than Q, and the correct order is P \lt Q \lt R.

BTW, if your curious, according to Wolfram Alpha, the sum of the factors of 36000, not including 36000, is Q = 91,764, and the sum of all the prime number less than 36000 is R = 64,711,067.

Answer = A

Q3. Here, we have to “un-simplify” the square-roots to get a sense of their relative size.

I. 2\sqrt{5} = \sqrt{4}\sqrt{5} = \sqrt{20}

II. 3\sqrt{2} = \sqrt{9}\sqrt{2} = \sqrt{18}

From this, we see that II is less than I. From this alone, we can eliminate (A) & (B).

The trickier item on the list is III. Without a calculator, it would be nearly impossible compute an exact value for the fourth-root of 401. But consider this: the fourth root of a number is the number to the power of \dfrac{1}{4}, and \dfrac{1}{4} = \bigg(\dfrac{1}{2}\bigg)*\bigg(\dfrac{1}{2}\bigg), so the fourth root is the square root of a square root. Now, of course, 401 is not itself a perfect square, but it is very close to a perfect square.

III. \sqrt[4]{401} = \sqrt{\sqrt{401}} \gt \sqrt{\sqrt{400}} = \sqrt{20}

This demonstrates that III. is slightly larger than I. Therefore, the order from least to greatest is II, I, III.

Answer = C

Q4. First, let's start with E. We know two 2-digit numbers sum to a three-digit number. If we add a two digit number and another two digit number together, what is their sum? Well, 99 + 99 = 198. The two largest two-digit numbers have a sum not bigger than 200. So I know that E is not equal to 2 or more.

Could E be a 1? Sure! It worked with 99 and 99. With the numbers we have in the problem, we could have 6 + 4 = 10. We might also add 5 + 4 = 9 and add a 1 that is carried over from the digits column, but we are getting ahead of ourselves. For now, we know that E is equal to 1 because the only possible digit that could carry over from the addition of A + C is a 1: E = 1.

While solving for E, we were thinking about A + C. So let's move on to this part of the problem. We determined that A + C equals 10 or more. From the remaining digits {0, 2, 3, 4, 5, 6}, we have three possibilities for A + C:

(1) 6 + 5 = 11

(2) 5 + 4 or 6 + 3 (with a 1 carried over from the digit column) = 10

(3) 6 + 4 = 10

Also, I am not concerned about the order (i.e. "Is A = 6 or 5?") because it is not crucial information to solving the problem. I just need to know G.

(1) 6 + 5 = 11

We can eliminate this possibility right away. We were told that we can't repeat digits. Adding 6 + 5, results in 11. That would make E = 1 and F =1. This is not allowed by the rules set forth in the problem.

You might think, "Wait! What if we carry over a 1. Then F would equal 2." This cannot happen because B + D would have to equal at least 10. With 6 and 5 already used, we are left with {0, 2, 3, 4}, and none of these numbers can be added together to make 10 or more.

(2) 5 + 4 or 6 + 3 (with a 1 carried over from the digit column) = 10

This scenario will not work for the same reason that the previous one did not work. If you use 5 and 4 or 6 and 3, there won't be numbers left over for B and D to equal 10 or more. So, we can eliminate this possibility.

(3) 6 + 4 = 10

In this case, A + C = 4 + 6 = 10. We have a 1 for E and don't need to worry about carrying over a 1 from the digits column (B + D), so F = 0. This works!

So, A + C = 4 + 6 = 10.

(Remember: we don't care if A is 4 or 6; the order doesn't matter in terms of solving the problem).

Also, as explained in the video, the 0 could not be B or D because that would lead to the repetition of a digit (2 + 0 = 2), and we already know we can't repeat a digit. So F = 0 is our only possibility.

Last part: we have {2, 3 and 5}. G is one of these numbers. Since G is the sum of B and D (B + D = G), the answer should be clear. 2 + 3 = 5, so B and D, in some order, equal 2 and 3, and G = 5.

Answer = D

FAQ: What if B = 6 and D = 4? Then G = 0.

A: That's totally correct! If B=6 and D=4, then yes, G would equal 0. However, the answer explanation in the video says that "G cannot equal zero because none of our answer choices is zero. This is true. Remember: our goal here is ultimately to identify the value of G. Since there is no answer choice of 0, if G were to equal zero, we wouldn't have an answer choice! So we have to use the answer options -- 2 through 6 -- for some guidance. :)

Could this kind of question actually show up on the test? How can I study for these types of questions?

FAQ: What if B = 6 and D = 4? Then G = 0.

A: This is in fact a kind of problem that you may see on test day, which is why we've included it in our set of practice questions. It's definitely not a common question type -- my educated guess will be you'll see 1 problem of this type, if you see it at all!

We don't have a lesson exactly on this type of problem, because it illustrates the importance of critical thinking and strategic use of plugging in numbers. The idea is that anyone can learn a bunch of basic math rules and formulas, but not everyone can apply them in interesting and abstract ways. That's what makes this test so difficult! This is also why Magoosh tries to not emphasize rote memorization, but rather mathematical thinking and number sense.

The basic idea is that you should try as hard as possible to get used to mental math and make your default mental math as opposed to, say, the calculator. Why? Because essentially when you're doing mental math, you're flexing your math muscle in your brain and you're building it up. Working with number sense allows you to develop a sense of how numbers work together.