What do all world-class athletes have in common? Great fundamentals. You can’t win Wimbledon with uncoordinated footwork, and you can’t become the NBA’s best shooter with a sloppy follow-through.
The same is true for GMAT math. You won’t be able to solve a complex overlapping probability question if you don’t understand overlapping sets. Good fundamentals make the extraordinary feats possible. That’s why the best athletes practice the fundamentals of their sport everyday, and it’s also why you should practice and master the math fundamentals that underly many of the GMAT’s math questions.
These GMAT math formulas are the basis for countless questions on the exam; expect to have to employ them while solving both easy and difficulty questions.
Nothing advanced here – you’ve seen distance/rate/time questions since middle school. Expect to see problems on the GMAT asking you to find the speed of a car, the distance traveled by a train or the time it takes to bike to the store. However, don’t expect the math to be as straightforward as they were in 9th grade algebra.
Here’s an example of how a GMAT math question may use the distance/rate/time formula:
Three planes, A, B, and C, consume $10, $9, and $19 worth of fuel per mile respectively. Planes A, B, and C travel at a speed of 300, 250, and 600 mph respectively. If a pilot is hired at a rate of $20 per hour, which plane(s) would be the most cost-efficient to travel a distance of 1,200 miles?
A) Plane A
B) Plane B
C) Plane C
D) Planes B and C
E) Planes A and B
In order to find which plane will be the cheapest, we need to compare the total cost of the journey for each plane. The total cost is going to be the total cost of fuel (1200 x cost of fuel per mile) plus the total cost of the pilot (20 x time it takes to travel 1200 miles).
To find the time it takes for each plane to the travel the 1,200 miles, use the distance/rate/time formula and solve for time: t = d/r. So for plane A, it takes 4 hours to travel 1,200 miles (1200/300 = 4), which at a rate of $20 per hour costs $80.
Now add in the cost of fuel (1200 x $10 = $12000) to get a total cost journey cost of $12,080.
Repeat this process for the other planes:
Plane B: (1200 x 9) + (1200/250) x 20 = $10,896
Plane C: (1200 x 19) + (1200/600) x 20 = $22,840
The correct answer is B, plane B.
Like , the Pythagorean theorem has likely been etched into the back of your brain since middle school. This equation crops up in a host of GMAT questions, so it’s a good idea to practice many different situations in which it can be applied.
Here’s an example:
A square is inscribed in a circle with a radius of 2 inches. Find the perimeter of the square.
D) 12 inches
It may help to draw a picture for this question. Since the edges of the square touch the circle, the diagonal of the square is equal to the diameter of the circle: 2r or 4.
The diagonal splits the square into two right triangles, so we can use the Pythagorean theorem to solve for the side length.
a2 + b2 = 42, but since a and b are sides of a square, we know that they are equal, so we can rewrite the equation as:
2a2 = 16
a2 = 8
a = or
Don’t stop here and fall for the trap in answer choice A. Remember you want the perimeter of the square, so multiply by 4 to get , the correct answer B.
The work rate theorem is actually a form of the distance/rate/time formula. However, instead of distance changing at a certain rate, an amount of something is being produced at a certain rate. For example, the amount of washing machines made, rooms painted, books written or cars washed.
Often work-rate problems ask you to find the total time it takes two people or machines working together to do a job. This is where the modified formula above comes into play.
For example, a question may ask you to find the total time it takes for two painters to paint a room working together given their different rates working alone. In this case, A is the time it takes the first painter to paint the room working alone and B is the time it takes for the second painter working alone.
Here’s an example of the formula in action:
Mack and Ed are building the frame of a new house. If Ed can build a frame in 5 days working alone and Mack can build a frame in 8 days working alone, approximately how long will it take to build the frame working together, but independently?
A) 3 days
B) 4 days
C) 5 days
D) 6 days
E) 7 days
The task is building the frame of a house, which it takes Ed 5 days to do working alone and Mack 8. This is all we need to know in order to find the total time it will take both of them working together.
Plugging in the numbers we get:
13 goes into 40 a little over 3 times (3.08 to be exact). The correct answer is A, 3 days.
Remember factorials? Even if you don’t, don’t be intimidated by this formula; it’s much simpler than it looks, which is good because it’s a GMAT math formula you’ll need to know inside and out.
Combinations are used to calculate how many ways you can select an unordered group from a larger set. For example, how many different 3-topping pizzas can you make from a selection of 10 toppings, or how many different 4-person teams can you make with 20 people?
An unordered group means that it doesn’t matter in which order you select the members of the group. For example, an onion, sausage and anchovy pizza is the same as a sausage, anchovy and onion pizza (and equally delicious) — changing the selection order of the ingredients has no impact. (Compare this to selecting 3 letters for a password; you could have the same three letters but each ordering of the letters would result in a different password.)
In the equation, n stands for the larger set (the 10 toppings or the 20 people) while k stands for the smaller selection you make (the 3 toppings or the teams of 4 people).
Let’s use the pizza example to demonstrate a basic combination problem you may encounter on the GMAT:
A pizzeria allows you to create a custom pizza by selecting exactly 3 toppings from a list of 10 available toppings. How many different custom pizzas could be created?
You want to know how many different 3-topping combinations you make from 10 available toppings, so n = 10 and k = 3. Plug these figures into the formula to get
Remember the ! means factorial. For example 3! = 3 x 2 x 1 = 6 and 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5,040, but in combination problems, there’s no need to multiply the factorials out. Instead cancel out numbers.
After simplifying the equation by canceling out 7! you have a simple multiplication and division problem that results in 120. The correct answer is C.
Overlapping sets are emblematic of GMAT math because they require solid mathematical reasoning rather than complicated arithmetic. You may know these problems as Venn diagram problems, in which you’re presented with multiple overlapping groups and tasked to find either the total number of items in all the groups or the total number of items in a single group.
Let’s use an example to show exactly what we mean and how this formula is applied.
A certain high school has 800 students. Of these students, 257 play sports and 100 students are in the drama club. If 38 students play sports and are in the drama club, how many students at the school neither play sports nor participate in the drama club?
In this question we have 2 different groups that overlap: students who play sports and students in the drama club. These are groups A and B, 257 students and 100 students respectively, and Both represents the overlap, the students who a part of both groups, 38.
We can plug these numbers into the equation along with the total number of students, 800, in order to find the answer – the number of students who neither play sports or participate in drama.
257 + 100 – 38 + Neither = 800
Neither = 481. The correct answer is D.
You can also solve overlapping set questions such as the one above by drawing a Venn diagram. If you’re a visual learner, you may want to try this method.
Not surprisingly, there’s more to these formulas than just these forms. If you have these formulas committed to memory, start to work on related concepts. Here are some of the ways the GMAT will complicate these formulas:
If you know the basic formulas well, these more complicated problems won’t be nearly as challenging. It’s simply a matter of good fundamentals.