The post How to Solve GMAT Motion Problems appeared first on Magoosh Blog — GMAT® Exam.
]]>Word Problems make up a majority of the quantitative section of the GMAT (almost 60 percent). Of the word problems they’ll face, students tend to need the most help with GMAT motion problems. This type of problem centers around the “dust” formula, which is short for Distance equals Speed multiplied by Time, or \(\text{D}\times\text{S}=\text{T}\). But there are many varieties of motion problem, and we will discuss techniques for each of them. At the end of this article, you’ll also find motion word problems with solutions for you to test your knowledge!
Like we mentioned above, Distance equals Speed multiplied by Time, or \(\text{D}\times\text{S}=\text{T}\). If you learn this basic equation well, you’ll be able to dust your math troubles away! (Insert rimshot.)
We can rearrange this formula to determine that Speed is equal to Distance divided by Time.
And Time is equal to Distance divided by Speed.
Let’s demonstrate this. Walking at a constant rate of 160 meters per hour, Monroe can cross a bridge in 2 hours. What is the length of the bridge?
Here, the length of the bridge is the distance Monroe must cross. Using Distance equals Speed multiplied by Time, we get:
\((160\text{meters per hour})\times(2\text{hours})=320\text{meters}\)
Seems simple enough so far, right? Let’s check out a few more GMAT motion problems.
Oftentimes one traveler will be dividing their trip into multiple segments. In these situations, you need a separate “dust” equation for each segment. You will usually have two equations, which requires a two-variable system of equations to solve. You will often perform substitution.
Let’s try this out. Julianne drives the first 360 miles of a trip at 60 miles per hour. How fast does he have to drive, in miles per hour, on the final 210 miles of the trip if the total time of the trip is to equal nine hours?
First, we identify all of our parameters. We will call the variables for the first leg of the trip distance D_{1}, speed S_{1}, and time T_{1}; we’ll call the variables for the second leg D_{2}, S_{2}, and T_{2}.
From the given information, we know that
\(\text{D}_1=360\text{miles}\text{S}_1=60\text{miles per hour}\text{D}_2=210\text{miles}\)
\(\text{T}_1=\frac{\text{D}_1}{\text{S}_1}\)
\(\text{T}_1=\frac{360\text{miles}}{60\text{miles per hour}}=6\text{hours}\)
Because the total time of the trip is to equal nine hours, T_{2} must equal nine minus T_{1} hours.
\(\text{T}_2=3\text{hours}\)
\(\text{S}_2=\frac{\text{D}_2}{\text{T}_2}\)
\(\text{S}_2=\frac{210\text{miles}}{3\text{hours}}=70\text{miles per hour}\)
As we discussed previously, Speed is equal to Distance divided by Time. Let’s take that a step further and talk about average speed. Average speed is defined as total distance traveled divided by the total time period spent traveling. This means that if you have a trip with multiple segments, you’ll want to take the sum of the distances of each segment and divide that by the sum of the times of each segment. Average speed captures the constant speed needed to travel the total distance in the total time.
Let’s demonstrate this. Koki drove 16 miles in 10 minutes, and then drove an additional 6 miles in 5 minutes. What is Koki’s average speed for the entire trip in miles per hour?
Well, average speed is the total distance divided by the total time.
\(\text{D}_\text{Total}=16\text{miles}+6\text{miles}=22\text{miles}\)
\(\text{T}_\text{Total}=10\text{minutes}+5\text{minutes}=15\text{minutes}\)
\(\text{S}_\text{Average}=\frac{\text{D}_\text{Total}}{\text{T}_\text{Total}}\)
\(\text{S}_\text{Average}=\frac{22\text{miles}}{15\text{minutes}}\times\frac{60\text{minutes}}{1\text{hour}}=88\text{miles per hour}\)
That’s straightforward enough, but what if we are not given any distances or times? It is possible to solve an average speed problem, even if all you are given are the different speeds in each segment of the trip. You might then think that average speed would just be the average of all of the speeds, but that is not correct.
Let’s say that Nathaniel drove from Gwenville to Samton at an average speed of 24 miles per hour. He then drove the same route on the return trip back from Samton to Gwenville at an average speed of 36 miles per hour. If you were asked to find Nathaniel’s average speed, it would not just be 30 miles per hour (the average of 24 and 36).
To see this, let’s go back to our MVP dust formula. Since there are two legs of the trip, we will have two equations. D_{1}, S_{1}, T_{1}; D_{2}, S_{2}, T_{2}. Because Nathaniel’s trip is a round trip, we can assume that D_{1}and D_{2} are the same, so we will set both of them equal to D.
\(\text{D}_1=\text{D}\text{S}_1=24\text{miles per hour}\text{T}_1=\frac{D}{24\text{miles per hour}}\)
\(\text{D}_2=\text{D}\text{S}_2=36\text{miles per hour}\text{T}_2=\frac{D}{36\text{miles per hour}}\)
\(\text{D}_\text{Total}=\text{2D}\)
\(\text{T}_\text{Total}=\frac{D}{24\text{miles per hour}}+\frac{D}{36\text{miles per hour}}\)
\(\text{S}_\text{Average}=\frac{\text{D}_\text{Total}}{\text{T}_\text{Total}}\)
\(\text{S}_\text{Average}=\frac{\text{2D}}{\frac{D}{24\text{miles per hour}}+\frac{D}{36\text{miles per hour}}}\)
We can factor a D out of this \fraction.
\(\text{S}_\text{Average}=\frac{2}{\frac{1}{24}+\frac{1}{36}}\text{miles per hour}\)
We can find a common denominator between 24 and 36.
\(\text{S}_\text{Average}=\frac{2}{\frac{3}{72}+\frac{2}{72}}\text{miles per hour}\)
\(\text{S}_\text{Average}=\frac{2}{\frac{5}{72}}\text{miles per hour}=\frac{2}{1}\times\frac{72}{5}=\frac{144}{5}\text{miles per hour}=28.8\text{miles per hour}\)
In summary, whenever you want to find the average speed of a round trip, and you are given the two segment speeds, you can put it in the form \(\text{S}_\text{Average}=\frac{\text{2}}{\frac{1}{\text{S}_1}+\frac{1}{\text{S}_2}}\). You can also use this formula to find one of the segment speeds, given the other segment speed and the average speed.
As we discussed in the last section, when one traveler divides their trip into multiple segments, you will need a separate “dust” equation for each segment.
Similarly, when we have multiple travelers, we need a separate “dust” equation for each traveler. And, like in the last segment, we will need a system of equations for these problems, too, since the two travelers will have some relationship to each other in distance, speed, and/or time.
Let’s demonstrate this. Car X and Y are traveling from A to B on the same route at constant speeds. Car X is initially behind Car Y, but Car Y’s speed is 0.8 times Car X’s speed. Car X passes Car Y at noon. At 1:45 pm, Car X reaches B, and at that moment, Car Y is still 35 miles away from B. What is the speed of Car X in miles per hour?
Like last time, we’ll start by defining our parameters: D_{X}, S_{X}, T_{X}, and D_{Y}, S_{Y}, T_{Y}. We’ll consider the point at which Car X passes Car Y to be our initial time, and we’ll consider the point at which Car X reaches B to be our final time. This means that both T_{X} and T_{Y} are one hour and 45 minutes, or 1.75 (\(\frac{7}{4}\)) hours.
In that timeframe, Car X travels all the way to B, but Car Y is still 35 miles away from B.
\(\text{D}_\text{X}=\text{D}\text{D}_\text{Y}=\text{D}-35\)
Car Y’s speed is 0.8 (\(\frac{4}{5}\)) \times Car X’s speed.
\(\text{S}_\text{Y}=\frac{4}{5}\times\text{S}_\text{X}\)
Now we have defined all of our parameters; let’s set up the two equations.
Car X: \(\text{D}=\text{S}_\text{X}\times(\frac{7}{4}\text{hour})\)
Car Y: \(\text{D}-35=\frac{4}{5}\times\text{S}_\text{X}\times(\frac{7}{4}\text{hour})\)
We can substitute the D in Car X’s equation into Car Y’s equation.
\((\text{S}_\text{X}\times(\frac{7}{4}\text{hour}))-35=\frac{4}{5}\times\text{S}_\text{X}\times(\frac{7}{4}\text{hour})\)
Multiply \(\frac{4}{5}\) by \(\frac{7}{4}\).
\(\frac{7}{4}\text{S}_\text{X}-35=\frac{7}{5}\text{S}_\text{X}\)
\(\frac{7}{4}\text{S}_\text{X}-\frac{7}{5}\text{S}_\text{X}=35\)
\(\frac{35}{20}\text{S}_\text{X}-\frac{28}{20}\text{S}_\text{X}=35\)
\(\frac{7}{20}\text{S}_\text{X}=35\)
\(\frac{20}{7}\times\frac{7}{20}\text{S}_\text{X}=\frac{35}{1}\times\frac{20}{7}\)
\(\text{S}_\text{X}=100\text{miles per hour}\)
Imagine that you see your friend in the park after the pandemic is over. If you start excitedly running towards each other, you would reach each other faster than you would if your friend had remained stationary. If instead you run away from each other, the distance between you would grow faster than if one of you had remained stationary. In both cases, the two of you are traveling in opposite directions, and your combined speed is faster than each of your separate speeds.
Now, imagine that you start running toward your friend, but your friend turns around and runs away from you. You would catch your friend more slowly than you would if your friend had remained stationary. And if you passed your friend after catching up to them, it would take you longer to grow your lead than if they had stopped running. In both cases, the two of you are traveling in the same direction, and your combined speed would be slower than each of your separate speeds.
In each of these scenarios, there is a gap between you and your friend that is either shrinking or expanding. If two bodies are traveling in opposite directions, you need to add their speeds together to find their gap speed. The gap speed is faster than each of the separate speeds. If they are traveling in the same direction, you need to take the positive difference of the speeds. The gap speed is slower than each of the separate speeds.
Instead of using two separate “dust” equations for two different travelers, you can just use one for the shrinking or expanding of the gap between the travelers. The distance is the length of the gap, the time is how long it takes for the gap to fully shrink or expand, and the speed is either the sum or the difference of the two travelers’ speeds, depending on their respective directions. This technique can greatly simplify and expedite problems like these. To help you keep track of all the parameters in the problem, you can use a diagram.
Let’s say that a car and truck are moving in the same direction on the same highway. The truck is moving at 50 miles an hour, and the car is traveling at a constant speed. At 3 pm, the car is 30 miles behind the truck and at 4:30 pm, the car overtakes and passes the truck. What is the speed of the car?
The car and truck are moving in the same direction, and the car is gaining on the truck. This means that the gap between the vehicles is shrinking and that the gap rate is the difference of the two vehicles’ respective speeds.
\(\text{S}_\text{G}=\text{S}_\text{C}-\text{S}_\text{T}\)
The distance of the gap is initially 30 miles.
\(\text{D}=30\text{miles}\)
The time frame we are given for the closing of the gap is from 3 pm to 4:30 pm.
\(\text{T}=1.5\text{hours}=\frac{3}{2}\text{hours}\)
\(\text{S}_\text{G}=\frac{\text{D}}{\text{T}}\)
\(\text{S}_\text{G}=\frac{30\text{miles}}{\frac{3}{2}\text{hours}}=30\times\frac{2}{3}=20\text{miles per hour}\)
\(20\text{miles per hour}=\text{S}_\text{C}-50\text{miles per hour}\)
\(20\text{miles per hour}+50\text{miles per hour}=\text{S}_\text{C}=70\text{miles per hour}\)
You may also encounter problem solving in motion in the con\text of a data sufficiency question. These GMAT motion problems may ask you to identify one of the “dust” elements, i.e., distance, speed, or time. To achieve data sufficiency, you will need to have both of the other elements.
The question may ask you to determine the average speed of a person making a round trip. Referring back to the round-trip average speed formula, \(\text{S}_\text{Average}=\frac{\text{2}}{\frac{1}{\text{S}_1}+\frac{1}{\text{S}_2}}\), in order to achieve sufficiency, you will need to have the person’s speeds on each leg of their trip.
Multiple traveler problems may also be featured in data sufficiency problems. In these situations, just follow the classic data sufficiency rule: you need as many equations as you have unknowns. For example, if you have two travelers, and thus two “dust” equations, you cannot achieve sufficiency unless you only have two unknowns (e.g., distance and speed of one of the travelers.)
OK, now let’s practice! The following GMAT motion problems deals with one or more of the motion topics discussed above. Happy solving!
As a result of the “Dust” formula, Speed is equal to Distance divided by Time. The object’s speed is the 374 miles it travels divided by the 11 hours in which it travels. 374 divided by 11 is 34, so the object’s speed is 34 miles per hour. The correct answer choice is A.
Since there are two legs of the trip in this average speed question, we will have two equations. D_{1}, S_{1}, T_{1}, and D_{2}, S_{2}, T_{2}.
Our S_{1} is 120 miles per hour. Our T_{1} is 5 hours. Therefore, we can find that our D1_{1} is 120 times 5, or 600 miles. Our S_{2} is 180 miles per hour. We don’t know our D_{2} or T_{2}.
However, we are solving for the time length of the entire trip, which we can call T. We know that our T_{1} is 5 hours, so we can write T_{2} as T-5 hours. Furthermore, D_{2} can be described as S_{2} times T_{2}, or \(180(T-5)\text{miles}\). If we plug the necessary values into the average speed formula, we will end up with \(\frac{(600+180(T-5))}{T}=170\).
We can multiply both sides of the equation by T. \(600+180(T-5)=170T\). Distribute the 180. \(600+180T-900=170T\). Subtract 900 from 600, and subtract 180T from both sides of the equation. \(-300=-10T\). Divide both sides by −10. \(T=30\text{hours}\).
The correct answer choice is E.
First, we will define our parameters. We’ll have D_{J}, S_{J}, T_{J}, and D_{R}, S_{R}, T_{R}. We’ll consider the point at which Rosalia passes Jacob to be our initial time. Rosalia and Jacob both eventually travel the same distance from their meeting point to Betaville, so we will set both D_{J} and D_{R} equal to D.
We are told that Jacob’s speed is 60 miles per hour; \(\text{S}_\text{J}=60\text{miles per hour}\). It takes Rosalia two hours to get from their meeting point to Betaville, so \(\text{T}_\text{R}=2\text{hours}\). Jacob takes 50 minutes or \(\frac{5}{6}\) of an hour longer, so this means that \(\text{T}_\text{J}=\frac{17}{6}\text{hours}\).
Since we are solving for Rosalia’s speed, we will call S_{R} \(x\). For Jacob, we have \(D=60\times\frac{17}{6}\). From this equation, we can determine D is 170 miles. Then for Rosalia, we have \(170=2x\). We can divide both sides of the equation by 2, and we determine that \(x\) equals 85 miles per hour.
The correct answer choice is B.
In this problem, Car M and Car N are traveling in opposite directions, and we have a shrinking gap. The gap must be shrinking at a rate that is the sum of the speeds of Car M and Car N: \(\text{S}_\text{G}=\text{S}_\text{M}+\text{S}_\text{N}\).
Car N is driving at twice the speed of Car M. \(\text{S}_\text{N}=2\text{S}_\text{M}\). Isolate S_{M}. \(\text{S}_\text{M}=\frac{1}{2}\text{S}_\text{N}\). Substitute S_{M} into the gap speed formula. \(\text{S}_\text{G}=\frac{1}{2}\text{S}_\text{N}+\text{S}_\text{N}\) Combine like terms. \(\text{S}_\text{G}=\frac{3}{2}\text{S}_\text{N}\).
The distance of the gap is the 480 miles between City J and City K. \(D=480\text{miles}\). The time frame we are given for the closing of the gap is from 10 am to 2 pm, or \(T=4\text{hours}\). Our “dust” equation is ultimately \(480=(\frac{3}{2}\text{S}_\text{N})(4)\). Multiply \(\frac{3}{2}\) by 4. \(480=6\text{S}_\text{N}\). Divide both sides by 6. \(\text{S}_\text{N}=80\text{miles per hour}\).
The correct answer choice is B.
Throughout the entire problem, Cars P and Q are still traveling in opposite directions, so their gap speed does not change. It remains 47 plus 63 or 110 miles per hour. The total distance traveled by both cars is the length of their initial 121 mile gap plus the extra 44 miles after they pass each other, or 165 miles. The total time spent traveling would then be 165 miles divided by 110 miles per hour, or 1.5 hours. This means that they will be moving away from each other and are 44 miles apart at 3:30 pm. The correct answer choice is A.
Let us know how you did on these practice questions in the comments below. If you’re looking for more GMAT motion problems, try out one of Magoosh’s GMAT plans, which comes with practice tests, video lessons, and study schedules. Good luck![crp limit=”4″ post_thumb_op=”after”]
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]]>In fact, rates are just ratios in disguise. Here are a four GMAT practice problems exploring rates and ratios. Remember: no calculator!
1) Someone on a skateboard is traveling 12 miles per hour. How many feet does she travel in 10 seconds? (1 mile = 5280 feet)
(A) 60
(B) 88
(C) 120
(D) 176
(E) 264
2) At 12:00 noon, a machine, operating at a fixed rate, starts processing a large set of identical items. At 1:45 p.m., the twenty-first item has just been processed, and 15 have not yet been processed. At what time will all 36 items be processed?
(A) 2:25 pm
(B) 3:00 pm
(C) 3:27 pm
(D) 4:13 pm
(E) 5:15 pm
3) An importer wants to purchase N high quality cameras from Germany and sell them in Japan. The cost in Germany of each camera is E euros. He will sell them in Japan at Y yen per camera, which will bring in a profit, given that the exchange rate is C yen per euro. Given the exchange rate of D US dollars per euro, and given that profit = (revenue) – (cost), which of the following represents his profit in dollars?
(A) N(YC – DE)
(B) ND(YC – E)
(C) ND((Y/C) – E)
(D) N((Y/C) – DE)
(E) ND(Y – E)/C
4) Machine A and machine B process the same work at different rates. Machine C processes work as fast as Machines A & B combined. Machine D processes work three times as fast as Machine C; Machine D’s work rate is also exactly four times Machine B’s rate. Assume all four machines work at fixed unchanging rates. If Machine A works alone on a job, it takes 5 hours and 40 minutes. If all four machines work together on the same job simultaneously, how long will it take all of them to complete it?
(A) 8 minutes
(B) 17 minutes
(C) 35 minutes
(D) 1 hour and 15 minutes
(E) 1 hours and 35 minutes
Solutions will come to these at the end of the article. Can’t contain your excitement? Click here to skip to the explanations.
Ratios are fractions. When we have an equation of the form fraction = fraction, that’s called a proportion. By far, the hardest part of dealing with a proportion is what you CAN and what you CAN’T cancel in a proportion. Many students are quite confused on this issue.
First of all, let’s be clear that cancelling is simply division. If I start with the fraction 24/32, and I “cancel the 8’s” to get 3/4, what I have really done is divide both the numerator and the denominator by 8. Similarly, if I have 5/35, and I cancel the 5’s, in the numerator, I am left with 1: the simplified version is 1/7. Too many student have the naïve view that canceling means “going away” or some other fairy-godmother operation. Instead, cancelling is a card-carrying legitimate mathematical operation, the operation of division.
Clearly, it’s always legitimate to cancel in the numerator and denominator of the same fraction, the same ratio, so of course we can do that on each side in a proportion.
We might call that “vertical canceling” in a proportion: that’s 100% legal. The one that often surprises folks is what we might call that “horizontal canceling” in a proportion, which looks like this:
Canceling a common factor from a & c would simply involve dividing both sides of the equation by the same number, a 100% legal move. Similarly, canceling a common factor from b & d would simply involve multiplying both sides of the equation by the same number, another totally legal move. Even though “horizontal canceling” across the equal sign may look suspect, it’s totally valid.
Now, the one that causes real problems is what we might call “diagonal canceling,” because so many students seem to be under the impression that is this OK, but in fact, it’s 100% illegal and incorrect.
I suspect people confuse this with “cross-canceling” in the process of multiplying fractions. I actually abhor that uses term, “cross-canceling”: I think this term causes dozens of times more harm than good. If we were to perform the canceling of a with d, that would essential be equivalent to dividing one side of an equation by a number and multiply the other side of the equation by the same number! That’s not allowed! We always have to do the same thing to both sides! This is why this kind of “diagonal canceling” in a proportion is always disastrously incorrect.
OK, that’s the relevant mathematics without the real world stuff involved!
Rates are ratios, that is, fractions. Any fraction with different units in the numerator and in the denominator is a rate: miles per hour, $ per pound, grams per cubic centimeter, etc. Most rate questions can be solved by setting up a proportion. One common proportion type involves a (part)/(whole) on each side: for example, part of the job over all of the job, and part of the price or time over all of the price or time. In setting up any rate proportion, we have to make sure that units match: the same units in the two numerators, and the same units in the two denominators. The GMAT will expect you to know a few common unit changes (e.g. 1 hour = 60 min; 1 dozen items = 12 items, etc.); because some test-takers are familiar with metric and other are familiar with English, the GMAT most often would specify the conversion, as in #1 above. In any case, a GMAT rate problem often involves reconciling units differences in some way before we can do the math.
Another pertinent topic is that of work rates. Suppose Machine P does a job in 3 hours and Machine Q can do the same job in 6 hours. How fast would it take both machines working together? You see, we can’t add or subtract the times it takes to perform jobs. What we can add are the work rates! It doesn’t matter that these work rates would have the ambiguous units of “job/hour”—it doesn’t matter as long as every rate in the problem has the same units. The rate of P, job per time, would be 1/3, which means either one job every three hours or one-third of a job every hour: either is correct. Similarly, the rate of Q would be 1/6. We can’t add or subtract times, but we can add individual work rate to find a combined work rate. Adding fractions, we get (1/6) + (1/3) = (1/6) + (2/6) = 3/6 = 1/2. The combined rate of P & Q working together is 1/2, or one job per 2 hours. Thus, if P & Q were working together, it would take them just two hours to get the job done. That is the basic logic of work rates.
If you understand the rules of fractions and the concept of work rate, there’s nothing about rates and ratios you can’t understand. If you had any “aha” moments while reading this article, give the practice problems above another look before jumping in the solutions below.
1) The speed is 12 mph. To change this to feet/second, we need to multiply by (5280 ft/mile), to cancel miles, and to multiply by (1 hour/3600 second) to cancel seconds.
So, in 10 seconds, the skateboarder moves 176 feet. Answer = (D)
2) At 1:45, that is, 105 minutes after starting, the machine has completed 21/36 = 7/12 of the job. Let T be the whole time in minutes. For the total time, set up a proportion
Remember, with proportions, we can cancel a common factor in the two numerators; cancel the factor of 7.
Now, cross-multiply, and use the doubling & halving shortcut for multiplying.
T = 15*12 = 30*6 = 180
Now, 180 minutes = 3 hours, so the task finishes 3 hours later, at 3 p.m.
Answer = (B)
3) All the other currencies are related to euros, so we should focus on getting everything to euros and then changing it all at once to dollars.
Remember that profit = revenue – cost. For one camera, cost is E euros. The revenue of one camera is Y yen: let’s change that to euros, so that we can express cost, revenue, and profit all in euros.
We have an exchange rate of C yen/euro, with yen in the numerator and euros in the denominator. If we were to multiply this, we could cancel euros and wind up with yen. We don’t want that. We want to cancel yen and wind up with euros, so we need to divide by C. Y/C is the revenue of one camera in euros.
This means that ((Y/C) – E) is the profit in euros of one camera.
Now, the other exchange rate is D dollars/euro, with dollars in the numerator and euros in the denominator. If we multiply this, we cancel euros and get dollars. That’s exactly what we want. Thus, D((Y/C) – E) is the profit, in dollars, of one camera.
Now, just multiply by the number of cameras: ND((Y/C) – E)
Answer = (C)
4) Let A, B, C, and D be the rates of Machines A, B, C, and D respectively. We know that
(i) C = A + B
(ii) D = 3C
(iii) D = 4B
Starting with (ii) and (iii), equate the two expressions equal to D, and then substitute in the expression from (i) equal to C.
4B = 3C = 3(A + B) = 3A + 3B
B = 3A
Then, C = A + 3A = 4A, and D = 3*(4A) = 12A
The combined rate,
A + B + C + D = A + 3A + 4A + 12A = 20A
Since the combined rate is 20 times faster than Machine A alone, the combined time should be divided by 20.
Machine A alone takes 5 hr 40 min, or 340 minutes for the whole job. Divide this by 20:
340/20 = 17
The combination of the four machines will take 17 minutes to complete the job.
Answer = (B)
Editor’s Note: This post was originally published in August, 2014 and has been updated for freshness, accuracy, and comprehensiveness.[crp limit=”4″ post_thumb_op=”after”]
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]]>Statement #1: x < 1
Statement #2: x > –1
5) If A is an integer, what is the value of A?
This is a tricky set of topics. Here are some previous blogs I have written about these ideas:
2) Patterns of powers for different kinds of numbers
3) Adding & Subtracting Powers
4) Roots
If you read one of those articles and have some insights, you may want to give the problems above another glance. If you would like to clarify anything said here or in the solutions, please let us know the comments section.
1) For positive numbers less than one, when we raise them to positive integer powers (square, cube, etc.), then get smaller, further from one. When we take roots of them (square root, cube root), they get bigger, closer to one. The higher the root, the closer it is to one. Thus, for any x in the 0 < x < 1 range,
Now, for negative numbers, the mirror image happens. Of course, we can’t take even roots of negative numbers (square root, fourth root, etc.) but we can take odd roots. For numbers in the –1 < x < 0 range, negative numbers with an absolute value less than one, when we take a root, the value gets closer to –1, and the higher the root, the closer it is to –1. Thus, for these numbers
Thus, the rule works in opposite directions in these two regions, (0 < x < 1) vs. (–1 < x < 0). Both of these regions are included, even with combined statements, so we can give no definitive answer to the prompt question.
Answer = (E)
2) Statement #1: Given this statement, we could square 23,100, and that would be some number between 4 & 6 times some power of ten. We don’t actually have to perform that calculation. It’s enough to note that, from this statement, we can determine the value of both A and m. Unfortunately, n remains unknown, and without n, we can’t determine the value of T. This statement, alone and by itself, is insufficient.
Statement #2: This statement tells us nothing about A, and it doesn’t allow us to determine unique values for either n or m. We can’t determine anything with just this. This statement, alone and by itself, is insufficient.
Combined statements: The first statement allows us to determine unique values for A & m. Once we know m, we can use the second statement to find n. Once we know both A & n, we can determine the value of T. Combined, the statements are sufficient. Notice, we can arrive at this conclusion without a single calculation: that’s the ideal of GMAT Data Sufficiency!!
Answer = (C)
3) First, we need to know the formula for the square of a binomial:
Thus:
Now, we need to recognize that 5 to the x and 5 to the negative x are reciprocals, so their product is one:
Also, by the laws of exponents:
Similarly,
Thus,
Subtract two from both sides, and we have
Answer = (D)
Thus, a & b are in a ratio of 4-to-3. Thus, we could say that a = 4n and b = 3n, for some n. Thus, a + b = 4n + 3n = 7n = 56, so n = 8. Well, a – b = 4n – 3n = n = 8
Answer = (B)
Statement #1 tells us that, for some unknown positive integer, N is larger than 1,000,000 and smaller than 10,000,000. Could there be more than one power of 23 in that region? No, because even if N were equal to the lowest possible value in the range, 1,000,000, if we multiply another factor of 23, it would bring us up to 23,000,000, which is above the high end of the range. The ratio of the top end to the bottom end is just 10:1 = 10, which is smaller than 23, so more than one power of 23 cannot possibly fit in this range.
Thus, there can only be one integer value A that puts N in this range. We don’t have to calculate it: it’s enough to know that the information given unique determines the value of A. Thus, this statement, alone and by itself, is sufficient.
Statement #2 is a tautology. A tautology is a statement that is true by definition, and thus gives no information. The verbal statement “My book is a book” is a tautology: it tells us absolutely nothing useful about that book. The mathematical statement x + 5 = x + 5 is a tautology: that gives us absolute no mathematical information that would allow us to solve for x; in other words, it is true for every single number on the number line.
This statement is a more sophisticated tautology. To understand this, we need to know how to add and subtract powers.
This statement is true for all numbers on the number line, so it contains absolutely no useful information for determining the value of A. This statement, alone and by itself, is insufficient.
Answer = (A)
Cube both sides. The cube or fifth power of a negative remains negative, but the fourth power of a negative is positive.
Because we know that x cannot be zero, we can divide by x cubed. Notice that (positive) x-to-the-fourth divided by (negative) x-cubed equals (negative) x.
There’s no direct way to go from this to the product xy. We have to solve for y in terms of x.
Now, we can simply take the fifth root of both sides.
Now, we are in a position to find the product.
Answer = (C)
7) For this one, we need to know how to simplify square roots. We can use the formula
Suppose we cleverly factor the number under the square root so that Q is a perfect square: then we could simplify that factor.
Notice that all three numbers under the radical sign are of the form: two times a perfect square. That gives us a big clue to simplifying them.
Answer = (A)
8) Start with the given equation
Cube both sides.
Now, square both sides.
Now, one solution is x = 0, but that’s not listed as an answer. We are looking for a non-zero answer, so if x is not zero, we can divide by x.
Now, take the fifth-root of both sides.
Answer = (D)
9) With this one, it’s helpful to rewrite the roots as a fractional exponents:
Because x ≠ 0, we can divide all three terms by . This leaves:
This is now an ordinary quadratic. To see this, use the substitution
(u + 7)(u – 3) = 0
u + 7 = 0 OR u – 3 = 0
u = –7 OR u = +3
We have to be careful here. Here, –7 is a solution for u, but not for x. Now that we have values for u, we have to solve for the corresponding values of x.
The first is not an available answer choice, but the second is answer choice (E).
10) Start with the first equation
Square both sides:
Now, multiply this by the second equation:
Answer = (C)
To find out where powers and roots sit in the “big picture” of GMAT Quant, and what other Quant concepts you should study, check out our post entitled:
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]]>6) Suppose you have access to a large vat of distilled water, several gallons large. You have two precise measuring pipettes, one to measure exactly 1/3 of an ounce and one to measure exactly 1/4 of an ounce. You can pour precisely measured amounts into a beaker, which initially is empty. You can use either pipette to remove distilled water from the vat or from the beaker and use either pipette to dispense water into either of those receptacles, but you cannot use either pipette to take any quantity of distilled water other than the amount for which it is designed. Which of the following represents, in ounces, a precise amount of distilled water you can transfer from the vat to the beaker?
I have already written a few blogs that would be germane to these topics.
1) Fractions: some of the basic ideas of fractions
2) Terminating and Repeating Decimals: an assortment of decimal ideas
3) Advanced Factoring: some sophisticated techniques for factoring ungainly looking decimals
More advanced ideas will be discussed the explanations to these problems, below.
1) First, let’s separate out the denominators and simplify them.
and
We are adding the reciprocals of these two fractions:
Answer = (E).
2) Let’s handle the numerator and denominator separately to begin. The number is the cube root of a decimal. The first thing we have to recognize is that 4 cubed is 64, so the cube root of 64 is 4. There are six total decimal places, so when we take a cube root, that will get divided by 3, down to only two decimal places. Thus,
The denominator is a little easier.
Those are the two numbers we have to divide. When divide decimals, we move both decimals an equal number of places to the right until the denominator is a whole number. Here, after we set up the fraction, we will have to move both decimals four places to the right, because the denominator starts with four decimal point.
Answer = (D).
3) There are a few different ways to think about this. First, I will multiply the entire inequality by positive 6. This will leave the direction of the inequality unchanged, and I can multiply right through the absolute value signs. This will eliminate any fractions.
|3y – 1| < 4
Well, the only way a thing can have an absolute value less than 4 is if it’s true value is between –4 and +4. Thus
–4 < 3y – 1 < 4
Add one to each term.
–3 < 3y < 5
Now, divide by +3. Because we are dividing by a positive, the direction of the inequalities stay the same.
So, y could be any positive or negative fraction between –1 and +1, so (B) & (C) & (D) are all allowed, and choice (E) is less than 5/3, so that’s also allowed. The only one that is not allowed is (A), and that’s the answer.
4) Notice that all the denominators are multiples of 30, so factor out a factor of 1/30:
Answer = (B).
5) It’s actually better to change the decimals into fractions:
Answer = (B).
6) If you fill the 1/3 oz pipette and put this into the beaker. Then use the other pipette to remove 1/4 oz from the beaker. This 1/4 oz can be put back in the vat.
Thus, there would be 1/12 oz left in the beaker: that’s the amount that would have been transferred from the vat to the beaker.
If we repeat this same procedure, we will transfer another 1/12 oz from the vat to the beaker, and 2/12 = 1/6. Therefore, we could transfer either 1/12 or 1/6 from the vat to the beaker.
There is no way to transfer 1/7 to the beaker. No combination of arithmetic involving 1/3 and 1/4 will produce 1/7.
Answer = (C).
7) Let’s think about this is in stages. First, call the entire denominator D; then (0.2)/D = 4. From this, we must recognize that D must be 1/4 of 0.2, or D = 0.05.
Now, set that denominator equal to 0.05.
0.3 – x = 0.05
x = 0.3 – 0.05 = 0.25 = 1/4
Answer = (A).
8) Let’s make things easier by breaking this into two fraction. First, let’s work with 3.3 divided by 0.015. We will begin by sliding the decimals to the right a couple spaces:
Now, notice that the denominator is 3/2, so we will replace that decimal with the fraction:
That’s the first piece of the fraction. Now, consider the rest:
Multiply the pieces
(220)*(200) = 44,000
Answer = (D).
9) For this one, we need to use some advanced factoring. Notice that
0.9996 = 1 – 0.0004
Thus, we can express this as a difference of two squares, and use that to factor it
Now, consider that 0.98 = 1 – 0.02; then
When we subtract 1, we get 0.02, which equals 1/50.
Answer = (A).
10) Think of this in stages. Call the denominator D. If 3/D = 12, then D must equal 1/4.
Now, look at the denominator. One minus thing equals 1/4, so that thing must equal 3/4.
Well, 6/c = 3/4, so c = 8.
Answer = (D).
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]]>1) In the figure above, AC = BC = 8, angle C = 90°, and the circular arc has its center at point C. Find the area of the shaded region.
2) Employees at a company will vote for an executive team of five people from eight qualified candidates. The executive team consists of a president, a treasurer, and three warrant officers. If an executive team is considered different if any of the same people hold different offices, then how many possible executive teams could be selected from the eight candidates?
3) In the above diagram, the 16 dots are in rows and columns, and are equally spaced in both the horizontal & vertical direction. How many triangles, of absolutely any shape, can be created from three dots in this diagram? Different orientations (reflections, rotations, etc.) and/or positions count as different triangles. (Notice that three points all on the same line cannot form a triangle; in other words, a triangle must have some area.)
What all three of these problems have in common is: relying on formula alone will not help you. Complete explanations will come at the end of this article.
Clearly, it’s a good idea to know some GMAT math formulas —– many of the best formulas to know are listed on that blog. Formulas can be wonderful shortcuts, very efficient time-savers. Clearly it would be a mistake to walk into the GMAT Math section without knowing a single formula.
Formulas are useful to a point, but one of the worst mistakes folks make when they study for GMAT Math is to think that all they have to do to master the GMAT Math section is to memorize all the formulas. Similarly, the worst way to approach a GMAT math section is to approach each question asking only, “What’s the best formula to use for this question?” That’s roughly the same as thinking that all I need to be the world’s best lion tamer is a fancy hat that says “lion tamer”! In fact, the GMAT regularly designs math questions that predictably punish folks who blindly memorize formulas. Most of the harder math problems on the GMAT have this property: if you approach the problem with the perspective of “what’s the best formula to use to solve this question?”, then you very likely will miss what the question is really about and get it wrong. Here are some tips you can use to escape these traps.
What do I mean by this? Yes, you have to remember some formulas for the GMAT. One option would be to memorize each one separately as an isolate mathematical factoid. Some people are exceptionally good memorizers, but for most people, this is not too helpful: in particular, if, in the stress of the real GMAT, you forget what you memorized, then you are out of luck. Memorizing allows you no recourse if you forget, and it gives you no insight into mathematical logic.
Instead, I would recommend: understand the derivation of every formula: that is, understand the logical argument that underlies the formula and from which the formula arises. Think in terms of “why is it true?”: don’t stop at simply “what is true?” For example, both the permutation and combination formulas can be derived directly from the Fundamental Counting Principle. In another blog, I talked about the derivation of the formula for the area of an equilateral triangle. For most of the formulas you need to know for the GMAT, there is a context of logical argument that is also helpful to know. One exception is Archimedes‘ formula for the area of a circle:
Yes, the great master Archimedes had a brilliant argument for why that formula is true, but that’s well beyond the kind of logic you need to employ on the GMAT. For this one, you can take a pass and simply memorize the formula, if you haven’t already. For most formulas, though, you should explore the logical of the context in depth.
Problem-solving in mathematics is a subtle and sophisticated subject, and even in the limited context of the GMAT math section, the harder questions sometimes demand insightful approaches. Formulas are very cookie-cutter: if you have a formula to do something, it’s very good at doing that one thing, but that one thing is all it can do. That’s quite different from the flexible analyses characteristic of mathematical thinking.
Think about it this way. When you approach the GMAT math section, you should have a well-equipped tool box. The individual formulas you know are tools, but each one is a highly specific tool, useful for only one thing. Other more widely applicable tools are skills such as estimation or backsolving or picking numbers. It’s important to have in your “tool box” a mix of tools, some more general and some more specific; what’s more important, though, it knowing when and how to use them.
In the post How to do GMAT Math Faster, I talk in depth about left-brain vs. right-brain thinkers. Left-brain dominant thinkers want to know “what to do,” and they love formulas, because it’s always very clear what to do with a formula. A more right-brain perspective focuses on “how to look at the problem“, on the question of the best way to frame a problem, the best problem-solving perspective to adopt. Often, on a challenging GMAT problem, when one adopts the best perspective, what to do becomes obvious. Given the correct perspective, many test-takers could solve the problem, but most folks get that problem wrong because it’s hard to come up with that right perspective on one’s own. Coming up with the best perspective for a problem, the best way to frame a problem, takes time. It’s a right-brain pattern-matching skill: these always take time & experience to master. It’s very important to read solutions carefully: left-brain thinkers will want to jump ahead to the formulas, to “what to do“, and may be frustrated because they knew all those parts already. To get the most from solutions, it’s very important to study the very beginning, any reasons given for making one perceptual choice rather than another. Sometimes, it may be helpful to solve the problem in more than one way, to experience the difference in different solution routes. Sometimes, the perceptual choices are hard to get from the solution, and the only way to find an answer is to post a question to an expert in the forums, asking them: “why do we get the solution this way but not in that way?” Through asking questions and investigating, over time one develops more of a sense for problem-solving perspective.
Yes, know the GMAT formulas, but don’t think they are a magic bullet. You will get considerably more from understanding the logic behind each formula, than you will from blind memorization of only the formula. Finally, it’s most important to develop the problem-solving perspective, which will help you to see which formulas to use when. Here’s another practice question from inside the Magoosh product:
4) http://gmat.magoosh.com/questions/124
If you would like to share your own experience with formulas, or would like to clarify something I have said, let us know in the comments section.
1) There’s not a single formula we can use to get the answer, but by combining a few formulas, we can calculate this.
First, think about the circle. The circle has radius r = 8, so its total area would be
This entire figure is a quarter of the circle, so that area would be
Now, the shaded area (technically known as a circular segment), would have an area of
(circular segment) = (quarter circle) – (triangle ABC)
Well, we already have the area of the quarter circle. The triangle would have an area of (1/2)bh = 32. Therefore, the area of the segment is
Answer = (B)
2) We can use a quick formula for this. We have to go back to the Fundamental Counting Principle and think this through.
Choice #1: for the president, we have eight choices
Choice #2: for the treasurer, we have seven remaining choices
Choice #3: we have to pick 3 warrant officers from the remaining 6. This would be
The FCP tells us to multiply the number of choices in each selection:
8*7*20 = 56*20 = 1120 choices — Answer = (D)
3) This is a very subtle counting problem. This post on difficult counting problems may provide some insights.
The basic strategy will be to calculate the total number of sets of three points we can select in this diagram, and then subtract the small number of sets of three collinear points.
The first step: in how many ways can we select three points from a set of 16 points?
There are 560 sets of three points we can select. Right away, notice, we can eliminate answer choices (C), (D), and (E) as too big.
How many sets of three points are collinear? Well, first look at the column. In any column, we have four collinear points, so we could eliminate any one point, and we would be left with a unique set of three collinear points. That means we have four collinear sets in each column, or 16 vertical sets of collinear points. By symmetry, we must have 16 horizontal sets of collinear point. That’s 32 together.
But that’s not all. Notice there are two long diagonals, of four points each: much in the same way as in the argument above, there must be four unique sets of three collinear points along each diagonal. For the two diagonals, that’s 8 more. We up to 40 now.
Now, notice the short diagonals of only three points:
That’s the last four. There are a total of 44 unique sets of three collinear points. Since any of these would not constitute valid triangles, we have to subtract this from the total number of sets of three points.
560 – 44 = 516 —- answer = (A)
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]]>2) In the month of August, Pentheus Corporation made $200,000 in profit. Pentheus made 6% of that profit on the second Wednesday of August. If the profits that day were approximately 14.5% of the revenue for that day, then what was Pentheus’s revenue on the second Wednesday of August?
3) Harold needs to buy a ticket to attend a conference for work. His own department contributes $4 less than half the price of the ticket. The HR department will contributes $1 more than a third of the price of the ticket. With these two contributions, Harold has to pay only $10 out of his own pocket to cover the cost of the ticket. What was the price of the ticket?
4) Jackson invested $300,000, dividing it all unequally between Account P and Account Q. At the end of the year, it turned out that Account P had earned 12% interest and Account Q had earned 25% interest. If Jackson earned a total of $60,000 in interest between the two accounts, which of the following is approximately the amount he put in account P?
Solutions will come at the end of this article.
When all five answers are numerical, that puts us in an excellent position to use backsolving. Backsolving is an alternative to the algebraic method of solution that your wizened Algebra Two teacher would have deemed the only correct way. Backsolving means starting with a numerical answer, and working through with that number to see if it fits the requirements of the scenario.
In a previous post, I recommended the #1 backsolving strategy: start with answer (C). You see, if all five answer are integers, the GMAT always lists them from smallest to largest. Answer choice (C) will always be in the middle. If we plug in (C) and it comes out to the correct value, then we are done. If it is too big, then we can eliminate (C) & (D) & (E); if it is too small, then we can eliminate (A) & (B) & (C). Choice (C) is often the best place to start, because whether it turns out to be too high or too low, there’s more than one answer to eliminate.
Make it easy!
Of course, backsolving is a strategy that should be used to make things easier. Suppose a problem presents a scenario, and then has these answer choices:
(A) 14,641
(B) 15,000
(C) 16,384
(D) 17,711
(E) 19,683
Well, hmmm. Whatever the calculations in the problem may be, starting with (C) doesn’t seem like it would be a whole lot of fun without a calculator. Also, notice the answers are too close together to estimate. BUT, notice that of the five answer choices, here answer choice (B) is a nice neat round number. In a way, the design of the answer choices presents (B) on a silver platter as the best possible choice for backsolving. If you decide to backsolve, don’t simply go on automatic pilot and choose (C). Scope out the answers, and see if one stands out as a much easier choice for backsolving calculations: if there is such answer choice available, chances are very good that the question-writer placed it for just that purpose.
The GMAT math section demands flexibility, and backsolving is an excellent alternative approach to have up your sleeve. If you had any insights reading this post, you may want to give the problems at the top another look before reading the explanations below. Here’s another practice problem that lends itself to backsolving.
5) http://gmat.magoosh.com/questions/29
If you have any questions about what I have said here, or if you would like to add anything, please use our comments section below.
1) We’ll use backsolving. Start with choice (C). If x = 4, then
That’s just larger than 80, so this may well be the answer. We have to verify that x = 3 is below 80:
That’s well below 80, so x = 4 is the lowest integer for which the expression is greater than 80. Answer = (C).
2) First of all, 1% of $200,000 is $2000, and six times this means 6% is $12,000. That’s the easy part of the problem. Now we know 14.5% of revenue is $12,000.
As we discussed in this post, starting with (C) in this problem would not be fun. Instead, we’ll backsolving starting with the round number, choice (D). Suppose revenue = $80,000.
(a) 1% of $80,000 = $800
(b) 10% of $80,000 = $8,000
(c) Multiply (a) by 4: 4% of $80,000 = $3,200
(d) Divide (a) by half: 0.5% of $80,000 = $400
(e) add (b) & (c) & (d): 14.5% of $80,000 = $11,600
That’s a bit shy of the required $12,000, so revenue must be larger than $80,000. That leads up immediately to the last choice. Answer = (E)
3) We’ll use backsolving. Start with choice (C). If price = 48, then
department pays $4 less than half = $24 – $4 = $20
HR pays $1 more than a third = $16 + $1 = $17
Together, they pay $37, leaving Harold to pay the remaining $11. Harold pays a bit too much, so the price must be a bit less. We strongly suspect that (B) will work, but we have to verify that assumption.
Try (B). If price = $42, then
department pays $4 less than half = $21 – $4 = $17
HR pays $1 more than a third = $14 + $1 = $15
Together, they pay $32, leaving Harold to pay the remaining $10. This works.
Answer = (B).
4) As we discussed in this post, starting with (C) in this problem would not be fun. We’ll backsolving starting with a round number. The answers give us two round numbers, choice (B) and choice (E). Let’s think strategically about this. If we try (E) and it works, then we have the answer, but if it doesn’t work, we would know the amount would have to be less, but we can already see that. Backsolving with (E) does not necessarily give us a lot of information.
Instead, backsolving with (B). Let’s say Jackson puts $120,000 in P and $180,000 in Q.
In P, Jackson earns 12% of $120,000
10% of $120,000 = $12,000
1% of $120,000 = $1,200
2% of $120,000 = $2,400
12% of $120,000 = $14,400
In Q, Jackson earns 25% of $180,000
50% of $180,000 = $90,000
half of that: 25% of $180,000 = $45,000
Total interest earned = $14,400 + $45,000 = $59,400
Not quite enough interest. To earn that extra $600 of interest that would bring the total interest up to $60,000, Jackson would have to have taken money out of P, which earns less interest, and have put it in Q, which earns more interest. That means, the amount in Account P must be less than $120,000. That leaves only one answer.
Answer = (A)
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]]>Full solutions will come at the end of this article.
First of all, let’s review what should be more familiar — the arithmetic of equations. Suppose A = B and P = Q. The soundbyte is: you can combine them in almost any way imaginable to get a new valid equation. You can add them, in either order (A + P = B + Q) or (A + Q = B + P). You can subtract then, either one from the other, in either order (four subtraction equations: e.g. A – P = B – Q). You can multiply them in either order (A* P = B * Q) or (A * Q = B * P). You can divide either one by either other (assuming you are not dividing by zero), in either order (four division equations: e.g. A/P = B/Q). You can raise either one to the power of the other, in either order (four exponentiation equation: e.g. A^P = B^Q).
(Note: for the GMAT, you are responsible for know what happens when you raise a base, say 3, to the power of integers or to the 1/2. Technically, you can raise any positive base to any power — say, 3 to the pi — but that’s more complicated than the GMAT expects you to know. For negative bases, things get dicier. For the GMAT, just worry about negative bases to integer powers, (–2)^3 or (–2)^(–1). Non-integral powers of negatives leads into complex numbers — again, beyond what the GMAT expects you to know).
Finally, the above may seem simple, and if A & B & P & Q are just individual numbers, then those equations are pretty much “duh”-simple. Things get considerably more interesting if some or all of those letters are not individual numbers but algebraic expressions. Even with four different algebra expressions, all this still holds for equations.
Everything gets trickier with inequalities. First of all, an equation, such as A = B, is inherently symmetrical and “two-sided”, but an inequality is more a one-sided, unidirectional thing. With any arithmetic of inequalities, we must consider the direction of the inequality. Furthermore, equations are very intuitive, but some of the arithmetic of inequalities is quite unintuitive.
Adding inequalities is not so bad: you can add inequalities with the same direction of inequality. Thus, if A > B, and P > Q, then it must be true that A + P > B + Q. That always works, and it is in many ways what you’d expect.
This is the one that’s much trickier. If A > B, and P > Q, then naïvely one might expect that (A – P) would be greater than (B – Q), but that’s not necessarily the case. For example, suppose we have 10 > 5 and 2 > 1 — then we could subtract them in the same direction of inequality, and we’d get 8 > 4, which still works. BUT, suppose 10 > 5, and 100 > 1, both true — now, if we subtract in the same direction of inequality, we find that (–90) is not greater than 4. If we subtract two inequalities in the same direction of inequality, we may get another true statement, but we are not guaranteed that this will lead to a true statements, so it’s a very unwise move in problem solving.
Here’s the real head-scratcher: we can’t subtract inequalities with the same direction of inequality, but we CAN subtract inequalities with the opposite direction of inequality — in other words, if A > B, and P > Q, then it must be true that (A – Q) > (B – P). That may be perplexing symbolically, so think about it in real world terms. Suppose Ann has more money than Bob. Suppose I take less from the richer person, Ann, and more from the poorer person, Bob. When I am done, Ann still will have more than Bob — in fact, a greater difference than existed between them before!
Everything gets much hairier with multiplying or dividing inequalities when you consider — one or both sides could be negative. Hmmm. If we multiply or divide both sides of inequality by a single negative number, that’s perfectly legal, but we must remember to reverse the direction of inequality. What happens if we were to multiply or divide inequalities and negatives were involved? For example, if we know that x > –6 and y > –7, then what can we say about the product xy? As it turns out, we could pick an x and a y that would satisfy the original inequalities and make the product xy equal absolutely any number on the number line. This mindboggling bit of math is just to demonstrate why the GMAT is not going to touch multiplying or dividing general inequalities with a ten-foot pole!
Let focus, though, on a special case that could, in very advance problems, appear on the GMAT. Suppose we know that all four numbers or expressions are positive: A > B > 0 and P > Q > 0. Then, as with addition, we can multiply inequalities with the same direction: A*P > B*Q must be true. And, as with subtraction, we can divide inequalities with the opposite direction: A/Q > B/P. Again, remember the caveat: everything must be positive for these patterns to work. If anything can be negative, things get much more complicated, so complicated that the GMAT won’t ask about them.
Absolute value inequality is a sizeable topic, with some mind-bending twists and turns, but chances are very good the GMAT is not going to probe this topic all that deeply. In fact, probably most of the absolute value inequalities on the GMAT will be of the form: |x – p|, where p is some given fixed number.
Here, we must remember a few key mathematical facts. First of all, subtraction gives us distance on the number line. Technically, subtraction gives signed distance. What do I mean by that? Well, 5 – 2 = +3, indicating that it’s a signed distance of positive 3, i.e. three units to the right, from 2 to 5; by contrast, 2 – 5 = –3 indicating that it’s a signed distance of negative 3, i.e. three units to the left, from 5 to 2.
For ordinary distance, distance in the geometric sense of the word, we don’t care about sign or direction — the distance between two points is just a positive number and is the same, whether from A to B or from B to A. That’s where absolute value comes in. The expression |p – q| is the distance between numbers p & q on the number line. That is a HUGE idea.
Thus, |x – 5| is the distance between variable point x and fixed point 5. The expression |x – 5| < 2 indicates the set of all points x that have a distance to the point 5 of less than two. Immediately, just thinking about this logic, and without any further calculations, we can see that |x – 5| < 2 is entirely equivalent to 3 < x < 7. Recognize that this is exactly the kind of math the GMAT adores: with simple logic, you can jump to conclusions without having to do any calculations. The GMAT goes crazy for math of this sort. I can’t emphasize enough how important this particular set of logically interconnected ideas is. If the GMAT asks you anything at all about absolute value inequalities, it is highly likely it would be something of this genre, and unlikely that it would be anything else in this extensive topic.
If you had some insights reading this article, I encourage you to take another look at the practice questions at the top before reading the solutions below. Here’s a similar question from inside Magoosh.
5) http://gmat.magoosh.com/questions/960
If you have anything to add, or any questions, please let us know in the comment section below.
1) In this problem, the tempting incorrect answer would be (A), or somehow would involve statement #1 as sufficient, but it’s not. This gets into an idea discussed in this blog —- we can’t subtract inequalities with the same direction of inequality, but we can subtract inequalities with the opposite direction of inequality.
Statement #1: So the prompt inequality and this statement’s inequality would be true for a + b = 15, c + d = 7, b = 6, d = 2
prompt inequality: 15 > 7 — true
statement #1 inequality: 6 > 2 — true
prompt question: 9 > 5 — a > c, an answer of “yes”
But, both inequalities would still be true for a + b = 20, c + d = 18, b = 15, d = 3
prompt inequality: 20 > 18 — true
statement #1 inequality: 15 > 3 — true
prompt question: 5 < 15 — a < c, an answer of “no”
We can make different choices consistent with all the given statements that would produce either a “yes” or “no” answer to the prompt question. This means that, given this statement, we are unable to determine a unique definitive answer to the prompt question. This statement, alone and by itself, is insufficient.
Statement #2: Here, we are allowed legitimately to subtract the inequalities, because the directions of inequality are in opposite order. When we subtract (b < d) from (a + b > c + d), we get a > c, a definitive “yes” answer.
Another way to think about it: (a + b) = LARGER, and (c + d) = SMALLER, so of course, the former is greater than the latter. Now, say that b = tiny, and d = bigger, so of course, b < d. Now, think about the two differences we will compare:
(i) a = (a + b) – b = LARGER – tiny
(ii) c = (c + d) – d = SMALLER – bigger
Obviously, it start with something LARGER, and subtract something tiny, the result will be greater than starting with something SMALLER and subtracting something bigger. Therefore, a > c, a definitive answer to the prompt question.
This statement allows us to give a definitive “yes” answer to the prompt, so this statement, alone and by itself, is sufficient.
Answer = B
2) Statement #1: if p = 2 and q = 1, then this statement’s equation is true, 4 > 1, and p > q, so the answer to the prompt is “yes”.
But, if p = –2 and q = –1, then it’s still true that the square of p is larger than the square of q, 4 > 1, but now it’s true that p < q, so the answer to the prompt question is “no.”
We can make different choices consistent with all the given statements that would produce either a “yes” or “no” answer to the prompt question. This means that, given this statement, we are unable to determine a unique definitive answer to the prompt question. This statement, alone and by itself, is insufficient.
Statement #2: if p = 2 and q = 1, then this statement’s equation is true, 8 > 1, and p > q, so the answer to the prompt is “yes”.
Switching to negatives won’t make a difference, because the cube of a negative is still negative. The values p = –2 and q = –3 satisfy the statement inequality, and it’s still true that p > q
What about fractions? If p = 1/2 and q = 1/3, then p cubed is still larger than q cubed, (1/8) > (1/27), and p is still greater than q. No matter what numbers we pick, the inequality in statement #2 directly implies the prompt inequality. The mathematical way to say this is: cubing, or taking a cube-root, preserves the order of any inequality.
This statement gives us a definitive answer of “yes” to the prompt question, so this statement, alone and by itself, is sufficient.
Answer = B
3) Think about the distance interpretation of absolute value inequalities. We want to know: is x further than two units away from 6 on the number line?
Statement #1: the x-values allowed by this statement are x’s that are more than three units from 4. Here’s a picture of these values, in green:
Notice, the endpoints, 1 & 7, are not allowed, because they are exactly three units from 4, and exactly 3 is not greater than 3. Most of these points are further than two from the point 6, but some, such as the point 7, is closer than two units to six. Thus, given this constraint, we could find many points that produce a “yes” answer to the prompt, but some that produce a “no” answer. Different choices give different answers. This means that, given this statement, we are unable to determine a unique definitive answer to the prompt question. This statement, alone and by itself, is insufficient.
Statement #2: the x-values allowed by this statement are x’s that are more than one unit from 8. Here’s a picture of these values, in green:
Again, notice the endpoints are not included. Many of these points are further than two from the point 6, but some, such as the point 7, is closer than two units to six. In fact, the point 6 itself — which is a distance of zero units from 6 — is allowed by this statement! Thus, given this constraint, we could find many points that produce a “yes” answer to the prompt, but some that produce a “no” answer. Different choices give different answers. This means that, given this statement, we are unable to determine a unique definitive answer to the prompt question. This statement, alone and by itself, is insufficient.
Combined statements: Now, we combine the constraints of the individual statements. Now, the allowed points must be both more than three units from 4 and more than one unit from 8. Shown in green are the points that simultaneously satisfy both constraints:
Now, all the green points are more than two units away from 6, and it absolutely impossible to pick a value of x that simultaneously satisfies the constraints of both statements and is closer than two units to 6. The combined statements allow for a definitive “yes” answer to the prompt. Together, the statements are sufficient.
Answer = C
4) Statement #1: this gets at the fundamental meaning of the inequality. If y is one more than x, it must be greater than x. To add one to a number means to move it one unit to the right on the number line, so y must be one unit right of x, which means it is greater than x. This allows us to determine a definite “yes” to the prompt question. This statement, alone and by itself, is sufficient.
Statement #2: There are a couple ways to think about this one. One is to treat different categories of numbers.
(i) if x is negative, then by squaring, y will be positive, and y > x.
(ii) if x = 0, y = 1, and y > x
(iii) if x is a fraction between 0 and 1, then its square will also be between 0 and 1, and adding one to this will produce a number greater than 1, between 1 and 2. Therefore, y > x.
(iv) if x = 1, y = 2, and y > x
(v) if x > 1, then squaring x makes it bigger, and adding one makes that even bigger, so y > x
Thus, for all possible values of x, y > x. Thus, the prompt gives a definitive “yes” answer.
A totally different way to think about it: using coordinate geometry. The graph of y = (x^2) is a parabola that passes through the origin. The graph of y = (x^2) + 1 is this same parabola shift up one in the y-direction, passing through (0, 1) instead of the origin. Now, compare this shifted parabola to the line y = x. One of the special properties of the line y = x is that all points above this line, regardless of quadrant, have the property y > x. Think about the graph:
The parabola is always above the line y = x, so every point on the parabola must satisfy y > x.
Either way, this allows us to determine a definite “yes” to the prompt question. This statement, alone and by itself, is sufficient.
Both statements are individually sufficient.
Answer = D
To find out where inequalities sit in the “big picture” of GMAT Quant, and what other Quant concepts you should study, check out our post entitled:
What Kind of Math is on the GMAT? Breakdown of Quant Concepts by Frequency[crp limit=”4″ post_thumb_op=”after”]
The post GMAT Quant: Arithmetic with Inequalities appeared first on Magoosh Blog — GMAT® Exam.
]]>The post GMAT Quantitative: Ratio and Proportions appeared first on Magoosh Blog — GMAT® Exam.
]]>1) A certain zoo has mammal and reptiles and birds, and no other animals. The ratio of mammals to reptiles to birds is 11: 8:5. How many birds are in the zoo?
Statement (1): there are twelve more mammals in the zoo than there are reptiles
Statement (2): if the zoo acquired 16 more mammals, the ratio of mammals to birds would be 3:1
2) In a certain large company, the ratio of college graduates with a graduate degree to non-college graduates is 1:8, and ratio of college graduates without a graduate degree to non-college graduates is 2:3. If one picks a random college graduate at this large company, what is the probability this college graduate has a graduate degree?
3) Dan’s car gets 32 miles per gallon. If gas costs $4/gallon, then how many miles can Dan’s car go on $50 of gas?
4) For a certain concert, the price of balcony tickets was exactly half the price of orchestra tickets. The ratio of balcony to orchestra tickets sold was 3:2. What was the price of one orchestra ticket?
Statement (1): the total revenue taken in from tickets of both kinds was $4200
Statement (2): the difference between the number of balcony tickets sold and the number of orchestra tickets sold was 40
5) At a certain high school, there are three sports: baseball, basketball, and football. Some athletes at this school play two of these three, but no athlete plays in all three. At this school, the ratio of (all baseball players) to (all basketball players) to (all football players) is 15:12:8. How many athletes at this school play baseball?
Statement (1): 40 athletes play both baseball and football, and 75 play football only and no other sport
Statement (2): 60 athletes play only baseball and no other sport
6) In a certain class, the ratio of girls to boys is 5:4. How many girls are there?
Statement (1): If four new boys joined the class, the number of boys would increase by 20%.
Statement (2): If the number of girls increases by 50%, then after such an increase, the probability that a randomly chosen student would be a boy would be 8/23
7) If 28 passes to a show cost $420, then at the same rate, how much will 42 passes cost?
8) Metropolitan Concert Hall was half full on Tuesday night. How many seats are in the Hall?
Statement (1): If the number of people in the Hall increased by 20% from Tuesday night to Wednesday night, then the Hall would be 60% full on Wednesday night.
Statement (2): If 20 more people showed up on Tuesday night, that would have increased the number of people in the Hall by 4%.
Answers will come at the end of the article.
Probably ratios and proportions have been on your radar since some time in grade school or middle school, when they are introduced. You remember there are a lot of “mathy” facts related to these things, but it’s a bit blurry. Here are a few quick facts as reminders.
1. What is a ratio? Fundamentally, a ratio is a fraction, and is subject to all the laws of fractions. “Ratios” and “fractions” are mathematically identical.
2. What is a proportion? Is it the same as a ratio? No, a proportion is NOT the same as a ratio. Whereas a ratio is single fraction by itself (e.g. 1/3), a proportion is an equation that sets two ratios equal to each other (e.g. 1/3 = 4/12). See the fraction post for a refresher on what you can and can’t cancel in a proportion.
3. When geometric figures are similar, the sides are proportional. Geometric similarity is a topic rife with ratios and proportions. One helpful idea discussed in that post is the idea of a scale factor, which, it turns out, is helpful in many many proportional situations well beyond anything having to do with geometry.
4. Percents and probabilities are specialized cases of ratios, and either gives you very much the same kind of information.
Those are four simple ideas, and there’s one more, but it takes a little setting up to express. Let’s divide the mathematical information that can appear in a problem into two categories. The first category is “ratio information”, and this includes any statements about percents or probability information. The second I’ll call, for lack of a better term, “count information” — not a percent or ratio, but an actual count — i.e. how many people or animals or whatever; it could be how many in any particular group, the sum or difference of multiple groups, or how many are in the whole population. This leads us the final important simple idea:
5. To get count information as an output, you need some count information as an input. If you all you have is ratio information as an input, it is impossible to get count information as an output.
Suppose, in some population, there are three kinds of things, A & B & C, and they are in a proportion of 3:8:4. That’s ratio information. Suppose we want to know either the percent that A makes of the whole, or the ratio of A to the whole. That’s also ratio information, so we should be able to calculate it from that given ratio.
It can be very helpful to understand ratios in terms of “parts”. For every 3 parts of A, there are 8 parts of B and 4 parts of C. To get the whole in the same ratio, we simply add up the parts — 3 + 8 + 4 = 15 parts. Keep in mind, we have zero information about the actual size of the population — we have no count information. This simply indicates the size of the population in the same ratios, so we could say A to the whole is 3/15 = 1/5, and B to the whole is 8/15, and C to the whole is 4/15. This means that A is 1/5 of the whole, or 20%.
Similarly, suppose a class is made up of people with brown eyes and people with blue eyes. 4/7 of the class have brown eyes. Those with brown eyes are four parts, and the whole is seven parts, so those with blue eyes must be the other three parts. Proceeding, we see that those with blue eyes must be 3/7 of the class, the ratio of blue eyes to brown eyes must be 3:4. Thinking about “parts” and portioning can be a powerful way to expand the information you get from any given ratio.
If you had any realizations while reading this blog, you may want to go back and give the practice problems another glance, before proceeding to the solutions. Here’s another practice question, involving two less-than-lovable baseball teams:
9) http://gmat.magoosh.com/questions/60
If you have any questions about this article, please let us know in the comment section at the bottom!
1) A short way to do this problem. The prompt gives us ratio information. Each statement gives use some kind of count information, so each must be sufficient on its own. From that alone, we can conclude: answer = D. This is all we have to do for Data Sufficiency.
Here are the details, if you would like to see them.
Statement (1): there are twelve more mammals in the zoo than there are reptiles
From the ratio in the prompt, we know mammals are 11 “parts” and reptiles are 8 “parts”, so mammals have three more “parts” than do reptiles. If this difference of three “parts” consists of 12 mammals, that must mean there are four animals in each “part.” We have five bird “parts”, and if each counts as four animals, that’s 5*4 = 20 birds. This statement, alone and by itself, is sufficient.
Statement (2): if the zoo acquired 16 more mammals, the ratio of mammals to birds would be 3:1
Let’s say there are x animals in a “part” — this means there are currently 11x mammals and 5x birds. Suppose we add 16 mammals. Then the ratio of (11x + 16) mammals to 5x birds is 3:1. —-
(11x + 16)/(5x) = 3/1 = 3
11x + 16 = 3*(5x) = 15x
16 = 15x – 11x
16 = 4x
4 = x
So there are four animals in a “part”. The birds have five parts, 5x, so that’s 20 birds. This statement, alone and by itself, is sufficient.
Both statements sufficient. Answer = D.
2) We are given ratio information, and we are asked for ratio information: a probability. That’s fine. For simplicity, let’s use the abbreviations:
P = college graduates with a graduate degree
Q = college graduates without a graduate degree
R = non-college graduates
The two ratios we are given is
P:R = 1:8
Q:R = 2:3
We have combine the ratios, by making the common term the same. The non-college graduates, R, are the common member, accounting for 8 parts in the first ratio, and 3 parts in the second, so we have to multiply the first ratio by 3/3, and the second by 8/8.
P:R = 1:8 = 3:24
Q:R = 2:3 = 16:24
P:Q:R = 3:16:24
Now, the probability about which the question asks is about only college graduates, so ignore the non-college graduates, and just focus on the ratio among college graduates:
P:Q = 3:16
There are 3 + 16 = 19 parts in total, and of those, 3 are the folks in P, so that’s a probability of 3/19. Answer = (D).
3) To get from dollars to gallons, we have to start with the $50, and divide by ($4/gallon) — that cancels the unit of dollars, and leaves us with gallons —– 50/4 gallons. (For the moment, I’ll leave it as that un-simplified fraction).
Now, we need to get from gallons to distance. We multiply by (32 miles/gallon), to cancel the units of gallons, and leave only miles —-
(50/4) gallons*(32 miles/gallon) = (32*50)/4 miles = (8*50) miles = 400 miles
Notice, we didn’t do anything with the four in the denominator until we got a 32 in the numerator with which we could cancel.
Answer = (C)
4) We know the price of a balcony ticket, B = (1/2)*C, or 2B = C, where C = the price of orchestra ticket. The 3:2 ratio tells us that, for some n, the concert sold 3n balcony tickets and 2n orchestra tickets.
Statement (1): the total revenue taken in from tickets of both kinds was $4200
We know total revenue would be (3n) balcony ticket plus (2n) orchestra tickets:
(3n)*B + (2n)*C = (3n)*B + (2n)* 2B = 3nB + 4nB = 7nB = $4200, or nB = 600. The problem with this — we have two variables, n and B. In one extreme case, we could say n = 1 (sold 3 balcony ticket and 2 orchestra tickets), and a balcony ticket cost $600. At another extreme, we could say n = 600 (sold 1800 balcony ticket and 1200 orchestra tickets) and a balcony ticket cost $1. This information alone does allow us to determine a definitive answer to the prompt question. This statement, alone and by itself, is insufficient.
Statement (2): the difference between the number of orchestra tickets sold and the number of balcony tickets sold was 40
Well, balcony tickets are 3 parts, and orchestra tickets are two parts, so there’s one part of difference between them. The statement lets us know: one part = 40, so that allows us to figure out: we sold 3*40 = 120 balcony tickets, and 2*40 = 80 orchestra tickets. That’s great, but unfortunately, with this statement, we get absolutely no financial information, so we can’t solve for a price. This statement, alone and by itself, is insufficient.
Combined statements: now, put this altogether. From the first statement, we got down to the equation nB = 600, and the second statement, in essence, tells us n = 40. Therefore, B = 600/40 = 60/4 = $15. That’s the price of a balcony ticket. The price of an orchestra ticket is twice that, $30. With both pieces of information, we were able to solve for this. Together, the statements are sufficient.
Statement are sufficient together but not individually. Answer = C
5) This is a tricky one. The ratio 15:12:8 double-counts some students. In terms of a triple Venn diagram:
So, in that diagram
a = folks who play baseball only
b = folks who play football only
c = folks who play basketball only
d = folks who play baseball and football only
e = folks who play baseball and basketball only
f = folks who play football and basketball only
g = folks who play all three sports
We know from the prompt that g = 0, but at the outset, that’s still six unknowns!!
Now, notice:
everyone who plays baseball = a + d + e
everyone who plays basketball = c + e + f
everyone who plays football = b + d + f
So, the ratio given in the problem is:
(a + d + e):(c + e + f):(b + d + f) = 15:12:8
Of course, these are all fractions, so we can’t simplify: there is no way to simplify.
We would like to find the total number of baseball players, a + d + e.
Statement #1 tells us that d = 40 and b = 75. Thus
total number of baseball players = a + 40 + e
We don’t have enough information to calculate this, and we don’t have enough ratio information to solve. This statement, alone and by itself, is insufficient.
Statement #2 tells us a = 60. Thus
total number of baseball players = 60 + d + e
We don’t have enough information to calculate this. This statement, alone and by itself, is insufficient.
Combined statements: Now we know a = 60, b = 75, and d = 40.
total number of baseball players = 60 + 40 + e= 100 + e
total number of football players = 75 + 40 + f = 115 + f
We know the ratio of these two quantities, baseball to football, is 15:8. Unfortunately, that would give us only one equation for two unknowns, e & f. If we can’t solve for these, then we can solve for the total number of baseball players. Even with both statements combined, we cannot determine the answer.
Both statements combined are insufficient. Answer = E
6) Statement (1): If four new boys joined the class, the number of boys would increase by 20%. This means, those four new boys count as 20% of the original boys, which means 2 new boys would be 10% of the original boys, which means there must have been 20 original boys. If we know how many boys, we can use the ratio to calculate how many girls. This statement, alone and by itself, is sufficient.
Statement (2): If the number of girls increases by 50%, then after such an increase, the probability that a randomly chosen student would be a boy would be 8/23. The prompt gives us “ratio information”, and this statement also gives us more “ratio information”. We have absolutely no “count information”, so we can’t figure out the count or number of anything. This statement, alone and by itself, is insufficient.
First statement sufficient, second insufficient. Answer = A
7) We can solve this question by setting up a proportion:
Divide the left fraction, numerator and denominator, by 7:
Now, cancel the factor of 4:
For this multiplication, use the doubling & halving trick — double 15 is 30, and half of 42 is 21:
x = (15)*(42) = 30*21 = $630
Answer = C
8) The prompt gives us “ratio information”.
Statement #1 also gives us “ratio information”, so there is no way we can calculate a count, such as total number of seats in the hall. This statement, alone and by itself, is insufficient.
Statement #2: 20 people would be 4% of the audience. Divide by two — 10 people would have be 2%; now multiply by five — 50 people would have been 10% of the audience; now multiply by ten — 500 people would be 100% of the audience on Tuesday night. Since the concert hall was half full, there must be 1000 seats in total. This piece of information allows us to solve for the answer to the prompt question. This statement, alone and by itself, is sufficient.
First statement insufficient, second sufficient. Answer = B
To find out where ratios and proportions sit in the “big picture” of GMAT Quant, and what other Quant concepts you should study, check out our post entitled:
What Kind of Math is on the GMAT? Breakdown of Quant Concepts by Frequency[crp limit=”4″ post_thumb_op=”after”]
The post GMAT Quantitative: Ratio and Proportions appeared first on Magoosh Blog — GMAT® Exam.
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]]>Priority #1: P = Parentheses
Priority #2: E = Exponents
Priority #3: MD = Multiplication & Division
Priority #4: AS = Addition and Subtraction
A few things to notice about this. First of all, within the parentheses, you can have any combination of symbols, so the order of mathematical prioritization would be repeated inside the parentheses (exponents inside the parentheses, then multiplication & division inside the parenthesis, then addition & subtraction inside the parenthesis); once the entire contents of the parentheses have been simplified to a single number, the entire parenthetical expression, parentheses and all, get replaced by this single number, and then outside-the-parentheses operations can proceed according to their prioritization levels. If we have nested parentheses, then this process is extended, with the innermost parentheses getting the highest prioritization.
Notice also: multiplication and division are at the same level of mathematical prioritization. Some mathematicians would say that, really, division doesn’t exist: when we think we are “dividing by 7”, we are really multiplying by 1/7. To some extent, that’s a semantic distinction, but it does underscore the profound relationship between multiplication and division. As long as we don’t change which factors are multiplied and by which we divide, we can swap around the order in any way we like. For example:
Each of those is 100% correct, and they are all perfectly valid reorganizations of each other. One extremely practical implication of this concerns how you might deal with, for example, the following fraction:
Suppose you need to simplify this in a GMAT problem. Many people, doing math on automatic pilot, will begin by multiplying 48 times 91. Since you get no calculator on the GMAT Quant section, this beginning is just about the most boneheaded move one could possibly make. Starting with that multiplication is the mathematical equivalent of driving from NYC to Boston by route of Albuquerque, NM. Now, Albuquerque, NM is an truly lovely town in its own right, but if you start in NYC and just want to get to Boston, Albuquerque is more than a little bit out of the way! If you multiply 48 time 91, to get whatever four-digit monstrosity you get, you have gone way out of your way and you have made the problem 15 times harder than it has to be. Instead, CANCEL BEFORE YOU MULTIPLY. For example
ON the GMAT, there is almost NEVER a reason you will have to do any kind of significant arithmetic with anything larger than a two-digit number. If you wind up with a four-digit number that you have to multiply or divide, know that you inadvertently have made a choice somewhere along the way to make the problem much harder than it had to be.
The foregoing section address the aggressive and productive cancelling you are 100% allowed to employ if everything is either multiplied or divided — if there’s no addition or subtraction mucking up the works. What kind of cancelling is allowed if there is addition or subtraction involved? Consider the following question. (This is not a standard GMAT Quant question format: this “multiple answer” format never appears on the GMAT, but does appear on the GRE).
In my teaching career, I have seen analogs of each of the nine of these appear, at various times, in a student’s work as something that the student believes is a valid form of cancelling. In fact, only (H) is correct, and the other eight are hideously incorrect, yet folks arrive at many of those others time and time again. What’s going on here?
You see, that long horizontal “fraction bar” that divide the numerator from the denominator —- that is something mathematically know as a grouping symbol. Parentheses & brackets are the most common grouping symbol, but the horizontal fraction bar is another. For example, suppose we wanted to write this original fraction in plaintext. Many Magoosh users, when they send us support emails, have to write complicated fractions in plaintext. First of all, here is the wrong way to do it.
That is an unholy abomination. Technically, the right side of that equation say: take 12x by itself, add it to 36 divided by 4x, and then add that to 6. That’s what the right says, which is completely different from what the left side says. To say that these two are equal is a mathematical obscenity! And yet, many students rewrite the left side as the right side, without any idea of the problem.
The long horizontal fraction bar of a properly written fraction, a fraction with the numerator vertically above the denominator — that fraction bar is a grouping symbol. We can’t express that grouping symbol in plaintext, and the simple slash, /, is NOT a grouping symbol. That’s part of the problem — the left side, the true fraction, has a bonafide grouping symbol, and the red expression on the right is devoid of one. Therefore, we need to substitute the only grouping symbol allowed in plaintext: parentheses.
Now, that’s 100% mathematically correct. First of all, this is important for any to appreciate who has to write about a fraction as part of asking a question in an email, in a forum post, etc. BUT notice: this is not merely a matter of persnickety bookkeeping. In fact, this difference makes explicit a profound mathematical issue.
Think about the cancelling question above. If we rewrite the vertical fraction as a plaintext fraction, we have to include the parentheses, and that makes explicitly visible the way elements are groups. Order of operations tells us that operations inside parentheses must take priority, so we can’t cancel, say, the 12x with the 4x, because they are included in separate calculations.
Rewriting the fraction in the form involving the parentheses makes clearer how order of operations comes into play, but of course, the way we write the fraction doesn’t change what legitimately can be cancelled. No matter what form the fraction is in, we can’t cancel that 12x with that 4x, nor that 36 with that 6. Technically, order of operations is not PEMDAS but GEMDAS, where the G = grouping symbols, which includes not only parentheses and brackets but also structures such as horizontal fraction bars. All of these things take precedence over the other arithmetic operations.
Now, having said that, it may seem to some readers that no cancelling is ever possible if there are algebraic expressions in the numerator and denominator. While the possibilities for cancelling are far fewer than the mathematically naïve would suppose, there is one possible route. Suppose we can factor out something from both the numerator and denominator: for example, with this fraction —
Now, as discussed in the first section, we can rearrange multiplication and division in any order, so we can cancel those factored-out numerical factors.
This is precisely how we arrived at answer H above.
Another grouping symbol is the “exponent slot” (that’s not an official term — I just don’t know what else to call it!!)
The “x+3” is “grouped” in the exponent. Writing the ordinary superscript exponents makes this grouping clear. When we have to write something like this in plaintext, as with fractions, we lose the original grouping symbol. Many students make atrocious mistakes when writing something like this in plaintext. For example:
This too is an unholy abomination. Literally, the right side now says: raise 2 to the power of x, and when you are done with that, add 3 to the result. That’s what the right side says, which of course, is completely different from what the left side says. As in the fraction case, we have lost a grouping symbol, and we need to replace it with the only grouping symbol allowed in plaintext: parentheses. BUT, we must be very careful where we put the parentheses. For example:
This is an unholy abomination in parentheses. This change is like choosing a more attractive font with which to write an obscene word: it’s a cosmetic difference that doesn’t address the core issue. We only fix the problem if we precisely replace one grouping symbol with another grouping symbol:
This is now 100% correct. Again, more than a matter of simple notation, these difference demonstrate something quite profound about the underlying mathematical relationships. The plaintext format of emails and forums is annoying in some ways, but in some ways it makes explicit some mathematical ideas of which some students might have remained unaware.
I hope this post gave you some insight into mathematical notation and its logical consequences. Here’s an easy question that employs some related ideas:
2) http://gmat.magoosh.com/questions/50
If you would like to ask any questions about this post, please let us know in the comments section below.
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Strange as it may seem, there is no generally accepted formal definition of the word mathematics. Most of the dictionary definitions, for example, cover the more familiar branches of mathematics, but not, say, typical Ph.D. dissertation topics in math. One aspect of math has to do with logical relationships and logical deductions — we could call this the more left brain part of math. Another aspect of math has to do with identifying formal abstract patterns. The words “logic” and “patterns” give two very important clues. For the first, you really understand a mathematic topic when you understand all the logical connections —- for example, knowing not just the rule for adding fractions, but understanding why fractions have to be added that way. The second is relevant because several out-of-the-box GMAT questions will throw some new situation at you, and you will have to identify what is the best way to parse the situation into recognizable components. Perhaps even more important is understanding what math is not.
I believe if one were to ask many a GMAT test taker, “What math does one need to know for the GMAT?”, most people would indicate the content of the OG, the section entitled “Math Review.” This notion is simultaneously very helpful but slightly misleading. Yes, those are the math facts you need to know — and if you’re rusty on them, as many folks are at the beginning of studying, it’s good to get very clear on them — but the math facts are not the same as mathematics. What’s the difference?
Well, think of the difference between, say, the content of a French dictionary vs. the language of French in living dialogue. As helpful as a dictionary may be, no one goes from no knowledge of the language to fluency simply by relying on the dictionary. In addition to knowing what each word means in isolation, one also has to know how it is used, in combination, in the context of real live communication. In this analogy, the “Math Review” communicates the meaning of each individual math factoid, and knowing these is essential, but the mathematics one really has to know is how to use these in problem-solving. If this is what you need to learn, you only find it by solving problem after problem, and carefully reading the solutions of each problem.
When you read the solutions to a problem that baffled you, pay attention to (a) the logic of the argument connecting one step to the next — not just the math factoids, but the strategies employed, and (b) the patterns the solution points out and employs. Any logic that is new to you or any patterns you didn’t see on your own — make a note of these, and return to them until you are confident you would be able to use these in the solution of a new problem on your own.
Much in the same vein, a student can easily lean on formulas too much, and mistake this reliance on formulas for mathematical thinking. Don’t memorize the formula for the sum of the first n integers. Don’t memorize the formula for the area of an equilateral triangle. Don’t memorize the formula for distance in the coordinate plane. Yes, those are all things you may need to calculate in a GMAT question, but instead of memorizing a formula, I highly recommend understanding, step by step, the logical argument that underlies the formula. Knowing the formula, by itself, is not mathematical thinking, but understanding the logic, the argument, that leads to the formula — that most certainly is mathematical thinking.
There are very few formulas you should simply memorize. One that leaps to mind is Archimedes‘ formula for the area of a circle.
Archimedes was one of the greatest mathematical geniuses of all time, and he concocted a truly brilliant argument for why this formula had to be true. That argument, fascinating as it may be, is a bit more math than I would recommend for the average GMAT preparer. On this one, I would say: take a pass — just memorize this formula by rote, if you don’t already have it memorized from school. Having said that, I can’t think of any other GMAT formula about which I unreservedly would say the same thing. For every other formula, do not settle for rote memorization. Strive to understand the underlying logic.
St. Paul wrote “the letter killeth but the spirit giveth life” (2 Cor. 3:6). While this is arguably an over-dramatic analogy in context here, the sense of it holds. Mathematical thinking is a vital fluency with math, and you won’t find it printed on the page of the GMAT “Math Review.” What’s on those pages is necessary but not sufficient. Here are a few problems, for practicing mathematical thinking.
1) http://gmat.magoosh.com/questions/880
2) http://gmat.magoosh.com/questions/59
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