Rate Problems

to sum up:

Remember: average speed is not necessarily the average of the speeds: the average speed is closer to the speed we spend more time in. Use Ballparking and POE to quickly unravel a question and avoid the trap answer - always try to guesstimate before you try serious calculations or formulas.

You probably know what speed (or velocity) means on an intuitive level. A typical GMAT test contains 1-2 "speed" questions. These questions come in all levels of difficulty. As usual, these questions combine general GMAT skills with the specific topic at hand.

Speed (or velocity) is the rate of movement. It tells us what distance is covered in a time unit. It always carries units of length per time. For example: 0.3 centimeters per second, or 200 miles per hour. Those units are important - don't neglect them.

Work work work...

There are few more prominent mathematical cliches than that of the "two pipes filling a pool".

The Lazy Elfin Spider advances 840 centimeters in 7 minutes. What is its speed? 0.2 centimeters per second 0.5 centimeters per second 2 centimeters per second 4 centimeters per second

C Correct! Here is the unit conversion according to the "replace with an equivalent" principle: Speed...............Time.............Distance ..?..................7 mins............840 cms ..?................7ₓ60sec............840 cms The speed is 2 Cms / Sec

Now, consider the Rough Frenzy Spider. It leaps 60 centimeters in 1.2 seconds. What is its speed? 0.05 meters per second 0.5 meters per second 5 meters per second 50 meters per second

C Incorrect. Because all the answers are expressed in meters per second, you have to convert the centimeters to meters. Try replacing the word "centimeters" with the equivalent expression "0.01 meters". B Correct. It is OK to replace the expression "60 centimeters" by the equivalent "0.6 meters", or, alternatively, replace the word "centimeters" with "0.01 meters". The speed is 0.5 m/s

The red kangaroo hops across 120 meters in 5 seconds. What is the speed of the kangaroo in kilometers per minute? 1.44 14.4 4 1.25

A Correct. First of all, assign each of the values in the question to the appropriate cell in the speed table. Then, convert the units using the 'replace with an equivalent' principle: Replace the word "meters" with the equivalent "×(1/1000) kilometers"; Go on to replace the word "seconds" with the equivalent "×(1/60) minutes". The speed is 1.44 Km/ Min

To sum up: Combined rates:

Identification: Questions with more than one worker. First step is always to find the rate of each worker Then combine the rates (and only rates, never combine time or work). Add the rates of those who work towards the target. Subtract the rate of those who work against the common target. Now use the combined rate in the Work Table to find the desired value.

A GMAT test will contain between 1 and 3 rate problems. Included in this number are work problems as well as velocity problems, which as we'll see, are in fact a certain kind of work problems.

We'll start by focusing on the fundamentals of rate and work. As you'll soon learn, a better grip of the basics helps you avoid common misconceptions and sets the foundations for great performance on advanced questions.

Here is a summary of what we've learned:

You may have to convert time or length units. Often this is imposed by the units found in the answers. Using the "replace with an equivalent" principle you can replace the unit (the actual word in the expression) by an equivalent expression. For example, you can replace "kilometers" by "1000 meters". You can replace other expressions as well, as long as the substitute is equivalent. Such was the case when we interchanged "2 minutes" and "120 seconds". Using this principle is not limited to one replacement. You may need to convert both the time units and the distance units.

A rule of thumb that helps eliminate answer choices and that results from the rate equation, refers to the units of the correct answer:

If the answer does not come in the form: [result] work unit per time unit (e.g. 3 tables per hour) then it doesn't describe rate.

Let's summarize: Use the table to organize!

1. Fill the table: Assign each of the values in the question to the appropriate cell in the table. Use a new row whenever there is more than one journey, more than one traveler or more than one segment. 2. Highlight the value you are asked about 3. Plug in the answer choices until one of them fits in with all the data in the table.

Summary:

A combined rates question can be identified by the presence of more than one worker, working simultaneously. Some of the data refer to the general pace of each worker - we start by calculating the rate of each worker out of that data. The rest of the data refer to the task that is performed when all are working together - sum all the rates of the workers who work towards that aim and use the combined rate in the relevant cell in the work table.

Some questions present a situation that starts with one worker, and gradually more workers are added. You have all the knowledge needed to solve such questions but let's take a look at the right approach. Read the following question:

Jerry and Seline are sipping a huge glass of margarita through two straws. Had Jerry sipped the margarita alone it would have taken him 4 minutes. Had Seline sipped alone it would have taken her 5 minutes. When the margarita was served, Seline first tried it, sipping the drink for 30 seconds. Then Jerry joined and they sipped together until they finished the drink. How long did it take to empty the margarita glass altogether? How do I even begin to solve this? First of all, before we relate to the changes in the number of workers (in rate questions even the people sipping cocktails are workers), this is another rate question with a invisible variable that calls for plugging-in. What is the size of the margarita glass? In school you were trained not to ask that question, and use 1 glass as the work unit in the question. But this leads to dealing with fractions, and complicates the calculations which exposes you to risk of careless mistakes. Therefore plug-in a volume for the glass. The volume should be divisible by 4 and 5 being the times mentioned in the question. Plug-in 20 oz. as the volume of the glass. Now let's look at the question again:

Well, that wasn't so bad. Is that all we need to know about combined rates?

Not quite. Do two people who work together always join forces in an attempt to achieve the common aim? The answer is of course no. Not in real life, and not in the GMAT. Here, take a look at the following situation: Melissa builds an average of 8.5 model airplanes per hour. Cheryl destroys Melissa's completed model airplanes at an average rate of 4.9 models per hour. If both Melissa and Cheryl progress at their respective rates over a full work day of 7.5 hours, How many completed model airplanes are present in Melissa's workshop at the end of the day? 2 16 27 58 100.5 C Very good. To solve combined rates with opposite targets we work in the same orderly manner. The first step is always the same - calculate the rate for each worker: In our basic question the rate is given: Melissa's rate = 8.5 model airplanes per hour Cheryl's rate = 4.9 model airplanes per hour Next combine the rates, giving a negative sign to the rate of the worker(s) who interfere with the mission: The work is to build model airplanes. Melissa works towards that aim while Cheryl's work is in the opposite direction. Combined Rate = 8.5 - 4.9 = 3.6 model airplanes per hour Now simply assign the rate and time in the relevant cells of the work table: Rate..................Time................Work 3.6.....................7.5...................... Work = 3.6*7.5 = ? Work with good numbers: 7.5 is three quarters of 10, so 3.6*7.5 is 3.6*10*3/4 = 36*3/4 = 9*3/1 = 27 model airplanes per hour.

Now that we have defined rate, let's see how to use the formula in the most convenient way.

The definition of rate as Rate = Work / Time can be represented as a the multiple: Rate ₓ Time = Work Based on this multiple we can represent the data from the question in a table that looks like this: ..............Rate.....Time......Work The solution for every rate question, even the most complicated ones, starts with the same initial step. Although it may seem simple, it is crucial: Assign each of the values in the question to the appropriate cell in the table. Take a look at the following question: If it takes the rising pop-star Erwin Enchilada 6 hours to finish recording 5 songs, then what is the rate at which Enchilada records songs? 1 hour and 20 minutes per song 84 minutes per song 1 hour and 12 minutes per song 50 minutes per song Approximately 0.84 songs per hour E Very good. Let's use this basic question to review the work order: 1. Assign a value from the question to the appropriate cells in the Work table: .......Rate.......Time.......Work ......................6............5 2. Identify the value that the question wishes to calculate. In this case it's easy - we want to know what the rate is. 3. Calculate the value of the missing cell in the table. Pay attention to units. R=W/T = 5 Songs/6 Hours = 5/6 Songs/Hour Only answer E is presented as songs per time unit so in this question there's no need to spend anymore time on calculations. Just POE answers A through D and rush to the next question

Henry eats X scones in X percent of the time it takes Rachel to eat Y scones. If Rachel eats four scones in ten minutes, then the number of minutes it takes Henry to eat 8 scones must be equal to which of the following? Y / 5 320 / Y 100Y / (5X) XY / 250 Y / (5X)

A Correct. Variables in the answer choices? Plug in numbers and POE! Plug in good numbers to clear the fog and see what's going on. Plug in Y=4, so that Rachel eats Y=4 scones in 10 minutes. A good number for X will be 10, so that Henry eats 10 scones in 10% of the time it takes Rachel to eat Y=4 scones. Arrange the information in a table as you go: .............Rate.....................Time....Work Rachel.........................10 min..Y=4Sc Henry...10s/m....10%·10=1 min...10Sc Henry (8 scones).10S/m...???.........8 Now that you have Henry's rate, find the time it takes him to eat 8 scones: 8/10, or 4/5 min. That's your Goal. Now plug in Y=4 and X=10 into the answer choices, and POE all those which do not match your Goal. Remember to check all 5 answer choices. Y/5 = 4/5, which matches your Goal. All the other answer choices do not equal 4/5 for Y=4 and X=10, so this is the right answer choice.

Now solve another one - Tom chased Jerry down a long corridor. When the chase began, Tom was 24 meters behind Jerry. After a while, Tom reached Jerry and caught him. If Tom was running at a constant speed of 8 meters per second, and Jerry was escaping Tom at a constant speed of 4 meters per second, then the duration of the chase was - 2 seconds 3 seconds 4 seconds 6 seconds 8 second

A Incorrect. Again, this question involves two moving objects, and the issue is the time it takes to close the gap between them. Draw a sketch of the situation, --------------------------------> ......................................-----------> TJ [--------------------] .....gap=24m................. Treat the 24 meter gap as Work. To find the rate of closing of the gap ask yourself "How did the gap between Tom and Jerry change during one second?" Every Second T--8m----> J--4m----> [----------------------------] T..................................................J Every second, Tom reduced the gap by 8 meters, while Jerry enlarged the gap by 4 meters. Therefore, the Rate of closing the gap was 8-4 = 4m per second. The original problem now becomes a Rate problem Rate problem - "How long does it take to close the 24 meter gap at a rate of 4 meters per second?" Find the Time so - In both the Fanny-Alexander question and the Tom-Jerry question the gap is being closed. In either question, the closing rate of the gap is actually a combined rate. In the Fanny-Alexander case, both work to reduce the gap, and therefore the combined rate is the sum of their individual speeds. In the Tom-Jerry case, Tom works to reduce the while Jerry works to build up the gap, and therefore the combined rate is the difference between their individual speeds. In fact, a common feature of gap problems is adding the individual speeds or subtracting them, in order to find the combined rate of closing (or opening) of the gap.

In many rate problems the work is not given in absolute numbers but in units - 1 pool, 1 bucket etc. This is a case of a Hidden Plug-In: PI the volume of the pool, number of worms in the bucket etc. The criterion for choosing a good number to PI is a number that would be divisible by all the bottoms of the fractions in the question. The time(s) mentioned in a rate question are also considered bottoms of fractions Plug-In a number that is a multiple of the bottoms of all fractions and of times mentioned in the question.

A hole drained 5/8 of a pool in 2 hours. How long will it take the hole to finish emptying the pool at the same rate? 3 hours and 20 minutes 1 hour and 10 minutes 1 hour and 20 minutes 3 hours and 12 minutes 1 hour and 12 minutes D Wrong answer. You calculated the time it takes to empty the whole pool. Was that what you were asked to do? A Wrong answer. Were you trying to calculate the entire time it takes to empty the pool? Well, that's not the value that the question sought. E Very good. This question demonstrates how you can use plugging-in to simplify rate questions. It takes the hole two hours to empty 5/8 of a pool. What is the volume of the pool? We don't know. So instead of using 1 pool as a unit, choose a number that's easy to work with as the volume of the pool. What number would you pick to represent the size of the pool in, say, liters? My response does not match any of the above Since the question talks about 5/8 of the pool, a good number choice would be a multiple of 8. Very well. Let's solve the question while plugging in a value of 800 Liters for the volume of a full pool. It takes 2 hours to empty 5/8 of the 800 Liter pool (which is 500 liters): Rate..................Time................Work ......................2 hrs.............500 Ltrs R= 500/2 = 250 Liters per hour Now that we have the rate as a convenient number, Let's move on to calculate the time it takes to complete emptying the pool. 500 Liters have already been emptied out of the total 800. The work now is to empty the remaining 300 Liters. Rate..................Time................Work 250 lts/hr............................300 Ltrs T = W/R = 300/250 = 6/5 hours Now let's convert the result to hour+minutes: 6/5 hrs = 1 1/5 hrs = 1 hr + 1/5 * 60mins= 1 hour and 12 minutes.

On a second thought, how about you do it yourself? Ask yourself: If Kate loses a guy in 10 days - what portion of the guy does she lose every day? A...1/10 B...Some information is missing

A... Whatever answer you chose, you were right. If you asked "How do we know that Kate loses the guy at a constant rate, maybe she loses half of him in the first day and the rest in the remaining 9 days?", well that's a good question! The answer is that we (as well as the GMAT writers) assume that there's a constant pace at which the work is done. Unless of course it is stated explicitly otherwise. So, if losing 1 guy takes 10 days, Kate loses 1 Guy/10 Days which means a portion of 1/10 Guy/Day In the GMAT this would be presented as 1/10 guy per day. So rate is the portion of work that is done in a time unit. And thus we get the rate equation: Rate = Work / Time

Mr. Wayne bungee jumps from the top of a building straight to the ground at a constant speed. 3 seconds after he starts plummeting he passes the 20th floor. 15 seconds after he starts plummeting he passes the 5th floor. Each floor is 3 meters high. What is Mr. Wayne's speed in meters per second? 3 3.75 4 4.25 5

B Correct. First of all, assign each of the values in the question to the appropriate cell in the table. speed.....time........distance ?.......12 seconds...15×3 meters The relevant distance is 15 floors long (from floor 20 to floor 5.) In meters, this is 15x3 meters. The relevant time (or duration) is 12 seconds (from the 3rd till the 15th second.) So, according to the definition, Speed=15ₓ3 meters/12 Seconds = 3.75 meters/seconds This question is actually more about extracting relevant data from a pile of numbers than about speed. The GMAT offers no partial credit for accurate calculation with wrong numbers.

Consider the following question: The Wahoo fish swims 2.40 kilometers in 2 minutes. What is its speed in meters per second? 1.2 20 40 120 200

B Correct. Well, the solution looks so: Speed...............Time.............Distance ..?...........2 minutes.....2.40 kilometers ..?......120 seconds....2.40×1000 meters And the speed is 2400 Meters / 120 Seconds = 20 m/s How did we do that? We made two separate changes to the original expression: We replaced the "kilometers" with the equivalent expression "1000 meters". We independently replaced "2 minutes" with 120 seconds. (Two minutes are 2x60 seconds.) Replacing one expression with an equivalent expression is always allowed and will not change the value of the whole expression. You can replace more than one expression, either at once, or in separate stages. Let's do some more unit conversions.

If pipe B fills a container at a constant rate in 100 minutes, how many minutes does it take pipe A, working at is own constant rate, to fill the same container? (1) Pipe A and pipe B together fill the container at (1/4) the time it takes pipe A alone. (2) Pipe A and pipe B together fill the container at (3/4) the time it takes pipe B alone.

B Incorrect. This is a "what is the value of..." DS question. In this type of question, a statement will be sufficient only if it leads to a single value of the variable (or expression) you're asked about. The issue is combined rate. This problem presents a case of two workers (pipe A and pipe B), working together, and asks for the time it takes pipe A alone to do a fill a certain container. Since the two pipes are working together, the combined rate is calculated by adding the rate of pipe A with the rate of pipe B. Organize the information in a rate table, adding rows as needed. It takes B 100 min to fill the container, so B's individual rate is 1/100 container/min. Assign a variable x to the time it takes A to fill the container alone. Enter the information in the statements into the table and see if you can construct an equation to find x. Yes, Stat.(2) is Sufficient. But what about Stat.(1)? D Correct. Stat. (1): Place the information in the rate table: ............Rate..................Time........Work Pump B..1/100.....100 min.............1 Pump A..1/x.........x min.................1 Combined rate A+B 1/100+1/x stat. (1): x/4 min 1 If A takes x minutes alone to fill the container, stat. (1) says that A and B working together take x/4 minutes to fill the container. Form an equation: 1/ x/4 = 1/100 + 1/x --> 4/x = 1/100 + 1/x From this equation it is possible to find a single value of x, so Stat.(1)->S->AD. Stat. (2): there's no real reason why this should be any different than (1), but check to be sure.

A vacuum pump reduces the number of gas molecules in a vacuum chamber by approximately 6.5·1024 molecules per minute. Which of the following best approximates the time (in minutes) it takes to reduce the number of gas molecules in the vacuum chamber by 30·1021? 0.005 0.03 0.05 2.9 27

C Incorrect. Organize the information in a rate box. Form the fraction, reduce, then solve. Be careful when counting places in power with scientific notation. A Correct. Rate..................Time................Work 6.5·10²⁴ mol/min...1 min.....6.5·10²⁴ 6.5·10²⁴ mol/min...???......30·10^21 Use the given work and time (1 min) to find the rate of 6.5·10²⁴/1 molecules/min. At this rate, it will take the pump 30·10^21 / 6.5·10²⁴ minutes to reduce the number of molecules by the required amount. Reduce the fraction: 30·10^21 / 6.5·10²⁴ = --> 30 / 6.5·10³ = Ballpark 30 / 6.5 ≈ 30/6 = 5. --> 5 / 10³ = --> 5 ·10^-3 = --> 0.005

Having understood the basics of Ballparking and POE, you can now benefit from combining these two in rate problems. Juan goes out cycling outdoors. He travels at an average speed of 15km/h, going uphill. On the return trip down hill, Juan travels at an average speed of 20km/h. Which of the following is the closest approximation of Juan's average speed, in kilometers per hour, for the round trip? What do you intend to do in order to solve the problem? A..Put all the data in the average speed formula, and solve. B..Is there a better, faster, more accurate way?

B... Good question. GMAT Math is more about understanding concepts than remembering formulas from fourth grade. Before trying to use needless and time consuming formulas in the GMAT, let's see how ballparking and a little understanding of the concept of speed can help. Q....Juan goes out cycling outdoors. He travels at an average speed of 15km/h, to the top of the hill where the midpoint of the trip is. Going down hill, Juan travels at an average speed of 20km/h. Which of the following is the closest approximation of Juan's average speed, in kilometers per hour, for the round trip? First off, there are two answer choices here that can be eliminated on sight. Which are they? Click on each of the answer choices, to see which ones you can POE. A Juan's average speed must be higher since he also rode at a speed of 20 mph. POE and move on. E Juan's average speed must be lower since he also rode at a speed of 15 mph. POE and move on. B,C and D These one's possible, since it's between 15-20 mph. Keep it. Now you know that the ballpark for Juan's average speed for the round trip must be between 15 and 20 mph. Having used POE to get rid of some of the answer choices, let's try to ballpark further... What is the most important factor in determining an average speed? Time Speed Distance A Correct. The fact that the distance up and down the hill is equal is not important. The average speed is determined according to how much time Juan spent at each speed. If Juan had spent an equal amount of time (say one hour) riding at each speed, his average speed would be the average of the speeds: 1 hour at 20 km/hour and 1 hour at 15 km/hour comes down to two hours at an average speed of 17.5 hours. However, since Juan traveled the same distance at different speeds, the time he rode at each speed is different. To be more exact, Juan spent more time riding at the slower speed. The slower speed of 15 km/hour carries more weight in the average, and the average speed should be closer to the 15 than to 20. Now try to guesstimate Juan's average speed in miles per hour from the remaining answer choices 17.1 17.5 17.9 A Correct. Since the time Juan spent in the lower speed (15mph) is greater, his average speed has to be closer to the lower bound - closer to 15 mph than to 20 mph. If we spend the same time at both speeds (say one hour at 15 and one hour at 20), then our average speed will 17.5 - straight down the middle. But when averaging speeds, time is the deciding factor: the average speed should be closer to the speed we spend more time in. Understanding and remembering this concept not only allows you to quickly choose B without calculating anything, but also helps you avoid the obvious trap answer C. Remember: average speed is not necessarily the average of the speeds.

Barbara walked up and down a 25 meter long corridor. When she walked one way, she walked at 4 meters per second, and when she walked the other way, her speed was 2 meters per second. What was her average speed in meters per second? 1 1/3 1 2/3 2 1/3 2 2/3 3

C Incorrect. Organize the information in a table and use the formula, Avg Speed = total Dist / total time . D Correct. Your table should look something like this: The last row is used for summing up the times and distances, and for the average speed. Now, either calculate the average speed (total distance / total time), ballpark it, or plug in the answers to see which answer choice fits the third row, Average Speed × Total Time = Total Distance.

What is the slope of a line that passes through the points A and B, whose coordinates are (x,y) and (y,x) respectively? x/y y/x −1 1 x−y

C The slope is given in the formula Variables in the answer choices? Make the algebra go away - plug in good numbers and eliminate. Correct. Plug in x=4 and y=3. (x, y) becomes (4, 3), and (y, x) is then (3, 4). The slope of a line joining points (4, 3) and (3, 4) is (4 - 3) / (3 - 4) = 1/-1 = -1. Since no other answer choice will give you -1 for x=4 and y=3, this is the correct answer.

Q 2*.Carl and Mark run in opposite directions around a stadium. Carl runs at a constant speed, and so does Mark. Carl's speed is how many times greater than Mark's speed? (1) Mark completes a whole round every 60 seconds. (2) Carl and Mark meet every 24 seconds.

C Incorrect. This is a "what is the value of..." DS question. In this type of question, a statement will be sufficient only if it leads to a single value of the variable (or expression) you're asked about. The issue is combined rates. You need the ratio between Carl's and Mark's speed. Organize the information in a rate box to see what's going on. Stat. (2) tells you that Mark and Carl take 24 seconds to cover the stadium at their combined rate. Think of the problem in terms of stadium parts/sec: they could each cover half a stadium in those 24 seconds, and then their speeds would be equal. But stat. (2) still allows other cases: for example, If mark covers a quarter of a stadium and Carl covers the remaining 3/4 in the same time, their speed ratio would be quite different. Alone, Stat.(2)->IS. C Correct. Stat. (1): That's nice to know, but tells you nothing about Carl's speed. Stat.(1)->IS->BCE. Stat. (2) tells you that Mark and Carl take 24 seconds to cover the entire circuit of the stadium at their combined rate - Carl from his end, Mark from his. Think of the problem in terms of stadium parts/sec: they could each cover half a stadium in those 24 seconds, and then their speeds would be equal. But stat. (2) still allows other cases: for example, If mark covers a quarter of a stadium and Carl covers the remaining 3/4 in the same time, their speed ratio would be quite different. Alone, Stat.(2)->IS->CE. Stat. (1+2): draw a figure to help you see what's going on. Mark completes a full stadium in 60 seconds. Thus, in 24 seconds he completes 24/60 = 2/5 of the stadium. Stat. (2) tells you that together Mark and Carl complete a full stadium in those 24 seconds, thus Carl has to complete the remaining 3/5 of the stadium in 24 seconds. Since you now have both Carl's and Mark's rates, in terms in Stadium part/sec, it is possible to find the ratio between the two. Stat.(1+2)->S->C.

A squirrel hops 0.055 kilometers in 11 seconds. What is the speed of the squirrel expressed in centimeters per second? 0.5 5 50 500

Correct. First of all, assign each of the values in the question to the appropriate cell in the speed table: Replace the word "kilometers" with the equivalent "1000 meters"; Go on to replace the word "meters" with the equivalent "100 centimeters": The speed is 500 Cms / Sec

Consider the following question, Bridget runs 210 yards in 30 seconds. What is her speed? 210/30 7 yards 7 feet per second 7 yards per second

D Correct. First of all, assign each of the values in the question to the appropriate cell in the table. Remember to keep track of the time and length units as you make calculations: speed..............time..............distance ?..............30 seconds........210 yards In order to find the speed, calculate Speed = 210 yards/30 Seconds = 7 Yards/Seconds Usually, when you encounter speed in a GMAT question, it is written in this form: 7...........yards...........per........second x....(distance unit).....per........(time unit) In the process of calculating speed you often get a slightly different expression: From "Bridget runs 210 yards in 30 seconds," it follows that Speed=210 yards/30 Seconds or Speed=numberₓ(Distance Unit) / numberₓ(time unit) It is possible to go from the calculation form to the answers' form by separating the numbers from the units: Speed=210 yards/30 Seconds =210/30 yards/Seconds which can then be simplified to 7 yards per second. However, not everything that is possible is also necessary. Finding the speed may sometimes be a superfluous step, as the following problem shows:

Q27...If the coordinates of points A, B and C are (2,4), (8,4) and (2,2) respectively, what is the area of triangle ABC? 12 10 8 6 4

D Correct. Remember the area of a triangle is AREA= ½×BASE×HEIGHT. Use the coordinates of the three points given to calculate the base and the height of the triangle shown in the figure. Based on the information given in the question, the length of line AB is 8 - 2 = 6 and the length of line AC 4 - 2 = 2. Hence, the area of triangle ABC = (1/2) x 2 x 6 = 6.

Q37...Home Excel: If O is the center of the circle in the figure above and the area of the unshaded sector is 5, what is the area of the shaded region? 25/√π 30/√π 20 25 30

D Correct. Thus, The ratio between the inscribed angles of the two sectors in this circle is also the ratio between the areas of the two sectors. Since the small (white) sector 'has' an inscribed angle of 60º, the unshaded sector constitutes 60/360 = 1/6 of the total area. The unshaded region must constitute the remaining 5/6 of the area, or 5 times the area of the unshaded region. Thus, the shaded region is 5×unshaded region, or 5×5=25.

Here are some common replacements

Distance Units: "kilometers" = "1000 meters" "meters" = "0.001 kilometers" "meters" = "100 centimeters" "centimeters" = "0.01 meters" "meters" = "1000 millimeters" "millimeters" = "0.001 meters" Time Units: "day" = "24 hours" "hour" = "1/24 days" "hour" = "60 minutes" "minute" = "1/60 hour" "minute" = "60 seconds" "second" = "1/60 minute" American units (such as inches and feet) are rarely used, and the GMAT does not assume that you remember the ratios between them. Any required ratios for American units will be supplied by the question.

A certain pump pumps one third of a gallon of water every 2.5 minutes. If the pump starts pumping from a completely full 12 gallon container, and pumps for one hour, the amount of water pumped would be what percent of the water left in the tank? 25% 50% 80% 100% 200%

E Correct. Rate is calculated as When solving a question, translate it into the Work table. Rate....Time......Work The table represents the formula Rate×Time=Work so that the multiple of the two left columns equals the right column. A third of a gallon every 2.5 minutes equals a rate of: Rate..................................Time..Work (1/3)/(5/2)=2/15 gln/min..2.5..1/3 Now find the amount pumped out in 60 minutes: Rate................Time.........Work 2/15 gln/min..60 min...(2/15)·60 = 8 8 gallons are pumped out, thus 12-8=4 are left in. Translate the statement 8 gallons are what percent of 4 gallons?" to get (8/4)·100 = 200%

Q36...On Goal Sept15 Excel Sheet... In the regular hexagon above, what is x? 15° 20° 22.5° 30° 60°

E Incorrect. Draw a circle around the hexagon. An n sided regular polygon circumscribed in a circle divides the circle into n equal segments. The size of any inscribed angle is half the size of the arc segment defined by its rays. D Great job! Draw the hexagon inside a circle. The hexagon divides the circle into six arcs, each of which measures 360°/6 or 60°. Each inscribed angle equals half the arc segment defined by its rays. Therefore, ∠BDA, or x, is half of arc segment AB, or 30°.

Bruno and Sacha are running in the same direction around a stadium. Sacha runs at a constant speed of 6 meters per second, and Bruno runs at a constant speed of 5 meters per second. At a certain point Sacha overtakes Bruno. If one minute afterward, Sacha stops and waits for Bruno to reach him, then how many seconds does he have to wait? 12 24 36 60 72

E Incorrect. This is a rate problem with two objects. Since the question asks for the time it takes Bruno to overtake Sacha, you need the gap between them and the rate at which this gap is closed. Once you have these two pieces, use the rate box to find the time. Deal with the problem in two stages: Opening the Gap, and closing the Gap. A Correct. Opening the Gap: First, find the distance, which is the gap opened by Sacha by running for 1 minute ahead of Bruno. Every Second, Sacha runs 6 meters, while Bruno runs 5 meters. Thus, the gap between them is opened at 6-5=1 meters per second. Place the numbers in the rate box: Stage Rate Time Work Opening the gap 6-5=1 m/sec 1 min = 60 sec 1×60=60 meters Now deal with the second stage: closing the gap. At this stage, Sacha stands still, while Bruno advances at a constant rate of 5 meters per second. Thus, the gap between them is closed at a rate of 5 meters per second. The gap between them is the 60 meters calculated in the first stage. Place the numbers in the rate box: Stage Rate Time Work Opening the gap 6-5=1 m/sec 1 min = 60 sec 1×60=60 meters Closing the gap 5 m/sec 60 / 5 = 12 sec 60 meters

Often, when we describe motion, we are interested only in the ratio between the total time and total distance traveled. This is exactly what average speed means: Avg Speed = Total Distance / Total Time When you make your way from A to B, you may not move at a constant speed. You may even stop for a while. The average speed gives you an overview of the total distance divided by the total duration of the trip. Let us see what happens when "average speed" meets the GMAT:

Gill drives 120 miles from Los Angeles to San Diego to fetch a package. On her way there she drives at 40 miles per hour. On her way back she drives 50% faster. What is Gill's average velocity for the round trip? 24 miles per hour 48 miles per hour 50 miles per hour 53 1/3 miles per hour 68 miles per hour B Correct. The way to San Diego is 120 miles long, and it takes 3 hours (120/40). On the way back, the speed is 1.5×40 miles per hour, or 60 miles per hour. The way back takes 2 hours (120/60). The average speed is therefore: (120+120 miles) / (3+2 hours) = 240/5 miles per hour = 48 miles per hour. What makes this question tricky for some is the temptation to calculate the average of the two speeds, 40 and 60 miles per hour. Average speed is NOT the plain average of speeds. When calculating the average speed please refer to the formula Avg Speed = Total Distance / Total Time

So far we have dealt with rate problems that had one worker performing one task. This can be complicated by adding a few additional workers. They could be working on the same task, a different task, or even have one working against the other workers.

If you're worried because the two pipes filling a pool and one hole emptying it scenario still comes back to haunt you sometimes - well you shouldn't be. You have mastered the rate equation tasks and that means your foundations are solid. Now, as you'll soon see, it's all about going back to the basics over and over again. If you start with the correct initial step, the rest will follow.

We hit the road with a very basic question - What is rate? The intuitive answer, not mathematically formulated yet, is that rate is the pace at which work (of some sort) is done. Therefore rate relates to pace - which brings in the time factor - and to work. We all know that time is measured in seconds, minutes, hours etc. In what units do we measure work?

In the GMAT, as opposed to formal physics, there are no specific units for work, because anything could be defined as work. And by anything, we mean anything. From painting walls, through doing the dishes, dating guys or sipping cocktails. Every action, every situation can be described in terms of work. Try reading this sentence as a rate problem: Kate loses a guy in ten days. Mathematically speaking, this sentence is no different than the sentence: Kate paints one wall in 10 days. There is one unit of Work in the sentence: be it losing 1 guy or painting 1 wall. The Time it takes to finish that work is also given - 10 days. We said that rate is the pace at which work is done. Now, let's rephrase it in a way that's easier to use when solving a problem.

Let's summarize:

Rate is the work that is done in a unit of time. It is calculated as R = W/T When solving a question we translate it into the Work table. .......Rate.......Time.......Work The table represents the formula Rate*Time=Work so that the multiple of the two left columns equals the right column. 1. Assign a value from the question to each appropriate cell in the work table. 2. Calculate the value that of the empty cell based on the equation. Make sure you pay attention to units - do they match? Can they help you eliminate answer choices?

The equation form of the above definition is

Speed = Distance / Time The definition of speed as can be represented as a product: Based on this equation we can represent the data from the question in the following table: speed.....time........distance Most GMAT speed problems are all about organizing the information given in the question in a clear form. Therefore, the solution for every speed question, even the most complicated ones, starts with the same initial step. Although it may seem simple, it is crucial: Assign each of the values in the question to the appropriate cell in the table. In order to get used to the table, use it from now on.

Let's summarize:Speed

Speed, the rate of movement, is the distance travelled during a time unit. As always in the GMAT test, you will have to handle the data in the question efficiently. This is where the table comes in. The table represents the formula speed × time = distance so that the multiple of the two left columns equals the right column. 1. Assign a value from the question to each appropriate cell in the speed table. speed.....time........distance 2. Calculate the value that of the empty cell based on the formula speed × time = distance. Make sure you pay attention to units - do they match? Can they help you eliminate answer choices?

To summarize -Gap Problems

When a Speed Problem involves two moving objects and two speeds, concentrate on the gap between the two objects. Draw a sketch to help find the gap between the two objects. Find the rate of closing (or opening) of the gap. Ask yourself "How does the gap change every hour (or second, etc.)?" Turn the original problem into a Rate Problem where the gap is the Work, and the rate of closing/opening of the gap, is the Rate. Use the rate Box to organize the new information and find the required data.

You may be aware that there is more than one unit system. Distances can be measured in meters, centimeters and kilometers, as well as in feet, inches, yards and miles. This causes a lot of headaches in the world outside of the GMAT. Well, we have some good news for you. The only unit conversions you will have to make are within the metric system such as from meters to kilometers, centimeters to meters etc. Another common type of conversion you may need to do is between different time units. You must be able to interchange seconds, minutes, hours or even days.

When is unit conversion needed? You have to convert between units when you combine data with different units. Often this happens when the answers are expressed in units that are different from those given in the question.

Consider the following problem - Since 07:00, Fanny and Alexander are driving on the same road toward each other. At 08:00, they were 240 km apart. If Fanny drives at a constant speed of 120 km per hour, and Alexander drives at a constant speed of 60 km per hour, at what time will they meet?

Usually, the issue of Speed Problems is using two of the given values (Distance, Time, or Speed) to find the third. The situation in Rate Problems is similar - you have to find one of the three values, Work, Time, or Rate, based on the other two. The problem with the problem above, is that there are two moving objects, and two speeds. Luckily, there is a way to turn this problem into a Rate Problem, with just one value for the work, and just one value for the rate. It helps to draw a sketch of the situation presented in the problem. F---------> <-----------A [-------------------------------] .........................240 Km...................... The issue of the problem is the Time it takes to close the gap between Fanny and Alexander. The trick is to treat this gap as Work in a rate problem. The above problem asks for the Time. To find the Time, you need both the Work and the Rate. The Work is the gap of 240 km. The Rate is the rate of closing the gap. To find the rate of closing the gap, ask yourself "How does the gap change during one hour?" Every Hour F--120Km----> <----60Km----A [-------------------------------] .........................240 Km...................... Every hour, the distance between Fanny and Alexander is decreased on one side by 120 km, and on the other by 60 km. Therefore, the rate of closing the gap is (120+60) km per hour. This is the Rate in the following Rate problem: "How much time does it take to close a gap of 240 km at a rate of 180 km/hour?" All that remains is to calculate the Time based on the Work and Rate. Since 07:00, Fanny and Alexander are driving on the same road toward each other. At 08:00, they were 240 km apart. If Fanny drives at a constant speed of 120 km per hour, and Alexander drives at a constant speed of 60 km per hour, at what time will they meet? 08:45 09:00 09:15 09:20 14:00 D Correct. At 08:00 the gap between Fanny and Alexander is 240 km. The rate of closing the gap is 180 km per hour. Find the Time it takes to close the gap - It takes 4/3 hours = one hour and 20 minutes to close the gap. Thus, they meet at 09:20

Jerry and Seline are sipping a huge glass of margarita through two straws. Had Jerry sipped the margarita alone it would have taken him 4 minutes. Had Seline sipped alone it would have taken her 5 minutes. When the margarita was served, Seline first tried it, sipping the drink for 30 seconds. Then Jerry joined and they sipped together until they finished the drink.How long did it take to empty the margarita glass altogether?

Why all these colors? The colors help break the question to its three stages. Stage 1: The first (orange colored) data uses a language that presents a hypothetical situation - "had Jerry..". It is used to calculate the rates of Saline and Jerry. Jerry's rate: Rate..........Time............Work ....................4...................20 Rate= 20/4 = 5 oz. per minute Seline's rate: Rate..........Time............Work ....................4...................20 Rate= 20/5 = 4 oz. per minute Stage 2: This is the stage where Saline sips the drink alone for 30 seconds = 0.5 minute. Rate..........Time............Work 4.................0.5................... Work = 4*0.5 = 2 oz. Seline drank 2 oz. alone. Now at the end of this stage there are 20-2=18 oz. of margarita left in the glass. Stage 3 Now Jerry joins Seline in drinking from the glass. The drinking rate would change from Seline's rate to the combined rate. A change in the number of workers changes the rate and therefore separates between stages of the calculation. Combined rate= 4+5 = 9 oz. per minute Rate..........Time............Work 9.......................................18 Time = 18/9 = 2 minutes Altogether: 30 seconds of Saline alone, then 2 minutes until both finished the drink = 2 minutes and 30 seconds.

m and n are both positive. Pump A pumps n liters of water per minute, and pump B pumps m liters of water per minute. Is m < n ? (1) x is the number of minutes it takes pump B to fill a 100 liter container alone, and 3x is the number of minutes it takes pump B to fill a 100 liter container while pump A pumps water out of that container. (2) It takes (3x/2) minutes for pump A to fill a 100 liter container alone. statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked; statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked; BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked; EACH statement ALONE is sufficient to answer the question asked; statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

A Correct. This is a Yes/No DS question. Answering a definite "Yes" or a definite "No" means Sufficient. If the answer is sometimes "Yes" and sometimes "No", it means Maybe which means Insufficient. The issue is rates. n and m are the rates of the two pumps. the question asks whether m < n, so the real issue is whether A pumps more water per minute than B does. Stat. (1): No need for complicated rate boxes here. The information tells us that pump B is faster than pump A, since the 100 liter container gets filled even though A pumps water out of it. This means that m is not less than n, but that's a definite "No", which is sufficient. Stat.(1)->S->AD. Stat. (2): Since x is only given in statement (1), stat. (2) alone tells you nothing about pump B. Thus, there is no way to compare the rates of the two pipes, and Stat.(2)->IS->A.

Sari and Ken climb up a mountain. At night, they camp together. On the day they are supposed to reach the summit, Sari wakes up at 06:00 and starts climbing at a constant pace. Ken starts climbing only at 08:00, when Sari is already 700 meters ahead of him. Nevertheless, Ken climbs at a constant pace of 500 meters per hour, and reaches the summit before Sari. If Sari is 50 meters behind Ken when he reaches the summit, at what time did Ken reach the summit? 13:00 13:30 14:00 15:00 15:30

A Correct. This is a rate problem with two objects, Sari and Ken. Deal with the problem in two stages: Stage I: Sari climbs 700 meters alone from 06:00 to 08:00 while Ken sleeps. 700 meters in two hours places Sari's rate at 700/350 m/hour. Stage II: Ken starts climbing at 500 meters/hour, closes the 700 meter gap from Sari and reaches the top 50 meters ahead of him. Use the following figure to see what's going on: Every hour, Ken climbs 500 meters, but Sari moves 350 meters from him. Thus, their combined rate of closing the Gap is 500-350 = 150 meters/hour. Use this to find the time it takes Ken to reach the summit. Organize the information in a rate box. Rate Time Work Sari 700/2 = 350 m/hr 2 hr 700 Ken 500 m/hr Combined rate 500-350 = 150 m/hr ??? 700 + 50 = 750 m Ken has to close a gap of 700 meters, and open a new gap of 50 meters, all at the same combined rate of 150 m/hr. Thus, the total gap is 750 meters, and the time it takes him is 750/150 = 5 hours. If Ken started climbing at 08:00, he will reach the summit five hours later at 13:00.

In a certain factory two mills are grinding sesame seeds. The old machine grinds 250 lb. every 10 minutes, and the new machine grinds 500 lb. every 9 minutes. Which of the following is closest to the number of minutes it takes the new machine and the old machine together to grind 750 lb. of sesame? 9 12 15 19 29

A Correct. This problem looks like a combined rate problem: find the individual rates of the old and new machines, add to get their combined rate, find the time it takes them to grind 750 lb. of Sesame together at their combined rate. However, while the method described above will work, it's the long way to go about solving this question. There really is no need to calculate their combined rate, as the numbers presented for the work done by the old and new machine, 250 lb. and 500 lb. respectively, conveniently add up to the required 750 lb. Since the question requires an approximation, there's no need to calculate all the way - just think: How long does it take each machine to finish one batch. After 9 minutes, the new machine finishes its 500 lb. After one more minute, the old machine will have worked enough to grind its given 250 lb, bringing the final load to 750 lb. So an approximation of the required time is 10 minutes. Note that the new machine will continue grinding during the extra minute, so the 750 pounds will be ready before the 10 minutes are up. Since all the other answer choices are greater than 10, A must be the right answer.

Prof. Zinc walks along a corridor. He walks 35 meters in 14 seconds . What distance will he walk in 28 seconds? 56 meters 70 meters 98 meters 140 meters

B Correct. First of all, assign each of the values in the question to the appropriate cell in the table. speed.....time........distance .........14 seconds...35 meters If Zinc passes 35 meters every 14 seconds, he will pass 35 plus 35 meters, in two identical periods of 14 seconds. Simplifying the fraction 35/14 meters per second is an unnecessary calculation and should be avoided. Organizing the data in the table helps you focus on the necessary calculations. Here is a detailed solution: speed.....time........distance .........14 seconds...35 meters .........28 seconds....70 meters The times and distances add to 70 meters in 28 seconds. Of course, you don't have to actually duplicate the first line on your erasable noteboard, but you may. This question is easier than an average GMAT speed question. Organizing the data in the table will prove more and more helpful as the questions become harder.

Martin and Wood are hired to perform a copywriting job. Working alone, Martin could finish 2/3 of the job in 15 days, and if Wood would work alone, he could finish half the job in 9 days. If Martin and Wood work together, how many days would it take them to finish the entire job? 9 10 11 12 13

B Correct. This is a combined rate question, with two workers working together towards a combined goal of finishing the job. Find the rate of each individual worker, add the two rates to find the combined rate, then find the required time to finish the job. Note that the job itself is given in general terms - "the job". This is a classic sign of an invisible plug in in rate problems - Plug In a good number for the job to ease calculations and reduce the need to work with fractions. The question presents two fractions (2/3 and half) and two times (15 and 9 days). A good number would be a number divisible by all four bottoms of the fractions, such as 2·9·15 = 2·135 = 270. Therefore, assume that the job includes 270 pages, and place the information in a rate box. ..........Rate..................Time................Work Martin......................15 days..2/3×270 = 180 Wood.......................9 days...1/2×270 = 135 Combined rate 12+15=27 pg/day ??? 270 Therefore, it will take the copywriting duo 270/27 = 10 days to complete the job together.

Two pumps are connected to a certain empty container at the same time. Pump X fills the container water at a constant rate, while pump Y drains water out of the container at a constant rate. The two pumps finish filling the container in four times the duration it would take pump X alone to fill the container. If pump Y alone can empty a whole container in 48 minutes, then how many minutes does it take pump X alone to fill the container? 24 36 48 50 64

B Correct. This problem presents a case of two workers (pump X and pump Y), working against each other, and asks for the time it takes pump X alone to do a certain job. Since the two pumps are working against each other, the combined rate is calculated by subtracting the rate of pump Y from the rate of pump X. As an added difficulty, the work in the question is not defined in absolute numbers but rather as a single item - a container. This is a surefire sign of a Hidden Plug-In - Plug in a number for the work (think of it as "the number of liters in the container"). A good number would be 48 liters, as it is divisible by both 48 minutes and the "times four factor" of the combined rate to Xs individual rate. Organize the information in a rate table, adding rows as needed. It takes y 48 min to empty the container, so Y's individual rate is 48/48=1 lit/min. Assign a variable a to the time it takes X to fill the container alone. The question states that the time it takes both pumps working together is four times the time it takes X to fill the pump alone, so write 4a in the time box for combined rate. ...............Rate............Time..........Work Pump X..48/a............a................48 Pump Y..48/48 = 1..48...............48 Combined rate X-Y 48/4a = 12/a 4a 48 Now build an equation: combined rate = rate X - rate Y, so: 12/a = 48/a - 1 /·a Solve for a: --> 12 = 48 - a --> a = 36

So what is that correct initial step that will make things easy for me? Excellent question! Glad you asked. And like a good (yet quite annoying) teacher, I'll answer it with a question:

Bob the builder builds 3 walls in 2 hours. Tom the constructor builds 6 walls in 3 hours. Working together, how many walls do they build in 5 hours? We're not going to solve this question. Not yet. There's just one thing I want to know - what in your opinion is the first step in the solution? A...First I'll combine the work of Bob and the work of Tom. 3 walls + 6 walls = 9 walls B...First I'll combine the time it takes Bob and the time it takes Tom to complete the task. 2 hours + 3 hours = 5 hours C...First I'll find the rate of Bob and the rate of Tom, so that on the next step I can combine them. C That's precisely the point! When combining two or more workers working together, it's always the rates that are combined. Make sure your first step is always finding the rate of each worker in the question. Bob: Rate..................Time................Work ...........................2.....................3 Rate: 3/2 = 1.5 walls per hour Tom: Rate..................Time................Work ............................3....................6 Rate = 6/3 = 2 walls per hour Thus the first part of the question is only destined to bring us to the point where we have the rate for each worker seperately. The next sentence in the question describes the task they perform together. The task is building walls, They're given 5 hours to do that. We need their combined rate. Combined rate: Sum all the rates of the workers that work towards completing the task. Bob's rate + Tom's rate = 1.5+2 = 3.5 walls per hour Now we can assign that combined rate in the work table: Rate..................Time................Work ..........................3.5...................5 Work = 5×3.5 = 17.5 walls

The speed of a sprinting cheetah is 108 kilometers per hour. What is the speed of the cheetah in meters per second? 300 180 30 18

C Correct. First of all, assign each of the values in the question to the appropriate cell in the speed table. Then convert the units using the 'replace with an equivalent' principle: Replace the word "kilometers" with the equivalent "×1000 meters"; Replace the word "hour" with the equivalent "60 minutes"; And finally, replace the word "minutes" with the equivalent "×60 seconds". The speed is 10800 / 60 * 60 = 30 m/s

Q28...If P, a vertex of an equilateral triangle of side 6, is the center of a circle, as shown above, what is the length of arc ACB? π 3+π 2π 6+2π 6π

C Correct. In order to find the length of an arc in a circle, you need two pieces of data: the circle's radius and a central or inscribed angle lying on the said arc. Here you have central angle APB, which is also an angle in an equilateral triangle, and thus equals 60º. The side of the triangle = the radius of the circle = 6. Thus, the circumference of the circle = 2πr = 12π. Now plug in the circumference and the central angle to find the arc length: 60/360 = arc/12π --> arc =2π

A certain communication line transmits 250 megabytes in 60 seconds. Which of the following is the approximate number of seconds it takes this line to transmit 60 megabytes? 1 second 11 seconds 14 seconds 16 seconds 250 seconds

C Correct. Rate is calculated as When solving a question, translate it into the Work table. Rate....Time......Work The table represents the formula Rate×Time=Work so that the multiple of the two left columns equals the right column. First calculate the line's rate using the rate table: Rate.....................Time....Work 250/60 = 25/6....60......250 Now find the needed time: Rate....Time.............................Work 25/6...60/(25/6) = (60·6)/25..60 --> (60·6)/25 = (12·6)/5 = 72/5 ≈ 14

Q35...If the perimeter of the square above is 3, what is the length of arc BD, centered at A? π/8 3π/16 3π/8 3π/6 3π/4

C Correct. Since the right angle of the square is also a central angle lying on the arc, the arc is a quarter of a circle (90/360 = 1/4). The radius of the arc is a side of the square, or a quarter of the perimeter=3/4. The arc is therefore 1/4*(2πr)=1/4*(2*π*3/4)=3π/8.

Alice and Bob run at constant speeds around a stadium whose perimeter is 400 meters. Alice runs at 8 m/s, and Bob runs 10% slower. If they start running together in the same direction, how long will it take Alice to be 200 meters ahead of Bob? 2 minutes 50 seconds 4 minute 4 minutes 10 second 8 minutes 25 minutes

C Correct. This is a rate problem with two objects. Since the question asks for the time it takes Alice to be 200 meters ahead of Bob, you need the gap between them and the rate at which this gap is opened. Once you have these two pieces, use the rate box to find the time. The Gap is already given in the question - 200 meters. To find the rate, ask yourself - "what happens after one second?" Use the following diagram to see what's going on: Every second, Alice opens the gap by 8 meters, while Bob closes it by 90%*8 = 7.2 meters. The combined rate of opening the gap is therefore 8-7.2=0.8 meters per second. Place the information in a Rate box: Stage Rate Time Work Opening the gap 0.8 m/sec 200/0.8 = 200×10/8 seconds 200 meters Time = 200×10/8 = 200×5/4 = 1000/4 = 250. Now, remember that these are 250 seconds, while the answer choices are in minutes. Divide 250 seconds by 60 to get 4 minutes and 10 seconds.

Two machines produce optical fibers. Working at a uniform rate, it takes one machine 3 seconds to produce one meter of fiber, and it takes the other machine 5 seconds to produce one meter of fiber. The two machines start working simultaneously and work together until they produce a total of 1200 meters of fiber. Then the faster machine stops to allow the slower machine to catch up. How many more seconds would it take the slower machine to match the amount of fiber produced by the faster machine? 60 600 1500 2000 2400

C Correct. This problem presents a case of two machines, working together, each at its own rate. Deal with the problem in two stages. In the first stage, both machines work together to reach a common goal of 1200 meters. Find the individual rate of each machine, add together to find the combined rate, find the time it takes them to reach the goal, then use rate and time to find the individual work each of them contributed towards that goal. In the second stage, the faster machine stops, and the slower machine continues alone to match the amount of work done by the faster machine in stage I. Find the difference between the amount of fibers for each machine, and use it together with the individual rate of the slow machine to find the required time. Organize the information in a rate box, adding rows as needed. Stage I: both machines work together at a combined rate. ......Rate..........Time......Work Fast machine 1/3 3 1 Slow machine 1/5 5 1 Combined rate 1/3+1/5 = 8/15 ??? 1200 The combined rate is 1/3+1/5 = 5/15+3/15 = 8/15. The time it takes both machines is 1200 / (8/15) = 1200×15/8 = (1200/8)×15 = 150×15 sec. During that time: Working at a rate of 1/3 meter/sec, the fast machine will do 1/3 × 150·15 = 150·5 = 750 meters Working at a rate of 1/5 meter/sec, the slow machine will do 1/5 × 150·15 = 150·3 = 450 meters Stage II: the slow machine works alone to catch up. The difference in work is 750-450 = 300 meters. Enter the final row of the box: Rate Time Work Slow machine (stage II) 1/5 ??? 300 meters It will take the slow machine 300 / 1/5 = 300×5 = 1500 sec to match the fast machine.

Sometimes, Speed problems are just an excuse to check your ability to handle a lot of data. It might look like this: Gustav ran 32 meters uphill at a constant speed, then he ran 36 meters downhill at a faster constant speed, so that his downhill speed was faster by 2 meters per second than his uphill speed. Running uphill took Gustav 2 seconds more than running downhill. Gustav's speed running downhill was how many meters per second? 2 3 4 6 8

C Incorrect. If you feel confused by all the details in this problem, don't blame yourself. The writers of this problem try to confuse. However, you can tackle this complication, if you organize the data in the "speed table": Gustav ran 32 meters uphill at a constant speed, then he ran 36 meters downhill at a faster constant speed, so that his downhill speed was faster by 2 meters per second than his uphill speed. Running uphill took Gustav 2 seconds more than running downhill. Gustav's speed running downhill was how many meters per second? .........speed............time.........distance Up.......v................t+2.............32 Down...v+2............t................36 Assign each of the values in the question to the appropriate cell in the table. Use a new row whenever there is more than one journey, more than one traveler, or more than one segment of the road. A new segment is determined by a change in conditions, usually a change of speed. Here, for example, the speed changes between the way up and the way down. In the table above you can see the relevant excerpts next to each row. Highlight the value you are asked about - In this case it is the downhill speed (in the table above it is in highlighted in red. In the GMAT where you only have a black marker, you'll just circle the relevant cell). What are you going to do now? A..Each row can be translated into an equation according to speed×time=distance. There are two equations and two unknown variables, t and v. I am ready to solve them! B..I'd prefer to plug the answer choices into the table (i.e., into v+2) and eliminate the ones that don't fit in the table. B Way to go! Don't be tempted to solve equations. There is a faster, better option: Plugging In the Answers. Pick up the answer choices one by one and plug them into v+2. Eliminate (POE) the ones that do not fit the rest of the data. Start with answer choice A: v+2=2. Well, that means that v=0, which is absurd. Eliminate this choice. Go on to answer choice B: v+2=3. This means that t=36/3=12 seconds. Now check whether these values also fit the first row, which now reads Speed=v=1, Time=(12+2), Distance=32. Clearly, the values don't fit the first row. POE B and go on to C. When you check answer choice D, everything fits in: v+2=6. This means that v=4 meters per second, and that t=36/6=6 seconds. In the first row the Speed=v=4, Time=t+2=8, and the Distance is 4×8=32 indeed. The correct answer is D.

Q 1...Jasmin and Ron drive at constant speeds toward each other on a certain route. Jasmin drives at a constant speed of 25 miles per hour. At a certain time they pass by each other, and then keep on driving away from each other, maintaining their speeds. If Jasmin is 105 miles away from Ron at 8am, and also 105 miles away from Ron later, at 11am, then Ron is driving at how many miles per hour? 10 15 20 35 45

C Incorrect. This is a rate problem with two objects, Jasmin and Ron. Draw a figure of the problem to see what's going on (initial standpoint on top, final situation on bottom): Every hour, Jasmin moves 25 miles and Ron moves at his own unknown speed. Together they close the gap at 25+[Ron's speed] miles/hour, and use the same speed to open a new gap once they pass each other. They travel at the same combined rate for 3 hours, from 8 to 11. In that time, the close the initial gap of 105 miles, and open an new gap of another 105 miles. Thus, the total gap is 105+105 = 210 miles. Organize this information in a rate box and use it to find the combined rate. Since Combined rate = Jasmin+Ron, use the combined rate and Jasmin's speed to find Ron's speed. E Correct. ........Rate..................Time................Work Jas..25 miles/hr................................... Ron....???.............................................. Combined rate (J+R) 210/3 = 70 mph 3 hours 105+105=210 Jasmin and Ron's total Gap at their combined rate is 105+105=210. Their combined rate is therefore 210/3=70 mph. Since Jasmin travels at 25 mph, it follows that Ron's speed is 70-25 = 45 mph.

Every day Daniel drives 80 miles back from work. On Sunday, Daniel drove all the way back from work at a constant speed of x miles per hour. On Monday, Daniel drove the first 32 miles back from work at (2x) miles per hour, and the rest of the way at (x/2) miles per hour. The time it took Daniel to drive back from work on Monday is longer than the time it took him to drive back from work on Sunday by what percent? 10% 20% 30% 40% 50%

D Correct. This is a speed problem. In order to find the percent difference between the time on Monday and the time on Sunday, you need to find both times. Organize the information in speed box, and calculate what you need. Note that the speed is given in terms of the variable x. Rather than working with an uncomfortable variable, plug in a good number for x. In this case, x=8 is a good number, as the distances in the question (80 miles for the entire road, 32 for the first section at high speed) are both divisible by 8. ...............Speed.........Time.........Distance Sunday...x=8.....80/8 = 10......80 miles Monday..2x = 16.. 32/16 = 2hr..32 miles High Speed Monday..x/2 = 4..48/4 = 12..80-32 = 48 miles Low Speed Thus, on Monday the trip back took a total of 12+2=14 hours. This is 4 hours longer than the 10 hours the trip took on sunday, or 4/10·100 = 40% longer.

Laura and Sue are walking in the same direction along the coast road. Sue is 120 feet ahead of Laura. If Sue is walking at a constant speed of 10 feet per second, and Laura is walking at a constant speed of 15 feet per second, then how many seconds will it take Laura to overtake Sue? 4 12 20 24 30

D Correct. This looks like a Speed Problem, but it involves two moving objects, and two speeds. Turn it into a Rate Problem by focusing on what happens to the gap between the walkers. Draw a sketch of situation in the question to help you determine the rate at which the gap is closed. Then ask yourself "What happens to the gap if Laura and Sue walk for one second?" Finally, organize the information in a ratio box, and find the Time based on the Rate and Work. During the relevant part of the travel, the gap of 120 feet between Laura and Sue is closed. The question is, how long does this take. Thus, the Work is 120 feet. Now determine the rate at which the gap is closed. If Laura and Sue walked for one second, then Laura walks 15 feet that reduce the gap, while Sue walks 10 feet that enlarge the gap. Therefore, the rate of closing the gap is 15−10=5 feet per second. Once you've found the rate of closing the gap, the problem has turned into a standard Rate problem. Put the information in ratio box: Rate Time Work 5 ft/sec ?? 120 feet

A certain car uses one gallon of gasoline every 30 miles when it travels on highway, and one gallon of gasoline every 20 miles when it travels in the city. When a car travels 4 miles on highway and 4 additional miles in the city, it uses what percent more gasoline than if it travels 8 miles on the highway? 15% 20% 22.5% 25% 50%

D Correct. This question compares gasoline consumptions over the same distance in two modes of travel - 8 miles on a highway Vs. 8 miles, half on the highway and half in the city. The difficulty is that 4 and 8 miles do not lend themselves to easy calculations with the given 20 and 30 miles per gallon given in the question. Bypass that difficulty by plugging in a good number for the distance traveled - since the question only asks about percent of use of gasoline, that percent will remain the same over 8 miles or 800 miles, provided that the ratio of of the use of the roads is maintained. Plug in a good number for the distance: a number that is divisible by both 20 and 30, such as 60 miles. The new model of the question is thus: compare 120 miles on the highway to a combination of 60 miles on the highway and 60 miles in the city. Use a slightly modified rate box to organize the information: Consumption (mile/gallon) × gallons = miles So a 120 mile drive on the highway consumes 4 gallons, while a mixed highway/city drive of the same length consumes 2+3=5 gallons. The car therefore uses 1 gallon more, which is 25% more than it will consume on the highway alone.

Before we even begin to discuss the subject of this entity - take a look at the following question: A bucket full of worms kept for Mike's next fishing trip was discovered by the rat that resides in the basement. The rat eats a sixth of the total number of worms in 5 hours. How much time will it take the rat to finish a quarter of the bucket? And our question is - How many worms are in the bucket? Yuck gross. It's hard to think of anything with this image of rat eating worms in my mind. It doesn't say, maybe you ask me again after I solve the problem and I'd be able to calculate it I don't know and I don't care. I'm just using 1 bucket as the total work in the problem. It's undetermined, so I can plug in any number that suits my purposes.

D Exactly. Any number of worms would work. How do you choose the best number of worms to plug-in? A...I was taught to multiply the bottoms. But what would be the bottoms in a rate question? B...What difference does it make? C...I'd multiply 4*5*6 . C Exactly. When calculating rate you divide Work/Time. So the times mentioned in the question are also bottoms of fractions. Thus in our question the bottoms of the fractions are: 6 - since a sixth of the bucket have been eaten. 5 - as the time for that was 5 hours. 4 - since we are asked how long it will take to eat a quarter of the bucket Thus 4*5*6=120 is a good Plug In for the number of worms. The rest follows quite easily so why don't you go ahead and finish the question. 4 hours 6 hours 7.5 hours 8 hours 30 hours c Correct. First determine the rate of the rat: Work: The rat eats sixth of the number of worms. 120/6=20 worms Time: 5 hours Rate= W/T = 20/5 = 4 worms per hour Then calculate the time for eating a quarter of the bucket Time= 30/4=7.5 hours

Dawn chases her young brother Paul. She runs at a constant speed, six times as fast as he does. After Dawn overtakes Paul, she goes on running at the same speed for two more minutes, and then stops and waits for Paul to reach her. If Paul maintains his constant speed throughout the chase, how many minutes does Dawn wait for Paul? 5 6 8 10 12

E Incorrect. This is a rate problem with two objects. Since the question asks for the time it takes Paul to overtake Dawn, you need the gap between them and the rate at which this gap is closed. Once you have these two pieces, use the rate box to find the time. An additional difficulty presented by the question is that it provides no information regarding the actual speed of the two siblings - only a ratio. Overcome this difficulty by treating the question as an invisible unknown - plug in a speed for Paul, and a corresponding speed for Dawn. Then work the problem in two stages: Opening the Gap, and closing the Gap. Do not forget that at the first stage of opening the gap, Paul is still running, thus slowing down the rate of opening the gap between him and Dawn. B Incorrect. D Correct. Plug in 1 meter/min (Paul is very slow, as you can see) for Paul's speed, making Dawn's speed at 6 meter/min. Opening the Gap: First, find the distance, which is the gap opened by Dawn by running for 2 minute ahead of Paul. Every minute, Paul runs 1 meter, while Dawn runs 6 meters. Thus, the gap between them is opened at 6-1=5 meters per minute. Place the numbers in the rate box: Stage Rate..........Time............Work Opening the gap 6-1=5 m/min 2 minutes 2×5=10 meters Now deal with the second stage: closing the gap. At this stage, Dawn stands still, while Paul advances at a constant rate of 1 meters per minute. Thus, the gap between them is closed at a rate of 1 meter per minute. The gap between them is the 10 meters calculated in the first stage. Place the numbers in the rate box: Stage Rate..........Time............Work Opening the gap 6-1=5 m/min 2 minutes 2×5=10 meters Closing the gap 1 m/min 10 / 1 = 10 minutes 10 meters

An astronaut who is freely floating in space throws an expensive camera at a certain direction. Consequently, the astronaut and the camera move in opposite directions. If the astronaut moves at 0.5 meters per second, and the camera moves 10 times faster, how many seconds pass from the moment the astronaut throws the camera to the moment the camera is 55 meters away from the astronaut? 5 5.5 9 10 11

E Incorrect. This looks like a Speed Problem, but it involves two moving objects, and two speeds. It asks about the time it takes to form a 55 meter gap between the astronaut and the camera, so actually this is a "gap" problem. Turn it into a Rate Problem by focusing on what happens to the gap between the astronaut and the camera. Draw a sketch of situation in the question to help you determine the rate at which the gap is closed. Then ask yourself "What happens to the gap when the astronaut and the camera move for one second?" Finally, use the rate equation to solve for the Time based on the Work (=the gap), and the Rate (=closing rate). D Correct. During the relevant part of the space float, the gap between the astronaut and the camera opens up to 55 meters. Thus, the Work is 55 meters. The question is, how long does this take. To find that, you need to determine the rate at which the gap is opened. When the astronaut and the camera move for one second, then the astronaut enlarges the gap by 0.5 meters, and the camera enlarges the gap by 10×0.5=5 meters. Therefore, the rate of opening the gap is 0.5+5=5.5 meters per second. By now, the problem has turned into a standard Rate problem of finding the Time based on the values of the Work and the Rate: "How long does it take to open a 55 meter gap at a rate of 5.5 meters per second?" The answer is pretty obvious, 10 seconds. If you are not sure, use the rate equation to find the Time so - 55m / 5.5m/s = 10 secs