{"id":18371,"date":"2019-12-07T04:28:16","date_gmt":"2019-12-07T04:28:16","guid":{"rendered":"https:\/\/www.manhattanprep.com\/gmat\/?p=18371"},"modified":"2019-12-07T04:28:45","modified_gmt":"2019-12-07T04:28:45","slug":"gmat-rate-problems","status":"publish","type":"post","link":"https:\/\/www.manhattanprep.com\/gmat\/blog\/gmat-rate-problems\/","title":{"rendered":"GMAT Rate Problems"},"content":{"rendered":"

\"gmat<\/span><\/p>\n

If this post is 1500 words long, and you can process 120 words per minute, then how long will it take you to read this whole post? If you could read 20% faster, then what effect would that have on how long it takes you to read the whole thing? If I were adding 80 words per minute to the blog post, then how long (at your original speed) would it take for you to reach the end?<\/span><\/p>\n

Those questions were a taste of the often daunting world of GMAT Rate problems. Before we get any deeper, we should acknowledge that Rate problems do not seem to be tested as frequently on <\/span>GMAT Quant<\/span><\/a> nowadays as they once were. So while you\u2019ll see plenty of Rate problems in the Official Guides and on Manhattan Prep\u2019s practice GMATs (<\/span>take a free one<\/span><\/a>), you might not see many or any of these on your real GMAT.<\/span><\/p>\n

<\/p>\n

Nevertheless, Rate problems are one of the most common requests I get from tutoring students. Because these problems usually come in <\/span>Word Problem format<\/span><\/a> and take on many different flavors, students frequently feel like there is too much density or variety for them to handle. Indeed, there are a variety of moves we might employ, and a variety of formulas\/relationships that would be useful to know, depending on the situation. Let\u2019s discuss.<\/span>
\n<\/span>
\n<\/a>GMAT Rate Problems: Useful Formulas<\/b><\/p>\n

    \n
  1. Rate * Time = Work (or Distance) <\/span><\/li>\n
  2. Time for Two People to do a Job Together = (Product of their separate Times) \/ (Sum of their separate Times)
    \n<\/span><\/li>\n
  3. \u00bd (smaller time)\u00a0 < Time it takes to do a job together\u00a0 < \u00bd (larger time)<\/span><\/li>\n
  4. Rate of 2 or more things working together =\u00a0 Sum of those individual rates<\/li>\n
  5. Average Speed for Entire Trip = (Total Distance) \/ (Total Time)<\/li>\n<\/ol>\n

    At their cores, all Rate problems are testing that 1st equation:<\/span>
    \n<\/span> Rate * Time = Work (or Distance) <\/span>
    \n<\/span>
    \n<\/span>Let\u2019s look back at our initial questions:<\/span>
    \n<\/span>If this post is 1500 words long, and you can process 120 words per minute, then how long will it take you to read this whole post?\u00a0 <\/span><\/i>
    \n<\/span><\/i>
    \n<\/span>One great habit to develop is actually writing the R * T = W or R * T = D formula on your page. Even though you know you know that formula, dumping it on the page allows your brain to take one number or variable at a time from the problem and then place it in the right spot.<\/span>
    \n<\/span>
    \n<\/span>First, I\u2019ll write\u00a0 <\/span>
    \n<\/span>R * T = W<\/span><\/i>
    \n<\/span>
    \n<\/span>Then I\u2019ll think, where does that <\/span>1500 words <\/span><\/i>go?<\/span>
    \n<\/span>R * T = W<\/span><\/i>
    \n<\/span><\/i> \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 1500<\/span><\/i>
    \n<\/span>
    \n<\/span>Where does that <\/span>120 words per minute <\/span><\/i>go? <\/span>
    \n<\/span>R\u00a0 \u00a0 \u00a0 * T = W<\/span><\/i>
    \n<\/span><\/i>120 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 1500<\/span><\/i>
    \n<\/span>
    \n<\/span>Now it\u2019s easier to see that what I need to solve is<\/span>
    \n<\/span>120 * T = 1500<\/span><\/i>
    \n<\/span>
    \n<\/span>So T = 1500\/120, or 150\/12, or 75\/6, or 12.5 minutes.<\/span><\/p>\n

    \n

    Let\u2019s look at another problem:<\/span><\/p>\n

    Romeo can finish a task in 10 hours.\u00a0 Juliet can complete the same task in 8 hours. Working at their respective rates, and not complicated or slowed down by their torrid yet star-crossed love, how long would it take them to complete the task together?<\/span><\/i>
    \n<\/span><\/i><\/p>\n

    This is the type of task for which we could take advantage of our 2<\/span>nd<\/span> and 3<\/span>rd<\/span> formulas.\u00a0<\/span>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/span><\/p>\n

    Time Together =\u00a0 (Product of separate times) \/ (Sum of separate times) = (10 * 8) \/ (10 + 8) =\u00a080 \/ 18 = 40 \/ 9 hrs<\/span><\/p>\n

    If Romeo can do a job in 10 hrs and Juliet can do it in 8 hrs, we\u2019d be tempted to think that it would take them 9 hours working together. But wait\u2014that\u2019s nonsensical!\u00a0 If Juliet can do it in 8 hours by herself, then having Romeo\u2019s help would mean the job would take <\/span>less <\/span><\/i>than 8 hrs. If we want to APPROXIMATE working together, then we do it by thinking about cloning each person. That\u2019s what our 3<\/span>rd<\/span> formula was all about.<\/span><\/p>\n

    \u00bd (smaller time)\u00a0 < Time it takes to do a job together\u00a0 < \u00bd (larger time)<\/span>
    \n<\/span><\/p>\n

    If Romeo were working with his clone, another Romeo, then instead of taking 10 hrs it would take 5 hrs.<\/span><\/p>\n

    If Juliet were working with her clone, another Juliet, then instead of taking 8 hrs it would take 4 hrs.<\/span>
    \n<\/span>
    \n<\/span>So when Juliet and Romeo work together, it will take somewhere between 4 and 5 hrs.\u00a0 It\u2019s a weighted average that always leans a little to the left of the midpoint. We\u2019d approximate a little under 4.5 hrs, and the actual answer (40\/9), was 4 4\/9<\/span>.\u00a0<\/span><\/p>\n

    \n

    When Romeo and Juliet work together, their collective rate is just the sum of their individual rates. <\/span>
    \n<\/span>
    \n<\/span>If it takes Romeo 10 hrs to complete a task, his rate is 1\/10 of the job per hour.<\/span>
    \n<\/span>If Juliet takes 8 hrs, her rate is 1\/8 of the job per hour.<\/span>
    \n<\/span>Working together, their rate is (1\/10 + 1\/8) of the job per hour, or 9\/40 of the job per hour.<\/span>
    \n<\/span>
    \n<\/span>Most of us aren\u2019t great about thinking in terms of fractions, so frequently on Rate problems, we can make life easier by <\/span>Making Up Numbers<\/b>.<\/span><\/p>\n

    On Work problems, we usually make up a total number of boxes to make. If the problem involves pumps filling up a tank, then we make up a total number of gallons. For Distance problems, we usually make up a total number of miles for the distance. In both cases, the easiest number to work with is the least common multiple of the times or rates involved in the problem.<\/span><\/p>\n

    Since Romeo takes 10 hrs and Juliet takes 8 hrs, we would choose 40 boxes.<\/span><\/p>\n

    If it takes Romeo 10 hrs to make 40 boxes, then he makes 4 box \/ hr.<\/span><\/p>\n

    If it takes Juliet 8 hrs to make 40 boxes, then she makes 5 box \/ hr.<\/span><\/p>\n

    Working together, they are making 9 box \/ hr.<\/span><\/p>\n

    Rate of 2 or more things working together =\u00a0 Sum of those individual rates<\/span><\/p>\n

    \n

    Here\u2019s another problem to take a look at applying Rate formulas:<\/span><\/p>\n

    Lebron drives 50 mph on the way to Kobe\u2019s house and then drives home along the same route at 30mph. What is his average speed for the entire trip?<\/span><\/i><\/p>\n

    If I go 50mph one way and 30mph the other, it\u2019s tempting to think that my average speed would be 40mph. But in these \u201cround trip\u201d stories, we have to remember that we spend more time driving the slower speed, so the slower speed is more represented in the average. Hence, we can always approximate that the average speed will always be a little lower than the simple average <\/i>of the two speeds. We can Approximate<\/b> that Lebron\u2019s average speed will be a little below 40mph.<\/span><\/p>\n

    If we Make Up Numbers<\/b>, we can solve pretty easily for it.\u00a0 We just want a common multiple of 30 and 50, so 150 miles or 300 miles or anything like that.<\/p>\n

    If it\u2019s 150 miles each way, then the 50mph trip takes 3 hours, and the 30mph trip takes 5 hours.<\/p>\n

    \"gmat<\/span><\/h3>\n

    GMAT Rate Problems: Going Beyond the Basics<\/b><\/p>\n

    Let’s remind ourselves that Rate is actually a <\/span>ratio<\/b>. It will always be expressed as \u201csome unit of work\/distance PER some unit of time\u201d<\/span><\/p>\n

    We usually think of 30 miles per hour as 30mph, but it would behoove us to remember that this is a fraction:<\/span><\/p>\n

    R = 30 miles \/ 1 hour<\/span><\/i>
    \n<\/span>
    \n<\/span>Sometimes we are given clunky looking rates, like \u201cBen takes 3 minutes to stamp 7 envelopes\u201d.\u00a0 If we write that as a fraction, we want to make sure time is on the bottom, as we\u2019re used to seeing it.<\/span>
    \n<\/span>Rate = 7 env \/ 3 min<\/span><\/i>
    \n<\/span>
    \n<\/span>If we wanted to know how many envelopes Ben does per hour, we could do a <\/span>Unit Conversion<\/b> from minutes to hours. We could solve for his Rate per minute (he makes 7\/3 envelopes per minute) and then multiply by 60. But, we could also make use of the idea of <\/span>Scaling Up Ratios<\/b>.<\/span>
    \n<\/span>
    \n<\/span>If he does 7 envelopes in 3 minutes, then he\u2019ll do ___ envelopes in 6 minutes?<\/span>
    \n<\/span>14 of course. Twice as much time, twice as many envelopes.\u00a0<\/span><\/p>\n

    We can use that simple logic with any sort of multiplier.<\/span>
    \n<\/span><\/p>\n

    If he does 7 envelopes in 3 minutes, then he\u2019ll do ___ envelopes in 60 minutes?<\/span>
    \n<\/span>140 envelopes.\u00a0 Twenty times as many minutes, twenty times as many envelopes.<\/span>
    \n<\/span><\/p>\n

    Writing Rates horizontally as a ratio, and then scaling them up or down as needed, is often a quicker\/easier way to get to your destination.<\/span>
    \n<\/span>Our second initial question:<\/span>
    \n<\/span>If you could read 20% faster, then what effect would that have on how long it takes you to read the whole thing?\u00a0 <\/span><\/i>
    \n<\/span><\/i>
    \n<\/span>I was originally asking this question in the context of the 1500 word blog post and 120 words\/minute reading speed. But in reality, the question doesn\u2019t need any other information beyond the 20% faster. This is the realm of <\/span>Reciprocal Thinking<\/b>.<\/span><\/p>\n

    We\u2019re all familiar with the common sense reciprocal idea that \u201cIf I could read TWICE as fast, it would only take me HALF as long to read this.\u201d\u00a0 That truism is playing off the idea that that 2\/1 and 1\/2 are reciprocals.<\/span>
    \n<\/span>
    \n<\/span>In order to do this with 20% faster, we need to already be fluent with our percentage and fraction conversions.<\/span>
    \n<\/span>20% more than x = 120% of x<\/span>
    \n<\/span>20% less than x = 80% of x<\/span>
    \n<\/span>
    \n<\/span>For the GMAT, we\u2019re supposed to get really good at switching from percents into simplified fractions. When a GMAT student sees 20%, she thinks 1\/5. <\/span>
    \n<\/span>1\/5 more than x = 6\/5 of x<\/span>
    \n<\/span>1\/5 less than x = 4\/5 of x<\/span><\/p>\n

    So when we think about someone reading 20% faster, we think about them reading 120% as fast, or 6\/5 as fast.<\/span>
    \n<\/span>When I read twice as fast (2\/1) it takes me 1\/2 the time.<\/span><\/p>\n

    When I read 6\/5 as fast, it takes me 5\/6 the time.<\/span>
    \n<\/span><\/p>\n

    So, <\/span>If you could read 20% faster, then what effect would that have on how long it takes you to read the whole thing?\u00a0\u00a0<\/span><\/i><\/p>\n

    It would take you 5\/6 as long. You would save 1\/6 the time of what you previously had to spend.<\/span><\/p>\n

    <\/i>This comes into play on tricky DS questions.<\/span><\/p>\n

    How long did it take for Kanye to drive to the White House?<\/i>
    \n<\/i>(1)\u00a0 The White House was 240 miles away<\/i>
    \n<\/i>(2)\u00a0 Had Kanye driven 30% faster, the trip would have taken 30 minutes less<\/i>
    \n<\/i>
    \n<\/i>The 1st statement is insufficient, since only knowing distance doesn\u2019t suffice. We would need to know Kanye\u2019s rate to calculate his time. The 2nd statement doesn\u2019t feel like enough information, but it is.<\/span><\/p>\n

    Driving 30% faster means driving 130% as fast as he actually did.
    \nDriving 3\/10 faster means driving 13\/10 as fast as he actually did. Thus it will only take 10\/13 as much time.
    \nKanye\u2019s trip would take 3\/13 less time. Since statement 2 told us it would take 30 minutes less time, we know that 30 mins = 3\/13 (his actual time)<\/p>\n

    So apparently, his actual time was 130 minutes.\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0
    \n<\/span><\/p>\n

    \n

    Last of the initial three questions:<\/span><\/p>\n

    If I were adding 80 words per minute to the blog post, which started at 1500 words, then how long would it take for you to reach the end, if you can read 120 words per minute?<\/span><\/i>
    \n<\/span><\/i>
    \n<\/span>This sort of problem is similar to those in which you have water leaking out of a tank at the same time that you\u2019re adding water into the tank.\u00a0 It\u2019s similar to problems in which Person A is trying to catch up with Person B, even while Person B is driving away from them.<\/span><\/p>\n

    What we need to consider for this sort of task is the <\/span>Net Rate<\/b>. If you\u2019re reading 120 words per minute while I\u2019m adding 80 words per minute, then you\u2019re getting 40 words closer to catching up with me every minute. If a tank is leaking 3 gallons per hour while a hose is filling it at a rate of 8 gallons per hour, then 5 gallons per hour is being added to the tank.<\/span>
    \n<\/span>If Person A is chasing Person B at a speed of 20mph while Person B is running away at a speed of 12mph, then Person A is getting 8 miles closer each hour.<\/span><\/p>\n

    Let\u2019s talk about ways to solve the original one:<\/span>
    \n<\/span>If I were adding 80 words per minute to the blog post, which started at 1500 words, then how long would it take for you to reach the end, if you can read 120 words per minute?<\/span><\/i>
    \n<\/span><\/i>
    \n<\/span>We could write out our normal equation and think about the variables:<\/span>
    \n<\/span> R\u00a0 \u00a0 * T\u00a0 \u00a0 \u00a0 = W<\/span>
    \n<\/span>
    \n<\/span>Do we already know the rate of reading, the time it will take to read, or the total work (word count) of reading?<\/span>
    \n<\/span>
    \n<\/span>We know the rate of reading is 120.<\/span><\/p>\n

    R\u00a0 \u00a0 * T\u00a0 \u00a0 \u00a0 = W<\/span>
    \n<\/span>120<\/span><\/p>\n

    We don\u2019t know time, because that\u2019s what we\u2019re solving for. We almost know Work, 1500 words, but there\u2019s that wrinkle that more words will be added. How many words?\u00a0 We don\u2019t know. But it will be 80 words per minute for whatever period of minutes it takes. So I guess we could say this:<\/span>
    \n<\/span> \u00a0 R \u00a0 * T\u00a0 \u00a0 \u00a0 = W<\/span>
    \n<\/span>120 \u00a0 * T\u00a0 \u00a0 \u00a0 = 1500 + (80 * T)<\/span><\/p>\n

    That\u2019s hard for me to reliably come up with, in the pressure cooker of a 2 minute GMAT problem. If I think of it instead as a net rate of 40 words per minute, then it\u2019s more straightforward:<\/span>
    \n<\/span>\u00a0 R \u00a0 \u00a0 * T \u00a0 = W<\/span>
    \n<\/span>40 \u00a0 \u00a0 * T \u00a0 \u00a0 = 1500<\/span><\/p>\n

    If we\u2019re not confident in our abilities to pull off the algebra, then we can always go to an <\/span>Hour by Hour Table. <\/b>In this case, it would be minute by minute.<\/span><\/p>\n

    \"gmat<\/p>\n

    Some problems need that \u201c0 hr\u201d or \u201c0 mins\u201d line because you need to record some initial conditions. Then you just go up one unit of time and calculate what\u2019s going on.<\/span>
    \n<\/span>
    \n<\/span>We know that in 1 minute, you read 120 and I add 80. So we could think 1500 \u2013 120 = 1380 words left, but then Patrick annoyingly adds 80 words, so there are 1460 left.<\/span>
    \n<\/span>
    \n<\/span>We might need to only do one more line to grasp the pattern here.<\/span><\/p>\n

    \"gmat<\/p>\n

    There were 1500 words. You\u2019ve read 240, so that would leave 1260. But I\u2019ve added 160, so it\u2019s really left at 1420.<\/span><\/p>\n

    Once we see how that rightmost column is trending …. 1500, 1460, 1420 … then we\u2019ll stumble upon the <\/span>Net Rate<\/b> inference that we\u2019re basically getting 40 words closer every minute. At that point, we can stop calculating all three columns and just worry about the one we care about.<\/span><\/p>\n

    \"gmat<\/p>\n

    In the example I came up with, it would take unrealistically long to work our way down from 1340 to 0, minute by minute. But in actual GMAT problems, you can usually get there within 5 to 10 lines. You can also start to move forward by chunks.\u00a0 If you know that we do 40 words per minute, you could take a bigger jump and think \u201c10 minutes later, I\u2019ll be 400 words farther\u201d.<\/span><\/p>\n

    \"gmat<\/p>\n

    One last example:<\/span>
    \n<\/span>Hobbes & Shaw are 360 miles apart. Hobbes heads due east at 20mph and Shaw heads due west at 40mph. How long will it take until they meet?<\/span><\/i><\/p>\n

    I realize that 20mph and 40mph are neither Fast nor Furious, but that\u2019s okay. They already have a bunch of points on their license for knocking a helicopter out of the sky with their cars.<\/span>
    \n<\/span>
    \n<\/span>We could do an <\/span>Hour by Hour Table<\/b> for this kind of problem.<\/span>
    \n<\/span><\/p>\n

    \"gmat<\/p>\n

    Again, by this point we would see the trend with our final column:\u00a0 it keeps decreasing by 60 miles. So we could just work that column until we get our answer:\u00a0 3 hrs = 180mi, 4 hrs = 120 mi, 5 hrs = 60 mi, 6 hrs. = 0 mi.<\/span>
    \n<\/span>
    \n<\/span>Only 6 lonely hours until they can consummate their bromance with a tender fist bump.<\/span>
    \n<\/span>
    \n<\/span>It\u2019s possible to set up an R * T = D equation for each person and then write an equation that relates their distances together so that they add up to 360.<\/span>
    \n<\/span>Hobbes: \u00a0 20 * T<\/span>h<\/sub><\/span> = D<\/span>h<\/sub><\/span>
    \n<\/span>Shaw:\u00a0 40 * T<\/span>s<\/sub><\/span> = D<\/span>s<\/sub><\/span>
    \n<\/span>D<\/span>h<\/sub><\/span> + D<\/span>s<\/sub><\/span> = 360<\/span>
    \n<\/span>
    \n<\/span>But, an easier path involves putting this all into one SHORTCUT FORMULA, by thinking about their aggregate rate and distance.<\/span>
    \n<\/span>
    \n<\/span>R\u00a0 \u00a0 \u00a0 * T \u00a0 = D<\/span>
    \n<\/span>
    \n<\/span>The aggregate distance they\u2019ll cover is 360 miles, so D = 360.<\/span>
    \n<\/span>What is their aggregate rate?\u00a0 i.e. \u201cIn one hour, how much closer to they get to each other?\u201d\u00a0 60 mph.<\/span>
    \n<\/span>
    \n<\/span>So we could just do<\/span>
    \n<\/span>R \u00a0 * T \u00a0 = D<\/span>
    \n<\/span>60\u00a0 * T\u00a0 = 360<\/span><\/p>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0T = 6 hrs<\/span>
    \n<\/span><\/p>\n

    GMAT Rate Problems: Takeaways<\/b><\/h3>\n