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GRE Quant - Average, Mixture, Rate, and Work Problems Theory
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14 Jun 2020, 03:38
14 Jun 2020, 03:38
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Frequency of the concepts tested: Medium
- Word Problems Made Easy
- ‘Work’ Word Problems Made Easy
- 'Distance/Speed/Time' Word Problems Made Easy
- Word Problems - Translating Statements into Algebraic Relationships
- Mixture Problems
Attachment:
- This is an introductory post to word problems.
- It deals primarily with the translation of word problems into equations.
- Discussions relating to specific types of word problems will be dealt with separately (see end of post).
The Following Points Outline a General Approach to Word Problems:
1) Read the entire question carefully and get a feel for what is happening. Identify what kind of word problem you're up against.
2) Make a note of exactly what is being asked.
3) Simplify the problem - this is what is usually meant by 'translating the English to Math'. Draw a figure or table. Sometimes a simple illustration makes the problem much easier to approach.
4) It is not always necessary to start from the first line. Invariably, you will find it easier to define what you have been asked for and then work backwards to get the information that is needed to obtain the answer.
5) Use variables (a, b, x, y, etc.) or numbers (100 in case of percentages, any common multiple in case of fractions, etc.) depending on the situation.
6) Use SMART values. Think for a moment and choose the best possible value that would help you reach the solution in the quickest possible time. DO NOT choose values that would serve only to confuse you. Also, remember to make note of what the value you selected stands for.
7) Once you have the equations written down it's time to do the math! This is usually quite simple. Be very careful so as not to make any silly mistakes in calculations.
8) Lastly, after solving, cross check to see that the answer you have obtained corresponds to what was asked. The makers of these GMAT questions love to trick students who don’t pay careful attention to what is being asked. For example, if the question asks you to find ‘what fraction of the remaining...’ you can be pretty sure one of the answer choices will have a value corresponding to ‘what fraction of the total…’
Translating Word Problems
These are a few common English to Math translations that will help you break down word problems. My recommendation is to refer to them only in the initial phases of study. With practice, decoding a word problem should come naturally. If, on test day, you still have to try and remember what the math translations to some English term is, you haven’t practiced enough!
ADDITION: increased by ; more than ; combined ; together ; total of ; sum ; added to ; and ; plus
SUBTRACTION: decreased by ; minus ; less ; difference between/of ; less than ; fewer than ; minus ; subtracted from
MULTIPLICATION: of ; times ; multiplied by ; product of ; increased/decreased by a factor of (this type can involve both addition or subtraction and multiplication!)
DIVISION: per ; out of ; ratio of ; quotient of ; percent (divide by 100) ; divided by ; each
EQUALS: is ; are ; was ; were ; will be ; gives ; yields ; sold for ; has ; costs ; adds up to ; the same as ; as much as
VARIABLE or VALUE: a number ; how much ; how many ; what
Some Tricky Forms:
'per' means 'divided by'
Jack drove at a speed of 40 miles per hour OR 40 miles/hour.
'a' sometimes means 'divided by'
Jack bought twenty-four eggs for $3 a dozen.
'less than'
In English, the ‘less than’ construction is reverse of what it is in math.
For example, ‘3 less than x’ means ‘x – 3’ NOT ‘3 – x’
Similarly, if the question says ‘Jack’s age is 3 less than that of Jill’, it means that Jacks age is ‘Jill’s age – 3’.
The ‘how much is left’ construction
Sometimes, the question will give you a total amount that is made up of a number of smaller amounts of unspecified sizes. In this case, just assign a variable to the unknown amounts and the remaining amount will be what is left after deducting this named amount from the total.
Consider the following:
A hundred-pound order of animal feed was filled by mixing products from Bins A, B and C, and that twice as much was added from Bin C as from Bin A.
Let "a" stand for the amount from Bin A. Then the amount from Bin C was "2a", and the amount taken from Bin B was the remaining portion of the hundred pounds: 100 – a – 2a.
In the following cases, order is important:
‘quotient/ratio of’ construction
If a problems says ‘the ratio of x and y’, it means ‘x divided by y’ NOT 'y divided by x'
‘difference between/of’ construction
If the problem says ‘the difference of x and y’ it means |x – y|
Now that we have seen how it is possible, in theory, to break down word problems, lets go through a few simple examples to see how we can apply this knowledge.
Example 1.
The length of a rectangular garden is 2 meters more than its width. Express its length in terms of its width.
Solution:
Key words: more than (implies addition); is (implies equal to)
Thus, the phrase ‘length is 2 more than width’ becomes:
Length = 2 + width
Example 2.
The length of a rectangular garden is 2 meters less than its width. Express its length in terms of its width.
Solution:
Key words: less than (implies subtraction but in reverse order); is (implies equal to)
Thus, the phrase ‘length is 2 less than width’ becomes:
Length = width - 2
Example 3.
The length of a rectangular garden is 2 times its width. Express its length in terms of its width.
Solution:
Key words: times (implies multiplication); is (implies equal to)
Thus, the phrase ‘length is 2 times width’ becomes:
Length = 2*width
Example 4.
The ratio of the length of a rectangular garden to its width is 2. Express its length in terms of its width.
Solution:
Key words: ratio of (implies division); is (implies equal to)
Thus, the phrase ‘ratio of length to width is 2’ becomes:
Length/width = 2 → Length = 2*width
Example 5.
The length of a rectangular garden surrounded by a walkway is twice its width. If difference between the length and width of just the rectangular garden is 10 meters, what will be the width of the walkway if just the garden has width 6 meters?
Solution:
Ok this one has more words than the previous examples, but don’t worry, lets break it down and see how simple it becomes.
Key words: and (implies addition); twice (implies multiplication); difference between (implies subtraction where order is important); what (implies variable); is, will be (imply equal to)
Since this is a slightly more complicated problem, let us first define what we want.
'What will be the width of the walkway' implies that we should assign a variable for width of the walkway and find its value.
Thus, let width of the walkway be ‘x’.
Now, in order to find the width of walkway, we need to have some relation between the total length/width of the rectangular garden + walkway and the length/width of just the garden.
Notice here that if we assign a variables to the width and length of either garden+walkway or just garden, we can express every thing in terms of just these variables.
So, let length of the garden+walkway = L
And width of garden+walkway = W
Thus length of just garden = L – 2x
Width of just garden = W - 2x
Note: Remember that the walkway completely surrounds the garden. Thus its width will have to be accounted for twice in both the total length and total width.
Now let’s see what the question gives us.
‘Garden with width 6 meters’ translates to:
Width of garden = 6
W – 2x = 6
Thus, if we know W we can find x.
‘Length of a rectangular garden surrounded by walkway is twice its width’ translates to:
Length of garden + length of walkway = 2*(width of garden + width of walkway)
L = 2*W
‘Difference between the length and width of just the rectangular garden is 10 meters’ translates to:
Length of garden – width of garden = 10
(L – 2x) – (W – 2x) = 10
L – W = 10
Now, since we have two equations and two variables (L and W), we can find their values. Solving them we get: L = 20 and W = 10.
Thus, since we know the value of W, we can calculate ‘x’
10 – 2x = 6
2x = 4
x = 2
Thus, the width of the walkway is 2 meters.
Easy wasn't it?
With practice, writing out word problems in the form of equations will become second nature. How much you need to practice depends on your own individual ability. It could be 10 questions or it could be 100. But once you’re able to effortlessly translate word problems into equations, more than half your battle will already be won.
Re: GRE Quant - Average, Mixture, Rate, and Work Problems Theory
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14 Jun 2020, 09:52
14 Jun 2020, 09:52
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What is a ‘Work’ Word Problem?
Sounds simple enough doesn’t it? Well it is!
There is just one simple concept you need to understand in order to solve any ‘work’ related word problem.
The ‘Work’ Problem Concept
STEP 1: Calculate how much work each person/machine does in one unit of time (could be days, hours, minutes, etc).
How do we do this? Simple. If we are given that A completes a certain amount of work in X hours, simply reciprocate the number of hours to get the per hour work. Thus in one hour, A would complete \(\frac{1}{X}\) of the work. But what is the logic behind this? Let me explain with the help of an example.
Assume we are given that Jack paints a wall in 5 hours. This means that in every hour, he completes a fraction of the work so that at the end of 5 hours, the fraction of work he has completed will become 1 (that means he has completed the task).
Thus, if in 5 hours the fraction of work completed is 1, then in 1 hour, the fraction of work completed will be (1*1)/5
STEP 2: Add up the amount of work done by each person/machine in that one unit of time.
This would give us the total amount of work completed by both of them in one hour. For example, if A completes \(\frac{1}{X}\) of the work in one hour and B completes \(\frac{1}{Y}\) of the work in one hour, then TOGETHER, they can complete \(\frac{1}{X}+\frac{1}{Y}\) of the work in one hour.
STEP 3: Calculate total amount of time taken for work to be completed when all persons/machines are working together.
The logic is similar to one we used in STEP 1, the only difference being that we use it in reverse order. Suppose \(\frac{1}{X}+\frac{1}{Y}=\frac{1}{Z}\). This means that in one hour, A and B working together will complete \(\frac{1}{Z}\) of the work. Therefore, working together, they will complete the work in Z hours.
Advice here would be: DON'T go about these problems trying to remember some formula. Once you understand the logic underlying the above steps, you will have all the information you need to solve any ‘work’ related word problem. (You will see that the formula you might have come across can be very easily and logically deduced from this concept).
Now, lets go through a few problems so that the above-mentioned concept becomes crystal clear. Lets start off with a simple one :
Example 1.
Jack can paint a wall in 3 hours. John can do the same job in 5 hours. How long will it take if they work together?
Solution:
This is a simple straightforward question wherein we must just follow steps 1 to 3 in order to obtain the answer.
STEP 1: Calculate how much work each person does in one hour.
Jack → (1/3) of the work
John → (1/5) of the work
STEP 2: Add up the amount of work done by each person in one hour.
Work done in one hour when both are working together = \(\frac{1}{3}+\frac{1}{5}=\frac{8}{15}\)
STEP 3: Calculate total amount of time taken when both work together.
If they complete \(\frac{8}{15}\) of the work in 1 hour, then they would complete 1 job in \(\frac{15}{8}\) hours.
Example 2.
Working, independently X takes 12 hours to finish a certain work. He finishes 2/3 of the work. The rest of the work is finished by Y whose rate is 1/10 of X. In how much time does Y finish his work?
Solution:
Now the only reason this is trickier than the first problem is because the sequence of events are slightly more complicated. The concept however is the same. So if our understanding of the concept is clear, we should have no trouble at all dealing with this.
‘Working, independently X takes 12 hours to finish a certain work’ This statement tells us that in one hour, X will finish \(\frac{1}{12}\) of the work.
‘He finishes 2/3 of the work’ This tells us that \(\frac{1}{3}\) of the work still remains.
‘The rest of the work is finished by Y whose rate is (1/10) of X’ Y has to complete \(\frac{1}{3}\) of the work.
‘Y's rate is (1/10) that of X‘. We have already calculated rate at which X works to be \(\frac{1}{12}\). Therefore, rate at which Y works is \(\frac{1}{10}*\frac{1}{12}=\frac{1}{120}\).
‘In how much time does Y finish his work?’ If Y completes \(\frac{1}{120}\) of the work in 1 hour, then he will complete \(\frac{1}{3}\) of the work in 40 hours.
So as you can see, even though the question might have been a little difficult to follow at first reading, the solution was in fact quite simple. We didn’t use any new concepts. All we did was apply our knowledge of the concept we learnt earlier to the information in the question in order to answer what was being asked.
Example 3.
Working together, printer A and printer B would finish a task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A?
Solution:
This problem is interesting because it tests not only our knowledge of the concept of word problems, but also our ability to ‘translate English to Math’
‘Working together, printer A and printer B would finish a task in 24 minutes’ This tells us that A and B combined would work at the rate of \(\frac{1}{24}\) per minute.
‘Printer A alone would finish the task in 60 minutes’ This tells us that A works at a rate of \(\frac{1}{60}\) per minute.
At this point, it should strike you that with just this much information, it is possible to calculate the rate at which B works: Rate at which B works = \(\frac{1}{24}-\frac{1}{60}=\frac{1}{40}\).
‘B prints 5 pages a minute more than printer A’ This means that the difference between the amount of work B and A complete in one minute corresponds to 5 pages. So, let us calculate that difference. It will be \(\frac{1}{40}-\frac{1}{60}=\frac{1}{120}\)
‘How many pages does the task contain?’ If \(\frac{1}{120}\) of the job consists of 5 pages, then the 1 job will consist of \(\frac{(5*1)}{\frac{1}{120}} = 600\) pages.
Example 4.
Machine A and Machine B are used to manufacture 660 sprockets. It takes machine A ten hours longer to produce 660 sprockets than machine B. Machine B produces 10% more sprockets per hour than machine A. How many sprockets per hour does machine A produce?
Solution:
The rate of A is \(\frac{660}{t+10}\) sprockets per hour;
The rate of B is \(\frac{660}{t}\) sprockets per hour.
We are told that B produces 10% more sprockets per hour than A, thus \(\frac{660}{t+10}*1.1=\frac{660}{t}\) --> \(t=100\) --> the rate of A is \(\frac{660}{t+10}=6\) sprockets per hour.
As you can see, the main reason the 'tough' problems are 'tough' is because they test a number of other concepts apart from just the ‘work’ concept. However, once you manage to form the equations, they are really not all that tough.
And as far as the concept of ‘work’ word problems is concerned – it is always the same!
- It involves a number of people or machines working together to complete a task.
- We are usually given individual rates of completion.
- We are asked to find out how long it would take if they work together.
Sounds simple enough doesn’t it? Well it is!
There is just one simple concept you need to understand in order to solve any ‘work’ related word problem.
The ‘Work’ Problem Concept
STEP 1: Calculate how much work each person/machine does in one unit of time (could be days, hours, minutes, etc).
How do we do this? Simple. If we are given that A completes a certain amount of work in X hours, simply reciprocate the number of hours to get the per hour work. Thus in one hour, A would complete \(\frac{1}{X}\) of the work. But what is the logic behind this? Let me explain with the help of an example.
Assume we are given that Jack paints a wall in 5 hours. This means that in every hour, he completes a fraction of the work so that at the end of 5 hours, the fraction of work he has completed will become 1 (that means he has completed the task).
Thus, if in 5 hours the fraction of work completed is 1, then in 1 hour, the fraction of work completed will be (1*1)/5
STEP 2: Add up the amount of work done by each person/machine in that one unit of time.
This would give us the total amount of work completed by both of them in one hour. For example, if A completes \(\frac{1}{X}\) of the work in one hour and B completes \(\frac{1}{Y}\) of the work in one hour, then TOGETHER, they can complete \(\frac{1}{X}+\frac{1}{Y}\) of the work in one hour.
STEP 3: Calculate total amount of time taken for work to be completed when all persons/machines are working together.
The logic is similar to one we used in STEP 1, the only difference being that we use it in reverse order. Suppose \(\frac{1}{X}+\frac{1}{Y}=\frac{1}{Z}\). This means that in one hour, A and B working together will complete \(\frac{1}{Z}\) of the work. Therefore, working together, they will complete the work in Z hours.
Advice here would be: DON'T go about these problems trying to remember some formula. Once you understand the logic underlying the above steps, you will have all the information you need to solve any ‘work’ related word problem. (You will see that the formula you might have come across can be very easily and logically deduced from this concept).
Now, lets go through a few problems so that the above-mentioned concept becomes crystal clear. Lets start off with a simple one :
Example 1.
Jack can paint a wall in 3 hours. John can do the same job in 5 hours. How long will it take if they work together?
Solution:
This is a simple straightforward question wherein we must just follow steps 1 to 3 in order to obtain the answer.
STEP 1: Calculate how much work each person does in one hour.
Jack → (1/3) of the work
John → (1/5) of the work
STEP 2: Add up the amount of work done by each person in one hour.
Work done in one hour when both are working together = \(\frac{1}{3}+\frac{1}{5}=\frac{8}{15}\)
STEP 3: Calculate total amount of time taken when both work together.
If they complete \(\frac{8}{15}\) of the work in 1 hour, then they would complete 1 job in \(\frac{15}{8}\) hours.
Example 2.
Working, independently X takes 12 hours to finish a certain work. He finishes 2/3 of the work. The rest of the work is finished by Y whose rate is 1/10 of X. In how much time does Y finish his work?
Solution:
Now the only reason this is trickier than the first problem is because the sequence of events are slightly more complicated. The concept however is the same. So if our understanding of the concept is clear, we should have no trouble at all dealing with this.
‘Working, independently X takes 12 hours to finish a certain work’ This statement tells us that in one hour, X will finish \(\frac{1}{12}\) of the work.
‘He finishes 2/3 of the work’ This tells us that \(\frac{1}{3}\) of the work still remains.
‘The rest of the work is finished by Y whose rate is (1/10) of X’ Y has to complete \(\frac{1}{3}\) of the work.
‘Y's rate is (1/10) that of X‘. We have already calculated rate at which X works to be \(\frac{1}{12}\). Therefore, rate at which Y works is \(\frac{1}{10}*\frac{1}{12}=\frac{1}{120}\).
‘In how much time does Y finish his work?’ If Y completes \(\frac{1}{120}\) of the work in 1 hour, then he will complete \(\frac{1}{3}\) of the work in 40 hours.
So as you can see, even though the question might have been a little difficult to follow at first reading, the solution was in fact quite simple. We didn’t use any new concepts. All we did was apply our knowledge of the concept we learnt earlier to the information in the question in order to answer what was being asked.
Example 3.
Working together, printer A and printer B would finish a task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A?
Solution:
This problem is interesting because it tests not only our knowledge of the concept of word problems, but also our ability to ‘translate English to Math’
‘Working together, printer A and printer B would finish a task in 24 minutes’ This tells us that A and B combined would work at the rate of \(\frac{1}{24}\) per minute.
‘Printer A alone would finish the task in 60 minutes’ This tells us that A works at a rate of \(\frac{1}{60}\) per minute.
At this point, it should strike you that with just this much information, it is possible to calculate the rate at which B works: Rate at which B works = \(\frac{1}{24}-\frac{1}{60}=\frac{1}{40}\).
‘B prints 5 pages a minute more than printer A’ This means that the difference between the amount of work B and A complete in one minute corresponds to 5 pages. So, let us calculate that difference. It will be \(\frac{1}{40}-\frac{1}{60}=\frac{1}{120}\)
‘How many pages does the task contain?’ If \(\frac{1}{120}\) of the job consists of 5 pages, then the 1 job will consist of \(\frac{(5*1)}{\frac{1}{120}} = 600\) pages.
Example 4.
Machine A and Machine B are used to manufacture 660 sprockets. It takes machine A ten hours longer to produce 660 sprockets than machine B. Machine B produces 10% more sprockets per hour than machine A. How many sprockets per hour does machine A produce?
Solution:
The rate of A is \(\frac{660}{t+10}\) sprockets per hour;
The rate of B is \(\frac{660}{t}\) sprockets per hour.
We are told that B produces 10% more sprockets per hour than A, thus \(\frac{660}{t+10}*1.1=\frac{660}{t}\) --> \(t=100\) --> the rate of A is \(\frac{660}{t+10}=6\) sprockets per hour.
As you can see, the main reason the 'tough' problems are 'tough' is because they test a number of other concepts apart from just the ‘work’ concept. However, once you manage to form the equations, they are really not all that tough.
And as far as the concept of ‘work’ word problems is concerned – it is always the same!
Re: GRE Quant - Average, Mixture, Rate, and Work Problems Theory
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14 Jun 2020, 09:53
14 Jun 2020, 09:53
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What is a ‘D/S/T’ Word Problem?
The ‘D/S/T’ Formula: Distance = Speed x Time
I’m sure most of you are already familiar with the above formula (or some variant of it). But how many of you truly understand what it signifies?
When you see a ‘D/S/T’ question, do you blindly start plugging values into the formula without really understanding the logic behind it? If then answer to that question is yes, then you would probably have noticed that your accuracy isn’t quite where you’d want it to be.
My advice here, as usual, is to make sure you understand the concept behind the formula rather than just using it blindly.
So what’s the concept? Lets find out!
In the next section we will learn how to do just that.
Resolving the Components
Warning: Make sure that the units for time and distance agree with the units for the rate. For instance, if they give you a rate of feet per second, then your time must be in seconds and your distance must be in feet. Sometimes they try to trick you by using the wrong units, and you have to catch this and convert to the correct units.
A Few More Points to Note
Now that we know the concept in theory, let us see how it works practically, with the help of a few examples.
Note for tables : All values in black have been given in the question stem. All values in blue have been calculated.
Example 1.
To qualify for a race, you need to average 60 mph driving two laps around a 1-mile long track. You have some sort of engine difficulty the first lap so that you only average 30 mph during that lap; how fast do you have to drive the second lap to average 60 for both of them?
Solution:
Let us first start with a problem that has already been introduced. You will see that by clearly listing out the given data in tabular form, we eliminate any scope for confusion.
In the first row, we are given the distance and the speed. Thus it is possible to calculate the time.
Time(1) = Distance(1)/Speed(1) = 1/30
In the second row, we are given just the distance. Since we have to calculate speed, let us give it a variable 'x'. Now, by using the 'D/S/T' relationship, time can also be expressed in terms of 'x'.
Time(2) = Distance(2)/Speed(2) = 1/x
In the third row, we know that the total distance is 2 miles (by taking the sum of the distances in row 1 and 2) and that the average speed should be 60 mph. Thus we can calculate the total time that the two laps should take.
Time(3) = Distance(3)/Speed(3) = 2/60 = 1/30
Now, we know that the total time should be the sum of the times in row 1 and 2. Thus we can form the following equation :
Time(3) = Time(1) + Time(2) ---> 1/30 = 1/30 + 1/x
From this, it becomes clear that '1/x' must be 0.
Since 'x' is the reciprocal of 0, which does not exist, there can be no speed for which the average can be made up in the second lap.
Example 2.
An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. The executive boarded the helicopter and flew to the corporate offices at an average speed of 60 mph. The entire distance was 150 miles; the entire trip took three hours. Find the distance from the airport to the corporate offices.
Solution:
Since we have been asked to find the distance from the airport to the corporate office (that is the distance he spent flying), let us assign that specific value as 'x'.
Thus, the distance he spent driving will be '150 - x'
Now, in the first row, we have the distance in terms of 'x' and we have been given the speed. Thus we can calculate the time he spent driving in terms of 'x'.
Time(1) = Distance(1)/Speed(1) = (150 - x)/30
Similarly, in the second row, we again have the distance in terms of 'x' and we have been given the speed. Thus we can calculate the time he spent flying in terms of 'x'.
Time(2) = Distance(2)/Speed(2) = x/60
Now, notice that we have both the times in terms of 'x'. Also, we know the total time for the trip. Thus, summing the individual times spent driving and flying and equating it to the total time, we can solve for 'x'.
Time(1) + Time(2) = Time(3) --> (150 - x)/30 + x/60 = 3 --> x = 120 miles
Answer : 120 miles
Note: In this problem, we did not calculate average speed for row 3 since we did not need it. Remember not to waste time in useless calculations!
Example 3.
A passenger train leaves the train depot 2 hours after a freight train left the same depot. The freight train is traveling 20 mph slower than the passenger train. Find the speed of the passenger train, if it overtakes the freight train in three hours.
Solution:
Since this is an 'overtaking' problem, the first thing that should strike us is that the distance traveled by both trains is the same at the time of overtaking.
Next we see that we have been asked to find the speed of the passenger train at the time of overtaking. So let us represent it by 'x'.
Also, we are given that the freight train is 20 mph slower than the passenger train. Hence its speed in terms of 'x' can be written as 'x - 20'.
Moving on to the time, we are told that it has taken the passenger train 3 hours to reach the freight train. This means that the passenger train has been traveling for 3 hours.
We are also given that the passenger train left 2 hours after the freight train. This means that the freight train has been traveling for 3 + 2 = 5 hours.
Now that we have all the data in place, we need to form an equation that will help us solve for 'x'. Since we know that the distances are equal, let us see how we can use this to our advantage.
From the first row, we can form the following equation :
Distance(1) = Speed(1) * Time(1) = x*3
From the second row, we can form the following equation :
Distance(2) = Speed(2) * Time(2) = (x - 20)*5
Now, equating the distances because they are equal we get the following equation :
3*x = 5*(x - 20) --> x = 50 mph.
Answer : 50 mph.
Example 4.
Two cyclists start at the same time from opposite ends of a course that is 45 miles long. One cyclist is riding at 14 mph and the second cyclist is riding at 16 mph. How long after they begin will they meet?
Solution:
Since this is a 'meeting' problem, there are two things that should strike you. First, since they are starting at the same time, when they meet, the time for which both will have been cycling will be the same. Second, the total distance traveled by the will be equal to the sum of their individual distances.
Since we are asked to find the time, let us assign it as a variable 't'. (which is same for both cyclists)
In the first row, we know the speed and we have the time in terms of 't'. Thus we can get the following equation :
Distance(1) = Speed(1) * Time(1) = 14*t
In the second row, we know the speed and again we have the time in terms of 't'. Thus we can get the following equation :
Distance(2) = Speed(2) * Time(2) = 16*t
Now we know that the total distance traveled is 45 miles and it is equal to the sum of the two distances. Thus we get the following equation to solve for 't' :
Distance(3) = Distance(1) + Distance(2) --> 45 = 14*t + 16*t --> t = 1.5 hours
Answer : 1.5 hours.
Example 5.
A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in calm water?
Solution:
Since this is a question on round trip, the first thing that should strike us is that the distance going and coming back will be the same.
Now, we are required to find out the boats speed in calm water. So let us assume it to be 'b'. Now if speed of the current is 3 mph, then the speed of the boat while going downstream and upstream will be 'b + 3' and 'b - 3' respectively.
In the first row, we have the speed of the boat in terms of 'b' and we are given the time. Thus we can get the following equation :
Distance(1) = Speed(1) * Time(1) = (b + 3)*3
In the second row, we again have the speed in terms of 'b' and we are given the time. Thus we can get the following equation :
Distance(2) = Speed(2) * Time(2) = (b - 3)*4
Since the two distances are equal, we can equate them and solve for 'b'.
Distance(1) = Distance(2) --> (b + 3)*3 = (b - 3)*4 --> b = 21 mph.
Answer : 21 mph.
- Usually involve something/someone moving at a constant or average speed.
- Out of the three quantities (speed/distance/time), we are required to find one.
- Information regarding the other two will be provided in the question stem.
The ‘D/S/T’ Formula: Distance = Speed x Time
I’m sure most of you are already familiar with the above formula (or some variant of it). But how many of you truly understand what it signifies?
When you see a ‘D/S/T’ question, do you blindly start plugging values into the formula without really understanding the logic behind it? If then answer to that question is yes, then you would probably have noticed that your accuracy isn’t quite where you’d want it to be.
My advice here, as usual, is to make sure you understand the concept behind the formula rather than just using it blindly.
So what’s the concept? Lets find out!
- The Distance = Speed x Time formula is just a way of saying that the distance you travel depends on the speed you go for any length of time.
If you travel at 50 mph for one hour, then you would have traveled 50 miles. If you travel for 2 hours at that speed, you would have traveled 100 miles. 3 hours would be 150 miles, etc.
If you were to double the speed, then you would have traveled 100 miles in the first hour and 200 miles at the end of the second hour.
- We can figure out any one of the components by knowing the other two.
For example, if you have to travel a distance of 100 miles, but can only go at a speed of 50 mph, then you know that it will take you 2 hours to get there. Similarly, if a friend visits you from 100 miles away and tells you that it took him 4 hours to reach, you will know that he AVERAGED 25 mph. Right?
- All calculations depend on AVERAGE SPEED.
Supposing your friend told you that he was stuck in traffic along the way and that he traveled at 50 mph whenever he could move. Therefore, although practically he never really traveled at 25 mph, you can see how the standstills due to traffic caused his average to reduce. Now, if you think about it, from the information given, you can actually tell how long he was driving and how long he was stuck due to traffic (assuming; what is false but what they never worry about in these problems; that he was either traveling at 50 mph or 0 mph). If he was traveling constantly at 50 mph, he should have reached in 2 hours. However, since he took 4 hours, he must have spent the other 2 hours stuck in traffic!
- Now lets see how we can represent this using the formula.
We know that the total distance is 100 miles and that the total time is 4 hours. BUT, his rates were different AND they were different at different times. However, can you see that no matter how many different rates he drove for various different time periods, his TOTAL distance depended simply on the SUM of each of the different distances he drove during each time period?
E.g., if you drive a half hour at 60 mph, you will cover 30 miles. Then if you speed up to 80 mph for another half hour, you will cover 40 miles, and then if you slow down to 30 mph, you will only cover 15 miles in the next half hour. But if you drove like this, you would have covered a total of 85 miles (30 + 40 + 15). It is fairly easy to see this looking at it this way, but it is more difficult to see it if we scramble it up and leave out one of the amounts and you have to figure it out going "backwards". That is what word problems do.
Further, what makes them difficult is that the components they give you, or ask you to find can involve variable distances, variable times, variable speeds, or any two or three of these. How you "reassemble" all this in order to use the d = s*t formula takes some reflection that is "outside" of the formula itself. You have to think about how to use the formula.
So the trick is to be able to understand EXACTLY what they are giving you and EXACTLY what it is that is missing, but you do that from thinking, not from the formula, because the formula only works for the COMPONENTS of any trip where you are going an average speed for a certain amount of time. ONCE the conditions deal with different speeds or different times, you have to look at each of those components and how they go together. And that can be very difficult if you are not methodical in how you think about the components and how they go together. The formula doesn't tell you which components you need to look at and how they go together. For that, you need to think, and the thinking is not always as easy or straightforward as it seems like it ought to be.
In the case of your friend above, if we call the time he spent driving 50 mph, T1; then the time he spent standing still is (4 - T1) hours, since the whole trip took 4 hours. So we have 100 miles = (50 mph x T1) + (0 mph x [4 - T1]) which is equivalent then to: 100 miles = 50 mph x T1
So, T1 will equal 2 hours. And, since the time he spent going zero is (4 - 2), it also turns out to be 2 hours.
- Sometimes the right answers will seem counter-intuitive, so it is really important to think about the components methodically and systematically.
There is a famous trick problem: To qualify for a race, you need to average 60 mph driving two laps around a 1 mile long track. You have some sort of engine difficulty the first lap so that you only average 30 mph during that lap; how fast do you have to drive the second lap to average 60 for both of them?
I will go through THIS problem with you because, since it is SO tricky, it will illustrate a way of looking at almost all the kinds of things you have to think about when working any of these kinds of problems FOR THE FIRST TIME (i.e., before you can do them mechanically because you recognize the TYPE of problem it is). Intuitively it would seem you need to drive 90, but this turns out to be wrong for reasons I will give in a minute.
The answer is that NO MATTER HOW FAST you do the second lap, you can't make it. And this SEEMS really odd and that it can't possibly be right, but it is. The reason is that in order to average at least 60 mph over two one-mile laps, since 60 mph is one mile per minute, you will need to do the whole two miles in two minutes or less. But if you drove the first mile at only 30, you used up the whole two minutes just doing IT. So you have run out of time to qualify.
To see this with the d = s*t formula, you need to look at the overall trip and break it into components, and that is the hardest part of doing this (these) problem(s), because (often) the components are difficult to figure out, and because it is hard to see which ones you need to put together in which way.
In the next section we will learn how to do just that.
Resolving the Components
- When you first start out with these problems, the best way to approach them is by organizing the data in a tabular form.
Use a separate column each for distance, speed and time and a separate row for the different components involved (2 parts of a journey, different moving objects, etc.). The last row should represent total distance, total time and average speed for these values (although there might be no need to calculate these values if the question does not require them).
- Assign a variable for any unknown quantity.
If there is more than one unknown quantity, do not blindly assign another variable to it. Look for ways in which you can express that quantity in terms of the quantities already present. Assign another variable to it only if this is not possible.
- In each row, the quantities of distance, speed and time will always satisfy d = s*t.
- The distance and time column can be added to give you the values of total distance and total time but you CANNOT add the speeds.
Think about it: If you drive 20 mph on one street, and 40 mph on another street, does that mean you averaged 60 mph?
- Once the table is ready, form the equations and solve for what has been asked!
Warning: Make sure that the units for time and distance agree with the units for the rate. For instance, if they give you a rate of feet per second, then your time must be in seconds and your distance must be in feet. Sometimes they try to trick you by using the wrong units, and you have to catch this and convert to the correct units.
A Few More Points to Note
- Motion in Same Direction (Overtaking): The first thing that should strike you here is that at the time of overtaking, the distances traveled by both will be the same.
- Motion in Opposite Direction (Meeting): The first thing that should strike you here is that if they start at the same time (which they usually do), then at the point at which they meet, the time will be the same. In addition, the total distance traveled by the two objects under consideration will be equal to the sum of their individual distances traveled.
- Round Trip: The key thing here is that the distance going and coming back is the same.
Now that we know the concept in theory, let us see how it works practically, with the help of a few examples.
Note for tables : All values in black have been given in the question stem. All values in blue have been calculated.
Example 1.
To qualify for a race, you need to average 60 mph driving two laps around a 1-mile long track. You have some sort of engine difficulty the first lap so that you only average 30 mph during that lap; how fast do you have to drive the second lap to average 60 for both of them?
Solution:
Let us first start with a problem that has already been introduced. You will see that by clearly listing out the given data in tabular form, we eliminate any scope for confusion.
In the first row, we are given the distance and the speed. Thus it is possible to calculate the time.
Time(1) = Distance(1)/Speed(1) = 1/30
In the second row, we are given just the distance. Since we have to calculate speed, let us give it a variable 'x'. Now, by using the 'D/S/T' relationship, time can also be expressed in terms of 'x'.
Time(2) = Distance(2)/Speed(2) = 1/x
In the third row, we know that the total distance is 2 miles (by taking the sum of the distances in row 1 and 2) and that the average speed should be 60 mph. Thus we can calculate the total time that the two laps should take.
Time(3) = Distance(3)/Speed(3) = 2/60 = 1/30
Now, we know that the total time should be the sum of the times in row 1 and 2. Thus we can form the following equation :
Time(3) = Time(1) + Time(2) ---> 1/30 = 1/30 + 1/x
From this, it becomes clear that '1/x' must be 0.
Since 'x' is the reciprocal of 0, which does not exist, there can be no speed for which the average can be made up in the second lap.
Example 2.
An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. The executive boarded the helicopter and flew to the corporate offices at an average speed of 60 mph. The entire distance was 150 miles; the entire trip took three hours. Find the distance from the airport to the corporate offices.
Solution:
Since we have been asked to find the distance from the airport to the corporate office (that is the distance he spent flying), let us assign that specific value as 'x'.
Thus, the distance he spent driving will be '150 - x'
Now, in the first row, we have the distance in terms of 'x' and we have been given the speed. Thus we can calculate the time he spent driving in terms of 'x'.
Time(1) = Distance(1)/Speed(1) = (150 - x)/30
Similarly, in the second row, we again have the distance in terms of 'x' and we have been given the speed. Thus we can calculate the time he spent flying in terms of 'x'.
Time(2) = Distance(2)/Speed(2) = x/60
Now, notice that we have both the times in terms of 'x'. Also, we know the total time for the trip. Thus, summing the individual times spent driving and flying and equating it to the total time, we can solve for 'x'.
Time(1) + Time(2) = Time(3) --> (150 - x)/30 + x/60 = 3 --> x = 120 miles
Answer : 120 miles
Note: In this problem, we did not calculate average speed for row 3 since we did not need it. Remember not to waste time in useless calculations!
Example 3.
A passenger train leaves the train depot 2 hours after a freight train left the same depot. The freight train is traveling 20 mph slower than the passenger train. Find the speed of the passenger train, if it overtakes the freight train in three hours.
Solution:
Since this is an 'overtaking' problem, the first thing that should strike us is that the distance traveled by both trains is the same at the time of overtaking.
Next we see that we have been asked to find the speed of the passenger train at the time of overtaking. So let us represent it by 'x'.
Also, we are given that the freight train is 20 mph slower than the passenger train. Hence its speed in terms of 'x' can be written as 'x - 20'.
Moving on to the time, we are told that it has taken the passenger train 3 hours to reach the freight train. This means that the passenger train has been traveling for 3 hours.
We are also given that the passenger train left 2 hours after the freight train. This means that the freight train has been traveling for 3 + 2 = 5 hours.
Now that we have all the data in place, we need to form an equation that will help us solve for 'x'. Since we know that the distances are equal, let us see how we can use this to our advantage.
From the first row, we can form the following equation :
Distance(1) = Speed(1) * Time(1) = x*3
From the second row, we can form the following equation :
Distance(2) = Speed(2) * Time(2) = (x - 20)*5
Now, equating the distances because they are equal we get the following equation :
3*x = 5*(x - 20) --> x = 50 mph.
Answer : 50 mph.
Example 4.
Two cyclists start at the same time from opposite ends of a course that is 45 miles long. One cyclist is riding at 14 mph and the second cyclist is riding at 16 mph. How long after they begin will they meet?
Solution:
Since this is a 'meeting' problem, there are two things that should strike you. First, since they are starting at the same time, when they meet, the time for which both will have been cycling will be the same. Second, the total distance traveled by the will be equal to the sum of their individual distances.
Since we are asked to find the time, let us assign it as a variable 't'. (which is same for both cyclists)
In the first row, we know the speed and we have the time in terms of 't'. Thus we can get the following equation :
Distance(1) = Speed(1) * Time(1) = 14*t
In the second row, we know the speed and again we have the time in terms of 't'. Thus we can get the following equation :
Distance(2) = Speed(2) * Time(2) = 16*t
Now we know that the total distance traveled is 45 miles and it is equal to the sum of the two distances. Thus we get the following equation to solve for 't' :
Distance(3) = Distance(1) + Distance(2) --> 45 = 14*t + 16*t --> t = 1.5 hours
Answer : 1.5 hours.
Example 5.
A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in calm water?
Solution:
Since this is a question on round trip, the first thing that should strike us is that the distance going and coming back will be the same.
Now, we are required to find out the boats speed in calm water. So let us assume it to be 'b'. Now if speed of the current is 3 mph, then the speed of the boat while going downstream and upstream will be 'b + 3' and 'b - 3' respectively.
In the first row, we have the speed of the boat in terms of 'b' and we are given the time. Thus we can get the following equation :
Distance(1) = Speed(1) * Time(1) = (b + 3)*3
In the second row, we again have the speed in terms of 'b' and we are given the time. Thus we can get the following equation :
Distance(2) = Speed(2) * Time(2) = (b - 3)*4
Since the two distances are equal, we can equate them and solve for 'b'.
Distance(1) = Distance(2) --> (b + 3)*3 = (b - 3)*4 --> b = 21 mph.
Answer : 21 mph.
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