The post Intro to GMAT Word Problems, Part 2: Assigning Variables appeared first on Magoosh GMAT Blog.
]]>1) Each month, after Jill pays for rent, utilities, food, and other necessary expenses, she has one fifth of her net monthly salary left as discretionary income. Of this discretionary income, she puts 30% into a vacation fund, 20% into savings, and spends 35% on eating out and socializing. This leaves her with $96 dollar, which she typically uses for gifts and charitable causes. What is Jill’s net monthly salary?
(A) $2400
(B) $3200
(C) $6000
(D) $6400
(E) $9600
2) Right now, Al and Eliot have bank accounts, and Al has more money than Eliot. The difference between their two accounts is 1/12 of the sum of their two accounts. If Al’s account were to increase by 10% and Eliot’s account were to increase by 20%, then Al would have exactly $22 more than Eliot in his account. How much money does Eliot have in his account right now?
(A) $110
(B) $120
(C) $180
(D) $220
(E) $260
3) A pool, built with one edge flush against a building, has a length that is 5 meters longer than its width. The short width is against the building. A 4 meter wide path is built on three side around the pool, as shown in the diagram (the path is yellow). If the area of the path is 216 sq m, what is the width of the pool in meters?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20
4) Four friends, Saul, Peter, Quirinal, and Roderick, are pooling their money to buy a $1000 item. Peter has twice as much money as Saul. Quirinal has $60 more than Peter. Roderick has 20% more than Quirinal. If they put all their money together and spend the $1000, they will have $20 left. How much money does Peter have?
(A) $120
(B) $160
(C) $180
(D) $200
(E) $240
Full solutions will appear at the end of this article.
Most GMAT word problem concern real world quantities and are stated in real world terms, and we need to assign algebraic variables to these real world quantities.
Sometimes, one quantity is directly related to every other quantity in the problem. For example:
“Sarah spends 2/5 of her monthly salary on rent, 1/12 of her monthly salary on auto costs including gas and insurance, and 1/10 of her monthly salary automatically goes into saving each month. With what she has left each month, she spend she spends $800 on groceries and …”
In that problem, everything is related to “monthly salary,” so it would make a lot of sense to introduce just one variable for that, and express everything else in terms of that variable. Also, please don’t always use the boring choice of x for a variable! If we want a variable for salary, you might use the letter S, which will help you remember what the variable means! If we are given multiple variables that are all related to each other, it’s often helpful to assign a letter to the variable with the lowest value, and then express everything else in terms of this letter.
If there are two or more quantities that don’t depend directly on each other, then you may well have to introduce a different variable for each. Just remember that it’s mathematically problematic to litter a problem with a whole slew of different variables. You see, for each variable, you need an equation to solve it. If we want to solve for two different variables, we need two different equations (this is a common Word Problem scenario). If we want to solve for three different variables, we need three different equations (considerably less common). While the mathematical pattern continues to extend upward from there, more than three completely separate variables is almost unheard of on GMAT math.
When you assign variables, always be hyper-vigilant and over-the-top explicit about exactly what each variable means. Write a quick note to yourself on the scratch paper: T = the price of one box of tissue, or whatever the problem wants. What you want to avoid is the undesirable situation of solving for a number and not knowing what that number means in the problem!
Here’s an easier-than-the-GMAT word problem as an example:
“Andrew and Beatrice each have their own savings account. Beatrice’s account has $600 less than three times what Andrew’s account has. If Andrew had $300 more dollars, then he would have exactly half what is currently in Beatrice’s account. How much does Beatrice have?”
The obvious choices for variables are A = the amount in Andrew’s account and B = the amount in Beatrice’s account. The GMAT will be good about giving you word problems involving people whose names start with different letter, so that it’s easier to assign variables. We can turn the second & third sentences into equations.
second sentence: B = 3A – 600
Both equations are solved for B, so simply set them equal.
3A – 600 = 2(A + 300)
3A – 600 = 2A + 600
A – 600 = 600
A = 1200
We can plug this into either equation to find B. (BTW, if you have time, an excellent check is to plug it into both equations, and make sure the value of B you get is the same!)
B = 3000
Thus, Andrew has $1200 in his account, and Beatrice, $3000 in hers.
If the foregoing discussion gave you any insights into assigning variables, it may well be worthwhile to look at those four practice problems again before preceding to the explanations below. If you join Magoosh, you can watch our 20+ video lessons on Word Problems.
1) Everything is in terms of Jill’s discretionary income, which is one-fifth of the net monthly rent. It makes sense to assign a variable to the former, solve for it, and then compute the latter. I will assign the letter D, to remind us that this represents the monthly discretionary income, not the answer to the question. We will not yet have the answer when we find the value of D.
vacation = 30% of D
savings = 20% of D
eating out & socializing = 35% of D
Together, those account for 85% of her monthly discretionary income. That leaves 15%. This 15% equals $96.
15% of D = $96
Divide by sides by 3.
5% of D = $32
Double.
10% of D = $64
Now, multiply by 10.
100% of D = D = $640
Remember, this is the value of D, the monthly discretionary income, not what the question asked. The question wanted monthly salary, which is five times this. Well, ten times D is $6400, so five times D would be half of that, $3200.
Answer = (B)
2) Names in this problem from a famous Al and a famous Eliot. Let’s start with two variables, A and E. The difference (A – E) is 1/12 the sum (A + B).
12(A – E) = A + E
12A – 12E = A + E
11A = 13E
Now, since we have related these variables, it doesn’t make sense to move through the rest of problem with two different variables. We could express E = (11/13)*A, and express everything in terms of A, but 11/13 is an especially ugly fraction. Here’s an alternative, using a little number sense. Clearly A equals 13 parts of something, and E equals 11 parts of something. Let’s say that P = the “part” in this ratio; then A = 13P and E = 11P. We can express everything in terms of P.
Al’s account increases by 10%:
New Al = 1.10*(13*P) = 14.3*P
Eliot’s account increases by 20%:
New Eliot = 1.20*(11*P) = 13.2*P
Difference = 14.3*P – 13.2*P = 1.1*P = $22
Multiply both sides by 10 to clear the decimal.
11*P = $220
We could solve for P at this point, but notice that what we want, Eliot’s amount, is already equal to 11*P. This is the answer! Eliot has $220 in his account.
Answer = (D)
3) Call the width W. Then the length is L = W + 5. The section of path to the left of the pool in the diagram is a rectangle L tall and 4 wide, so it’s area is
A1 = 4L = 4(W + 5)
In the upper-left hand corner of the path, there’s a 4 x 4 square, with area:
A2 = 16
Above the pool is a rectangle with a height of 4 and width of W, with an area:
A3 = 4W
Another 4 x 4 square in the upper right-hand corner:
A4 = 16
And finally, another rectangle on the right, equal to the one on the left
A5 = A1 = 4(W + 5)
All these pieces add up to 216.
Total = A1 + A2 + A3 + A4 + A5
Total = (4W + 20) + 16 + 4W + 16 + (4W + 20)
Total = 12W + 72 = 216
12W = 216 – 72 = 144
W = 12
The pool has a width of 12 m and a length of 17 m.
Answer = (A)
4) Saul appears to have the least money, so we will put everything in terms of his amount.
P = 2*S
Q = P + 60 = 2*S + 60
R = 1.2*Q = 1.2*(2*S + 60) = 2.4*S + 72
Total = S + P + Q + R
Total = S + 2*S + 2*S + 60 + 2.4*S + 72
Total = 7.4*S + 132 = 1020
7.4*S = 888
74*S = 8880
37*S = 4440
At this point, it’s very helpful to know that 3*37 = 111. This means that 12*37 = 444, and 120*37 = 4440. Thus, S = 120. Saul has $120. Notice, though, the question is not asking for what Saul has: it is asking for what Peter has. Peter has twice Saul’s amount, so Peter has $240.
Answer = (E)
This is beyond what you need to know for the test, but in this problem there’s a pattern encrypted in the names. The abbreviation of the four names spells out SPQR, which was the abbreviation in Latin for the name of the Roman Empire (Senatvs Popvlvsqve Romanvs = “The Senate and the People of Rome”). The four names are folks associated with the city of Rome in one way or another. In the Christian tradition, Saul (who became St. Paul) and St. Peter are believed to have lived and died in Rome. The somewhat obscure male name Quirinal was the name of a son of the god Mars, and it is also the name of the one of the seven hills of Rome. The name Roderick is an inside-joke from a Monty Python film set during Roman times.
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]]>1) Seven more than a number is 2 more than four times the number. What is the number?
(A) 1
(B) 2/3
(C) 3/5
(D) 4/7
(E) 5/3
2) If $40,000 less than John’s salary is $5,000 more than 25% more than half his salary, then what is John’s salary?
(A) $80,000
(B) $100,000
(C) $120,000
(D) $135,000
(E) $140,000
3) Twice a number is 3 times the square of the number less than one. If the number is positive, what is the value of the number?
(A) 1
(B) 1/2
(C) 1/3
(D) 2/3
(E) 3/2
4) The original price of an item is discounted 20%. A customer buys the item at this discounted price using a $20-off coupon. There is no tax on the item, and this was the only item the customer bought. If the customer paid $1.90 more than half the original price of the item, what was the original price of the item?
(A) $61
(B) $65
(C) $67.40
(D) $70
(E) $73
It’s one thing to understand algebra in the abstract, and quite another to think about where the rubber meets the road. The reason human beings created algebra was to solve problems about real world situations, and the GMAT loves asking math problems about numbers and about real world situations, a.k.a. word problems! Even folks who can do algebra in the abstract sometimes find word problems challenging. In this blog and the next, we present a rough-and-ready guide to what you need about word problems.
Suppose we have the following sentence in a word problem:
“Three-fifths of x is 14 less than twice y squared.”
How do we change words to math? Here’s a quick guide
1) the verb “is/are” is the equivalent of an equal sign; the equal sign in an equation is, in terms of “mathematical grammar,” the equivalent of a verb in a sentence. Every sentence has a verb and every equation has an equal sign.
2) The word “of” means multiply (often used with fractions and percents). Ex. “26% of x” means (0.26x)
3) The words “more than” or “greater than” mean addition. Ex. “5 greater than x” means (x + 5) and “7 more than y” means (y + 7)
4) The words “less than” means subtraction. Ex. “8 less than Q” means (Q – 8). Notice that the first element is always subtracted: in other words, “J less than K” means (K – J).
With that in mind, let’s go back to the sentence from the hypothetical problem above.
“three fifths of x” means [(3/5)*x]
“is” marks the location of the equal sign
“twice y squared” means 2(y^2)
“14 less than twice y squared” means 2(y^2) – 14
Altogether, the equation we get is:
Using this strategy, it’s straightforward to translate from a verbal statement about numbers to an equation.
When all the answer choices are numerical, one further strategy we have at our disposal is backsolving. Using this strategy, we can pick one answer, plug it into the problem, and see whether it works. If this choice is too big or too small, it guides us in what other answer choices to eliminate. Typically, we would start with answer choice (C), but if another answer choice is a particularly convenient choice, then we would start there.
If the strategies discussed here gave you any insights, you may want to give the problems above another look before turning to the solutions below. Look for the second article on Assigning Variables in Word Problems.
1) Translate this one step at a time. Let N be the number we seek.
“seven more than a number” = N + 7
The “is” is the equal sign.
“two more than four times the number” = 4N + 2
Answer = (E)
2) We will say that S is John’s salary. “$40,000 less than John’s salary” is (S – 40000). The second part is tricky: “25% more than half his salary” is (1.25*(1/2)*S), so “$5,000 more than 25% more than half his salary” would be that plus 5000. We can write the whole first part of the prompt sentence as
Multiply both sides by 8 (we don’t have to perform the numerical multiplication yet)
8S – 8*45000 = 5S
3S = 8*45000
S = 8*15000 = 4*30000 = 120,000
Notice the use of the doubling and halving trick to perform the multiplication in the last line. John’s salary must be $120,000.
Answer = (C)
3) Call the number x. Of course, “twice the number” equals 2x. The part after the word “is” can be tricky. Remember that the information of the form “J less than K” takes the mathematical form (K – J), in which the first part is the part that’s subtracted. What we have here is “3 times the square of the number less than one.” That would be one minus 3 times x squared. Now we can translate that entire first sentence of the prompt into math:
This is a quadratic. We need to get all the terms on one side, equal to zero, and then factor.
Because the prompt tells us that the number must be positive, we can reject the negative root. The number must be +1/3.
Answer = (C)
4) We will show two solutions for this: (i) backsolving, and (ii) the full algebraic solution.
For the backsolving solution, notice that (C) is an ugly number. (B) and (D) are nicer numbers. Let’s start with (D).
Original price = $70
10% of price = $7, so
20% of price = $14.
After discount, the price is 70 – 14 = $56. The customer then uses a $20-off coupon, so this customer pays $36.
How does this price compare to half the original price? Well, half the original was $35, so the customer paid exactly $1 more than half the original price.
First of all, we know that answer choice (D) does not work. We know we need a bigger difference, so we need a bigger price. The only price bigger is (E). This must be the answer.
Answer = (E)
Now, a full algebra solution. Let the original price be P. Then, 20% off would be 0.8*P. Then, if we subtract $20, that’s (0.8*P – 20). That is the price the customer paid, which equals “$1.90 more than half the original price,” or (0.5P + 1.9). We will set these equal.
08*P – 20 = 0.5*P + 1.9
0.3*P = 21.9
3P = 219
P = 73
The original price was $73. Answer = (E)
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]]>1) On a ferry, there are 50 cars and 10 trucks. The cars have an average mass of 1200 kg and the trucks have an average mass of 3000 kg. What is the average mass of all 60 vehicles on the ferry?
2) In Plutarch Enterprises, 70% of the employees are marketers, 20% are engineers, and the rest are managers. Marketers make an average salary of $50,000 a year, and engineers make an average of $80,000. What is the average salary for managers if the average for all employees is also $80,000?
3) At Didymus Corporation, there are just two classes of employees: silver and gold. The average salary of gold employees is $56,000 higher than that of silver employees. If there are 120 silver employees and 160 gold employees, then the average salary for the company is how much higher than the average salary for the silver employees?
4) In a company of only 20 employees, 10 employees make $80,000/yr, 6 employees make $150,000/yr, and the 4 highest-paid employees all make the same amount. If the average annual salary for the 20 employees is $175,000/yr, then what is the annual salary of each highest-paid employee?
5) In a certain apartment building, there are one-bedroom and two-bedroom apartments. The rental prices of the apartment depend on a number of factors, but on average, two-bedroom apartments have higher rental prices than do one-bedroom apartments. Let R be the average rental price for all apartments in the building. If R is $5,600 higher than the average rental price for all one-bedroom apartments, and if the average rental price for all two-bedroom apartments is $10,400 higher that R, then what percentage of apartments in the building are two-bedroom apartments?
6) At a certain upscale restaurant, there just two kinds of food items: entrees and appetizers. Each entrée item costs $30, and each appetizer item costs $12. Last year, it had a total of 15 food items on the menu, and the average price of a food item on its menu was $18. This year, it added more appetizer items, and the average price of a food item on its menu dropped to $15. How many appetizer items did it add this year?
Solutions will appear at the end of the article.
For the purposes of the GMAT, the weighted average situations occurs when we combine groups of different sizes and different group averages. For example, suppose in some parameter, suppose the male employees in a company have one average score, and the females have another average score. If there were an equal number of males and females, we could just average the two separate gender averages: that would be ridiculously easy, which is precisely why the GMAT will never present you with two groups of equal size in such a question. Instead, the number of male employees and number of female employees will be profoundly different, one significantly outnumbering the other, and then we will have to combine the individual gender averages to produce a total average for all employees. That’s a weighted average.
That example, included just two groups, which is common, but sometimes there are three, and conceivably, on a very hard problem, there could be four groups. Each group is a different size, each has its own average, and the job is to find the average of everyone all together. Or, perhaps the question will give us most of the info for the individual groups, and give us the total average for everyone, and then ask us to find the size or average of a particular group.
We have three basic approaches we can take to these question.
Remember that, even with ordinary average questions, thinking in terms of the sum can often be helpful. We can’t add or subtract averages, but we can add or subtract sums! Right there is the key to one approach to the weighted average situation. If we calculate the sums for each separate group, we can simply add these sums to get the sum of the whole group. Alternately, if we know the size of the total group and the total average for everyone, we can figure out the total sum for everyone, and simply subtract the sums of the individual groups in order to find what we need.
Some weighted average problems give percents, not actual counts, of individual groups. In that case, we could simply pick a convenient number for the size of the population, and use the sums method from there. For example, if group A has 20% of the population, group B has 40% of the population, and group C also has 40% of the population, then we could just pretend that group A has one person, groups B & C each have two people, and total population has five people. From this, we could calculate all our sums.
This method always works, although is not always the most convenient in more advanced problems. This will be demonstrated in a few of the answers below.
Sometimes the information about the sizes of individual groups is given, not in absolute counts of members, but simply in percents or proportions. In the problems above, question #2 simply gives percents, and question #5 asks for a percent. Yes, we could use Approach I, but there’s a faster way.
In Approach II, we simply multiply each group average by the percent of the population, expressed as a decimal, which that group occupies. When we add these products up, the sum is magically the total average for everyone. Suppose we have three groups, groups J and K and L which together constitute the entire population. Suppose this summarizes their separate information:
The percent have to be expressed in decimal form, so that:
Then, the total average is simply given by
This approach will be demonstrated in #2 below.
This final method can be hard to understand at first, but if you appreciate it, it is an incredibly fast time-saver. This approach only works if there are exactly two groups, no more.
Suppose there are two groups in a population, group 1 and group 2, and suppose that group 1 is bigger than group 2. Let’s also suppose that group 1 has a lower group average, and group 2 has the higher group average. Of course, the total combined average of the two groups together will be between the two individual group averages. In fact, because group 1 is bigger, the combined average will have to be closer to group 1’s average, and further away from group 2’s average.
Now, think about a number line, with the two individual group averages and the total averages indicated on the number line.
On that number line, I have labeled d1, the distance from the average of group #1 to the combined average, and d2, the distance from the combined average to the average of group 2. The ratio of these two distances is equal to a ratio of the size of the two individual groups. Let’s think about this. The bigger group, here group 1, will have more of an effect on the combined average and therefore will be closer to the combined average—a smaller distance. Therefore, the ratio of the distances must equal the reciprocal of the same ratio of the sizes of the groups:
Let’s say group 1 is 3 times larger than group 2. This means d2, the distance from group 2’s average to the combined average, would have to be 3 times bigger than d1, the distance from group 1’s average to the combined average. If the latter is x, then the former is 3x, and the total distance is 4x. If we know the averages of the two groups individually, we would simply have to divide the difference between those group averages by 4: the combined average would be one part away from group 1’s average, or three parts away from group 2’s average.
This approach is hard to explain clearly in words. You really have to see it demonstrated in the solutions below to understand it fully. Once you understand it, though, this is an extremely fast method to solve many problems.
If the above article gave you any insights, you may want to give the practice problems another look. Remember, in your practice problems on weighted averages, practice all three of these methods. The more ways you have to understand any problems, the more options you will have on test day!
1) Method I: using sums
We will divide the two masses by 1000, 1.2 and 3 respectively, to simplify calculations. Note the use of the Doubling and Halving trick in the first multiplication.
sum for cars = 50*1.2 = 100*0.6 = 60
sum for trucks = 10*3 = 30
total sum = 60 + 30 = 90
To find the total average, we need to divide this total sum by the total number of vehicles, 60.
Since we divided masses by 1000 earlier, we need to multiply by 1000 to get the answer. Total average = 1500 kg. Answer = (B)
Method II: proportional placement of the total average
Cars to trucks is 5:1, so if the distance between the car’s average and truck’s average were divided into 6 parts, the car’s average is 1 part away from the total average, and the truck’s average is 5 parts away.
Well, the difference in the two group averages is 3000 – 1200 = 1800 kg. Divide that by six: each “part” is 300 kg. Well, the total average must be 300 kg bigger than 1200 kg, or 5*300 kg smaller than 3000 kg. Either way, that’s 1500 kg. Answer = (B)
2) We will approach this use the proportion & percents approach. Divide all dollar amounts by 1000 for smaller numbers. Multiply each group average by the percent expressed as a decimal:
marketers = 0.70*50 = 35
engineers = 0.20*80 = 16
managers = 0.10*x = 0.1x
where x is the average salary for the managers. These three should add up to the average for all employees:
35 + 16 + 0.1x = 80
0.1x = 80 – 35 – 16
0.1x = 29
x = 290
Now, multiply the 1000 again, to get back to real dollar amounts. The average salary for managers is $290,000. Answer = (D)
3) Method I: using sums
We can use this if we pick a value for the average salary for the silver employees. It actually doesn’t matter what value we pick, because averages will fall in the same relative places regardless of whether all the individual values are slid up and down the number line. The easiest value by far to pick is zero. Let’s pretend that the silver folks make $0, and the gold folks make $56. (I divided dollars by 1000 for simplicity).
Now, we also have to simplify the numbers of employees. We could reduce the number of employees as long as we preserve the relative ratio.
silver : gold = 120:160 = 12:16 = 3:4
So everything would be the same if we just had 3 silver employees and 4 gold employees. OK, now find the sums.
silver = 3*0 = 0
gold = 4*56
I am not even going to bother to multiply that out, because we know that the next step is to divide by 7, the total number of employees.
total average = 4*56/7 = 4*8 = 32
The average salary is $32,000, which is $32,000 higher than the average for the silver employees. Answer = (C)
Method II: proportional placement of the total average
The ratio of silver employees to gold employees is
silver : gold = 120:160 = 12:16 = 3:4
If we divide the distance between the two averages by 7, then silver will be “four parts” away from the total average, and gold will be “three parts” away.
Well, the difference is $56,000, so that divided by 7 is $8000. That’s one part. Four parts would be $32,000, which has to be the distance from the silver average to the total average.
Answer = (C)
4) We will approach this using sums. The individual employee numbers are small. We will divide all dollar amounts by 1000, for easier calculations. Call the highest-paid employee salary x. Then the sums are
lowest = 10*80 = 800
middle = 6*150 = 3*300 = 900
highest = 4x
Individual sums must add up to the total sum.
800 + 900 + 4x = 3500
4x = 1800
2x = 900
x = 450
The salary of each of those four highest paid employees is $450,000.
Answer = (E)
5) This question is designed for an analysis involving proportional placement of the mean. First, observe that R is much closer to the average for one-bedroom apartments, so there must be more one-bedroom apartments and fewer two-bedroom apartments.
The ratio of the distances to R is
5600:10400 = 56:104
Cancel a factor of 8 from both 56 and 104
56:104 = 7:13
One-bedroom apartments are “13 parts” of the building, and two-bedroom apartments are “7 parts.” That’s a total of 7 + 13 = 20 parts in the building. Two-bedroom apartments constitute 7/20 of the apartments in the building. Since 1/20 = 5%, 7/20 = 35%.
Answer = (B)
6) Method I: using sums
First, last year. Let x be the number of entrees. Then (15 – x) is the number of appetizers. The sums are:
entrees = 30x
appetizers = (15 – x)*12 = 12*15 – 12x = 6*30 – 12x = 180 – 12x
total = 15*18 = 30*9 = 270
Notice the use of the Doubling and Halving trick in the second and third lines. The two individual sums should add up to the total sum.
30x + 180 – 12x = 270
18x = 90
x = 5
They start out with 5 entrees and 10 appetizers.
Let N be the number of appetizers added, so now there are 5 entrees and (10 + N) appetizers. We need to solve for N. Again, the sums:
entrees = 5*30 = 150
appetizers = (10 + N)*12 = 120 + 12N
total = (15 + N)*15 = 225 + 15N
Again, the two individual sums should add up to the total sum.
150 + 120 + 12N = 225 + 15N
270 = 225 + 3N
45 = 3N
15 = N
They added 15 more appetizers. Answer = (E)
Method I was do-able, but we had to solve for many values.
Method II: proportional placement of the total average
Originally, the entrée price was 30 – 18 = 12 from the total average, and the appetizer price was 18 – 12 = 6. This means there must have been twice as many appetizers as entrees. Therefore , with 15 items, there must have been 10 appetizers and 5 entrees.
The number of entrees doesn’t change. The average drops to $15, so the distance from the entrée price is now 30 – 15 = 15, and the distance from the appetizer price is now 15 – 12 = 3. That’s a 5-to-1 ratio, which means there must be 5x as many appetizers as entrees. Since there still are 5 entrees, there must now be 25 appetizers, so 15 have been added.
Answer = (E)
If you know how to employ this method, it is much more elegant.
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]]>The post Counting Practice Problems for the GMAT appeared first on Magoosh GMAT Blog.
]]>(A) 720
(B) 1,512
(C) 2,520
(D) 6,400
(E) 12,600
2) A librarian has a set of ten different books, including four different books about Abraham Lincoln. The librarian wants to put the ten books on a shelf with the four Lincoln books next to each other, somewhere on the shelf among the other six different books. How many different arrangements of the ten books are possible?
3) Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?
(A) 22!
(B) 63!
(C) 63!/3!
(D) 66!/3
(E) 66!/3!
4) A bag contains ten marbles of the same size: 3 are identical green marbles, 2 are identical red marbles, and the other 5 are five distinct colors. If 5 marbles are selected at random, how many distinct combinations of five marbles could be drawn?
(A) 41
(B) 51
(C) 62
(D) 72
(E) 82
5) In a bag, there are five 6-sided dice (numbered 1 to 6), three 12-sided dice (numbered 1 to 12), and two 20-sided dice (numbered 1 to 20). If four of these dice are selected at random from the bag, and then the four are rolled and we find the sum of numbers showing on the four dice, how many different possible totals are there for this sum?
(A) 61
(B) 106
(C) 424
(D) 840
(E) 960
6) A chessboard is an 8×8 array of identically sized squares. Each square has a particular designation, depending on its row and column. An L-shaped card, exactly the size of four squares on the chessboard, is laid on the chessboard as shown, covering exactly four squares. This L-shaped card can be moved around, rotated, and even picked up and turned over to give the mirror-image of an L. In how many different ways can this L-shaped card cover exactly four squares on the chessboard?
(A) 256
(B) 336
(C) 424
(D) 512
(E) 672
7) A shipping company has four empty trucks that will head out in the morning, all four to the same destination. The clerk has four different boxes to ship to that same destination. All four boxes could go on any one of the trucks, or the boxes could be split up into any groupings and given to the trucks in any combinations (ie. two to one truck, one to another, and one to another). In how many different ways could the boxes be put on the four trucks?
(A) 16
(B) 64
(C) 256
(D) 576
(E) 4096
8) Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one green marble. Suppose we are going to put them into three cups: a black cup, a white cup, and a purple cup. We could put all six in any cup and leave two cups empty; or we could put marbles in two cups and leave one cup empty; or we could put some marbles in each of the three cups. How many combinations are possible?
(A) 90
(B) 180
(C) 360
(D) 540
(E) 720
9) Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one green marble. Suppose we are going to put them into three cups: a black cup, a white cup, and a purple cup. The only restriction is that the two red marbles can’t be in the same cup. We could put as many as five (all except one of the reds) in any cup. We could leave one cup empty, or put some in each of the three cups. All combinations are allowed that don’t involve the two red marbles in the same cup. How many combinations are possible?
(A) 90
(B) 180
(C) 360
(D) 540
(E) 720
10) In a certain mathematical activity, we start with seven cards, each with a different prime number written on it. These seven cards are randomly put into three boxes in the following way: one box must get four cards, one must get two, and one gets only one. Then, for each box, we find the product of all the cards in the box, and that’s the “number” of the box. Then, we put those three numbers in order, from lowest to highest, and that is our set. How many different sets can be created from this process?
(A) 35
(B) 105
(C) 210
(D) 420
(E) 630
11) In a certain mathematical activity, we have five cards with five different prime numbers on them. We will distribute these five cards among three envelope: all could go in any envelope, or they could be broken up in any way among the envelopes. Then in each envelop, we find the product of all the cards in that envelope: that is the “number” of the envelope. An envelope containing no cards has the number 1. We then put the three envelope numbers in order, from lowest to highest, and that is our set. How many different sets can be produced by this process?
(A) 41
(B) 89
(C) 125
(D) 243
(E) 512
12) In the diagram below, points A and B are on opposite corners of a lattice consisting of 12 segments. A “true path” from A to B is a path on which no segment is traversed more than once. How many “true paths” from A to B are there?
(A) 6
(B) 10
(C) 12
(D) 14
(E) 18
Solutions will appear at the end of this article.
The following blogs cover some of the basic ideas of counting problems:
1) The Fundamental Counting Principle
2) Permutations and Combinations
4) Difficult Counting Problems
The first three just cover the basic rules. The fourth, in addition to four hard practice problems, has a discussion of some of the reasons why counting is such a challenging topic. I would refer you to that discussion, and would simply add: if you found the problems above challenging, then read the solutions very careful. The hardest part of a counting problem is often how to frame the problem, how to parse the situation, how to imagine it in do-able stages: once these choices are made, the application of the rules is often quite straightforward. How do you learn how to frame a counting problem? Well, first of all, pay attention to how solutions do this, starting with the solutions below. For more on right brain skills, see this post.
If the related blogs and the solutions here gave you some new insights, please let us know about this in the comments section!
1) If the ten books were all different, the arrangements would be (10!), a very big number. Because we have repeats of identical copies, not all 10! arrangements will be unique. For a subset of k identical members, those k members could be interchanged in k! orders, and the resulting arrangements would be the same, so we have to divide by k! to remove repetitions. In this example, we need to divide 10! by 4! and 3! and 2!:
Answer = (E)
2) Start out by thinking of the four books about Lincoln as one big book. This is one book, and we have six others, so we start by figuring out the arrangements of these seven books. That’s 7!
Now, for each one of those 7! arrangements, we can put the four Lincoln books in four different orders. Thus, for each 7! arrangements of the Lincoln slot with the six other books, we have 4! variants because of the order of the Lincoln books. That’s a total of
N = (7!)(4!)
Answer = (C)
3) If the constraints with these three individuals didn’t matter, there would be 66! arrangements. Any one arrangement will have those three in some particular order. If we kept the other 63 participants in the same places, and just re-arranged these three, there would be 3! = 6 possibilities:
We could group all (66!) arrangements into groups of 6 like this, and in each case, only 1 of the 6 would have the correct order for these three individuals. Thus, we need to divide (66!) by 3! = 6.
N = (66!)/(3!)
Answer = (E)
4) Here, we have to do a combination of listing and calculating. We have to think about several individual scenarios. First, the “three green” scenarios. By “other color”, I mean the colors other than green and red; there are five of these.
Scenario #1: 3 green, 2 red = 1 possibility
Scenario #2: 3 green, 1 red, 1 other color = 5 possibilities
Scenario #3: 3 green, 2 other colors = 5C2 = 10 possibilities
Now, the “two green” scenarios:
Scenario #4: 2 green, 2 red, 1 other color = 5 possibilities
Scenario #5: 2 green, 1 red, 2 other colors = 5C2 = 10 possibilities
Scenario #6: 2 green, 3 other colors = 5C3 = 10 possibilities
Now, the “one green” scenarios:
Scenario #7: 1 green, 2 red, 2 other colors = 5C2 = 10 possibilities
Scenario #8: 1 green, 1 red, 3 other colors = 5C3 = 10 possibilities
Scenario #9: 1 green, 4 other colors = 5C4 = 5 possibilities
Now, the “no green” scenarios:
Scenario #10: 2 red, 3 other colors = 5C3 = 10 possibilities
Scenario #11: 1 red, 4 other colors = 5C4 = 5 possibilities
Scenario #12: 5 other colors = 1 possibility
Now add these.
(10 + 10 + 10 + 10 + 10 + 10) + (5 + 5 + 5 + 5) + 1 + 1 = 60 + 20 + 2 = 82
Answer = (E)
5) This is not really a counting question, in that it doesn’t involves any of the standard counting techniques. We just have think about this logically. No matter what four dice we pick, the lowest roll we could get is a “1” on each of the four dice, for a total of 4. We could get any integer value from 4 up to the highest value. The highest value would occur if we picked the two 20-sided dice and two of the 12-sided dice, and got the highest value on each die: 20 + 20 + 12 + 12 = 64. We could get any integer from 4 to 64, inclusive. For this, we simply need inclusive counting. 64 – 4 + 1 = 61.
Answer = (A)
6) Consider the L in its current orientation, and start with it in the lower left corner.
We can move this up so that the top square is in any of the five empty squares above it: those, plus this original, is 6 positions. Now, we can move any of these six to the right one space, then again, then again, until we have moved it to the sixth new space, when the right side of the L will come up flush against the rightmost boundary. That’s seven total columns, each with 6 positions, for a total of 42 while it is in this orientation.
Clearly, we can rotate by 90° clockwise, and we would have 42 new positions for that orientation. Then we can rotate by 90° clockwise again, and again. Four orientations, each with 42 positions, for 42*4 = 168 positions.
All of this is for the “forward L.” Now, if we pick up the card, and put it down flipped over, to get a “mirror image L,” this again will have 42 positions in each of 4 rotated orientations, for another 168 position.
The total is 2*168 = 336
Answer = (B)
7) Where we put one box has absolutely no bearing on where we put any of the other boxes. The placement of the four boxes is completely independent of one another. For each box, we have four choices.
N = 4*4*4*4 = 16*16 = 256
Answer = (C)
8) First, consider possibilities the 3 blue marbles.
Blue Case I: all three in one cup = 3 possibilities
Blue Case II: two in one cup, one in another è three choices for the cup with two, then two choices for the cup with one = 6 possibilities
Blue Case III: one in each cup = 1 possibility
total = 3 + 6 + 1 = 10 possibilities for the blue marbles
Now, the 2 red marbles.
Red Case I: two in one cup = 3 possibilities
Red Case II: one in cup, one in another = 3 possibilities for which cup is empty
total = 6 possibilities for the red marbles
Now, the green marble can simply go in one of the three cups = 3 possibilities.
By the Fundamental Counting Principle, multiple the blue & red & green possibilities:
N = 10*6*3 = 180
Answer = (B)
9) First, consider possibilities the 3 blue marbles.
Blue Case I: all three in one cup = 3 possibilities
Blue Case II: two in one cup, one in another è three choices for the cup with two, then two choices for the cup with one = 6 possibilities
Blue Case III: one in each cup = 1 possibility
total = 3 + 6 + 1 = 10 possibilities for the blue marbles
Now, the 2 red marbles.
Red Case I: two in one cup = FORBIDDEN!
Red Case II: one in cup, one in another = 3 possibilities for which cup is empty
total = 3 legal possibilities for the red marbles
Now, the green marble can simply go in one of the three cups = 3 possibilities.
By the Fundamental Counting Principle, multiple the blue & red & green possibilities:
N = 10*3*3 = 90
Answer = (A)
10) Part of the logic of the problem involves recognizing that every unique combination of prime numbers produces a unique product. There is no way that two different groupings cards will produce the same set of numbers.
First of all, the cards to go into the cup holding 4 cards.
;
Once we have placed 4 card in the first cup, we have three cards left, which means we have three choices of a single card to put into the third cup. Three choices. Once we place that, the remaining two cards must go into the second cup: no choice there.
N = 35*3 = 105
Answer = (B)
11) We have to consider different groupings. First, all the cards together.
Case I: all five cards in one envelope (2 envelopes empty) = one possibility
Now, two envelopes used, one empty.
Case II: four in one, one in another, one empty = five choices for the one by itself = 5 possibilities
Case III: three in one, two in another, one empty = 5C3 = 10 possibilities
Now, no empty envelopes, all three used:
Case IV: 3-1-1 split = 5C3 = 10 possibilities
Case V: 2-2-1 split = 5C2 = 10 for the first pair, then 3 possibilities for the single one, leaving the other two for the pair. This would be 10*3 = 30, but that double-counts the two pairs, so we need to divide by two. 15 possibilities.
N = 1 + 5 + 10 + 10 + 15 = 41
Answer = (A)
12) First of all, consider the minimum paths, the paths of just four segments from A to B. Each one of them will consist of, in some order, two horizontal segments and two vertical segments. How many of these are there? Well, how many ways can we distribute two horizontal segments among four slots? 4C2 = 6. We put the horizontal segments in two slots, and then the two vertical segments must go in the remaining slots. There are six minimum slots.
Now, we have to consider paths that take more than four segments to get from A to B. Consider the possibilities that begin: (down) then (right) then (up). The two possibilities are
Now, consider the possibilities that begin (down)(down) then (right) then (up). The two possibilities are:
Those are all the possibilities for path that start (down). We also could start with a first step of (right), but those possibilities are just the reflections of these over an imaginary mirror line from A to B. There are four path that start (down), and four mirror images that start (right). That’s 8 longer-than-necessary paths, to add to the 6 minimum paths, for a total of 14 paths.
Answer = (D)
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]]>Yesterday we solved this problem:
Div’s bicycle tour consists of three legs of equal length. For the first leg Div averaged 16 kilometers per hour. For the second leg he averaged 24 kilometers per hour. What speed must Div average for the final leg in order to average 24 kilometers per hour for the entire tour?
A) 20 kilometers per hour
B) 28 kilometers per hour
C) 32 kilometers per hour
D) 40 kilometers per hour
E) 48 kilometers per hour
We made a pretty good use of the RTD table, but we still took quite a while to arrive at our answer, and we did a lot of computation and algebra, every step inviting some sort of error.
There are two shortcuts that could save us some time and energy while reducing opportunities for computational error.
Notice that this problem doesn’t specify the lengths of the legs, but that every answer is a constant. That implies that so long as the legs are all of the same length as required by the problem, any leg length will yield the same answer.
Let’s take advantage of that fact to avoid some unnecessary algebra. Let’s stipulate an easy leg length. What leg length would be easiest to work with? 48 kilometers, since 48 is the least common multiple of 16 and 24.
Now we can put constants in the cells for leg lengths as well as in some of the cells for rates.
We can also determine a time for the bottom row—a combined time for all three legs—by dividing the combined distance by the average rate.
That allows us to determine the time for the third leg, since the sum of the times for the three legs is the combined time for the tour.
Finally, we can determine the rate for the third leg by dividing the distance by the time.
The little shortcut isn’t as powerful as the big one, and it’s peculiar to average problems, so you won’t get much use out of it, but here it is.
Since the rate for the second leg is equal to the average rate, we can ignore that leg. Because Div averages 24 kph for the entire tour and 24 kph for the second leg, he must also average 24 kph for the balance of the tour. So we could leave the second leg out altogether.
What would the RTD table for that amended problem look like? Well, if we use the big shortcut, we’ll still probably make each leg 48 kilometers, since 48 is the least common denominator of the rate in kph of the first leg and the rate in kph of the entire tour.
We can complete the top and bottom rows by dividing distance by the time.
Next, we can complete the “time” column.
Finally, we can divide the length of the third leg, 48 kilometers, by the time for the third leg, 1 hour, to determine the rate for the third leg.
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]]>The post One More RTD Table Problem: Average Rates Part 1 appeared first on Magoosh GMAT Blog.
]]>Here’s today’s problem:
Div’s bicycle tour consists of three legs of equal length. For the first leg Div averaged 16 kilometers per hour. For the second leg he averaged 24 kilometers per hour. What speed must Div average for the final leg in order to average 24 kilometers per hour for the entire tour?
A) 20 kilometers per hour
B) 28 kilometers per hour
C) 32 kilometers per hour
D) 40 kilometers per hour
E) 48 kilometers per hour
I usually start with the bare outline of an RTD table, and then fill it in bit-by-bit. Today, though, I’m going to start with a warning. Every problem on average rates has at least one attractive wrong answer that is intuitively appealing and seems promise a quick solution. Can you spot it here?
You might have said 20 kilometers per hour, because that is the average of the given rates, 16 and 24 kilometers per hour. Fair enough. That is wrong and will attract some people who read too quickly. I’m more concerned about a different trap, though.
Average-rate problems encourage you to assume that spending the same distance at two (or more) different rates allows you simply average those rates. For instance, in this problem, we might suppose that traveling one leg at 16 kph, one at 24 kph, and one at 32 kph would yield an average speed for the whole tour of 24 kph, because 24 is the simple mean of 16, 24, and 32 . And there’s 32 kilometers per hour waiting for you!
In fact, though, speeds are weighted according to how much time you spend at each, not how much distance you spend at each. This means that average speeds are generally less than you’d guess based on their component speeds, since it takes you more time to travel a given distance when you travel it slowly. (This neat fact deserves a post of its own, and I’ll write one soon.)
As a practical matter, what this means for average speed problems is that you will usually need to approach them not as simple averages or even as weighted averages. Instead, you need to view average speed problems through this formula:
Let’s devote one row of the table to each leg, and a fourth row to the total.
For an average-rate problem, you’ll need a wall between the rates of the various legs and the rate for the whole trip. I’ve represented that wall by drawing a heavy black line above the bottom-left cell. The only way to fill in that cell is from the other information in the bottom row, not directly from the cells above it in the same column.
The red lines below show the direction in you will usually add each of the columns and divide the bottom row.
Today’s problem is a bit strange though, so we’ll do things in a slightly different way.
Okay, let’s get rid of those red lines and add the rate information we’ve been given: the rates for the first two legs and the average rate.
Those are the only constant values we’ve been given, but we can add a little more information even so. We don’t know the distance, but we know that it’s the same for every leg, so let’s call that distance d, and then sum the legs to get a combined distance.
Now that we’ve represented both rate and distance for three of our four rows, we can determine times for those rows by division;
Notice that the column for “time” is almost complete. Since the total time, d/8, must be the sum of the other times, we can determine the missing time for the third leg.
Multiply through by 48, the least common denominator.
Transpose.
Divide.
Let’s add that to the table.
Now that we have the distance and time for the third leg, we can divide to determine the rate;
Okay, even using the RTD table, we still had to do an awful lot of work. It turns out, though, that a couple of shortcuts could have saved us most of that work. Come back tomorrow to see how.
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]]>The post Using the RTD Table for a Complicated Problem appeared first on Magoosh GMAT Blog.
]]>If you’d like to review before moving on, check out my previous posts on the topic:
We used the RTD (Rate, Time, and Distance) table to manage these problems, but some test-takers probably found that method to be overkill. Those readers judged that they could translate the problems directly from English to algebra without benefit of any table.
Today I’m going to give you a more complicated problem, one in which the travelers don’t move simultaneously:
Jan’s house and Cindy’s house are joined by a straight road 24 miles long. Jan and Cindy agree to meet at a restaurant along that road, twice as far from Jan’s house as from Cindy’s. Beginning at noon, Cindy walks to the restaurant at a constant speed of 3 miles per hour. Later, Jan drives to the restaurant at a constant speed of 30 miles per hour. If they arrive at the restaurant simultaneously, at what time did Jan begin her drive?
A) 12:32
B) 1:25
C) 1:52
D) 2:08
E) 2:40
This is likely to be a frustrating problem. The algebra—if we can get to the algebra—probably won’t be too complicated, but the translation into algebra looks daunting, if only because we haven’t likely seen a very close model of this problem before. Let’s try our good friend the RTD table, and see if it eases the translation.
We’ll set the table up as we have before, with columns for rate, time, and distance, and with rows for Jan, Cindy, and their combined distance:
We’re given Jan’s and Cindy’s rates directly, so let’s put those in.
We are given the combined distances for Jan and Cindy, 24 miles, so we can include that. We’re told that the restaurant is “twice as far from Jan’s house as from Cindy’s.” If we call the distance from Cindy’s house d, then the distance from Jan’s house must be 2d.
d+2d=24
3d=24
d=8.
So the restaurant is 8 miles from Cindy’s house and 16 miles from Jan’s.
Let’s complete the rows for Cindy and Jan. Since we know that Cindy’s rate is 3 mph and that her distance is 8 miles, we can conclude that her time was 8/3 hours. Since we know that Jan’s rate is 30 mph and that her distance is 16 miles, we can conclude that her time was 16/30 hours.
But now what? Well, if Cindy took 8/3 of an hour, then she took hours, or 2 hours and 40 minutes. Since she started at noon, she arrived at the restaurant at 2:40.
Since Cindy and Jan arrived simultaneously, Jan too arrived at 2:40. At what time did she leave? Well, she took 16/30 of an hour. That’s 32/60, or 32 minutes. Since she arrived at 2:40 after 32 minutes of travel, she must have started her drive at 2:08.
If you’ve read my other blogs on the use of this table, then you know that the answer is “yes,” at least for some people. The table is an aid to translation, and a few people are able to translate even very complicated rate problems directly into algebra without benefit of the table.
Still, I think that the RTD table is especially useful for this problem, even though as we used it here it turned out to just be two equations stacked one on the other, and even though it left us with a fair bit of work still to do. Disciplined well-practiced use of the RTD table allows us to chip away at the familiar parts of the problem and to reserve our cognitive resources for the less familiar parts.
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]]>We wanted to solve for Mary’s time, t.
In every row the relationship among rate, time, and distance is the same: RT=D. In this diagram the bottom row looks the most promising, since it alone contains only the variable for which we’re solving. (Why mess around with d if we don’t need to?) So let’s look at the equation implicit in that bottom row:
Cross-multiplying to get the difference between those fractions yields:
or:
Yes. I wanted to show an efficient use of the table, but I didn’t want to insist on the optimal use.
You could, for instance, have represented Mary’s rate as 200 rather than as 1000/5, and Kate’s rate as 500/3 rather than as 1000/6. Doing that in the diagram or simplifying the moment you’d pull the equation out of the diagram would have yielded instead of That would have allowed you to multiply through by 3 rather than to cross-multiply:
That definitely saves some time and effort, but if our first, efficient-but-not-optimal use of the table comes more easily, that’s OK. Sometimes there’s a little trade-off between the ease of translation (English-into-algebra) and the ease of solution (algebraic manipulation).
By the way, I wouldn’t take it a step further and represent Kate’s rate as . Mixed numbers are usually more difficult to clear than are improper fractions.
For instance, many people build RTD tables with no bottom row for the sum or the difference of the rates and distances. This is just inviting complexity and computational error. Such a table often yields a system of two equations and two variables rather than a single equation with a single variable.
Consider what our table would have looked like with no bottom roappw:
This would have yielded a system of equations:
If you distribute those fractions right away you’re in for a lot of work. It might occur to you to instead isolate t in each equation, then to solve for d, and finally to solve for t. It turns out that that, too, is a lot of work:
But 3750 isn’t one of our answers! That’s because we’re solving for t rather than for d. We have to return to one of our equations that related t to d, substitute 3750 for d, and solve for t:
Bottom-line? Always include an extra row for simultaneous-movement problems. You won’t need to use it every time, but it will you save you a lot of trouble when you do need it.
Hey, I think that I could answer this problem without the table!
Well, then you probably could.
You might remember this formula from my earlier post on the RTD table: (combined rate)(time)=(combined distance). We used that for travelers moving simultaneously in opposite directions.
Here’s a similar, and similarly simple, formula for travelers moving simultaneously in the same direction: (difference in rates)(time)=(difference in distances).* Applying that formula to this problem directly yields our equation:
You’ve still got to do the math, of course.
*In fact, some people point out that the second formula is just a special case of the first, since subtraction is one way to combine values. I find that confusing, since for non-mathematicians like us “combined” usually means “added.”
Just in case you’ve patiently read these posts on the RTD table and you still haven’t seen a problem that you couldn’t translate directly from English to Algebra, my next post will feature a more complicated rate problem.
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]]>Today and tomorrow we’re going to use the table just a tiny bit differently to solve another common sort of rate problem, one in which two travelers move in the same direction simultaneously. We’ll set up the table today and use it to solve the problem tomorrow.
Many problems in which two travelers move in the same direction simultaneously involve a faster traveler starting behind a slower traveler, and catching up to or passing him.
Mary and Kate are running clockwise around a circular track with a circumference of 500 meters, each at her own constant speed. Mary runs 1000 meters every five minutes and Kate runs 1000 meters every six minutes. If Mary and Kate start opposite one another on the circular track, how many minutes must Mary run in order to pass Kate and catch her again?
We’ll begin with our familiar layout: three columns (for rate, time, and distance) and three rows (for Mary, Kate, and the difference between them). A lot of people mark the bottom row in such a table as “C,” for “combined,” just as they would if Mary and Kate were traveling in opposite directions, though “combined” in this case doesn’t mean “added together.” Just to be clear, I’m going to label that bottom row with a delta, to remind myself to determine the difference between Mary’s and Kate’s rates and distances. I’m also going to circle the cell representing Mary’s time, since that’s what we’re solving for.
Before we talk about how best to manage this problem and table, go ahead and copy the table above, then add the information from the problem.
Now let’s add the information that we’ve been given. Begin with the rates. Mary’s rate is not 5 minutes and Kate’s rate is not 6 minutes. Speeds are always D/T, so Mary’s and Kate’s rates are 1000/5 and 1000/6 respectively.
Similarly, Mary’s time and distance are not 5 minutes and 1000 meters, and Kate’s time and distance are not 6 minutes and 1000 meters. Those figures are given to you only to determine Mary’s and Kate’s rates. NEVER ASSUME THAT THE TIME AND DISTANCE THAT FIX THE RATE ARE THE APPROPRIATE FIGURES FOR THE TIME AND DISTANCE COLUMNS!
In fact, Mary’s and Kate’s distances are so far unknown to us. Let’s represent them with variables. Or better, let’s represent them with a single variable, using d for Kate’s distance. If Kate ran d meters, how far would Mary run? Well, she’d have to run an additional 250 meters to catch Kate, and then an additional 500 meters to lap her. That means that if Kate’s distance were d meters, Mary’s distance would be d+750 meters.
Since Kate and Mary run simultaneously, a single time t will suffice for all three rows.
Finally, subtract Kate’s rate and distance from Mary’s to fill in the bottom row.
As we saw in our earlier posts on the RTD table, diagrams generally produce equations which we then solve algebraically. Diagrams are usually tools to help us translate word problems into algebra; they are not usually substitutes for manipulating equations.
What equation can you draw from this table that would allow you to solve for t?
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]]>The post Using Diagrams to Solve GMAT Rate Problems: Part 2 appeared first on Magoosh GMAT Blog.
]]>There’s more than one correct way to use the table.
If you keep in mind a few simple truth about the RTD table and it’s application to simultaneous travel problems, your correct approach might look different than my correct approach, but we should both get the same right answer. If we also keep the number of variables to a minimum, we should get that right answer very efficiently.
Let’s suppose that we started as we did in Part 1, by circling the cell for whose value we’re solving, and by recording the time (2/3 of an hour), total distance (100 miles) and the rates for the two trains (r for X and r-30 for Y).
For last post’s solution we added r and r-30 to get a combined rate of 2r-30. That’s probably the easiest way to approach this, but suppose that instead of completing the R column, we instead completed the bottom, “combined” row.
What we did there was to divide the combined distance (100) by the time (2/3) to get the combined rate (100/(2/3) or 150). Since the rates must sum to the combined rate, we can now solve for r using the equation,
r+(r-30)=150
2r-30=150
2r=180
r=90
As in Part 1, we still have one more step, to solve for (2/3)r=60.
Part 1 and today’s Part 2 both I represented X’s rate as r, just because we needed to eventually solve for X’s distance. Might as well keep X’s row simple, right? But notice that we still would have got the same answer if we’d called Y’s rate r, called X’s rate r+30, and solved for (2/3)( r+30).
One table consistent with that choice would look like this:
Add the rates to complete the first column.
Using the formula DR=T with the bottom row yields,
(2r+30)(2/3)=100
(4r/3)+20=100
(4r/3)=80
4r=240
r=60
Substitute 60 for r in the top row, and solve for 2/3 of 90, X’s distance.
The post Using Diagrams to Solve GMAT Rate Problems: Part 2 appeared first on Magoosh GMAT Blog.
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