This article is continued from the first, on “Translating from Words to Math.” First of all, here are four word problems that present issues with assigning variables.
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.
Assigning Variables
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.
Summary
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.
Explanations to Practice 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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