Algebraic solutions for the Word Problems

In this lesson we will talk about word problems in which you can see not certain numbers but letters. They seem to be hard, but really are approximately as easy as such with fixed values. Let’s look at some problems.

1.    A group of 7 fishermen chartered a boat for a day to fish for flounder. The boat costs x dollars per day to rent. If the group can find 3 more fishermen on the docks who are willing to come aboard and share the rental costs, how much less will the rental cost be per person in terms of x?

A. x/70
B. x/35
C. 3x/70
D. 3x/10
E. 3x/7

At the very beginning, each fisherman has to pay 1/7 of the total amount of rent, or simply x/7. If the group can find 3 more fishermen, the amount of rent will be divided between 10 men, and each one will have to pay x/10. The difference in cost is the difference between x/7 and x/10:
x/7 - x/10 = (10x - 7x)/70 = 3x/70. So, the answer is C.

2.    If Finn was 18 months old one year ago, how old was he in months, x months ago?

A. x – 30
B. x – 12
C. 18 – x
D. 24 – x
E. 30 – x

If Finn was 18 months old 1 year (or 12 months) ago, then he is now 18 + 12 = 30 months old. Then 30 – x represents his age x months ago. The answer is E.

3.    Billy picks apples, and he is able to pick one apple per minute. How many apples will he have picked in ten minutes from now if he would have picked x apples total in y minutes from now?

A.    x + y + 10.
B.    x – y – 10.
C.    x – 10y.
D.    x + 10y.
E.    x – y + 10.

In y minutes from now, Billy will have picked x apples. As he picks one apple per minute, in y minutes he will have picked y additional apples. If we call the current amount of apples he has z, then in y minutes he will have picked a total of z + y apples, which we know is x apples: z + y = x.
We are looking for the amount he will have picked 10 minutes from now, which will be the amount which he has already picked, plus 10, which he will pick in the next ten minutes. So our solution is z + 10. However, we cannot give our solution in terms of z, so we can take our original expression, z + y = x and get z = x – y. So, z + 10 = x – y + 10, which is answer E.

4.    When ticket sales began, Pat was the nth customer in line for a ticket, and customers purchased their tickets at the rate of x customers per minute. Of the following, which best approximates the time, in minutes, that Pat had to wait in line from the moment ticket sales began?
A (n - 1) x
B n + x –1
C(n-1/x)
D x/(n-1)
E n/(x-1)
If Pat was nth customer in the line, there were (n-1) customers in front of him. Each customer takes x minutes to purchase a ticket, so, for (n-1) person total amount of time will be (n-1)x. The answer is A.

Resume:
This lesson shows some typical word problems in which you can see not certain numbers but letters and the most effective ways to solve them in 2 minutes or even faster and add some points to your GMAT total score.
    
Material prepared by Ksenia Zueva,
GMAT consultant at MBA Strategy