# The Importance of Organization for Word Problems

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Molly worked at an amusement park over the summer.Every two weeks she was paid according to the following schedule:at the end of the first two weeks she received \$160.At the end of each subsequent two week period she received \$1, plus an additional amount equal to the sum of all the payments she had received in the previous weeks.How much money was Molly paid during full 10 weeks of summer?

A. 644
B. 1288
C. 1770
D. 2575
E. 3229

To start, let’s create a table just to lay out the general information first:

After 2 weeks —> \$160

After 4 weeks —-> \$1 + (all previous)

After 6 weeks —-> \$1 + (all previous)

After 8 weeks —-> \$1 + (all previous)

After 10 weeks —-> \$1 + (all previous)

Then fill in what the “all previous” would mean:

After 2 weeks —> \$160

After 4 weeks —-> \$1 + \$160 = \$161

After 6 weeks —-> \$1 + (\$160 + \$161) = \$161 + \$161

After 8 weeks —-> \$1 + (\$160 + \$161 + \$161 + \$161) = \$161 + \$161 + \$161 + \$161

After 10 weeks —> \$1 + (all previous) = eight \$161’s

At this point we can see that “all previous” is going to equal the initial \$160 + seven \$161 and then we’re adding \$1 to it, so it’s really just \$161 x 8 = \$1288

We have an interesting sequence:

After 2 weeks —> \$160

After 4 weeks —-> one \$161

After 6 weeks —-> two \$161

After 8 weeks —-> four \$161

After 10 weeks —-> eight \$161

Basically the “number” of \$161’s just doubled from 4 weeks to 10 weeks.

Now to add them all up, again, we’re just going to count up all of the number of \$161’s.

We have our initial \$160 + 15*(\$161) = \$2575. The correct answer is (D).

You will see over and over again that it’s not the Math calculations in Word Problems that will trip you up, but the inability to be super organized on your scratch paper. 🙂

# A Sequence Question or an Interest Question?

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Sometimes the concept being tested on the GMAT is not clear from the first read. A question that might look a little intimidating could be easier than it initially appears! Take, for example, this word problem:

Torres invested \$P in a long term investment plan for 24 years at an interest compound annually. But due to some financial issues he had to withdraw money at the end of 6 years.How much more amount could he have gotten after 24 years if he received \$3P after 6 years , if it’s known that there were no pre-closing charges or deductions in the plan?

A. 9P
B. 12P
C. 72P
D. 78P
E. 81P

This question is tricky at first since we don’t know the compounding interest rate, and feels like perhaps we would need a compound interest formula to solve, but let’s consider.

It started at P and then compounded 6 times and tripled itself to 3P. So this is more like a Geometric sequence question!

The rule is “P triples every 6 years.”

Year 1 – P
Year 6 – 3P
Year 12 – 9P
Year 18 – 27P
Year 24 – 81P

How much more means we subtract 3P from 81P:
81P-3P = 78P

The answer is (D). What is interesting is that we don’t need to know ANYTHING about interest to get this question correct!

The skill required is simply good scratch-work organization!

# Math Things to Memorize: the Combinations Formula

For some questions on the GMAT, you will need to know the Combinations formula.

Combinations formula = n! / k! (n-k)!

n = the bigger number (what we’re choosing from)
k = the smaller number (how many we’re choosing)

Check out this question from Math Revolution:

How many committees can be formed comprising 2 male members selected from 4 men, 3 female members selected from 5 women, and 3 junior members selected from 6 juniors?

A. 900
B. 1200
C. 1500
D. 1800
E. 2400

We could almost rephrase this as three different questions:

How many ways to choose 2 from 4?
How many ways to choose 3 from 5?
How many ways to choose 3 from 6?

Let’s start with the first question: how many ways to choose 2 from 4?

4! / 2! (4-2)!
4! / 2!2!
(4 x 3 x 2 x 1) / (2 x 1)(2 x 1)
24/4 = 6

But if we counted this and said the four males were ABCD, we could just list AB, AC, AD, BC, BD, and CD, and you can see it also equals 6.

Anyway, finding the other two:

How many ways to choose 3 from 5?

5! / 3! (5-3)!
5! / 3! 2!
We can cancel out 3! from both numerator and denominator.
5 x 4 / 2
20/2 = 10

How many ways to choose 3 from 6?

6! / 3! (6-3)!
6! / 3! 3!
We can cancel out one 3! from both numerator and denominator.
6 x 5 x 4 / 3 x 2
120 / 6 = 20

Back to our original questions:

How many ways to choose 2 from 4? 6
How many ways to choose 3 from 5? 10
How many ways to choose 3 from 6? 20

Now, multiply all those numbers together!

6 x 10 x 20 = 1200

# Backsolving Problem Solving “Work” Questions

One way to do these type of question is to Backsolve, or essentially “try out” the answer choices. Let’s look at a problem:

Working together, each at his or her own constant rate, Jeff and Ashley painted their apartment in 6 hours. Working at his constant rate, Jeff could have painted the whole apartment in 10 hours. How many hours would it have taken Ashley, working at her constant rate, to paint the apartment?

A. 4
B. 12
C. 15
D. 16
E. 20

Let’s say it takes Ashley 15 hours (choosing answer choice (C)). Ashley would have a 1/15 rate and Jeff’s rate is 1/10 each hour. Working together they would do 1/15 + 1/10 each hour, or 2/30 + 3/30 = 5/30 = 1/6 in one hour. Then, yes, it does make sense that together the would do the job in 6 hours.

We got lucky here that (C) ended up being the correct answer, but if our answer had not matched the given information, then we probably could have discerned the correct answer based on how “too big” or “too small” we were.

While not required, sometimes we forget how useful leveraging the answer choices can be!