The GMAT Bootcamp » jake becker

Weighted Averages On The GMAT

jake becker — Fri, 26 Feb 2010 17:29:24 +0000

This post will introduce weighted average questions you’ll see on the GMAT. There is one main formula you need to solve simple GMAT Average questions:

Average = SUM / # of observations

Note that this can be rearranged to read:

SUM = Average x (# of obs)
# obs = SUM / Average

So, if you are given ANY 2 of the 3 values, you should be able to find the 3rd. For example:

John drinks an average of 1.5 cups of water/day. After how many days has he drank 3 gallons of water? (1 gallon = 16 cups.)

In this case, we are looking for the number of days (or observations) such that we total 48 cups (3 gallons) of water.

# = SUM / Average

# days = 48 cups / 1.5 cups/day
# days = 32 days

NEVER AVERAGE AVERAGES!

Class A has 15 students and an average height of 60”. Class B has 20 students. What is class B’s average height if the average height of both classes is 65”.

One might say: (A + B) / 2 = 65”; A = 60”; so B must be 70”. However, keep in mind:

TOTAL AVERAGE = TOTAL SUM / TOTAL OBS

CLASS A + CLASS B = BOTH

15 students + 20 students = 35 students
60” average + 68.75” average = 65” average
900” total in A + 1375” total in B = 2275” total in Both

The given information is in black. The necessary intermediate steps are in blue, and the red is your answer. Note that the average of the averages ≠ total average. We must calculate each average separately, and to do this we need the SUM and # of observations for each category. This brings us to the idea of WEIGHTED AVERAGES.

A WEIGHTED AVERAGE is needed when you are taking average of a large group in which there are subgroups with a different number of observations in each. Take a look at this generalized formula, assuming there are 3 groups A, B and C.

(Average of A x Obs in A) + (Average of B x Obs in B) + (Average of C x Obs in C)

(Obs in A) + (Obs in B) + (Obs in C)

Think of weighted averages like a tug of war between numbers. The “stronger” one side (dog) is, the more that weighted average (tennis ball) will be “pulled” in that direction.

In the previous question, we had:

CLASS A + CLASS B = BOTH

15 students + 20 students = 35 students
60” average + 68.75” average = 65” weighted average

Note that the weighted average is CLOSER to B’s average than it is to A’s. This is because there are 20 students in Class B compared to only 15 students in Class A.

Two More Examples

At a certain restaurant, the average (arithmetic mean) number of customers served for the past x days was 75. If the restaurant serves 120 customers today, raising the average to 90 customers per day, what is the value of x?

A. 2

B. 5

C. 9

D. 15

E. 30

WITHOUT using the formula, we can see that today the restaurant served 30 customers above the average. The total amount ABOVE the average must equal total amount BELOW the average. This additional 30 customers must offset the “deficit” below the average of 90 created on the x days the restaurant served only 75 customers per day.

30/15 = 2 days. Choice (A).

WITH the formula, we can set up the following:

90 = (75x + 120)/(x + 1)
90x + 90 = 75x + 120
15x = 30

x = 2 Answer Choice (A)

Use whichever makes more sense to you!

Anita spent a total of $780 on 52 bottles of wine for her wedding. She then decided to buy 8 bottles of sparkling wine for the toasts, as well. Was the average (arithmetic mean) price per bottle of wine less than $20?

(1) Each bottle of sparkling wine cost more than $15.

(2) Each bottle of sparkling wine cost less than $40.

Take another look at what exactly the question is calling for: the TOTAL average price of all the wine at the wedding. We should look at the suggested average ($20) and use that as our threshold amount.

60 bottles * $20/bottle = $1200 total
$1200 total – $780 (given) = $420 (left for sparkling wine)
$420 / 8 bottles = $52.50/bottle of sparkling wine (for the total average to equal $20)

Which of the answer choices are conclusively above or below $52.50/bottle of sparkling wine? Only (2). With (1), we can be below OR above the threshold, so (1) is not sufficient.

Answer Choice (B)

Now, you’re score will be above average! Please visit the Grockit forum or leave a comment here if you have more questions on weighted averages.

Good luck!

Estimation Strategies For GMAT Problem Solving

jake becker — Tue, 19 Jan 2010 20:32:51 +0000

The GMAT is not testing who is the fastest at long division. It is a test that seeks to measure problem solving skills that are not necessarily the “textbook” ways to discover solutions. Let’s discuss some estimation strategies, which are not used as often as they should be.

1. Round Up AND Round Down When Multiplying

Be aware of the direction in which you are altering the result. If you want to estimate a product of two “ugly” numbers, you can move one up and one down, which is an attempt to minimize the error in your estimation. For example:

658*436 = 286,888

If we round 658 UP to 700 and 436 DOWN to 400, we can approximate using:

700*400 = 280,000

2. Round In The Same Direction When Dividing

When you want to approximate a fraction, you can either adjust only the numerator (or denominator) or move both in the same direction. For example:

8/19 = .4210526…

8/20 = 0.4 (Note that increasing the denominator, will decrease the fraction.)

9/20 = 0.45 (Note that increasing both top and bottom will increase the fraction.)

Your estimate is somewhere between .40 and .45.

3. Remember These Other Helpful Tips

Peek at your answer choices: If your answer choices are relatively far apart, this could be hint that approximation is helpful. If the answers are very tight together, you may still estimate, but you have to be more careful and do due diligence.
Geometry shortcut 1: √2 =~ 1.4 and √3 =~ 1.7. Try to commit these to memory, as they are very common.
Geometry shortcut 2: Be careful when using = 3. Recognize that you are using a smaller number, so your result will be smaller too. Test makers love to give tempting answer choices that assume = 3. It’s not.
Geometry shortcut 3: Even though you cannot assume charts are drawn to scale, they can still be a resource. Obtuse/acute angles are typically shown as much, and angles can be approximated in many circumstances. That’s not to say “if it looks like a right angle, it must be 90.” But you can use the drawing as a guide to your estimation.
Use the extremes: If you are given a range, it helps to plug in those extremes to see between which values your answer falls. This will focus your attention on the cases that are above (or below) those endpoints.

Two Examples

If a square has a perimeter of 80 inches, what is the approximate length of its diagonal, in inches?

A. 20

B. 28

C. 40

D. 56

E. 112

This question uses the word “approximate,” so that should be a very big hint that you will need to find a number “close enough.” If P = 80, then s = 20. The diagonal is essentially a hypotenuse of a 45-45-90 triangle, so d = 20√2.

Two strategies:

1) 20√1 = 20 and 20√4 = 40. Therefore 20 < 20√2 < 40. (B) 28 is the only option.

2) Since we remember that √2 =~1.4, we can simply multiply 20*1.4 = 28. (B).

Addison High School’s senior class has 160 boys and 200 girls. If 75% of the boys and 84% of the girls plan to attend college, what percentage of the total class plan to attend college?

A. 75

B. 79.5

C. 80

D. 83.5

E. 84

84 is an obscure number. When you see obscure numbers, that is another sign that you may want to look for an approximating shortcut.

Firstly, we should eliminate the overtly incorrect choices. This will be (A) 75 (since that’s the low extreme) and (E) 84 and (D) 83.5 (since they are both essentially equal to the high extreme).

Secondly, find the average of the given percents. Since there are more girls than boys, we know that the weighted average will be closer to the girls’ percent than the boys’ percent. By finding 79.5% as the mean of 75% and 84%, we are given the low extreme. Again, we recognize the weight placed on 84%, making the answer higher than 79.5. (C) 80 it is!

(For similar questions in the future where we actually need to calculate, we could drop the extra “0” from 160 and 200. The ratio of 16:20 is the same (4:5), and the calculation is much easier.)

Sequences On The GMAT Math Section

jake becker — Mon, 28 Dec 2009 19:10:54 +0000

There are four main categories of Sequences that appear GMAT Quantitative Section: Arithmetic, Geometric, Repeating and Sums. Think of sequences as a simple pattern, and detecting this pattern is probably the most difficult part.

Arithmetic Sequences

An Arithmetic Sequence is when subsequent terms in a sequence increase (or decrease) by a constant amount. Here’s the standard formula:

= + (n – 1)d, where is the value at term n, and d is the constant change.

Note that – = d

If given the following sequence, we can derive both and d, to solve for any term.

8, 11, 14, 17, 20, 23….. ( = 8 and d = 3)

So, if asked what term number 86 is, we can just plug in to the formula:

= + (n – 1)d

= 8 + (86 – 1)3 = 8 + 85*3 = 8 + 255 = 263

Geometric Sequences

The same principle applies to Geometric Sequences, in which each subsequent term is multiplied by a certain constant. Compound interest in an example of a geometric sequence. Here’s the standard formula.

= * , where is the value at term n, and r is multiplicative rate of increase.

Note that / = r

Similarly, when given the rate of increase and the value of any term, we can find any other term. For example:

John originally put $8 in his piggy bank in 2001. If his parents double the money in the piggy bank once a year (without John adding anymore himself), how much money will he have in 2009?

r = 2

= *

= 8

= 8*

n = 8

= $1,024

Repeating Patterns

Many times, repeating patterns will yield a remainder question.

The first term of a sequence is -2 and the second term is 2. Each subsequent odd term is found by adding 2 to the previous term, and each subsequent even terms is found by multiplying the previous term by -1. What is the sum of the first 669 terms?

Clearly, we are not looking to enter all 669 terms and see what the last one is. But, we can do a few and check out the pattern.

n = 1, = -2

n = 2, = 2

n = 3, = 4

n = 4, = -4

n = 5, = -2

n = 6, = 2

Since we can see that the pattern will repeat every 4 terms, we can solve for the remainder after dividing 669/4. Since 4 goes into 668 evenly, we know that the value will be equivalent to that of the first term, which = -2.

Sums

With sequence questions involving sums, identify the pattern. This is the most important first step toward finding the solution. Check out the following example.

A set of consecutive integers begins at -19. After which term will the sum of all the terms equal 41?

Again, we don’t want to write all these out. BUT, we can start, and try to see if we can detect the pattern. In this particular example, we will be adding integers in the negative space until 0, after which we will be adding the positive pair of the same integers all the way up to +19. The sum after +19 equals zero, equals 20 after +20 and equals 41 after +21.

That means 19 negative terms + 1 (zero) + 21 positive terms = 41 total terms.