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.

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.)

Any other estimation tricks? Just post in the comment field or check out Grockit’s forums for more strategies on GMAT math.

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.

Remainders are the NUMERATOR of a fraction from a mixed number that results from division. For example, 19/3 leaves a remainder of 1, since 19/3  = 6 1/3.

Some quick tips:

1. Your remainder can only range from zero to the denominator of the fraction. For example, when dividing by 9, your remainder options are 0-8, since a remainder of 9 leaves you a new whole number (with a remainder of 0).
2. Look for the closest whole number and count up or down from there. For example, when trying to find the remainder of 146/15, you can see that 15 would go into 150 evenly. You then count down four from 150 to 146, so your remainder is (15 -4) = 11. This is easier than recognizing 135/15 is a whole number and counting up.
3. Become familiar with common trends or patterns. For example, multiples of even numbers are even, so 167/(even #) must have an ODD remainder.
4. The remainder should NOT be reduced. 18/4 = 4 2/4. The remainder stays equal to 2, even though you can reduce 4 2/4 to  4 1/2.

I recently came across this question, which I think is a good introduction:

What is the remainder of 3^(4n+3) divided by 5, assuming n is a positive integer?

Firstly, when dividing by 5, we are looking for the remainder above a one’s digit of either 0 or 5. In this scenario, we only care about the one’s digit, so we only need to look at the one’s digit while multiplying.

We can break 3^(4n+3) into 3^4n * 3^3 by the rules of exponents.

If n = 1, 3^4n = 3^4 = 81 = one’s digit of 1.

If n = 2, 3^4n = 3^8 = 81*81 = one’s digit of 1.

We detect the pattern that regardless the value of n, we will be multiplying a term with a one’s digit of 1 with a term with a one’s digit of 7 (3³), so the result will have a one’s digit of 7. When and number with a one’s digit of 7 dividing by 5, we are left with a remainder of 2.

Pattern questions with division are many times Remainder questions at their core

The 4 members of the Jones Family rotate who takes out the trash on a daily basis. The order goes as follows: Mom, Dad, Brother, Sister. If Dad takes out the trash on January 18th, who takes out the trash on March 26th? (There are 31 days in January and 28 days in February.)

It’s clear that we don’t want to whip out our calendars and start counting. (A general rule of thumb is that if you think it’s taking too long, it probably is….)

Instead, we see how many days pass between January 18 and March 26:

January 19-31: 13 +

February 1-28: 28 +

March 1 – 26:  26  = 67 days.

67/4 leaves you will a remainder of 3, so we count 3 from Dad, leaving us with Mom on March 26th.

Fractions and Decimals are the same thing

You should be familiar with common decimals, mainly:

1/2 = .5

1/3 = .33 repeating

1/4 = .25

1/5 = .20

1/6 = .166 repeating

1/8 = .125

1/9 = .11 repeating

Note that multiplying these by constants will leave similarly instructive results, such as:

3/8 = 3*1/8 = 3*0.125 = 0.375

The more familiar with these you become, the quicker you can eliminate answer choices are clearly wrong. For example:

If x is an integer, which of the following is a possible value of (x² +2x – 7)/9?

A. 0.268

B. 4.555 repeating

C. -2.4

D. 1.166 repeating

E. 8.125

We don’t have to start plugging in. We know that when divided by 9, the remainder will be a number repeating to the right of the decimal place. Only choice (B) fits that description. ((C) is divided by a factor of 5, (D) by a factor of 6, and (E) by a factor of 8.)

Join a Grockit game for more GMAT math practice with Jake!

There is one very important equation that guides all rate and work questions: r = d/t, “rate equals distance over time.” If given any two of the three, you should be able to find the third, if the units remain constant. Problems involving “work” are essentially rate problems, where a worker’s “efficiency” is calculated by amount completed in a given period of time.

Working Together

In questions where individuals work at different speeds, we typically need to add their separate rates together. Make sure you keep your units straight. This doesn’t mean wasting time and writing each and every one out, but rather simply recognizing their existence. Note that when working together, the total time to complete the same task will be less than BOTH of the individual rates, but not necessarily in proportion. Nor, are you averaging or adding the given times taken. You must add rates.

A worker can load 1 full truck in 6 hours. A second worker can load the same truck in 7 hours. If both workers load one truck simultaneously while maintaining their constant rates, approximately how long, in hours, will it take them to fill 1 truck?

A. 0.15

B. 0.31

C. 2.47

D. 3.23

E. 3.25

The rate of worker #1 is 1 truck/6 hours. This can also be 1/6 trucks/1 hour. The rate of worker #2 is 1/7. When together, they will complete 1/6 + 1/7 trucks/ 1 hour.

1/6 + 1/7 = 6/42 + 7/42 = 13/42 trucks/1 hour. Remember the question is asking for the number of hours to fill 1 truck, NOT the number of trucks completed in 1 hour. To find this, we find the reciprocal of 13/42.

42/13 hours/truck = 3 3/13 hours/truck.

At this point, we may not be able to decide between (D) or (E). However, the decimal is important. Because the denominator is 13, we know the decimal cannot equal .25. We can also see that 3/12 will yield .25, so 3/13 will be slightly lower. Choice (E).

Relative Velocity

Planes, trains and automobiles. Sometimes walking. Objects moving at given speeds on the GMAT usually travel toward or away from each other. When moving at an angle, we may be looking at a geometry question. If moving toward or away from each other, we can add their speeds to see their relative velocities. If moving in the same direction, we instead subtract their speeds to find the relative velocity. Again, be careful of units.

Train A traveling at 60 m/hr leaves New York for Dallas at 6 P.M. Train B traveling at 90 m/hr also leaves New York for Dallas at 9 P.M. Train C leaves Dallas for New York at 9 P.M. If all three trains meet at the same time between New York and Dallas, what is the speed of Train C if the distance between Dallas and New York is 1260 miles?

A. 60 m/hr

B. 90 m/hr

C. 120 m/hr

D. 135 m/hr

E. 180 m/hr

Relative to Train A, Train B’s velocity is 30 m/hr. This means that B will gain on A at a rate of 30 miles every hour. In the three hours from 6pm to 9pm, A gets to mile marker 180. To catch up the 180 miles, it will take Train B 6 hours. So when they all meet up, the time will be 3am, and they will be at mile marker 540. Notice that:

Train A = 9 hours at 60 miles/hour = 540 miles

Train B = 6 hours at 90 miles/hour = 540 miles

We can now tackle Train C, which has traveled the same time as B (6 hours), and traveled (1260 – 540) miles.

Rate of Train C = 720 miles/ 6 hours = 120 miles/hour. Choice (C).

Man Hours

Many times you may be asked to calculate the number of workers would be need to complete a certain task. Keep in mind that the number of workers (at the same efficiency) is inversely proportional to the amount of time it takes one to complete a given task. It may help consider the unit man-hours as the multiplication between workers and time, which is then compared to the work completed. For example:

Three plows working at identical constant rates can clear 123 ft of snow per minute. At this rate, how much snow could 8 plows remove in 5 minutes?

A. 328

B. 984

C. 1,640

D. 16,400

E. 131,200

Instead of man-hours, here we want to interact plow-minutes. Feet and minutes are already compared, so all we have to is add “plows” to the expression. If we divide 123 ft/min by 3 plows, we get:

123 ft/minute/3 plows = 41 ft/plow-minute

At this rate, if we want to increase minutes to 5 and plows to 8, we can simply insert these into the existing rate. Note the absolute rate does not change, since we are multiplying top and bottom by 40, so the value is constant.

41*40 feet / 40 plow-minutes = 1640 feet / 40 plow-minutes. Choice (C).

There are LOADS more rate questions; some are much more difficult. This is by no means exhaustive. Please visit the Grockit forum or leave a comment here to discuss further.