# Breaking Down the Scale Method for Weighted Averages

Before you dive into this post, make sure you are are familiar with the Scale Method for weighted averages, which we have discussed in previous posts.

We know that the scale formula of weighted averages is the following:

w1/w2 = (A2 – Aavg)/(Aavg – A1)

One point of confusion for many test takers regarding this formula is figuring out what A1, A2, w1 and w2 actually are.

Here is the simple answer: they can be anything. You can choose to set up the solution as you want. The only thing is that it must be consistent across. A1 and w1 could be the parameters of either solution; A2 and w2 will be the parameters of the other solution. We could also work with the concentration of either ingredient of the solution. We will illustrate this point with an example GMAT question:

A container holds 4 quarts of alcohol and 4 quarts of water. How many quarts of water must be added to the container to create a mixture that is 3 parts alcohol to 5 parts water by volume?

(A) 4/3
(B) 5/3
(C) 7/3
(D) 8/3
(E) 10/3

Now, we have been given two solutions that we have to mix:

1. A container holding 4 quarts of alcohol and 4 quarts of water
2. Water (which means it has no alcohol in it)

When these solutions are mixed together, they give us a mixture that is 3 parts alcohol to 5 parts water by volume.

So, what are A1, w1, A2, w2 and Aavg? We can work with the concentration of either alcohol or water. Let’s first see how we can work with the concentration of water:

Method 1:
A1 is the concentration of water in the solution of 4 quarts of alcohol and 4 quarts of water. So A1 = 4/8.
w1 is the volume of this solution.
A2 is the concentration of water in the solution of water only. So A2 = 8/8 (we want to write this in the same format that we write A1 in.)
w2 is the volume of this solution.
Aavg is the concentration of water in the final solution i.e. 5/8

w1/w2 = (A2 – Aavg)/(Aavg – A1)
w1/w2 = (8/8 – 5/8)/(5/8 – 4/8)
w1/w2 = 3/1

So 3 parts of the solution with alcohol and water should be mixed with 1 part of pure water.

Method 2:
A1 is the concentration of water in pure water. So A1 is 8/8
w1 is the volume of this solution.
A2 is the concentration of water in the solution of 4 quarts alcohol and 4 quarts water. So A2 is 4/8
w2 is the volume of this solution.
Aavg is the concentration of water in the final solution i.e. 5/8

w1/w2 = (A2 – Aavg)/(Aavg – A1)
w1/w2 = (4/8 – 5/8)/(5/8 – 8/8)
w1/w2 = 1/3

So 1 part of water should be mixed with 3 parts of the solution with alcohol and water (same result as above).

Now we will see how to work with the concentration of alcohol. Of course the result will be the same.

Method 3:
A1 is the concentration of alcohol in the solution of 4 quarts alcohol and 4 quarts water. So A1 is 4/8.
w1 is the volume of this solution.
A2 is the concentration of alcohol in the solution of water only. So A2 is 0/8 (to write in the same way as above)
w2 is the volume of this solution.
Aavg is the concentration of alcohol in the final solution i.e. 3/8

w1/w2 = (A2 – Aavg)/(Aavg – A1)
w1/w2 = (0/8 – 3/8)/(3/8 – 4/8)
w1/w2 = 3/1

So 3 parts of the solution with alcohol and water should be mixed with 1 part of pure water (same as above).

Method 4:
A1 is the concentration of alcohol in pure water. So A1 is 0/8
w1 is the volume of this solution.
A2 is the concentration of alcohol in the solution of 4 quarts alcohol and 4 quarts water. So A2 is 4/8.
w2 is the volume of this solution.
Aavg is the concentration of alcohol in the final solution i.e. 3/8

w1/w2 = (A2 – Aavg)/(Aavg – A1)
w1/w2 = (4/8 – 3/8)/(3/8 – 0/8)
w1/w2 = 1/3

So 1 part of pure water should be mixed with 3 parts of the solution with alcohol and water (same result as above).

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Quarter Wit, Quarter Wisdom: Try to Answer This GMAT Challenge Question!

Today, we will give you a GMAT challenge question. The challenge of reviewing this question is not that the question is hard to understand – it is that you will need to solve this official question within a minute using minimum calculations.

Let’s take a look at the question stem:

 Date of Transaction Type of Transaction June 11 Withdrawal of \$350 June 16 Withdrawal of \$500 June 21 Deposit of x dollars

For a certain savings account, the table shows the three transactions for the month of June. The daily balance for the account was recorded at the end of each of the 30 days in June. If the daily balance was \$1,000 on June 1 and if the average (arithmetic mean) of the daily balances for June was \$1,000, what was the amount of the deposit on June 21?

(A) \$1,000
(B) \$1,150
(C) \$1,200
(D) \$1,450
(E) \$1,600

The average of daily balances = (Balance at the end of June 1 + Balance at the end of June 2 + … + Balance at the end of June 30) / 30 = 1000

Now we have been given the only three transactions that took place:

• A withdrawal of \$350 on June 11 – so on June 11, the account balance goes down to \$650.
• A withdrawal of \$500 on June 16 – so on June 16, the account balance goes down to \$150.
• A deposit of \$x on June 21 – So on June 21, the account balance goes up to 150 + x.

Now we can plug in these numbers to say the average of daily balances = [1000 + 1000 + …(for 10 days, from June 1 to June 10) + 650 + 650 + … (for 5 days, from June 11 to June 15) +  150 + … (for 5 days, from June 16 to June 20) + (150 + x) + (150 + x) + … (for 10 days, from June 21 to June 30)] / 30 = 1000

One might then end up doing this calculation to find the value of x:

[(1000 * 10) + (650 * 5) + (150 * 5) + ((150 + x) * 10)] / 30 = 1000
x = \$1,450

But this calculation is rather tedious and time consuming. Can’t we use the deviation method ? After all, we are dealing with large values here! How?

Note that we are talking about the average of certain data values. Also, we know the deviations from those data values:

• The amount from June 11 to June 30 is 350 less.
• The amount from June 16 to June 30 is another 500 less.
• The amount from June 21 to June 30 is x in excess.

Through the deviation method, we can see the shortfall = the excess:

350 * 20 + 500 * 15 = x * 10
x = 1,450 (D)

This simplifies our calculation dramatically! Though saving only one minute on a question like this may not seem like a very big deal, saving a minute on every question by using a more efficient method could be the difference between a good Quant score and a great Quant score!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Quarter Wit, Quarter Wisdom: Using the Deviation Method for Weighted Averages

We have discussed how to use the deviation method to find the arithmetic mean of numbers. It is very useful in cases where the numbers are huge, as it considerably brings down the calculation time.

The same method can be applied to weighted averages, as well. Let’s look at an example very similar to the one we examined when we were working on deviations in the case of arithmetic means:

What is the average of 452, 452, 453, 460, 467, 480, 499,  499, 504?

What would you say the average is here? Perhaps, around 470?

Shortfall:
We have two 452s – 452 is 18 less than 470.
453 is 17 less than 470.
460 is 10 less than 470.
467 is 3 less than 470.

Overall, the numbers less than 470 are (2*18) + 17 + 10 + 3 = 66 less than 470.

Excess:
480 is 10 more than 470.
We have two 499s – 499 is 29 more than 470.
504 is 34 more than 470.

Overall, the numbers more than 470 are 10 + (2*29) + 34 = 102 more than 470.

The shortfall is not balanced by the excess; there is an excess of 102-66 = 36.

So what is the average? If we assume that the average of these 9 numbers is 470, there will be an excess of 36. We need to distribute this excess evenly among all of the numbers, and hence, the average will increase by 36/9 = 4.

Therefore, the required mean is 470 + 4 = 474. (If we had assumed the mean to be 474, the shortfall would have balanced the excess.)

This method is used in exactly the same way when we have a simple average as when we have a weighted average. The reason we are reviewing it is that it can be very handy in weighted average questions involving more than two quantities.

We often deal with questions on weighted averages involving two quantities using the scale method. Let’s see how to use the deviation method for more than 2 quantities on an official GMAT question:

Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x+y+z gallons of a 1.5 percent grade, what is x in terms of y and z?

(A) y + 3z
(B) (y +z) / 4
(C) 2y + 3z
(D) 3y + z
(E) 3y + 4.5z

Grade 1 milk contains 1% fat. Grade 2  milk contains 2% fat. Grade 3 milk contains 3% fat. The mixture of all three contains 1.5% fat. So, grade 1 milk provides the shortfall and grades 2 and 3 milk provide the excess.

Shortfall = x*(1.5 – 1)
Excess = y*(2 – 1.5) + z*(3 – 1.5)

Since 1.5 is the actual average, the shortfall = the excess.

x*(1.5 – 1) = y*(2 – 1.5) + z*(3 – 1.5)
x/2 = y/2 + 3z/2
x = y + 3z

And there you have it – the answer is A.

We easily used deviations here to arrive at the relation. It’s good to have this method – useful for both simple averages and weighted averages – in your GMAT toolkit.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# What to Do When You Find a Weighted Average Question In the Verbal Section of the GMAT

Weighted averages show up everywhere on the GMAT. Most test-takers are prepared to see them on the Quantitative Section, but they’ll show up on the Integrated Reasoning and Verbal Sections, as well.  Because it is such an exam staple, we want to make sure that we have a thorough, intuitive understanding of the concept.

In class, I’ll typically start with a simple example. Say you have two solutions, A and B. A is 10% salt and B is 20% salt. If we combine these two solutions to get a composite solution that is 14% salt, do we have more A or B in this composite solution? Most students eventually see that we’ll have more of solution A, but it doesn’t always feel instinctive. If we had had equal quantities of both solutions, the combined solution would have been 15% salt – equidistant from 10% and 20%. So, if there is 14% salt, the average skews closer to A than B, and thus, there must be more of solution A.

I’ll then give another example. Say that there is an intergalactic party in which both humans and aliens are present. The humans, on average, are 6 feet tall. The aliens, on average, are 100 feet tall. If the average height at the party is 99 feet, who is dominating the party? It isn’t so hard to see that this party is packed with aliens and that the few humans present would likely spend the evening cowering in some distant corner of the room. The upshot is that it’s easier to feel the intuition behind a weighted average question when the numbers are extreme.

Take this tough Critical Reasoning argument for example:

To be considered for inclusion in the Barbizon Film Festival, a film must belong either to the category of drama or of comedy. Dramas always receive more submissions but have a lower acceptance rate than comedy. All of the films are either foreign or domestic. This year, the overall acceptance rate for domestic films was significantly higher than that for foreign films. Within each category, drama and comedy, however, the acceptance rate for domestic films was the same as that for foreign films.

From the cited facts it can be properly concluded that

(A) significantly fewer foreign films than domestic films were accepted.
(B) a higher proportion of the foreign than of the domestic films submitted were submitted as dramas.
(C) the rate of acceptance of foreign films submitted was the same for drama as it was for comedies.
(D) the majority of the domestic films submitted were submitted as comedies.
(E) the majority of the foreign films submitted were submitted as dramas.

Okay. We know that dramas had a lower acceptance rate than comedies, and we know that the overall acceptance rate for domestic films was significantly higher than the acceptance rate for foreign films. So, let’s assign some easy numbers to try and get a handle on this information:

Say that the acceptance rate for dramas was 1% and the acceptance rate for comedies was 99%.

We’ll also say that the acceptance rate for domestic films was 98% and the acceptance rate for foreign films was 2%.

The acceptance rate within both domestic and foreign films is a weighted average of comedies and dramas. If only dramas were submitted, clearly the acceptance rate would be 1%. If only comedies were submitted, the acceptance rate would be 99%. If equal amounts of both were submitted, the acceptance rate would be 50%.

What do our numbers tell us? Well, if the acceptance rate for domestic films was 98%, then almost all of these films must have been comedies, and if the acceptance rate for foreign films was 2%, then nearly all of these films must have been dramas. So, domestic films were weighted towards comedies and foreign films were weighted towards drama. (An unfair stereotype, perhaps, but this is GMAC’s question, not mine.)

We can see that answer choice A is out, as we only have information regarding rates of acceptance, not absolute numbers. C is also out, as it violates a crucial premise of the question stem – we know that the acceptance rate for dramas is lower than for comedies, irrespective of whether we’re talking about foreign or domestic films.

That leaves us with answer choices B, D and E. So now what?

Let’s pick another round of values, but see if we can invalidate two of the three remaining options.

What if the acceptance rate for domestic films was 3% and the acceptance rate for foreign films was still 2%? (We’ll keep the acceptance rate for dramas at 1% and the acceptance rate for comedies at 99%.) Now domestic films would be mostly dramas, so option D is out – the majority of domestic films would not be comedies, as this  answer choice states.

Similarly, what if the acceptance rate for domestic films was 98% and the rate for foreign films was 97%? Now the foreign films would be mostly comedies, so option E is also out – the majority of foreign films would not be dramas, as this answer choice states.

Because the acceptance rate is lower for dramas than it is for comedies, and foreign films have a lower acceptance rate than do domestic films, the foreign films must be weighted more heavily towards dramas than domestic films are. This analysis is perfectly captured in option B, which is, in fact, the correct answer.

Takeaway: certain concepts, such as weighted averages, are such exam staples that will appear in both Quant and Verbal questions. If you see one of these examples in the Verbal Section, assigning extreme values to the information you are given can help you get a handle on the underlying logic being tested.