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Re: Tough airthmetic [#permalink]
10 Dec 2012, 23:40

VeritasPrepKarishma wrote:

gettinit wrote:

Karishma can you please explain your method a bit more? I don't understand how you solved the problem using it? thanks

What is weighted average? It is average when each value has a different weight. e.g. a group of friends has 10 boys and 20 girls. Average age of boys is 20 years and average age of girls is 17 years. What is the average age of the group?

Here, the average is weighted since we have different number of boys and girls. We calculate it as follows: \(W Avg = \frac{20*10 + 17* 20}{10 + 20}\)

What we are doing instinctively here is using weighted average formula which as given below: \(C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}\)

You need to find the average of C and W is the weight. In the example above, C is age and W is number of boys and girls.

The alligation method, or the scale method as we call it, is based on the weighted averages formula itself:

\(C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}\) If I re-arrange the formula, I get \(\frac{W_1}{W_2} = \frac{C_2 - C_{avg}}{C_{avg} - C_1}\) So I get that weights will be in the same ratio as difference between higher value of C and average value of C and difference between average value of C and lower value of C.

How does this help? Knowing this, we can directly make a diagram and get the answer. e.g. A group of friends has 10 boys and some girls. Average age of boys is 20 years and average age of girls is 17 years. The average age of the group is 18 years. How many girls are there? Draw:

Attachment:

Ques1.jpg

On a scale (number line), mark 17 years as age of girls, 18 years as average and 20 years as age of boys. Now, distance between 17 and 18 is 1 and distance between 18 and 20 is 2, The ratio of W1/W2 will be 2:1 (Note, the numbers 1 and 2, give a ratio of 2:1 for girls:boys as seen by the formula) Since there are 10 boys, there will be 20 girls.

This method is especially useful when you have the average and need to find the ratio of weights.

[/quote]

Karishma, this was very helpful and your way of doing the scale method makes sense because MGMAT doesn't explain it clearly in the foundation book. My only question is your equation

\(C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}\) If I re-arrange the formula, I get \(\frac{W_1}{W_2} = \frac{C_2 - C_{avg}}{C_{avg} - C_1}\)

How did you re-arrange the formula to get that? I don't see it. Thanks in advance!

Re: Tough airthmetic [#permalink]
11 Dec 2012, 23:39

Expert's post

dcastan2 wrote:

Karishma, this was very helpful and your way of doing the scale method makes sense because MGMAT doesn't explain it clearly in the foundation book. My only question is your equation

\(C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}\) If I re-arrange the formula, I get \(\frac{W_1}{W_2} = \frac{C_2 - C_{avg}}{C_{avg} - C_1}\)

How did you re-arrange the formula to get that? I don't see it. Thanks in advance!

Re: Tough airthmetic [#permalink]
27 Aug 2013, 14:52

VeritasPrepKarishma wrote:

anish319 wrote:

karishma

how did you derive this: Since the weights are in the ratio 3:2, the distance on the scale will be in the ratio 2:3. So x = 82.5%

I followed the whole weighted average discussion. what is C2 for the above question

Weights are the number of boys and number of girls. Here we need to find the number of girls so we do not have the ratio of weights. We need to find it. What we do have is C1, CAvg and C2. As shown in the diagram above, ratio of distance between C2 and CAvg and distance between CAvg and C1 gives the ratio of weights.

C1 = 17 yrs, CAvg = 18 yrs and C2 = 20 yrs Hence distance on the scale between 17 and 18 is 1 and distance on the scale between 18 and 20 is 2. This gives us a ratio of 2:1 for the weights i.e. for the number of girls:number of boys.

Remember C is what you want to find the average of. Here average age is 18 yrs. So age is C. Weights is the number of boys/girls.

Is there a document which can explain the Mixture concept in detail. I seem to be having issues with it.

Re: The final exam of a particular class makes up 40% of the [#permalink]
24 Aug 2014, 13:17

VeritasPrepKarishma wrote:

anish319 wrote:

VeritasPrepKarishma wrote:

Since the weights are in the ratio 3:2, the distance on the scale will be in the ratio 2:3. So x = 82.5%

If you have time, it would be a good idea to be comfortable with the scale method. It will save you a lot of time and energy.

I was actually about the above solution....sorry for the confusion. how did you get "Since the weights are in the ratio 3:2, the distance on the scale will be in the ratio 2:3. So x = 82.5%"

Oh ok... I thought I was missing something! In the question above, you have the average of marks. So your C is marks. Here, the difference is that weights are given to you and one of the C (i.e. C2) is missing. We know that ratio of weights will be the distance on the number line.

So if you look at the diagram above, the ratio of weights is 3:2 (because weightage of mid terms is 60% and weightage of finals is 40%, so 60:40 = 3:2). This means that the distance on the number line should be in the ratio 2:3 (The ratio on the number line flips). So distance between 45 and 60 is 2 units. This means 1 unit is 15/2 = 7.5 on the number line. Now we need to find what 3 units distance is because C2 (i.e. x in the diagram) will be 3 units away from 60. Since 1 unit is 7.5, 3 units will be 22.5. Adding 22.5 to 60, we get 82.5. So x, the missing extreme right value must be 82.5

Hi karishma,

I thought that I understood your line method until I read this statement. Whenever we look at the line method (You usually have an average weight and the two ends(weights) of the data given). So in the example with ages, there are 17yr old girls, 20yr old boys and the group is 18, you have a clear number that's above the average and below the average -- which makes complete sense.

Two questions: 1) Line method: If I were to use the line method in this example -- we are already TOLD that the middle weight is 40. In addition to that, i'm a little confused because the "middle weight" is actually not in the middle? I'm not sure if I'm explaining this properly but if we are looking to score a 60% (that should be the average of the two ends, correct?) -- we know that we have scored 40% but we don't have the other end (a number higher than 60). How do we find a ratio without knowing that number, which incidentally, is the same number that the question is asking for.

We know that the weight of the final is 40% which means that the weight of everything but the exam is 100-40 = 60%. So we have w1=40 and w2 = 60

Additionally, we know that we are looking to score a 60% in the class, which is different than the 60% I outlined above. So we have that the Average. Avg = 60.

We also know that currently, he has scored 45%, which is on the non final part, therefore, we are looking to find what he needs to score on the Final. Correct? Meaning, we need to find A1 in the equation above. Correct?

This seems to get me to 82.5 but it was extremely difficult for me to draw the gap between what weights correspond to what grades. I've read your "quarter wit" documents but it still eludes me a little. Any help would be appreciated.

Re: The final exam of a particular class makes up 40% of the [#permalink]
27 May 2015, 12:24

VeritasPrepKarishma wrote:

russ9 wrote:

I thought that I understood your line method until I read this statement. Whenever we look at the line method (You usually have an average weight and the two ends(weights) of the data given). So in the example with ages, there are 17yr old girls, 20yr old boys and the group is 18, you have a clear number that's above the average and below the average -- which makes complete sense.

Two questions: 1) Line method: If I were to use the line method in this example -- we are already TOLD that the middle weight is 40. In addition to that, i'm a little confused because the "middle weight" is actually not in the middle? I'm not sure if I'm explaining this properly but if we are looking to score a 60% (that should be the average of the two ends, correct?) -- we know that we have scored 40% but we don't have the other end (a number higher than 60). How do we find a ratio without knowing that number, which incidentally, is the same number that the question is asking for.

We know that the weight of the final is 40% which means that the weight of everything but the exam is 100-40 = 60%. So we have w1=40 and w2 = 60

Additionally, we know that we are looking to score a 60% in the class, which is different than the 60% I outlined above. So we have that the Average. Avg = 60.

We also know that currently, he has scored 45%, which is on the non final part, therefore, we are looking to find what he needs to score on the Final. Correct? Meaning, we need to find A1 in the equation above. Correct?

This seems to get me to 82.5 but it was extremely difficult for me to draw the gap between what weights correspond to what grades. I've read your "quarter wit" documents but it still eludes me a little. Any help would be appreciated.

Thanks.

Scale method:

w1/w2 = (C2 - Cavg)/(Cavg - C1)

Cavg is the thing we need to average - marks here w1 and w2 are the weights allotted to marks - 40% to finals and 60% to rest of the tests

Question: The final exam of a particular class makes up 40% of the final grade, and Moe is failing the class with an average (arithmetic mean) of 45% just before taking the final exam. What grade does Moe need on his final exam in order to receive the passing grade average of 60% for the class?

"The final exam of a particular class makes up 40% of the final grade" - this tells us that weight allotted to final exam is 40% and hence 60% is allotted to exams before finals. So we have w1 and w2. This is the tricky part.

"Moe is failing the class with an average (arithmetic mean) of 45%" - This means in exams before finals, Moe has 45% marks.

"What grade does Moe need on his final exam in order to receive the passing grade average of 60% for the class" - To pass, Moe needs average 60% marks i.e. Cavg should be 60%. So what we need is his marks in finals i.e. C2

This is still quite unclear to me. I too understand the boys and girls problem just fine, but like the previous poster metioned, I don't understand how we get that upper limit. Any chance you can show it on a number line too?

Re: The final exam of a particular class makes up 40% of the [#permalink]
27 May 2015, 22:11

Expert's post

The question can be interpreted in an another way

Total marks needed to pass We are given that out of total marks(let's assume it to be \(y\)) Moe should receive 60% to pass i.e. 60% of \(y\) = \(0.6y\)

Final exam marks Since the final exam makes up for 40% of the total marks, it constitutes 40% of \(y\) \(= 0.4y\) marks. Let's assume he should get \(x\)% of marks in his final exam to pass the exam i.e. he should get \(x\)% of \(0.4y\)

Rest of the marks The rest 60% of total marks constitute of 60% of \(y\) \(= 0.6y\) marks

Out of these \(0.6y\) marks, Moe has got only 45% i.e. 45% of \(0.6y = 0.45 * 0.6y\)

Writing the equation We can write the equation for marks of Moe as

Total marks needed by Moe to pass = Rest of the marks + Final exam marks

\(0.6y = 0.45 * 0.6y +\) \(x\)% \(* 0.4y\)

Solving this would give us \(x = 82.5\)% of marks he needs in his final exam to pass the class

Re: The final exam of a particular class makes up 40% of the [#permalink]
28 May 2015, 07:46

EgmatQuantExpert wrote:

The question can be interpreted in an another way

Total marks needed to pass We are given that out of total marks(let's assume it to be \(y\)) Moe should receive 60% to pass i.e. 60% of \(y\) = \(0.6y\)

Final exam marks Since the final exam makes up for 40% of the total marks, it constitutes 40% of \(y\) \(= 0.4y\) marks. Let's assume he should get \(x\)% of marks in his final exam to pass the exam i.e. he should get \(x\)% of \(0.4y\)

Rest of the marks The rest 60% of total marks constitute of 60% of \(y\) \(= 0.6y\) marks

Out of these \(0.6y\) marks, Moe has got only 45% i.e. 45% of \(0.6y = 0.45 * 0.6y\)

Writing the equation We can write the equation for marks of Moe as

Total marks needed by Moe to pass = Rest of the marks + Final exam marks

\(0.6y = 0.45 * 0.6y +\) \(x\)% \(* 0.4y\)

Solving this would give us \(x = 82.5\)% of marks he needs in his final exam to pass the class

Hope this helps

Regards, Harsh

Quite simple way to interpret & arrive at the final solution.

gmatclubot

Re: The final exam of a particular class makes up 40% of the
[#permalink]
28 May 2015, 07:46

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