Problem Solving

A class is given a 20-question exam in which each question is equally weighted. Of the students in the class, 20% answer zero questions incorrectly, 40% answer (x - 1) questions incorrectly, and 40% answer 3x questions incorrectly. What is the value of x if the average (arithmetic mean) of the students' scores on the exam is 70%?


(A) 2

(B) 3

(C) 4

(D) 5

(E) 6

easy to plug choices. 20 students score 20/20 so total correct score = 400 (20*20)
To make all class avg = 70% correct answers, avg no of correct answers out of 20 = 14 (70% of 20)
and total correct answers = 100*14 = 1400 (assuming 100 students in class)

remaining 80% of class population (40% each) mark (x-1) and 3x incorrect answers and they together should score 1400-400 = 1000 correct answers score.

Let those who answer (x-1) Qs incorrect have A Correct Answers
Let those who answer 3x Qs incorrect have B Correct Answers
And 40*A + 40*B = 1000 so A+B = 25

plugging answer choices tells us only C accomplishes the A+B=25
Ans C

I didnt understand ur concept, can you bit elaborate (one of weak area is to interpret weighted average problems)

srcc25anu wrote:easy to plug choices. 20 students score 20/20 so total correct score = 400 (20*20)
To make all class avg = 70% correct answers, avg no of correct answers out of 20 = 14 (70% of 20)
and total correct answers = 100*14 = 1400 (assuming 100 students in class)

remaining 80% of class population (40% each) mark (x-1) and (x-3) incorrect answers and they together should score 1400-400 = 1000 correct answers score.

Let those who answer (x-1) Qs incorrect - A
Let those who answer (x-1) Qs incorrect - B
And 40*A + 40*B = 1000 so A+B = 25

plugging answer choices tells us only C accomplishes the A+B=25
Ans C

karthikpandian19 wrote:A class is given a 20-question exam in which each question is equally weighted. Of the students in the class, 20% answer zero questions incorrectly, 40% answer (x - 1) questions incorrectly, and 40% answer 3x questions incorrectly. What is the value of x if the average (arithmetic mean) of the students' scores on the exam is 70%?


(A) 2

(B) 3

(C) 4

(D) 5

(E) 6

Let the total number of students = 10.
Since each student could earn 20 points, the total number of possible points = 10*20 = 200.
For the average to be 70%, the total number of points earned = .7(200) = 140.
Thus, the total number of MISSED POINTS = 200-140 = 60.
We can plug in the answers, which represent the value of x.

Answer choice C: 4
40% answer (x-1) questions incorrectly.
Thus, 4 students each answer 4-1=3 questions incorrectly.
Number of points missed by these 4 students = 4*3 = 12.
40% answer 3x questions incorrectly.
Thus, 4 students each answer 3*4=12 questions incorrectly.
Number of points missed by these 4 students = 4*12 = 48.
Total missed = 12+48 = 60.
Success!

The correct answer is C.

karthikpandian19 wrote: one of weak area is to interpret weighted average problems

First off: percentage of a thing in a mixture or weighted problems all boil down to one basic equation.

If we have 5 things A, B, C, D , E etc. And each contains a, b, c, d, e percent of some thing X, then the overall amount of X in the mixture is:
(a*A+b*B+c*C+d*D+e*E). Oh, I wrote complicated algebra. Let's look at more recognizable data.

Here's an example.
If the following thing is mixed together, let's find the overall amount and percentage.
1) Scotch bottle A contains 10% water. The volume of Scotch in the bottle is 1 liter.
2) Scotch bottle B contains 20% water. The volume of Scotch in the bottle is 2 liters.
3) Scotch bottle C contains 30% water. The volume of Scotch in the bottle is 3 liters.
4) Scotch bottle D contains 40% water. The volume of Scotch in the bottle is 4 liters.
5) Scotch bottle e contains 50% water. The volume of Scotch in the bottle is 5 liters.

If all the bottles are mixed together in a big vat, what is:
A) total volume of water in the mixture.
B) percentage of water in the mixture.

If you understand the fundamental behind this problem, you should be able to formulate any word problem with weighted averages or mixtures.

Let's find out.
A) Total volume of water in the mixture is 10%*1 + 20%*2 + 30%*3 + 40%*4 + 50%*5 = 5.5 liters.
B) The percentage of water in the mixture = volume of water * 100 / (total volume of scotch) =
= 5.5*100/15 = 36.7% etc.

Key take-away from this straight forward example is:
a) Identify which one is the percentage and which one is the amount.
b) Set up the equation for the total amount.

You can then divide the amount by the total volume (and multiply by 100) to change the equation to find the overall percentage.

With this in mind, let's look at the problem at hand.

karthikpandian19 wrote:A class is given a 20-question exam in which each question is equally weighted. Of the students in the class, 20% answer zero questions incorrectly, 40% answer (x - 1) questions incorrectly, and 40% answer 3x questions incorrectly. What is the value of x if the average (arithmetic mean) of the students' scores on the exam is 70%?

This is a disguised problem. They are telling you that overall 70% of questions were answered correctly. But the data is given for incorrect answers. We should mold the data to a common thing. Since more stuff is written about incorrect numbers, I have chosen to think about the incorrect number of questions.

So we need to note that if 70% of questions were answered correctly, 100-70 = 30% of questions were answered incorrectly. Once you know this, the problem is anything, but, over. We know that the percentages are 20%, 40%, and 40% for the quantities of 0, x-1, and 3x (Same as our scotch problem and pretty much any problem out there).

Let's set up the equation then. We have overall percent of incorrect percentage as 30% and overall questions are 20. So our equation becomes:

20*0 + 40*(x-1) + 40*(3x) = 30*20.
=> 40x-40+120x = 600
=> 160x = 640 => x=4.

Let's rehash the key takeaways in this problem:
a. Converting data to a usable/consistent form
b. Identifying which one is the percentage (or the fraction) and which is the quantity
c. Setting up that equation.

Of course, for some of the problems, there are other methods, such as balancing (for two variable ones) and allegation etc. and they should be used when warranted, but this is the crux of handling these problems.

If you still have questions, let me know.

Let me know if this helps

That's one real easy and quick way to understand the concept. Thanks.

Eventhough i was bit confused when you multiply 30 by 20 on the Right hand side of eqn

"20*0 + 40*(x-1) + 40*(3x) = 30*20. "

eagleeye wrote:

karthikpandian19 wrote: one of weak area is to interpret weighted average problems

First off: percentage of a thing in a mixture or weighted problems all boil down to one basic equation.

If we have 5 things A, B, C, D , E etc. And each contains a, b, c, d, e percent of some thing X, then the overall amount of X in the mixture is:
(a*A+b*B+c*C+d*D+e*E). Oh, I wrote complicated algebra. Let's look at more recognizable data.

Here's an example.
If the following thing is mixed together, let's find the overall amount and percentage.
1) Scotch bottle A contains 10% water. The volume of Scotch in the bottle is 1 liter.
2) Scotch bottle B contains 20% water. The volume of Scotch in the bottle is 2 liters.
3) Scotch bottle C contains 30% water. The volume of Scotch in the bottle is 3 liters.
4) Scotch bottle D contains 40% water. The volume of Scotch in the bottle is 4 liters.
5) Scotch bottle e contains 50% water. The volume of Scotch in the bottle is 5 liters.

If all the bottles are mixed together in a big vat, what is:
A) total volume of water in the mixture.
B) percentage of water in the mixture.

If you understand the fundamental behind this problem, you should be able to formulate any word problem with weighted averages or mixtures.

Let's find out.
A) Total volume of water in the mixture is 10%*1 + 20%*2 + 30%*3 + 40%*4 + 50%*5 = 5.5 liters.
B) The percentage of water in the mixture = volume of water * 100 / (total volume of scotch) =
= 5.5*100/15 = 36.7% etc.

Key take-away from this straight forward example is:
a) Identify which one is the percentage and which one is the amount.
b) Set up the equation for the total amount.

You can then divide the amount by the total volume (and multiply by 100) to change the equation to find the overall percentage.

With this in mind, let's look at the problem at hand.

karthikpandian19 wrote:A class is given a 20-question exam in which each question is equally weighted. Of the students in the class, 20% answer zero questions incorrectly, 40% answer (x - 1) questions incorrectly, and 40% answer 3x questions incorrectly. What is the value of x if the average (arithmetic mean) of the students' scores on the exam is 70%?

This is a disguised problem. They are telling you that overall 70% of questions were answered correctly. But the data is given for incorrect answers. We should mold the data to a common thing. Since more stuff is written about incorrect numbers, I have chosen to think about the incorrect number of questions.

So we need to note that if 70% of questions were answered correctly, 100-70 = 30% of questions were answered incorrectly. Once you know this, the problem is anything, but, over. We know that the percentages are 20%, 40%, and 40% for the quantities of 0, x-1, and 3x (Same as our scotch problem and pretty much any problem out there).

Let's set up the equation then. We have overall percent of incorrect percentage as 30% and overall questions are 20. So our equation becomes:

20*0 + 40*(x-1) + 40*(3x) = 30*20.
=> 40x-40+120x = 600
=> 160x = 640 => x=4.

Let's rehash the key takeaways in this problem:
a. Converting data to a usable/consistent form
b. Identifying which one is the percentage (or the fraction) and which is the quantity
c. Setting up that equation.

Of course, for some of the problems, there are other methods, such as balancing (for two variable ones) and allegation etc. and they should be used when warranted, but this is the crux of handling these problems.

If you still have questions, let me know.

Let me know if this helps

karthikpandian19 wrote:That's one real easy and quick way to understand the concept. Thanks.

Eventhough i was bit confused when you multiply 30 by 20 on the Right hand side of eqn

"20*0 + 40*(x-1) + 40*(3x) = 30*20. "

Karthik:

Could you please detail a little more about what exactly is confusing about the RHS? I will try to explain it better once I find what is troubling you.

ee!

Quantity * Percentage is the rule that you had mentioned.....

40% of (x-1)
40% of 3x
and hence, 30% of 20....(but 70% is correct and 30% is wrong, so it should be 30% * 6)???????

(i might have deeply messed somewhere here, explain pls)

eagleeye wrote:

karthikpandian19 wrote:That's one real easy and quick way to understand the concept. Thanks.

Eventhough i was bit confused when you multiply 30 by 20 on the Right hand side of eqn

"20*0 + 40*(x-1) + 40*(3x) = 30*20. "

Karthik:

Could you please detail a little more about what exactly is confusing about the RHS? I will try to explain it better once I find what is troubling you.

ee!

karthikpandian19 wrote:Quantity * Percentage is the rule that you had mentioned.....

40% of (x-1)
40% of 3x
and hence, 30% of 20....(but 70% is correct and 30% is wrong, so it should be 30% * 6)???????

(i might have deeply messed somewhere here, explain pls)

Oh, I see what you are doing. You are doing the thing twice. Let me give you an example which might explain it a little better.

If you have 50% salt in a sand and salt combo of 4 kg and you mix it with 50% mix of salt and sand combo of 2kg, what do you expect to get? You expect 6 kg mixture with 50% salt in it, right?.

So the way we would equate it in this case is :

50*4 + 50*2 = 50*6
What you are doing is:
50*4 + 50*2 = 50* (50% of 6).

So just to rehash the earlier problem.

We take the individual percentages and multiply with their individual quantities. What we get overall is the overall percentage and overall quantity. So the equation is:

(%age1)*(quantity1) + (%age2*quantity2) + (%age3*quantity3) = (%ageTOTAL)*(quantityTOTAL).

Let me know if this helps

Silly mistake....but i understood now ............thanks

eagleeye wrote:

karthikpandian19 wrote:Quantity * Percentage is the rule that you had mentioned.....

40% of (x-1)
40% of 3x
and hence, 30% of 20....(but 70% is correct and 30% is wrong, so it should be 30% * 6)???????

(i might have deeply messed somewhere here, explain pls)

Oh, I see what you are doing. You are doing the thing twice. Let me give you an example which might explain it a little better.

If you have 50% salt in a sand and salt combo of 4 kg and you mix it with 50% mix of salt and sand combo of 2kg, what do you expect to get? You expect 6 kg mixture with 50% salt in it, right?.

So the way we would equate it in this case is :

50*4 + 50*2 = 50*6
What you are doing is:
50*4 + 50*2 = 50* (50% of 6).

So just to rehash the earlier problem.

We take the individual percentages and multiply with their individual quantities. What we get overall is the overall percentage and overall quantity. So the equation is:

(%age1)*(quantity1) + (%age2*quantity2) + (%age3*quantity3) = (%ageTOTAL)*(quantityTOTAL).

Let me know if this helps