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On a recent test, Kevin scored m percent higher than the class average

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On a recent test, Kevin scored m percent higher than the class average [#permalink]

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New post 25 May 2018, 03:27
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On a recent test, Kevin scored m percent higher than the class average, while Katherine scored n percent higher than the class average. What was the class average?

(1) n – m = 7
(2) m = 12

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Re: On a recent test, Kevin scored m percent higher than the class average [#permalink]

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New post 25 May 2018, 03:49
Bunuel wrote:
On a recent test, Kevin scored m percent higher than the class average, while Katherine scored n percent higher than the class average. What was the class average?

(1) n – m = 7
(2) m = 12


Say class avg =x,
Kavins score= x(1+m) & Katherine score= x(1+n)

(1) n – m = 7 gives the difference between the extra %points of Katherine & Kevin than the class avg... no real value of x... Thus Insufficient
(2) m =12 ....Kevin's score = x*1.12............................no real value of x........................................................... Thus Insufficient
(1)+(2) n=12+7=19, Katherine's Score = 1.19*x & Kevin's score = x*1.12..................Again no real value of x................... Thus Insufficient

Hence I would go for option E.
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Re: On a recent test, Kevin scored m percent higher than the class average [#permalink]

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New post 09 Jun 2018, 05:15
Could someone please tell me if my approach described below is valid?

I assume that I have 5 different variables (m, n, average, Kevin's score, Kathrerine's score) and only two equations.
Statement (1) gives me one more equation - INSUFF
Statement (2) gives me one more equation - INSUFF
Statement (1) & (2), combined, give me two more equation
Then I still have 5 different variables for 4 different equations, hence INSUFF

Does this approach makes sense?
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Re: On a recent test, Kevin scored m percent higher than the class average [#permalink]

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New post 09 Jun 2018, 05:24
PedroBodnar wrote:
Could someone please tell me if my approach described below is valid?

I assume that I have 5 different variables (m, n, average, Kevin's score, Kathrerine's score) and only two equations.
Statement (1) gives me one more equation - INSUFF
Statement (2) gives me one more equation - INSUFF
Statement (1) & (2), combined, give me two more equation
Then I still have 5 different variables for 4 different equations, hence INSUFF

Does this approach makes sense?


Hey PedroBodnar ,

Unfortunately, your approach isn't valid. You don't have any two separate values as Kevin's or Katherine's score. There are nothing but a relation of A and m/n.

We are given the Kevin's Score = A + m % of A (where A is class Average).

Similarly, Katherine's score = A + n % of A

Question is asking you to find out A.

Statement 1: n - m = 7 ==> You can't find out anything here.
Statement 2: m = 12 ==> No information about n or A.

Combining,

m = 12 => n - 12 = 7

or n = 19

Now, you have m and n but you don't have relationship between Katherine's and Kevin's score. Thus, you can't find out A. Hence, E.

Does that make sense?
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Re: On a recent test, Kevin scored m percent higher than the class average [#permalink]

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New post 09 Jun 2018, 05:53
abhimahna wrote:
PedroBodnar wrote:
Could someone please tell me if my approach described below is valid?

I assume that I have 5 different variables (m, n, average, Kevin's score, Kathrerine's score) and only two equations.
Statement (1) gives me one more equation - INSUFF
Statement (2) gives me one more equation - INSUFF
Statement (1) & (2), combined, give me two more equation
Then I still have 5 different variables for 4 different equations, hence INSUFF

Does this approach makes sense?


Hey PedroBodnar ,

Unfortunately, your approach isn't valid. You don't have any two separate values as Kevin's or Katherine's score. There are nothing but a relation of A and m/n.

We are given the Kevin's Score = A + m % of A (where A is class Average).

Similarly, Katherine's score = A + n % of A

Question is asking you to find out A.

Statement 1: n - m = 7 ==> You can't find out anything here.
Statement 2: m = 12 ==> No information about n or A.

Combining,

m = 12 => n - 12 = 7

or n = 19

Now, you have m and n but you don't have relationship between Katherine's and Kevin's score. Thus, you can't find out A. Hence, E.

Does that make sense?


I see.

I have spent some time trying to figure out if this approach will work in every case, and I concluded then it will not.

It has to do with what you mentioned.

If I have 5 variables and only 4 equations, but one equation gives me the relationship between two variables, then I might be able to solve it, like you mentioned in this case.

In other scenario, for instance, if I have to find n and I have the equations:
(1) n+x+y = 1
(2) x+y = 3

Even though I have only two equations for 3 variables, I'm still able to solve it.

Conclusion: In any given case that I have a n variables for m equations, I'll have to find out if:
1 - any of those equations are exactly the same (say, (1) x+y = 5 and (2) 2x+2y=10), or;
2 - any of those equations give me any relationship between variables that will allow me to eliminate two variables from one equation

Does this new approach makes sense?
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Re: On a recent test, Kevin scored m percent higher than the class average [#permalink]

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New post 09 Jun 2018, 07:38
1
PedroBodnar wrote:
I see.

I have spent some time trying to figure out if this approach will work in every case, and I concluded then it will not.

It has to do with what you mentioned.

If I have 5 variables and only 4 equations, but one equation gives me the relationship between two variables, then I might be able to solve it, like you mentioned in this case.

In other scenario, for instance, if I have to find n and I have the equations:
(1) n+x+y = 1
(2) x+y = 3

Even though I have only two equations for 3 variables, I'm still able to solve it.

Conclusion: In any given case that I have a n variables for m equations, I'll have to find out if:
1 - any of those equations are exactly the same (say, (1) x+y = 5 and (2) 2x+2y=10), or;
2 - any of those equations give me any relationship between variables that will allow me to eliminate two variables from one equation

Does this new approach makes sense?


Hey PedroBodnar ,

Yes, this approach works perfectly fine :-)
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Re: On a recent test, Kevin scored m percent higher than the class average   [#permalink] 09 Jun 2018, 07:38
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