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A candidate who gets 20% marks fails by 10 marks but another

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A candidate who gets 20% marks fails by 10 marks but another [#permalink] New post 08 Jul 2005, 12:07
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A candidate who gets 20% marks fails by 10 marks but another candidate who gets 42% marks gets 12% more than the passing marks. Find the maximum marks. :oops: :oops:

1. 50%
2. 100%
3. 150%
4. 200%
5. 250%
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 [#permalink] New post 08 Jul 2005, 12:53
Indeed a wierd Q...

Reason why it so is - the question asks the 'total marks' but the answer choices are expressed as %ages. That sounds wierd to me... pls. correct me if I am not getting this right...

In any case: 2 equations will be formed for this.

Assuming 't' is the total marks & 'p' being the passing marks.

(1) 0.20t = p -10
(2) 0.42t = 1.12p

Solving for t yields, t = 560/9.8 , which is approx. 57 or 58. Also, p = approx. 21.

So, the question would make more sense if it were asking "what percent of the total marks is the passing marks" - which would be 21/57*100 = 37%. Since that is nowhere near any of the choices ... I resign...:)
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 [#permalink] New post 08 Jul 2005, 13:36
himjhamb wrote:
Indeed a wierd Q...

Reason why it so is - the question asks the 'total marks' but the answer choices are expressed as %ages. That sounds wierd to me... pls. correct me if I am not getting this right...

(1) 0.20t = p -10
(2) 0.42t = 1.12p



Well, 100% actually will mean 100 marks, i believe. I used your approach himjhamb, and i ended up with about 57%. However, when you backsolve (like i would have done if it were matchday), you get 100% (B) as the answer. However, i am only wondering what is the problem with our approach and why we are not getting 100%.
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 [#permalink] New post 08 Jul 2005, 17:34
However, when you backsolve (like i would have done if it were matchday), you get 100% (B) as the answer. However, i am only wondering what is the problem with our approach and why we are not getting 100%.

The reason why you are getting 100% using backsolving is due to the interpretation of "but another candidate who gets 42% marks gets 12% more than the passing marks"

In my equation # 2, I have taken the above underlined part to mean 12% of passing marks & you have assumed it to be 12% of the total marks.

From my answer, passing marks = approx. 21. So, 12% more of 21 means, passing marks = 23.4 approx.... which is 42% of 57.

Hence the difference.

Thanks
- Himanshu
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 [#permalink] New post 08 Jul 2005, 20:36
Let the total marks be X

Now

(20/100) *X + 10 = (42/100)*X - 12/100*[(20/100)*X + 10] should be the equation....

Solve it and I got it as close to 50% ...SO I will pick A
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 [#permalink] New post 09 Jul 2005, 07:25
Hi,
The equation you came up with, yields the value of x=56, which is the same as what I got for the variable 't' in my equation.

However, thats the total marks - my issue with the question was its asking for total marks but giving %ages in the choices.

Would you agree that the question's wording is a bit obscure?

Thanks
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 [#permalink] New post 09 Jul 2005, 07:49
We are getting 100 as max marks...which is 100%

Consider my equation

(20/100) *X + 10 = (42/100)*X - 12/100*[(20/100)*X + 10]

Now on the left hadn side we have 10 marks which indicates 10% of the total marks

This is because...the difference between the 1st and the second guys score are 22%(42 -20)...We can do this because max marks is same for both the guys

Now the second guy has 12% more than cutoff

22/100 = 12/100 + y

y = 10% = 10 marks(which the first guy fell short off)

So 1 mark is 1% which menas 100 marks is 100%


Answer is B
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 [#permalink] New post 09 Jul 2005, 12:13
its B). the 2nd candidate with 42 % failed by 12 %. so the passing is at 30 %. the difference between the 1st candidate with 20 % to the passing mark is 10%. so its 10%*x=10. x=100.
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 [#permalink] New post 10 Jul 2005, 07:17
The reason why you are getting 100% using backsolving is due to the interpretation of "but another candidate who gets 42% marks gets 12% more than the passing marks"

In my equation # 2, I have taken the above underlined part to mean 12% of passing marks & you have assumed it to be 12% of the total marks.


Thanks Christoff... it makes sense now - I think my assumption of "12% of passing marks" was incorrect... its actually, "12% of total marks".

So, my 2nd equation becomes

0.42t = p+0.12t

Solving the two equations yields, t=100.
  [#permalink] 10 Jul 2005, 07:17
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