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Director
Joined: 13 Mar 2007
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Schools: MIT Sloan

There were 36.000 hardback copies of a certain novel sold [#permalink]
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02 May 2007, 21:57
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There were 36.000 hardback copies of a certain novel sold before the paperback version was issued. From the time the first paperback copy was sold until the last copy of the novel was sold. 9 times as many paperback copies as hardback copies were sold. If a total of 441.000 copies of the novel were sold in all, how many paperback copies were sold?
(A) 45.000
(B) 360.000
(C) 364.500
(D) 392.000
(E) 396.900
If y ≠ 3 and (3x/y) is a prime integer greater than 2, which of the following must be true?
Ⅰ. x = y
Ⅱ. y = 1
Ⅲ. x and y are prime integers.
(A) None
(B) Ⅰ only
(C) Ⅱonly
(D) Ⅲonly
(E) Ⅰand Ⅲ



Senior Manager
Joined: 19 Feb 2007
Posts: 325

Re: 1000 series  dont agree with the OA ! [#permalink]
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02 May 2007, 22:23
Quote: There were 36.000 hardback copies of a certain novel sold before the paperback version was issued. From the time the first paperback copy was sold until the last copy of the novel was sold. 9 times as many paperback copies as hardback copies were sold. If a total of 441.000 copies of the novel were sold in all, how many paperback copies were sold?
(A) 45.000 (B) 360.000 (C) 364.500 (D) 392.000 (E) 396.900 Total copies sold = 36000 + 9x + x = 441000
=> x = 40500
So no of Paperback copies = 9x = 364500
Ans CQuote: If y ≠ 3 and (3x/y) is a prime integer greater than 2, which of the following must be true? Ⅰ. x = y Ⅱ. y = 1 Ⅲ. x and y are prime integers.
(A) None (B) Ⅰ only (C) Ⅱonly (D) Ⅲonly (E) Ⅰand Ⅲ
Ans B



Director
Joined: 13 Mar 2007
Posts: 543
Schools: MIT Sloan

Yes, the answer for (2) must be B.
However, the reference key gives A !
For the 1st problem, I followed the below approach,
Hard copies  36000
# Hard copies sold after paper back was introduced  x
hence total# hard copies sold = (36000 + x)
hence total paperbacks = 9(36000 + x)
=> 10(36000 + x) = 441000 => paper copies = 396900 ! E
What is it I am missing here ?



Senior Manager
Joined: 19 Feb 2007
Posts: 325

grad_mba wrote: Yes, the answer for (2) must be B. However, the reference key gives A !
For the 1st problem, I followed the below approach,
Hard copies  36000
# Hard copies sold after paper back was introduced  x
hence total# hard copies sold = (36000 + x)
hence total paperbacks = 9(36000 + x)
=> 10(36000 + x) = 441000 => paper copies = 396900 ! E
What is it I am missing here ?
From the time the first paperback copy was sold until the last copy of the novel was sold. 9 times as many paperback copies as hardback copies were sold.
The statement says from the time the first paperback copy was sold............this implies papaerback copies are 9 times the hardcopies sold after the first paperback cpy was introduced.........i.e 9x and not 9 (36000 + x)...................now u can figure out the solution.



Director
Joined: 13 Mar 2007
Posts: 543
Schools: MIT Sloan

oh cool .. thanks sid



Manager
Joined: 04 Oct 2006
Posts: 76

hardback = h
paperback = p
36 000 + h + p = 441 000
p= 9h <=> h = p/9
36 000 + p/9 + p = 441 000
p = 405 000 * 9/10
Answer C.



VP
Joined: 08 Jun 2005
Posts: 1145

My way:
441,000 = all books
441,000  36,000 = 405,000 (we don't care about hardback)
ratio p9:h1 = 9+1 = 10
405,000/10 = 40,500
we need only paperback
40,500*9 = 364,500
the answer is (C)



Intern
Joined: 25 Apr 2007
Posts: 3

i got A (none) for the 2nd problem..
If y ≠ 3 and (3x/y) is a prime integer greater than 2, which of the following must be true?
x = y
y = 1
x and y are prime integers.
looking at the question i made the example, 3 * 14 / 6 = 7, so x=14 y=6 and it equals a prime integer (7)
in this 1 example, x doesn't equal y, y doesn't equal 1, and neither x nor y is prime.. so none must be true
what are you guys thinking that i'm missing?



Director
Joined: 25 Oct 2006
Posts: 635

thanks uhmno, dat's right, I missed out the same too.
A is OK. Thanx for the example.



Director
Joined: 25 Oct 2006
Posts: 635

Can anybody explain this? [#permalink]
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03 May 2007, 14:39
Month Average Price per Dozen
April $1.26
May $1.20
June $1.08
The table above shows the average (arithmetic mean) price per dozen of the large grade A eggs sold in a certain store during three successive months. If as many dozen were sold in April as in May, and twice as many were sold in June as in April, what was the average price per dozen of the eggs sold over the threemonth period?
(A) $1.08
(B) $1.10
(C) $1.14
(D) $1.16
(E) $1.18
OA is D, I got E



VP
Joined: 08 Jun 2005
Posts: 1145

Re: Can anybody explain this? [#permalink]
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03 May 2007, 14:53
priyankur_saha@ml.com wrote: Month Average Price per Dozen April $1.26 May $1.20 June $1.08 The table above shows the average (arithmetic mean) price per dozen of the large grade A eggs sold in a certain store during three successive months. If as many dozen were sold in April as in May, and twice as many were sold in June as in April, what was the average price per dozen of the eggs sold over the threemonth period? (A) $1.08 (B) $1.10 (C) $1.14 (D) $1.16 (E) $1.18
OA is D, I got E
April  number of dozen eggs sold  x  average 1.26$
May  number of dozen eggs sold  x  average 1.20$
June  number of dozen eggs sold  2x  average 1.08$
average for three months:
(1.26x+1.2x+1.08*2x)/4x = (total price for dozen eggs/number of dozen eggs sold)
4.62x/4x = 4.62/4 = 1.155 ~ 16
so the answer is (D)



Intern
Joined: 01 Feb 2007
Posts: 17

Yeah D is correct . Same method



Director
Joined: 25 Oct 2006
Posts: 635

Thanks for the explanation. I messed up with the question stem. It was pretty easy one



Director
Joined: 25 Oct 2006
Posts: 635

I need explanation for this too...
At a certain diner, a hamburger and coleslaw cost $3.59, and a hamburger and french fries cost $4.40. If french fries cost twice as much as coleslaw, how much do french fries cost?
(A) $0.30
(B) $0.45
(C) $0.60
(D) $0.75
(E) $0.90
Suppose H + C = 3.59; H + F = 4.40
So F  C = 4.40  3.59 = 0.81
Again, F = 2C.
Then, 2C  C = 0.81 & C = 0.81, and F = 2C = 1.62
Is that right?? But not in answer choice??



VP
Joined: 08 Jun 2005
Posts: 1145

priyankur_saha@ml.com wrote: I need explanation for this too... At a certain diner, a hamburger and coleslaw cost $3.59, and a hamburger and french fries cost $4.40. If french fries cost twice as much as coleslaw, how much do french fries cost? (A) $0.30 (B) $0.45 (C) $0.60 (D) $0.75 (E) $0.90 Suppose H + C = 3.59; H + F = 4.40 So F  C = 4.40  3.59 = 0.81 Again, F = 2C. Then, 2C  C = 0.81 & C = 0.81, and F = 2C = 1.62 Is that right?? But not in answer choice??
hi priyankur_saha@ml.com  can you please post new topics as new subject ? so every question will be separate from the others.
you will get more viewers that way.
Thanks










