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There were 36.000 hardback copies of a certain novel sold

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There were 36.000 hardback copies of a certain novel sold [#permalink] New post 02 May 2007, 20:57
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
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Re: 1000 series - dont agree with the OA ! [#permalink] New post 02 May 2007, 21:23
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 C


Quote:
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
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 [#permalink] New post 02 May 2007, 21:50
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 ?
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 [#permalink] New post 02 May 2007, 22:01
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.
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 [#permalink] New post 02 May 2007, 22:39
oh cool .. thanks sid :)
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 [#permalink] New post 03 May 2007, 00:11
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.
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 [#permalink] New post 03 May 2007, 02:00
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)
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 [#permalink] New post 03 May 2007, 12:17
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?
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thanx [#permalink] New post 03 May 2007, 13:36
thanks uhmno, dat's right, I missed out the same too.

A is OK. Thanx for the example.
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Can anybody explain this? [#permalink] New post 03 May 2007, 13:39
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 three-month period?
(A) $1.08
(B) $1.10
(C) $1.14
(D) $1.16
(E) $1.18

OA is D, I got E
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Re: Can anybody explain this? [#permalink] New post 03 May 2007, 13:53
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 three-month 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) :-D
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 [#permalink] New post 03 May 2007, 16:27
Yeah D is correct . Same method
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thanx [#permalink] New post 03 May 2007, 18:34
Thanks for the explanation. I messed up with the question stem. It was pretty easy one :(
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 [#permalink] New post 03 May 2007, 18:41
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?? :roll:
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 [#permalink] New post 04 May 2007, 01:35
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?? :roll:


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
  [#permalink] 04 May 2007, 01:35
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