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# An equal number of desks and bookcases are to be placed

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An equal number of desks and bookcases are to be placed [#permalink]

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17 Nov 2008, 16:25
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An equal number of desks and bookcases are to be placed along a library wall that is 15 meters long. Each desk is 2 meters long, and each bookshelf is 1.5 meters long. If the maximum possible number of desks and bookcases are to be placed along the wall, then the space along the wall that is left over will be how many meters long?

A. 0.5
B. 1
C. 1.5
D. 2
E. 3
[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Jan 2012, 11:46, edited 1 time in total.
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17 Nov 2008, 17:56
Let x be the number of desks and bookcases that are placed along the library wall.

2x + 1.5x < 15
3.5x < 15

Since x is a non negative integer, the largest number x can be is 4.
When x is 4, the desks and bookcases take up 3.5 * 4 = 14m, leaving 1m of empty space.

Thus, I believe the answer is B) 1
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17 May 2011, 14:12
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IEsailor wrote:
Can somebody pls help here. I did not do this algebrically but by actually picking the values.
IF we have 7 Book cases and 2 desks then the maximum no of them are kept.
So,
(1.5 * 7 ) + ( 2 * 2 ) = 10.5 + 4 = 14.5.
So the area left would be 0.5

Now to further maximize then only other option is to take 10 Bookcases which would leave the area as zero.

Although i am quite convinced by the algebric solution given above by sfwon i somehow cannot seem to find the error in my own approach to this question.

Can somebody pls help !!!

You overlooked the restriction that there must be EQUAL number of desks and bookcases.
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17 May 2011, 22:05
each will be 4,an integer number.

hence total = 4*3.5 = 14
1m left.
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19 May 2011, 21:53
haichao wrote:
An equal number of desks and bookcases are to be placed
along a library wall that is 15 meters long. Each desk is
2 meters long, and each bookshelf is 1.5 meters long. If
the maximum possible number of desks and bookcases
are to be placed along the wall, then the space along the
wall that is left over will be how many meters long?

A.0.5
B.1
C.1.5
D.2
E.3

The way I approached this problem is much more simplistic than the answers otherwise listed. We know that the length available to be filled is 15. I then chose to see how many desks/book shelfs can be placed:

1.5Bookshelf = 15
Bookshelf = 10

2 Desks = 15
Desks = 7 (the answer is 7.5, but you can not place half a desk there)

Max bookshelves that can be placed = 15 (assuming no desks) and maximum desks that can be placed = 7 (assuming no bookshelves). Since the value of max bookshelves > max desks, we want to maximize the # of bookshelves by using only 1 desk.

2(1) + 1.5(x)= 15
1.5x = 13
x = 8 (rounded down to the nearest whole number)

So we know that we will have 1 desk and 8 bookshelves. Now you just calculate 2(1) + 1.5(8) = x.
and then 15 - x = answer, which happens to be the value 1, aka B.
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27 Jan 2012, 10:58
trial & error. Each will be 4
4 x 2 = 8
4 x 1.5 =6
8+6 = 14
Left 15-14 = 1
Ans. B
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27 Jan 2012, 11:45
haichao wrote:
An equal number of desks and bookcases are to be placed along a library wall that is 15 meters long. Each desk is 2 meters long, and each bookshelf is 1.5 meters long. If the maximum possible number of desks and bookcases are to be placed along the wall, then the space along the wall that is left over will be how many meters long?

A. 0.5
B. 1
C. 1.5
D. 2
E. 3

Given: $$d=b$$ and $$2d+1.5b\leq{15}$$ --> $$2d+1.5d\leq{15}$$ --> $$3.5d\leq{15}$$ --> $$d=4$$ --> $$3.5d=14$$ --> 15-14=1.

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Re: An equal number of desks and bookcases are to be placed [#permalink]

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15 Oct 2014, 22:19
lets place x desks and x book shelves , then we have 2x + 1.5x metres covered we need 2x+1.5 x <= 15 => 3.5 x<= 15 x<= 30/7 hence max value of x =4 hence length covered = 14 m remaining =1m
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Re: An equal number of desks and bookcases are to be placed [#permalink]

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16 Jan 2016, 07:13
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Re: An equal number of desks and bookcases are to be placed   [#permalink] 16 Jan 2016, 07:13
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