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haichao
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An equal number of desks and bookcases are to be placed Updated on: 27 Jan 2012, 11:46
Originally posted by haichao on 17 Nov 2008, 16:25.
Last edited by Bunuel on 27 Jan 2012, 11:46, edited 1 time in total.
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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
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B
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An equal number of desks and bookcases are to be placed 17 Nov 2008, 17:56
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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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An equal number of desks and bookcases are to be placed 17 May 2011, 14:12
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IEsailor
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 !!!
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You overlooked the restriction that there must be EQUAL number of desks and bookcases.
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An equal number of desks and bookcases are to be placed 17 May 2011, 22:05
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each will be 4,an integer number.

hence total = 4*3.5 = 14
1m left.
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An equal number of desks and bookcases are to be placed 19 May 2011, 21:53
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haichao
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
Show more

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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An equal number of desks and bookcases are to be placed 27 Jan 2012, 10:58
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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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An equal number of desks and bookcases are to be placed 27 Jan 2012, 11:45
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haichao
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
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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.

Answer: B.
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An equal number of desks and bookcases are to be placed 15 Oct 2014, 22:19
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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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An equal number of desks and bookcases are to be placed 31 May 2019, 05:41
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Given:

Length of each desk = 2m

Length of each bookshelf = 1.5m

Length of the wall = 15m

The combined length of 1 desk and 1 bookshelf = 2m + 1.5m = 3.5m

Maximum number of desk and bookshelf that can be placed along the wall = 15/3.5 = 4.2

Since the number of desk and bookshelf will be a natural number. Therefore the Maximum number of desk and bookshelf that can be placed along the wall is 4.

The combined length of 4 desks and 4 bookshelves = 3.5*4 = 14m

Leftover space = 15m - 14m = 1m

The correct answer is B
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An equal number of desks and bookcases are to be placed 02 Jun 2019, 17:19
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haichao
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
Show more

Let n = the number of desks and also the number of bookcases to be placed along the wall. Thus we can create the equation:

2n + 1.5n ≤ 15

3.5n ≤ 15

n ≤ 4.29

Since n is a positive integer, the maximum value of n is 4, and thus we have 15 - 3.5(4) = 15 - 14 = 1 meter of space left over along the wall that is not occupied by a desk or bookcase.

Alternate Solution:

Let’s group a bookcase and a desk together to form a 2 + 1.5 = 3.5 meter long desk-bookcase unit. We see that we can place a maximum of 4 such groups on the library wall, which will take up a space of 3.5 x 4 = 14 meters. Therefore, when the maximum number of desks and bookcases are placed on the wall, 15 - 14 = 1 meter of space will be left over.

Answer: B
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An equal number of desks and bookcases are to be placed 23 Apr 2022, 11:10
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