An equal number of desks and bookcases are to be placed

each will be 4,an integer number.

hence total = 4*3.5 = 14
1m left.

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.

trial & error. Each will be 4
4 x 2 = 8
4 x 1.5 =6
8+6 = 14
Left 15-14 = 1
Ans. B

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.

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

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

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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