SOLUTION: A wall is to be constructed 31 ft. in length. 4 different lengths of stone to be used; 2.2 ft., 2.5 ft., 3 ft., and 4.1 ft. In laying a course, determine number of each. Not su

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Question 1177510: A wall is to be constructed 31 ft. in length. 4 different lengths of stone to be used; 2.2 ft., 2.5 ft., 3 ft., and 4.1 ft. In laying a course, determine number of each.
Not sure how to solve.

Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

A wall: L=31ft
if the length of stone is x, then 4 different lengths of stone to be used;
2.2x ft.
2.5x ft.
3x ft.
4.1x ft
and their sum must be 31

2.2x%2B2.5x%2B3x%2B4.1x=31
11.8x=31
x=31%2F11.8
x=2.62712+
so, the lengths of stone are:
2.2x=2.2%2A2.62712+=5.77
2.5x=2.5%2A2.62712=6.56
3x=3%2A2.62712=7.9
4.1x=4.1%2A2.62712=10.77

check:
5.77%2B6.56%2B7.9%2B10.77=31
31=31



Answer by greenestamps(13208) About Me  (Show Source):
You can put this solution on YOUR website!


The other tutor has a VERY strange interpretation of the question....!

We have stones of lengths 2.2 ft, 2.5 ft, 3 ft, and 4.1 ft; we are to make a wall 31 feet long using those stones. We are to find out how many of each length stone we need.

So we need to find whole number solutions to the equation

2.2a%2B2.5b%2B3c%2B4.1d=31

That is a single equation with 4 unknowns; an algebraic solution is not possible. We need to find solution(s) using logical reasoning and trial and error.

Note the statement of the problem implies that we need to use at least one stone of each length.

The stones of lengths 2.5 ft and 3 ft are both multiples of 0.5ft, as is the total required length of 31 feet. That means the total length of the stones of lengths 2.2 ft and 4.1 ft must be a multiple of 0.5 ft.

So find combinations of the stones of length 2.2 feet and 4.1 feet that make a total length that is a multiple of 0.5 feet.

Solution #1:

Two stones of length 2.2 feet and one of length 4.1 feet make a total of 8.5 feet.
That leaves 22.5 feet to be made using the stones of lengths 2.5 and 3 feet.
Since the remaining length is not a whole number, the number of stones of length 2.5 feet must be odd.
Trying different odd numbers of stones of length 2.5 feet finds a solution with three stones of length 2.5 and five of length 3 feet.

SOLUTION #1: 2(2.2)+1(4.1)+3(2.5)+5(3) = 4.4+4.1+7.5+15 = 31

Solution #2:

One stone of length 2.2 feet and three of length 4.1 feet make a total of 14.5 feet.
That leaves 16.5 feet to be made using the stones of length 2.5 feet and 3 feet.
Again the number of stones of length 2.5 feet must be odd.
Trying different odd numbers of stones of length 2.5 feet finds another solution using three stones of length 2.5 feet and three of length 3 feet.

SOLUTION #2: 1(2.2)+3(4.1)+3(2.5)+3(3) = 2.2+12.3+7.5+9 = 31

Solution #3:

Four stones of length 2.2 feet and two of length 4.1 feet make a total of 17 feet.
That leaves 14 more feet; that means the number of stones of length 2.5 feet must be even; that essentially means we need to make the remaining 14 feet using multiples of 5 and 3. That's easy -- one 5 (i.e., two stones of length 2.5 feet) and three 3's.

SOLUTION #3: 4(2.2)+2(4.1)+2(2.5)+3(3) = 8.8+8.2+5+9 = 31

There are no more solutions.