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Kay [80]
3 years ago
12

Current law requires that wheelchair ramps have a height to length ratio of 1:12. An old ramp has a height to length ratio of 2:

15. It's height is 12 inches. Workers will replace this ramp with a new one that has the same height. How much longer must the new ramp be to meet the ramp ratio required by law? Please Help!!!
Mathematics
1 answer:
crimeas [40]3 years ago
8 0

Answer:

The new ramp must be 54 inches long than the old ramp to meet the requirement of the new law

Step-by-step explanation:

Old height to length ratio is:

h : l =  2 : 15.

Height of old ramp h = 12

Length of old ramp can be find out by putting h  = 12 in the above ratio

\frac{h}{l}  = \frac{2}{15}\\\\h = 12\\\\\frac{12}{l} = \frac{2}{15}\\\\l = 90

Length of old ramp = 90

Height to length ratio by current law:

h : l = 1 : 12

Height will be the same i.e. h = 12

\frac{h}{l}  = \frac{1}{12}\\\\h = 12\\\\\frac{12}{l} = \frac{1}{12}\\\\l = 144

Length of new ramp = 144

How much longer the new ramp should be to meet the new ratio law:

New length - Old length = 144 - 90 = 54 inches

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Answer:

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John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m
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Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

Let's start by saying that

x = is the initial number of fishes

John:

When John arrives:

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x-1

  • divides the remaining fish into three.

\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}

  • takes a third for himself.

\dfrac{x-1}{3} + \dfrac{x-1}{3}

the remaining fish are expressed by the above expression. Let's call it John

\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}

and simplify it!

\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}

When Joe arrives:

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\text{John} -1

  • divides the remaining fish into three

\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

  • takes a third for himself.

\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

the remaining fish are expressed by the above expression. Let's call it Joe

\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

and simiplify it

\text{Joe}=\dfrac{2}{3}(\text{John}-1)

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}

When James arrives:

We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

14=\dfrac{8x}{27} - \dfrac{38}{27}

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

14=\dfrac{8x}{27} - \dfrac{38}{27}

x=52

This is minimum possible amount of fish before John threw out the first fish

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