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3 years ago

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Six machines, each working at the same constant rate, together can complete a certain job in 12 days. How many additional machin

es, each working at the same constant rate, will be needed to complete the job in 8 days?

1 answer:

Schach [20]3 years ago

7 0

Answer:

3 machine

Step-by-step explanation:

It is given that 6 machine each working at the same rate can complete the work in 12 days

We the has to complete in 8 days

Let we need x extra machine to complete the work in 8 days

So total number of machines =x+6

Now according to man work day equation $M_1D_1=M_2D_2$

$6\times 12=(x+6)\times 8$

$8x+48=72$

$8x=24$

x=3 machine

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Given a Venn diagram showing the number of students that like blue uniform only as 32, the number of students that like gold uniform only as 25, the number of students that like blue and gold uniforms as 12 and the number of students that like neither blue nor gold uniform as 6.

Thus, the total number of students interviewed is 75.

Recall that relative frequency of an event is the outcome of the event divided by the total possible outcome of the experiment.

From the relative frequency table, a represent the relative frequency of the students that like gold but not blue.
From the Venn, diagram, the number of students that like gold uniform only as 25, thus the relative frequency of the students that like gold but not blue is given by
$\frac{25}{75} = \frac{1}{3}=0.33=33\%$

Therefore, a = 33% to the nearest percent.

Similarly, from the relative frequency table, b represent the relative frequency of the students that like blue but not gold.
From the Venn, diagram, the number of students that like blue uniform only as 32, thus the relative frequency of the students that like gold but not blue is given by
$\frac{32}{75} = 0.427=42.7\%$

Therefore, b = 43% to the nearest percent.

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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

Dmitry [639]

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:

he throws away one fish from the bunch

$x-1$

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:

he throws away one fish from the remaining bunch

$\text{John} -1$

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

8 0

2 years ago