I'll give you a medal and be a fan if you HELP PLEASE
1. The pump used to fill the water tank on a fire truck takes 40 minutes to fill the tank. If the pump sending water to the hose is on, it takes 60 minutes to fill the tank. If the pump filling the water tank is off, how long does the pump sending water to the hose take to empty the tank?
A. 120 minutes
B. 60 minutes
C. 30 minutes
D. 24 minutes
2. Annmarie can plow a field in 240 minutes. Gladys can plow a field 80 minutes faster. If they work together, how many minutes does it take them to plow the field?
A. 96 minutes
B. 160 minutes
C. 400 minutes
D. 480 minutes
3.Brody can fill a bowl with candy in 3 minutes. While Brody fills the bowl, Hudson takes the candy out of the bowl. With Hudson taking candy out of the bowl, it takes 5 minutes for Brody to fill the bowl.
Which of the following can be used to determine the amount of time it takes for Hudson to empty the bowl if Perry does not add candy?
A. 1 over 3 minus 1 over x equals 1 over 5
B. 1 over 3 minus 1 over 5 equals 1 over x
C. 1 over 5 minus 1 over x equals 1 over 3
D. 1 over 3 minus 1 over x equals x over 5
4. Leanne can clean a fish tank in 30 minutes. Karl can clean a fish tank in 20 minutes. If they work together, how many minutes does it take them to clean a fish tank?
A. 6 minutes
B.10 minutes
C. 12 minutes
D. 50 minutes
5.Genna can build a cabinet in 4 hours. Claude can build a cabinet in only 3 hours.
Which of the following can be used to determine the amount of time it would take for Genna and Claude to build a cabinet together?
A. 1 over 4 plus 1 over x equals 1 over 3
B. 1 over 4 plus 1 over 3 equals 1 over x
C. 1 over x plus 1 over 3 equals 1 over 4
D. 1 over 3 plus 1 over 4 equals x over 7

- anonymous

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- anonymous

@ganeshie8

- anonymous

@iGreen

- anonymous

@Michele_Laino

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## More answers

- anonymous

@TheSmartOne

- Michele_Laino

Question 5)
the working rate of Genna is W/4, the working rate of Claude is W/3, where W is the work to do.
When Gwenna and Claude work together, then the working rate is:
W/4 + W/3
so we can write:
\[\Large \left( {\frac{W}{4} + \frac{W}{3}} \right)t = W\]
where t is time needed to Genna and Claude to do the work W
Please solve that equation for T, what do you get?

- Michele_Laino

oops.. for t, what do you get?

- anonymous

3 hrs?

- Michele_Laino

hint:
what is:
\[\frac{1}{4} + \frac{1}{3} = ...?\]

- anonymous

\[\frac{ 7 }{ 12 }\]

- Michele_Laino

ok! so, from preceding equation, we can write:
\[\frac{{7W}}{{12}}t = W\]
or:
\[\frac{7}{{12}}t = 1\]

- anonymous

ok

- Michele_Laino

namely:
\[t = \frac{{12}}{7}\]

- anonymous

so the answer is D then?

- Michele_Laino

yes!

- anonymous

Thanks ! can you help with the others
?

- Michele_Laino

ok!
Question 4)

- Michele_Laino

the working rate of Leanne is W/30, where the working rate of Karl is W/20, where as usual, W is the work to do.
So when Leanne and Karl work together the working rate is:
W/30+ W/20, and we can write:
\[\Large \left( {\frac{W}{{30}} + \frac{W}{{20}}} \right)t = W\]
where t is time neeeded to Leanne and Karl to finish the work W.
Please solve for t

- Michele_Laino

oops...whereas the working rate of Karl

- Michele_Laino

as before, we can write:
\[\Large \left( {\frac{1}{{30}} + \frac{1}{{20}}} \right)t = 1\]

- Michele_Laino

what is t?

- anonymous

\[\frac{ 5 }{ 15 } ?\]

- Michele_Laino

we have:
\[\frac{1}{{30}} + \frac{1}{{20}} = \frac{{2 + 3}}{{60}} = ...?\]

- anonymous

i entered the wrong thing my bad!

- anonymous

\[\frac{ 1 }{ 12 }\]

- Michele_Laino

ok!
So we have:
\[\frac{t}{{12}} = 1\]

- Michele_Laino

and therefore:
\[t = 12 \times 1 = ...?\]

- anonymous

12?

- Michele_Laino

that's right!

- Michele_Laino

it is 12 minutes

- anonymous

Ok Thanks!

- anonymous

Have time for another one?

- Michele_Laino

yes!

- Michele_Laino

question 3)

- anonymous

ok

- Michele_Laino

I'm pondering....

- Michele_Laino

the working rate of Brody is:
W/3, when he works alone
whereas the working rate of Brody is W/5, when Brody and Hudson work together
W is the work to be done
Now the working rate of Hudson is:
W/x, where x is the unknown quantity. So, we can write:
\[\Large \frac{W}{3} - \frac{W}{x} = \frac{W}{5}\]
or:
\[\Large \frac{1}{3} - \frac{1}{x} = \frac{1}{5}\]
so, what is the right option?

- Michele_Laino

hint:
\[\Large \frac{1}{x} = \frac{1}{3} - \frac{1}{5}\]

- Michele_Laino

@amielli

- anonymous

2/15

- Michele_Laino

yes! What is the right option?

- anonymous

D?

- Michele_Laino

I think that option D is wrong

- Michele_Laino

please look at this equation:
\[\Large \frac{1}{3} - \frac{1}{x} = \frac{1}{5}\]
do you recognize to which option corresponds to?

- anonymous

yes that would be A

- Michele_Laino

yes! correct!

- anonymous

I dont get it though

- Michele_Laino

since Brody is filling the bowl, he is adding candies to that bowl.
Hudson is subtracting candies to that bowl.
Here is why I used the minus sign in the subsequent equation:
\[\large \frac{W}{3} - \frac{W}{x} = \frac{W}{5}\]

- Michele_Laino

now, is it clear?

- anonymous

Oh yes I do

- anonymous

Thanks

- Michele_Laino

ok!

- Michele_Laino

question 2)

- Michele_Laino

here we can write:
working rate of Annmarie is W/240, whereas the working rate of Gladies is W/80.
Now when Annmarie and Gladies work together, the working rate is:
W/240 + W/80.
As usual W is the work to be done.
So we can write:
\[\Large \left( {\frac{W}{{240}} + \frac{W}{{80}}} \right)t = W\]
or:
\[\Large \left( {\frac{1}{{240}} + \frac{1}{{80}}} \right)t = 1\]
Please solve for t

- Michele_Laino

hint:
\[\Large \frac{1}{{240}} + \frac{1}{{80}} = \frac{{1 + 3}}{{240}} = ...?\]

- Michele_Laino

please wait a moment, I think to have made an error

- anonymous

ok

- Michele_Laino

the working rate of Gladies is:
\[\Large \frac{W}{{240 - 80}} = \frac{W}{{160}}\]

- Michele_Laino

so the right equation is:
\[\Large \left( {\frac{W}{{240}} + \frac{W}{{240 - 80}}} \right)t = W\]

- Michele_Laino

or:
\[\Large \left( {\frac{1}{{240}} + \frac{1}{{160}}} \right)t = 1\]

- Michele_Laino

please solve for t, keep in mind that:
\[\Large \frac{1}{{240}} + \frac{1}{{160}} = \frac{{2 + 3}}{{480}} = \frac{5}{{480}}\]

- Michele_Laino

after a substitution, we have:
\[\Large \frac{5}{{480}}t = 1\]
what is t?

- anonymous

just simplify it?

- Michele_Laino

hint:
\[\Large t = \frac{{480}}{5} = ...?\]

- anonymous

96

- anonymous

I get that one

- anonymous

So the last one

- Michele_Laino

ok! question 1)

- Michele_Laino

this question is similar to the question of candies, namely question 3).
So the rate of working for the pump is W/40, when the pump works alone
the working rate of the firehose is W/x, where x is the unknown quantity
and
W/60 is the working rate of the pump, when the pump and the firehose work together.
So we can write:
\[\Large \frac{W}{{40}} - \frac{W}{x} = \frac{W}{{60}}\]
or:
\[\Large \frac{1}{{40}} - \frac{1}{x} = \frac{1}{{60}}\]

- anonymous

so the answer is 60?

- Michele_Laino

no, since we have to compute the value of x

- Michele_Laino

hint:
\[\Large \frac{1}{x} = \frac{1}{{40}} - \frac{1}{{60}} = \frac{{3 - 2}}{{120}} = ...\]

- Michele_Laino

please continue my computation

- anonymous

1/120

- Michele_Laino

ok! so we have:
\[\Large x = \frac{{120}}{1} = ...{\text{minutes}}\]

- anonymous

thanks very much I got 80%

- anonymous

number 5 was wrong

- Michele_Laino

are you sure?

- Michele_Laino

I'm very sorry, the correct answer is option B.
Sorry again

- anonymous

That is totally fine

- anonymous

Thanks!

- Michele_Laino

since we can write this:
\[\Large \left( {\frac{W}{4} + \frac{W}{3}} \right)t = W\]
and then:
\[\Large \frac{1}{4} + \frac{1}{3} = \frac{1}{t}\]
Sorry again

- Michele_Laino

thanks for your comprehension!

- anonymous

Thanks for helping me

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