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A standard machine fills paint cans at a rate of 1 gallon every 4 minu

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A standard machine fills paint cans at a rate of 1 gallon every 4 minu  [#permalink]

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New post 16 Jun 2018, 09:16
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A standard machine fills paint cans at a rate of 1 gallon every 4 minutes. A deluxe machine fills gallons of paint at twice the rate of a standard machine. How many hours will it take a standard machine and a deluxe machine, working together, to fill 135 gallons of paint?

(A) 1
(B) 1.5
(C) 2
(D) 2.5
(E) 3

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Re: A standard machine fills paint cans at a rate of 1 gallon every 4 minu  [#permalink]

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New post 16 Jun 2018, 10:47
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A standard machine will fill 1/4 gallon per minute
A deluxe machine works twice the rate of a standard machine so 1/2 gallon per minute.

Working together - 3/4 gallon per minute.

Total hours required to fill 135 gallons would be:
(3/4) * X = 135

(135*4)/(3*60) = 3 hours.

OA - E.
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A standard machine fills paint cans at a rate of 1 gallon every 4 minu  [#permalink]

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New post 16 Jun 2018, 21:33
Bunuel wrote:
A standard machine fills paint cans at a rate of 1 gallon every 4 minutes. A deluxe machine fills gallons of paint at twice the rate of a standard machine. How many hours will it take a standard machine and a deluxe machine, working together, to fill 135 gallons of paint?

(A) 1
(B) 1.5
(C) 2
(D) 2.5
(E) 3

If the denominator of a rate in minutes is a factor of 60, multiply bottom and top of the rate by the number that will yield 60 minutes. Then change "60 minutes" to "1 hour."

Standard machine rate:
\(\frac{1gal}{4mins}*\frac{15}{15}=\frac{15gal}{60mins}=\frac{15gal}{1hour}\)

Deluxe machine rate is 2 * Standard rate:
30 gallons in one hour

Combined rate: (15 + 30) = 45 gallons in one hour

Work = 135 gallons to fill
Working together, rate: 45 gallons per hour
Time needed to complete work?

\(t=\frac{W}{r}\)
\(t=\frac{135}{45}=3\) hours

Answer E
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Re: A standard machine fills paint cans at a rate of 1 gallon every 4 minu  [#permalink]

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New post 25 Jun 2018, 04:43
generis I did it as per your method, however can you also present as per standard algebric method ?
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A standard machine fills paint cans at a rate of 1 gallon every 4 minu  [#permalink]

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New post 25 Jun 2018, 16:28
Bunuel wrote:
A standard machine fills paint cans at a rate of 1 gallon every 4 minutes. A deluxe machine fills gallons of paint at twice the rate of a standard machine. How many hours will it take a standard machine and a deluxe machine, working together, to fill 135 gallons of paint?

(A) 1
(B) 1.5
(C) 2
(D) 2.5
(E) 3

hero_with_1000_faces wrote:
generis I did it as per your method, however can you also present as per standard algebric method ?

hero_with_1000_faces - Sure. I am not certain about what "standard" means. If my reply is not "standard," please PM me, explain what you mean by "standard," and I will change this answer. :-)

If units are an issue (rates in minutes vs. time in hours, or "twice the rate"), see II.
If not (if you just want to see "minutes" carried all the way through), see I.

I. Without units

• RATES (in # of gallons per # of minutes)

Standard machine's rate: \(R_{s}=\frac{1}{4}\)

Deluxe machine's rate is twice the rate of the standard machine, so
Deluxe machine's rate: \(R_{d}=(\frac{1}{4}*2)=\frac{2}{4}\)

Combined rate of machines: \((\frac{1}{4}+\frac{2}{4})=\frac{3}{4}\)

• TIME needed for machines working together to fill 135 gallons of paint?
(Rates are in minutes, hence calculated time will be in minutes)

\(R_{(s+d)}*T=W\)

\((\frac{3}{4}*T)=135\)

\((\frac{4}{3}*\frac{3}{4})*T=135*\frac{4}{3}\)

\(T=(135*\frac{4}{3})=\frac{540}{3}=180\)
minutes

Time in hours: Use a conversion ratio written so that minutes will cancel

\((180 mins*\frac{1hour}{60 mins})=(\frac{180}{60}*1hour)=(3*1hour)=\) \(3\) hours

Answer E

II. With units

• RATES

Standard machine's rate: \(R_{s}=\frac{1gal}{4mins}\)

Deluxe machine's rate? Twice the rate* of the standard machine =
twice the amount of work in the same amount of time.
Hence, for the deluxe machine's rate, double the standard machine's
amount of work (# of gallons of paint)

Deluxe machine's rate: \(R_{d}=(\frac{1gal*2}{4mins})=\frac{2gals}{4mins}\)

Combined rate**: \((\frac{1gal}{4mins}+\frac{2gals}{4mins})=\frac{3gals}{4mins}\)

TIME to finish? "How many hours will it take [the machines] to fill 135 gallons of paint?"

Issue: minutes vs. hours. Rates in minutes will yield time needed in minutes.
Adjust at the end. Use RT=W, find # of minutes needed, and convert to number of hours

Time, T in minutes: At their combined rate of \(\frac{3gals}{4mins}\)

how many minutes do the machines need to fill 135 gallons of paint?

\(R_{(s+d)}*T=W\) ("gallons" will cancel, leaving only minutes)

\((\frac{3gals}{4mins}*T)=135gallons\)

\((\frac{4mins}{3gals}*\frac{3gals}{4mins}*T)=(135gals*\frac{4mins}{3gals})\)

\(T=(135gals)*(\frac{4mins}{3gals})\)

\(T=\frac{135*4mins}{3}=\frac{540mins}{3}=180\)
minutes

Time in hours: 180 minutes = how many hours? ("minutes" will cancel)

\(180mins*\frac{1hour}{60mins}\) \(=3\) hours

Answer E

Hope that helps!

*"Twice the rate" can be confusing, especially if the denominator is not in single units of time (denominator, e.g., is not ONE minute)
Use logic. Work rates = Amount of work/Amount of time. Twice the rate?

• If X can fill ONE bucket in 1 minute, and Y's rate is "twice the rate of X," then Y can fill TWO buckets in 1 minute
• If X can fill ONE bucket in 4 minutes, and Y's rate is "twice the rate of X," then Y can fill TWO buckets in 4 minutes

That is, if the amount of time is the same, for the faster machine's rate, double the slower machine's amount of work (# of gallons filled), whether the denominator is 1 minute, 4 minutes, 1 hour, 4 days . . . Double the numerator. Leave the denominator (# of units of time) alone. (Algebraic method is in Part I. Multiply slower machine's rate by 2. Not very helpful if units are at issue.)


**Shortcut - From this combined rate we can convert to gallons per hour easily.
(And we could do so in Part I as soon as we have the combined rate.)

Combined rate: \(\frac{3gals}{4mins}\)

Find combined rate in gallons per hour. Multiply combined rate by minutes per hour (minutes will cancel)

Combined rate in gals/hour: \((\frac{3gals}{4mins}*\frac{60mins}{1hour})=\frac{3*15gals}{1hour}=\frac{45gals}{1hr}\)

Time needed to fill cans with 135 gallons of paint? \(T = \frac{W}{r}\)

\(T = \frac{135gals}{(\frac{45gals}{1hour})}=(135gals*\frac{1hour}{45gals})=3\) hours

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Re: A standard machine fills paint cans at a rate of 1 gallon every 4 minu  [#permalink]

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New post 28 Jun 2018, 18:03
Bunuel wrote:
A standard machine fills paint cans at a rate of 1 gallon every 4 minutes. A deluxe machine fills gallons of paint at twice the rate of a standard machine. How many hours will it take a standard machine and a deluxe machine, working together, to fill 135 gallons of paint?

(A) 1
(B) 1.5
(C) 2
(D) 2.5
(E) 3


The rate of the standard machine is 1/4 and the rate of the deluxe machine is 2 x 1/4 = 2/4 = 1/2.

Thus, the combined rate is 1/4 + 1/2 = 1/4 + 2/4 = 3/4.

So, it takes 135/(3/4) = (135 x 4)/3 = 45 x 4 = 180 minutes = 180/60 = 3 hours.

Answer: E
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Re: A standard machine fills paint cans at a rate of 1 gallon every 4 minu &nbs [#permalink] 28 Jun 2018, 18:03
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