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

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

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

Hi, I am restarted my studies. Please help me with this silly doubt
I tried using the LCM of t1 and t2 method but didn't get the right answer.
For standard machines
1 gallon is filled in 4 min
Therefore t1 = 4min

For Deluxe machines
1 gallon is filled in 2 min(rate is twice)
Therefore t2 = 2min

Now LCM(4,2) = 4
Rate of Standard machine will be 4/4 = 1
Rate of Deluxe machine will be 4/2 = 2

Combined rate => 1+2=3
Total time = Work/Rate = 135/3 = 45 min
Final time (in hours)= 45/60

I know the rates are given to use. But normally we use this method when the time periods are given. Hence I should get the same answer when using this method.
Can somebody please tell me where I am wrong?
Bunuel generis VeritasKarishma chetan2u

Rate of standard machine = 1/4 gallons/min
Rate of deluxe machine = 2/4 gallons/min

Combined rate = 1/4 + 2/4 = 3/4 gallons/min

Time = Work/Rate = 135 gallons / (3/4) gallons/min = 180 mins = 3 hrs

Answer (E)

I am not sure what you mean by rate for standard machine = 1

Keep the units and then see if it makes sense. Rate must be in gallons per minute.
Rate of standard machine = 1 gallon per 4 mins = (1/4) gallon/min

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