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Sub 505 Level|   Work and Rate Problems|                     
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Bunuel
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Each machine can produce (270/6) = 45 bottles per minute

45 * 10 = 450 bottles per minute

450 * 4 = 1800 bottles in 4 minutes.
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Let's take the approach that uses the answer choices to eliminate wasted time.

270/6 is a little less than 300/60 = 50 bottles per minute per machine. 10 machines = 500 per minute. 4 minutes worth = 2000 bottles as being slightly HIGHER than the answer we need.

Looking at the answers it is clear...we can only choose (B) 1800 bottles starting with an upper bound of 2000.
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270/6 = 45 BPM X 10 Machines = 450 BPM X 4 MIN = 1,800.

Kudos if this is easier!
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Bunuel
Running at the same constant rate, 6 identical machines can produce a total of 270 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes?

(A) 648
(B) 1,800
(C) 2,700
(D) 10,800
(E) 64,800

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Question: 23
Page: 155
Difficulty: 600

6 identical machine produce 270bottles/min then 1machine - 45bottles/min

10 machine - 450 bottles/min

for 4 minutes--> 1800
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better as fractions:

270*10/6*4/1=1800

B
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Answer = B

Equation would be

\(= \frac{270}{6} * 10 * 4 = 1800\)
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Bunuel
Running at the same constant rate, 6 identical machines can produce a total of 270 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes?

(A) 648
(B) 1,800
(C) 2,700
(D) 10,800
(E) 64,800

To solve this problem, we first want to determine the rate of each machine. Since each machine works at the same constant rate, each machine’s rate is 270/6 = 45 bottles per minute. Now since there are 10 such machines, the rate of the 10 machines together is 10 x 45 = 450 bottles per minute.

Finally, we are asked to determine the number of bottles these 10 machines can produce in 4 minutes. Thus:

rate x time = work

450 x 4 = 1,800 bottles

Answer: B
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Bunuel
Running at the same constant rate, 6 identical machines can produce a total of 270 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes?

(A) 648
(B) 1,800
(C) 2,700
(D) 10,800
(E) 64,800

Practice Questions
Question: 23
Page: 155
Difficulty: 600

6 machines can produce 270 bottles, so 10 machines can produce 270*10/6

This can be produced in 1 minute. Total bottles that can be produced in 4 minutes :-
270*10*4/6= 1800 bottles

B is the answer
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Bunuel
Running at the same constant rate, 6 identical machines can produce a total of 270 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes?

(A) 648
(B) 1,800
(C) 2,700
(D) 10,800
(E) 64,800

Practice Questions
Question: 23
Page: 155
Difficulty: 600

r*t=w
we have 6 machine here so

6*(r*1)=270
r*1=45===> r=45

r*t=w

we got 10 machine
10*(r*t)=w
10*(45*4)=1800

t was 4 according to the question
B
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Looking at the answer choices we can approximate and save a lot of time



Rate of each bottle \(= \frac{270}{6}\) = This is somewhere between 40 and 50 bottles per minute

10 bottles will produce between 400 and 500 bottles per minute
in 4 minutes the same 10 bottles will produce between 1600 and 2000 bottles.

The only answer that lies in this range is B
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Hi All,

We're told that 6 identical machines can produce a total of 270 bottles per minute. We're asked for the number of bottles that 10 such machines would produce in 4 minutes when working at this same rate.

Since this is just a big 'rate' question, you can approach the math in a couple of different ways. From the information in the first sentence, since 6 divides evenly into 270, you could figure out the rate/minute for 1 machine...

270/6 = 45 bottles/minute for 1 machine

Since we have 10 machines working for 4 minutes each, that would give us...

(10)(4)(45) = 1800 bottles

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This is one too easy. 6 machines can produce 270 / minute.

270 / 6 = 45 per machine.
45 * 10 machines = 450 per minute.
450 * 4 minute = 1800.

You don't want to miss this on the test.

Another simple way to do this is....

270 * 4 minutes = (250 + 20) * 4 = 1,080
1,080 = 60% * X
1,080 = (6/10) * X
10,800x = 6x
10,800 / 6 = 1800

Cheers
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First, we will keep the time constant at 1 minute and use unitary method to connect machines and bottles.

1 minute: 6 machines —— 270 bottles
1 minute: 1 machine —— 270/6 = 45 bottles

So, 1 minute: 10 machines —- 45(10) = 450 bottles —- (A)

Now, we will keep number of machines constant at 10 and use unitary method to connect minutes and candles.

We can rewrite relationship (A) as:
10 machines: 1 minute —— 450 bottles

Finally,
10 machines: 4 minutes —— (450)(4) = 1800 bottles
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