galaxyblue wrote:

Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then Machine Y, operating alone for 3 hours, filled the rest of this lot. How many hours would it have taken Machine X operating alone to fill the entire production lot?

(1) Machine X produced 30 bottles per minute.

(2) Machine X produced twice as many bottles in 4 hours as Machine Y produced in 3 hours.

My work:

Bottles = (rateX)4 + (rateY)3 *Did not forget to convert minutes to hours

1) RateX = 30 b/min = 1800 b/hr

Xbottles = 1800(4) = 7200 bottles

Insuff

2) Bottles Y = 1/2 Bottles X

I went with putting them together, getting the total number of bottles produced and backing out the rate for the answer. WRONG.

I have the answer, I just don't understand it.

There are several important things you should know to solve work problems:

1. Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance in rate problems.time*speed=distance <-->

time*rate=job \ done. For example when we are told that a man can do a certain job in 3 hours we can write:

3*rate=1 -->

rate=\frac{1}{3} job/hour. Or when we are told that 2 printers need 5 hours to complete a certain job then

5*(2*rate)=1 --> so rate of 1 printer is

rate=\frac{1}{10} job/hour. Another example: if we are told that 2 printers need 3 hours to print 12 pages then

3*(2*rate)=12 --> so rate of 1 printer is

rate=2 pages per hour;

So, time to complete one job = reciprocal of rate. For example if 6 hours (time) are needed to complete one job --> 1/6 of the job will be done in 1 hour (rate).

2. We can sum the rates.If we are told that A can complete one job in 2 hours and B can complete the same job in 3 hours, then A's rate is

rate_a=\frac{job}{time}=\frac{1}{2} job/hour and B's rate is

rate_b=\frac{job}{time}=\frac{1}{3} job/hour. Combined rate of A and B working simultaneously would be

rate_{a+b}=rate_a+rate_b=\frac{1}{2}+\frac{1}{3}=\frac{5}{6} job/hour, which means that they will complete

\frac{5}{6} job in one hour working together.

3. For multiple entities: \frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+...+\frac{1}{t_n}=\frac{1}{T}, where T is time needed for these entities to complete a given job working simultaneously.For example if:

Time needed for A to complete the job is A hours;

Time needed for B to complete the job is B hours;

Time needed for C to complete the job is C hours;

...

Time needed for N to complete the job is N hours;

Then:

\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+...+\frac{1}{N}=\frac{1}{T}, where T is the time needed for A, B, C, ..., and N to complete the job working simultaneously.

For two and three entities (workers, pumps, ...):

General formula for calculating the time needed for two workers A and B working simultaneously to complete one job:Given that

t_1 and

t_2 are the respective individual times needed for

A and

B workers (pumps, ...) to complete the job, then time needed for

A and

B working simultaneously to complete the job equals to

T_{(A&B)}=\frac{t_1*t_2}{t_1+t_2} hours, which is reciprocal of the sum of their respective rates (

\frac{1}{t_1}+\frac{1}{t_2}=\frac{1}{T}).

General formula for calculating the time needed for three A, B and C workers working simultaneously to complete one job:T_{(A&B&C)}=\frac{t_1*t_2*t_3}{t_1*t_2+t_1*t_3+t_2*t_3} hours.

BACK TO THE ORIGINAL QUESTION:Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then Machine Y, operating alone for 3 hours, filled the rest of this lot. How many hours would it have taken Machine X operating alone to fill the entire production lot?You can solve this question as Karishma proposed in her post above or algebraically:

Let the rate of X be

x bottle/hour and the rate of Y

y bottle/hour.

Given:

4x+3y=job. Question:

t_x=\frac{job}{rate}=\frac{job}{x}=?(1) Machine X produced 30 bottles per minute -->

x=30*60=1800 bottle/hour, insufficient as we don't know how many bottles is in 1 lot (job).

(2) Machine X produced twice as many bottles in 4 hours as Machine Y produced in 3 hours -->

4x=2*3y, so

3y=2x -->

4x+3y=4x+2x=6x=job -->

t_x=\frac{job}{rate}=\frac{job}{x}=\frac{6x}{x}=6 hours. Sufficient.

Answer: B.

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