Plug in a number that is a multiple of all the times given: the LCM (lowest common multiple) of 12, 15, and 18 is 180.lawri wrote:Farooq, the pick a number strategy is quite elegant. My only question is how can one easily and quickly pick a mumber, in this case 180. Is there a method to picking a number?farooq wrote:Lets say total no. of pages that each printer prints are 180.resilient wrote:working alone, printers x,y, and z can do a certain printing job, consisitning of a large number of pages, 12, 15, and 18 hours, respectively. What is the ratio of the time it takes printer x to do the job, working at its rate, to time it takes printers y and z to do the job, working together at their individual rates?
a. 4/11
b.1/2
c. 15/22
d.22/15
e.11/4
qa is d. I dont see why C is wrong. I dont see why the solution flips the combined rate of y and z working together. help stuart?
X prints 180 pages in 12 hours. 15 pages per hour.
Y prints 180 pages in 15 hours. 12 Pages per hour.
Z prints 180 pages in 18 hours. 10 pages per hour.
If Y and Z both work together, they will print 12+10 = 22 pages per hour.
Therefore total no. of hours to print 180 pages by Y and Z will be 180/22.
Time taken by X to print total no. of pages/Time taken by (Y and Z together)to print total no. of pages = 12/(180/22) = 22/15
og math # 130
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there is something I don't understand:
we have RT=W
For printer 1 : xt1=w1
For printer 2 : yt2=w2
For printer 3 : zt3=w3
Since job 1 is the same than job 2 + job 3
we have w1=w2+w3
as printer 2 and 3 finish together, they work together and logically finish at the same time: t2=t3=t
so we should have : 12t1=15t + 18t, thus t1/t=11/4
why it is not correct???
thanks,
we have RT=W
For printer 1 : xt1=w1
For printer 2 : yt2=w2
For printer 3 : zt3=w3
Since job 1 is the same than job 2 + job 3
we have w1=w2+w3
as printer 2 and 3 finish together, they work together and logically finish at the same time: t2=t3=t
so we should have : 12t1=15t + 18t, thus t1/t=11/4
why it is not correct???
thanks,
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12, 15, and 18 are the respective times for the 3 printers, not the respective rates.mazwaz wrote:there is something I don't understand:
we have RT=W
For printer 1 : xt1=w1
For printer 2 : yt2=w2
For printer 3 : zt3=w3
so we should have : 12t1=15t + 18t, thus t1/t=11/4
why it is not correct???
thanks,
Thus:
12t1 = time*time.
15t = time*time.
18t = time*time.
Since work = rate*time, you need to use the given times to determine the rate for each printer.
The algebra needed to determine the three rates is a bit messy. I think that the easiest way to determine the three rates is to plug in a value for the work, as I did in my original post.
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Time taken by x to complete 1 job =12 hr . Therefore rate = 1/12 per hrsayanchakravarty wrote:D. 22/15
Similarly y,z rates are 1/15 and 1/18 per hr.
combined rate = 1/15 + 1/18 => 15+18/15*18 and therfore time = total work/combined rate
=1/15+18/15*18 => 15*18/15+18 => 22/15. Ans D
Though i feel, after solving it, plug in would have been an easier option
LCM => 180
So, individual rates are x=180/12 = 15 per hr
y=180/15 = 12 per hr
z= 180/18 = 10 per hr
combined rate of y and z = 22 per hr
time by y and z -> 180/22 =90/11
ratio => 12/90/11 => 12*11/90 => 22/15 hr
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x-12, y-15, z-18working alone, printers x,y, and z can do a certain printing job, consisitning of a large number of pages, 12, 15, and 18 hours, respectively. What is the ratio of the time it takes printer x to do the job, working at its rate, to time it takes printers y and z to do the job, working together at their individual rates?
a. 4/11
b.1/2
c. 15/22
d.22/15
e.11/4
y&z together = (1/(1/15 + 1/18)) = 90/11
thus ratio = 12:(90/11) = 22/15
IMO D
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SUPPOSE 'W' BE THE JOB TO BE DONE.
FOR PRINTER 'X' --> THE RATE OF THE JOB IS W/12
FOR PRINTER 'Y' --> THE RATE OF THE JOB IS W/15
FOR PRINTER 'Z' --> THE RATE OF THE JOB IS W/18
PRINTERS Y AND Z WORK TOGETHER --> THE RATE OF THE JOB IS (W/15 + W/18) = 11W/90 --> W/(90/11)
--> COMBINEDLY Y AND Z TOOK 90/11 HOURS.
SO THE RATION REQUIRED IS (12)/(90/11) = 22/15
THANKS,
DEEPAK
FOR PRINTER 'X' --> THE RATE OF THE JOB IS W/12
FOR PRINTER 'Y' --> THE RATE OF THE JOB IS W/15
FOR PRINTER 'Z' --> THE RATE OF THE JOB IS W/18
PRINTERS Y AND Z WORK TOGETHER --> THE RATE OF THE JOB IS (W/15 + W/18) = 11W/90 --> W/(90/11)
--> COMBINEDLY Y AND Z TOOK 90/11 HOURS.
SO THE RATION REQUIRED IS (12)/(90/11) = 22/15
THANKS,
DEEPAK
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If you want to bipass the algebra, you can just find really simple ways to approximate the rates and you'll be able to quickly identify that only one option comes close.
If Y and Z complete the job in 15 and 18 hours respectively, the time it will take them to complete it combined is going to be very close to half of the midpoint between those two values.
12 : 8.25 is the approximate ratio.
and looking at the options available, you don't even really need to do the algebra to see which one the answer will be.
just keep in mind that the only reason this works is because the rates of Y and Z are close to begin with. if one was significantly faster than the other, it's combined rate would be weighted more heavily towards the faster one. ie: printer Y took 10 hours and printer Z took 20 hours. in this case, the combined rate would complete the job in 6.66 and not 7.5.
If Y and Z complete the job in 15 and 18 hours respectively, the time it will take them to complete it combined is going to be very close to half of the midpoint between those two values.
12 : 8.25 is the approximate ratio.
and looking at the options available, you don't even really need to do the algebra to see which one the answer will be.
just keep in mind that the only reason this works is because the rates of Y and Z are close to begin with. if one was significantly faster than the other, it's combined rate would be weighted more heavily towards the faster one. ie: printer Y took 10 hours and printer Z took 20 hours. in this case, the combined rate would complete the job in 6.66 and not 7.5.
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For me it was simpler to start discarding options.
A, B and C can be discarded directly because the Y & Z printers together will print the job faster than X so the ratio should have a bigger number on the left side.
Now assuming that Y & Z print at the fastest speed, to take the extreme at 15 hours/job, both together will print the job in, at least, 7.5 hours. Actually it´ll take more than that.
X on the other side, we know it will print in 12 hours.
12/7.5 will be 1.??? (will be closer to 1 because actually Y & Z will print slower than our assumption).
Option D is 22/15 which is 1.??? --> CLOSER TO OUR ASSUMPTION
Option E is 11/4 which is 2.???
If the difference between the two ratios (D & E) was less, then the other calculation would be needed but in this case, it was much easier this way.
Regards,
A, B and C can be discarded directly because the Y & Z printers together will print the job faster than X so the ratio should have a bigger number on the left side.
Now assuming that Y & Z print at the fastest speed, to take the extreme at 15 hours/job, both together will print the job in, at least, 7.5 hours. Actually it´ll take more than that.
X on the other side, we know it will print in 12 hours.
12/7.5 will be 1.??? (will be closer to 1 because actually Y & Z will print slower than our assumption).
Option D is 22/15 which is 1.??? --> CLOSER TO OUR ASSUMPTION
Option E is 11/4 which is 2.???
If the difference between the two ratios (D & E) was less, then the other calculation would be needed but in this case, it was much easier this way.
Regards,
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Dont know if its the best way but i always follow this in Work/Time problems..
Took the LCM of 12, 15 and 18 as 180.
Now Rate of X = 180/12 = 15
Rate of Y = 180/15 = 12
Rate of Z = 180/18 = 10
Time taken by X to finish 180 units = 12
Time taken by Y nd Z to finish 180 units = 180/ (12+10) = 180/22
So, ratio = 12/(180/22) = 12 * 22 /180 = 22/15
Hence d) is the correct answer.
Took the LCM of 12, 15 and 18 as 180.
Now Rate of X = 180/12 = 15
Rate of Y = 180/15 = 12
Rate of Z = 180/18 = 10
Time taken by X to finish 180 units = 12
Time taken by Y nd Z to finish 180 units = 180/ (12+10) = 180/22
So, ratio = 12/(180/22) = 12 * 22 /180 = 22/15
Hence d) is the correct answer.
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rate x = 1/12resilient wrote:working alone, printers x,y, and z can do a certain printing job, consisitning of a large number of pages, 12, 15, and 18 hours, respectively. What is the ratio of the time it takes printer x to do the job, working at its rate, to time it takes printers y and z to do the job, working together at their individual rates?
a. 4/11
b.1/2
c. 15/22
d.22/15
e.11/4
qa is d. I dont see why C is wrong. I dont see why the solution flips the combined rate of y and z working together. help stuart?
rate y = 1/15
rate z = 1/18
time x = 1/1/12 = 12
time y+z = 1/ 1/15+1/18 = 1/ 33/270 = 270/33 = 90/11
ratio = 12/90/11 = 12*11/90 = 44/30 = 22/15
Set up problem as follows:
i)it takes x 12 hours
ii)it takes y and z together (15*18)/15+33. (simplifying to 90/11)
Now that you have the rates for the two required scenarios set them up as the ratio required in the question; i.e 12 : 90/11
We know from our ratio bascis that we can rewrite this as the fraction 12 divided by 90/11
OR 12*11/90 --simplify this you get 22/15
And 22/15 is choice D
i)it takes x 12 hours
ii)it takes y and z together (15*18)/15+33. (simplifying to 90/11)
Now that you have the rates for the two required scenarios set them up as the ratio required in the question; i.e 12 : 90/11
We know from our ratio bascis that we can rewrite this as the fraction 12 divided by 90/11
OR 12*11/90 --simplify this you get 22/15
And 22/15 is choice D