qr word problem

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keibee82

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how would you go about solving this problem?

Tom has to paint his aunt's garage, which will normally take him
10 hours. However, after working half an hour, Tom persuades
a friend to paint while he rests. Each half hour later he persuades
another friend to join in, and this pattern continues until the job
is done, Assuming everyone works at the same rate as Tom, how
long will it take until the garage is competely painted?
 
"Tom persuades a friend to paint while he rests."

There arent 2 people painting, my guess would be 10 hours.
 
I won't do the calculations, but I can give you the general gist:

1st half hour: 1 guy works
2nd half hour: 1 guy works (Tom's friend)
3rd half hour: 2 guys work together, each at the same rate, at a combined rate of... (calculate this)... then, find the difference between the time it takes them to do this half hour versus the standard half hour it would've taken Tom... that is to say: Tom can do the job in half an hour... Two "Toms" can do the job in... (?)... the difference between the two is the time saved.
4th half hour: 3 guys work together... calculate the time saved here as well

Tom can do the job in 10 hours. The combined rest can do it much quicker. It's a long problem, but it gives you incredibly great practice on these rate questions.

aranjuez
 
aranjuez said:
I won't do the calculations, but I can give you the general gist:

1st half hour: 1 guy works
2nd half hour: 1 guy works (Tom's friend)
3rd half hour: 2 guys work together, each at the same rate, at a combined rate of... (calculate this)
4th half hour: 3 guys work together

Tom can do the job in 10 hours. The combined rest can do it much quicker. It's a long problem, but it gives you incredibly great practice on these rate questions.

aranjuez

i knew that set up, but that as far as i can go. how do i calculate
the rate of each half hour and how do i do that in less than a minute???
 
1/(rate of the first guy) + 1/(rate of the second guy) = 1/(rate of both combined)

Go for it.

aranjuez
 
keibee82 said:
i knew that set up, but that as far as i can go. how do i calculate
the rate of each half hour and how do i do that in less than a minute???

10hrs / 1 entire garage = .5hrs / (1/20 of the garage)
 
hmmm...so can anyone do the problem and come up with the
answer? i'm still confused...... gosh i hate qr and rc section.
 
What are the choices? In hours or in number of friends?
 
keibee82 said:
how would you go about solving this problem?

Tom has to paint his aunt's garage, which will normally take him
10 hours. However, after working half an hour, Tom persuades
a friend to paint while he rests. Each half hour later he persuades
another friend to join in, and this pattern continues until the job
is done, Assuming everyone works at the same rate as Tom, how
long will it take until the garage is competely painted?


don't DO the problem. list it out!!! don't try to set up an equation and solve it!

first, there are 20 half-hours (hr/2) worth of work to be done. with every ACTUAL half-hour that passes, you're adding an extra half-hr man power for every half-hr (time):

1st half hr: 1 (hr/2), .......1 total
2nd: +1 .....2 total <--here, tom stops work, so you're adding one only
3rd:+2 ......4 total
4th:+3 ....7 total
5th:+4 .....11 total
6th: +5 ......16 total.
7th: +6 ......22 total. <---this exceeds the 20 units of work2b done already!

Therefore, a shade before the 7th half-hour, the work's done. find the answer a shade before 3.5 hrs.

this question should've taken you 15-20 seconds to solve.
 
I'm going to have to agree. I finally sat down and thought about it. It's just a bit under 3.5 hours.

aranjuez
 
jkh1886 said:
don't DO the problem. list it out!!! don't try to set up an equation and solve it!

first, there are 20 half-hours (hr/2) worth of work to be done. with every ACTUAL half-hour that passes, you're adding an extra half-hr man power for every half-hr (time):

1st half hr: 1 (hr/2), .......1 total
2nd: +1 .....2 total <--here, tom stops work, so you're adding one only
3rd:+2 ......4 total
4th:+3 ....7 total
5th:+4 .....11 total
6th: +5 ......16 total.
7th: +6 ......22 total. <---this exceeds the 20 units of work2b done already!

Therefore, a shade before the 7th half-hour, the work's done. find the answer a shade before 3.5 hrs.

this question should've taken you 15-20 seconds to solve.
Yep, that's about right. It's wise to list them out. Now, what if I changed it to ...say 20 hours or 1000 hours instead of 10 hours? :scared:
 
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luder98 said:
Yep, that's about right. It's wise to list them out. Now, what if I changed it to ...say 20 hours or 1000 hours instead of 10 hours? :scared:

Then this would become a more complicated number series question. In that case, to solve it, you take advantage of the fact that you're adding one more each time. Solving THAT problem would be very complex, requires clever thinking, and would be prone to errors. In other words, don't worry about it at that stage. It'll never appear on the DAT. Only math competitions throughout the country.

...btw, since you're adding one each time, you can track the total number of additions you've made. Look at the sequence....

1, 2, 3, 4, 5, 6... n.... Note that the first number added to the last number is the same as the 2nd number plus the second to last number:

1+6 = 2+5 = 3+4.

Therefore, the sum of the numbers is... Sum=(n+1)(n/2)
I'll let you think about how to solve the question when you have to deal with 1000 hrs. It involves this kind of thinking, and I'll only hint to it for now....

and beware, there's one trap when you apply it to this problem!
 
I got 3 hours and 20 minutes.

I figure that Tom knocks out 5% per half hour.

Half hour : Percent of total done
1st: 5% (tom)
2nd: 5% (1st friend)
3rd: 10% (2 friends)
4th: 15% (3 friends)
5th: 20% (4 friends)
6th: 25% (5 friends)
after 3 hours, 80% of the work is done

in the 7th half hour, they can knock out 30% of the work, or 10% per 10 minutes. They would only need 20 minutes to finish the job.

3 hours and 20 minutes... but that question is worded kind of screwy so I could be wrong.
 
aggie-master said:
I got 3 hours and 20 minutes.

I figure that Tom knocks out 5% per half hour.

Half hour : Percent of total done
1st: 5% (tom)
2nd: 5% (1st friend)
3rd: 10% (2 friends)
4th: 15% (3 friends)
5th: 20% (4 friends)
6th: 25% (5 friends)
after 3 hours, 80% of the work is done

in the 7th half hour, they can knock out 30% of the work, or 10% per 10 minutes. They would only need 20 minutes to finish the job.

3 hours and 20 minutes... but that question is worded kind of screwy so I could be wrong.

Just note that any percentage you calculated before the 7th half-hour is unnecessary. It's good practice, but you wouldn't do it on the real test. you just need to note that at the 7th half-hr, 16 hrs of work was completed, leaving only 4 hr/2's left. So, on that last hr/2, you have 4 hrs of work, and 6 people doing them. So the final calculation is:

4 hr work x (30min/6) = 20 min. Your answer is correct.

My only point here.... only do the calculations where you must.
 
yea the answer is 3 hours and 20 mins.
hey so i don't have to worry about these kinds of problems?
i keep getting 15-17 on qr and rc on practice tests.
Is there a chance for me to get 20s in qr and rc in the real one?
 
jkh1886 said:
Therefore, the sum of the numbers is... Sum=(n+1)(n/2)
Yep. That's it.

jkh1886 said:
I'll let you think about how to solve the question when you have to deal with 1000 hrs. It involves this kind of thinking, and I'll only hint to it for now....
and beware, there's one trap when you apply it to this problem!
😡 😀 😛

OP: Sorry if I panicked you. I was just trying to bring up a way that helps solve the problem under a general condition. The ways jkh1886 and aggie did are perfectly fine. As the poster mentioned, quite often, you don't really have to do the calculation. This also applies to probability. To be able to score 20 or higher on QR, you need to be fast and determined to skip questions. If you see something that you think will take you more than 1 minute, just mark and skip. There are many questions later on that are extremely simple. Good Luck!
 
keibee82 said:
yea the answer is 3 hours and 20 mins.
hey so i don't have to worry about these kinds of problems?
i keep getting 15-17 on qr and rc on practice tests.
Is there a chance for me to get 20s in qr and rc in the real one?


You don't have to worry about the more complicated situation like, 20 hrs, or 1000 hrs. The 10 hr question's fair game, althou a bit towards the tricky side. It'll be the hardest of the hard questions. The thing to learn from this is... you don't have to calculate every question. Do what it takes to get the correct answer.
 
I've never missed a QR question on a real or practice DAT test and I've never come close to running out of time.

I'd rather go through all the steps and know that I'm right then take shortcuts or estimate and possibly get it wrong.

That's just me, I guess.
 
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