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working together printer A and B can finish a job in 24 min. [#permalink]
06 Oct 2005, 18:23

1

This post received KUDOS

working together printer A and B can finish a job in 24 min. Printer A working alone can do the job in 60 mins. How many pages does the task contain, if printer B can print 5 pages a minute more than A.

600
800
1000
1200
1500

Folks please provide explanation...just posting answers isnt helpful...

Let x be the total number of pages.
The number of pages A prints in 1 min is x/60
--->the number of pages B prints in 1 min is x/60+5
A and B work together in 24 min to finish the task, thus:
24( x/60+x/60+5) = x ----> x=600 pages.

Re: tricky work problem from challenge 23 [#permalink]
06 Oct 2005, 18:48

A can do the same job in 60 minuets.
A can do 1/60 job in 1 minuet.

B can do the same job in 40 minuets.
B can do 1/40 job in 1 minuet.

so, B can do (1/40)-(1/60) or 1/120 job in 1 minuet
B can do 1/120 of whole job, which is 5 more pages paint, in 1 minuet
so the whole job contain 600 pages.

Verification:
A can paint 10 (600/60) pages in 1 minuet.
B can paint 15 (600/40) pages in 1 minuet.

so the difference between B's work and A's work in 1 minuet is 5 (15-10) pages.

Re: tricky work problem from challenge 23 [#permalink]
06 Oct 2005, 18:59

HIMA
could you explain this in more detail, why are you subtracting As rate from B?

so, B can do (1/40)-(1/60) or 1/120 job in 1 minuet
B can do 1/120 of whole job, which is 5 more pages paint, in 1 minuet
so the whole job contain 600 pages.

i

HIMALAYA wrote:

A can do the same job in 60 minuets. A can do 1/60 job in 1 minuet.

B can do the same job in 40 minuets. B can do 1/40 job in 1 minuet.

so, B can do (1/40)-(1/60) or 1/120 job in 1 minuet B can do 1/120 of whole job, which is 5 more pages paint, in 1 minuet so the whole job contain 600 pages.

Verification: A can paint 10 (600/60) pages in 1 minuet. B can paint 15 (600/40) pages in 1 minuet.

so the difference between B's work and A's work in 1 minuet is 5 (15-10) pages.

Re: tricky work problem from challenge 23 [#permalink]
06 Oct 2005, 19:24

we know: the comparision is of apple with apple and orange with orange, right.

so to have correct comparsion, we need to find the rate of work done by B in 1 minuet with the same of the A in the same minuet. so find the rate of work/minuet for both, A and B.

so find the rate of work A and B can do in 1 minuet.

lets suppose the whole job = x.

A can do the whole job (x) in 60 minuets.
or, it takes 60 minuets for A to complete the job (x).
if so, then in 1 inuet A can do 1/60 part of the whole job (x).

Same to B:
B can do the same job (x) in 40 minuets.
B can do 1/40 job (x) in 1 minuet.

here, we know that B does more work in 1 inuet than A does. So now get the excess work done by B, i.e. the difference between the work done by A and B in 1 minuet:

= {(1/40)-(1/60)} of the whole job (x)
= 1/120 of the whole job (x)

as given, B can paint 5 more pages than A in 1 minuet and we have from the above solution that B can do (1/120) (x) more work in 1 minuet than A. so,

5 pages = (1/120) x
x = 600 pages.
hope helpssssss................

fresinha12 wrote:

HIMA, could you explain this in more detail, why are you subtracting As rate from B?

so, B can do (1/40)-(1/60) or 1/120 job in 1 minuet B can do 1/120 of whole job, which is 5 more pages paint, in 1 minuet so the whole job contain 600 pages.

HIMALAYA wrote:

A can do the same job in 60 minuets. A can do 1/60 job in 1 minuet.

B can do the same job in 40 minuets. B can do 1/40 job in 1 minuet.

so, B can do (1/40)-(1/60) or 1/120 job in 1 minuet B can do 1/120 of whole job, which is 5 more pages paint, in 1 minuet so the whole job contain 600 pages.

Verification: A can paint 10 (600/60) pages in 1 minuet. B can paint 15 (600/40) pages in 1 minuet.

so the difference between B's work and A's work in 1 minuet is 5 (15-10) pages.

Re: tricky work problem from challenge 23 [#permalink]
06 Oct 2005, 22:14

fresinha12 wrote:

working together printer A and B can finish a job in 24 min. Printer A working alone can do the job in 60 mins. How many pages does the task contain, if printer B can print 5 pages a minute more than A.

600 800 1000 1200 1500

Folks please provide explanation...just posting answers isnt helpful...

I chose the answer from the answer choice provided. Since A can complete in 24 min and B in 60 min, I know the answer should be a multiple of 6, so eliminated 800 and 1000. I plucked in 1200 and found the combined effort to be greater than 24 minutes. So the answer has to be A, and verified this by substituting in the question as well.

Re: tricky work problem from challenge 23 [#permalink]
07 Oct 2005, 08:30

Brilliant....I was stuck there...thanks!

OA is 600

HIMALAYA wrote:

we know: the comparision is of apple with apple and orange with orange, right.

so to have correct comparsion, we need to find the rate of work done by B in 1 minuet with the same of the A in the same minuet. so find the rate of work/minuet for both, A and B.

so find the rate of work A and B can do in 1 minuet.

lets suppose the whole job = x.

A can do the whole job (x) in 60 minuets. or, it takes 60 minuets for A to complete the job (x). if so, then in 1 inuet A can do 1/60 part of the whole job (x).

Same to B: B can do the same job (x) in 40 minuets. B can do 1/40 job (x) in 1 minuet.

here, we know that B does more work in 1 inuet than A does. So now get the excess work done by B, i.e. the difference between the work done by A and B in 1 minuet:

= {(1/40)-(1/60)} of the whole job (x) = 1/120 of the whole job (x)

as given, B can paint 5 more pages than A in 1 minuet and we have from the above solution that B can do (1/120) (x) more work in 1 minuet than A. so,

5 pages = (1/120) x x = 600 pages. hope helpssssss................

fresinha12 wrote:

HIMA, could you explain this in more detail, why are you subtracting As rate from B?

so, B can do (1/40)-(1/60) or 1/120 job in 1 minuet B can do 1/120 of whole job, which is 5 more pages paint, in 1 minuet so the whole job contain 600 pages.

HIMALAYA wrote:

A can do the same job in 60 minuets. A can do 1/60 job in 1 minuet.

B can do the same job in 40 minuets. B can do 1/40 job in 1 minuet.

so, B can do (1/40)-(1/60) or 1/120 job in 1 minuet B can do 1/120 of whole job, which is 5 more pages paint, in 1 minuet so the whole job contain 600 pages.

Verification: A can paint 10 (600/60) pages in 1 minuet. B can paint 15 (600/40) pages in 1 minuet.

so the difference between B's work and A's work in 1 minuet is 5 (15-10) pages.

Let x be the total number of pages. The number of pages A prints in 1 min is x/60 --->the number of pages B prints in 1 min is x/60+5 A and B work together in 24 min to finish the task, thus: 24( x/60+x/60+5) = x ----> x=600 pages.

thanks. this was very quick. _________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Re: tricky work problem from challenge 23 [#permalink]
17 Dec 2011, 10:57

together in 1 min = 1/24 a in 1 min = 1/60 b in 1 min = 1/24 - 1/60 = 1/40 b did more than a in 1 minute = 1/40-1/60 = 1/120 so 1/120 part = 5 pages 1 part or whole part = 120x5 = 600 pages ans. A _________________

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