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3 employers rake the beach each day. working together, employees A and B can rake the beach in 3 hours, whereas A and C can rake the beach in 2.5 h. working together, can A,B, and C rake the beach in less than 2 h?
1. B rakes faster than A
2. working alone, C can rake the beach in less than 5 h
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I am getting C . I know I am wrong . But please correct me.
A B C are hours which they can do raking alone.
1/A + 1/B = 1/3 1/A + 1/C = 1/2.5 = 2/5
Stmnt 1. B is faster means 1/A < 1/6 Adding both the questions we have 1/A + 1/A + 1/B + 1/C = 1/3 + 2/5 1/A + 1/B + 1/C = 1/3 + 2/5 - 1/6 = ( 10 + 12 - 5 ) / 30 = 17/30 Work which can be done by all three 30/17 < 2 hours . So not suff .
you are right. "(10+12-5)/30 = 17/30" instead of "(15+6-5)/30 = 16/30"
ashkrs
I am getting C . I know I am wrong . But please correct me.
A B C are hours which they can do raking alone.
1/A + 1/B = 1/3 1/A + 1/C = 1/2.5 = 2/5
Stmnt 1. B is faster means 1/A < 1/6 Adding both the questions we have 1/A + 1/A + 1/B + 1/C = 1/3 + 2/5 1/A + 1/B + 1/C = 1/3 + 2/5 - 1/6 = ( 10 + 12 - 5 ) / 30 = 17/30 Work which can be done by all three 30/17 < 2 hours . So not suff .
Stmnt 2 - suff.
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1/t=1/A + 1/B + 1/C > 1/3 + 2/5 - 1/6 = ( 10 + 12 - 5 ) / 30 = 17/30 1/t>17/30 ==> t<30/17<2 working together, can A,B, and C rake the beach in less than 2 h? Yes. t<30/17
1. B rakes faster than A. if A and B were equal, then they would be 6 and 6. thus, B<6 and A >6. at the same time if A and C were equal they would be 5 (theri sum is 10). if A<6, C<4. let's say A and B together x. we know that XC/X+C is the total amount of time needed by the three. substitute any value <4 for C and 3 for x. we obtain only values <2. suff
2. C<5. it means A>5 and B<7. let's say A and C together y: yB/y+B is the formula for the three. substitute . we obtain only values <2. suff.
1/3 = 1/ta + 1/tb ( ta and tb are times of A and B) 1/2.5 = 1/ta + 1/tc
we need to get if tabc < 2hr
statement 1 : tb<ta assume any value for tb and find ta from above equation , putting that value in second equation we get tc .we can find tabc from 1/tabc = 1/ta + 1/tb +1/tc sufficient
3 employers rake the beach each day. working together, employees A and B can rake the beach in 3 hours, whereas A and C can rake the beach in 2.5 h. working together, can A,B, and C rake the beach in less than 2 h?
1. B rakes faster than A 2. working alone, C can rake the beach in less than 5 h
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1/A+1/B=1/3
1/A+1/C=2/3
The best way to work these problems is to take the "worst scenario or most extreme scenario" case.
1: B is faster than A. This means that B dsnt equal A. Lets make B equal to A just to get a bearing. (Wel make them equal b/c we can say B is just slightly bigger than A, but the difference is so small its negligable, hence our extreme scenario).
So 1/6+1/6=1/3 --->
Lets find C: Lets say A is 1/6 ---> 2/3 - 1/6: So C is 1/2. This is our worst case scenario: we can see that C will be around 1/2 or larger. Thus we can definitively say that the job will be done in less than 2 hours.
2: Lets say that C is 1/5.
thus A is 7/15. Again we can say that the job will be done in less than 2 hours.
D
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.