Machine A and machine B process the same work at different rates. Mac

That is better approach to calculate work done by machine instead of capacity of machine.. Kudos.

Answer = B = 17 Minutes

A .............................. B ........................... C ............................ D

\(\frac{1}{a}\) .......................... \(\frac{1}{b}\) ......................... \(\frac{a+b}{ab}\) ..................... \(\frac{3(a+b)}{ab}\) ............... Rates

Given that \(4 * \frac{1}{b} = \frac{3(a+b)}{ab}\)

a = 3b

Re-writing all rates in terms of b

\(\frac{1}{3b}\) ........................... \(\frac{1}{b}\) ....................... \(\frac{4}{3b}\) ............................ \(\frac{4}{b}\)

Let the work done by A = 1

\(\frac{1}{3b} * 340 = 1\)

\(b = \frac{340}{3}\) ....................... (1)

Combined rate of all machines\(= \frac{1}{3b} + \frac{1}{b} + \frac{4}{3b} + \frac{4}{b} = \frac{20}{3b}\)

Placing value of b from (1)

Combined rate \(= \frac{1}{17}\)

Time required = 17 Minutes

The answer is B

A,B,C=A+B, D=3(A+B) orD=4B

A =Ra*340=1
Ra=1/340
D=3A+3B=4B
3A=B
3/340=B
C=1/340+3/340=4/340
D=4B=4(3/340)=12/340

so working together =A+B+C+D*T=1
1/340+3/340+4/340+12/340*T=1
T=340/20=17 minutes

Great and simple solution.

I had trouble putting all the rates into terms of A, this made it simple.

Posted from my mobile device

hi bunnuel can you please explain the same with rate of a taken as 1/a

then am getting 3b=a

Machine A processes work in 5 hours 40 mins = 5x60 + 40 = 340 Mins

i.e. Machine A processes in 1 mins = 1/340 work

Let Machine B processes in 1 min = 1/b work

i.e. The Work of Machine C in 1 Min = Work of A and B together in 1 min = (1/340)+(1/b)

i.e. The Work of Machine D in 1 Min = 3*Work of C in 1 min = 3*[(1/340)+(1/b)]

But, Machine D’s work rate is also exactly four times Machine B’s rate

i.e. The Work of Machine D in 1 Min = 4* (1/b) = 3*[(1/340)+(1/b)]

i.e. 4/b = 3*[(b+340)/340*b]

i.e. (4/b) * (340*b) = 3b + 1020

i.e. 1360 = 3b + 1020
i.e. b = 340/3

i.e. B's 1 Min work = 3/340
i.e. C's 1 Min work = (3/340)+(1/340) = 4/340
i.e. D's 1 Min work = 3*(4/340) = 12/340

Work of A, B, C and D combined in 1 min = (1/340) + (3/340) + (4/340) + (12/340) = (20/340)

i.e. Time taken by A, B, C and D working together = 340/20 mins = 17 mins

Answer: option

P.S.Your Mistake is you are getting 3b = a whereas the CORRECT relation is b=3a

a pretty basic question, yet with a lot of information...

C=A+B
D=4B=3(A+B) = 3A+3B -> B=3A.
thus, we have:
A
B=3A
C=4A
D=12A
total: 20A - combined rate.

time taken by A alone to finish: 5h40m -> i think it's better to work with minutes...340 minutes.
Rate for A is thus 1/340
total rate A+B+C+D = 20A
20/340 or 2/34 or 1/17
this means that all machines, combined, do 1 job in 17 minutes.

the answer must be B.

Since Machine A’s time to complete the job is 5 hours and 40 minutes, or 5 ⅔ = 17/3 hours, its rate is 3/17. We can let Machines B’s rate = b and thus, Machine C’s rate = 3/17 + b and Machine D’s rate = 3(3/17 + b) = 9/17 + 3b. We are also given that Machine D’s rate = 4b. Thus we have:

9/17 + 3b = 4b

9/17 = b

Now, we can determine that C’s rate = 3/17 + 9/17 = 12/17 and D’s rate = 4(9/17) = 36/17.

Thus, if we let x = the time, in hours, for the 4 machines to work together to complete the job, we have:

(3/17)x + (9/17)x + (12/17)x + (36/17)x = 1

[(3 + 9 + 12 + 36)/17]x = 1

(60/17)x = 1

x = 1/(60/17) = 17/60

Thus it takes 17/60 hours, or 17 minutes to complete the job when 4 machines work together.

Answer: B

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