Machine A, Machine B, and Machine C can complete a task in 5, 10, and

Question

Machine A, Machine B, and Machine C can complete a task in 5, 10, and 20 days, respectively. How many Machine Cs must assist 2 Machine As and 4 Machine Bs for them to complete the task in 1 day?

A. 1
B. 2
C. 3
D. 4
E. 8

Alamin4you · Answer

Total= 60 unit
Lcm of 5, 10, 20

Efficiency of
A= 60/5 = 12 unit
B = 6 unit
C= 3 unit

Atq, (2*12+4*6)+x*3 =60
X=4✅✅

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gmatophobia · Answer

Let's assume that the total task = 20 units.

Rate of Machine A = 20/5 = 4 units / day 
 Rate of Machine B = 20/10 = 2 units / day 
 Rate of Machine C = 20/20 = 1 units / day

As the task needs to be completed in 1 day, 20 units of work must be completed by the three machines in 1 day.

Units of work that can be done by 2 Machine As and 4 Machine Bs = 2*4 + 4*2 = 16 units.

The remaining (20 - 16 =) 4 units of work must be done by Machine C. As each Machine C does 1 unit of work per day, we need 4 Machine Cs to complete the work.

Option D

Archit3110 · Answer

combined rate
1/5 + 1/10+1/20 = 7/20
with count of machines
2/5+4/10+x/20 = 1
x = 20-16 ; 4
option D

coucoudanger · Answer

2 machines A + 4 machines B + x machines C will complete the task in 1 day.

machine A rate :  \frac{1}{5}  ; 2 machines A rate :  \frac{2}{5}  
machine B rate :  \frac{1}{10}  ; 4 machines B rate :  \frac{4}{10}  
machine C rate :  \frac{1}{20}  ; x machines C rate :  \frac{x}{20}

Working together they will complete the job in one day :

\frac{2}{5}+\frac{4}{10}+\frac{x}{20} = 1

x=4

Answer : D

Abhishek009 · Answer

Let the total work be 20 units (LCM of 5, 10 & 20)

Efficiency of machine A = 4 Units/Day ; Efficiency of machine B = 2 Units/Day & Efficiency of machine C = 1 Units/Day
Work done by 2 Machine As and 4 Machine Bs in a day is -

2*4 + 4*2 + c*1= 20

Or,  8 + 8 + c = 20

Or,  c = 4 , Answer must hence be (D)

JeffTargetTestPrep · Answer

Rate of Machine A = 1/5 task/day
Rate of Machine B = 1/10 task/day
Rate of Machine C = 1/20 task/day

2 Machines A, 4 Machines B, and k Machines C together:

Rate = 2/5 + 4/10 + k/20 = (16 + k)/20 task/day
Time = 1 day
Work = 1 task

Rate × Time = Work

(16 + k)/20 × 1 = 1
16 + k = 20
k = 4

Answer: D

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