A factory has two machines. Machine X can produce 200 pins per hour an

Question

A factory has two machines. Machine X can produce 200 pins per hour and machine Y can produce 500 pins per hour. At least one machine is in operation throughout the 8-hour work day. The factory must produce 5,000 pins every day. What is the least possible number of hours that machines X and Y must work together on each day?

A. 4 
B. 5 
C. 6 
D. 7 
E. 8

MidhilaMohan · Accepted Answer

Machine x- 200 pins/hr
Machine y-500 pins/hr

they will run least together when machine with high output works the most

therefore, in 8 hrs Machine y will produce 500x8=4000 pins
 remaining pins= 5000- 4000=1000pins

Machine x will have to work for= 1000/200=5 hrs.

hence option B.

jaykpatel21 · Answer

A+B working together = 700 pins/hr
We need minimum number of both machines working together, so we need to maximize the period of 8 hours and produce as close to as 5000 pins.
Option 1 : 7 hours work : 7*700= 4900 + MachineX/Y any to complete anonther 100
Option 2 : 6 hours work 
7*600 = 4200 + Machine Y for another 800
Option 3 : 5 hours work : 
7*500 = 3500 + Machine Y for another 1500 (3 hrs) 
This 5 hours is the correct answer. If we go to 4 hours we can maximum 4800 and not 5000

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Archit3110 · Answer

question needs to be understood well before solving as i found it to be a bit confusing;

given ; At least one machine is in operation throughout the 8-hour work day
since we need to find the least together working hours so so let the machine Y with highest rate be in operation for 8 hrs ; it will produce ; 8*500; 4000 pins
left with 5000-4000; 1000 pins ; since contribution of both machines is required so machine x will take ; 1000/200; 5 hrs 
now see wording 
what is the  least possible number of hours that machines X and Y must work together on each day 
the least hours worked by both machine will be 5 hrs ; but total hours would be 13 .. 
IMO B

MathRevolution · Answer

=>

In order to minimize the number of hours that the two machines work together, the more efficient machine  Y  should work for the entire day.

Thus,  Y  works for  8  hours and produces  8 * 500 = 4,000  pins.

There are  1,000  pins remaining for machine  X  to produce. This takes  \frac{1,000}{200} = 5 hours.

Therefore, B is the answer. 
Answer: B

manasmridulmitra · Answer

700*n + 500*(8-n) = 5000
700n + 4000 - 500n = 5000
200n = 1000
n=5

p.s i used this method to solve the question within 20 seconds.

Shwetabh54 · Answer

Better to approach through options as we have to find least time and options are in ascending order.

Rate X - 200, Rate Y - 500, Rate X+Y= 700. 
Total work - 5000

Option 1 - 4 hours
Total pins in 4 hours working together = (700 * 4)= 2800
Remaining pins = 5000 - 2800 = 2200
Remaining hours = 4 
we can't produce 2200 pins by working X & Y alone for 4 hours (X in 4 hours will produce 2000 and Y will 800)

Option 2 - 5 hours
Total pins in 5 hours working together = (700 * 5)= 3500
Remaining pins = 5000 - 3500 = 1500
Remaining hours = 3
We can produce 1500 pins by working Y for 3 hours. Hence the option is perfect.

A factory has two machines. Machine X can produce 200 pins per hour an

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A factory has two machines. Machine X can produce 200 pins per hour an

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