10 factories, working for h hours per day, can produce P units of cars

Question

10 factories, working for h hours per day, can produce P units of cars in every 5 days. Each factory produces car at the same constant rate. What is the additional time taken by the 10 factories to produce 2P units of cars, if they work for h/2 hours per day instead of h hours per day?

A. 0
B. 2
C. 5
D. 10
E. 20

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sj296 · Answer

Assuming the rate of producing the car = k

10*5*h*k = p ........(1)

10*(h/2)*x*k = 2P.....(2)

Equiation (1)/(2)

x = 20.

time taken = 20

Additional time = 20-10 = 10

bv8562 · Answer

# of factories * Rate * Time = Work

# of factories = 10
Rate = r
Total Time = h  \frac{hours}{day}  * 5 days = 5h hours
Work = P

10*r*5h = P
 r=\frac{P}{50h}

Let the additional time be = x days

Since it takes 10 factories 5 days to produce P cars, it must take them 10 days (twice) to produce 2P cars at the same rate each working h hours per day.

# of factories * Rate * Time = Work

10 * \frac{P}{50h} * \frac{h}{2}(10+x) = 2P

x = 10 days (D)

singh.00 · Answer

From where did we get the value of 10 in {(h/2)(10+x)}?

Posted from my mobile device

bv8562 · Answer

singh.00  Hey, (10+x) is correct because we are asked for 2P cars and NOT P cars. Since it takes 10 factories 5 days to produce P cars, it must take them 10 days (twice) to produce 2P cars at the same rate working h hours per day. This is how we got 10 in (10+x).

10 factories, working for h hours per day, can produce P units of cars

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10 factories, working for h hours per day, can produce P units of cars

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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My Rewards