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I like the solution - it’s helpful.
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I like the solution - it’s helpful.
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I've come up with a different solution.

First, we convert the task to hours and assign a target (value) of 1200 items to be completed by each machine. I chose 1200 because it's a multiple of both 600 and 300:

Machine A: 1200 items in 10 hours → 1200 items in 600 minutes → 120 items per hour or 2 items per minute

Machine B: 1200 items in 5 hours → 1200 items in 300 minutes → 240 items per hour or 4 items per minute

Now, let's break down the timeline:

1st hour (only A working): 120 items

2nd hour (only A working): 120 items

3rd hour (both A and B working): A produces 120 items, B produces 240 items

4th hour (both A and B working): A produces 120 items, B produces 240 items

At this point, the total number of items produced is:

120 + 120 + (120 + 240) + (120 + 240) = 960 items

Now, during the additional 15 minutes when A, B, and C are all working, we calculate the output of each machine:

A → 15 minutes × 2 items/minute = 30 items

B → 15 minutes × 4 items/minute = 60 items

So, the total after 15 more minutes is:

960 items + 30 (A) + 60 (B) = 1050 items

To reach the target of 1200 items, we still need:

1200 - 1050 = 150 items

This means that C must produce 150 items in 15 minutes, which is equivalent to:

150 ÷ 15 = 10 items per minute

Which means 600 items per hour

So, to produce 1200 items, C alone would take 2 hours.

Answer: The value of C is 2.

I hope this helps and makes the logic easier to understand — it’s much clearer to me when I think about it this way.
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This is a great question that’s helpful for learning and I like the solution - it’s helpful.
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Great question. What is the fastest way to do these type of questions?
How can we use Answer options in this case to eliminate the wrong one?
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1111fate
Great question. What is the fastest way to do these type of questions?
How can we use Answer options in this case to eliminate the wrong one?

The fastest way for me is exactly the one shown in the official solution. Other people may find a different method faster, and several good approaches are already posted in this thread.
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I like the solution - it’s helpful.
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