Sirish123 wrote:
grenico wrote:
Answer is A.
The time it takes Machine A to make \(w\) widgets is 6 days, so to make \(2w\) widgets it would take 12 days.
I used algebra to solve this one and it's a pain to do, but will write it out if someone needs help with explanation
Can you please explain this?
So this boils down to systems of equations and manipulating them.
We are given the times it takes both the machines to make \(w\) widgets. Let's assign some variables:
Let \(x\) be the rate of Machine A
Let \(y\) be the rate of Machine B
Let \(t\) be the time it takes Machine B to create \(w\) widgets
If it takes 2 days longer for Machine A to produce \(w\) widgets, then it takes Machine A \(t+2\) days to create them.
Putting these together, we get:
\(w = x(t+2)\)
\(w = yt\)
Additionally, we know that working together, they create \(\frac{5}{4}\)\(w\) widgets in 3 days, so that means:
\(\frac{5}{4}\)\(w = (x+y)(3)\)
Let's isolate the right side of this equation:
\(\frac{5}{12}\)\(w = (x+y)\)
What we should do now is take advantage of the first two equations we have so we can make the entire equation consist of only one variable, namely \(x\), since we're looking for the time it takes Machine A to create widgets. In other words, we're looking for \(t+2\), so lets replace \(w\) and \(y\) with \(x\) wherever we can.
So let's look at the first two equations:
\(w = x(t+2)\)
\(w = yt\)
Since both are equal to \(w\), let's set them equal to each other and solve for \(y\) so we can sub in the value for it in the third equation above:
\(yt = x(t+2)\)
\(y = \)\(\frac{{x(t+2)}}{t}\)
Subbing this value for \(y\) into the third equation, we get:
\(\frac{5}{12}\)\(w = x +\)\(\frac{{x(t+2)}}{t}\)
Now we know that:
\(w = x(t+2)\)
So let's sub this in for \(w\):
\(\frac{5}{12}\)\(x(t+2) = x +\)\(\frac{{x(t+2)}}{t}\)
Now we have everything in terms of \(x\), so we can begin to solve.
It may not be obvious at first glance because this has become one giant mess, but notice how we have \(x(t+2)\) on both sides. Let's put them on the same side:
\(\frac{5}{12}\)\(x(t+2) - \)\(\frac{{x(t+2)}}{t}\) = \(x\)
We can now factor out the \(x(t+2)\) from the left side:
\(x(t+2)(\)\(\frac{5}{12}\)\(-\)\(\frac{1}{t}\)\()\) = \(x\)
Now we can divide both sides by \(x\):
\((t+2)(\)\(\frac{5}{12}\)\(-\)\(\frac{1}{t}\)\()\) = \(1\)
And simplify the fraction on the left side:
\((t+2)(\)\(\frac{{5t-12}}{12t}\)\()\) = \(1\)
Multiply both sides by \(12t\):
\((t+2)(5t-12) = 12t\)
Now we solve for \(t\):
\(5t^2 -12t + 10t - 24 = 12t\)
\(5t^2 -14t - 24 = 0\)
Factoring this results in:
\((5t + 6)(t - 4)\)
Which means \(t = -\)\(\frac{6}{5}\) or \(t = 4\)
Negative days don't make sense, so we know that \(t = 4\).
Now that we know \(t\), we can solve the problem.
Recall that:
\(w = x(t+2)\)
Subbing in 4 for \(t\) gives:
\(w = x(4+2)\)
\(w = x(6)\)
Multiply both sides by 2:
\(2w = x(12)\)
So it takes Machine A 12 days to make \(2w\) widgets.
Therefore the answer is A