Running at their respective constant rates, Machine X takes 2 days

Back tracking method:
We have to find the number of days it takes Machine X alone to produce 2w widgets.
Answer options are as below:

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12

We are given that X and Y produce (5/4)w in 3 days and also given that X take 2 days more than Y.
So to find corresponding value of X and Y use the answer options.
First half the value of X to find number of days it takes Machine X to complete *W* widgets and find the value of Y by reducing 2 from all answer options.

--> X to complete W widgets
(A) X=2 and Y= 2-2 =0
(B) X=3 and Y= 3-2 =1
(C) X=4 and Y= 4-2 =2
(D) X=5 and Y= 5-2 =3
(E) X=6 and Y= 6-2 =4

Now we have all the values of X and Y. Find out the combination X and Y, whose values will produce (5/4)w for 3 days of work.

3(1/6+1/4)=>5/4-->Option (E) gives this value so correct answer is (E)-->12.

Hi,

I am new here and I have solved this math correctly. However, here's my problem:

If I consider the number of days required to produce 'W' widgets as 'X' and 'X-2' for X and Y respectively, I end up with the equation 5t^2 - 34t + 24 which yields me 2 solutions, one of which is correct (x=6)

However, when I take 'X' and 'Y' as 'X+2' and 'X', which are the same as above, because either ways, X is taking 2 days longer, I end up with the equation 5x^2 - 14x - 24 which has no solutions (I hope I'm not making some stupid mistake here)

Can someone please clarify this and correct me where I'm wrong.

Refer my post above; I did in the same way.....

As far as solving the equation, refer below.....

\(5x^2 - 14x - 24 = 0\)

\(5x^2 - 20x + 6x - 24 = 0\)

\(5x(x-4) + 6(x-4) = 0\)

(x-4)(5x+6) = 0

x = 4 (Ignore the -ve equation)

x+2 = 6

It doesn't matter whether you take the number of days as
Days taken by machine X = X and
Days taken by machine Y = X-2
or
Days taken by machine X = X+2
Days taken by machine Y = X

Either way, you will get one valid value for X and you will need to manipulate that to get your answer. You need to find the number of days that will be taken by machine X to make 2w widgets.

In first case, you get the equation as \(5X^2 - 34X + 24 = 0\) and get X = 6 as the only valid value (X = 4/5 gives X-2 as negative but number of days cannot be negative).
Machine X takes X days i.e. 6 days to make w widgets. So it will take 12 days to make 2w widgets.

In the second case, you get \(5x^2 - 14x - 24 = 0\)
which is \(5x^2 - 20x + 6x - 24 = 0\)
5x(x - 4) + 6(x - 4) = 0
(x-4)(5x + 6) = 0
x = 4
This is the number of days taken by machine Y to make w widgets. So machine X takes 4+2 = 6 days to make w widgets. It will take 12 days to make 2w widgets.

Either way, your answer will be the same.

Just wanted to point out that this problem can be solved quite easily without solving a quadratic.

Our equation for x is given by:

\(\frac{1}{x} + \frac{1}{x-2} = \frac{5}{12}\).

\(\frac{x-2 + x} {x(x-2)} = \frac{5}{12}\).

\(\frac{2x-2}{x(x-2)} = \frac{5}{12}\).

\(\frac{2(x-1)}{x(x-2)} = \frac{5}{12}\).

\(\frac{(x-1)}{x(x-2)} = \frac{5}{24}\).

We see immediately that by letting \(x=6\) we find our solution of x. Therefore, since we're looking for the value of 2x, we see that the answer must be 12.

Answer: E

can someone pelase explain how they got from (1/t) + (1/((t-2)) = 5/12
TO
5t^2-34t+24=0

You can also give the quadratic route a miss.
Once you get (1/t) + (1/(t-2)) = 5/12, you can try out some values for t. t is an integer since it is one of the given 5 options.

To get 12 in denominator on the right hand side, t*(t-2) should give you 12 or a multiple.
12 = 4*3 or 6*2 (not of the form t*(t-2))
24 = 6*4 (this is possible)

Check 1/6 + 1/4 = 10/24 = 5/12
You get the value of t.

Or you can also try substituting the value of t from each option.

Tough one! My recommended solution is plugging in 1 for w. Attached is a visual that should help.

1. Do all the calculations for 2w widgets
2. It takes x 4 days more to produce 2w widgets'
3. Together they take 4.8 days to produce 2w widgets
4. 1/y + 1/x=1/4.8 or 1/y + 1/(y+4) =5/24, y=8 . Therefore y+4=12.
X takes 12 days to produce 2w widgets

Running at their respective and constant rates, machine x takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4w widgets in 3 days, hoe many days would it take machine X aline to produce 2w widgets.

So far I have
X's rate = w/(d+2)
Y's rate = w/d

[w/d + w/(d+2)] * 3 = 5w/4

Not sure what to do with those fractions - how do you get a common denominator??

Strictly speaking, one doesn't need a formula for rates problems. The well-known formula works by expressing the amount of work each machine does in the *same* amount of time- one hour. If you know this, you can do any rates problem without a formula; just get the same time (rather than the same number of widgets) for each machine. Algebraically, this produces the same equation as the formula (though without the fractions), but is more flexible in case you encounter some kind of twist. Applied to this problem, it's still quite abstract, but it works:

Y produces w widgets in d days
X produces w widgets in d+2 days
X+Y produce w widgets in 12/5 days

Getting the same time for each:

Y produces dw + 2w widgets in d(d+2) days
X produces dw widgets in d(d+2) days
X+Y produce (5/12)*d*(d+2)*w widgets in d(d+2) days

So:
dw + 2w + dw = (5/12)*d*(d+2)*w
(24/5)w(d+1) = dw(d+2)
24d + 24 = 5d^2 + 10d
5d^2 - 14d - 24 = 0
(5d + 6)(d - 4) = 0

And since d > 0, we find d = 4.

The question asks us to find 2d+4, which is 12.

Let # days required for machine Y to produce w widgets be y days

Hence rate of machine Y = w/y widgets per day

Hence rate of machine X = w/y+2 widgets per day

given that both machines together can produce 5w/4 widgets in 3 days

Hence rate at which both machines working together = (5w/4)/3

We get w/(y+2) + w/y = (5w/4)/3

solving this for y we get, y = 4 days

Hence # of days machine X takes to produce w widgets = 4 + 2 = 6 days

Therefore # of days machine X takes to produce 2w widgets = 12 days


Answer E.



Thanks,
GyM

Not sure if it is weird, but I tried a different approach:

Rate of doing work by X = x widgets per day
Rate of doing work by Y = y widgets per day

According to the question,
Time taken by machine X to produce w widgets is 2 days more than time taken by machine Y to produce w widgets

time = (items produced)/(rate) { "time = distance/speed" analogy}

Therefore,
(w/x) = (w/y) + 2 ...... (1)

Also, working simultaneously, X and Y produce 5w/4 widgets in 3 days

i.e. 3(w/x) + 3(w/x) = 5w/4 ...... (2) {distance = speed * time}

Substituting (1) in (2):
(w/x) +(w/x) -2 = 5w/12
2(w/x) = (5w/12) + 2

Note that (2(w/x)) is the time taken by X to produce 2w widgets, which is what we need to find out.

Since all answers are in integers, we know that (5w/12) + 2 must be an integer
Therefore, w = {12, 24, 36, ...}

w = 12; 2(w/x) = 7 - Does not match with any option
w = 24; w(w/x) = 12 - Matches with (E)
All other options are less than 12, so not possible. Therefore, the correct answer is (E)

avigutman is there any way that we can solve this Question using Reasoning, all the procedures were an algebraic ways of solving

Because the data given says that one machine takes t days and another t + 2 days, there will be a variable and equation involved. When we have rates in multiples (say time taken is t days and 2t days), we can often make-do without any variable and equation.

kskumar Video solution from Quant Reasoning:
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Maybe its a super ignorant doubt but what am I doing wrong if I do the following -

If Y does W work - its rate is 1/W so the other one is 1/(W+2)

Giving us

1/W + 1/(W+2) = 5W/12

Not sure why this is throwing me off