Machine A and Machine B - XYZ

[asharma8080]

Machine A, working alone at a constant rate, can complete a certain production lot in x hours. Machine B, working alone at a constant rate, can complete 1/5 of the same production lot in y hours. Machines A and B, working together, can complete 1/2 of the same production lot in z hours. What is the value of y in terms of x and z?

A. (5x - 10z)/ 2xz
B (2xz) / (5x - 10z)
C. (5xz)/ (x + z)
D. xz/ x + z
E. xz/ x + 2z

I started with the algebra approach and it started to get ugly and then paused but then I could not decide whether to plug-in here or not and ended up wasting valuable time in trying to make that decision. Is there something in this problem that tells us that plugging is going to suck?

I ended up seeing the algebraic solution afterwards and have no problems with that approach.

[jnelson0612]

Machine A, working alone at a constant rate, can complete a certain production lot in x hours. Machine B, working alone at a constant rate, can complete 1/5 of the same production lot in y hours. Machines A and B, working together, can complete 1/2 of the same production lot in z hours. What is the value of y in terms of x and z?

A. (5x - 10z)/ 2xz
B (2xz) / (5x - 10z)
C. (5xz)/ (x + z)
D. xz/ x + z
E. xz/ x + 2z

I started with the algebra approach and it started to get ugly and then paused but then I could not decide whether to plug-in here or not and ended up wasting valuable time in trying to make that decision. Is there something in this problem that tells us that plugging is going to suck?

I ended up seeing the algebraic solution afterwards and have no problems with that approach.

See, I don't think that this problem has to suck by plugging numbers. :-) I just did it with a tutoring student.

Here's what we did:
1) We decided that the job would be 30 widgets.
2) We decided that x=6, so A completes the job in six hours, or builds 5 widgets per hour.
3) We decided that y=2, so B completes 1/5 of the job in 2 hours. 1/5 of the job is 6 widgets, so 6 widgets/2 hours = 3 widgets.
4) We now know that together A + B complete 8 widgets per hour. The problem says they can do half the lot (15 widgets) in z hours. Thus rate (8 widgets) * time = work (15 widgets). Time is 15/8. z=15/8
5) Now, the question asks what is y in terms of x and z? y=2 and x=6 and z=less than 2.
6) Estimate. C is too big. A and E are too small. Run the actual numbers in B and D. Run D first; it's easier and if it doesn't work you can just pick B and move on. B is in fact the answer and checks out perfectly.

It's certainly a little more time consuming as a plug in question, but not impossible. :-)

[nocheivyirene]

Machine A, working alone at a constant rate, can complete a certain production lot in x hours. Machine B, working alone at a constant rate, can complete 1/5 of the same production lot in y hours. Machines A and B, working together, can complete 1/2 of the same production lot in z hours. What is the value of y in terms of x and z?

A. (5x - 10z)/ 2xz
B (2xz) / (5x - 10z)
C. (5xz)/ (x + z)
D. xz/ x + z
E. xz/ x + 2z

My approach:
Let production = 1
Rate of A + Rate of B = Rate Together

1/x + 1/5y = 1/2z

Separate 1/5y since we want y.

1/5y = 1/2z - 1/x
1/5y = (x-2z)/2xz

Multiply both sides by 5

1/y = 5x - 10z / 2xz
y = 2xz / (5x - 10z)

Answer: B

[tim]

:)

[krishnagmat]

It took me about 3 minues to solve this problem but I thought it can be simply done in few seconds by just substituting values.

Eg, let say Z=X , that means only A performs the entire task. The remaining time should be Zero.

In some questions we may consider few other scenarios like this.

[tim]

Z=X won't work here given the specific conditions of the problem (see if you can figure out why!), but in general this type of approach is good to use on problems with variables in the answer choices.

[JbhB682]

Another approach -- took me 3.5 mins though

Assuming 1 production unit

Time for A : x for 1 production unit

Time for B :
-- y for 1/5 production unit
-- 5 y for 1 production unit

so time for B for 1 production unit -- 5 y

Combined times for 1 production unit : product of times of A and B / sum of times of A and B

So thats (x times 5y) / (x + 5y) for 1 production unit

Now z is half a production unit so we need to halve the above equation

z = (1/2) [5xy / (x+5y)]

re-arrange and isolate Y - you get B

Takes some time though and potential for error given the movement ...but right answer

[Sage Pearce-Higgins]