Machine A and Machine B can produce 1 widget in 3 hours work

It should be the second D that I think is supposed to be E. I've attached my work in a spreadsheet

the quickest way to solve this problem is to know the following shortcuts ..

If machine A and B work together, then:
1 hour = (A+B)/AB of work done ..... (1)
AB/(A+B) hour = 1 job done ..... (2)

the questions discusses time, so we'll use (1) equation. plug in the values.

(a+b)/ab = 3
(a/2+b)/(a/2*b) = 2 ....... [the speed is doubled so the time is halved]

solve the equations and you'll get a=6 hrs

My way of doing it:
Check all the times given in Question-3 hr and 2 hr - take LCM = 6; SO 6 is the total units of work to be done.
W=6 units
now , a+b = 6units/3hr= 2u/hr -(I) (work done by a and b together in 1 hr)

with double speed of a:

2a+b=6u/2hr= 3u/hr (II)

by I & II a=1 units per hour -> so total time taken to complete the full work is 6*1 (6 units * 1 unit per hour) = 6 hours is the answer.

NOTE: This method helps to solve the problem orally !

My attempt:

Given rate at which A & B works at normal pace to complete 1 widget is (1/3).

Hence A's rate = B's rate = half of (1/3).

Hence A's rate is (1/6), so to complete 1 widget A requires 6 hours.

Any thoughts ?????

lol I put it in a spreadsheet... what a nerd I am. I forgot that I did that.

hi
I dont think it wil give u a correct result everytime ..
I dont think 1/6+1/6= 1/3 ( where in A and B rate of work is same )

however these speeds may vary ad yet the totalmay be 1/3....not sure If i have explained u??

I understand your point. A's rate could be 1/12 and B's rate be 1/4 but still working together they could end up with a combined rate of 1/3.

I believe the key mistake of my approach is not understanding the key part of the question -- "[highlight]working together at their respective constant rates[/highlight]"

Thanks for pointing this and correcting me. +1 from me.

I just wanted to point out that answers A, B, and C don't even make sense.

Bunuel

can you please help vvith this problem I could not understand this

please explain

crack700 already gave you the correct answer (6 is correct, so the answer is E and not D).

The two equations are:

\(\frac{1}{A}+\frac{1}{B}=\frac{1}{3}\)

\(\frac{2}{A}+\frac{1}{B}=\frac{1}{2}\)

Subtract the first equation from the second. You obtain \(\frac{1}{A}=\frac{1}{6}\) , so \(A=6.\)

Answer E

\(\frac{1}{A}+\frac{1}{B}=\frac{1}{3}\)
\(\frac{2}{A}+\frac{1}{B}=\frac{1}{2}\)

Combine the two eq:

\(\frac{2}{A}-\frac{1}{A}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{A}=\frac{1}{6}\)

\(t=6\)

Hi All,

This question is a bit more complex than a typical Work question, but you can still use the Work Formula to solve it.

Work = (A)(B)/(A+B) where A and B are the speeds of the two individual machines

From the prompt, we know that Machine A and Machine B, working together, can produce 1 widget in 3 hours. This is the same as saying "it takes 3 hours to complete 1 job."

Using the Work Formula, we have....

(A)(B)/(A+B) = 3

AB = 3A + 3B

Next, we're told that if Machine A's speed were DOUBLED, then the two machines would need 2 hours to produce 1 widget. Mathematically, doubling Machine A's speed means that we have to refer to it as A/2 (if the original speed is 1 widget every 10 hours, then DOUBLING that speed means 1 widget every 5 hours.....thus A becomes A/2).

Using the Work Formula, we have....

(A/2)(B)/(A/2 + B) = 2

(AB)/2 = A + 2B
AB = 2A + 4B

Now we have two variables and two equations. Both equations are set equal to "AB", so we have....

3A + 3B = 2A + 4B
A = B

This tells us that the original speeds of both machines are the SAME. Going back to the original formula, we can substitute in the value of "B" which gives us....

AB = 3A + 3B

A(A) = 3A + 3(A)

A^2 = 6A

A^2 - 6A = 0
A(A-6) = 0

Since a machine cannot have a rate of 0, Machine A's rate must be 1 unit per 6 hours.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hi guys, here is my solution, please take a look and let me know if this is a correct way to think:

Let A,B be rates of machines A,B

3 hours*A + 3 hours*B = 1 widget
or
3A + 3B = 1

2 hours*2*A + 2 hours*B = 1 widget
or
4A + 2B = 1

Subtract the two equations:

A-B = 0 => A = B

Plug back in:

3A + 3A = 1, A = 1/6

Therefore A takes 6 hours working at its rate of 1/6 to make 1 widget

Test answer choices, three of which are out immediately. A and B at their normal rates take 3 hours to finish one job together.

Eliminate answers A, B, and C.
They mean that machine A works faster alone than it does with Machine B.*

Given the times A and B take together to finish (3 and 2 hours), Answer E, a multiple of both times, makes more sense to test first.

Answer E) 6 hours

Per (E), machine A currently takes 6 hours to finish the job on its own. A's rate = \(\frac{1}{6}\). B's rate?

\((\frac{1}{A}+\frac{1}{B})=\frac{1}{3}\)

\((\frac{1}{6}+\frac{1}{B})=\frac{1}{3}\)

\(\frac{1}{B}=(\frac{1}{3}-\frac{1}{6})=\frac{1}{6}\)

A's rate = B's rate = \(\frac{1}{6}\)? Use the second scenario

A's speed doubles. Rate IS speed. A's original rate, doubled:
\((\frac{1job}{6hrs}*2)=\frac{2jobs}{6hrs}=\frac{1job}{3hrs}\)

A's rate now (A\(_2\)) = \(\frac{1}{3}\)
B's rate still = \(\frac{1}{6}\)
Together, given faster rate A\(_2\), they should = \(\frac{1}{2}\)

\((\frac{1}{3}+\\
\frac{1}{6})=\frac{3}{6}=\frac{1}{2}\)

That's correct

Answer E

* At A's current rate, A and B working together take 3 hours to finish. The first three answers mean that A takes \(\leq{3}\) hours by itself. Not possible. B cannot make a negative number of widgets. Nor, per option C, can B make 0 widgets. When A's speed doubles, A's time is cut in half. If B = 0, then A would finish in \(\frac{3}{2}\) hours -- not, as prompt says, in 2 hours.

We can let a and b, respectively, be the time, in hours, it takes machines A and B to produce 1 widget on their own. The current rate equation is::

1/a + 1/b = 1/3

If Machine A’s rate is doubled, the new rate equation is:

2/a + 1/b = 1/2

Subtracting the first equation from the second, the 1/b terms cancel, so we have:

2/a - 1/a = 1/2 - 1/3

1/a = 1/6

a = 6

Answer: E

Solution

Method 1: Using a for rate (Faster), Need: 1/a

Name, R, T, W
A, a, 1/a, 1
B, b, 1/b 1
A+B, 1/3, 3, 1
2A+B, 1/2, 2, 1

Eq 1: a + b = 1/3 => 3a + 3b = 1 [x2]
Eq 2: 2a + b = 1/2 => 4a + 2b = 1 [x3]
Using elimination: -6a = -1 => a = 1/6 => 1/a = 6

Method 2: Using a for time (Slower), Need: a

Name, R, T, W
A, 1/a, a, 1
B, 1/b, b 1
A+B, 1/3, 3, 1
2A+B, 1/2, 2, 1

Eq 1: 1/2 = 2/a + 1/b => 1/2 = (2b+a)/ab => ab = 4b + 2a
Eq 2: 1/3 = 1/a + 1/b => 1/3 = (b + a)/ab => ab = 3b + 3a
4b + 2a = 3b + 3a => a = b
a^2 = 3a + 3a => a^2 = 6a => a(a-6) = 0 => a=6 (a can’t be 0).

Veritas Prep Official Solution

Tricky, eh? It is a little cumbersome if you get into variables. If you just try to reason it out, it could be done rather quickly and easily. Let’s see!

Machine A and B together complete 1 work in 3 hrs i.e. together, they do 1/3rd work every hour.

If machine A’s speed were double, they would do 1/2 work in 1 hour together. How come they do (1/2 – 1/3 =) 1/6th work extra in 1 hour now? Because machine A’s speed is double the previous speed. The extra speed that machine A has allows it to do 1/6th work extra. This means, at normal speed, machine A used to do 1/6 work in an hour (its speed had doubled so work had doubled too). Hence, at usual speed, it will take 6 hrs to produce 1 widget.

ANSWER: E