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

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by BTGmoderatorDC » Sun Sep 23, 2018 2:51 am
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Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?

A. 1/2
B. 2
C. 3
D. 5
E. 6

OA E

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by GMATGuruNY » Sun Sep 23, 2018 3:06 am
BTGmoderatorDC wrote:Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?

A. 1/2
B. 2
C. 3
D. 5
E. 6
We can PLUG IN THE ANSWERS, which represent A's time to produce 1 widget.
Since A and B working together take 3 hours to produce 1 widget, A working alone must take MORE than 3 hours.
Eliminate A, B and C.

Let the value of each widget = 30 units.
Since A and B together take 3 hours to produce 1 widget, their combined rate = w/r = 30/3 = 10 units per hour.

Answer choice D: 5 hours for A alone
Since A alone takes 5 hours to produce 1 widget, A's rate alone = 30/5 = 6 units per hour.
Thus, B's rate alone = (combined rate for A and B) - (A's rate alone) = 10-6 = 4 units per hour.
When A's rate is doubled, the new combined rate for A and B = 2*6 + 4 = 16 units per hour.
Time to produce 1 widget when A's rate is doubled = w/r = 30/16 = 15/8 hours.
Since the required time is 2 hours, eliminate D.

The correct answer is E.

Answer choice E: 6 hours for A alone
Since A alone takes 6 hours to produce 1 widget, A's rate alone = 30/6 = 5 units per hour.
Thus, B's rate alone = (combined rate for A and B) - (A's rate alone) = 10-5 = 5 units per hour.
When A's rate is doubled, the new combined rate for A and B = 2*5 + 5 = 15 units per hour.
Time to produce 1 widget when A's rate is doubled = w/r = 30/15 = 2 hours.
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by GMATGuruNY » Sun Sep 23, 2018 3:30 am
Algebra:

Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates.
(A+B)(3) = 1 widget.
If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates.
(2A+B)(2) = 1 widget.
Since the amount of work is the same in each case, we get:
(A+B)(3) = (2A+B)(2)
3A+3B = 4A+2B
B=A.

Since A=B, A+B = 2A.
Since A+B take 3 hours to produce 1 widget, and A+B = 2A, the time for 2 type A machines to produce 1 widget is 3 hours.
Since one machine A will work 1/2 as fast, the time for one machine A must double to 6 hours.

The correct answer is E.
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Machine A and Machine B can produce 1 widget in 3 hours work

by fskilnik@GMATH » Sun Sep 23, 2018 7:32 am
BTGmoderatorDC wrote:Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?

A. 1/2
B. 2
C. 3
D. 5
E. 6

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$${A_{{\text{usual}}}}\,\,\, - \,\,\,1\,\,Job\,\,\, - \,\,\,\boxed{? = A}\,\,{\text{h}}$$
$${{\text{A}}_{{\text{doubled}}}} - 1\,\,Job\,\,\, - \,\,\,\frac{A}{2}\,\,{\text{h}}$$
$${B_{{\text{usual}}}}\,\,\, - \,\,\,1\,\,Job\,\,\, - \,\,\,B\,\,{\text{h}}$$
Let´s use the "work/rate shortcut":
$${1 \over {{T_{x \cup y}}}} = {1 \over {{T_x}}} + {1 \over {{T_y}}}$$
$$\left\{ \matrix{
\,\left( {\rm{I}} \right)\,\,\,\,{1 \over 3} = {1 \over A} + {1 \over B} \hfill \cr
\,\left( {{\rm{II}}} \right)\,\,\,{1 \over 2} = {2 \over A} + {1 \over B} \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {{\rm{II}}} \right) - \left( {\rm{I}} \right)} \,\,\,\,\,{1 \over 6} = {1 \over 2} - {1 \over 3} = {2 \over A} - {1 \over A} = {1 \over A}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = A = 6\,\,\,\,\left[ {\rm{h}} \right]$$

This solution follows the notations and rationale taught in the GMATH method.

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Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by [email protected] » Sun Sep 23, 2018 7:36 pm
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: E

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by Scott@TargetTestPrep » Thu Sep 27, 2018 4:23 pm
BTGmoderatorDC wrote:Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?

A. 1/2
B. 2
C. 3
D. 5
E. 6

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

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