A factory has three types of machines – A, B, and C – each of which wo

Target question: How many widgets could one machine A, one Machine B, and one Machine C produce in one 8-hour day?

Statement 1: 7 Machine A's and 11 Machine B's can produce 250 widgets per hour
No information about Machine C
NOT SUFFICIENT

Statement 2: 8 Machine A's and 22 Machine C's can produce 600 widgets per hour
No information about Machine B
NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1: 7 Machine A's and 11 Machine B's can produce 250 widgets per hour
Statement 2: 8 Machine A's and 22 Machine C's can produce 600 widgets per hour

Hmmm, I see that we're given info about 11 Machine Bs and 22 Machine Cs. Perhaps we might gain some useful information, if we create an EQUIVALENT statement such that Machines B and C produce the SAME number of widgets.

Take statement 2 and HALVE everything to get: 4 Machine As and 11 Machine Cs can produce 300 widgets per hour
Combine this new (equivalent) info with statement 1 to see we have two useful pieces of info:

7 Machine A's and 11 Machine B's can produce 250 widgets per hour
4 Machine As and 11 Machine Cs can produce 300 widgets per hour

So, if we ADD all of the machines we get:
11 Machine A's, 11 Machine B's and 11 Machine C's can produce 550 widgets per hour

Now divide everything by 11 to get: 1 Machine A, 1 Machine B and 1 Machine C can produce 50 widgets per hour
So, 1 Machine A, 1 Machine B and 1 Machine C can produce 400 widgets in 8 hours
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer:

Require both the options.
(7A+11B) = 250 widgets per hr
Multiple by 2
14A+22B = 500 widgets per hr ---(1)
Using 2nd statement
8A+22C= 600 widgets per hr ---(2)

Add (1) &(2)
22A+22B+22C=1100
A+B+C= 50 widgets per hr.

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Hello Friends,
Bunuel, Vyshak, VeritasPrepKarishma, msk0657 and others
Shouldn't this question mention: Working together how many widgets could one machine A, one Machine B, and one Machine C produce in one 8-hour day?

Please clarify,
thanks in advance.

This is exactly what the question is asking. No need to over-complicate...
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Thanks Bunuel for clarifying.

Could we possibly create a system of simultaneous equations?
7A + 11B = 250
8A + 22C = 600

So 56A + 88B = 2000 --> equation 1
56A + 154C = 4200 --> equation 2

Subtracting equation 2 from equation 1 gives us 88B - 154C = -2200 --> equation 3

Three equations are enough for us to solve A, B, and C. And therefore solve A+B+C

Hi, From the looks of other solutions, it is the right method. However, I have a doubt if you could clarify. How did you write 7A+11B=250 I mean does A here denote the A's rate ? Sorry for asking a silly question

\(7A + 11B = 250\)

\(8A + 22C = 600\)

\(4A + 11C = 300\)

\(4A + 7A + 11C + 11B = 550\)

\(11A + 11C + 11B = 550\)

\(A + B + C = 50\)

\(50 * 8 hours = 400 widgets\)

Answer is C.
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