from the given, we can deduce that each large machine dis 4 loads, and each small machine did 6 loads during the two hours.
so the question is what is the value of \(4m + 6n = ??\)

from statement (1), the ratio between the large and small machines are given, but this can't provide the exact numerical --> insufficient

from statement (2), if \(2m + 3n = 55\), then \(4m + 6n = 110\) --> sufficient

official solution

During the two-hour period each large washer does four loads (the number of 30-minute periods in two hours is four) and each small washer does six loads (the number of 20-minute periods in two hours is six). Therefore, the total number of loads done during the two-hour period by m large washers and n small washers is 4m + 6n.

If m = 1 and n = 3, then n = 3m and the total number of loads of laundry done in the two-hour period is 4(1) + 6(3) = 22. However, if m = 2 and n = 6, then n = 3m and the total number of loads of laundry done in the two-hour period is 4(2) + 6(6) = 44; NOT sufficient.
Since 2m + 3n = 55, it follows that 4m + 6n = 110. Therefore, by the remarks above, the total number of loads done during the two-hour period is 110; SUFFICIENT.

1. If there is only 1 large and 1 small machine then, in 2 hours, the large washing machine will process 4 loads \((120/30 = 4)\) and the small washing machine will process 6 loads \((120/20 = 6)\) of laundry
2. Hence, if there were \(m\) large washing machines and \(n\) small washing machines, then in a 2 hour period these would process \(4m + 6n\) loads of laundry
3. We need the value of \(4m + 6n\) to answer the question

Option A
\(4m + 6 * 3m = 4m + 18m = 22m\)

Option B
\(2m + 3n = 55\)

Multiplying the equation by 2 yields,

\(4m + 6n = 110\)

Ans B
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Large washer = 30 minutes per load
small washer = 20 minutes per load

In a 2 hour period, a large washer can complete 4 loads. A small washer can complete 6 loads.

We can use the equation 4m+ 6n for our total loads.

(1) n = 3m

We can't obtain the total loads from this equation. Insufficient.

(2) 2m + 3n = 55

4m + 6n = 110

SUFFICIENT

Answer is B.
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KNOW: m Large machines --> 30 minutes each --> 4 loads in 2 hrs
n Small machines --> 20 minutes each --> 6 loads in 2 hrs

Find: 4m + 6n

STATEMENTS
(1) n = 3m

INSUFFICIENT

(2) 2m + 3n = 55
Multiple both sides by 2
=> 4m + 6n = 110

SUFFICIENT

Thus, the answer is B.

Hope this helps.

NOTE: Make sure you read the question carefully and write down the equations and expressions on the scratch paper.
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Classic reducing equation trap option, in 2nd condition, forcing students to look for another equation. But if its proportional to the question stem, then you dont need the 2nd equation. The trick in this question is basically translating the question into this equation -> 4m + 6n

Each m laundry can do 2 washes in an hour and thus 4 washes in the 2 hour period. Similarly, n can do 6n.

Thus, Statement 2 is adequate, since we can just multiply the given info by 2 and arrive at 4m + 6n

n = the number of small machines
m = the number of large machines

4 = the number of loads in two hrs from a large machine
6 = the number of loads in two hrs from a small machine

(2) gives us an equation for the number of machines, however this equation can be solved in more than one way:

A) if m = 20 (large machines), then n = 5 (small machines)
2*20 + 3*5 = 55 machines

B) if m = 5 (large machines), then n = 15 (small machines)
2*5 + 3*15 = 55 machines

A) gives us a total of 20*4 + 5*6 = 190 loads in two hours
B) gives us a total of 5*4 + 15*6 = 110 loads in two hours

The correct answer must be C?

Now I get it.

We know that the number of loads will be 4m + 6n in two hours. Therefore we dont need the number of machines if we know the solution to 4m + 6n.

4m + 6n = 110 (number of machines AND number of loads)

This was intuitively upside down for me first.

If (2) would have been like 2m + 4n = 55 we would not be able to find an answer.

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