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19 Sep 2025, 11:56
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1st hour: 17
8th hour: 38
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
60% (03:01) correct
40%
(03:48)
wrong
based on 200
sessions
History
Date
Time
Result
Not Attempted Yet
A factory is working on a production order using identical machines. The production begins with a certain number of machines operating in the first hour. At the start of the second hour, and at the start of each hour thereafter, the same fixed number of additional machines is added, thus increasing the units produced in each subsequent hour. By the end of 8 hours, the machines have produced exactly 220 units.
Select for 1st hour the number of units produced in the first hour, and select for 8th hour the number of units produced in the eighth hour that would be jointly consistent with the given information. Make only two selections, one in each column.
Select for 1st hour the number of units produced in the first hour, and select for 8th hour the number of units produced in the eighth hour that would be jointly consistent with the given information. Make only two selections, one in each column.
| 1st hour | 8th hour | |
| 17 | ||
| 27.5 | ||
| 32 | ||
| 38 | ||
| 40 | ||
| 55 |
ShowHide Answer
Official Answer
1st hour: 17
8th hour: 38
27 Sep 2025, 01:00
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Bunuel
In 8 hours, the machines produced a total of 220 units.
So the average production per hour is 220/8 = 27.5 units.
Because the output increases each hour, the production in the first hour must be less than the average. Among the choices, the only value below 27.5 is 17.
Thus, in the first hour, 17 units were produced.
Since the same number of additional machines is added each hour, the hourly outputs form an arithmetic sequence. For such a sequence, the average equals (first hour + eighth hour)/2.
So:
(17 + eighth-hour output)/2 = 27.5
17 + eighth-hour output = 55
Eighth-hour output = 38
I29-170
General Discussion
19 Sep 2025, 13:42
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I tried to solve this using work and rate. That did not work. It was so simple once I understood we can assign variables to the number of units produced.
Say there were \(N\) machines in the first hour. They produce \(x\) units in one hour.
Say \(M\) machines were added each hour and they produce \(y\) units on their own.
In the first hour, \(N\) machines produce \(x\) units.
In the second hour, \(N+M\) machines produce \(x+y\) units.
In the third hour, \(N+2M\) machines produce \(x+2y\) units and so on.
In the eighth hour, \(N+7M\) machines produce \(x+7y\) units.
Adding all the units -> \((x)+(x+y)+(x+2y)+........+(x+7y) = 220\)
\(8x+28y = 220\)
\(2x+7y=55\)
\(y=\frac{55-2x}{7}\)
Now plug in values for \(x\)
1. \(x=17\)
\(y = \frac{55-34}{7}\)
\(y=3\)
Units produced in the eighth hour \(= x+7y = 17+7(3) = 17+21 = 38\)
First column -> \(17\)
Second column -> \(38\)
Say there were \(N\) machines in the first hour. They produce \(x\) units in one hour.
Say \(M\) machines were added each hour and they produce \(y\) units on their own.
In the first hour, \(N\) machines produce \(x\) units.
In the second hour, \(N+M\) machines produce \(x+y\) units.
In the third hour, \(N+2M\) machines produce \(x+2y\) units and so on.
In the eighth hour, \(N+7M\) machines produce \(x+7y\) units.
Adding all the units -> \((x)+(x+y)+(x+2y)+........+(x+7y) = 220\)
\(8x+28y = 220\)
\(2x+7y=55\)
\(y=\frac{55-2x}{7}\)
Now plug in values for \(x\)
1. \(x=17\)
\(y = \frac{55-34}{7}\)
\(y=3\)
Units produced in the eighth hour \(= x+7y = 17+7(3) = 17+21 = 38\)
First column -> \(17\)
Second column -> \(38\)
