SOLUTION: The sample of work produced by lathe is 10000. a. If the machine doubled this production every 20 minutes, how much will it produce in one hour? b. How long will it take the mac

Question 1169051: The sample of work produced by lathe is 10000.
a. If the machine doubled this production every 20 minutes, how much will it produce
in one hour?
b. How long will it take the machine to produce 8 million?
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
Let the initial production of the lathe be $P_0 = 10000$.
**a. Production in one hour:**
One hour has 60 minutes. The machine doubles its production every 20 minutes. So, in one hour, the production will double $60 / 20 = 3$ times.
* After the first 20 minutes: $P_1 = P_0 \times 2 = 10000 \times 2 = 20000$
* After the next 20 minutes (total 40 minutes): $P_2 = P_1 \times 2 = 20000 \times 2 = 40000$
* After the final 20 minutes (total 60 minutes or one hour): $P_3 = P_2 \times 2 = 40000 \times 2 = 80000$
Alternatively, we can use the formula:
$P(t) = P_0 \times 2^{t/d}$
where:
* $P(t)$ is the production after time $t$
* $P_0$ is the initial production
* $t$ is the total time
* $d$ is the doubling time
For one hour (60 minutes):
$P(60) = 10000 \times 2^{60/20}$
$P(60) = 10000 \times 2^3$
$P(60) = 10000 \times 8$
$P(60) = 80000$
So, the machine will produce **80,000** units in one hour.
**b. Time to produce 8 million:**
We want to find the time $t$ when the production $P(t)$ reaches 8,000,000.
$P(t) = 10000 \times 2^{t/20} = 8000000$
Divide both sides by 10000:
$2^{t/20} = \frac{8000000}{10000}$
$2^{t/20} = 800$
To solve for $t$, we can take the logarithm of both sides (using base 2 or natural logarithm):
Using base 2 logarithm:
$\log_2(2^{t/20}) = \log_2(800)$
$\frac{t}{20} = \log_2(800)$
We know that $2^9 = 512$ and $2^{10} = 1024$. So, $\log_2(800)$ is between 9 and 10.
$\log_2(800) = \log_2(8 \times 100) = \log_2(2^3 \times 100) = 3 + \log_2(100)$
Since $2^6 = 64$ and $2^7 = 128$, $\log_2(100)$ is between 6 and 7 (approximately 6.64).
$\log_2(800) \approx 3 + 6.64 = 9.64$
Now, solve for $t$:
$t = 20 \times \log_2(800)$
$t \approx 20 \times 9.64$
$t \approx 192.8$ minutes
Alternatively, using natural logarithm:
$\ln(2^{t/20}) = \ln(800)$
$\frac{t}{20} \ln(2) = \ln(800)$
$t = 20 \times \frac{\ln(800)}{\ln(2)}$
$t \approx 20 \times \frac{6.6846}{0.6931}$
$t \approx 20 \times 9.644$
$t \approx 192.88$ minutes
So, it will take approximately **192.88 minutes** for the machine to produce 8 million units.
To express this in hours and minutes:
$192.88 \text{ minutes} = 3 \text{ hours and } 12.88 \text{ minutes}$
$0.88 \text{ minutes} \times 60 \text{ seconds/minute} \approx 53 \text{ seconds}$
So, it will take approximately 3 hours, 12 minutes, and 53 seconds.
Final Answers:
a. The machine will produce **80,000** units in one hour.
b. It will take approximately **192.88 minutes** (or about 3 hours, 12 minutes, and 53 seconds) for the machine to produce 8 million units.
Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.

        The " solution " by @CPhill is out of logic and out of elementary Arithmetic.

        Here I will give my solution for question (a).

The sample amount is 10,000. An hour is three times 20 minutes. Production for the first 20 minutes 2*10,000 = 20,000. Production for the next 20 minutes 2*20,000 = 40,000. Production for the last 20 minutes 2*40,000 = 80,000. Total production for one hour is 20,000 + 40,000 + 80,000 = 140,000. Answer to question (a)

RELATED QUESTIONS
Ace Machine Works estimates that the probability its lather tool is properly adjusted is... (answered by ikleyn)
Of the parts produced by a particular machine, 1.2% are defective. If a random sample of... (answered by rothauserc)
A pacer pencil manufacturer determines that 3 out of every 10000 pacers produced are... (answered by stanbon)
An old machine can do a certain unit of work in 12 hours. To increase production, a more... (answered by josmiceli)
60(m-15)this is the expression to solve the problems. I sure as heck cant get it to work. (answered by scott8148)
Suppose that it is known that 5% of the items produced by a particular machine are... (answered by stanbon)
It is known that 10% of the items produced by a certain machine are defective. Find the... (answered by mathmate)
Please could you kindly help me solve this question.Kindly show me all the working... (answered by lynnlo)
``60(m-15)this is the expression to solve the problems. I sure as heck cant get it to... (answered by ReadingBoosters)