Bunuel wrote:
If a machine consumes k/5 kilowatts of power every t hours, how much power in kilowatts, will three such machines consume in 10 hours?
A. 6t/k
B. t/k
C. 30kt
D. k/t
E. 6k/t
Rate and algebraThe
fractional number of kilowatts in the prompt might obscure the obvious: "___ kilowatts every ___ hours" is a
rate.
What is the rate of consumption for one machine?
Number of kilowatts used: k/5
Per time unit: every t hours
Rate of consumption (no units after first term):
\(\frac{\frac{k}{5}kws}{t_{hours}}=(\frac{k}{5}*\frac{1}{t})=\frac{k}{5t}\)At that rate, how much power will three such machines consume in 10 hours?
(Consumption rate) * (time spent consuming, 10 hrs) * (# of consumers, 3) = Total consumption
= Amount of power used
\((\frac{k}{5t} * 10 * 3) = \frac{30k}{5t}=\frac{6k}{t}\) kilowatts
Answer E
Assign valuesAssign values that result in integers.
Let k = 5
Let t = 1
Rate of power consumptionOne machine consumes k/5 kilowatts of power every t hours
\(\frac{k}{5}kw=\frac{5}{5}kw = 1\) kilowatt is consumed . . .
Every
\(t = 1\) hours
Consumption rate is
\(\frac{1kw}{1hr}\) How much power, in kilowatts, will three such machines consume in 10 hours?
(Consumption rate) * (# of machines) * (Time) = Total consumption
\(\frac{1kilowatt}{1hr} * 3 * 10hrs = 30\) kilowatts
With k = 5, t = 1, find the answer choice that yields 30.
At a glance with mental math, eliminate B and D immediately (much too small), and C (huge). That leaves A and E
A. 6t/k:
\(\frac{(6*1)}{5} = \frac{6}{5}\). NO
E. 6k/t:
\(\frac{(6*5)}{1} = \frac{30}{1} = 30\). MATCH
Answer E
*
The eliminated options, which are indeed small or huge compared to 30:
B. t/k: \(\frac{1}{5}\). NO
C. 30kt: \((30)*(5)*(1) = 150\). NO
D. k/t: \(\frac{5}{1} = 5\). NO
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