IMAT 2012 Q16 [Light Bulb]

Past Papers Solutions Archives
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1

The 100W light bulb (cost €0.60) is not going to be used anymore and is being replaced with the 20W (cost €3) low energy light bulb.

If electricity is charged at €0.15 per kWh, for how many hours must the low energy bulb be used in order for the lower cost of running it to exactly compensate for its higher initial cost?

A. 0.25
B. 250.00
C. 200.00
D. 160.00
E. 720.00

2

Simple steps to solve word problems:

  • Underline key information
  • Determine what they are trying to ask, and what you will need to solve it
    Eliminate any non-essential information
  • Draw a picture, graph, or equation
    In moments of high stress like exam taking, always work with the paper they give you to avoid careless mistakes.
  • Solve.

To solve this question, we need to calculate the cost of running each light bulb at different times. The difficulty in this question is that the cost is in kWh and we are giving watts. The problem with this is that we will be working with a lot of decimals, so it would be advised to practice using uncomfortable numbers like these before exam day.

  1. Convert the units so we are working all in the same

100W = 0.1kW

20W = 0.02kW

  1. Find an appropriate interval and then test different numbers of hours.

To find the interval, we can look at the answers. Notice how there is a fairly steep increase between options? We can start with an interval of 100 hours, and we can narrow it if we overshoot our target.

100 Hours

100W Bulb

(€0.15kWh)(0.1kW)(100 hours) + €0.60= €2.10

20W Bulb

(€0.15kWh)(0.02kW)(100 hours) + €3.00= €3.30

200 Hours

100W Bulb

(€0.15kWh)(0.1kW)(200 hours) + €0.60= €3.60

20W Bulb

(€0.15kWh)(0.02kW)(200 hours) + €3.00= €3.60

After 200 hours they are the same price, and we know that the low energy bulb is cheaper per hour, it is just the initial cost that is more. So we can conclude that after 200 hours, the low energy bulb will have compensated for its higher overall cost.

\fcolorbox{red}{grey!30}{Therefore C is the answer.}
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