Walkabout wrote:
A dealer originally bought 100 identical batteries at a total cost of q dollars. If each battery was sold at 50 percent above the original cost per battery, then, in terms of q, for how many dollars was each battery sold?
(A) 3q/200
(B) 3q/2
(C) 150q
(D) q/100
(E) 150/q
We are given that 100 batteries cost a TOTAL of q dollars. We are also given that EACH battery was sold at 50% above the original cost. The first thing we must do is create an equation for q. Remember that q is the TOTAL COST. So if we make b = the original cost per battery we can say:
100 x b = q
b = q/100
We now have the original cost per battery in terms of q. Next, we determine the selling price when we increase the cost by 50%. To calculate this increase we simply multiply q/100 by 1.5. We have:
(q/100) x 1.5
(q/100) x 3/2 = 3q/200
Answer: A
If you don't like working with variables, you could instead substitute a convenient number for q. Normally I would not suggest the plugging-in method in a problem such as this; however, since we have only one variable, the method will be sufficient.
Let's say we make the total cost of the 100 batteries q = 200. This is a convenient number that will work well with the numbers presented in the problem. We can now set up a similar equation to what we did above, where b = the original price per battery.
100 x b = 200
b = 200/100
b = 2
The cost per battery is $2. Now we need to show the selling price per battery by increasing $2 by 50%.
2 x 1.5 = 3
The answer 3 is the selling price of the battery.
The last step is to now plug our value of q = 200 into each answer choice to see which one provides us with a value of 3. This will yield the correct answer.
(A) 3q/200
(3 x 200)/200 = 3
This IS equal to 3.
(B) 3q/2
(3 x 200)/2 = 600/2 = 300
This IS NOT equal to 3.
(C) 150q
150 x 200 = 30,000
This IS NOT equal to 3.
(D) q/100
200/100 = 2
This IS NOT equal to 3.
(E) 150/q
150/200 = 15/20 = ¾
This IS NOT equal to 3.
Answer choice A is the only one that is equal to 3.