IMAT 2014 Q16 [Coffee Jars]

There is a special offer on jars of coffee this week. Customers who buy two jars at the normal price have the option to buy up to four more jars at one quarter of the normal price.

For customers buying coffee this week, which of these bar graphs shows how the average cost per jar (as a percentage of the normal cost) varies with the number of jars bought?

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

For this question, we are given percentages, but to picture the problem better we can think of them as euros.

The first two jars cost €100, but the next 4 jars are a quarter of this, so €25 each. We can take the averages every time we add a jar and then see which graph best represents this.

We know the first two jars are full price so the first two bars in the graph should be 100%.

When you buy 3 jars….

€100 + €100 + €25 = €225

Find the average:

225 / 3 = €75

So the third bar should be at 75%

When you buy 4 jars….

€100 + €100 + €25 + €25 = €250

Find the average:

250 / 4 = €62.5
(the math seems difficult but 240/4 = 60 and then do 10/4 = 2.5 and then add the two answers together).

So the fourth bar should be at 62.5%

With two calculations we can already eliminate all options but C and E, and C looks more reasonable because E has a decrease that is too consistent and linear.

When you buy 5 jars….

€100 + €100 + €25 + €25 + €25= €275

Find the average:

275 / 5 = €55
(this is a multiple of 5 so it should be fairly easy, if not, take 300 divided by 5 and subtract 5 from the answer because you are adding an extra 25, which is equal to 5 when divided by 5

So the fifth bar should be at 55%

So we can now make our final decision, the fifth bar should be barely above the 50%. In this case, we can eliminate E and confirm that C is the correct answer.