   SEARCH HOME Math Central Quandaries & Queries  Question from Armstrong, a student: Given that 50 litres of kerosine costs 6000 nira. Draw a graph and read off; A: The costs of 15 and 34 liters. B: The number of litres that can be bought for 2300 nira. First, note that this is not a particularly practical way to solve this problem; it is intended to give practice in using graphs. I mention this because students are sometimes puzzled if a problem liekl this seems to take an unrealistic length of time.

Secondly, there is an unstated assumption here that the only cost involved is per liter; you are not paying a service fee, buying a jerrycan, or getting a volume discount. Thus 0 litres of kerosine cost 0 nira, and all other (volume,cost) points lie on a straight line through (0,0) and (50,6000).

You can now draw this line. Use accurately labelled graph paper and do not try to make 1 liter on the horizontal axis the same length as 1 nira on the vertical axis! You might use 1 liter increments and 100 nira increments instead.

Starting with 15 liters on the horizontal axis, follow straight up till you hit the line; then go left till you hit the vertical axis where you should find 1800 nira. Do the same for 34 liters; and for part B, find 2300 nira on the vertical axis, measure right to the line, and down to the horizontal axis.

Do note that if you merely wanted the numerical answers you would do better to divide 6000 by 50 to get 120 nira per liter, then multiply the number of liters (or divide the number of nira) by 120 to get the answers.

Good Hunting!
RD     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.