This task is a somewhat more complicated version of "Accurately weighing pennies I'' as a third equation is needed in order to solve part (a) explicitly. Instead, students have to combine the algebraic techniques with some additional problem-solving (numerical reasoning, informed guess-and-check, etc.) Part (b) is new to this task, as with only two types of pennies the weight of the collection determines how many pennies of each type are in the collection. This is no longer the case with three different weights but in this particular case, a collection of 50 is too small to show any ambiguity. This is part of the reason for part (c) of the question where the weight alone no longer determines which type of pennies are in the roll. This shows how important levels of accuracy in measurement are as the answer to part (b) could be different if we were to measure on a scale which is only accurate to the nearest tenth of a gram instead of to the nearest hundredth of a gram.