Question

The math club sells candy bars and drinks. 60 candy bars and 110 drinks will sell for $265. 120 candy bars and 90 drinks will sell for $270. How much does each candy bar sell for?

Answer 1

OK, we are in the land of simultaneous equations here. They are fun to do, but need some careful steps, including checking at the end.

Explanation:

Let’s call the number of candy bars, c and the number of drinks, d.

We are told that:

60c + 110d = $265.12 (equation 1)

And:

120c + 90d = $270 (equation 2)

We now set off to eliminate one of these factors (c or d) so that we can solve it for the other factor. Then we substitute our new value back into one of the original equations.

If we multiply equation 1 by 2 I’ve spotted that the factor c could be eliminated by subtraction:

(1) x 2 = 120c + 220d = $530.24. (Eqn. 3)

Eqn 2 - 3 = -130d = -$260.24 (Eqn. 4) [makes me wonder if the original question was copied correctly, suspect this was meant to be $260]

So d = $-260.24/-130 = $2.002

Put this “back into” equation (1), i.e. substitute for d in that equation to give:

60c + (110 x $2.002) = $265.12

So 60c = $265.12 - 220.22

c = $44.9 / 60 = $0.748

Pretty sure now the original question was wrong, but still .... you must now check your answer by substituting your two values back into either of the original equations to check it works

I’ll use Eqn 1.

60 x 0.748 + 110 x 2.002 = 265.12

TA DAAH!