8/13/10

Test 6 Section 4 - #13 (page 716)

This would be a great backsolve if it were multiple choice. But it isn't, so we have to use another way. Blech.

We know the total cost was $600. We also know we have twice as many $1 bulbs as $2 bulbs. Think about that part for a second. If we have twice as many of the bulb that costs half as much, aren't we saying basically we paid the same for each? Shouldn't we have paid $300 for each kind of bulb? That's gonna mean 150 $2 bulbs, and 300 $1 bulbs. Total: 450 bulbs.

You could also approach this algebraically, if you want. Say x = # of cheap bulbs, and y = # of expensive ones. Solve these two equations:

x + 2y = 600 (this equation represents the cost)
x = 2y (this equation represents the fact that we had twice as many cheap ones, and if you tried this algebraically and got it wrong, this is likely where you made your mistake. Let me guess: your answer was 360?)

Solve it, and you get x = 300, y = 150. Total bulbs: 450, just like above.

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