Math problems in BB similar to this one? Concept?

Test Preparation SAT Preparation
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<p>p.716 #13.
A school ordered $600 worth of lightbulbs. Some of the lightbulbs cost $1 each and the others cost $2 each. If twice as many $1 bulbs as 2$ bulbs were ordered, how many lightbulbs were ordered together?</p>

<p>What kind of problem is this? I remember while taking the oct sat, i wasted a lot of time trying to figure out this problem. I wanna master this sorta problem, so i won't get stumped next time, so can anyone provide me with questions similar to this one? If not, then can you just explain the concept?</p>

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<p>Ok, so this can be solved with a set of equations. The trickiest part is converting what we have given in the problems into equations.</p>

<p>In this case we’ll use two variables. x for the number of $1 lightbulbs and y for 2$ lightbulbs.
So we know that the school spend 600$ total. </p>

<p>We know that the cost of all the 1$ lightbulbs is given by the total number of lightbulbs time 1$ for each one. The total cost of 2$ lightbulbs is given by 2$ times the number of 2$ lightbulbs [y].</p>

<p>So from that we get:
600 = x + 2y</p>

<p>The next equation we get from the line “twice as many 1$ bulbs as 2$ bulbs were ordered”.
So from that we get:
x = 2y</p>

<p>So by subbing in 2y for x in the top equation we get:
600 = 2y + 2y
600 = 4y
so y = 150</p>

<p>The next thing that we can do is plug that y = 150 into x = 2y
so:
x = 2*150
x=300</p>

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<p>ya this is a set of two equations with two variables</p>

4

<p>Can also use guess and check instead of algebra…</p>

<p>You know that they ordered twice as many $1 bulbs. So take a guess to get started.</p>

<p>For example:</p>

<p>Suppose they ordered 50 $2 bulbs and 100 $1 bulbs.<br>
That would have cost 50 x 2 + 100 x 1 = $200</p>

<p>That’s too small. In fact, it’s 1/3 of the cost we wanted. So triple our guess:</p>

<p>150 $2 bulbs, 300 $1 bulbs…check and see that it comes to $600. So now we know that they ordered a total of 450 bulbs.</p>

<p>The SAT often has these “two equation, two variable” problems. For many people, trial and error is quicker and easier than algebra. Really, it’s good to know both methods so that during the test, you can do whichever method comes to you more easily for that particular problem.</p>

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<p>Thanks guys! Ima work on two equations with two variables more! figured out thats one of my weaknesses.</p>