SOLUTION: A mathematics teacher prepared a test consisting of 50 problems worth 2 points, 5 points, and 10 points each. If the number of 2-point problems is thrice the number of 10-point pro

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Question 1181290: A mathematics teacher prepared a test consisting of 50 problems worth 2 points, 5 points, and 10 points each. If the number of 2-point problems is thrice the number of 10-point problems, and the number of 5-point problems is 10 less than twice the number of 2-point problems, how many points is the entire test?
Found 3 solutions by Solver92311, greenestamps, ikleyn:
Answer by Solver92311(821)   (Show Source):
You can put this solution on YOUR website!

Let represent the number of 2-point questions, represent the number of 5-point questions, and represent the number of 10-point questions.

Solve the 3X3 system for , , and

Then calculate

John

My calculator said it, I believe it, that settles it

From
I > Ø

Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!

You can certainly solve the problem as the other tutor suggested, using three unknowns, leading to having to solve a system of three equations in those three unknowns to solve the problem.

Your time is far better used if you take the time to set up the problem using a single variable, which means you will only need to solve a single equation.

x = # of 10-point problems
3x = # of 2-point problems (thrice the number of 10-point problems)
6x-10 = # of 5-point problems (10 less than twice the number of 2-point problems)

It should be easy to finish the problem from there....

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Using a single variable might increase the time required for setting up the problem by 10-20%; but it will decrease the amount of time and effort required to solve the problem by 80-90%.

Answer by ikleyn(52817)   (Show Source):
You can put this solution on YOUR website!
.
A mathematics teacher prepared a test consisting of 50 problems worth 2 points, 5 points, and 10 points each.
If the number of 2-point problems is thrice the number of 10-point problems,
and the number of 5-point problems is 10 less than twice the number of 2-point problems,
how many points is the entire test?
~~~~~~~~~~~~~~~

Let x be the number of the 10-point problems. Then the number of the 2-point problems is 3x, according to the condition, and the number of the 5-point problems is (2*(3x) - 10) = (6x-10). Now we write an equation for the total number of problems x + 3x + (6x-10) = 50 problems. Simplify and find x x + 3x + 6x = 50 + 10 10x = 60 x = 60/10 = 6. So, the number of the 10-point problems is 6; the number of the 2-point problems is 3*6 = 18, and of the 5-point problems is (6x-10) = 6*6-10 = 26. As a check, 6 + 18 + 26 = 50 problems, in all. Now the last step of the solution is to calculate the number of points. It is 6*10 + 18*2 + 26*5 = 226. ANSWER. The entire test is 226 points.

Solved.

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Couple of post-notes.

From the first glance, this problem is for 3 unknowns and 3 equations. But it is not so. It easily can be solved using one single equation in one single unknown. Moreover, this problem and many other similar problems, are DESIGNED, are INTENDED and are EXPECTED to be solved PRECISELY in this way. Solving such problems in this way has important educational goal to train and to prepare a mathematically literate student.