# algebra word problem

• Mar 16th 2010, 11:40 AM
donnagirl
algebra word problem
Mike got a score of 56 on a recent 25-question multiple choice test. The scoring for the test was +6 for each correct answer, -2 for each incorrect answer and 0 for each unanswered question. What is the maximum number of questions Mike could have answered correctly?

My book says the answer is 13 but I'm not sure how to calculate this.
• Mar 16th 2010, 01:04 PM
Henryt999
well
the maximum point is 150 (6*25)
He got $56$
that is $150-94.$
so he had somehow minus 94 points from the maximum.

Each wrong answer gives you minus 8 points from the maximum points.
(-6 for not getting it correct and the minus 2 for answering wrong).
So the question is now.
How many questions can go wrong and still get 56 points?
The answer is $\frac{94}{8} = 12,\frac{6}{8}$
so 12 questions can go wrong and u can still get 56 points.
That leaves you with 25-12 = 13

Understandable?
Regards Henry
• Mar 16th 2010, 01:05 PM
Soroban
Hello, donnagirl!

Quote:

Mike got a score of 56 on a recent 25-question multiple choice test.
The scoring was: +6 for each correct answer, -2 for each incorrect answer,
and 0 for each unanswered question.

What is the maximum number of questions Mike could have answered correctly?

Let: . $\begin{Bmatrix}R &=& \text{number correct} \\ W &=& \text{number wrong} \\ B &=& \text{number blank} \end{Bmatrix}$

We have: . $R + W + B \:=\:25$ .[1]

and: . $6R - 2W + 0B \:=\:56 \quad\Rightarrow\quad 3R - W \:=\:28$ .[2]

Add [1] and [2]: . $4R + B \:=\:53 \quad\Rightarrow\quad R \:=\:\frac{53-B}{4}$

Since $R$ is an integer, $53-B$ must be divisible by 4.

The first time this happens is when $B = 1.$

. . Hence: . $R \:=\:\frac{53-1}{4} \:=\:\frac{52}{4} \:=\:13$

There are other values of $B\!:\;5,9,14, \hdots$
. . but they produce smaller values of $R.$