Track / Stage 1 / 260 of 320 #260 of 2604

Problem 260

Algebra Difficulty 1.6 Multiple choice CEMC Gauss (Grade 8) · Canada · 2014

On a science test, Janine got 80% of the 10 multiple choice questions correct and 70% of the 30 short answer questions correct. What percentage of the 40 questions on the test did she answer correctly?

Pick one

Official solution

Solution 1

Since $PQR$ is a straight line segment, then $\angle PQR=180\degree$ .

Since $\angle SQP+ \angle SQR=180\degree$ , then $\angle SQR=180\degree -\angle SQP=180\degree-75\degree=105\degree$ .

The three angles in a triangle add to $180\degree$ , so $\angle QSR+\angle SQR+\angle QRS=180\degree$ , or
$\angle QSR=180\degree-\angle SQR-\angle QRS=180\degree-105\degree-30\degree=45\degree$ .

Solution 2

The exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles of the triangle.

Since $\angle SQP$ is an exterior angle of $\triangle SQR$ , and the two opposite interior angles are $\angle QSR$ and $\angle QRS$ , then $\angle SQP=\angle QSR+\angle QRS$ .
Thus, $75\degree=\angle QSR + 30\degree$ or $\angle QSR=75\degree-30\degree=45\degree$ .

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.