Maths Olympiad Prep

Track / Stage 1 / 184 of 240 #184 of 2444

Problem 184

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

Next problem →

Official solution

Solution 1

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

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

The three angles in a triangle add to 180°180\degree, so QSR+SQR+QRS=180°\angle QSR+\angle SQR+\angle QRS=180\degree, or
QSR=180°SQRQRS=180°105°30°=45°\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 SQP\angle SQP is an exterior angle of SQR\triangle SQR, and the two opposite interior angles are QSR\angle QSR and QRS\angle QRS, then SQP=QSR+QRS\angle SQP=\angle QSR+\angle QRS.
Thus, 75°=QSR+30°75\degree=\angle QSR + 30\degree or QSR=75°30°=45°\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.