Track / Stage 1 / 185 of 240 #185 of 2444

Problem 185

Geometry Difficulty 1.4 Multiple choice CEMC Gauss (Grade 8) · Canada · 2015

In the graph shown, which of the following represents the image of the line segment $PQ$ after a reflection across the $x$ -axis?

Hide/Reveal Description of Line Segments

The segment PQ is in the first quadrant. The point P is 3 units right and 3 units up from the origin. The point Q is 4 units right and 3 units up from the point P.
The segment PS is in the first quadrant. The point P is 3 units right and 3 units up from the origin. The point S is 4 units right and 3 units down from the point P.
The segment MN is in the second quadrant. The point M is 3 units left and 3 units up from the origin. The point N is 4 units left and 3 units up from the point M.
The segment FG is in the third quadrant. The point F is 3 units left and 3 units down from the origin. The point G is 4 units left and 3 units down from the point F.
The segment TU is in the fourth quadrant. The point T is 3 units right and 3 units down from the origin. The point U is 4 units right and 3 units down from the point T.
The segment WV is in the fourth quadrant. The point W is 3 units right and 6 units down from the origin. The point V is 4 units right and 3 units up from the point W.

Pick one

Official solutions — 2

Solution 1

To find the image of $PQ$ , we reflect points $P$ and $Q$ across the $x$ -axis, then join them.

Since $P$ is 3 units above the $x$ -axis, then the reflection of $P$ across the $x$ -axis is 3 units below the $x$ -axis at the same $x$ -coordinate.

That is, point $T$ is the image of $P$ after it is reflected across the $x$ -axis.

Similarly, after a reflection across the $x$ -axis, the image of point $Q$ will be 6 units below the $x$ -axis but have the same $x$ -coordinate as $Q$ .

That is, point $U$ is the image of $Q$ after it is reflected across the $x$ -axis.
Therefore, the line segment $TU$ is the image of $PQ$ after it is reflected across the $x$ -axis.

Solution 2

Solution 1

The multiples of 5 between 1 and 99 are: $5,10,15,20,25,30,35,40,45,50,55,60,65,70, 75,80,85,90,95.$ Of these, only $10,20,30,40,50,60,70,80,$ and $90$ are even.
Therefore, there are 9 even whole numbers between 1 and 99 that are multiples of 5.

Solution 2

To create an even multiple of 5, we must multiply 5 by an even whole number (since 5 is odd, multiplying 5 by an odd whole number creates an odd result).

The smallest positive even multiple of 5 is $5\times2=10$ .

The largest even multiple of 5 less than 99 is $5\times18=90$ .

That is, multiplying 5 by each of the even numbers from 2 to 18 results in the only even multiples of 5 between 1 and 99.
Since there are 9 even numbers from 2 to 18 (inclusive), then there are 9 even whole numbers between 1 and 99 that are multiples of 5.

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