# Solution to 2014 MMC Division Finals-Orals 4th Year

This blog post will outline the solution to last year’s MMC division finals-orals team category. We will start for the first ten questions, the next set of questions will be added very soon.

15-second mental questions worth 1 point and their solutions.

1. Find the remainder when is divided by .

Solution: This can be done in many ways. But I prefer to use substitution.

Let , then is the remainder.

The answer is 1.

2. Find the exact radian measure of

Solution: we need to know that , hence

3. Find the domain of

Solution: To answer the question, take note that in order for logarithmic curve to exist, in its form , b>1 and a>0.

Going back to our problem, the base is fine since it is 4. We just have to focus on x-4. Since it must be positive real number, we have

The domain is **{x|x>4}** or **(4,+∞)**

4. Find the 12th term of an arithmetic series whose first term is 8 and third term is 2.

Solution: Observe that the progression is decreasing and there is one term between 2 and 8. With the 15-second cramming fever I guess we don’t have much time to use the series to find the common difference. Just observe the following numbers.

8 , x, 2, . . . . What do you think is the number between them to make it an arithmetic series?

yes! it is 5. Therefore, the common difference is -3.

Solving for 12th term:

Therefore, the 12th term is -25.

5. Find the vertex of the parabola

Solution: in the vertex form of parabola , the vertex is at (h,k). Therefore, in , the vertex is at (-3,8). don’t get confused with the negative sign. It is just telling us that the graph is opening downward.

6. Determine all x in so that the ordered pairs (1,3), (2,1), and (x,3) are on the graph of the function.

Solution: first to think is that a function passed the vertical line test. This means that there should be no the same values for the x-coordinates. If x=1, the point will become (1,3) which is the same as the first point. We can still consider the points are on the graph of a function.

If x=2, the set of points will become (1,3), (2,1), and (2,3). Points (2,1) and (2,3) form a vertical line making the graph to not pass the vertical line test. But the question here is to determine all x so that the ordered pairs are on the graph of the function. The answer here are all real numbers except 2 or .

7. Find the x-intercept of the line

Solution: x-intercept is a point at which the given line intersects the x-axis. Thus, by letting y=0, we can get the value of x-intercept.

$latex y=7x-4$

Therefore the x-intercept is 4/7.

8. Find a so that the line is perpendicular to a line with slope 3.

Solution: This problem can be solved quickly if we know that the product of the slope of perpendicular lines is equal to -1. So the value of a is the negative reciprocal of a to attain such condition. Thus, a=-1/3.

9. The parabola passes through the point (1,b). Find b.

Solution: If (1,b) is on the graph of the equation, it must satisfy the equation. Thus, by simple substitution, we can solve for the value of b.

10. Suppose u varies directly as the square root of v. If u=3 when v=4, find u when v=10.

Solution: If we know already that this is a direct variation, we can just use ratio and proportion as follows.

### Dan

#### Latest posts by Dan (see all)

- 2016 MMC Schedule - November 4, 2015
- 2014 MTAP reviewer for Grade 3 - September 30, 2015
- 2015 MTAP reviewer for 4th year solution part 1 - August 22, 2015