1- In the system of equations below, a is a constant and (x, y) is a solution, where x = 3.
What is the value of a?
3y = 12 - 3y
y + a = x - 1
3y = 12 - 3y
3y + 3y = 12
6y = 12
y =
12
/
6
y = 2
Substitute y = 2 and x = 3 in equation 2
2 + a = 3 - 1
2 + a = 2
a = 0
2- The graph shown below is that of a linear function ƒ whose expression is given by ƒ(x) = cx + d, where c and d are constants. Which of the following is true about c and d?
c is the slope, and the line is decreasing, therefore c < 0.
d is the y-intercept, and the y-intercept is on the positive side of the y-axis, therefore d > 0.
We conclude then that c < d as c cannot be greater than d, nor equal to 0 or to d.
3- The given graph shows the speed v in meters per second (m/s) of Sam and Daniel, as they do their morning jogs, as a function of time t in seconds (s).
By how much less is the difference in the speed of the two boys at t = 0.2s compared to the difference in speed of the boys initially?
Initially (t=0 s),
Sam’s velocity is 0 m/s (from the graph).
Daniel’s velocity is 0.6 m/s (from the graph).
the difference in speeds = 0.6 - 0 = 0.6 m/s.
At t = 0.2 s,
Sam’s velocity is 0.6 m/s (from the graph).
Daniel’s velocity is 0.3 m/s (from the graph).
the difference in speeds = 0.6 - 0.3 = 0.3 m/s.
The difference in speeds initially and after two seconds = 0.6 - 0.3 = 0.3 m/s.
4- Number of Covid-19 cases according
to age.
The histogram shown summarizes the number of people who got infected by the Covid-19 virus according to their age in Lebanon. The survey was done over 11,200 people.
Which of the following is closest to the ratio of the number of infected people below the age of 10 to those above the age of 80?
The number of infected people below the age of 10 is 400 (according to the histogram).
The number of infected people above the age of 80 is 200 (according to the histogram).
The correct ratio is
400
/
200 = 2
From all the alternatives, alternative C is the closest to the correct ratio as
410
/
215 = 1.9069
5- The scatterplot below shows the Bee population in a certain farm for every year since 2012.
A line of best fit and its equation are also shown.
Which of the following is the best interpretation of the value -56.429 in the equation of the line of best fit?
The scatterplot is decreasing, which means alternative A should be eliminated.
The number 525.71 represents the y-intercept, which means when x = 0, y = 525.71 and therefore its unit is “Bee Population” and not “years”. Alternative B should be then eliminated.
Alternative D is eliminated because logically the number of bees can’t be negative.
Proof that alternative C is correct:
The scatterplot is decreasing, and the slope of the equation is negative.
As well, the slope shows the direction and the steepness of the line, and in our case it will show the average decrease in the number of bees each year.
6- If the average (arithmetic mean) of three numbers a, b, and c is 10, what is the average of a and b in terms of c?
a + b + c
/
3
= 10
a + b + c = 30
a + b = 30 - c
The average of a + b is:
a + b
/
2
=
30 - c
/
2 = 15 - 0.5c
7- If p(x) = x2 - 7x + 5 and
q(x) = -3x3 - 7x2 + 2x - 5,
which of the following expressions is equal to the difference p(x) – q(x)?
p(x) - q(x) =
Horizontal method:
(x2 - 7x + 5) - (-3x3 - 7x2 + 2x - 5)
= x2 - 7x + 5 + 3x3 + 7x2 -2x + 5
= 3x3 + 8x2 - 9x + 10
OR:
Vertical method:
8- The graph (C) of the function ƒ(x) = 2(x + 2)(x - 6) is a parabola.
If the line x = k is the axis of symmetry of the parabola, what is the value of k?
f(x) = 2(x + 2)(x - 6)
f(x) = 2(x2 - 6x + 2x - 12)
f(x) = 2x2- 12x + 4x - 24
f(x) = 2x2 - 8x - 24
Equation of the axis of symmetry:
x = -
b
/
2a
= -
(-8)
/
2(2)
=
8
/
4
= 2
OR:
The axis of symmetry passes through the midpoint of the two x-intercepts of the parabola (x = -2 and x = 6).
xmidpoint =
-2 + 6
/
2 = 2 so k = 2
9- John owns a drone that has a radio range of 55 meters; that is, the owner can control the drone only if it is within 55 meters from him. As John launches the drone, the drone flies off a distance D, measured in meters, given by the expression D = 4t2 + 20t, where t is the time in seconds after the drone is launched. Assuming John stays where he is, at least how many seconds, after being launched, does the drone get out of range?
D = 4t2 + 20t
55 meters from the owner is the maximum range where he can control the drone. So the maximum distance to control the drone is when D = 55 meters.
4t2 + 20t = 55
4t2 + 20t - 55 = 0
∆ = b2 - 4ac = (20)2 - 4(4)(-55)
∆ = 1280 > 0. Then we have two solutions.
t =
-b ± √∆
/
2a
=
-20 ± 16√5
/
2(4)
t = 1.97 seconds (accepted) ≅ 2
(because for t > 1.97, the drone gets out of range 55 meters)
t = -6.97 (rejected because time should be positive)
10- A right triangle has an area of 96 cm2. If the shorter leg is 4 cm less than the longer leg, what is the length of the hypotenuse?
20
A =
1
/
2
bh
Let longer leg = x (base)
And shorter leg = x - 4 (height)
A =
1
/
2
(x)(x - 4) = 96
1
/
2
(x2 - 4x) = 96
x2 -4x - 192 = 0
∆ = b2 -4ac = (-4)2
- 4(1)(-192)
∆ = 784 > 0. Then we have two distinct real solutions.
x =
-b ± √∆
/
2a
=
4 ± 28
/
2(1)
x = 16 (accepted)
x = -12 (rejected because the dimension of a side in a triangle cannot be negative)
Therefore base = 16 and height = 12
Pythagorean theorem:
c2 = a2 + b2
c2 = 162 + 122
c2 = 400
c = ±√400
c = 20
We took the positive value since we are finding the dimension of the hypotenuse and it cannot be negative.
11- If x and y are positive measures of acute angles, and sin (x - 20°) = cos (y + 12°), what is a possible value of x + y? (Disregard the degree sign when gridding in your answer).
98
Recall that sin θ = cos θ if θ = 45° (acute angle).
x - 20 = 45
x = 65
y + 12 = 45
y = 33
Then x + y = 65 + 33 = 98
12-
In the given figure, F is the center of the circle, and J,H, and I are points on the circle, and FI = 4 If ∠JHI = ∠JIH = 45°, what is the area of the shaded region?
Since m∠JHI = m∠JIH = 45°, then triangle JHI is a right isosceles triangle (m∠HJI = 180 - 45 - 45 = 90°).
To find the dimension of JH and JI:
JI = JH (legs in a right isosceles triangle)
c2 = a2 + b2
(4 + 4)2 = JI2 + JI2
82 = 2JI2
64 = 2JI2
JI2 = 32
JI = 4√2
Area of the semi-circle:
πR2
/
2
=
π(4)2
/
2
= 8π
Area of triangle JHI:
1
/
2
bh =
1
/
2
(4√2)(4√2) = 16
Area of the shaded region:
8π - 16
/
2
= 4π - 8
We divided by 2 since we have two equal regions from the left side and the right side of the right isosceles triangle. However, the shaded part is for one of the two regions only.