2018 MTAP Grade 10 Division Orals Solutions

 2018 GRADE 10 DIVISION ORALS SOLUTIONS:

15-second question

1.) Difference = 4-1 = 3. Thus, 4 + 2(3) = 10

2.) x² + x - 12 = 0. (x+4)(x-3) = 0. Thus, roots are -4 and 3

3.) 4! = (4)(3)(2)(1) = 24

4.) 10! / 7! = (10)(9)(8) = 720

5.) 2/4 = 3/x. Thus, x = 6

6.) √(2² + 3²) = √13

7.) 3/15 or 1/5

8.) sin²A + cos²A = 1
9/49 + cos²A = 1
cos²A = 40/49
cosA = 2√10 / 7

9.) x-1 = 0, x = 1
(1)² + 1 + 1 = 3

10.) Ratio = cbrt(24/3) = cbrt(8) = 2
3 * 2² = 12

11.) Discriminant should be zero
b² - 4ac = 0
81 - 4k = 0
k = 81/4 or 20.25

30-second question

1.) Ratio = 7/3 / 7/2 = 2/3
Sum = a_1 / (1-r)
Sum = 7/2 / (1/3) = (7/2)(3) = 21/2 or 10.5

2.) x+4 = 0, x = -4
2(-4)³ + 6(-4)² + 5(-4) + 2
-128 + 96 - 20 + 2 = -50

3.) 7C3 * 5C2
(7! / 4!3!)(5! / 3!2!) = (35)(10) = 350

4.) radius² = 87 + (14/2)² + (16/2)²
radius² = 87+49+64 = 200
radius = 10√2

5.) Arc AD + Arc CB = 2 * Angle AED
55° + Arc CB = 2(40°)
Arc CB = 25°

6.) Difference = 2.5
Sum = (a_1 + a_n)(n/2)
a_n = a_1 + (n-1)d
Sum = (a_1 + a_1 + 2.5(n-1))(n/2)
374 = (2 + 2 + 2.5(n-1))(n/2)
748 = 2.5n² + 1.5n
5n² + 3n - 1496 = 0
(5n+88)(n-17) = 0
Thus, answer is 17

1-minute question

1.) Distance of (6,3) and (12,-7) is √136, which is also the diameter
Midpoint is (9,-2), which is also the center
Note that new circle has twice the radius of the old, thus, the diameter of the old is the radius of the new.
Thus, (x-9)² + (y+2)² = 136

2.) By synthetic division trial and error, roots are -1, ±2, 3

3.) Let x be the first term of arithmetic
Let x+3 be the second term of arithmetic
Let x+6 be the third term of arithmetic
Let x-2 be the first term of geometric
Let 2x+6 be the second term of geometric
Let 5x+30 be the third term of geometric

(2x+6) / (x-2) = (5x+30) / (2x+6)
(2x+6)² = (5x+30)(x-2)
4x² + 24x + 36 = 5x² + 20x - 60
x² - 4x - 96 = 0
(x-12)(x+8) = 0, x = 12
Thus, 12,15,18

4.) Since Arc BC = 120°, meaning, angle BDC and angle CAB are 60° , and since arc AD is 60°, meaning, angle ACD and angle ABD is 30°. We will notice that the center are 90°, and this forms 4 right triangles with hypotenuse as the sides of the quadrilateral.
Let E be the point of intersection, meaning, by 30-60-90 triangle, BE is 6√3, AE is 6, CE is 9√3, and DE is 9.
BD = BE + DE = 9 + 6√3
AC = AE + CE = 6 + 9√3
By observation, we will notice that 9 + 6√3 is shorter.

5.) Maximum sum is 4*6 = 24. Question is "at least 23", thus, 23 and 24.
To get a sum of 23, the dice must be (6,6,6,5). Thus, it can be arranged in 4!/3! ways or 4 ways
To get a sum of 24, there is only 1 possible way, and that is if all dices show 6.
Thus, (4+1) / 6⁴ = 5 / 1296

6.) a = 1 and c = -8
-8 can be 4 ways:
(8)(-1) , (-8)(1) , (4)(-2) , (-4)(2).
If (8)(-1) , b = 7
If (-8)(1) , b = -7
If (4)(-2) , b = 2
If (-4)(2) , b = -2
By trial and error, b = 2

Clincher question

1.) (4² + 1) - (3² + 1) = 7

2.) 2(4x - 10) = 6x + 10
4x - 10 = 3x + 5
x = 15

3.) Since P(1) = P(2) = P(3) = P(4) = 0, then, they are zeros of the function. Thus, P(x) = a(x-1)(x-2)(x-3)(x-4)
Solve for a by substituting 5
P(5) = a(4)(3)(2)(1)
120 = 24a , thus, a = 5
Substitute 6 to get P(6)
P(6) = 5(5)(4)(3)(2)(1)
P(6) = (5)(5!) = (5)(120) = 600

Do-or-Die question

Notice that (9)(2018)² = (3²)(2018²) = (3*2018)² = 6054²
Last number of 6054² is 6. Thus, the integer must be ending in 4 or 6. By trial and error, you will get 6056