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TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

Find the value of a:

D

a 2 2a 5 0

PLOD: M

CONCEPT/ COMPETENCY TESTED/ BLOOM’S TAXONOMY/ EXPLANATION/ REF. CONCEPT: Algebra: Quadratic Functions COMPETENCY TESTED: solves quadratic equations by: (a) extracting square roots; (b) factoring; (c) completing the square; and (d) using the quadratic formula.

A.

a 1 6

B.

a 1 6

C.

a 1 5

BT: Application

D.

a 1 6

EXPLN: By completing squares

a 2 2a 5 0 a 2 2a 5 a 2 2a 1 5 1 (a 1) 2 6 a 1 6 a 1 6 2.

Find the discriminant and characteristic of the root of the equation:

PLOD: E

4x2 + 20x – 53 = 0 A. B. C. D.

D

-1248; imaginary roots 0; exactly one root 612; real roots 1248; real roots

CONCEPT: Quadratic Equations COMPETENCY TESTED: characterizes the roots of a quadratic equation using the discriminant BT: Remembering, Analyzing EXPLN: The discriminant of a quadratic equation is given by b 4ac . Thus, the discriminant of the given equation is 2

20 2 (4 4 53) 400 848 1248 Hence, the quadratic equation has discriminant 1248 and has real roots.. CONCEPT: Variation

3. X

1

5

9

Y

1.5

.7

.61

Z

3

7

11

Refer to the table above. Which variation statement satisfies the values in the table? A. Y kX Z

ADMATH – TG 2015

D PLOD: E

COMPETENCY TESTED: translates into variation statement a relationship between two quantities given by: (a) a table of values; (b) a mathematical equation; (c) a graph, and vice versa. BT: Applying

1

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1.

ANS./ PLOD

QUESTION

NO.

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

kX Z k C. Y XZ kZ D. Y X

EXPLN: (Trial and Error) The first step should be to find the constant k. By substituting the values of x, y and z in the ordered pairs to the equations in the choices,

B. Y

the answer will be D. Y

CONCEPT: Exponents Simplify:

(3x 4 y 7 z 12 ) 3 (5 x 9 y 3 z 4 ) 2 A.

675 x 18 y 9 z 13

B.

675 x 30 y 15 z 28

B

C. 675 x 15 y 27 z 30 D. 675 x

15

y

30

z

COMPETENCY TESTED: applies the laws involving positive integral exponents to zero and negative integral exponents.

PLOD: E

28

BT: Applying EXPLN: Applying the laws of exponents, the given expression can be solved as follows.

(3x 4 y 7 z 12 ) 3 (5 x 9 y 3 z 4 ) 2 =

5.

(27 x 12 y 21 z 36 )(25 x 18 y 6 z 8 ) 675 x 30 y 15 z 28

CONCEPT: Radicals

Simplify the given expression:

COMPETENCY TESTED: simplifies radical expressions using the laws of radicals.

8x 3 y 3 3 2 x 5 y 5 18 x 7 y 7 A.

B.

C.

2 xy 3 2 3x 2 y 2 2 3 2 2 xy x y

2 2

3x y D.

2

1 xy

BT: Applying, Analyzing C PLOD: M

EXPLN: Using the Laws of radicals and exponents, the expression can be simplified as follows:

8x 3 y 3 3 2 x 5 y 5 18 x 7 y 7

3x 2 y 2 2 3 xy

ADMATH – TG 2015

2

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4.

kz where k=1/2. x

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

=

2 xy 2 xy 3 x 2 y 2 2 xy 3 x 3 y 3 2 xy 2 xy 3 x 2 y 2 3x 3 y 3 2 3 xy 3x 2 y 2 2 2

3x y 6 6.

2

1 xy

CONCEPT: Geometry COMPETENCY TESTED: uses properties to find measures of angles, sides and other quantities involving parallelograms. BT: Remembering, Analyzing A PLOD: E

7.

A quadrilateral is inscribed in a circle (see figure above). What is angle θ? A. 113o B. 117o C. 130o D. 243o 3 5 s and s = t , find the ratio of t to r. If r = 4 6 A. 5:8 B. 8:5 C. 9:10 D. 10:9

EXPLN: For quadrilaterals inscribed in circles, opposite angles supplement each other. Hence, 180o = θ + 67o θ = 180o - 67o = 113°

CONCEPT: Proportions B PLOD: E

COMPETENCY TESTED: applies the fundamental theorems of proportionality to solve problems involving proportions. BT: Application

EXPLN: Substitution r=

ADMATH – TG 2015

3 4

s,s=

5 t 6

3

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TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

r=

B. 3 C. 6 D.

9.

2

6 2

Find the sixth term of an arithmetic sequence whose second and tenth terms are 14 and 58, respectively. A. 20 B. 36 C. 44 D. 72

5

C PLOD: E

B PLOD: E

COMPETENCY TESTED: finds the trigonometric ratios of special angles. BT: Remembering, Applying EXPLN: Each leg of a 45-45-90 triangle is 1/√2 of the hypotenuse. Each leg is then √9 = 3. The sum of two legs is 6 . CONCEPT: Sequences COMPETENCY TESTED: determines arithmetic means and nth term of an arithmetic sequence.*** BT: Application EXPLN: Notice that the middle of 2nd and 10th term is the 6th term, thus, the 6th term is the average;

14 58

10.

divided by (x 3) .

COMPETENCY TESTED: performs division of polynomials using long division and synthetic division.

2 x 9

2 x 9 2 C. x 3x 9 2 D. x 3x 9

72

36 2 2 CONCEPT: Algebra

3 Find the quotient when P(x) x 27 is

A.

B.

D PLOD: M

BT: Analysis EXPLN: Sum of Two Cubes 3 3 2 2 a b (a b)(a ab b )

3 2 x 27 (x 3)(x 3x 9) Or Synthetic division x 3 x 3

ADMATH – TG 2015

4

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A 45-45-90 triangle has a hypotenuse of length √18. How long is the sum of the lengths of its two legs? A. 3

t

46

t 8 Thus, r : t = 5:8 t : r = 8:5 CONCEPT: Geometry r=

8.

35

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS -3 1 0 0 27 -3 9 -27 1 -3 9 -27

2 x 3x 9 11.

B PLOD: E

CONCEPT: Geometry COMPETENCY TESTED: solves problems on circles. BT: Remembering, Analyzing

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EXPLN: Inscribed Angle theorem: Central angle = 2θ Arc length: C = 2R

12.

Refer to the figure above. If R is the radius of the circle, what is the length of C? A. Rθ B. 2Rθ C. πR D. 2πR Consider a family of four standing side by side for a family portrait, in how many ways can they arrange themselves? A. 24 B. 28 C. 30 D. 32

A PLOD: A

CONCEPT: Permutations COMPETENCY TESTED: solves problems involving permutations. BT: Applying EXPLN: n Pr

n! , (n r )!

where n = total number of objects; r = number of objects chosen (want)

n Pr

13.

A Four coins are flipped. What are the chances of NOT getting heads? A. 1/16 B. 1/4 C. 1/2 D. 15/16

PLOD: E

4! 4! 24 (4 4)!

CONCEPT: Probability COMPETENCY TESTED: solves problems involving probability. BT: Creating, Understanding, Evaluating EXPLN:

ADMATH – TG 2015

5

TEACHER’S GUIDE SET A

The following are the scores of 20 students on their 30 item exam: 4,5,6,6,7,8,10,10,11,16,17,17,18,19,20,20,21, 23,25,30 Determine the 90thpercentile of the given data. A. B. C. D.

C PLOD: E

20 22 24 26

The probability of not getting heads is equal to the probability of getting four tails Each flip has a 1/2 chance to get a tails. Do it four times, the chance of getting four tails is (1/2)4 = 1/16 CONCEPT: Measure of Position COMPETENCY TESTED: calculates a specified measure of position (e.g. 90th percentile) of a set of data. BT: Creating, Understanding, Evaluating For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

14.

ACET 2016 MATHEMATICS

EXPLN: Make sure first that all data are arranged from least to greatest. The formula for percentile:

X PK

(

nk ) 100

X (

nk 1) 100

2

where n=number of observations and k=percentile.

nk (20)(90) 18 100 100 (20)(90) nk 1 1 19 100 100

So

Note that X1=4, X2=5, X3=6 and so on… So X18=23 and X19=25 and substituting,

PK 15.

CONCEPT: Quadratic equations

Determine the quadratic equation with roots 3/2 and 5. 2 A. 15 x 7 x 2 2 B. 7 x 2 x 15 2 C. 2 x 15 x 7

D.

X 18 X 19 23 25 24 2 2

D PLOD: E

COMPETENCY TESTED: describes the relationship between the coefficients and the roots of a quadratic equation. BT: Application

2 x 7 x 15 2

EXPLN: Given the roots, the quadratic equation can be solved as follows:

x

3 2

or x 5

(2 x 3)( x 5) 0 2 x 2 7 x 15 0 ADMATH – TG 2015

6

TEACHER’S GUIDE SET A 16.

ACET 2016 MATHEMATICS

If y varies directly as x2 and inversely as z, and y=8 when x=1 and z=2, find y when x=3 and z=6. A. 8 B. 16 C. 24 D. 32

C

CONCEPT: Variation

PLOD: E

COMPETENCY TESTED: solves problems involving variation BT: Application EXPLN:

kx 2 z k (1) 2 8 2 k 16, then

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y

16(3) 2 16(9) 6 6 y 24

y

17.

1

Simplify:

A.

B.

1 2

x 3 y 2

C PLOD: E

3 x2

COMPETENCY TESTED: simplifies expressions with rational exponents.

EXPLN:

3 2

x

CONCEPT: Rational Expressions

BT: Understanding, Analyzing, Applying

5 y2

y

C.

x2 y

1

5 2

x2 y

1 2

x 3 y 2

3 y2

x

1 1 ( 3) ( 2 ) 2 y 2

1

5 3 ( ) ( ) x 2 y 2

y x

3 ( ) 2 5 ( ) 2

5

x2 5

D.

y2

x

18.

5 2

Find the solution/s to the following equation:

1 1 x2 2 x 4 x 4 x 16 A. B. C. D.

x=0,2 x=2,4 x=0,4 x=4,-4

ADMATH – TG 2015

A PLOD: M

CONCEPT: Geometry COMPETENCY TESTED: solves equations transformable to quadratic equations (including rational algebraic equations). BT: Remembering, Analyzing EXPLN:

7

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS 1 1 x2 2 x 4 x 4 x 16 The LCD is 𝑥 2 − 16, so the equation will become,

x4 x 2 16

x4 x 2 16

x2 x 2 16

x 4 x 4 x2 x( x 2) 0 x 0 and x 2 19.

Simplify the expression:

C

CONCEPT: Radical Expressions

( 2 x 3 y )( 2 x 3 y )

A.

2x 3 y

B.

4x 2 9 y 2

PLOD: E

COMPETENCY TESTED: performs operations on radical expressions.***

C. ( 2 x )( 2 x 3 y ) ( 3 y )( 2 x 3 y )

BT: Application

D. ( 3 x )( 2 x 3 y ) ( 2 y )( 2 x 3 y )

EXPLN:

( 2 x 3 y )( 2 x 3 y ) ( 2 x )( 2 x 3 y ) ( 3 y )( 2 x 3 y )

20.

Find the values of x:

x 1 1 2 x 1 x 2 x 3x 2 A. B. C. D.

-1 2 -1 and 2 1 and 2

A PLOD: D

CONCEPT: Algebraic Equations COMPETENCY TESTED: solves problems involving quadratic equations and rational algebraic equations. BT: Analysis and Application EXPLN:

x 1 1 x 1 x 2 x 2 3x 2 x 1 1 x 1 x 2 ( x 1)( x 2) x ( x 2) 1( x 1) 1 ( x 1)( x 2) ( x 2)( x 1) ( x 1)( x 2) x( x 2) ( x 1) 1 x2 x 2 0 x 2 or x 1 But x=2 will make the equation undefined. So

ADMATH – TG 2015

8

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x 2 2x 0

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS only x= -1.

21.

Refer to the figure below:

B PLOD: M

CONCEPT: Midline Theorem COMPETENCY TESTED: proves the Midline Theorem.

EXPLN: Since line DE is a midline of triangle ABC, line DE || line AB. Note that ∠A+∠B+∠C=180°, so ∠A=180°-∠C-∠B ∠A=180°-45°-50°=85°

If ∠C=45°, ∠B=50°, find ∠BED + ∠CDE. A. B. C. D.

205° 215° 225° 235°

22.

A

PLOD: M

Since line DE || line AB, ∠A+∠ADE=180° ∠ADE=180°-∠A=180°-85° ∠ADE =95° And ∠B+∠BED=180° ∠BED=180°-∠B=180°-50° ∠BED=130° Now, ∠ADE + ∠CDE=180° (Supplementary Angles) ∠CDE=180° - ∠ADE =180° - 95° ∠CDE = 85° Hence, ∠BED + ∠CDE= 130°+85° ∠BED + ∠CDE= 215° CONCEPT: Similar Triangles COMPETENCY TESTED: solves problems that involve triangle similarity and right triangles.*** BT: Remembering, Analyzing

Find x and y. A. x =12, y = 15 B. x =12, y = 20 C. x =15, y = 15 D. x =15, y = 20

ADMATH – TG 2015

EXPLN: The triangles in the figure are all similar. From there x can be found: x 16 9 x x 2 16 9 x 12 Since x is 12, along with the other leg equal to 9 of the smallest triangle, these are recognized as the legs of a Pythagorean triple. Thus, 𝑦 = 15.

9

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BT: Analysis, Application

TEACHER’S GUIDE SET A A wooden plank leaning against a wall makes a 30° angle with the ground. If the point at which the plank touches the wall is 5 meters above the ground, what is the length of the wooden plank? A. 5 meters B. 5√3 meters C. 10 meters D. 10√3 meters

C PLOD: M

CONCEPT: Algebra COMPETENCY TESTED: uses trigonometric ratios to solve real-life problems involving right triangles. *** BT: Remembering, Analyzing EXPLN: Let l- length of the plank

l

5

30°

s l 5 l 10 meters 1 2 CONCEPT: Sequences sin 30

24.

An arithmetic sequence of seven numbers sums to 126. If the first number in the sequence is 6, what is the common difference? A. 3 B. 4 C. 5 D. 6

COMPETENCY TESTED: finds the sum of the terms of a given arithmetic sequence.*** BT: Remembering, Analyzing

B PLOD: M

ADMATH – TG 2015

EXPLN: Let d = common difference x = first number Set-up: x + (x+d) + (x+2d) + … + (x + 6d) = 126 Expect seven x’s and d + … + 6d = 21d 7x + 21d = 126 Since x = 6 42 + 21d = 126 21d = 84 d=4 OR Let Sn = sum of n numbers a1 = first number Formula: Sn = n(2a1 + (n-1)d)/2 Substitute n = 7, S7 = 126 126 = 42 + 21d 21d = 84

10

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23.

ACET 2016 MATHEMATICS

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS d=4

Consider the polynomial

x5 2x4 3x3 6x2 4x 8

D PLOD: E

Which of the following is/are factors of the given polynomial? I. II. III. A. B. C. D.

26.

EXPLN: Trial and Error. By Factor Theorem, we consider: I. (x-1) f(1)=1-2+3-6-4+8 = 0, so it IS a FACTOR II. (x+1) f(1)=-1-2-3-6+4+8 = 0, so it IS a FACTOR III. (x-2) f(2)=32-32+24-24-8+8=0, so it IS a FACTOR

I only I and II only II and III only I, II and III

CONCEPT: Circles

A circle with radius r is centered at (1,2). The point (x, y) = (5,-4) is on the circle. Find r.

13

B.

20

C.

2 13

D.

2 20

COMPETENCY TESTED: proves the Remainder Theorem and the Factor Theorem. BT: Remembering, Applying, Evaluating

(x-1) (x+1) (x-2)

A.

CONCEPT: Remainder Theorem/Factor Theorem

COMPETENCY TESTED: applies the distance formula to prove some geometric properties. C PLOD: E

BT: Analyzing, Applying EXPLN: Use the distance formula since r is the distance between the center of a circle and a point on the circle.

r (1 5) 2 (2 4) 2 16 36 52 2 13 27.

In how many different ways can the letters of the word “MISSISSIPPI” be arranged? A. 34,644 B. 34,646 C. 34,648 D. 34,650

CONCEPT: Combinations COMPETENCY TESTED: solves problems involving permutations and combinations. D BT: Application PLOD: M EXPLN: There are 11 letters: M-1, S-4, I-4, P-2

ADMATH – TG 2015

11

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25.

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS So,

Find the 75th percentile of the set of values below: {15,9,7,6,1,4,3,10,7,8,9,12} A. B. C. D.

B PLOD: E

9 9.5 10 11.25

CONCEPT: Measures of Position COMPETENCY TESTED: interprets measures of position. BT: Applying EXPLN: Arrange the data from lowest to highest: {1,3,4,6,7,7,8,9,9,10,12,15}

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

28.

11! 11.10.9.8.7.6.5.4.3.2.1 34,650 1!4!4!2! (4.3.2.1)(4.3.2.1)(2.1)

The formula for percentile:

X PK

(

nk ) 100

X (

nk 1) 100

2

where n=number of observations and k=percentile. Where n=12 and k=75 Compute for nk/100 = (12)(75)/(100) nk/100=9 nk/100 +1 = 10

PK

X (9) X (10 ) 2 9 10 2

, so the 75th percentile is 9.5 29.

Find a possible inequality whose solution set is given by 2 x 5 . A.

x 2 7 x 10 0

B.

x 2 7 x 10 0

C.

x 7 x 10 0

D.

x 2 7 x 10 0

B PLOD: M

CONCEPT: Algebra COMPETENCY TESTED: solves quadratic inequalities. BT: Analysis

2

EXPLN: Inequality Shortcut factor choices first, A. (x 2)(x 5) 0 B. (x 2)(x 5) 0 C. (x 2)(x 5) 0 D. (x+2)(x+5)≥ 0 Assuming there is no “negative x”, and the right-

ADMATH – TG 2015

12

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS hand side is 0, the solution set is given by Case 1: if 0 x s small critical # or x b big critical # Case 2: if ≤0 s x b

Find the solutions of:

x 3 x 3 A. B. C. D.

COMPETENCY TESTED: solves equations involving radical expressions.***

x=3 x=4 x=5 No solution

BT: Applying, Analyzing EXPLN:

x3 x 3 ( x 3 x ) 2 32 D PLOD: M

2 x 3 2 x 2 3x 9 2 x 12 2 x 2 3x ( x 6) 2 ( x 2 3x ) 2 x 2 12 x 36 x 2 3x 36 9 x 4x Check if the resulting value of x is a root of the given equation.

4 3 4 1 2 1 3 Therefore, x=4 is not a root of the equation and hence there are no solutions. CONCEPT: Quadrilaterals

31.

COMPETENCY TESTED: solves problems involving parallelograms, trapezoids and kites. B Refer to the figure above. The top base of the trapezoid is 8. A lateral side measures 10. What is the area of the trapezoid? A. 108 B. 112 C. 144 D. 160

ADMATH – TG 2015

PLOD: M

BT: Remembering, Applying, Analyzing EXPLN: Compute for length of the shortest leg of the right triangle at the left. Shortest leg length = √102 − 82 Shortest leg length = 6 (Or recognize that to make a Pythagorean triple,

13

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30.

Thus, answer is B. CONCEPT: Radical Equations

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS the leg must be 6) The triangles are congruent since they are SSS Method 1:□ + 2Δ Area of □ = 82 = 64 Total area of 2 Δs = 2(8×6/2) = 48

area of trapezoid 112

32.

CONCEPT: Geometry; Similar/Congruent Triangles

Three pairs of parallel segments make up two triangles, ∆𝑍𝑌𝑋 and ∆𝑊𝑉𝑈, shown below.

COMPETENCY TESTED: applies the theorems to show that given triangles are similar. BT: Evaluation EXPLN: C

Which of the following is/are ALWAYS true for∆𝑍𝑌𝑋and ∆𝑊𝑉𝑈? I. They are congruent. II. They are right triangles. III. They are AAA triangles. A. B. C. D. 33.

PLOD: M

WV = ZY, WU = ZX, UV = XY Therefore, angles are also the same. No information whether the angles form a right angle. I and III

II only III only I and III I, II, and III

Given the sequence 15,

15 15 , , ... 2 4

Find the 7th term (a7 ) and give the geometric mean (m) of the first and 7th term. A.

a7

15 15 ;m 32 8

B.

a7

15 15 ;m 32 16

ADMATH – TG 2015

C PLOD: E

CONCEPT: Geometric Sequences COMPETENCY TESTED: determines geometric means and nth term of a geometric sequence.*** BT: Applying, Analyzing EXPLN:

14

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Method 2: Trapezoid base1 = 20, base2 = 8, height = 8 area = (20+8)×8/2 = 28×8/2 = 28×4 area = 112

TEACHER’S GUIDE SET A

15 15 ;m 64 8

a 1 15 ; common ratio(r) a n a1 r ( n 1)

15 15 ;m D. a 7 64 16

a 7 (15)

1 ( 7 1) 15 2 64

m 15 34.

x 2x 4x 8 3

C. D.

225 15 64 8

2

COMPETENCY TESTED: factors polynomials.

x 4 16

B.

15 64

CONCEPT: Algebra

Simplify:

A.

1 2

1

BT: Application

x 4 1 x4 1 x2 1 2

EXPLN: Numerator: factored by grouping Denominator: difference of two squares (DOTS)

x 3 2x 2 4x 8 x 4 16

x 2 2

x 3 4x 2x 2 8 ( x 2 4)( x 2 4)

x( x 2 4) 2( x 2 4) ( x 2 4)( x 2 4)

Cancellation: C PLOD: D

x3 2 x 2 4 x 8

x 4 16 ( x 2)

( x 2)( x 2 4) ( x 2 4)( x 2 4)

( x 2 4)

Denominator: DOTS

x 3 2x 2 4x 8 x 16 4

( x 2) ( x 2)( x 2)

( x 2) ( x 2)( x 2)

Cancellation:

x 3 2x 2 4x 8 x 16 1 ( x 2) 4

35.

Find the center (x,y) and radius (r) of the circle 1 with equation: 2 x 5 y ( x 2 y 2 13) 0 2 A. (-5,2); r=4

ADMATH – TG 2015

C PLOD: M

is the hypotenuse of ΔBCF. is 9 since adjacent of 30 is the hypotenuse of ΔCDF. Thus, is the hypotenuse of ΔDEF. Thus, = . CONCEPT: Circles COMPETENCY TESTED: determines the center and radius of a circle given its equation and vice versa

15

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C. a 7

ACET 2016 MATHEMATICS

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

B. (-2,5); r=16 C. (-2,5); r=4 D. (5,-2); r=16

BT: Remembering, Analyzing EXPLN: Multiplying the equation by 2 will result to

4 x 10 y x 2 y 2 13 0 and rearranging the terms, the equation will be

x 2 4 x y 2 10 y 13 By completing the squares method, the resulting equation will be and by factoring the perfect square trinomials, will yield to

( x 2)2 ( y 5)2 16 Thus, the center is (-2,5) and the radius is

r 16 4 . 36.

Find the lower and upper quartile respectively, in the following data set: 1, 11, 19, 15, 20, 24, 28,34, 37, 47, 50, 57 A. 15 and 47 B. 17 and 42 C. 19 and 47 D. 20 and 50

B PLOD: E

CONCEPT: Quartiles, Decile, Percentiles COMPETENCY TESTED: solves problems involving measures of position. BT: Applying EXPLN: To get the lower quartile, make sure that the data set is arranged in increasing order. 1, 11, 15, 19, 20, 24, 28, 34, 37, 47, 50, 57 The median

24 28 26 .The lower 2

quartile (Q1) is the median of the lower half of the data set. Then the

15 19 17 . 2 37 47 The upper quartile 42 2 lower quartile

37.

A rectangular lot has an area of 560 square meters. The length of the lot is three more than twice its width. Find the length and width of the rectangular lot. A. L=30 meters, W=15 meters B. L=35 meters, W=15 meters C. L=32 meters, W=16 meters D. L=35 meters, W=16 meters

ADMATH – TG 2015

D PLOD: M

CONCEPT: Quadratic Functions COMPETENCY TESTED: models real-life situations using quadratic functions. BT: Application EXPLN: Let L - length of the lot; W – width of the lot

16

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( x 2 4 x 4) ( y 2 10 y 25) 13 4 25

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS Area of a rectangle: A=LW LW=560 (2W+3)(W)=560 2W 2+3W=560 2W 2+3W-560=0 Using the Quadratic Formula to solve for W:

W

b b 2 4ac 2a 3 3 2 4(2)( 560) 2(2) For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

3 4489 4 3 67 3 67 or 4 4 64 - 70 or 4 4 cannot be negative

64 16 meters 4 L = 2(16)+3=32+3= 35 meters

So, W

38. Evaluate:

48 27 2

A.

6 2

B.

6

C.

3 2

D.

3

A PLOD: E

CONCEPT: Radical Expressions COMPETENCY TESTED: solves problems involving radicals. BT: Applying, Evaluating EXPLN:

(16)(3) (9)(3)

4 3 3 3

2 Rationalizing, 39.

Given the quadratic equation:

x 2 24 x 45 y Which of the following statements is/are true? I. The graph of the equation is open downwards II. The vertex is at (12, -99)

ADMATH – TG 2015

D PLOD: M

2 3

2

3 2

6 2

2 2 CONCEPT: Quadratic Equation COMPETENCY TESTED: graphs a quadratic function: (a) domain; (b) range; (c) intercepts; (d) axis of symmetry; (e) vertex; (f) direction of the opening of the parabola. BT: Analyzing

17

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

III. The range is { y R | y 99} A. B. C. D.

EXPLN: The vertex form of the quadratic equation will be used to solve for the vertex. Rearranging terms in the equation will yield

II only I only I and II only II and III only

x 2 24 x y 45 And completing the squares will result to

x 2 24 x 144 y 99 ( x 12) 2 99 y

40.

X varies inversely as the cube of Z and

1 directly as . X will be 4 if Y=4 and Z=2. Find Y 1 𝑌

if X=5 and Z=3.

D

upwards. Therefore, only II and III are true. CONCEPT: Variation

PLOD: E

COMPETENCY TESTED: solves problems involving variation.

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

Hence, the vertex of the quadratic equation is at the point (12,-99). This implies that the range is { y R | y 99} and the graph is opening

BT: Remembering, Applying

32 A. 5 5 B. 32 32 C. 27 27 D. 32

EXPLN:

1 k X Y3 3 Z Z Y k 4 3 2 (15) k 160 k

1 160( ) Y X Z3 1 5(3)3 Y 160 1 135 27 Y 160 32 41.

A PLOD: E

CONCEPT: Similar Triangles COMPETENCY TESTED: applies the theorems to show that given triangles are similar. BT: Analysis EXPLN: Only I is true according to properties of similar triangles.

Given the triangle above, which of the

ADMATH – TG 2015

18

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

following statements is ALWAYS TRUE when triangle AEC is similar to triangle BCD? I. II. III.

42.

CBD BAE

AE BD CE CD BCD BAE AB BC

NOT BCD BAE III is false.

I only I and II only II and III only I, II and III CONCEPT: Algebra

Find the sum of the infinite sequence:

5 5 5 , , , ... 27 81 243

6 5 5 B. 9 5 C. 6 9 D. 5

COMPETENCY TESTED: finds the sum of the terms of a given finite or infinite geometric sequence.***

A.

BT: Remembering, Analyzing, Applying

C PLOD: D

EXPLN: Observe that the given sequence is a geometric sequence with common ratio of 1/3. This can be shown using the formula a n ra n 1 . Also, the first term is a

5 which can be found using the 9

formula a n ra n 1 . Since the common ratio r < 1, the formula is

i 1 a i 1 r a

Substituting values, the sum is

5 9 1

43. If 4a+b=3c and -8a+2b+6c=24, what is the value of 2b? A. 3 B. 6 C. 9 D. 12

1 3

5 . 6

D

. CONCEPT: Polynomial Equations

PLOD: A

COMPETENCY TESTED: solves polynomial equations. BT: Application EXPLN:

ADMATH – TG 2015

19

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

A. B. C. D.

II is false,

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS 8a 2b 6c 24 2 4a b 3c 12 4a b 3c 0

44.

Equ.1 Equ.2

C

2b = 12 CONCEPT: Circles

PLOD: D

COMPETENCY TESTED: solves problems on circles.

A kite is inscribed in a circle with center at C and the bottom of the kite has an interior angle of 30o. What is the area of the shaded region if the circle has a radius of 2? A. 2π - 4 B. 2π - 2 C. 4π – 4 D. 4π - 2

45.

In a family, boys and girls are equally likely to be born. Find the probability that in a family with three children, exactly one is girl. A. B. C. D.

3/8 1/8 5/8 7/8

ADMATH – TG 2015

A PLOD: M

EXPLN: The central angle with vertex at C is twice the inscribed angle. The central angle is 60o. Bisect the central angle, to get two triangles with 30o angles with vertices at C. Draw a chord from between unequal segments of the kite. Half of the chord can be found using the radius since opposite a 30o is half the radius which is 1. The kite’s smaller diagonal is 2. The longer diagonal is found since it is twice the radius which is 4. Area of kite: 2( 4) Ak= =4 2 Area of Circle: Ac = (2)2π = 4π Area of shaded region: As = Ac - Ak As = 4π-4 CONCEPT: Probability COMPETENCY TESTED: solves problems involving probability. BT: Analyzing, Applying EXPLN: Since the probability of having a boy or a girl is equal, then each has a probability of ½. Let B- event that a boy is born. G- event that a girl is born. There are 8 total combinations of having three children and these are: BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG. So, there are 3 combinations of having exactly one girl: BBG,

20

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

BT: Applying, Analyzing, Remembering

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS BGB, GBB. Hence, the probability of having exactly one girl is 3/8.

Using the set of values below:{26,5,5,8,20,2,3,24} find the second quartile (Q2) and the range (r). A. B. C. D.

C PLOD: M

Q2=5; r=24 Q2=5; r=26 Q2=6.5; r=24 Q2=6.5; r=26

CONCEPT: Measures of Position COMPETENCY TESTED: uses appropriate measures of position and other statistical methods in analyzing and interpreting research data. BT: Analyzing, Applying EXPLN: Arrange the data from lowest to highest: {2,3,5,5,8,20,24,26} Second Quartile (Q2) = 50th percentile The formula for percentile:

X (

PK

nk ) 100

X (

nk 1) 100

2

where n=number of observations and k=percentile. Where n=8 and k=50 Compute for nk/100 = (8)(50)/(100) nk/100=4, and nk/100+1=5

P50

X ( 4 ) X ( 5) 2

58 2 6 .5 so the 50th percentile is 6.5. The range can be computed by subtracting the highest value – lowest value = 26 - 2 =24 47.

A cannonball travels along the path with equation

3 4x 4x 2 0 Find the roots of the quadratic equation. A. B.

2 8 , 3 3 2 8 x , 3 3 x

ADMATH – TG 2015

C PLOD: E

CONCEPT: Quadratic Equations COMPETENCY TESTED: models real-life situations using quadratic functions. BT: Application

EXPLN:

21

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

46.

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

3 1 , 2 2 3 1 D. x , 2 2 C.

x

x

b b 2 4ac 2a

(4) (4) 2 4(4)(3) 2(4) 4 64 4 8 8 8 1 2 2 3 1 x , 2 2

48.

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

CONCEPT: Polynomials

Find the quotient if

f(x) x 3x 3x 1 3

2

COMPETENCY TESTED: solves problems involving polynomial functions.

is divided by (x - 1) . x 2 2x 1

A. B.

x 2 - 2x 1

C.

x 2 2x 1

B PLOD: A

2 D. x + 2x 1

BT: Application EXPLN: Synthetic Division

x 1 0 x 1 1 -3 3 -1 1 -2 1 1 -2 1 0 x

49.

2

- 2x 1

CONCEPT: Algebra

(-1,7)

(3,7) COMPETENCY TESTED: solves problems involving geometric figures on the coordinate plane.

(-1,3)

BT: Analyzing, Applying

(3,3) B

A square with corner coordinates (-1,3)(1,7)(3,3)(3,7) is circumscribed in a circle as shown in the diagram above. Find the area of the shaded region. A. B. C. D.

8π+16 square units 8π-16 square units 32π+16 square units 32π-16 square units

PLOD: D

EXPLN: Solving the length of the diagonal of a square will give the diameter of the circle. By distance formula or properties of a 45-45-90 triangle, the length of the diagonal can be found to be 4 √2.

D (3 1) 2 (7 3) 2 4 2 4 2 32 4 2 By 45-45-90 triangle, the base has length=4 so multiply by √2 to get the hypotenuse = diagonal = diameter 4 2

ADMATH – TG 2015

22

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS So, radius = 2 2 Solving for the area of the circle:

A r 2 (2 2 ) 2 8 square units Area of square = 42=16 square units Finally,

8 16 square units

50.

Find the sixth term of an arithmetic sequence whose second and tenth terms are 14 and 58, respectively. A. 20 B. 36 C. 44 D. 72

B PLOD: E

CONCEPT: Arithmetic COMPETENCY TESTED: solves problems involving sequences. BT: Application EXPLN: Notice that the middle of 2nd and 10th term is the 6th term, thus, the 6th term is the average; 14 58

51.

72

36 2 2 CONCEPT: Algebra

Find the solution set of the given polynomial equation:

COMPETENCY TESTED: solves problems involving polynomials and polynomial equations.

x 3 6 x 2 x 6. A. B. C. D.

{ -1, 1, -6 } { -1, 1, 6 } { -3, 2, 6 } { 3, -2 , 6 }

BT: Application

A PLOD: M

EXPLN: (Partly by Trial and Error) By the factor theorem, we will see that x+1 or x1 or x+5 are factors of the polynomial. Consider x+1. By factor or long/synthetic division,

x 3 6x 2 x 6 0 ( x 2 5 x 6)( x 1) 0 Furthermore,

( x 6)( x 1)( x 1) 0 So,

x 6 0; x 1 0 and x 1 0 Hence, the solution set is {-1,1,-6}

ADMATH – TG 2015

23

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

area of the shaded region =

TEACHER’S GUIDE SET A

52.

ACET 2016 MATHEMATICS

C PLOD: D

CONCEPT: Circles COMPETENCY TESTED: solves problems on circles.

EXPLN: (Area of the sector of circle) – (Area of triangle) = (Area of shaded region)

2 3 Area of the Sector (2 3 ) 2 4 2

If ̅̅̅̅ 𝐴𝐵 is the radius of the circle equal to 2√3 and ∠ABC=2π/3, what is the area of the shaded region? A.

̅̅̅̅ to ΔABC is an isosceles triangle. Bisect 𝐴𝐶 ̅̅̅̅ create two congruent right triangles. 𝐴𝐵 and ̅̅̅̅ 𝐵𝐶 are now hypotenuses of the two right triangles. The angles of the triangles, coinciding with the center, are just half of the central angle. They each measure π/3. BAC is then equal to π/6.

4 3

3 3 2 C. 4 3 3 B.

4

1 2

Bisector length = 2 3 ( ) 3

D. 4 4 3

̅̅̅̅ = 6 𝐴𝐶 Area of ABC

6 3 3 3 2

(Area of shaded region) = 4 3 3 53.

CONCEPT: Probability Three fair coins are tossed independently. Determine the probability of obtaining at most two heads.

1 4 3 B. 4 3 C. 8 7 D. 8

COMPETENCY TESTED: solves problems involving probability.

A.

BT: Analyzing, Applying

D PLOD: M

EXPLN: At most 2 heads means having 0 head, 1 head, or 2 heads. The combinations are as follows: TTT, HTT, THT, TTH, HHT, HTH, THH. Each combination has the probability

1 1 1 1 since the probability of obtaining a 2 2 2 8 head or a tail are equal. There are 7 combinations of having at most 2 heads. So 7

ADMATH – TG 2015

1 7 8 8

24

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

BT: Applying, Analyzing, Remembering

TEACHER’S GUIDE SET A

If the graph of the equation y=8(x-2)2+5 is changed to y=8(x-3)2+5, what will happen to the graph of the new equation compared to the original one? A. The graph will move one unit to the left. B. The graph will move one unit to the right. C. The graph will flip with respect to the xaxis. D. The graph will flip with respect to the yaxis.

55.

B

CONCEPT: Algebraic equations

PLOD: E

COMPETENCY TESTED: analyzes the effects of changing the values of a, h and k in the equation y = a(x – h)2 + k of a quadratic function on its graph.*** BT: Understanding, Analysis

B PLOD: M

EXPLN: The graph of y=8(x-2)2+5 is a parabola whose vertex is at x=2, while y=8(x-3)2+5 is a similar parabola but whose vertex is at x=3. CONCEPT: Similar triangles COMPETENCY TESTED: solves problems that involve triangle similarity and right triangles.*** BT: Remembering, Analyzing EXPLN: The smaller triangle is identified as a 30-60-90 right triangle since its hypotenuse is 18 and its leg is half of the hypotenuse. The angle opposite the smaller leg is ∠BCD= ∠ACE=30°. The leg adjacent to ∠BCD of the smaller triangle

In the triangle above, what is the length of ̅̅̅̅ 𝐶𝐷 ? A. 8√3 B. 9√3 C. 16 D. 17√3 56.

If tan

3 find sin sec . 5

is then √3/2 of 18, which is 9 3 . A PLOD: M

CONCEPT: Trigonometry COMPETENCY TESTED: uses trigonometric ratios to solve real-life problems involving right triangles. ***

A.

15 34 170

B.

49 34 170

BT: Remembering, Applying

34 15 170

EXPLN:

C. D.

34 49 170

tan

O 3 H 5

so finding the hypotenuse, H=

3 2 5 2 34

Hence,

ADMATH – TG 2015

25

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

54.

ACET 2016 MATHEMATICS

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS sin

O 3 34 3 34 H 34 34 34

sec

H 34 A 5

sin sec

Determine the equation described by the ordered pairs: (-4,18), (-2,6), (0,2), (1,3) and (3,11). A.

y x2

B.

y x2 2

C.

y x2 2

D.

y x 2 2x 2

B PLOD: E

49 34 170

CONCEPT: Geometry

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

57.

3 34 34 34 5

COMPETENCY TESTED: determines the equation of a quadratic function given: (a) a table of values; (b) graph; (c) zeros. BT: Applying EXPLN: (Trial and Error) By substituting the values of x and y in the ordered pairs to the equations in the choices, the answer will be B y x 2 2 .

58.

Which of the following is the graph of f (x) =–x3 ? A.

B

CONCEPT: Graphing Polynomials

PLOD: Moderat e

COMPETENCY TESTED: graphs polynomial functions. BT: Analyzing EXPLN: Plot arbitrary values for the function X Y -2 8 -1 1 0 0

ADMATH – TG 2015

26

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

B.

1 2 Answer is B.

-1 -8

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

REF: Graphs taken from Tsishchanka, Kiryl. Section 3.2 Polynomial Functions And Their Graphs. 1st ed. Web. 4 Nov. 2015.

C.

D.

59.

A square has coordinates (1,5) (1,1) (5,1)(5,5). Find the point of intersection of the diagonals of the square. A. B. C. D.

(2,4) (3,3) (4,2) (4,4)

B PLOD: M

CONCEPT: Trigonometry COMPETENCY TESTED: solves problems involving geometric figures on the coordinate plane BT: Remembering EXPLN: Plot the points. By the property of squares, the diagonals bisect each other, which means that they intersect at their midpoints. Using the midline theorem, the midpoint of the line with

1 5 1 5 , ) (3,3). 2 2

points (1,1) & (5,5) is (

Similarly, the midpoint of the line with points

1 5 1 5 , ) (3,3). Hence, the 2 2

(1,5) & (5,1) is (

ADMATH – TG 2015

27

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS point of intersection is (3,3).

60.

CONCEPT: Probability Three fair coins are tossed independently. Determine the probability of obtaining at most two heads. E.

BT: Analyzing, Applying

D PLOD: M

EXPLN: At most 2 heads means having 0 head, 1 head, or 2 heads. The combinations are as follows: TTT, HTT, THT, TTH, HHT, HTH, THH. Each combination has the probability

1 1 1 1 since the probability of obtaining a 2 2 2 8 head or a tail are equal. There are 7 combinations of having at most 2 heads. So 7

ADMATH – TG 2015

1 7 8 8

28

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

1 4 3 F. 4 3 G. 8 7 H. 8

COMPETENCY TESTED: solves problems involving probability.

View more...
ACET 2016 MATHEMATICS

Find the value of a:

D

a 2 2a 5 0

PLOD: M

CONCEPT/ COMPETENCY TESTED/ BLOOM’S TAXONOMY/ EXPLANATION/ REF. CONCEPT: Algebra: Quadratic Functions COMPETENCY TESTED: solves quadratic equations by: (a) extracting square roots; (b) factoring; (c) completing the square; and (d) using the quadratic formula.

A.

a 1 6

B.

a 1 6

C.

a 1 5

BT: Application

D.

a 1 6

EXPLN: By completing squares

a 2 2a 5 0 a 2 2a 5 a 2 2a 1 5 1 (a 1) 2 6 a 1 6 a 1 6 2.

Find the discriminant and characteristic of the root of the equation:

PLOD: E

4x2 + 20x – 53 = 0 A. B. C. D.

D

-1248; imaginary roots 0; exactly one root 612; real roots 1248; real roots

CONCEPT: Quadratic Equations COMPETENCY TESTED: characterizes the roots of a quadratic equation using the discriminant BT: Remembering, Analyzing EXPLN: The discriminant of a quadratic equation is given by b 4ac . Thus, the discriminant of the given equation is 2

20 2 (4 4 53) 400 848 1248 Hence, the quadratic equation has discriminant 1248 and has real roots.. CONCEPT: Variation

3. X

1

5

9

Y

1.5

.7

.61

Z

3

7

11

Refer to the table above. Which variation statement satisfies the values in the table? A. Y kX Z

ADMATH – TG 2015

D PLOD: E

COMPETENCY TESTED: translates into variation statement a relationship between two quantities given by: (a) a table of values; (b) a mathematical equation; (c) a graph, and vice versa. BT: Applying

1

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

1.

ANS./ PLOD

QUESTION

NO.

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

kX Z k C. Y XZ kZ D. Y X

EXPLN: (Trial and Error) The first step should be to find the constant k. By substituting the values of x, y and z in the ordered pairs to the equations in the choices,

B. Y

the answer will be D. Y

CONCEPT: Exponents Simplify:

(3x 4 y 7 z 12 ) 3 (5 x 9 y 3 z 4 ) 2 A.

675 x 18 y 9 z 13

B.

675 x 30 y 15 z 28

B

C. 675 x 15 y 27 z 30 D. 675 x

15

y

30

z

COMPETENCY TESTED: applies the laws involving positive integral exponents to zero and negative integral exponents.

PLOD: E

28

BT: Applying EXPLN: Applying the laws of exponents, the given expression can be solved as follows.

(3x 4 y 7 z 12 ) 3 (5 x 9 y 3 z 4 ) 2 =

5.

(27 x 12 y 21 z 36 )(25 x 18 y 6 z 8 ) 675 x 30 y 15 z 28

CONCEPT: Radicals

Simplify the given expression:

COMPETENCY TESTED: simplifies radical expressions using the laws of radicals.

8x 3 y 3 3 2 x 5 y 5 18 x 7 y 7 A.

B.

C.

2 xy 3 2 3x 2 y 2 2 3 2 2 xy x y

2 2

3x y D.

2

1 xy

BT: Applying, Analyzing C PLOD: M

EXPLN: Using the Laws of radicals and exponents, the expression can be simplified as follows:

8x 3 y 3 3 2 x 5 y 5 18 x 7 y 7

3x 2 y 2 2 3 xy

ADMATH – TG 2015

2

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

4.

kz where k=1/2. x

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

=

2 xy 2 xy 3 x 2 y 2 2 xy 3 x 3 y 3 2 xy 2 xy 3 x 2 y 2 3x 3 y 3 2 3 xy 3x 2 y 2 2 2

3x y 6 6.

2

1 xy

CONCEPT: Geometry COMPETENCY TESTED: uses properties to find measures of angles, sides and other quantities involving parallelograms. BT: Remembering, Analyzing A PLOD: E

7.

A quadrilateral is inscribed in a circle (see figure above). What is angle θ? A. 113o B. 117o C. 130o D. 243o 3 5 s and s = t , find the ratio of t to r. If r = 4 6 A. 5:8 B. 8:5 C. 9:10 D. 10:9

EXPLN: For quadrilaterals inscribed in circles, opposite angles supplement each other. Hence, 180o = θ + 67o θ = 180o - 67o = 113°

CONCEPT: Proportions B PLOD: E

COMPETENCY TESTED: applies the fundamental theorems of proportionality to solve problems involving proportions. BT: Application

EXPLN: Substitution r=

ADMATH – TG 2015

3 4

s,s=

5 t 6

3

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

r=

B. 3 C. 6 D.

9.

2

6 2

Find the sixth term of an arithmetic sequence whose second and tenth terms are 14 and 58, respectively. A. 20 B. 36 C. 44 D. 72

5

C PLOD: E

B PLOD: E

COMPETENCY TESTED: finds the trigonometric ratios of special angles. BT: Remembering, Applying EXPLN: Each leg of a 45-45-90 triangle is 1/√2 of the hypotenuse. Each leg is then √9 = 3. The sum of two legs is 6 . CONCEPT: Sequences COMPETENCY TESTED: determines arithmetic means and nth term of an arithmetic sequence.*** BT: Application EXPLN: Notice that the middle of 2nd and 10th term is the 6th term, thus, the 6th term is the average;

14 58

10.

divided by (x 3) .

COMPETENCY TESTED: performs division of polynomials using long division and synthetic division.

2 x 9

2 x 9 2 C. x 3x 9 2 D. x 3x 9

72

36 2 2 CONCEPT: Algebra

3 Find the quotient when P(x) x 27 is

A.

B.

D PLOD: M

BT: Analysis EXPLN: Sum of Two Cubes 3 3 2 2 a b (a b)(a ab b )

3 2 x 27 (x 3)(x 3x 9) Or Synthetic division x 3 x 3

ADMATH – TG 2015

4

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

A 45-45-90 triangle has a hypotenuse of length √18. How long is the sum of the lengths of its two legs? A. 3

t

46

t 8 Thus, r : t = 5:8 t : r = 8:5 CONCEPT: Geometry r=

8.

35

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS -3 1 0 0 27 -3 9 -27 1 -3 9 -27

2 x 3x 9 11.

B PLOD: E

CONCEPT: Geometry COMPETENCY TESTED: solves problems on circles. BT: Remembering, Analyzing

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

EXPLN: Inscribed Angle theorem: Central angle = 2θ Arc length: C = 2R

12.

Refer to the figure above. If R is the radius of the circle, what is the length of C? A. Rθ B. 2Rθ C. πR D. 2πR Consider a family of four standing side by side for a family portrait, in how many ways can they arrange themselves? A. 24 B. 28 C. 30 D. 32

A PLOD: A

CONCEPT: Permutations COMPETENCY TESTED: solves problems involving permutations. BT: Applying EXPLN: n Pr

n! , (n r )!

where n = total number of objects; r = number of objects chosen (want)

n Pr

13.

A Four coins are flipped. What are the chances of NOT getting heads? A. 1/16 B. 1/4 C. 1/2 D. 15/16

PLOD: E

4! 4! 24 (4 4)!

CONCEPT: Probability COMPETENCY TESTED: solves problems involving probability. BT: Creating, Understanding, Evaluating EXPLN:

ADMATH – TG 2015

5

TEACHER’S GUIDE SET A

The following are the scores of 20 students on their 30 item exam: 4,5,6,6,7,8,10,10,11,16,17,17,18,19,20,20,21, 23,25,30 Determine the 90thpercentile of the given data. A. B. C. D.

C PLOD: E

20 22 24 26

The probability of not getting heads is equal to the probability of getting four tails Each flip has a 1/2 chance to get a tails. Do it four times, the chance of getting four tails is (1/2)4 = 1/16 CONCEPT: Measure of Position COMPETENCY TESTED: calculates a specified measure of position (e.g. 90th percentile) of a set of data. BT: Creating, Understanding, Evaluating For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

14.

ACET 2016 MATHEMATICS

EXPLN: Make sure first that all data are arranged from least to greatest. The formula for percentile:

X PK

(

nk ) 100

X (

nk 1) 100

2

where n=number of observations and k=percentile.

nk (20)(90) 18 100 100 (20)(90) nk 1 1 19 100 100

So

Note that X1=4, X2=5, X3=6 and so on… So X18=23 and X19=25 and substituting,

PK 15.

CONCEPT: Quadratic equations

Determine the quadratic equation with roots 3/2 and 5. 2 A. 15 x 7 x 2 2 B. 7 x 2 x 15 2 C. 2 x 15 x 7

D.

X 18 X 19 23 25 24 2 2

D PLOD: E

COMPETENCY TESTED: describes the relationship between the coefficients and the roots of a quadratic equation. BT: Application

2 x 7 x 15 2

EXPLN: Given the roots, the quadratic equation can be solved as follows:

x

3 2

or x 5

(2 x 3)( x 5) 0 2 x 2 7 x 15 0 ADMATH – TG 2015

6

TEACHER’S GUIDE SET A 16.

ACET 2016 MATHEMATICS

If y varies directly as x2 and inversely as z, and y=8 when x=1 and z=2, find y when x=3 and z=6. A. 8 B. 16 C. 24 D. 32

C

CONCEPT: Variation

PLOD: E

COMPETENCY TESTED: solves problems involving variation BT: Application EXPLN:

kx 2 z k (1) 2 8 2 k 16, then

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

y

16(3) 2 16(9) 6 6 y 24

y

17.

1

Simplify:

A.

B.

1 2

x 3 y 2

C PLOD: E

3 x2

COMPETENCY TESTED: simplifies expressions with rational exponents.

EXPLN:

3 2

x

CONCEPT: Rational Expressions

BT: Understanding, Analyzing, Applying

5 y2

y

C.

x2 y

1

5 2

x2 y

1 2

x 3 y 2

3 y2

x

1 1 ( 3) ( 2 ) 2 y 2

1

5 3 ( ) ( ) x 2 y 2

y x

3 ( ) 2 5 ( ) 2

5

x2 5

D.

y2

x

18.

5 2

Find the solution/s to the following equation:

1 1 x2 2 x 4 x 4 x 16 A. B. C. D.

x=0,2 x=2,4 x=0,4 x=4,-4

ADMATH – TG 2015

A PLOD: M

CONCEPT: Geometry COMPETENCY TESTED: solves equations transformable to quadratic equations (including rational algebraic equations). BT: Remembering, Analyzing EXPLN:

7

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS 1 1 x2 2 x 4 x 4 x 16 The LCD is 𝑥 2 − 16, so the equation will become,

x4 x 2 16

x4 x 2 16

x2 x 2 16

x 4 x 4 x2 x( x 2) 0 x 0 and x 2 19.

Simplify the expression:

C

CONCEPT: Radical Expressions

( 2 x 3 y )( 2 x 3 y )

A.

2x 3 y

B.

4x 2 9 y 2

PLOD: E

COMPETENCY TESTED: performs operations on radical expressions.***

C. ( 2 x )( 2 x 3 y ) ( 3 y )( 2 x 3 y )

BT: Application

D. ( 3 x )( 2 x 3 y ) ( 2 y )( 2 x 3 y )

EXPLN:

( 2 x 3 y )( 2 x 3 y ) ( 2 x )( 2 x 3 y ) ( 3 y )( 2 x 3 y )

20.

Find the values of x:

x 1 1 2 x 1 x 2 x 3x 2 A. B. C. D.

-1 2 -1 and 2 1 and 2

A PLOD: D

CONCEPT: Algebraic Equations COMPETENCY TESTED: solves problems involving quadratic equations and rational algebraic equations. BT: Analysis and Application EXPLN:

x 1 1 x 1 x 2 x 2 3x 2 x 1 1 x 1 x 2 ( x 1)( x 2) x ( x 2) 1( x 1) 1 ( x 1)( x 2) ( x 2)( x 1) ( x 1)( x 2) x( x 2) ( x 1) 1 x2 x 2 0 x 2 or x 1 But x=2 will make the equation undefined. So

ADMATH – TG 2015

8

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x 2 2x 0

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS only x= -1.

21.

Refer to the figure below:

B PLOD: M

CONCEPT: Midline Theorem COMPETENCY TESTED: proves the Midline Theorem.

EXPLN: Since line DE is a midline of triangle ABC, line DE || line AB. Note that ∠A+∠B+∠C=180°, so ∠A=180°-∠C-∠B ∠A=180°-45°-50°=85°

If ∠C=45°, ∠B=50°, find ∠BED + ∠CDE. A. B. C. D.

205° 215° 225° 235°

22.

A

PLOD: M

Since line DE || line AB, ∠A+∠ADE=180° ∠ADE=180°-∠A=180°-85° ∠ADE =95° And ∠B+∠BED=180° ∠BED=180°-∠B=180°-50° ∠BED=130° Now, ∠ADE + ∠CDE=180° (Supplementary Angles) ∠CDE=180° - ∠ADE =180° - 95° ∠CDE = 85° Hence, ∠BED + ∠CDE= 130°+85° ∠BED + ∠CDE= 215° CONCEPT: Similar Triangles COMPETENCY TESTED: solves problems that involve triangle similarity and right triangles.*** BT: Remembering, Analyzing

Find x and y. A. x =12, y = 15 B. x =12, y = 20 C. x =15, y = 15 D. x =15, y = 20

ADMATH – TG 2015

EXPLN: The triangles in the figure are all similar. From there x can be found: x 16 9 x x 2 16 9 x 12 Since x is 12, along with the other leg equal to 9 of the smallest triangle, these are recognized as the legs of a Pythagorean triple. Thus, 𝑦 = 15.

9

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BT: Analysis, Application

TEACHER’S GUIDE SET A A wooden plank leaning against a wall makes a 30° angle with the ground. If the point at which the plank touches the wall is 5 meters above the ground, what is the length of the wooden plank? A. 5 meters B. 5√3 meters C. 10 meters D. 10√3 meters

C PLOD: M

CONCEPT: Algebra COMPETENCY TESTED: uses trigonometric ratios to solve real-life problems involving right triangles. *** BT: Remembering, Analyzing EXPLN: Let l- length of the plank

l

5

30°

s l 5 l 10 meters 1 2 CONCEPT: Sequences sin 30

24.

An arithmetic sequence of seven numbers sums to 126. If the first number in the sequence is 6, what is the common difference? A. 3 B. 4 C. 5 D. 6

COMPETENCY TESTED: finds the sum of the terms of a given arithmetic sequence.*** BT: Remembering, Analyzing

B PLOD: M

ADMATH – TG 2015

EXPLN: Let d = common difference x = first number Set-up: x + (x+d) + (x+2d) + … + (x + 6d) = 126 Expect seven x’s and d + … + 6d = 21d 7x + 21d = 126 Since x = 6 42 + 21d = 126 21d = 84 d=4 OR Let Sn = sum of n numbers a1 = first number Formula: Sn = n(2a1 + (n-1)d)/2 Substitute n = 7, S7 = 126 126 = 42 + 21d 21d = 84

10

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23.

ACET 2016 MATHEMATICS

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS d=4

Consider the polynomial

x5 2x4 3x3 6x2 4x 8

D PLOD: E

Which of the following is/are factors of the given polynomial? I. II. III. A. B. C. D.

26.

EXPLN: Trial and Error. By Factor Theorem, we consider: I. (x-1) f(1)=1-2+3-6-4+8 = 0, so it IS a FACTOR II. (x+1) f(1)=-1-2-3-6+4+8 = 0, so it IS a FACTOR III. (x-2) f(2)=32-32+24-24-8+8=0, so it IS a FACTOR

I only I and II only II and III only I, II and III

CONCEPT: Circles

A circle with radius r is centered at (1,2). The point (x, y) = (5,-4) is on the circle. Find r.

13

B.

20

C.

2 13

D.

2 20

COMPETENCY TESTED: proves the Remainder Theorem and the Factor Theorem. BT: Remembering, Applying, Evaluating

(x-1) (x+1) (x-2)

A.

CONCEPT: Remainder Theorem/Factor Theorem

COMPETENCY TESTED: applies the distance formula to prove some geometric properties. C PLOD: E

BT: Analyzing, Applying EXPLN: Use the distance formula since r is the distance between the center of a circle and a point on the circle.

r (1 5) 2 (2 4) 2 16 36 52 2 13 27.

In how many different ways can the letters of the word “MISSISSIPPI” be arranged? A. 34,644 B. 34,646 C. 34,648 D. 34,650

CONCEPT: Combinations COMPETENCY TESTED: solves problems involving permutations and combinations. D BT: Application PLOD: M EXPLN: There are 11 letters: M-1, S-4, I-4, P-2

ADMATH – TG 2015

11

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25.

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS So,

Find the 75th percentile of the set of values below: {15,9,7,6,1,4,3,10,7,8,9,12} A. B. C. D.

B PLOD: E

9 9.5 10 11.25

CONCEPT: Measures of Position COMPETENCY TESTED: interprets measures of position. BT: Applying EXPLN: Arrange the data from lowest to highest: {1,3,4,6,7,7,8,9,9,10,12,15}

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

28.

11! 11.10.9.8.7.6.5.4.3.2.1 34,650 1!4!4!2! (4.3.2.1)(4.3.2.1)(2.1)

The formula for percentile:

X PK

(

nk ) 100

X (

nk 1) 100

2

where n=number of observations and k=percentile. Where n=12 and k=75 Compute for nk/100 = (12)(75)/(100) nk/100=9 nk/100 +1 = 10

PK

X (9) X (10 ) 2 9 10 2

, so the 75th percentile is 9.5 29.

Find a possible inequality whose solution set is given by 2 x 5 . A.

x 2 7 x 10 0

B.

x 2 7 x 10 0

C.

x 7 x 10 0

D.

x 2 7 x 10 0

B PLOD: M

CONCEPT: Algebra COMPETENCY TESTED: solves quadratic inequalities. BT: Analysis

2

EXPLN: Inequality Shortcut factor choices first, A. (x 2)(x 5) 0 B. (x 2)(x 5) 0 C. (x 2)(x 5) 0 D. (x+2)(x+5)≥ 0 Assuming there is no “negative x”, and the right-

ADMATH – TG 2015

12

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS hand side is 0, the solution set is given by Case 1: if 0 x s small critical # or x b big critical # Case 2: if ≤0 s x b

Find the solutions of:

x 3 x 3 A. B. C. D.

COMPETENCY TESTED: solves equations involving radical expressions.***

x=3 x=4 x=5 No solution

BT: Applying, Analyzing EXPLN:

x3 x 3 ( x 3 x ) 2 32 D PLOD: M

2 x 3 2 x 2 3x 9 2 x 12 2 x 2 3x ( x 6) 2 ( x 2 3x ) 2 x 2 12 x 36 x 2 3x 36 9 x 4x Check if the resulting value of x is a root of the given equation.

4 3 4 1 2 1 3 Therefore, x=4 is not a root of the equation and hence there are no solutions. CONCEPT: Quadrilaterals

31.

COMPETENCY TESTED: solves problems involving parallelograms, trapezoids and kites. B Refer to the figure above. The top base of the trapezoid is 8. A lateral side measures 10. What is the area of the trapezoid? A. 108 B. 112 C. 144 D. 160

ADMATH – TG 2015

PLOD: M

BT: Remembering, Applying, Analyzing EXPLN: Compute for length of the shortest leg of the right triangle at the left. Shortest leg length = √102 − 82 Shortest leg length = 6 (Or recognize that to make a Pythagorean triple,

13

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30.

Thus, answer is B. CONCEPT: Radical Equations

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS the leg must be 6) The triangles are congruent since they are SSS Method 1:□ + 2Δ Area of □ = 82 = 64 Total area of 2 Δs = 2(8×6/2) = 48

area of trapezoid 112

32.

CONCEPT: Geometry; Similar/Congruent Triangles

Three pairs of parallel segments make up two triangles, ∆𝑍𝑌𝑋 and ∆𝑊𝑉𝑈, shown below.

COMPETENCY TESTED: applies the theorems to show that given triangles are similar. BT: Evaluation EXPLN: C

Which of the following is/are ALWAYS true for∆𝑍𝑌𝑋and ∆𝑊𝑉𝑈? I. They are congruent. II. They are right triangles. III. They are AAA triangles. A. B. C. D. 33.

PLOD: M

WV = ZY, WU = ZX, UV = XY Therefore, angles are also the same. No information whether the angles form a right angle. I and III

II only III only I and III I, II, and III

Given the sequence 15,

15 15 , , ... 2 4

Find the 7th term (a7 ) and give the geometric mean (m) of the first and 7th term. A.

a7

15 15 ;m 32 8

B.

a7

15 15 ;m 32 16

ADMATH – TG 2015

C PLOD: E

CONCEPT: Geometric Sequences COMPETENCY TESTED: determines geometric means and nth term of a geometric sequence.*** BT: Applying, Analyzing EXPLN:

14

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Method 2: Trapezoid base1 = 20, base2 = 8, height = 8 area = (20+8)×8/2 = 28×8/2 = 28×4 area = 112

TEACHER’S GUIDE SET A

15 15 ;m 64 8

a 1 15 ; common ratio(r) a n a1 r ( n 1)

15 15 ;m D. a 7 64 16

a 7 (15)

1 ( 7 1) 15 2 64

m 15 34.

x 2x 4x 8 3

C. D.

225 15 64 8

2

COMPETENCY TESTED: factors polynomials.

x 4 16

B.

15 64

CONCEPT: Algebra

Simplify:

A.

1 2

1

BT: Application

x 4 1 x4 1 x2 1 2

EXPLN: Numerator: factored by grouping Denominator: difference of two squares (DOTS)

x 3 2x 2 4x 8 x 4 16

x 2 2

x 3 4x 2x 2 8 ( x 2 4)( x 2 4)

x( x 2 4) 2( x 2 4) ( x 2 4)( x 2 4)

Cancellation: C PLOD: D

x3 2 x 2 4 x 8

x 4 16 ( x 2)

( x 2)( x 2 4) ( x 2 4)( x 2 4)

( x 2 4)

Denominator: DOTS

x 3 2x 2 4x 8 x 16 4

( x 2) ( x 2)( x 2)

( x 2) ( x 2)( x 2)

Cancellation:

x 3 2x 2 4x 8 x 16 1 ( x 2) 4

35.

Find the center (x,y) and radius (r) of the circle 1 with equation: 2 x 5 y ( x 2 y 2 13) 0 2 A. (-5,2); r=4

ADMATH – TG 2015

C PLOD: M

is the hypotenuse of ΔBCF. is 9 since adjacent of 30 is the hypotenuse of ΔCDF. Thus, is the hypotenuse of ΔDEF. Thus, = . CONCEPT: Circles COMPETENCY TESTED: determines the center and radius of a circle given its equation and vice versa

15

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C. a 7

ACET 2016 MATHEMATICS

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

B. (-2,5); r=16 C. (-2,5); r=4 D. (5,-2); r=16

BT: Remembering, Analyzing EXPLN: Multiplying the equation by 2 will result to

4 x 10 y x 2 y 2 13 0 and rearranging the terms, the equation will be

x 2 4 x y 2 10 y 13 By completing the squares method, the resulting equation will be and by factoring the perfect square trinomials, will yield to

( x 2)2 ( y 5)2 16 Thus, the center is (-2,5) and the radius is

r 16 4 . 36.

Find the lower and upper quartile respectively, in the following data set: 1, 11, 19, 15, 20, 24, 28,34, 37, 47, 50, 57 A. 15 and 47 B. 17 and 42 C. 19 and 47 D. 20 and 50

B PLOD: E

CONCEPT: Quartiles, Decile, Percentiles COMPETENCY TESTED: solves problems involving measures of position. BT: Applying EXPLN: To get the lower quartile, make sure that the data set is arranged in increasing order. 1, 11, 15, 19, 20, 24, 28, 34, 37, 47, 50, 57 The median

24 28 26 .The lower 2

quartile (Q1) is the median of the lower half of the data set. Then the

15 19 17 . 2 37 47 The upper quartile 42 2 lower quartile

37.

A rectangular lot has an area of 560 square meters. The length of the lot is three more than twice its width. Find the length and width of the rectangular lot. A. L=30 meters, W=15 meters B. L=35 meters, W=15 meters C. L=32 meters, W=16 meters D. L=35 meters, W=16 meters

ADMATH – TG 2015

D PLOD: M

CONCEPT: Quadratic Functions COMPETENCY TESTED: models real-life situations using quadratic functions. BT: Application EXPLN: Let L - length of the lot; W – width of the lot

16

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( x 2 4 x 4) ( y 2 10 y 25) 13 4 25

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS Area of a rectangle: A=LW LW=560 (2W+3)(W)=560 2W 2+3W=560 2W 2+3W-560=0 Using the Quadratic Formula to solve for W:

W

b b 2 4ac 2a 3 3 2 4(2)( 560) 2(2) For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

3 4489 4 3 67 3 67 or 4 4 64 - 70 or 4 4 cannot be negative

64 16 meters 4 L = 2(16)+3=32+3= 35 meters

So, W

38. Evaluate:

48 27 2

A.

6 2

B.

6

C.

3 2

D.

3

A PLOD: E

CONCEPT: Radical Expressions COMPETENCY TESTED: solves problems involving radicals. BT: Applying, Evaluating EXPLN:

(16)(3) (9)(3)

4 3 3 3

2 Rationalizing, 39.

Given the quadratic equation:

x 2 24 x 45 y Which of the following statements is/are true? I. The graph of the equation is open downwards II. The vertex is at (12, -99)

ADMATH – TG 2015

D PLOD: M

2 3

2

3 2

6 2

2 2 CONCEPT: Quadratic Equation COMPETENCY TESTED: graphs a quadratic function: (a) domain; (b) range; (c) intercepts; (d) axis of symmetry; (e) vertex; (f) direction of the opening of the parabola. BT: Analyzing

17

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

III. The range is { y R | y 99} A. B. C. D.

EXPLN: The vertex form of the quadratic equation will be used to solve for the vertex. Rearranging terms in the equation will yield

II only I only I and II only II and III only

x 2 24 x y 45 And completing the squares will result to

x 2 24 x 144 y 99 ( x 12) 2 99 y

40.

X varies inversely as the cube of Z and

1 directly as . X will be 4 if Y=4 and Z=2. Find Y 1 𝑌

if X=5 and Z=3.

D

upwards. Therefore, only II and III are true. CONCEPT: Variation

PLOD: E

COMPETENCY TESTED: solves problems involving variation.

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

Hence, the vertex of the quadratic equation is at the point (12,-99). This implies that the range is { y R | y 99} and the graph is opening

BT: Remembering, Applying

32 A. 5 5 B. 32 32 C. 27 27 D. 32

EXPLN:

1 k X Y3 3 Z Z Y k 4 3 2 (15) k 160 k

1 160( ) Y X Z3 1 5(3)3 Y 160 1 135 27 Y 160 32 41.

A PLOD: E

CONCEPT: Similar Triangles COMPETENCY TESTED: applies the theorems to show that given triangles are similar. BT: Analysis EXPLN: Only I is true according to properties of similar triangles.

Given the triangle above, which of the

ADMATH – TG 2015

18

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

following statements is ALWAYS TRUE when triangle AEC is similar to triangle BCD? I. II. III.

42.

CBD BAE

AE BD CE CD BCD BAE AB BC

NOT BCD BAE III is false.

I only I and II only II and III only I, II and III CONCEPT: Algebra

Find the sum of the infinite sequence:

5 5 5 , , , ... 27 81 243

6 5 5 B. 9 5 C. 6 9 D. 5

COMPETENCY TESTED: finds the sum of the terms of a given finite or infinite geometric sequence.***

A.

BT: Remembering, Analyzing, Applying

C PLOD: D

EXPLN: Observe that the given sequence is a geometric sequence with common ratio of 1/3. This can be shown using the formula a n ra n 1 . Also, the first term is a

5 which can be found using the 9

formula a n ra n 1 . Since the common ratio r < 1, the formula is

i 1 a i 1 r a

Substituting values, the sum is

5 9 1

43. If 4a+b=3c and -8a+2b+6c=24, what is the value of 2b? A. 3 B. 6 C. 9 D. 12

1 3

5 . 6

D

. CONCEPT: Polynomial Equations

PLOD: A

COMPETENCY TESTED: solves polynomial equations. BT: Application EXPLN:

ADMATH – TG 2015

19

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A. B. C. D.

II is false,

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS 8a 2b 6c 24 2 4a b 3c 12 4a b 3c 0

44.

Equ.1 Equ.2

C

2b = 12 CONCEPT: Circles

PLOD: D

COMPETENCY TESTED: solves problems on circles.

A kite is inscribed in a circle with center at C and the bottom of the kite has an interior angle of 30o. What is the area of the shaded region if the circle has a radius of 2? A. 2π - 4 B. 2π - 2 C. 4π – 4 D. 4π - 2

45.

In a family, boys and girls are equally likely to be born. Find the probability that in a family with three children, exactly one is girl. A. B. C. D.

3/8 1/8 5/8 7/8

ADMATH – TG 2015

A PLOD: M

EXPLN: The central angle with vertex at C is twice the inscribed angle. The central angle is 60o. Bisect the central angle, to get two triangles with 30o angles with vertices at C. Draw a chord from between unequal segments of the kite. Half of the chord can be found using the radius since opposite a 30o is half the radius which is 1. The kite’s smaller diagonal is 2. The longer diagonal is found since it is twice the radius which is 4. Area of kite: 2( 4) Ak= =4 2 Area of Circle: Ac = (2)2π = 4π Area of shaded region: As = Ac - Ak As = 4π-4 CONCEPT: Probability COMPETENCY TESTED: solves problems involving probability. BT: Analyzing, Applying EXPLN: Since the probability of having a boy or a girl is equal, then each has a probability of ½. Let B- event that a boy is born. G- event that a girl is born. There are 8 total combinations of having three children and these are: BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG. So, there are 3 combinations of having exactly one girl: BBG,

20

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BT: Applying, Analyzing, Remembering

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS BGB, GBB. Hence, the probability of having exactly one girl is 3/8.

Using the set of values below:{26,5,5,8,20,2,3,24} find the second quartile (Q2) and the range (r). A. B. C. D.

C PLOD: M

Q2=5; r=24 Q2=5; r=26 Q2=6.5; r=24 Q2=6.5; r=26

CONCEPT: Measures of Position COMPETENCY TESTED: uses appropriate measures of position and other statistical methods in analyzing and interpreting research data. BT: Analyzing, Applying EXPLN: Arrange the data from lowest to highest: {2,3,5,5,8,20,24,26} Second Quartile (Q2) = 50th percentile The formula for percentile:

X (

PK

nk ) 100

X (

nk 1) 100

2

where n=number of observations and k=percentile. Where n=8 and k=50 Compute for nk/100 = (8)(50)/(100) nk/100=4, and nk/100+1=5

P50

X ( 4 ) X ( 5) 2

58 2 6 .5 so the 50th percentile is 6.5. The range can be computed by subtracting the highest value – lowest value = 26 - 2 =24 47.

A cannonball travels along the path with equation

3 4x 4x 2 0 Find the roots of the quadratic equation. A. B.

2 8 , 3 3 2 8 x , 3 3 x

ADMATH – TG 2015

C PLOD: E

CONCEPT: Quadratic Equations COMPETENCY TESTED: models real-life situations using quadratic functions. BT: Application

EXPLN:

21

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46.

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

3 1 , 2 2 3 1 D. x , 2 2 C.

x

x

b b 2 4ac 2a

(4) (4) 2 4(4)(3) 2(4) 4 64 4 8 8 8 1 2 2 3 1 x , 2 2

48.

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

CONCEPT: Polynomials

Find the quotient if

f(x) x 3x 3x 1 3

2

COMPETENCY TESTED: solves problems involving polynomial functions.

is divided by (x - 1) . x 2 2x 1

A. B.

x 2 - 2x 1

C.

x 2 2x 1

B PLOD: A

2 D. x + 2x 1

BT: Application EXPLN: Synthetic Division

x 1 0 x 1 1 -3 3 -1 1 -2 1 1 -2 1 0 x

49.

2

- 2x 1

CONCEPT: Algebra

(-1,7)

(3,7) COMPETENCY TESTED: solves problems involving geometric figures on the coordinate plane.

(-1,3)

BT: Analyzing, Applying

(3,3) B

A square with corner coordinates (-1,3)(1,7)(3,3)(3,7) is circumscribed in a circle as shown in the diagram above. Find the area of the shaded region. A. B. C. D.

8π+16 square units 8π-16 square units 32π+16 square units 32π-16 square units

PLOD: D

EXPLN: Solving the length of the diagonal of a square will give the diameter of the circle. By distance formula or properties of a 45-45-90 triangle, the length of the diagonal can be found to be 4 √2.

D (3 1) 2 (7 3) 2 4 2 4 2 32 4 2 By 45-45-90 triangle, the base has length=4 so multiply by √2 to get the hypotenuse = diagonal = diameter 4 2

ADMATH – TG 2015

22

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS So, radius = 2 2 Solving for the area of the circle:

A r 2 (2 2 ) 2 8 square units Area of square = 42=16 square units Finally,

8 16 square units

50.

Find the sixth term of an arithmetic sequence whose second and tenth terms are 14 and 58, respectively. A. 20 B. 36 C. 44 D. 72

B PLOD: E

CONCEPT: Arithmetic COMPETENCY TESTED: solves problems involving sequences. BT: Application EXPLN: Notice that the middle of 2nd and 10th term is the 6th term, thus, the 6th term is the average; 14 58

51.

72

36 2 2 CONCEPT: Algebra

Find the solution set of the given polynomial equation:

COMPETENCY TESTED: solves problems involving polynomials and polynomial equations.

x 3 6 x 2 x 6. A. B. C. D.

{ -1, 1, -6 } { -1, 1, 6 } { -3, 2, 6 } { 3, -2 , 6 }

BT: Application

A PLOD: M

EXPLN: (Partly by Trial and Error) By the factor theorem, we will see that x+1 or x1 or x+5 are factors of the polynomial. Consider x+1. By factor or long/synthetic division,

x 3 6x 2 x 6 0 ( x 2 5 x 6)( x 1) 0 Furthermore,

( x 6)( x 1)( x 1) 0 So,

x 6 0; x 1 0 and x 1 0 Hence, the solution set is {-1,1,-6}

ADMATH – TG 2015

23

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area of the shaded region =

TEACHER’S GUIDE SET A

52.

ACET 2016 MATHEMATICS

C PLOD: D

CONCEPT: Circles COMPETENCY TESTED: solves problems on circles.

EXPLN: (Area of the sector of circle) – (Area of triangle) = (Area of shaded region)

2 3 Area of the Sector (2 3 ) 2 4 2

If ̅̅̅̅ 𝐴𝐵 is the radius of the circle equal to 2√3 and ∠ABC=2π/3, what is the area of the shaded region? A.

̅̅̅̅ to ΔABC is an isosceles triangle. Bisect 𝐴𝐶 ̅̅̅̅ create two congruent right triangles. 𝐴𝐵 and ̅̅̅̅ 𝐵𝐶 are now hypotenuses of the two right triangles. The angles of the triangles, coinciding with the center, are just half of the central angle. They each measure π/3. BAC is then equal to π/6.

4 3

3 3 2 C. 4 3 3 B.

4

1 2

Bisector length = 2 3 ( ) 3

D. 4 4 3

̅̅̅̅ = 6 𝐴𝐶 Area of ABC

6 3 3 3 2

(Area of shaded region) = 4 3 3 53.

CONCEPT: Probability Three fair coins are tossed independently. Determine the probability of obtaining at most two heads.

1 4 3 B. 4 3 C. 8 7 D. 8

COMPETENCY TESTED: solves problems involving probability.

A.

BT: Analyzing, Applying

D PLOD: M

EXPLN: At most 2 heads means having 0 head, 1 head, or 2 heads. The combinations are as follows: TTT, HTT, THT, TTH, HHT, HTH, THH. Each combination has the probability

1 1 1 1 since the probability of obtaining a 2 2 2 8 head or a tail are equal. There are 7 combinations of having at most 2 heads. So 7

ADMATH – TG 2015

1 7 8 8

24

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

BT: Applying, Analyzing, Remembering

TEACHER’S GUIDE SET A

If the graph of the equation y=8(x-2)2+5 is changed to y=8(x-3)2+5, what will happen to the graph of the new equation compared to the original one? A. The graph will move one unit to the left. B. The graph will move one unit to the right. C. The graph will flip with respect to the xaxis. D. The graph will flip with respect to the yaxis.

55.

B

CONCEPT: Algebraic equations

PLOD: E

COMPETENCY TESTED: analyzes the effects of changing the values of a, h and k in the equation y = a(x – h)2 + k of a quadratic function on its graph.*** BT: Understanding, Analysis

B PLOD: M

EXPLN: The graph of y=8(x-2)2+5 is a parabola whose vertex is at x=2, while y=8(x-3)2+5 is a similar parabola but whose vertex is at x=3. CONCEPT: Similar triangles COMPETENCY TESTED: solves problems that involve triangle similarity and right triangles.*** BT: Remembering, Analyzing EXPLN: The smaller triangle is identified as a 30-60-90 right triangle since its hypotenuse is 18 and its leg is half of the hypotenuse. The angle opposite the smaller leg is ∠BCD= ∠ACE=30°. The leg adjacent to ∠BCD of the smaller triangle

In the triangle above, what is the length of ̅̅̅̅ 𝐶𝐷 ? A. 8√3 B. 9√3 C. 16 D. 17√3 56.

If tan

3 find sin sec . 5

is then √3/2 of 18, which is 9 3 . A PLOD: M

CONCEPT: Trigonometry COMPETENCY TESTED: uses trigonometric ratios to solve real-life problems involving right triangles. ***

A.

15 34 170

B.

49 34 170

BT: Remembering, Applying

34 15 170

EXPLN:

C. D.

34 49 170

tan

O 3 H 5

so finding the hypotenuse, H=

3 2 5 2 34

Hence,

ADMATH – TG 2015

25

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

54.

ACET 2016 MATHEMATICS

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS sin

O 3 34 3 34 H 34 34 34

sec

H 34 A 5

sin sec

Determine the equation described by the ordered pairs: (-4,18), (-2,6), (0,2), (1,3) and (3,11). A.

y x2

B.

y x2 2

C.

y x2 2

D.

y x 2 2x 2

B PLOD: E

49 34 170

CONCEPT: Geometry

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

57.

3 34 34 34 5

COMPETENCY TESTED: determines the equation of a quadratic function given: (a) a table of values; (b) graph; (c) zeros. BT: Applying EXPLN: (Trial and Error) By substituting the values of x and y in the ordered pairs to the equations in the choices, the answer will be B y x 2 2 .

58.

Which of the following is the graph of f (x) =–x3 ? A.

B

CONCEPT: Graphing Polynomials

PLOD: Moderat e

COMPETENCY TESTED: graphs polynomial functions. BT: Analyzing EXPLN: Plot arbitrary values for the function X Y -2 8 -1 1 0 0

ADMATH – TG 2015

26

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS

B.

1 2 Answer is B.

-1 -8

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

REF: Graphs taken from Tsishchanka, Kiryl. Section 3.2 Polynomial Functions And Their Graphs. 1st ed. Web. 4 Nov. 2015.

C.

D.

59.

A square has coordinates (1,5) (1,1) (5,1)(5,5). Find the point of intersection of the diagonals of the square. A. B. C. D.

(2,4) (3,3) (4,2) (4,4)

B PLOD: M

CONCEPT: Trigonometry COMPETENCY TESTED: solves problems involving geometric figures on the coordinate plane BT: Remembering EXPLN: Plot the points. By the property of squares, the diagonals bisect each other, which means that they intersect at their midpoints. Using the midline theorem, the midpoint of the line with

1 5 1 5 , ) (3,3). 2 2

points (1,1) & (5,5) is (

Similarly, the midpoint of the line with points

1 5 1 5 , ) (3,3). Hence, the 2 2

(1,5) & (5,1) is (

ADMATH – TG 2015

27

TEACHER’S GUIDE SET A

ACET 2016 MATHEMATICS point of intersection is (3,3).

60.

CONCEPT: Probability Three fair coins are tossed independently. Determine the probability of obtaining at most two heads. E.

BT: Analyzing, Applying

D PLOD: M

EXPLN: At most 2 heads means having 0 head, 1 head, or 2 heads. The combinations are as follows: TTT, HTT, THT, TTH, HHT, HTH, THH. Each combination has the probability

1 1 1 1 since the probability of obtaining a 2 2 2 8 head or a tail are equal. There are 7 combinations of having at most 2 heads. So 7

ADMATH – TG 2015

1 7 8 8

28

For AHEAD teacher’s use only. Photocopying is not allowed. Please return this material after the review period.

1 4 3 F. 4 3 G. 8 7 H. 8

COMPETENCY TESTED: solves problems involving probability.

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