ACET Set A Math 111015.pdf

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For A H E A D t ea cher ’s use on ly . P hotoc o pyin g is not a ll owe d . P le a s e re turn th is m a te ri a l a ft e r th e re v ie w per io d.

NO. QUESTION ANS./

PLOD

CONCEPT/ COMPETENCY TESTED/ BLOOM’S TAXONOMY/ EXPLANATION/ REF.

1. Find the value of a:

0

5

2





a

A.

a



1 

6

B.

a



1 

6

C.

a



1 

5

D.

a



1 

6

D PLOD: M

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.

BT: Application EXPLN: By completing squares

6

1

6

1

6 )

1 (

1

5

1

2

5

2

0

5

2

2 2 2 2































a

2. Find the discriminant and characteristic of the

root of the equation:

4x2_{+ 20x – 53 = 0}

A. -1248; imaginary roots B. 0; exactly one root C. 612; real roots D. 1248; real roots

D PLOD:

E

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

2



4 ac

. Thus, the discriminant of the

given equation is 1248 848 400 ) 53 4 4 ( 202      

Hence, the quadratic equation has discriminant 1248 and has real roots..

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

D PLOD:

E

CONCEPT: Variation

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.

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For A H E A D t ea cher ’s use on ly . P hotoc o pyin g is not a ll owe d . P le a s e re turn th is m a te ri a l a ft e r th e re v ie w per io d. B.

Z

kX

Y



C.

XZ

k

Y



D.

X

kZ



Y

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, the answer will be D.

x

kz

Y



where k=1/2. 4. Simplify: 2 4 3 9 3 12 7 4 ₎ ₍₅ ₎ 3 ( x y z x y z A. 675x18y9z13 B. 675x30y15z28 C. 675x15y27z30 D. 675x15y30z28 B PLOD: E CONCEPT: Exponents

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

BT: Applying EXPLN:

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

2 4 3 9 3 12 7 4 ₎ ₍₅ ₎ 3 ( x y z x y z = 28 15 30 8 6 18 36 21 12

675 )

25 )(

27 (

z

y

x

z

y

x

z

y

x



 

5. Simplify the given expression:

7 7 5 5 3 3

18

2

3

8 y

x

y

x

y

x



A. 2 2

3

2

3

2 y

x

xy



B. 2 2

3

2 y

x

xy



C.

xy

y

x

1

3

2

2 2



D.

xy

y

x

3

2

3

2 2



C PLOD: M CONCEPT: Radicals

COMPETENCY TESTED: simplifies radical expressions using the laws of radicals. BT: Applying, Analyzing

EXPLN:

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

7 7 5 5 3 3

18

2

3

8 y

x

y

x

y

x



(3)

For A H E A D t ea cher ’s use on ly . P hotoc o pyin g is not a ll owe d . P le a s e re turn th is m a te ri a l a ft e r th e re v ie w per io d. =

6

1

3

2

3

2

3

2

3

2

3

2

2 2 2 2 3 3 2 2 3 3 2 2



















xy

y

x

y

x

xy

y

x

y

x

xy

y

x

xy

y

x

xy

6.

A quadrilateral is inscribed in a circle (see figure above). What is angle θ? A. 113o B. 117o C. 130o D. 243o A PLOD: E CONCEPT: Geometry

COMPETENCY TESTED: uses properties to find measures of angles, sides and other quantities involving parallelograms. BT: Remembering, Analyzing EXPLN:

For quadrilaterals inscribed in circles, opposite angles supplement each other. Hence,

180o_{= θ + 67}o θ = 180o_{- 67}o ₌_113° 7. If r = s 4 3 and s = t 6 5

, find the ratio of t to r. A. 5:8 B. 8:5 C. 9:10 D. 10:9 B PLOD: E CONCEPT: Proportions

COMPETENCY TESTED: applies the

fundamental theorems of proportionality to solve problems involving proportions.

BT: Application EXPLN: Substitution r = s 4 3 , s = t 6 5

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For A H E A D t ea cher ’s use on ly . P hotoc o pyin g is not a ll owe d . P le a s e re turn th is m a te ri a l a ft e r th e re v ie w per io d. r = t 6 5 4 3













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

8. 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 B.

3

2

C. 6 D.

6

2

C PLOD: E CONCEPT: Geometry

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 .

9. 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: Sequences

COMPETENCY TESTED: determines

arithmetic means and nth term of an arithmetic sequence.***

BT: Application EXPLN:

Notice that the middle of 2nd_{and 10}th_{term is the}

6th_{term, thus, the 6}th_{term is the average;}

36 2 72 2 58 14    10.

Find the quotient when P(x)x327 is

divided by (x3). A. x29 B.

x

2 

9

C.

x

2 

3x



9

D.

x

2 

3x



9

D PLOD: M CONCEPT: Algebra

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

BT: Analysis EXPLN:

Sum of Two Cubes

) 2 b ab 2 b)(a (a 3 b 3 a      9) 3x 2 3)(x (x 27 3 x      Or Synthetic division 3 x x3

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For A H E A D t ea cher ’s use on ly . P hotoc o pyin g is not a ll owe d . P le a s e re turn th is m a te ri a l a ft e r th e re v ie w per io d. -3 1 0 0 27 -3 9 -27 1 -3 9 -27

9 3x

2 x





11.

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 B PLOD: E CONCEPT: Geometry

COMPETENCY TESTED: solves problems on circles.

Inscribed Angle theorem: Central angle = 2θ Arc length:

C =2R



12. 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: )! ( ! r n n Pr n   ,

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

24 ! 4 )! 4 4 ( ! 4     r nP 13.

Four coins are flipped. What are the chances of NOT getting heads?

A. 1/16 B. 1/4 C. 1/2 D. 15/16 A PLOD: E CONCEPT: Probability

COMPETENCY TESTED: solves problems involving probability.

BT: Creating, Understanding, Evaluating EXPLN:

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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}

14. 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 90th_{percentile of the given data.}

A. 20 B. 22 C. 24 D. 26 C PLOD: E

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 EXPLN:

Make sure first that all data are arranged from least to greatest.

The formula for percentile:

2 P

) 1 100 ( ) 100 ( 





nk nk K

X

where n=number of observations and k=percentile. So

18

100 )

90 )(

20 (

100 



nk

19

1

100 )

90 )(

20 (

1

100 







nk

Note that X1=4, X2=5, X3=6 and so on…

So X18=23 and X19=25 and substituting,

24

2

25

23

2

19 18



_



_



X

P

_K

15. Determine the quadratic equation with roots

-3/2 and 5. A.

15 x

2



7 x



2

B.

7 x

2



2 x



15

C.

2 x

2



15 x



7

D.

2 x

2



7 x



15

D PLOD: E

CONCEPT: Quadratic equations

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

BT: Application

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

0

15

7

2

0 )

5 )(

3

2 (

5 or

2 2 3















x

(7)

16. 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 PLOD: E CONCEPT: Variation

COMPETENCY TESTED: solves problems involving variation BT: Application EXPLN: 24 6 ) 9 ( 16 6 ) 3 ( 16 y then , 16 2 ) 1 ( 8 kx y 2 2 2       y k k z 17. Simplify: ₃ ₂ 2 1 2 1 x   y x y A. 2 5 2 3 x y B. 2 5 2 3 x  y C. 2 5 2 3 x y D. 2 5 2 5  x y C PLOD: E

CONCEPT: Rational Expressions COMPETENCY TESTED: simplifies expressions with rational exponents. BT: Understanding, Analyzing, Applying EXPLN: ) 2 5 ( ) 2 3 ( ) 2 3 ( ) 2 5 ( ) 2 2 1 ( ) 3 2 1 ( 2 3 2 1 2 1 1 x x y y x y x y x y         

18. Find the solution/s to the following equation:

16

4

1

4

1

2 2









_x

x

A. x=0,2 B. x=2,4 C. x=0,4 D. x=4,-4 A PLOD: M CONCEPT: Geometry

COMPETENCY TESTED: solves equations transformable to quadratic equations (including rational algebraic equations).

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16

4

1

4

1

2 2









_x

x

The LCD is 𝑥2_{− 16, so the equation will}

become,

2 and

0

0 )

2 (

0

2

4

16

4

16

4

2 2 2 2 2 2



























x

19. Simplify the expression:

) 3 2 )( 3 2 ( x y x y A. 2x3y B. 4x29y2 C. ( 2x)( 2x3y)( 3y)( 2x3y) D. ( 3x)( 2x3y)( 2y)( 2x3y) C PLOD: E

CONCEPT: Radical Expressions

COMPETENCY TESTED: performs operations on radical expressions.*** BT: Application EXPLN: ) 3 2 )( 3 ( ) 3 2 )( 2 ( ) 3 2 )( 3 2 ( y x y y x x y x y x     

20. Find the values of x:

2

3

1

2

1

1 





2







x

A. -1 B. 2 C. -1 and 2 D. 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: 1 or 2 0 2 1 ) 1 ( ) 2 ( ) 2 )( 1 ( 1 ) 1 )( 2 ( ) 1 ( 1 ) 2 )( 1 ( ) 2 ( ) 2 )( 1 ( 1 2 1 1 2 3 2 1 2 1 1 2                                 x x x x x x x x x x x x x x x x x x x x x x x x x x

(9)

For A H E A D t ea cher ’s use on ly . P hotoc o pyin g is not a ll owe d . P le a s e re turn th is m a te ri a l a ft e r th e re v ie w per io d. only x= -1.

21. Refer to the figure below:

If ∠C=45°, ∠B=50°, find ∠BED + ∠CDE.

A. 205° B. 215° C. 225° D. 235° B PLOD: M

CONCEPT: Midline Theorem

COMPETENCY TESTED: proves the Midline Theorem.

BT: Analysis, Application 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° 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° 22. Find x and y. A. x =12, y = 15 B. x =12, y = 20 C. x =15, y = 15 D. x =15, y = 20 A PLOD: M

CONCEPT: Similar Triangles

COMPETENCY TESTED: solves problems that involve triangle similarity and right triangles.*** BT: Remembering, Analyzing

EXPLN:

The triangles in the figure are all similar. From there x can be found:

12 9 16 16 9 2     x x x x

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,

(10)

23. 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. ***

Let l- length of the plank

meters 0 1 2 1 5 30 sin     l l s

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 B PLOD: M CONCEPT: Sequences

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

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

5

30°

l

(11)

For A H E A D t ea cher ’s use on ly . P hotoc o pyin g is not a ll owe d . P le a s e re turn th is m a te ri a l a ft e r th e re v ie w per io d. d = 4

25. Consider the polynomial

8

4

6

3

2

4 3 2 5

_

_

_

_

x

Which of the following is/are factors of the given polynomial? I. (x-1) II. (x+1) III. (x-2) A. I only B. I and II only C. II and III only D. I, II and III

D PLOD:

E

CONCEPT: Remainder Theorem/Factor Theorem

COMPETENCY TESTED: proves the

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

26. A circle with radius r is centered at (1,2). The

point (x, y) = (5,-4) is on the circle. Find r.

A.

13

B.

20

C.

2

13

D.

2

20

C PLOD: E CONCEPT: Circles

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

13

2

52

36

16 )

4

2 (

)

5

1 (

2 2













r

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 D PLOD: M CONCEPT: Combinations

COMPETENCY TESTED: solves problems involving permutations and combinations. BT: Application

EXPLN:

There are 11 letters: M-1, S-4, I-4, P-2

(12)

For A H E A D t ea cher ’s use on ly . P hotoc o pyin g is not a ll owe d . P le a s e re turn th is m a te ri a l a ft e r th e re v ie w per io d. So,

34 ,

650 )

1 .

2 )(

1 .

2 .

3 .

4 )(

1 .

2 .

3 .

4 (

1 .

2 .

3 .

4 .

5 .

6 .

7 .

8 .

9 .

10 .

11 !

2 !

4 !

1 !

11 _

_

28. Find the 75th_{percentile of the set of values}

below: {15,9,7,6,1,4,3,10,7,8,9,12} A. 9 B. 9.5 C. 10 D. 11.25 B PLOD: E

CONCEPT: Measures of Position

COMPETENCY TESTED: interprets measures of position.

Arrange the data from lowest to highest: {1,3,4,6,7,7,8,9,9,10,12,15}

2 P

) 1 100 ( ) 100 ( 





nk nk K

X

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

2

10

9

2 P

_K (9) (10)









X

, 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

2



7 x



10 

0

D.

x

2



7 x



10 

0

B PLOD: M CONCEPT: Algebra

COMPETENCY TESTED: solves quadratic inequalities.

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

(13)

right-For A H E A D t ea cher ’s use on ly . P hotoc o pyin g is not a ll owe d . P le a s e re turn th is m a te ri a l a ft e r th e re v ie w per io d.

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 Thus, answer is B.

30. Find the solutions of:

3

3 





x

A. x=3 B. x=4 C. x=5 D. No solution D PLOD: M

CONCEPT: Radical Equations

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

BT: Applying, Analyzing EXPLN:

x































4

9

36

3

36

12 )

3 (

)

6 (

3

2

12

2

9

3

2

3

2

3 )

3 (

3

2 2 2 2 2 2 2 2 2

Check if the resulting value of x is a root of the given equation.

3

1

2

1

4

3

4 













Therefore, x=4 is not a root of the equation and hence there are no solutions.

31.

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 B PLOD: M CONCEPT: Quadrilaterals

COMPETENCY TESTED: solves problems involving parallelograms, trapezoids and kites. BT: Remembering, Applying, Analyzing EXPLN:

Compute for length of the shortest leg of the right triangle at the left.

Shortest leg length = √102_{− 8}2

Shortest leg length = 6

(14)

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

112 trapezoid

of

area



Method 2: Trapezoid

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

32. Three pairs of parallel segments make up two

triangles, ∆𝑍𝑌𝑋 and ∆𝑊𝑉𝑈, shown below.

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. II only B. III only C. I and III D. I, II, and III

C PLOD:

M

CONCEPT: Geometry; Similar/Congruent Triangles

COMPETENCY TESTED: applies the theorems to show that given triangles are similar.

BT: Evaluation EXPLN:

WV = ZY, WU = ZX, UV = XY

Therefore, angles are also the same. No information whether the angles form a right angle.

I and III

33. Given the sequence

. .. , 4 15 , 2 15 15,

Find the 7th_{term (a}₇_{) and give the geometric}

mean (m) of the first and 7th_term.

A. 8 15 ; 32 15 7  m a B. 16 15 ; 32 15 7  m a C PLOD: E

CONCEPT: Geometric Sequences COMPETENCY TESTED: determines geometric means and nth term of a geometric sequence.***

BT: Applying, Analyzing EXPLN:

(15)

For A H E A D t ea cher ’s use on ly . P hotoc o pyin g is not a ll owe d . P le a s e re turn th is m a te ri a l a ft e r th e re v ie w per io d. C. 8 15 ; 64 15 7  m a D. 16 15 ; 64 15 7  m a 8 15 64 225 64 15 15 64 15 2 1 ) 15 ( 2 1 ratio(r) common ; 15 ) 1 7 ( 7 ) 1 ( 1 1            m a r a a a n n 34. Simplify:

16

8

4

2

4 2 3





x

A.

4

1

2

_

x

B.

4

1 

x

C.

2

1 

x

D.

2

1

2

_

x

C PLOD: D CONCEPT: Algebra

COMPETENCY TESTED: factors polynomials. BT: Application

EXPLN:

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

)

4 )(

4 (

)

4 (

2 )

4 (

)

4 )(

4 (

8

2

4

16

8

4

2

2 2 2 2 2 2 2 3 4 2 3





















x

Cancellation:

)

4 (

)

2 (

)

4 )(

4 (

)

4 )(

2 (

16

8

4

2

2 2 2 2 4 2 3



















x

Denominator: DOTS

)

2 )(

2 (

)

2 (

16

8

4

2

4 2 3













x

Cancellation:

)

2 (

1 )

2 )(

2 (

)

2 (

16

8

4

2

4 2 3

















x

is the hypotenuse of ΔBCF. is 9 since adjacent of 30 is the hypotenuse of ΔCDF. Thus, is the hypotenuse of ΔDEF. Thus, = .

35. Find the center (x,y) and radius (r) of the circle

with equation: ( 13) 0 2 1 5 2x y x2y2  A. (-5,2); r=4 C PLOD: M CONCEPT: Circles

COMPETENCY TESTED: determines the center and radius of a circle given its equation and vice versa

(16)

For A H E A D t ea cher ’s use on ly . P hotoc o pyin g is not a ll owe d . P le a s e re turn th is m a te ri a l a ft e r th e re v ie w per io d. 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

0 13 x y 10 x

4   2y2  and rearranging the

terms, the equation will be

13 10 4

x2 xy2 y

By completing the squares method, the resulting equation will be 25 4 13 ) 25 10 ( ) 4 4 (x2 x  y2 y   

and by factoring the perfect square trinomials, will yield to 16 ) 5 ( ) 2 (x 2 y 2 

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

4 16 

r .

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.

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

26

2

28

24 

_



median

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

17

2

19

15 quartile

lower







. The

42

2

47

37 quartile

upper







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 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

(17)

Area of a rectangle: A=LW LW=560

(2W+3)(W)=560

2W2_+3W=560

2W2_+3W-560=0

Using the Quadratic Formula to solve for W:

negative be cannot 4 70 -or 4 64 4 67 3 or 4 67 3 4 4489 3 ) 2 ( 2 ) 560 )( 2 ( 4 3 3 2 4 2 2                    a ac b b W So, 16meters 4 64   W L = 2(16)+3=32+3= 35 meters 38. Evaluate: 2 27 48 A. 2 6 B. 6 C. 2 3 D. 3 A PLOD: E

CONCEPT: Radical Expressions

COMPETENCY TESTED: solves problems involving radicals. BT: Applying, Evaluating EXPLN: 2 3 2 3 3 3 4 2 ) 3 )( 9 ( ) 3 )( 16 (     Rationalizing, 2 6 2 2 2 3  

39. Given the quadratic equation:

y x

x224 45

Which of the following statements is/are true? I. The graph of the equation is open

downwards

II. The vertex is at (12, -99)

D PLOD:

M

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.

(18)

III. The range is

{

y



R

|

y





99 }

A. II only B. I only C. I and II only D. II and III only

EXPLN:

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

45 24 2 _ _ _ y x x

And completing the squares will result to

y

x

y

x













99 )

12 (

99

144

24

2 2

Hence, the vertex of the quadratic equation is at the point (12,-99). This implies that the range is

}

99 |

{

y



R

y





and the graph is opening

upwards. Therefore, only II and III are true.

40. X varies inversely as the cube of Z and

directly as

Y

1

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

1 𝑌 if X=5 and Z=3. A.

5

32

B.

32

5

C.

27

32

D.

32

27

D PLOD: E CONCEPT: Variation

COMPETENCY TESTED: solves problems involving variation. BT: Remembering, Applying EXPLN:

32

27

160

135

1

160 )

3 (

5

1 )

1 (

160

160 )

15 (

2

4

1

3 3 3 3 3



Y

Z

Y

X

k

Y

Z

k

Z

Y

k

X

41.

Given the triangle above, which of the

A PLOD:

E

CONCEPT: Similar Triangles

COMPETENCY TESTED: applies the theorems to show that given triangles are similar.

Only I is true according to properties of similar triangles.

(19)

following statements is ALWAYS TRUE when triangle AEC is similar to triangle BCD?

I.

CD

BD

CE

AE

_

II. BCDBAE III. ABBC A. I only B. I and II only C. II and III only D. I, II and III II is false,

BAE

BCD

NOT

E











CBD

BA

III is false.

42. Find the sum of the infinite sequence:

...

,

243

5 ,

81

5 ,

27

5

A.

5

6

B.

9

5

C.

6

5

D.

5

9

C PLOD: D CONCEPT: Algebra

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

BT: Remembering, Analyzing, Applying 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

9

5 

a

which can be found using the

formula

a

_n



ra

_n_₁.

Since the common ratio r < 1, the formula is

r

a

i i







 1

₁

Substituting values, the sum is

6

5

3

1

9

5 



. . 43.

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

CONCEPT: Polynomial Equations

COMPETENCY TESTED: solves polynomial equations.

(20)

For A H E A D t ea cher ’s use on ly . P hotoc o pyin g is not a ll owe d . P le a s e re turn th is m a te ri a l a ft e r th e re v ie w per io d. 12 3c b 4a 2 24 6c 2b 8a         Equ.1 0 3c b 4a   _Equ.2 2b = 12 44.

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 C PLOD: D CONCEPT: Circles

BT: Applying, Analyzing, Remembering 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: Ak= 2 ) 4 ( 2 = 4 Area of Circle: Ac = (2)2π = 4π

Area of shaded region:

As = Ac - Ak

As = 4π-4

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. 3/8 B. 1/8 C. 5/8 D. 7/8 A PLOD: M CONCEPT: 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,

(21)

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

46. Using the set of values

below:{26,5,5,8,20,2,3,24}

find the second quartile (Q2) and the range (r). A. Q2=5; r=24 B. Q2=5; r=26 C. Q2=6.5; r=24 D. Q2=6.5; r=26 C PLOD: M

CONCEPT: Measures of Position

COMPETENCY TESTED: uses appropriate measures of position and other statistical methods in analyzing and interpreting research data.

Arrange the data from lowest to highest: {2,3,5,5,8,20,24,26}

Second Quartile (Q2) = 50th_percentile

2 P

) 1 100 ( ) 100 (





nk nk K

X

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

5 .

6

2

8

5

2 P

₅₀ (4) (5)











X

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

0

4

3 

x



x

2



Find the roots of the quadratic equation. A.

3

8 ,

3

2 



x

B.

3

8 ,

3

2 _



x

C PLOD: E

CONCEPT: Quadratic Equations

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

(22)

For A H E A D t ea cher ’s use on ly . P hotoc o pyin g is not a ll owe d . P le a s e re turn th is m a te ri a l a ft e r th e re v ie w per io d. C.

2

1 ,

2

3 



x

D.

2

1 ,

2

3 _





x

2

1 ,

2

3

2

1

8

4

8

64

4 )

4 (

2 )

3 )(

4 (

4 )

4 (

)

4 (

2

4

2 2







































x

a

ac

b

x

48. Find the quotient if

1 3 3 x f(x) 3 x2  x is divided by (x-1). A. x22x1 B. x2-2x1 C. x22x1 D. x2+2x1 B PLOD: A CONCEPT: Polynomials

COMPETENCY TESTED: solves problems involving polynomial functions.

BT: Application

EXPLN: Synthetic Division

1 x

0

1 x









1 -3 3 -1 1 -2 1 1 -2 1 0 1 2x -2 x  49.

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. 8π+16 square units B. 8π-16 square units C. 32π+16 square units D. 32π-16 square units B PLOD: D CONCEPT: Algebra

COMPETENCY TESTED: solves problems involving geometric figures on the coordinate plane.

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.

2

4

32

4

4 )

3

7 (

)

1

3 (

2 2 2 2













D

By 45-45-90 triangle, the base has length=4 so multiply by √2 to get the hypotenuse = diagonal

= diameter

4

2 (-1,7)

(3,7)

(23)

For A H E A D t ea cher ’s use on ly . P hotoc o pyin g is not a ll owe d . P le a s e re turn th is m a te ri a l a ft e r th e re v ie w per io d. So, radius =

2

Solving for the area of the circle:

units

square

8 )

2

2 (

2 2





r

A

Area of square = 42_{=16 square units}

Finally,

area of the shaded region =

units

square

16

8 



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.

Notice that the middle of 2nd_{and 10}th_{term is the}

6th_{term, thus, the 6}th_{term is the average;}

36 2 72 2 58 14   

51. Find the solution set of the given polynomial

equation:

.

6

2 3

_

_x

_

_x

_

x

A. { -1, 1, -6 } B. { -1, 1, 6 } C. { -3, 2, 6 } D. { 3, -2 , 6 } A PLOD: M CONCEPT: Algebra

COMPETENCY TESTED: solves problems involving polynomials and polynomial equations. BT: Application

EXPLN:

(Partly by Trial and Error)

By the factor theorem, we will see that x+1 or x-1 or x+5 are factors of the polynomial.

Consider x+1.

By factor or long/synthetic division,

0 )

1 )(

6

5 (

0

6

2 2 3















x

Furthermore,

0 )

1 )(

6 (









x

So,

0

1 and

0

1 ;

0

6 











x

(24)

For A H E A D t ea cher ’s use on ly . P hotoc o pyin g is not a ll owe d . P le a s e re turn th is m a te ri a l a ft e r th e re v ie w per io d. 52.

If 𝐴𝐵̅̅̅̅ is the radius of the circle equal to 2√3

and ∠ABC=2π/3, what is the area of the shaded region? A.

4 



3

B.

2

3

4 



C.

4 



3

D.

4 



4

3

C PLOD: D CONCEPT: Circles

BT: Applying, Analyzing, Remembering EXPLN:

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



4 )

3

2 (

2

3

2 Sector

the

of

Area



2



ΔABC is an isosceles triangle. Bisect 𝐴𝐶̅̅̅̅ to

create two congruent right triangles. 𝐴𝐵̅̅̅̅ and 𝐵𝐶̅̅̅̅

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. BACis then equal to π/6.

Bisector length =

)

3

2

1 (

3

2 



𝐴𝐶 ̅̅̅̅ = 6 Area of

3

2

3

6 





ABC

(Area of shaded region) =

4 



3

53.

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

4

1

B.

4

3

C.

8

3

D.

8

7

D PLOD: M CONCEPT: Probability

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

8

1

2

1

2

1

2

1 _

_

_

since the probability of obtaining a head or a tail are equal. There are 7

combinations of having at most 2 heads. So

8

7

8

1

7 



(25)

54. 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

x-axis.

D. The graph will flip with respect to the y-axis.

B PLOD:

E

CONCEPT: Algebraic equations

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 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. 55.

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

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

is then √3/2 of 18, which is

9

3

.

56. If

5

3 tan





find sin



sec



.

A.

170

34

15

B.

170

34

49

C.

170

15

34

D.

170

49

34

A PLOD: M CONCEPT: Trigonometry

COMPETENCY TESTED: uses trigonometric ratios to solve real-life problems involving right triangles. *** BT: Remembering, Applying EXPLN:

5

3 tan



H

O



so finding the hypotenuse,

H =

3

2



5

2



34

(26)

170

34

49

5

34

3 sec

sin

5

34 sec

34

3

34

3 sin

















A

H

O

57. 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x22x2 B PLOD: E CONCEPT: Geometry

COMPETENCY TESTED: determines the equation of a quadratic function given: (a) a table of values; (b) graph; (c) zeros.

(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x2 2.

58. Which of the following is the graph of

f (x) =–x3 _? A. B PLOD: Moderat e

CONCEPT: Graphing Polynomials

COMPETENCY TESTED: graphs polynomial functions.

BT: Analyzing EXPLN:

Plot arbitrary values for the function

X Y

-2 8

-1 1

(27)

For A H E A D t ea cher ’s use on ly . P hotoc o pyin g is not a ll owe d . P le a s e re turn th is m a te ri a l a ft e r th e re v ie w per io d. B. C. D. 1 -1 2 -8 Answer is B.

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

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. (2,4) B. (3,3) C. (4,2) D. (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