acet set a math 111015.pdf

TEACHER’S GUIDE ACET 2016

SET A MATHEMATICS

ADMATH – TG 2015 1

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NO. QUESTION ANS./

PLOD

CONCEPT/ COMPETENCY TESTED/

BLOOM’S TAXONOMY/ EXPLANATION/ REF.

1. Find the value of a:

0522 aa

A. 61a

B. 61a

C. 51a

D. 61a

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

61

61

6)1(

1512

52

052

2

2

2

2

a

a

a

aa

aa

aa

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 ac4b2 . Thus, the discriminant of the

given equation is

1248848400)5344(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. ZkXY

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.

BT: Applying


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B. Z

kXY

C. XZ

kY

D. X

kZY

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

kzY where k=1/2.

4.

Simplify:

243931274 )5()3( zyxzyx

A. 13918675 zyx

B. 281530675 zyx

C. 302715675 zyx

D. 283015675 zyx

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.

243931274 )5()3( zyxzyx

=281530

8618362112

675

)25)(27(

zyx

zyxzyx

5. Simplify the given expression:

77

5533

18

238

yx

yxyx

A. 223

232

yx

xy

B. 22

32

yxxy

C. xyyx

1

3

222

D. xy

yx

32

3 22

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:

77

5533

18

238

yx

yxyx


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.

=

6

1

3

2

3

32

3

32

23

2322

22

22

33

22

33

22

xyyx

yx

xy

yx

yxxy

xyyx

xyyxxyxy

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.


EXPLN:

For quadrilaterals inscribed in circles, opposite

angles supplement each other. Hence,

180o = θ + 67o

θ = 180o - 67o = 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 = s4

3, s = t

6

5


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.

r = t6

5

4

3

r = t8

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. 23

C. 6

D. 26

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 10th term is the

6th term, thus, the 6th term is the average;

362

72

2

5814

10. Find the quotient when 27

3xP(x) is

divided by 3)(x .

A. 92

x

B. 92

x

C. 93x2

x

D. 93x2

x

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

bab2

b)(a(a3

b3

a

9)3x2

3)(x(x273

x

Or

Synthetic division

3x 3x


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-3 1 0 0 27

-3 9 -27

1 -3 9 -27

93x2

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.


EXPLN:

Inscribed Angle theorem:

Central angle = 2θ

Arc length:

C = R2

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:

)!(

!

rn

nPrn

,

where n = total number of objects;

r = number of objects chosen (want)

24!4)!44(

!4

rn P

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


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 90thpercentile 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:

2P

)1100

()100

(

nknk

K

XX

where n=number of observations and

k=percentile.

So 18100

)90)(20(

100

nk

191100

)90)(20(1

100

nk

Note that X1=4, X2=5, X3=6 and so on…

So X18=23 and X19=25 and substituting,

242

2523

2

1918

XX

PK

15. Determine the quadratic equation with roots -

3/2 and 5.

A. 2715 2 xx

B. 1527 2 xx

C. 7152 2 xx

D. 1572 2 xx

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:

01572

0)5)(32(

5or

2

23

xx

xx

xx


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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


involving variation

BT: Application

EXPLN:

24

6

)9(16

6

)3(16y

then,16

2

)1(8

kxy

2

2

2

y

k

k

z

17.

Simplify: 23

2

1

2

1

x

yx

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(

)22

1()3

2

1(

23

2

1

2

1

1

x

x

yyx

yx

yx

y

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

164

1

4

12

2

x

x

xx

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).


EXPLN:


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164

1

4

12

2

x

x

xx

The LCD is 𝑥2 − 16, so the equation will

become,

2 and 0

0)2(

02

44

1616

4

16

4

2

2

2

2

22

xx

xx

xx

xxx

x

x

x

x

x

x

19. Simplify the expression:

)32)(32( yxyx

A. yx 32

B. 22 94 yx

C. )32)(3()32)(x2( yxyyx

D. )32)(2()32)(x3( yxyyx

C

PLOD:

E

CONCEPT: Radical Expressions

COMPETENCY TESTED: performs operations

on radical expressions.***

BT: Application

EXPLN:

)32)(3()32)(2(

)32)(32(

yxyyxx

yxyx

20. Find the values of x:

23

1

2

1

1 2

xxxx

x

A. -1

B. 2

C. -1 and 2

D. 1 and 2

A

PLOD:

D

CONCEPT: Algebraic Equations


involving quadratic equations and rational

algebraic equations.

BT: Analysis and Application

EXPLN:

1or 2

02

1)1()2(

)2)(1(

1

)1)(2(

)1(1

)2)(1(

)2(

)2)(1(

1

2

1

1

232

1

2

1

1

2

xx

xx

xxx

xxxx

x

xx

xx

xxxx

x

xxxx

x

But x=2 will make the equation undefined. So


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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.***


EXPLN:

The triangles in the figure are all similar. From

there x can be found:

12

916

16

92

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,

𝑦 = 15.


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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. ***


EXPLN:

Let l- length of the plank

meters 01

2

1

5

30sin

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.***


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


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d = 4

25. Consider the polynomial

84632 2345 xxxxx

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. 132

D. 202

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.

132

523616)42()51( 22

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


involving permutations and combinations.

BT: Application

EXPLN:

There are 11 letters:

M-1, S-4, I-4, P-2


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So, 650,34)1.2)(1.2.3.4)(1.2.3.4(

1.2.3.4.5.6.7.8.9.10.11

!2!4!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.

BT: Applying

EXPLN:

Arrange the data from lowest to highest:

{1,3,4,6,7,7,8,9,9,10,12,15}


2P

)1100

()100

(

nknk

K

XX


k=percentile.

Where n=12 and k=75

Compute for nk/100 = (12)(75)/(100)

nk/100=9

nk/100 +1 = 10

2

109

2P

)10()9(

K

XX

, so the 75th percentile is 9.5

29. Find a possible inequality whose solution set

is given by 5x2 .

A. 01072 xx

B. 01072 xx

C. 01072 xx

D. 01072 xx

B

PLOD:

M

CONCEPT: Algebra

COMPETENCY TESTED: solves quadratic

inequalities.

BT: Analysis

EXPLN:

Inequality Shortcut factor choices first,

A. 05)2)(x(x

B. 05)2)(x(x

C. 05)2)(x(x

D. (x+2)(x+5)≥ 0

Assuming there is no “negative x”, and the right-


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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:

33 xx

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.***


EXPLN:

x

x

xxxx

xxx

xxx

xxx

xx

xx

4

936

33612

)3()6(

32122

93232

3)3(

33

22

222

2

2

22

Check if the resulting value of x is a root of the

given equation.

3121434

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


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 − 82

Shortest leg length = 6

(Or recognize that to make a Pythagorean triple,


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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 trapezoidof 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 (a7 ) and give the geometric

mean (m) of the first and 7th term.

A. 8

15;

32

157 ma

B. 16

15;

32

157 ma

C

PLOD:

E

CONCEPT: Geometric Sequences

COMPETENCY TESTED: determines

geometric means and nth term of a geometric

sequence.***


EXPLN:


SET A MATHEMATICS


Fo

r A

HE

AD

teach

er’

s u

se o

nly

.

Ph

oto

co

py

ing

is n

ot

allo

wed

. P

lease r

etu

rn t

his

mate

rial aft

er

the r

evie

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eri

od

.

C. 8

15;

64

157 ma

D. 16

15;

64

157 ma

8

15

64

225

64

1515

64

15

2

1)15(

2

1ratio(r)common ; 15

)17(

7

)1(1

1

m

a

raa

a

nn

34. Simplify:

16

8424

23

x

xxx

A. 4

1

2 x

B. 4

1

x

C. 2

1

x

D. 2

12 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(

824

16

842

22

22

22

23

4

23

xx

xxx

xx

xxx

x

xxx

Cancellation:

)4(

)2(

)4)(4(

)4)(2(

16

842

2

22

2

4

23

x

x

xx

xx

x

xxx

Denominator: DOTS

)2)(2(

)2(

16

8424

23

xx

x

x

xxx

Cancellation:

)2(

1

)2)(2(

)2(

16

8424

23

x

xx

x

x

xxx

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: 0)13(2

152 22 yxyx

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


SET A MATHEMATICS


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nly

.

Ph

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is n

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allo

wed

. P

lease r

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mate

rial aft

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the r

evie

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od

.

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

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

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


EXPLN:

Multiplying the equation by 2 will result to

013xy10x4 22 y and rearranging the

terms, the equation will be

13104x 22 yyx

By completing the squares method, the resulting

equation will be

25413)2510()44( 22 yyxx

and by factoring the perfect square trinomials,

will yield to

16)5()2( 22 yx

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

416 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


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 262

2824

median .The lower

quartile (Q1) is the median of the lower half of

the data set. Then the

172

1915 quartilelower

.

The 422

4737 quartileupper

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


SET A MATHEMATICS


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.

Ph

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wed

. P

lease r

etu

rn t

his

mate

rial aft

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the r

evie

w p

eri

od

.

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 becannot

4

70-or

4

64

4

673or

4

673

4

44893

)2(2

)560)(2(433

2

4

2

2

a

acbbW

So, meters 164

64W

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

38. Evaluate:

2

2748

A. 2

6

B. 6

C. 2

3

D. 3

A

PLOD:

E

CONCEPT: Radical Expressions


involving radicals.

BT: Applying, Evaluating

EXPLN:

2

3

2

3334

2

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

Rationalizing, 2

6

2

2

2

3

39. Given the quadratic equation:

yxx 45242

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.

BT: Analyzing


SET A MATHEMATICS


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.

Ph

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co

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is n

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wed

. P

lease r

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mate

rial aft

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the r

evie

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od

.

III. The range is }99|{ yRy

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

45242 yxx

And completing the squares will result to

yx

yxx

99)12(

9914424

2

2

Hence, the vertex of the quadratic equation is at

the point (12,-99). This implies that the range is

}99|{ yRy 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


involving variation.


EXPLN:

32

27

160

1351

160

)3(51

)1

(160

160

)15(24

1

3

3

3

33

Y

Y

Z

YX

k

k

YZ

k

Z

Yk

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.

BT: Analysis

EXPLN:

Only I is true according to properties of similar

triangles.


SET A MATHEMATICS


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r A

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lease r

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mate

rial aft

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the r

evie

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od

.

following statements is ALWAYS TRUE when

triangle AEC is similar to triangle BCD?

I. CD

BD

CE

AE

II. BAEBCD

III. BCAB

A. I only

B. I and II only

C. II and III only

D. I, II and III

II is false,

BAEBCD

NOT

E

BACBD

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 1 nn raa . Also, the

first term is 9

5a which can be found using the

formula 1 nn raa .

Since the common ratio r < 1, the formula is

r

aa

ii

11

Substituting values, the sum is 6

5

3

11

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.

BT: Application

EXPLN:


SET A MATHEMATICS


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r A

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lease r

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mate

rial aft

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the r

evie

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od

.

123cb4a

2

246c2b8a

Equ.1

03cb4a 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


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





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,


SET A MATHEMATICS


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r A

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.

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allo

wed

. P

lease r

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his

mate

rial aft

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the r

evie

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od

.

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.


EXPLN:

Arrange the data from lowest to highest:

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

Second Quartile (Q2) = 50th percentile


2P

)1100

()100

(

nknk

K

XX


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

85

2P

)5()4(

50

XX

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

0443 2 xx

Find the roots of the quadratic equation.

A. 3

8,

3

2x

B. 3

8,

3

2x

C

PLOD:

E

CONCEPT: Quadratic Equations

COMPETENCY TESTED: models real-life

situations using quadratic functions.

BT: Application

EXPLN:


SET A MATHEMATICS


Fo

r A

HE

AD

teach

er’

s u

se o

nly

.

Ph

oto

co

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is n

ot

allo

wed

. P

lease r

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his

mate

rial aft

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the r

evie

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od

.

C. 2

1,

2

3x

D. 2

1,

2

3x

2

1,

2

3

2

21

8

84

8

644

)4(2

)3)(4(4)4()4(

2

4

2

2

x

a

acbbx

48. Find the quotient if

133xf(x) 23 xx

is divided by 1)-(x .

A. 12x2

x

B. 12x-x 2

C. 12x2

x

D. 12x+2x

B

PLOD:

A

CONCEPT: Polynomials


involving polynomial functions.

BT: Application

EXPLN: Synthetic Division

1x01x 1 -3 3 -1

1 -2 1

1 -2 1 0

12x-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


involving geometric figures on the coordinate

plane.


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.

243244

)37()13(

22

22

D

By 45-45-90 triangle, the base has length=4 so

multiply by √2 to get the hypotenuse = diagonal

= diameter 24

(-1,7) (3,7)

(-1,3) (3,3)


SET A MATHEMATICS


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wed

. P

lease r

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mate

rial aft

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the r

evie

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eri

od

.

So, radius = 22

Solving for the area of the circle:

units square 8

)22( 2

2

rA

Area of square = 42=16 square units

Finally,

area of the shaded region =

units square 168

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


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;

362

72

2

5814

51. Find the solution set of the given polynomial

equation:

.66 23 xxx

A. { -1, 1, -6 } B. { -1, 1, 6 } C. { -3, 2, 6 } D. { 3, -2 , 6 }

A

PLOD:

M

CONCEPT: Algebra


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)(65(

066

2

23

xxx

xxx

Furthermore,

0)1)(1)(6( xxx

So,

01 and 01 ;06 xxx

Hence, the solution set is {-1,1,-6}


SET A MATHEMATICS


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wed

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mate

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the r

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od

.

52.

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

and ∠ABC=2π/3, what is the area of the

shaded region?

A. 34

B. 2

334

C. 334

D. 344

C

PLOD:

D

CONCEPT: Circles


circles.

BT: Applying, Analyzing, Remembering

EXPLN:

(Area of the sector of circle) – (Area of triangle)

= (Area of shaded region)

4)32(2

3

2

Sector theof Area 2

ΔABC is an isosceles triangle. Bisect 𝐴𝐶̅̅ ̅̅ to

create two congruent right triangles. 𝐴𝐵̅̅ ̅̅ and 𝐵𝐶̅̅ ̅̅

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.

Bisector length = 3)2

1(32

𝐴𝐶̅̅ ̅̅ = 6

Area of 332

36ABC

(Area of shaded region) = 334

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





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

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

17


SET A MATHEMATICS


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.

Ph

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. P

lease r

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mate

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the r

evie

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od

.

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

𝐶𝐷̅̅ ̅̅ ?

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.***


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 39 .

56. If

5

3tan find secsin .

A. 170

3415

B. 170

3449

C. 170

1534

D. 170

4934

A

PLOD:

M

CONCEPT: Trigonometry

COMPETENCY TESTED: uses trigonometric

ratios to solve real-life problems involving right

triangles. ***


EXPLN:

5

3tan

H

O

so finding the hypotenuse,

H = 3453 22

Hence,


SET A MATHEMATICS


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.

170

3449

5

34

34

343secsin

5

34sec

34

343

34

34

34

3sin

A

H

H

O

57. Determine the equation described by the

ordered pairs: (-4,18), (-2,6), (0,2), (1,3) and

(3,11).

A. 2xy

B. 22 xy

C. 22 xy

D. 222 xxy

B

PLOD:

E

CONCEPT: Geometry

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 22 xy .

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

0 0


SET A MATHEMATICS


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.

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


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

points (1,1) & (5,5) is ).3,3()2

51,

2

51(

Similarly, the midpoint of the line with points

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

51,

2

51(

Hence, the


SET A MATHEMATICS


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point of intersection is (3,3).

60.

Three fair coins are tossed independently.

Determine the probability of obtaining at most

two heads.

E. 4

1

F. 4

3

G. 8

3

H. 8

7

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

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

17

acet set a math 111015.pdf

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