Upload
christian-arranz
View
251
Download
0
Embed Size (px)
Citation preview
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 1
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
w p
eri
od
.
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
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 2
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
w p
eri
od
.
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
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 3
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
w p
eri
od
.
=
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.
BT: Remembering, Analyzing
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
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 4
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
w p
eri
od
.
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
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 5
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
w p
eri
od
.
-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.
BT: Remembering, Analyzing
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
COMPETENCY TESTED: solves problems
involving probability.
BT: Creating, Understanding, Evaluating
EXPLN:
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 6
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
w p
eri
od
.
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
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 7
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
w p
eri
od
.
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(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).
BT: Remembering, Analyzing
EXPLN:
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 8
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
w p
eri
od
.
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
COMPETENCY TESTED: solves problems
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
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 9
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
w p
eri
od
.
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
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.
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 10
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
w p
eri
od
.
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. ***
BT: Remembering, Analyzing
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.***
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
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 11
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
w p
eri
od
.
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
COMPETENCY TESTED: solves problems
involving permutations and combinations.
BT: Application
EXPLN:
There are 11 letters:
M-1, S-4, I-4, P-2
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 12
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
w p
eri
od
.
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}
The formula for percentile:
2P
)1100
()100
(
nknk
K
XX
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
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-
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 13
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
w p
eri
od
.
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.***
BT: Applying, Analyzing
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
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 − 82
Shortest leg length = 6
(Or recognize that to make a Pythagorean triple,
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 14
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
w p
eri
od
.
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.***
BT: Applying, Analyzing
EXPLN:
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 15
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
w p
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
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 16
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
w p
eri
od
.
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
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
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 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
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 17
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
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
COMPETENCY TESTED: solves problems
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
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 18
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
w p
eri
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
COMPETENCY TESTED: solves problems
involving variation.
BT: Remembering, Applying
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.
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 19
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
w p
eri
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:
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 20
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
w p
eri
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
COMPETENCY TESTED: solves problems on
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
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,
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 21
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
w p
eri
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.
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:
2P
)1100
()100
(
nknk
K
XX
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
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:
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 22
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
w p
eri
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
COMPETENCY TESTED: solves problems
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
COMPETENCY TESTED: solves problems
involving geometric figures on the coordinate
plane.
BT: Analyzing, Applying
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)
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 23
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
w p
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
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;
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
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)(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}
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 24
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
w p
eri
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
COMPETENCY TESTED: solves problems on
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
CONCEPT: Probability
COMPETENCY TESTED: solves problems
involving probability.
BT: Analyzing, Applying
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
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 25
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
w p
eri
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.***
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 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. ***
BT: Remembering, Applying
EXPLN:
5
3tan
H
O
so finding the hypotenuse,
H = 3453 22
Hence,
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 26
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
w p
eri
od
.
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
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 27
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
w p
eri
od
.
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
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
TEACHER’S GUIDE ACET 2016
SET A MATHEMATICS
ADMATH – TG 2015 28
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
w p
eri
od
.
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
CONCEPT: Probability
COMPETENCY TESTED: solves problems
involving probability.
BT: Analyzing, Applying
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