Upload
gilbert-joseph-abueg
View
294
Download
4
Embed Size (px)
Citation preview
Question 1
What is the smallest three-digit number that
leaves a remainder of 1 when divided by 2,
3, and 5?
set 1math-tanong CEER Supplement
(a) 121
(b) 151
(c) 181
(d) 211
Question 1 Solution
• Determine first the least common multiple
(LCM) of 2, 3, and 5 – which is 30.
• Now, find the smallest three-digit multiple of
the LCM – that’s 120.the LCM – that’s 120.
• The smallest 3-digit number that leaves a
remainder of 1 when divided by 2, 3, and 5 is
the smallest 3-digit number that leaves a
remainder of 1 when divided by 30, the LCM
– that’s 121 ☺
set 1math-tanong CEER Supplement
Question 2
The pie chart above shows the distribution of DVD
rentals from Dibidi Doo Bee shop for a single night. If
250 DVDs were rented that night, how many more
action movies were rented than horror movies?
set 1math-tanong CEER Supplement
(a) 10
(b) 20
(c) 22
(d) 25
Question 2 Solution
Those who rented action
movies are
TIP: For this type of problem, just subtract the rates
before applying the percentage – no need to multiply
first before subtracting.
movies are
22% — 12% = 10%
more than those who rented
horror movies. Thus, there are
250(10%) = 250(0.1) = 25
more action movies rented than
horror movies.
set 1math-tanong CEER Supplement
Question 3
Suppose that n and p are integers greater than 1. If 5n is
a perfect square and 75np is a perfect cube, what is the
smallest value of n + p?
set 1math-tanong CEER Supplement
(a) 5 (c) 8
(b) 6 (d) 15
Question 3 Solution
If 5n is a perfect square, the smallest possible value of n
would be 5. Now,275 5 3np np= i i
If 75np is a perfect cube and n = 5 is the smallest
set 1math-tanong CEER Supplement
If 75np is a perfect cube and n = 5 is the smallest
possible value of n, the smallest possible value for p is
32 = 9 since 2 2 3 375 5 3 5 3 5 3np = =i i i i
Hence, the smallest possible value for n + p is
5 + 9 = 14
Question 4
For real numbers a and b, define the operation
3 2a b b a∗ = −
If , what is the value of X? 2 3 3 X∗ = ∗
set 1math-tanong CEER Supplement
If , what is the value of ? 2 3 3 X∗ = ∗
( ) ( )
( ) ( )
11a 2 c
3
9b d 4
2
Question 4 Solution
By definition of the operation :
( ) ( ) ( )2 3 3 3 3 2 2 3 2 3
5 3 6
X X
x
∗ = ∗ ⇒ − = −
= −
" "∗
set 1math-tanong CEER Supplement
5 3 6
3 11
11
3
x
x
x
= −
=
=
Question 4 Solution
Of course, you can do substitution. By the definition,
( ) ( )
( ) ( )
2 3 3 3 2 2 5
3 3 2 3 3 6
and
X X X
∗ = − =
∗ = − = −
set 1math-tanong CEER Supplement
( ) ( )3 3 2 3 3 6X X X∗ = − = −
Q: Which of the choices will give a value of 5 when
substituted to X in 3X − 6? Choice (c).
Question 5
A test is composed of 25 questions. The score of an
examinee is obtained by giving him 4 points for each
correct answer and deducting 1 point for each wrong
answer. If an examinee answered all the questions and
obtained a score of 70, how many questions did he obtained a score of 70, how many questions did he
answer correctly?
set 1math-tanong CEER Supplement
(a) 17 (c) 19
(b) 18 (d) 20
Question 5 Solution
Let x = the number of correct answers
25 – x = the number of wrong answers
(since there are 25 questions in all)
:Equation
set 1math-tanong CEER Supplement
No. of questions answered
Points per
questionTotal points
Correct x 4 4x
Wrong 25 − x 11(25 − x) = 25 − x
( )4 25 70
4 25 70
5 95
9
:
1
Equatio
x x
x x
x
x
n
− − =
− + =
=
=
Question 6
In the figure below, . What is the
value of x?1 2� � �
(a) 40
set 1math-tanong CEER Supplement
(a) 40
(b) 50
(c) 60
(d) 70
`
Question 6 Solution
Draw an extra line passing through the vertex of
angle x and parallel to 1 2 and � �
set 1math-tanong CEER Supplement
m°
n°
Question 6 Solution
If m and n are the angles formed as in the figure, then,
using the ideas of angles formed by a transversal:
( )20 corresponding anglesm =
set 1math-tanong CEER Supplement
m°
n°
180 150 30
(alternate interior 150,
thenadjacent)
n = − =
50x m n∴ = + =
Question 7
If , what is the value of ? 6 5.43x− = 6 x+
(a) 6.43
set 1math-tanong CEER Supplement
(a) 6.43
(b) 6.57
(c) 17.43
(d) 17.51
Question 7 Solution
Warning: DO NOT ATTEMPT to solve
for x! Sayang ang effort!
set 1math-tanong CEER Supplement
6 5.43 5.43 6 0.57
0.57
6 6 0.57 6.57
x x
x
x
− = ⇒− = − = −
=
∴ + = + =
Question 9
Erin calculated the average of 5 numbers to be 38. Then
she found out that she had made an error and had
written 40 for one of the numbers when she should have
written 30. What is the average of the correct 5 numbers?
set 1math-tanong CEER Supplement
(a) 32
(b) 34
(c) 36
(d) 38
Question 9 Solution
Let S be the sum of the first 4 numbers that Erin has.
If the average of 5 numbers is 38, including the
erroneous 40, then
4038 40 190
5
Saverage S
+= = ⇒ + =
set 1math-tanong CEER Supplement
38 40 1905
average S= = ⇒ + =
But the last number should be 30 instead of 40, so
40 10 190 10 30 180S S+ − = − ⇒ + =
The average, then, should have been
30 18036
5 5
S+= =
Question 9 Solution
“Expert” shortcut solution: Start with
4038 40 190
5
Soriginal average S
+= = ⇒ + =
set 1math-tanong CEER Supplement
5
Then the correct average is
( )40 1030 190 10 18036
5 5 5 5
SS + −+ −= = = =
Question 10
In the diagram, FDCB is a rectangle. Segment ED is 6
units long, segment AB is 10 units long, and the measure
of angle ECD is 60°.What is the length of segment AE?
set 1math-tanong CEER Supplement
( ) ( )
( ) ( )
3a 20 c 20
2
3b d 20 4 3
2
−
−
Question 10 Solution
Fast facts about the figure:
1. Triangle CDE is a 30-60-90 triangle with angle DEC =
30 degrees.
2. Since DF is parallel
set 1math-tanong CEER Supplement
2. Since DF is parallel
to CB, and AE acts
as a transversal, then
angle AEF is 30
degrees. Hence,
triangle EAF is also a
30-60-90 triangle.
30°
x2
x
3
2
x
y
2
y
2
x
Question 10 Solution
Hence, we have the following:
1. By the properties of a 30-60-90 triangle:
set 1math-tanong CEER Supplement
6 12 32 4 3
3 3 3
12 3
2
EC
DC EC
= = =
= =
i
30°
x2
x
36
2
x=
Question 10 Solution
2. Since BCDF is a rectangle, DC = BF, so
2 3DC BF= =
set 1math-tanong CEER Supplement
3. Since AB = 10, then
10
10 2 3
AF BF= −
= −
30°
2 32 3
Question 10 Solution
4. Since EAF is a 30-60-90 triangle with angle AEF
= 30 degrees, then
2AE AF=
set 1math-tanong CEER Supplement
( )
2
2 10 2 3
20 4 3
AE AF=
= −
= −
30°
10 2 3−
2 3
( )2 10 2 3y = −
Question 10 Solution
Expert fast solution! Let x = EF and z = AE Based on the
established facts earlier, you have the figure shown.
3 12 36 4 3
2 3 3
xx= ⇒ = =i
set 1math-tanong CEER Supplement
30°
x2
x
36
2
x=
z
2
zy =
2
x
( )
4 32 3
2 2
10 10 2 32
10 2 3 2 10 2 32
20 4 3
x
xy
zy z
= =
= − = −
= = − ⇒ = −
= −