Under the amendment, qualified ALS & PEPT passers can avail of the following amount:
1. NCR = Php 22,500.00
2. Highly urbanized cities outside NCR = Php 20,000.00
3. Other locations = Php 17,500.00
Refer to the D.O. below for more information:
Anything about Alternative Learning System (ALS) Accreditation & Equivalency ( A & E) Program of the Department of Education (DepEd) of the Philippines in line with the K to 12 Basic Education Curriculum.
Wednesday, May 22, 2019
D.O. No. 009, s. 2019 - SHS Voucher System for ALS A&E and PEPT Passers
Under the amendment, qualified ALS & PEPT passers can avail of the following amount:
1. NCR = Php 22,500.00
2. Highly urbanized cities outside NCR = Php 20,000.00
3. Other locations = Php 17,500.00
Refer to the D.O. below for more information:
BAKIT MARAMING BUMAGSAK SA BATCH 2018 ALS A&E TEST?
Nakapanlulumong isipin na tila
mas maraming bumagsak sa nakaraang Alternative Learning System (ALS)
Accreditation & Equivalency (A&E) test na ibinigay ng Department of
Education (DepEd) noong ika-24 ng Pebrero, 2019 para sa Luzon at ika-3 ng Marso,
2019 para sa Visayas at Mindanao kahit hindi pa inilalabas ng DepEd ang opisyal
na bilang at poryento ng mga pasado. Maraming dahilan kung bakit hindi pumasa
sa pagsusulit ang mga mag-aaral ng ALS subali’t malaking bahagi nito ang dahil
sa kakulangan.
1. Kulang sa oras. Ito
ang malimit na dahilan kung bakit hindi pinalad ang isang ALS learners. Kulang
ang oras ng mga mag-aaral sa pagpasok sa mga araw na kailangang pumasok dahil
na rin sa pagtatrabaho at iba pang gawaing-bahay. Dahil kailangan munang kumita
at/o tapusin ang mga gawaing-bahay, limitado rin ang kanilang oras sa
pagrerepaso ng mga leksyon. Kulang din ang oras ng mga guro at coordinators ng
ALS upang pagtuunan ng pansin ang paparaming estudyante. Hindi sapat ang
kanilang oras upang masigurong nauunawaan ng mga mag-aaral ang bawat kasanayan
at karunungang dapat taglayin upang sila ay makapasa sa pagsusulit.
2. Kulang sa pokus at
konsentrasyon. Dahil maraming mga bagay ang dapat pagtuunan ng pansin ng
sabay-sabay, nagkukulang sa pokus at konsentrasyon ang mga mag-aaral ng ALS.
Kadalasan, mali pa ang pinagtutuunan nila ng pansin. Mas marami ang
nag-i-internet hindi upang maghanap ng mga review materials kundi upang
makipagsosyalan lamang sa Facebook at Twitter. Nawawaglit sa kanilang isipan na
dapat ang unahin ay kung paano pumasa sa ALS A&E kaysa kung paano
magkakaroon ng maraming kaibigan at/o nobyo/nobya. Nakasalalay sa pagpasa sa eksamin
ang katuparan ng kanilang mga pangarap kung kaya’t doon dapat sila naka-pokus.
3. Kulang sa tiwala. Marami ang hindi pumasa sa pagsusulit dahil
kulang sila sa tiwala sa sarili. Hindi pa man nagsisimula ang pagsusulit, sa
kanilang isipin, sila ay bagsak na. Hindi nila pinaghandaan o binigyang
solusyon ang kanilang kahinaan. Hindi nagtanong sa guro at mga kamag-aral kung
paano i-solve ang isang Math problem. Pinabayaan nilang lukobin sila ng takot
at ipagwalang-bahala ang kanilang mga pangarap. “Bahala na si Batman!”, ‘ika
nga nila.
4. Kulang sa load. Dahil sa kakulangan sa salapi, maraming mag-aaral
ang kulang sa load upang makahanap ng mga websites at social group na
makatutulong sa kanilang pag-aaral at pagrerepaso. Ito ang dahilan kung bakit hindi
nila nahahanap ang mga dapat nilang aralin at repasuhin. Gayunman, dapat ay
gumawa sila ng paraan upang kahit papaano ay maka-connect sila sa internet o
kung hindi man ay humingi ng tulong sa mga kamag-aral na gumagamit nito.
5. Kulang sa review materials. Isa ring kadahilanan ng pagbagsak ng
marami sa pagsusulit ay ang kakulangan ng mga review materials buhat sa kanilang
guro. Kung walang ideya ang mga guro kung anong karunungan at kasanayan ang
posibleng lumabas sa eksamin, mahihirapan pumasa ang mga mag-aaral dahil tila
nangangapa sila sa dilim sa dami ng dapat pag-aralan. Kung meron mang
reviewers, mali ang akala ng ibang estudyante na ang mga ito mismo ang lalabas
sa pagsusulit, words for words. Hindi nila batid na ang mga reviewers at giya
lamang. Konsepto ang dapat pag-aralan at hindi ang mismong sagot dahil iba ang
lalabas sa test kaysa sa nirepaso subali’t naroroon ang konsepto.
6. Kulang sa pansin. Dahil kulang sa pansin ng mga guro, magulang,
kamag-aral, at kaibigan, pa-bandying-bandying lamang ang ilan sa mga ALS
learners. Kulang ang mga nag-uudyok upang pagbutihin nila ang pag-aaral,
pagrerepaso, at pagkuha ng eksamin. Dahil kulang ang pumapansin sa kanila, hindi
sila nagpupunyagi nang husto dahil para sa mga mag-aaral na ito, pareho lamang
kung sila ay pumasa o bumagsak sa pagsusulit.
7. Kulang sa pagsagot. Isa ring dahilan ng pagbagsak ng mga mag-aaral
ng ALS ay ang katamaran nilang sumagot sa mga tanong na naka-post sa kanilang
grupo. Kalimitan, hanggang “SEEN” na lang sila. Hindi man lamang sinubukang
sagutin o i-solve ang isang Math problem. Kung hindi nila ito ginawa, paano
nila malalalaman na sapat na ang kanilang kaalaman, kasanayan, at kahandaan sa
pagsusulit? Mas mabuti nang magkamali sa pagsagot sa tanong bago pa ang
pagsusulit kaysa sa mismong eksamin.
Ikaw, pumasa ka ba sa 2018 ALS A&E test? Ano ang dahilan kung bakit
ka bumagsak? Ano ang kakulangan sa iyo? I-komento sa ibaba.
Ano man ang dahilan ang iyong pagbagsak sa A&E test, nararapat lamang na itama ang inyong pagkakamali dahil nakasalalay sa pagpasa rito ang iyong mga pangarap sa buhay. Bumangon at lumabang muli! Maging handa!
Tuesday, May 21, 2019
ALS A& E PASSERS - FEBRUARY 24, 2019
Region: NCR
Division: Quezon City
Level: Secondary/High School
Testing Center: Commonwealth Elementary School
Thursday, May 16, 2019
ALS A&E RESULTS
Likewise, guidelines for the availment of voucher system for ALS A&E passers are being processed.
For those who wish to have their names check against A&E passers once it was released, can comment their names, region, and division office in the following format:
FAMILY name, GIVEN names, MIDDLE name
Region No., Division Office
Secondary or Elementary
Example:
DELA CRUZ, JUAN PEDRO, SANTOS
NCR, Quezon City
Secondary
Monday, April 29, 2019
2019 ALS A&E Reviewer - SCIENCE
Alternative
Learning System (ALS)
Accreditation
& Equivalency (A&E)
Review Questions – SCIENCE
Choose the
letter of the correct answer.
1. Which of
the following statements describes the difference between active and inactive
volcano?
A. The
summit of an active volcano has higher elevation while an inactive volcano has
steeper sides.
B. The
summit of an active volcano is lower while an inactive volcano has less steep
sides.
C. An active
volcano had history of eruptions while an inactive volcano periodically erupts.
D. An active
volcano has a record of periodic eruption while an inactive volcano had
recorded history of eruptions.
2. Which of
the following explains the cause of volcanic eruptions?
I. Magma rises to the crater.
II. Magma is
moving in the chamber.
III. When built-up pressure is released, magma explodes to
the surface.
IV. Lava
solidifies and becomes part of the crust.
A. I only
B. I and III
C. III only
D. II and IV
3. Mariel
observes the patterns formed by prominent stars almost every night. Which of
the following statements may serve as her correct realization about
constellation?
A. Stars
assume different patterns every week.
B. Different
constellations appear in the sky every night.
C. A
constellation will appear in the same position in the sky throughout the year.
D. Different
patterns of stars appear in the night sky as the earth revolves around the sun.
4. Darlene observes that each day a
few stars are visible in the eastern region of the sky that were not visible
the night before. Which of the following statements explains her observation?
A. The Earth spins
on its axis as it revolves around the Sun.
B. The stars move around the Earth.
C. The sky moves around the Sun.
D. The Earth rotates on its axis.
5. John learned that the sun is a
star. Which of the following is a proof of his information?
A. Stars and the Sun, unlike planets,
become smaller and smaller after thousands of years.
B. Like the Sun,
stars produce light due to nuclear reactions.
C. Stars belong to the Solar System,
hence, it is a star.
D. Stars like the Sun do not move.
6. Which statement describes the cause
of plate movement?
A. Molten rock in
the mantle moves due to intense heat.
B. Molten rock beneath the Earth behaves
like liquid.
C. The plate moves due to the rotation
of the Earth.
D. The plate moves due to tidal waves.
7. Melai always looks at the expiry
dates of food items she buys at the supermarket in order to avoid food poisoning.
Which of the following is the cause of food expiration?
A. Cans and packaging deteriorate over
time making the food inside them not safe for consumption.
B. Canned and packed goods are treated
with chemicals which expire over time.
C. Foods are organic matters that naturally
degrade over time.
D. Food manufacturers
need to set an expiration date so that buyers will consume food items before
the expiration and buy new food items again.
8. Which of the following is an
evidence of Plate Tectonic Theory?
I. Shapes of continents fit like a
puzzle.
II. Identical rocks formed millions of
years ago are found in different continents.
III. Identical fossils are found in
different continents.
IV. Islands sunk and rise.
A. II and IV
B. I, II, and III
C. I and III
D. I, II, and IV
2019 Actual ALS & A&E Test - Mathematics with Answers and Solutions
Alternative Learning System (ALS)
Accreditation & Equivalency (A&E)
Review Questions
MATHEMATICS with Answers & Solutions
Read each item carefully and then select the letter of the correct
answer.
1. A factory packs 25 small shirts, 30 medium shirts, and 15 large
shirts in one box. If 550 boxes are packed with these shirts, how many shirts
are there in all?
A. 33,000
B. 38,500
C. 44,000
D. 49,500
Answer: B. 38,500
Solution: 25 + 30 + 15 = 70 x
550 = 38,500
2. The Nayon Homeowners Association generated Php 26,000.00 from the
sale of tickets for a fund raising show. Tickets for adults cost Php 150.00
while tickets for children cost Php 100.00 If 112 tickets for adults were sold,
how many tickets for children were sold?
A. 115
B. 100
C. 92
D. 88
Answer: C. 92
Solution: 150x + 100y = 26 000
150(112) + 100y = 26 000
16 800 + 100y = 26 000
100y = 26 000 – 16 800
100y = 9 200
y = 9 200/100
y = 92
3. Perform the indicated operations: 3 + 6 x 5 + 12 ÷ 3 – 7
A. 12
B. 30
C. 42
D. 44
Answer: B. 30
Solution: 3 + 6 x 5 + 12 ÷ 3 – 7 =
3 + (6 x 5) + (12 ÷ 3) – 7 =
3 + 30 + 4 – 7 =
37 – 7 = 30
4. Simplify: 150 + (20 + 5) x 2 – 100
A. 1, 000
B. 500
C. 200
D. 100
Answer: D. 100
Solution: 150 + (20 + 5) x 2 – 100 =
150 + (25) x 2 – 100 =
150 + 50 – 100 =
200 – 100 = 100
5. Lian has a daily allowance of Php 100.00. She spends Php 50.00 for
snacks, Php 25.00 for transportation, and saves the rest. To solve for Lian’s
saving’s S, which of the following number sentence is CORRECT?
A. S = (50 – 25) + 100
B. S = (50 + 25) – 100
C. S = 100 + (50 + 25)
D. S = 100 – (50 + 25)
Answer: D. S = 100 – (50 + 25)
6. Joan bought 2 pencils at Php 20.00 each and a ballpen for Php 30.00.
How much did she pay to the storeowner?
A. C = 2 + 20 + 30
B. C = 2 + 20(30)
C. C = 2(20) + 30
D. C = 2(20) – 30
Answer: C. C = 2(20) + 30
7. A piece of wire that is 7/8 m long will be cut into 16 pieces of
equal length. What is the length of each piece?
A. 128/7 m
B. 128/8 m
C. 7/2 m
D. 7/128 m
Answer: D. 7/128 m
Solution: 7/8 ÷ 16 = 7/8 x 1/16 = 7/128
(Rule on Dividing Fraction)
8. Mrs. Ruiz bought a cake for her son’s birthday. She cut 1/3 of the
cake and kept it. Then she divided the remaining part equally among her 4
children. What part of the cake did each child get?
A. 1/12
B. 1/6
C. 2/3
D. 8/3
Answer: B. 1/6
Solution: 2/3 ÷ 4 = 2/3 x ¼ = 2/12 = 1/6 (Rule on Dividing Fraction
& Reducing to Lowest Term)
9. Every month, the class treasurer collects Php 7.50 from each student
for the class fund. If there are 45 students in a class, how much is her
collection every month?
A. Php 33.75
B. Php 52.50
C. Php 337.50
D. Php 3,375.00
Answer: Php 337.50
Solution: 7.50 x 45 = 337.50
10. At the Science Centrum, Jefferson bought a T-shirt for Php 200.25,
a ballpen for Php 35.50, and a book for Php 105.75. If he gave Php 500.00 to
the cashier, how much change did he get?
A. Php 341.50
B. Php 158.50
C. Php 143.50
D. Php 185.50
Answer: B. Php 158.50
Solution: 500 – (200.25 + 35.50
+ 105.75) =
500 – 341.50 = 158.50
11. The temperature in a certain city is given in the table below. Find
the difference between the temperatures at 5 AM and 11 PM.
Time
|
Temperature (oF)
|
5 AM
|
-90
|
11 AM
|
100
|
5 PM
|
50
|
11 PM
|
-70
|
A. 20 oF
B. -20 oF
C. 10 oF
D. -10 oF
Answer: B. -20 oF
Solution: (-90) – (-70) = (-90) + 70 = -20 (Rule on Subtracting
Integers)
12. Mother lost 3 kg of her weight when she was sick. After recovering,
she gained 7 kg and then lost 2 kg. How much weight she finally lost or gained?
A. lost 2 k
B. lost 1 kg
C. gained 1 kg
D. gained 2 kg
Answer: gained 2 kg
Solution: x = – 3 + 7 – 2
x = -5 + 7 = 2
13. Cora lent 35 pesos to Lita. Then Cora received 50 pesos from her
uncle. Then Cora’s sister borrowed from her 20 pesos. Cora also paid 18 pesos
to a newsboy. Cora still had 43 pesos left. How much did she have at the start?
A. 80 pesos
B. 66 pesos
C. 26 pesos
D. 20 pesos
Answer: B. 66 pesos
Solution: x – 35 + 50 – 20 -18 = 43
x – 73 + 50 = 43
x = 43 + 73 – 50
x = 116 – 50
x = 66
14. The diagonal of the floor of a rectangular room is 10 m. The
shorter side of the room is 6 m. What is the floor area of the room?
A. 48 sq. m.
B. 60 sq. m.
C. 80 sq. m.
D. 480 sq. m.
Answer: A. 48 sq. m.
Solution: Use Pythagorean Formula to solve for the longer side of the
room
c 2 = a2 + b2
102 = 62 + b2
100 = 36 + b2
b2 = 100 – 36
b2 = 64
b = sqrt 64
b = 8
Area of rectangle = length x
width
Area = 8 x 6 = 48 sq. m.
15. If the radius of the circle is 10.5 m, what is its circumference?
A. 13.64 m
B. 16.49 m
C. 32.97 m
D. 65.94 m
Answer: D. 65.94 m
Solution: Circumference (C) = 2 π r
(Ï€ = pi = 3.14; r = radius)
C = 2 (3.14) (10.5)
C = 65.94 m
16. Andrew is staying at a hotel that has L-shaped swimming pool. He
wants to know the area of the pool. What is the area of the swimming pool given
the dimensions below:
A. 96 m2
B. 150 m2
C. 160 m2
D. 186 m2
Answer: D. 186 m2
Solution: Total area (TA) = Area of big rectangle + Area of small
rectangle
TA = (length x width) + (length x width)
TA = (15 x 10) + [ 6 x (16 -10)]
TA = 150 + ( 6 x 6)
TA = 150 + 36
TA = 186 m2
17. The park has a rectangular garden that is 18 m long and 9 m wide.
What is the area of the garden?
A. 324 sq. m.
B. 162 sq. m.
C. 54 sq. m.
D. 27 sq. m.
Answer: B. 162 sq. m.
Solution: Area of rectangle (A) = length x width
A = 18 x 9
A = 162 sq. m.
18. Compute for the surface area of a cube of side 6 m.
A. 216 cm2
B. 144 cm2
C. 108 cm2
D. 96 cm2
Answer: 216 cm3
Solution: Surface area of a cube (SA) = 6a2 (a = edge)
SA = 6(6)2
SA = 6(36)
SA = 216 cm2
19. A rectangular prism has a length of 20 cm, a width of 10 cm, and a
height of 30 cm. What is its volume?
A. 6,000 cm3
B. 3,000 cm3
C. 1,400 cm3
D. 60 cm3
Answer: A. 6,000 cm3
Solution: Volume (V) = length x width x height
V = 20 x 10 x 30
V = 6000 cm3
20. In a fruit basket, there are 4 papayas, 9 guavas, and 7 mangoes.
What percent of the fruits are papayas?
A. 2%
B. 4%
C. 8%
D. 20%
Answer: D. 20%
Solution: Percent (P) = (Number of papayas / Total fruits) x 100%
P = 4 / (4 + 9 + 7)
P = 4/20
P = 0.20 x 100%
P = 20%
21. A shirt costs Php 400.00. If it is on sale at 20% discount, how
much will a buyer pay for the shirt?
A. Php 200.00
B. Php 320.00
C. Php 340.00
D. Php 380.00
Answer: B. Php 320.00
Solution: x = 400 – (400 x 20%)
x = 400 – (400 x 0.20)
x = 400 - (80)
x = 400 – 80 = 320
For items 22-24. The graph
below shows Ada’s activities during Saturdays
22. About how many hours Ada has spent in marketing?
A. 17
B. 16
C. 6
D. 4
Answer: D. 4
Solution: Note: there are 24 hours in one day
x = 24 x 17%
x = 24 x 0.17
x = 4.08
x = 4 hours (round off)
23. How many percent does Ada spend in cooking?
A. 12%
B. 15%
C. 17%
D. 33%
Answer: B. 15% (See graph)
24. How many percent lesser does Ada spend in cleaning the house than
in sleeping?
A. 52%
B. 42%
C. 24%
D. 14%
Answer: D. 14%
Solution: Cleaning = 19% ; Sleeping =
33%
x = 33% - 19%
x = 14%
25. Jonas has made a scale model of his proposed new house as shown in
the drawing below. The house is 24 meters long. If the model scale is 1:4 in
centimeters, what is the length of the house in his drawing?
A. 6 cm
B. 10 cm
C. 14 cm
D. 28 cm
Answer: A. 6 cm
Solution: 24/ 4 = 6 cm
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