kuiz bulan vokasional

7/29/2019 KUIZ Bulan Vokasional

1/11

KUIZ BULAN TEKNIK DAN VOKASIONAL

APLIKASI MATEMATIK TAMBAHAN DI DALAM KEDIHUPAN HARIAN

MASA : 1 JAM

NAME : 1) ____________________ 3) ______________________

2) ____________________ 4) ______________________

TINGKATAN : 5SN _______ MARKAH : ________

ARAHAN: Kuiz ini dijalankan secara BERKUMPULAN (34 orang satu kumpulan). Soalan

boleh dijawab dalam Bahasa Inggeris atau Bahasa Malaysia. Tunjukkan langkah

penyelesaian dengan jelas dan teratur.

1) How many cows clear the field?Suppose you have a grassy field, and cows eat grass at a constant rate. Keep in mind, the grass

keeps growing continuously. 48 cows can clear all the grass off the field in 90 days. 120 cows can

clear all the grass off the field in 30 days. How many cows would be needed to clear all of the

grass in 16 days? (Round up to the nearest whole cow.)

Andaikan anda mempunyai satu padang berumput dan lembu-lembu makan rumput pada kadar

yang tetap. Ingat, rumput tumbuh secara berterusan. 48 ekor lembu boleh menghabiskan semua

rumput di padang dalam 90 hari. 120 ekor lembu boleh menghabiskan semua rumput di padang

dalam 30 hari. Berapakah ekor lembu diperlukan untuk menghabiskan kesemua rumput dalam

masa 16 hari? (Bundarkan kepada nombor bulat)

SOLUTION / PENYELESAIAN :
http://www.gottfriedville.net/mathprob/alg-cowsgrass.htmlhttp://www.gottfriedville.net/mathprob/alg-cowsgrass.html


2/11

2) How much money did they have?During the first week of their vacation trip, the Stanbert family spent $200 more than three fifths

of their vacation money and had more than $400 less than half of it left. If they started their trip

with a whole number of dollars, what was the greatest amount of vacation money they could have

had?Dalam minggu pertama percutian, keluarga Stanbert membelanjakan $200 lebih daripada

jumlah duit percutian dan ia melebihi $400 kurang daripada separuh bakinya. Jika mereka mula

bercuti dengan sejumlah wang dalam bentuk nombor bulat, cari nilai terbesar bagi duit

percutian yang mungkin.

http://www.gottfriedville.net/mathprob/alg-money.htmlhttp://www.gottfriedville.net/mathprob/alg-money.html


3/11

3) Dividing an estate equallyA man dies and leave his estate to his sons. The estate is divided as follows:

1st son gets 100 acres+ 1/10 of remainder of estate.

2nd son gets 200 acres + 1/10 of remainder of estate ...

(n)th son gets 100 (n) acres+ 1/10 of the remainderEach son receives same amount.

How many sons were there, what did each receive, and what was the estate?

Satu lelaki meninggal dunia dan meninggalkan harta-bendanya kepada anak-anak lelakinya.

Harta bendanya dibahagikan seperti yang berikut:

Anak lelaki pertama mendapat 100 ekar +

daripada baki harta bendanya.

Anak lelaki kedua mendapat 200 ekar +


Anak lelaki ke-n mendapat 100 x n ekar +


Setiap anak lelaki menerima jumlah yang sama.

Berapakah anak-anak lelaki yang ada? Setiap orang menerima berapa?

Cari jumlah kuantiti harta benda itu.

http://www.gottfriedville.net/mathprob/alg-will.htmlhttp://www.gottfriedville.net/mathprob/alg-will.html


4/11

4) Saving should start early!Consider twin sisters with two different retirement savings plan:

Plan 1: Ivy begins a retirement account at age 20. She starts with $2000 and then saves $2000 per

year at 7% interest compounded annually for 10 years. Then she stops contributing to the account

but keeps her savings invested at the same rate.

Plan 2: Sonia doesn't save any money in her twenties. When she turns 30 she starts with $2000and then saves $2000 per year at 7% interest compounded annually for 35 years.

Which one has more at age 65 ?

Andaikan sepasang saudara kembar perempuan mempunyai dua jenis rancangan simpanan

persaraan yang berbeza:

Rancangan 1: Ivy memulakan akaun persaraannya pada usia 20. Dia bermula dengan $2000 dan

kemudian menyimpan $2000 setiap tahun dengan faedah 7% setiap tahun selama 10 tahun.

Selepas itu, dia berhenti menambahkan duit tetapi mengekalkan

simpanannya melabur pada kadar yang sama.

Rancangan 2: Sonia tidak menyimpan sebarang duit pada usia 20-an. Apabila usianya

menjangkau 30, dia bermula dengan $2000 dan kemudian menyimpan $2000 setiap tahun

dengan faedah 7% setiap tahun selama 35 tahun.

Siapakah mempunyai lebih banyak simpanan pada umur 65?

SOLUTION / PENYELESAIAN:
http://www.gottfriedville.net/mathprob/misc-compound.htmlhttp://www.gottfriedville.net/mathprob/misc-compound.html


5/11

5) Gorilla transporting bananasA gorilla harvests 3,000 bananas and needs to carry them 1,000 kilometres to the supermarket. Hecan only carry 1,000 at a time. Since he is a gorilla he eats 1 banana every kilometre he goes in anydirection. He can (and will have to) leave bananas anywhere along the way. Once all his bananashave reached the end he DOES NOT need any to eat to get back. Remember he eats 1 banana

every kilometre he goes even if he is going back to pick up more bananas. What is the maximumnumber of bananas he can get to the market ?

Seekor gorila mendapat hasil 3000 pisang dan perlu membawanya sejauh

1000 kilometer ke pasar raya. Ia hanya membawa 1000 pisang pada satu masa. Disebabkan oleh

ia seekor gorila, ia memakan 1 pisang setiap kilometer dalam perjalanannya. Ia boleh

meninggalkan pisang di mana-mana sahaja sepanjang perjalanan. Sebaik sahaja semua pisangnya

sampai di pasar raya, ia TIDAK perlu makan lagi untuk kembali. Ingat bahawa ia makan 1 pisang

setiap kilometer dilaluinya walaupun ia kembali mengutip lebih banyak pisang. Apakah nilai

maksimum pisang yang boleh dibawanya sampai di pasar raya?

SOLUTION / PENYELESAIAN:


6/11

6) How far apart are the towns?A truck travels from X to Y. Going uphill, it goes at 56 km/h. Going downhill, it goes at72 km/h. On level ground, it goes at 63 km/h. If it takes 4 hours to travel from X to Y, and 5hours to come back, what is the distance between X and Y?

Sebuah lori bertolak dari X ke Y. Semasa menaiki cerun, lori itu berjalan dengan kelajuan 56km/j.

Semasa turun, lori itu berjalan pada kelajuan 72km/j. Di tanah rata, lori itu berjalan dengan

kelajuan 63km/j. Jika lori itu mengambil masa 4 jam untuk berjalan dari X ke Y dan 5 jam untuk

balik, apakah jarak antara X dan Y?


END of QUESTIONS
http://www.gottfriedville.net/mathprob/misc-truckxy.htmlhttp://www.gottfriedville.net/mathprob/misc-truckxy.html


7/11

ANSWER

1) A cow eats a certain amount of grass in one day, call it c.

The field grows by a certain amount each day, call it g.

The field has some initial amount of grass: i

i + 90 g - 48*90 c = 0

i + 30 g - 30*120 c = 0

1.2 i + 36 g - 4320 c = 0

i + 90 g - 4320 c = 0

0.2 i - 54 g = 0

0.2 i = 54 g

i = 270 g

The field starts with 270 days' growth

i + 30 g - 30*120 c = 0

300 g - 3600 c = 0

g = 12c

i + 90 g - 4320 c = 0

360 g - 4320 c = 0

g = 12c ... confirmed

It takes 12 cows to eat one day's growth in one day.

So 270 initial + 16 days' growth = 286 days' growth

In 16 days, n cows eat 16n / 12 days' growth.

16n/12 = 286

16n = 3432

n = 214.5 cows

Round to 215

215/12 = about 18 days' growth consumed each day

18 * 16 = 288 which exceeds the 286.


8/11

2) Let M = money they started with."During the first week of their vacation trip, the Gomez family spent

$200 more than three fifths of their vacation money"

First week spending = 3/5 M + 200,

leaving M - 3/5 M - 200 or 2/5 M - 200.

"had more than $400 less than half of it left."

2/5 M - 200 > M/2-400

Add 400

2/5 M + 200 > M/2

Convert to common denominator

4/10 M + 200 > 5/10 M

Subtract 4/10 M200 > 1/10 M

Mult by 10

2000 > M

The most they could have started with was $1999.

3) Answer: 9 sons, 8100 crowns,which we derive as follows:

Let Sn be the share of each son, and E be the entire estate.S1 = 100 + (E - 100)/10S2 = 200 + (E - S1 - 200)/10

S1 = S2

200 + (E - S1 - 200)/10 = 100 + (E - 100)/102000 + E - S1 - 200 = 1000 + E - 100

How handy: the E's cancel!

1800 - S1 = 900S1 = 900

S1 = 100 + (E - 100)/10(E - 100)/10 = 800E - 100 = 8000E = 8100So each got 900/8100 = 1/99 Sons.


9/11

S1 = 100 + 8000/10 = 900, Leaves 7200S2 = 200 + 7000/10 = 900, Leaves 6300S3 = 300 + 6000/10 = 900, Leaves 5400S4 = 400 + 5000/10 = 900,S5 = 500 + 4000/10 = 900,

S6 = 600 + 3000/10 = 900,S7 = 700 + 2000/10 = 900,S8 = 800 + 1000/10 = 900S9 = 900 .... done!

4) At the end of the 10 years, the first one has 2000 x (1.07^10 + 1.07^9 + 1.07^8 + ... + 1.07^2 + 1.07) =29567.20

This then is multiplied by 1.07^35 = 10.68 to yield 315,676.62

The other one gets 2000 x (1.07 + 1.07^2 + 1.07^3 + ... + 1.07^35) = 2000 x 137.4 = 274,800

She doesn't catch up.

5) GorillaWhen he has more than 2000 bananas, it costs him 5 bananas / mile to

transport them, since he makes 3 trips forward and 2 trips back.

When he has more than 1000 bananas, it costs him 3 bananas / mile, for 2

trips forward and 1 trip back.

And when he has 1000 or fewer, it's 1 banana / mile.

I think the way to do it is: see how far he can get 2000 bananas, then 1000

bananas, and then make one final trip.

200 miles @ 5 bananas / mile = 1000 bananas cost.

So if the first stopping point is at 200 miles,

he would carry 1000, eat 200, drop 600, and eat 200.

Then he would do that again.

On the third trip, he doesn't eat the second 200, so at the 200 mile

marker he has 600 + 600 + 800 = 2000 bananas.

If the second leg is 333 miles, it works out as:

carry 1000, eat 333, drop 334, eat 333.


10/11

carry 1000, eat 333, drop 667.

Now he has 1001 bananas, and it's just as well to throw one away. To

make it come out exact, we'd need to use fractional miles and fractional

bananas.

Now we have 1000 bananas at the 533 mile marker.

For the last leg, he just carries the last 1000 bananas,

eating 467 along the way, and arriving with 533 bananas.

If the initial legs are shorter, it works out the same.

If the initial legs are longer, he does more miles at higher banana /

mile rates, ending up with fewer bananas.

6) Fdsfds

When you reverse the direction,uphill becomes downhill, downhill becomes uphill, and flat stays flat.

Suppose when going from x to y we haveu miles of uphill,f miles of flatd miles of downhill(and the reverse coming back).

x to y: u / 56 + f / 63 + d / 72 = 4 hoursy to x: u / 72 + f / 63 + d / 56 = 5 hours

Subtracting top from bottom:u/72 - u/56 + d/56 - d/72 = 11/72 - 1/56 = 1/8*9 - 1/7*8 = -2/7*8*9

-2/7*8*9 u + 2/7*8*9 d = 1-2u + 2 d = 7*8*9 = 5042 (d - u) = 504d - u = 252

And we wantu/56 + f/63 + d/72 = 4u/72 + f/63 + d/56 = 5

Now suppose u = 0.Then we get d = 252x to y takes: 0/56 + f/63 + 252/72 = 3.5 + f/63 hoursf/63 = 0.5 hour (to bring total to 4)


11/11

f = 31.5and total distance = 283.5

For the return journey we have0/72 + 0.5 + 252/56 = 0.5 + 4.5 = 5 hours, as required.

Here is a chart of what happens with a few different values for u:

Umiles

Dmiles

Time U Time D Time FFmiles

Time X -> Y

Totaldistance

0 252 0 3.5 0.5 31.5 4 283.5

5.25257.25

0.09375

3.57292

0.33333

21 4 283.5

10.5 262.50.18750

3.64583

0.16667

10.5 4 283.5

15.75

267.75

0.28125

3.71875

0 0 4 283.5

For the return trip, the times workout as:

Time U Time D Time FTime Y -> X

0 4.5 0.5 5

0.07292

4.59375

0.33333

5

0.14583

4.68750

0.16667

5

0.21875

4.78125

0 5

15.75 is the maximum for U, since if it goes higher, F becomes negative.

In every case, u + f + d = 283.5 miles, and that is the answer to the problem.

Objektif: 1) Kerja berkumpulan saling bantu membantu

2) Meningkatkan kesedaran tentang kepentingan matematik

3) Mengaplikasikan pengetahuan matematik untuk menyelesaikan masalah harian

4) Memupuk kesedaran bahawa proses penyelesaian masalah adalah lebih penting daripada

hasil penyelesaian.

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