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Unit 33 · Selesaikan Masalah
Unit 33

Selesaikan Masalah

Setiap masalah boleh diselesaikan dengan empat langkah:
① Fahami soalan ② Fikir cara ③ Selesaikan ④ Semak

Selesaikan Masalah
解决问题 = selesaikan masalah 应用题可以用四个步骤来做：一、理解题目（Fahami soalan）：找出已知的资料和要求的东西；二、想方法（Fikir cara）：决定用加、减、乘、除还是画图；三、计算（Selesaikan）；四、检查（Semak）：用相反的运算或估计来验证答案。 例子：“有 50 本，送出 18 本，剩几本？”→ 已知 50 和 18，求剩下 → 用减法 → 32 → 32 + 18 = 50。 注意：每一题都要写出最后的答句。
problem solving = selesaikan masalah Word problems can be solved in four steps. 1 Understand the question (Fahami soalan): find the given facts and what is asked. 2 Plan (Fikir cara): choose add, subtract, multiply, divide or a diagram. 3 Solve (Selesaikan). 4 Check (Semak): use the opposite operation or an estimate. Example: 50 books, 18 given away, how many left? Given 50 and 18, find what is left, so subtract: 32; check 32 + 18 = 50. Watch out: always finish with an answer sentence.
Tambah dan bundar

Jumlah dan bundarkan

Buku di tiga perpustakaan: A 24 816, B 31 507, C 28 940. Jumlahkan buku di perpustakaan yang paling banyak dan paling sedikit. Bundarkan kepada ratus terdekat.
Fahami soalan
• paling banyak: B
• paling sedikit: A
• bundar ratus terdekat
Fikir cara
31 507 + 24 816
Selesaikan
= 56 323
→ 56 300
Semak
31 500 + 24 800 = 56 300 ✔
加法和取整 = tambah dan bundar 先比较找出最多的 B 和最少的 A，再相加：31 507 + 24 816 = 56 323。题目要取整到百，看十位是 2，比 5 小，百位不变，得 56 300。可以用 31 500 + 24 800 = 56 300 估计检查。 例子：18 250 + 12 610 = 30 860，取整到百是 30 900。 注意：先确定题目问的是哪两个数，再看要取整到哪一位。
add and round = tambah dan bundar First compare to find the most (B) and the fewest (A), then add: 31 507 + 24 816 = 56 323. The question asks for the nearest hundred; the tens digit is 2, less than 5, so the hundreds stay and the answer is 56 300. Check with 31 500 + 24 800 = 56 300. Example: 18 250 + 12 610 = 30 860, which is 30 900 to the nearest hundred. Watch out: first decide which two numbers the question is about, then which place to round to.
Pola nombor genap

Pilih kad genap, kemudian cari pola

Kad: 5 216, 5 239, 5 244, 5 201, 5 230, 5 257. Pilih kad nombor genap sahaja dan susun menaik. Jika pola diteruskan, apakah nombor keempat?
Fahami soalan
• pilih genap (sa: 0, 2, 4, 6, 8)
• susun menaik
• nombor keempat
Fikir cara
Genap: 5 216, 5 230, 5 244 → +14
Selesaikan
5 244 + 14
= 5 258
Semak
5 258 − 5 244 = 14 ✔
先挑偶数再找模式 = pilih nombor genap, kemudian cari pola 先看每张卡的个位，个位是 0、2、4、6、8 的才是偶数：5 216、5 244、5 230。把这三张按升序排好：5 216、5 230、5 244，相邻两数相差 14。第四个数就是 5 244 + 14 = 5 258。 例子：卡片 310、325、330、351、350，偶数排好是 310、330、350，每次加 20，第四个是 370。 注意：奇数卡不能混进来，不然就找不到规律。
pick the even numbers, then find the pattern = pilih nombor genap, kemudian cari pola Look at the ones digit of each card; only 0, 2, 4, 6 or 8 make an even number: 5 216, 5 244 and 5 230. Arrange them in ascending order: 5 216, 5 230, 5 244; neighbours differ by 14. The fourth number is 5 244 + 14 = 5 258. Example: from cards 310, 325, 330, 351, 350 the even ones are 310, 330, 350, going up by 20, so the fourth is 370. Watch out: keep the odd cards out, or the pattern cannot be seen.
Lebih daripada

Model bar

Januari, 18 460 tiket bas dijual. Februari, 2 750 lebih daripada Januari. Mac, 1 890 lebih daripada Februari. Berapakah tiket yang dijual pada bulan Mac?
Fahami soalan
• Feb = Jan + 2 750
• Mac = Feb + 1 890
Fikir cara
Mac = ?
18 460
2 750
1 890
Selesaikan
18 460 + 2 750 + 1 890
= 23 100
Semak
23 100 − 1 890 − 2 750 = 18 460 ✔
“多……”要加 = lebih daripada bermaksud tambah 二月比一月多 2 750，三月又比二月多 1 890，所以要连加：18 460 + 2 750 + 1 890 = 23 100。画条形图可以看出每个月都在前一个月的基础上加长。 例子：第一周 500 张，第二周比第一周多 100 张，第三周比第二周多 50 张 → 第三周 650 张。 注意：用减法倒回去检查。
more than means add = lebih daripada bermaksud tambah February is 2 750 more than January, and March is 1 890 more than February, so add them all: 18 460 + 2 750 + 1 890 = 23 100. A bar model shows each month growing from the one before. Example: week 1 sells 500, week 2 sells 100 more than week 1 and week 3 sells 50 more than week 2, so week 3 sells 650. Watch out: check by subtracting back.
Tolak berturut

Model bar

Tiga buah lori membawa 32 000 kg beras. Lori P membawa 12 850 kg dan lori Q membawa 9 475 kg. Berapakah kg beras yang dibawa oleh lori R?
Fahami soalan
• jumlah 32 000 kg
• lori P dan Q diketahui
• cari lori R
Fikir cara
32 000
12 850
9 475
?
P · Q · R
Selesaikan
32 000 − 12 850 − 9 475
= 9 675
Semak
12 850 + 9 475 + 9 675 = 32 000 ✔
Anggaran: 32 000 − 13 000 − 9 000 = 10 000 ✔
连减 = tolak berturut-turut 三辆罗厘一共载 32 000 公斤，P 和 Q 已知，R 是剩下的部分，所以用总数连减两个已知部分：32 000 − 12 850 − 9 475 = 9 675。检查有两种方法：三部分加起来要等于 32 000；或者用估计，32 000 − 13 000 − 9 000 = 10 000，和答案接近。 例子：总数 1 000，两部分 300 和 450，第三部分是 250。 注意：条形图里的“？”就是要求的部分。
subtract twice = tolak berturut-turut The three lorries carry 32 000 kg in all; P and Q are known and R is the rest, so subtract both known parts: 32 000 − 12 850 − 9 475 = 9 675. Check in two ways: the three parts must add to 32 000, or estimate: 32 000 − 13 000 − 9 000 = 10 000, which is close. Example: a total of 1 000 with parts 300 and 450 leaves 250. Watch out: the ? in the bar model is the part we need.
Tolak dan tambah

Tulis maklumat penting

Sebuah gudang ada 8 400 kotak jus. 6 925 kotak dihantar ke kedai. Kemudian gudang menerima 1 350 kotak baharu. Berapakah kotak jus dalam gudang sekarang?
Fahami soalan
• ada 8 400
• hantar keluar 6 925 (−)
• terima 1 350 (+)
Fikir cara
8 400 − 6 925 + 1 350
Selesaikan
1 475 + 1 350
= 2 825
Semak
8 400 + 1 350 = 9 750; 9 750 − 6 925 = 2 825 ✔
加减混合 = tambah dan tolak 先写重要资料：送出去是减少，收进来是增加。按事情发生的次序从左到右算：8 400 − 6 925 = 1 475，再 + 1 350 = 2 825。可以换个次序检查：先加后减，答案相同。 例子：仓库有 500 箱，运走 200 箱，又运来 100 箱 → 400 箱。 注意：每个符号都跟着它后面的数走。
add and subtract = tambah dan tolak Write the key facts: sending out makes it smaller and receiving makes it bigger. Work in the order things happen, left to right: 8 400 − 6 925 = 1 475, then + 1 350 = 2 825. Check in another order: add first, then subtract; the answer is the same. Example: a warehouse has 500 cartons, sends out 200 and receives 100, so it has 400. Watch out: each sign moves with the number after it.
Darab

Bilangan yang sama dalam setiap kumpulan

Sebuah sekolah ada 18 kelas. Setiap kelas ada 36 murid. Berapakah jumlah murid?
Fahami soalan
• 18 kelas
• setiap kelas 36
Fikir cara
18 × 36
Selesaikan
36 × 8 = 288
36 × 10 = 360
= 648
Semak
20 × 36 = 720, tolak 2 × 36 = 72 → 648 ✔
乘法应用 = masalah darab 每一班人数一样，所以用乘法：18 × 36。可以拆成 36 × 8 = 288 和 36 × 10 = 360，相加得 648。检查：20 × 36 = 720，多算的 2 班是 72，720 − 72 = 648。 例子：12 行、每行 25 张椅子 → 300 张。 注意：“每……都有”提示要用乘法。
a multiplication problem = masalah darab Every class has the same number, so multiply: 18 × 36. Split it into 36 × 8 = 288 and 36 × 10 = 360, which add to 648. Check: 20 × 36 = 720, and the 2 extra classes are 72, so 720 − 72 = 648. Example: 12 rows of 25 chairs make 300 chairs. Watch out: setiap usually tells us to multiply.
Kali ganda

12 kali bilangan

Buku A dicetak 2 150 naskhah. Buku B dicetak 12 kali bilangan buku A. Berapakah naskhah buku B?
Fahami soalan
• A = 2 150
• B = 12 kali A
Fikir cara
2 150 × 12
Selesaikan
2 150 × 2 = 4 300
2 150 × 10 = 21 500
= 25 800
Semak
25 800 ÷ 12 = 2 150 ✔
“几倍”用乘法 = kali ganda bermaksud darab B 是 A 的 12 倍，就是 2 150 × 12。拆成 2 150 × 2 = 4 300 和 2 150 × 10 = 21 500，相加得 25 800。用除法检查：25 800 ÷ 12 = 2 150。 例子：A 有 300，B 是 A 的 5 倍 → 1 500。 注意：“12 倍”不是加 12。
times as many means multiply = kali ganda bermaksud darab B is 12 times A, which is 2 150 × 12. Split it into 2 150 × 2 = 4 300 and 2 150 × 10 = 21 500, which add to 25 800. Check with division: 25 800 ÷ 12 = 2 150. Example: A is 300 and B is 5 times A, so B is 1 500. Watch out: 12 times is not the same as adding 12.
Kaedah kekisi

Kali ganda dengan kekisi

2 150 × 12

2150120204010205100000008520

2 150 × 12 = 25 800

格子乘法解应用题 = kaedah kekisi dalam masalah “12 倍”要算 2 150 × 12，也可以用格子乘法：上面写 2 150，右边写 1 和 2，每一格写两个数字的积，再沿斜线相加，读出 25 800，和拆分法的答案一样。 例子：“B 是 A 的 13 倍，A 是 120” → 120 × 13 = 1 560，也可以画格子来算。 注意：数字 0 那一列的格子都是 00，但仍然要画出来，位置才不会错。
lattice method in a word problem = kaedah kekisi dalam masalah 12 times means 2 150 × 12, which can also be worked with the lattice method: write 2 150 on top and 1 and 2 down the side, put the product of the two digits in each box, then add along the diagonals to read 25 800, the same as the splitting method. Example: B is 13 times A and A is 120, so 120 × 13 = 1 560, which can also be drawn as a lattice. Watch out: the column under 0 has 00 in every box, but it must still be drawn to keep the places right.
Tulis ayat matematik

Buku A dicetak 2 150 naskhah. Buku B dicetak 12 kali bilangan buku A. Ayat matematik manakah mencari jumlah naskhah buku A dan buku B?

写出算式 = tulis ayat matematik 先读清楚问题到底问什么：是其中一种的数量、两种相差多少，还是两种合起来的总数。再把题目里每一种东西的数量用数字或算式表示出来，最后按照问题选用加、减、乘或除，把它们连成一条算式。选好以后，按题意读一遍，看看这条算式算出来的是不是题目要的东西。 例子：绿球 45 个，黄球比绿球少 12 个，求黄球 → 45 − 12 = 33 个。 注意：先圈出问题里的关键词，例如 jumlah、beza、setiap、kali，再决定用哪一种运算。
write a number sentence = tulis ayat matematik First read exactly what the question asks: the amount of one kind, how many more one has than the other, or the total of both. Write the amount of each thing as a number or an expression, then join them with add, subtract, multiply or divide to match the question. Afterwards read it back against the story to check that it finds what is asked. Example: 45 green balls and 12 fewer yellow balls than green, so yellow is 45 − 12 = 33. Watch out: circle the key words in the question, such as jumlah, beza, setiap and kali, before choosing the operation.
Bahagi dengan baki

Tandakan maklumat penting

Sebuah pusat kitar semula mengumpul 25 000 tin. Setiap beg besar memuatkan 48 tin. Berapakah beg yang penuh? Berapakah tin yang berbaki?
Fahami soalan
• 25 000 tin
• 48 tin sebeg
• beg penuh & baki
Fikir cara
25 000 ÷ 48
Selesaikan
= 520 baki 40
520 beg, baki 40 tin
Semak
520 × 48 + 40 = 25 000 ✔
有余数的应用题 = masalah bahagi dengan baki 把 25 000 个罐子装进袋子，每袋 48 个，用除法：25 000 ÷ 48 = 520 余 40。商 520 是装满的袋数，余数 40 是剩下、装不满一袋的罐子。验算：520 × 48 + 40 = 25 000。 例子：100 颗糖每袋 8 颗 → 12 袋余 4 颗。 注意：答句里要分别写出商和余数的意思。
a division problem with a remainder = masalah bahagi dengan baki Put 25 000 cans into bags of 48: 25 000 ÷ 48 = 520 remainder 40. The quotient 520 is the number of full bags and the remainder 40 is the cans left over. Check: 520 × 48 + 40 = 25 000. Example: 100 sweets in bags of 8 make 12 bags with 4 left. Watch out: in the answer sentence, say what the quotient and the remainder each mean.
Darab, kemudian bahagi

Diberi dan dicari

96 kotak, setiap kotak ada 24 tin susu. Semua tin diagihkan sama banyak ke dalam 8 buah lori. Berapakah tin dalam setiap lori?
Fahami soalan
Diberi: 96 kotak × 24 tin, 8 lori
Dicari: tin setiap lori
Fikir cara
96 × 24 ÷ 8
Selesaikan
96 ÷ 8 = 12
12 × 24 = 288
Semak
288 × 8 = 2 304 = 96 × 24 ✔
先乘后除 = darab, kemudian bahagi 先求一共有多少罐：96 × 24；再平均分给 8 辆罗厘：÷ 8。可以先约简：96 ÷ 8 = 12，再 12 × 24 = 288。检查：288 × 8 = 2 304，刚好是 96 × 24。 例子：10 盒 × 6 个 ÷ 5 人 = 12 个。 注意：先写“已知”和“要求”，再列算式。
multiply, then divide = darab, kemudian bahagi First the total number of tins, 96 × 24, then share among 8 lorries: ÷ 8. Simplify first: 96 ÷ 8 = 12, then 12 × 24 = 288. Check: 288 × 8 = 2 304, which is 96 × 24. Example: 10 boxes × 6 ÷ 5 people = 12 each. Watch out: write what is given and what is asked before writing the number sentence.
Bahagi, kemudian darab

Cari untuk satu dahulu

5 basikalRM8 4501 basikalRM1 69012 basikalRM20 280RM20 280
先求一份 = cari satu unit dahulu 5 辆脚车一共 RM8 450，先求 1 辆多少钱：8 450 ÷ 5 = 1 690；再乘 12 辆：1 690 × 12 = 20 280，所以是 RM20 280。条形图把 RM8 450 分成 5 等份，每份就是一辆的价钱。 例子：4 本同样的书 RM60，1 本 RM15，7 本 RM105。 注意：不能直接拿总价乘 12。
find one unit first = cari satu unit dahulu 5 bicycles cost RM8 450 altogether, so one costs 8 450 ÷ 5 = 1 690 ringgit. Then 12 bicycles cost 1 690 × 12 = 20 280, which is RM20 280. The bar model splits RM8 450 into 5 equal parts, one for each bicycle. Example: 4 identical books cost RM60, so 1 costs RM15 and 7 cost RM105. Watch out: do not just multiply the total price by 12.
Bincang

Bolehkah ÷ 5 × 12 ditulis ÷ 60?

8 450 ÷ 5 × 12 dan 8 450 ÷ 60

✔ ÷ 5 × 12

8 450 ÷ 5 = 1 690
1 690 × 12 = 20 280

✘ ÷ 60

8 450 ÷ 60
= 140 baki 50

20 280 ≠ 140 baki 50

除后再乘不等于除以积 = ÷ 5 × 12 bukan ÷ 60 有人以为“÷ 5 × 12”可以合成“÷ 60”。其实除以 5 让数变小，乘 12 又让它变大；而除以 60 只会让数变小。两边算一算：8 450 ÷ 5 × 12 = 20 280，8 450 ÷ 60 只有 140 余 50，差很多。 例子：600 ÷ 3 × 10 = 2 000，但 600 ÷ 30 = 20，并不相等。 注意：只有“连续两次除”才能合成一次除，例如 ÷ 5 ÷ 12 = ÷ 60。
divide then multiply is not dividing by the product = ÷ 5 × 12 bukan ÷ 60 Some pupils think ÷ 5 × 12 can be joined into ÷ 60. But dividing by 5 makes the number smaller and multiplying by 12 makes it bigger again, while dividing by 60 only makes it smaller. Work both out: 8 450 ÷ 5 × 12 = 20 280, but 8 450 ÷ 60 is only 140 remainder 50. Example: 600 ÷ 3 × 10 = 2 000, but 600 ÷ 30 = 20, so they are not equal. Watch out: only two divisions in a row can be joined into one, for example ÷ 5 ÷ 12 = ÷ 60.
Anu dalam jadual

Susun maklumat dalam jadual

Jualan tiga jenis kuih berjumlah 30 000 biji. Kuih lapis 12 480, kuih talam ?, kuih bahulu 9 365. Berapakah kuih talam?
Fahami soalan
• jumlah 30 000
• dua jenis diketahui
Fikir cara
12 480 + ? + 9 365 = 30 000
Selesaikan
12 480 + 9 365 = 21 845
30 000 − 21 845 = 8 155
Semak
12 480 + 8 155 + 9 365 = 30 000 ✔
表格里的未知数 = anu dalam jadual 总数 30 000 已知，另外两种已知，要求的一种放在中间。先把已知的两种加起来：21 845，再用总数减：30 000 − 21 845 = 8 155。 例子：总数 500，两项 120 和 200，第三项 180。 注意：未知数在中间也一样做法。
an unknown in a table = anu dalam jadual The total 30 000 and two of the kinds are known; the unknown is in the middle. Add the two known kinds: 21 845, then subtract from the total: 30 000 − 21 845 = 8 155. Example: a total of 500 with parts 120 and 200 leaves 180 for the third. Watch out: it works the same way even when the unknown is in the middle.
Anu dalam tambah dan tolak

Ayat matematik dengan anu

Lim ada 45 guli. Adiknya ada k guli. Jumlah guli mereka 83 biji. Cari nilai k.
Fahami soalan
Diberi: 45 + k = 83
Dicari: nilai k
Fikir cara
83
45
k
Selesaikan
k = 83 − 45
k = 38
Semak
45 + 38 = 83 ✔
未知数应用题 = masalah anu 把故事写成算式：45 + k = 83。k 是加数，所以用减法求：k = 83 − 45 = 38。把 38 代回去：45 + 38 = 83，对了。 例子：m − 37 = 18（有 m 个，用掉 37 个，剩 18 个）→ m = 55。 注意：答案要写成“k 的值是 38”。
a problem with an unknown = masalah anu Write the story as a number sentence: 45 + k = 83. k is being added, so subtract to find it: k = 83 − 45 = 38. Put 38 back in: 45 + 38 = 83, which is right. Example: m − 37 = 18 (there were m, 37 were used and 18 are left), so m = 55. Watch out: write the answer as: the value of k is 38.
Cuba

Tangki ada 20 150 ℓ air. 8 795 ℓ digunakan. Berapakah air yang tinggal?

剩下多少 = berapa yang tinggal 水被用掉，就是减少，所以用减法：原来的水量 − 用掉的水量。五位数减四位数时，数位要对齐，从个位减起，不够减就向左边借 1。最后把剩下的和用掉的相加，看看是否回到原来的水量。 例子：有 5 000 升，用掉 1 200 升，剩 3 800 升。 注意：看到用掉、送出、卖出，一般要减；借位以后，左边那一位要记得少 1。
how much is left = berapa yang tinggal Water that is used is taken away, so subtract: water at first − water used. When taking a four-digit number from a five-digit number, line up the places, start at the ones and regroup from the left when needed. Then add what is left and what was used to check that you get the starting amount. Example: 5 000 litres with 1 200 litres used leaves 3 800 litres. Watch out: used, given away and sold usually mean subtract; after regrouping, remember the place on the left is one less.
Cuba

60 000 biji butang dimasukkan ke dalam kotak. Setiap kotak 35 biji. Berapakah kotak penuh?

整盒的数量 = bilangan kotak penuh 把纽扣装盒，每盒一样多，用除法：总数 ÷ 每盒的数量。商是装满的盒数；余数是剩下、装不满一盒的纽扣。题目问“装满的盒子”，所以只看商，不能把装不满的那一盒也算进去。 例子：50 支笔每盒 12 支 → 4 盒满，余 2 支。 注意：答案要配合情境想一想：装不满的不算满盒。
number of full boxes = bilangan kotak penuh Buttons go into boxes of the same size, so divide: total ÷ buttons in a box. The quotient is the number of full boxes and the remainder is the buttons that cannot fill a box. The question asks for full boxes, so use only the quotient and do not count the box that is not full. Example: 50 pens in boxes of 12 make 4 full boxes with 2 left. Watch out: think about the situation: a box that is not full is not a full box.
Genap dan ganjil

Kad: 5 216, 5 239, 5 244, 5 201, 5 230, 5 257. Berapakah beza nombor genap terbesar dengan nombor ganjil terkecil?

最大偶数与最小奇数的差 = beza genap terbesar dan ganjil terkecil 分三步：一、用个位把卡片分成偶数和奇数两组；二、在偶数组里找最大的，在奇数组里找最小的；三、用大的减小的求差。 例子：卡片 812、809、826、817，最大偶数 826，最小奇数 809，差是 17。 注意：不要拿全部卡片里最大和最小的数来减，要先分组再找。
difference between the largest even and smallest odd = beza genap terbesar dan ganjil terkecil Work in three steps: 1 use the ones digit to sort the cards into even and odd; 2 find the largest even card and the smallest odd card; 3 subtract the smaller from the larger. Example: from cards 812, 809, 826, 817 the largest even is 826 and the smallest odd is 809, a difference of 17. Watch out: do not subtract the largest and smallest of all the cards; sort them into groups first.
Cari nombor asal

Bilangan peserta sebuah larian ialah 46 000 apabila dibundarkan kepada ribu terdekat. Pilih semua nombor yang mungkin.

取整反推 = cari nombor asal daripada pembundaran 取整到千以后变成某个整千的数，原来的数一定在一个范围里：从“少 500”开始，到“多 499”为止。先写出这个范围的最小值和最大值，再逐个检查每个数是不是在范围里面。 例子：取整到千是 8 000 的数，范围是 7 500 到 8 499，例如 7 500、7 986、8 499 都可以。 注意：刚好在范围边上的数要特别小心：比范围多一点点，就会取整成下一个千。
work back from rounding = cari nombor asal daripada pembundaran A number that rounds to a whole thousand lies in a range: from 500 less up to 499 more. Write the smallest and largest numbers of that range first, then check whether each number is inside it. Example: numbers that round to 8 000 to the nearest thousand go from 7 500 to 8 499, so 7 500, 7 986 and 8 499 all work. Watch out: take care with numbers at the edge of the range; just one more rounds up to the next thousand.
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