Matematik T4

Matematik Tingkatan 4 Bab 9 129 (a) Seorang pekerja dipilih secara rawak dari sebuah hospital. A worker is chosen randomly from a hospital. • D ialah peristiwa pekerja yang dipilih ialah seorang doktor. D is an event where the chosen worker is a doctor. • E ialah peristiwa pekerja yang dipilih perlu membayar cukai pendapatan. E is an event where the chosen worker needs to pay an income tax. • F ialah peristiwa pekerja yang dipilih mempunyai pendapatan sebanyak RM2 000 sebulan. F is an event where the chosen worker earns a salary of RM2 000 a month. (i) D dan/ and E Peristiwa tidak saling eksklusif kerana peristiwa D dan E boleh berlaku bersama. Non-mutually exclusive event because events D and E can occur together. (ii) D dan/ and F Peristiwa saling eksklusif kerana peristiwa D dan F tidak boleh berlaku bersama. Mutually exclusive event because events D and F cannot occur together. (iii) E dan/ and F Peristiwa saling eksklusif kerana peristiwa E dan F tidak boleh berlaku bersama. Mutually exclusive event because events E and F cannot occur together. (b) Sebiji dadu adil dilambung. A fair dice is tossed. • F ialah peristiwa mendapat nombor ganjil. F is an event where an odd number is obtained. • G ialah peristiwa mendapat nombor yang melebihi 4. G is an event where the number obtained is more than 4. • H ialah peristiwa mendapat faktor bagi 4. H is an event where the number obtained is a factor of 4. (i) F dan/ and G Peristiwa tidak saling eksklusif kerana peristiwa F dan G boleh berlaku bersama. Non-mutually exclusive event because events F and G can occur together. (ii) F dan/ and H Peristiwa tidak saling eksklusif kerana peristiwa F dan H boleh berlaku bersama. Non-mutually exclusive event because events F and H can occur together. (iii) G dan/ and H Peristiwa saling eksklusif kerana peristiwa G dan H tidak boleh berlaku bersama. Mutually exclusive event because events G and H cannot occur together. (c) Seorang murid dipilih secara rawak dari sebuah kelas. A student is chosen randomly from a class. • J ialah peristiwa murid yang dipilih gagal ujian Sejarah. J is an event where the student chosen failed in History test. • K ialah peristiwa murid yang dipilih mendapat 80 markah. K is an event where the student chosen gets 80 marks. • L ialah peristiwa murid yang dipilih lemah subjek Sejarah. L is an event where the student chosen is weak in History subject. (i) J dan/ and K Peristiwa saling eksklusif kerana peristiwa J dan K tidak boleh berlaku bersama. Mutually exclusive event because events J and K cannot occur together. (ii) J dan/ and L Peristiwa tidak saling eksklusif kerana peristiwa J dan L boleh berlaku bersama. Non-mutually exclusive event because events J and L can occur together. (iii) K dan/ and L Peristiwa saling eksklusif kerana peristiwa K dan L tidak boleh berlaku bersama. Mutually exclusive event because events K and L cannot occur together. CONTOH

Matematik Tingkatan 4 Bab 9 130 2 Tentu sahkan rumus penambahan kebarangkalian bagi setiap peristiwa bergabung berikut dengan menyenaraikan semua kesudahan yang mungkin. SP: 9.3.2 TP3 Sederhana Verify the addition rule of probability for each of the following combined events by listing down all the possible outcomes. Diberi set semesta S = {x : 1 N x N 30, x ialah integer}, A ialah set yang mengandungi faktor bagi 30, B ialah set yang mengandungi gandaan bagi 3 dan C ialah set yang mengandungi gandaan bagi 4. Given the universal set S = {x : 1 N x N 30, x is an integer}, A is a set which consists of factors of 30, B is a set which consists of multiples of 3 and C is a set which consists of multiples of 4. (i) P(A atau/ or B) A = {1, 2, 3, 5, 6, 10, 15, 30} B = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30} A ∩ B = {3, 6, 15, 30} A ∪ B = {1, 2, 3, 5, 6, 9, 10, 12, 15, 18, 21, 24, 27, 30} P(A ∪ B) = 14 30 = 7 15 P(A) + P(B) – P(A ∩ B) = 8 30 + 10 30 – 4 30 = 14 30 = 7 15 ∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B) (ii) P(A atau/ or C) A = {1, 2, 3, 5, 6, 10, 15, 30} C = {4, 8, 12, 16, 20, 24, 28} A ∩ C = { } A ∪ C = {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 28, 30} P(A ∪ C) = 15 30 = 1 2 P(A) + P(C) = 8 30 + 7 30 = 15 30 = 1 2 ∴ P(A ∪ C) = P(A) + P(C) (a) Tujuh keping kad berlabel dengan huruf “G, E, M, B, I, R, A” dimasukkan ke dalam sebuah kotak. Sekeping kad dipilih secara rawak dari kotak tersebut. Seven cards labelled with letters “G, E, M, B, I, R, A” are put into a box. A card is chosen randomly from the box. • J ialah peristiwa mendapat kad berhuruf konsonan. J is an event of getting a card with consonant. • K ialah peristiwa mendapat kad berhuruf vokal. K is an event of getting a card with vowel. • L ialah peristiwa mendapat kad berhuruf R. L is an event of getting a card with letter R. (i) P(J atau/ or K) J = {G, M, B, R}, K = {E, I, A} J ∩ K = { } J ∪ K = {G, M, B, R, E, I, A} P(J ∪ K) = 7 7 = 1 ∴ P(J ∪ K) = P(J) + P(K) (ii) P(J atau/ or L) J = {G, M, B, R}, L = {R} J ∩ L = {R} J ∪ L = {G, M, B, R} P(J ∪ L) = 4 7 P(J) + P(L) – P(J ∩ L) = 4 7 + 1 7 – 1 7 = 4 7 ∴ P(J ∪ L) = P(J) + P(L) – P(J ∩ L) (iii) P(K atau/ or L) K = {E, I, A}, L = {R} K ∩ L = { } K ∪ L = {E, I, A, R} P(K ∪ L) = 4 7 ∴ P(K ∪ L) = P(K) + P(L) P(J) + P(K) = 4 7 + 3 7 = 7 7 = 1 P(K) + P(L) = 3 7 + 1 7 = 4 7 Contoh Tip Bestari Hukum penambahan kebarangkalian Addition rule of probability Peristiwa tidak saling eksklusif: Non-mutually exclusive event P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Peristiwa saling eksklusif: Mutually exclusive event P(A ∪ B) = P(A) + P(B) CONTOH

Matematik Tingkatan 4 Bab 9 131 (b) Sebiji dadu adil dan sekeping syiling adil dilambung secara serentak. A fair dice and a fair coin are tossed simultaneously. • F ialah peristiwa mendapat angka dan nombor genap. F is an event of getting a head and an even number. • G ialah peristiwa mendapat gambar dan nombor ganjil. G is an event of getting a tail and an odd number. • H ialah peristiwa mendapat angka dan gandaan 3. H is an event of getting a head and a multiple of 3. (i) P(F atau/ or G) F = {(Angka, 2), (Angka, 4), (Angka, 6)}/ F = {(Head, 2), (Head, 4), (Head, 6)} G = {(Gambar, 1), (Gambar, 3), (Gambar, 5)}/ G = {(Tail, 1), (Tail, 3), (Tail, 5)} F ∩ G = { } F ∪ G = {(Angka, 2), (Angka, 4), (Angka, 6), (Gambar, 1), (Gambar, 3), (Gambar, 5)} {(Head, 2), (Head, 4), (Head, 6), (Tail, 1), (Tail, 3), (Tail, 5)} P(F ∪ G) = 6 12 = 1 2 P(F) + P(G) = 3 12 + 3 12 = 6 12 = 1 2 ∴ P(F ∪ G) = P(F) + P(G) (ii) P(F atau/ or H) F = {(Angka, 2), (Angka, 4), (Angka, 6)}/ F = {(Head, 2), (Head, 4), (Head, 6)} H = {(Angka, 3), (Angka, 6)}/ H = {(Head, 3), (Head, 6)} F ∩ H = {(Angka, 6)}/ {(Head, 6)} F ∪ H = {(Angka, 2), (Angka, 3), (Angka, 4), (Angka, 6)}/ {(Head, 2), (Head, 3), (Head, 4), (Head, 6)} P(F ∪ H) = 4 12 = 1 3 P(F) + P(H) – P(F ∩ H) = 3 12 + 2 12 – 1 12 = 4 12 = 1 3 ∴ P(F ∪ H) = P(F) + P(H) – P(F ∩ H) (iii) P(G atau/ or H) G = {(Gambar, 1), (Gambar, 3), (Gambar, 5)}/ G = {(Tail, 1), (Tail, 3), (Tail, 5)} H = {(Angka, 3), (Angka, 6)}/ H = {(Head, 3), (Head, 6)} G ∩ H = { } G ∪ H = {(Gambar, 1), (Gambar, 3), (Gambar, 5), (Angka, 3), (Angka, 6)} {(Tail, 1), (Tail, 3), (Tail, 5), (Head, 3), (Head, 6)} P(G ∪ H) = 5 12 P(G) + P(H) = 3 12 + 2 12 = 5 12 ∴ P(G ∪ H) = P(G) + P(H) CONTOH

Matematik Tingkatan 4 Bab 9 132 3 Selesaikan setiap masalah yang berikut. SP: 9.3.3 TP3 TP4 Sederhana Solve each of the following problems. (a) Sebiji dadu adil dilambung sekali. Hitung kebarangkalian bahawa satu nombor ganjil atau satu nombor genap diperoleh. A fair dice is tossed once. Calculate the probability of getting an odd number or an even number. Nombor ganjil/ Odd numbers = {1, 3, 5} Nombor genap/ Even numbers = {2, 4, 6} P(nombor ganjil atau nombor genap): P(an odd number or an even number): P(A ∪ B) = P(A) + P(B) = 3 6 + 3 6 = 6 6 = 1 (b) Sebuah bekas mengandungi 30 biji gula-gula yang berlainan perisa. Lapan biji gula-gula berperisa epal, sembilan biji gula-gula berperisa anggur dan 13 biji gula-gula berperisa oren. Sebiji gula-gula dipilih secara rawak dari bekas tersebut. Hitung kebarangkalian gula-gula yang dipilih berperisa epal atau anggur. A container contains 30 sweets of different flavours. Eight are apple-flavoured sweets, nine are grapeflavoured sweets and 13 are orange-flavoured sweets. A sweet is chosen randomly from the container. Calculate the probability that the sweet chosen is apple-flavoured or grape-flavoured. P(gula-gula berperisa epal atau anggur): P(apple-flavoured sweet or grape-flavoured sweet): P(E ∪ A) = P(E) + P(A) = 8 30 + 9 30 = 17 30 (c) Kebarangkalian Ah Meng akan memenangi pertandingan menyanyi dan melukis masingmasing ialah 1 4 dan 1 3. Kebarangkalian Ah Meng akan memenangi kedua-dua pertandingan ialah 1 5. Hitung kebarangkalian Ah Meng akan memenangi salah satu pertandingan. The probability that Ah Meng will win a singing competition and drawing competition are 1 4 and 1 3 respectively. The probability that he will win in both competitions is 1 5. Calculate the probability that Ah Meng will win in either one competition. P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 1 4 + 1 3 – 1 5 = 15 + 20 – 12 60 = 23 60 (d) Dua kotak, X dan Y masing-masing mengandungi kad-kad berlabel dengan nombor dan huruf seperti yang ditunjukkan di bawah. Sekeping kad dipilih secara rawak masing-masing dari kotak X dan Y. Two boxes, X and Y contains cards labelled with numbers and letters respectively as shown below. A card is chosen randomly from boxes X and Y respectively. 4 5 7 9 M A S I N Kotak X/ Box X Kotak Y/ Box Y Dengan menyenaraikan semua kesudahan yang mungkin, hitung kebarangkalian mendapat nombor perdana dari kotak X atau huruf “N” dari kotak Y. By listing all the possible outcomes, calculate the probability of getting a prime number from box X or letter "N" from box Y. Kad nombor perdana/ Prime number cards = {5, 7} Kad huruf N/ Letter N card = {N} P(kad nombor perdana atau kad huruf N): P(a prime number card or letter N card): P(A ∪ B) = P(A) + P(B) = 2 4 + 1 5 = 10 + 4 20 = 14 20 = 7 10 CONTOH

Matematik Tingkatan 4 Bab 9 133 (e) Di dalam sebuah kelas yang mempunyai 40 orang murid, 22 orang murid ialah murid lelaki. Didapati 10 orang murid lelaki dan tujuh orang murid perempuan di dalam kelas tersebut memakai cermin mata. Hitung kebarangkalian seorang murid lelaki atau seorang murid perempuan yang tidak memakai cermin mata dipilih. In a class of 40 students, 22 of them are boys. It is found that 10 boys and seven girls in the class are wearing spectacles. Calculate the probability that a boy or a girl who does not wear spectacles is chosen. Memakai cermin mata Wearing spectacles Tidak memakai cermin mata Does not wear spectacles Lelaki Boys 10 12 Perempuan Girls 7 11 P(murid lelaki atau murid perempuan yang tidak memakai cermin mata): P(a boy or a girl who does not wear spectacles): P(LS' ∪ PS') = P(LS') + P(PS') = 12 40 + 11 40 = 23 40 (f) Kebarangkalian Muthu dan Mohan akan memenangi hadiah pertama dalam cabutan bertuah masingmasing ialah 2 7 dan 1 6. Hitung kebarangkalian The probability that Muthu and Mohan will win the first prize in a lucky draw are 2 7 and 1 6 respectively. Calculate the probability that (i) Muthu atau Mohan memenangi hadiah pertama, Muthu or Mohan will win the first prize, (ii) tiada yang memenangi hadiah pertama. none of them win the first prize. (i) P(Muthu atau Mohan menang hadiah pertama): P(Muthu or Mohan wins the first prize): P(Mu ∪ Mo) = P(Mu) + P(Mo) = 2 7 + 1 6 = 12 + 7 42 = 19 42 (ii) P(Muthu dan Mohan tidak memenangi hadiah pertama): P(Muthu and Mohan do not win the first prize): 1 – 19 42 = 23 42 (g) Sebuah syarikat menganjurkan perjalanan keluarga ke Pulau Redang. Kebarangkalian Danny dan Firdaus menyertai perjalanan keluarga ini masing-masing ialah 3 5 dan 5 8. Lakar satu gambar rajah pokok untuk menunjukkan semua kesudahan yang mungkin. Kemudian, hitung kebarangkalian hanya seorang daripadanya menyertai perjalanan keluarga ini. A company organises a family trip to Redang Island. The probability that Danny and Firdaus join the event are 3 5 and 5 8 respectively. Sketch a tree diagram to show all the possible outcomes. Then, calculate the probability that only one of them joins this event. Danny Firdaus Kesudahan Outcomes M M (M, M) (M, M�) (M�, M) (M�, M�) M M� M� M� 3 5 2 5 5 8 5 8 3 8 3 8 P(salah seorang menyertai)/ P(one of them joins): P(M , M�) + P(M� , M) = ( 3 5 × 3 8 ) + ( 2 5 × 5 8 ) = 9 + 10 40 = 19 40 M = Menyertai/ Join M� = Tidak menyertai/ Does not join CONTOH

Matematik Tingkatan 4 Bab 9 134 9.4 Aplikasi Kebarangkalian Peristiwa Bergabung Application of Probability of Combined Events Buku Teks m/s 262 – 264 1 Selesaikan setiap masalah yang berikut. SP: 9.4.1 TP4 TP5 KBAT Sukar Solve each of the following problems. (a) Kotak pensel A mengandungi 10 batang pen merah, lapan batang pen biru dan tujuh batang pen hijau. Kotak pensel B mengandungi lapan batang pen merah, sembilan batang pen biru dan lapan batang pen hijau. Sebatang pen akan dipilih secara rawak daripada kotak pensel A dan dimasukkan ke dalam kotak pensel B. Kemudian, sebatang pen dipilih secara rawak daripada kotak pensel B. Hitung kebarangkalian kedua-dua batang pen yang dipilih itu adalah sama warna. There are 10 red pens, eight blue pens and seven green pens in pencil case A. There are eight red pens, nine blue pens and eight green pens in pencil case B. A pen will be chosen randomly from pencil case A and placed into pencil case B. Then, a pen is chosen randomly from pencil case B. Calculate the probability that both chosen pens are of the same colour. Kesudahan Outcomes M M M M B B B (M, M) (M, B) (M, H) (B, M) (B, B) (B, H) (H, M) (H, B) (H, H) H H H H B 10 25 7 25 8 25 9 26 9 26 8 26 8 26 8 26 8 26 9 26 9 26 10 26 M = Merah/ Red B = Biru/ Blue H = Hijau/ Green P(M, M) + P(B, B) + P(H, H) = ( 10 25 × 9 26) + ( 8 25 × 10 26) + ( 7 25 × 9 26) = 90 + 80 + 63 650 = 233 650 (b) Sekeping syiling adil dan sebiji dadu adil dilambung. A fair coin and a fair dice are tossed. (i) Tunjukkan semua kesudahan yang mungkin. Show all the possible outcomes. (ii) Hitung kebarangkalian mendapat ‘ekor’ dan nombor genap. Calculate the probability that ‘tail’ and even numbers will be obtained. (iii) Hitung kebarangkalian mendapat ‘kepala’ dan nombor ‘3’ atau ‘ekor’ dan nombor ‘1’. Calculate the probability of obtaining ‘head’ and number ‘3’ or ‘tail’ and number ‘1’. (i) S = {(Kepala, 1), (Kepala, 2), (Kepala, 3), (Kepala, 4), (Kepala, 5), (Kepala, 6), (Ekor, 1), (Ekor, 2), (Ekor, 3), (Ekor, 4), (Ekor, 5), (Ekor, 6)} S = {(Head, 1), (Head, 2), (Head, 3), (Head, 4), (Head, 5), (Head, 6), (Tail, 1), (Tail, 2), (Tail, 3), (Tail, 4), (Tail, 5), (Tail, 6)} (ii) S = {(Ekor, 2), (Ekor, 4), (Ekor, 6)} S = {(Tail, 2), (Tail, 4), (Tail, 6)} P(ekor dan nombor genap) = 3 12 P(tail and an even number) = 1 4 (iii) S = {(Kepala, 3), (Ekor, 1)} S = {(Head, 3), (Tail, 1)} P[(Kepala, 3) ∪ (Ekor, 1)] P[(Head, 3) ∪ (Tail, 1)] = 1 12 + 1 12 = 2 12 = 1 6 Info Digital 9.4 CONTOH

Matematik Tingkatan 4 Bab 9 135 (c) Kotak A dan kotak B masing-masing mempunyai empat biji bola yang bernombor 1 hingga 4. Betty memilih sebiji bola daripada kotak A dan sebiji bola daripada kotak B secara rawak. Hitung kebarangkalian bahawa hasil tambah kedua-dua nombor tersebut ialah There are four balls labelled with numbers 1 to 4 in boxes A and B respectively. Betty chooses a ball randomly from boxes A and B respectively. Calculate the probability that the sum of the two numbers is (i) nombor ganjil dan melebihi 5, odd number and exceed 5, (ii) nombor ganjil atau melebihi 5. odd number or exceed 5. (i) 1 ✗ ✗ ✗ ✗ ✗ ✗ ✓ ✓ ✗ ✗ ✗ ✓ ✗ ✓ ✓ ✓ 1 3 2 4 2 3 4 A B Petunjuk/ Key: ✗ = Tidak melebihi 5/ Not exceed 5 ✓ = Melebihi 5/ Exceed 5 ✓ = Nombor ganjil dan melebihi 5 Odd numbers and exceed 5 P(nombor ganjil dan melebihi 5) = 2 16 P(an odd number and exceed 5) = 1 8 (ii) 1 ✗ ✗ ✗ ✗ ✗ ✗ ✓ ✓ ✗ ✗ ✗ ✓ ✗ ✓ ✓ ✓ 1 3 2 4 2 3 4 B A Petunjuk/ Key: ✗ = Tidak melebihi 5/ Not exceed 5 ✓ = Melebihi 5/ Exceed 5 =Nombor ganjil / Odd numbers P(nombor ganjil atau melebihi 5) P(an odd number or exceed 5) P(✓ ∪ ) = P(✓) + P( ) – P(✓ ∩ ) = 6 16 + 8 16 – 2 16 = 12 16 = 3 4 (d) Tiro dan Titus akan makan aiskrim atau biskut sebagai minum petang. Kebarangkalian Tiro dan Titus akan makan aiskrim masing-masing ialah 1 6 dan 2 7. Hitung kebarangkalian bahawa Tiro and Titus will eat ice cream or biscuits for high tea. The probability of Tiro and Titus will eat ice cream are 1 6 and 2 7 respectively. Calculate the probability that (i) Tiro akan makan biskut, Tiro will eat biscuits, (ii) Tiro akan makan aiskrim dan Titus akan makan biskut, Tiro will eat ice cream and Titus will eat biscuits, (iii) Tiro dan Titus makan makanan yang sama. Tiro and Titus will eat the same food. Tiro Titus Kesudahan Outcomes A A (A, A) (A, B) (B, A) (B, B) A B B B 1 6 5 6 2 7 2 7 5 7 5 7 A = Aiskrim/ Ice cream B = Biskut/ Biscuits (i) P(Tiro makan biskut)/ P(Tiro eats biscuits): 1 – 1 6 = 5 6 (ii) P(A, B) = 1 6 × 5 7 = 5 42 (iii) P(A, A) + P(B, B) = ( 1 6 × 2 7) + ( 5 6 × 5 7) = 2 + 25 42 = 27 42 = 9 14 CONTOH

Matematik Tingkatan 4 Bab 9 136 (e) Chloe dan Cindy berjaya memasuki pusingan akhir pertandingan catur. Mereka merupakan pemain yang berpengalaman dan pernah bertanding sebanyak 30 kali dalam pelbagai peringkat. Chloe menumpaskan Cindy sebanyak 10 kali dan Cindy menumpaskan Chloe sebanyak 14 kali, manakala permainan yang lain ditamatkan dalam keadaan seri. Jika Chloe dan Cindy berlawan sebanyak tiga pusingan dalam pertandingan ini, hitung kebarangkalian bahawa Chloe and Cindy managed to take part in the final round of a chess tournament. They both are experienced players and competed 30 times in multiple stages. Chloe defeated Cindy 10 times and Cindy defeated Chloe 14 times, while the rest of the games ended in a draw. If Chloe and Cindy were to compete for three rounds in this game, calculate the probability that (i) dua perlawanan adalah seri, two games are a draw, (ii) Chloe dan Cindy menang secara berselang-seli, Chloe and Cindy win alternately, (iii) Cindy menang sekurang-kurangya dua perlawanan. Cindy wins at least two games. P(Chloe) = 10 30 = 1 3 P(Cindy) = 14 30 = 7 15 P(S) = 6 30 = 1 5 (i) P(S, S, Chloe) + P(S, S, Cindy) = ( 1 5 × 1 5 × 1 3 ) + ( 1 5 × 1 5 × 7 15) = 1 75 + 7 375 = 4 125 (ii) P(Chloe, Cindy, Chloe) + P(Cindy, Chloe, Cindy) = ( 1 3 × 7 15 × 1 3 ) + ( 7 15 × 1 3 × 7 15) = 7 135 + 49 675 = 28 225 (iii) P(Cindy, Cindy, Chloe) + P(Cindy, Cindy, Cindy) + P(Cindy, Cindy, S): = ( 7 15 × 7 15 × 1 3 ) + ( 7 15 × 7 15 × 7 15) + ( 7 15 × 7 15 × 1 5 ) = 49 675 + 343 3 375 + 49 1 125 = 49 225 Kertas 1 Jawab semua soalan./ Answer all questions. 1 Austin melambung dua biji dadu yang mempunyai enam permukaan pada masa yang sama. Apakah kebarangkalian kedua-dua dadu itu menunjukkan nombor ganjil? Austin tossed two six-sided dice at the same time. What is the probability that both of the dice show odd numbers? A 1 2 C 1 6 B 1 4 D 1 8 2 Satu kajian mendapati bahawa 83% rakyat Malaysia suka minum teh tarik. Jika dua orang dipilih secara rawak, hitung kebarangkalian mereka berdua suka minum teh tarik. A survey found that 83% of Malaysians like to drink teh tarik. If two people were chosen randomly, calculate the probability that both of them like to drink teh tarik. A 0.59 B 0.68 C 0.69 D 0.96 3 Di dalam sebuah kelas yang terdiri daripada 20 orang murid lelaki dan 15 orang murid perempuan, dua orang murid akan dipilih untuk menyertai pertandingan bercerita. Hitung kebarangkalian dua orang murid lelaki akan dipilih menyertai Praktis Kendiri CONTOH

Matematik Tingkatan 4 Bab 9 137 pertandingan itu. In a class of 20 boys and 15 girls, two students are chosen to participate in a story telling competition. Calculate the probability that two boys will be chosen to participate the competition. A 73 245 C 30 119 B 38 119 D 12 49 4 Terdapat 10 biji epal dan lapan biji oren di dalam sebuah bakul. Sin Ling makan dua biji buah yang terdapat di dalam bakul tersebut. Hitung kebarangkalian Sin Ling makan dua biji buah yang berlainan. There are 10 apples and eight oranges in a basket. Sin Ling eats two fruits from that basket. Calculate the probability that she eats two different types of fruit. A 40 81 C 80 153 B 73 153 D 103 153 5 Diberi kebarangkalian bagi peristiwa P dan Q masing-masing ialah 1 3 dan 2 5. Jika P dan Q adalah saling eksklusif, cari P(P � Q). Given that the probability of events P and Q are 1 3 and 2 5 respectively. If P and Q are mutually exclusive, find P(P � Q). A 1 15 C 4 15 B 2 15 D 11 15 6 Kebarangkalian Joanne makan sebiji pai epal dan sebiji pai pisang masing-masing ialah 0.31 dan 0.45. Jika kebarangkalian Joanne makan kedua-dua pai ialah 0.09, tentukan kebarangkalian dia akan makan mana-mana satu jenis pai? The probability of Joanne eating an apple pie and a banana pie are 0.31 and 0.45 respectively. If the probability of Joanne eating both pies is 0.09, determine the probability that she will eat either type of pie? A 0.05 C 0.67 B 0.23 D 0.85 7 Sebiji dadu dan sekeping syiling dilambung secara serentak. Apakah kebarangkalian dadu tersebut menunjukkan nombor genap dan syiling menunjukkan ekor? A dice and a coin are tossed simultaneously. What is the probability that a dice shows an even number and the coin shows the tail? A 1 2 C 1 4 B 1 3 D 1 6 8 Sebuah kotak mengandungi 10 keping kad merah serta beberapa keping kad biru dan kad hijau. Jika sekeping kad dipilih secara rawak dari kotak itu, kebarangkalian bahawa sekeping kad biru dan sekeping kad hijau dipilih masing-masing ialah 2 5 dan 4 15. Kemudian, lima keping kad biru dimasukkan ke dalam kotak itu dan sekeping kad dipilih secara rawak. Hitung kebarangkalian bahawa sekeping kad merah dipilih. A box contains 10 red cards, some blue cards and green cards. If a card is drawn randomly from the box, the probability that a blue card and a green card drawn are 2 5 and 4 15 respectively. Then, five blue cards are put into the box and a card is drawn randomly. Calculate the probability that a red card will be drawn. A 4 15 C 1 3 B 2 7 D 2 5 K B A T Kertas 2 Jawab semua soalan./ Answer all questions. Bahagian A/ Section A 1 Jadual berikut menunjukkan bilangan bola pingpong dengan nombor yang berlainan di dalam sebuah kotak. The following table shows the number of ping pong balls with different numbers in a box. Nombor pada bola Number on the ball Bilangan bola pingpong Number of ping pong balls Jingga Orange Putih White 2 3 6 4 4 7 6 7 5 (a) Jika dua biji bola pingpong dipilih secara rawak dari kotak itu, hitung kebarangkalian bahawa kedua-dua bola pingpong berwarna jingga. If two ping pong balls are chosen randomly from the box, calculate the probability that both ping pong balls are orange. [2 markah/ marks] CONTOH

Matematik Tingkatan 4 Bab 9 138 (b) Jika dua biji bola pingpong berwarna putih dipilih secara rawak dari kotak itu, hitung kebarangkalian bahawa kedua-dua bola pingpong itu menunjukkan nombor 2. If two white ping pong balls are chosen randomly from the box, calculate the probability that both ping pong balls show number 2. [2 markah/ marks] (b) P(P, P) = 6 32 × 5 31 = 30 992 = 15 496 (a) P(J, J) = 14 32 × 13 31 = 182 992 = 91 496 2 Di dalam sebuah kelas yang mempunyai 40 orang murid, 18 orang daripada mereka ialah murid perempuan. Dua orang murid telah dipilih oleh guru kelas untuk menjadi ketua kelas. Hitung kebarangkalian dua orang murid lelaki telah dipilih dan tentukan sama ada peristiwa ini peristiwa bersandar atau peristiwa tak bersandar. In a class of 40 students, 18 of them are girls. Two students are chosen to become class monitors by the class teacher. Calculate the probability that two boys are chosen and determine whether it is a dependent event or an independent event. [3 markah/ marks] P(L, L) = 22 40 × 21 39 = 462 1 560 = 77 260 ∴ Peristiwa bersandar/ Dependent event P = Perempuan/ Girl L = Lelaki/ Boy Kesudahan Outcomes P P (P, P) (P, L) (L, P) (L, L) P L L L 18 40 17 39 18 39 21 39 22 39 22 40 3 Kebarangkalian Mary membeli oren dan tembikai di sebuah pasar masing-masing ialah 3 7 dan 2 3. The probability that Mary buys an orange and a watermelon at a market are 3 7 and 2 3 respectively. (a) Lengkapkan gambar rajah Venn di bawah untuk mewakili kebarangkalian Mary membeli oren dan tembikai di pasar tersebut. Complete the Venn diagram below to represent the probability that Mary buys an orange and a watermelon at the market. [3 markah/ marks] ξ O T 2 7 8 21 4 21 1 7 P(O ∩ T) = 3 7 × 2 3 = 2 7 P(O sahaja/ only) = 3 7 – 2 7 = 1 7 P(T sahaja/ only) = 2 3 – 2 7 = 14 – 6 21 = 8 21 (b) Hitung kebarangkalian Mary tidak membeli oren dan tembikai. Calculate the probability that Mary does not buy the orange and watermelon. [2 markah/ marks] P(Mary tidak membeli oren dan tembikai): P(Mary does not buy orange and watermelon): 1 – ( 1 7 + 8 21 + 2 7 ) = 1 – 17 21 = 4 21 CONTOH

Matematik Tingkatan 4 Bab 9 139 Bahagian B/ Section B 4 Terdapat 20 biji gula-gula berperisa laici dan 15 biji gula-gula berperisa oren di dalam sebuah balang. Mei Mei dibenarkan untuk mengambil tiga biji gula-gula secara rawak. Hitung kebarangkalian dia akan mendapat There are 20 lychee-flavoured candies and 15 orangeflavoured candies in a jar. Mei Mei is allowed to take three candies randomly. Calculate the probability that she will get (a) tiga biji gula-gula berperisa laici, three lychee-flavoured candies, [3 markah/ marks] (b) tiada sebiji gula-gula berperisa oren, none of the orange-flavoured candies, [3 markah/ marks] (c) sebiji gula-gula berperisa laici dan dua gula-gula berperisa oren. one lychee-flavoured candy and two orange-flavoured candies. [3 markah/ marks] (a) P(L, L, L) = 20 35 × 19 34 × 18 33 = 6 840 39 270 = 228 1 309 (b) P(L, L, L) = 20 35 × 19 34 × 18 33 = 6 840 39 270 = 228 1 309 (c) P(L, O, O) + P(O, L, O) + P(O, O, L) = ( 20 35 × 15 34 × 14 33) + ( 15 35 × 20 34 × 14 33) + ( 15 35 × 14 34 × 20 33) = 4 200 + 4 200 + 4 200 39 270 = 12 600 39 270 = 60 187 5 Diana dan Ezzaty menduduki suatu ujian Matematik. Kebarangkalian bahawa mereka lulus ujian tersebut masing-masing ialah 2 5 dan 2 3. Diana and Ezzaty sat for a Mathematics test. The probability that both of them passed in the test are 2 5 and 2 3 respectively. (a) Lakar satu gambar rajah pokok untuk menunjukkan semua kesudahan yang mungkin. Sketch a tree diagram to show all the possible outcomes. [3 markah/ marks] (b) Hitung kebarangkalian bahawa salah seorang gagal ujian tersebut. Calculate the probability that one of them failed in the test. [2 markah/ marks] (c) Hitung kebarangkalian bahawa sekurangkurangnya seorang lulus ujian tersebut. Calculate the probability that at least one of them passed in the test. [3 markah/ marks] (a) Diana Ezzaty Kesudahan Outcomes L L (L, L) (L, G) (G, L) (G, G) L G G G 2 5 2 3 2 3 1 3 1 3 3 5 L = Lulus/ Passed G = Gagal/ Failed (b) P(salah seorang gagal ujian): P(one of them failed in the test): P(L, G) + P(G, L) = ( 2 5 × 1 3 ) + ( 3 5 × 2 3 ) = 2 15 + 6 15 = 8 15 (c) P(sekurang-kurangnya seorang lulus ujian) P(at least one of them passed in the test) = P(L, G) + P(G, L) + P(L, L) = ( 2 5 × 1 3 ) + ( 3 5 × 2 3 ) + ( 2 5 × 2 3 ) = 2 + 6 + 4 15 = 12 15 = 4 5 CONTOH

140 Praktis Intensif 10.1 Perancangan dan Pengurusan Kewangan Financial Planning and Management Buku Teks m/s 272 – 289 1 Tentukan sama ada matlamat kewangan dalam situasi berikut memenuhi pendekatan SMART atau tidak. SP: 10.1.1 Determine whether the financial goal in the following situations fullfill the SMART approach. TP1 Mudah (a) Puan Siti ingin menyambut hari lahir anak perempuannya setengah tahun kemudian. Beliau memerlukan RM1 200 untuk sambutan tersebut. Puan Siti bercadang menyimpan RM200 daripada pendapatannya sebanyak RM2 400 setiap bulan untuk mencapai matlamat kewangan beliau. Mrs Siti wants to celebrate her daughter’s birthday half year later. She needs RM1 200 for this celebration. Mrs Siti plans to save RM200 from her income of RM2 400 every month to achieve her financial goal. • Khusus: Puan Siti ingin menyambut hari lahir anak perempuannya dalam tempoh setengah tahun kemudian. Specific: Mrs Siti wants to celebrate her daughter’s birthday half year later. • Boleh diukur: Puan Siti bercadang menyimpan RM200 sebulan daripada pendapatan bulanannya. Measurable: Mrs Siti plans to save RM200 every month from her monthly income. • Boleh dicapai: RM200 merupakan 8.33% daripada jumlah pendapatan bulanan Puan Siti, maka dia berkemampuan untuk menyimpannya. Attainable: RM200 is 8.33% of Mrs Siti’s monthly income, so she is able to save. • Realistik: Puan Siti mampu menyimpan RM200 setiap bulan dan dapat mencapai RM1 200 dalam setengah tahun. Realistic: Mrs Siti affords to save RM200 every month and able to achieve RM1 200 in half year. • Tempoh masa: Setengah tahun Time-bound: Half year Oleh itu, Puan Siti mengamalkan konsep SMART untuk mencapai matlamat tersebut. Therefore, Mrs Siti applied the SMART concept to achieve the goal. (b) Arjun merancang untuk bercuti ke Sabah dalam tempoh enam bulan. Percutian tersebut mengambil perbelanjaan sebanyak RM2 000. Beliau bercadang menyimpan RM250 sebulan daripada gaji bulanannya sebanyak RM3 500 untuk mencapai matlamat kewangan beliau. Arjun plans a trip to Sabah within six months. The trip expends RM2 000. He plans to save RM250 every month from his monthly salary of RM3 500 to achieve his financial goal. • Khusus: Arjun merancang untuk bercuti ke Sabah dalam tempoh enam bulan. Specific: Arjun plans a trip to Sabah within six months. • Boleh diukur: Arjun bercadang menyimpan RM250 sebulan daripada gaji bulanannya. Measurable: Arjun plans to save RM250 every month from his monthly salary. • Boleh dicapai: Arjun perlu lapan bulan untuk mencapai matlamat kewangannya sebanyak RM2 000. Simpanan selama enam bulan hanya mencapai 75% matlamat kewangannya. Attainable: Arjun needs eight months to achieve his financial goal of RM2 000. Six months' saving only reaches 75% of his financial goal. • Realistik: Matlamat kewangan Arjun untuk bercuti ke Sabah dalam tempoh enam bulan adalah tidak realistik dengan simpanan sebanyak RM250 sebulan. Beliau perlu menyimpan sebanyak RM334 sebulan untuk mencapai matlamatnya. Realistic: Arjun's financial goal to go for a trip to Sabah within six months is not realistic with a saving of RM250 per month. He needs to save RM334 per month to achieve his goal. • Tempoh masa: Enam bulan Time-bound: Six months Oleh itu, Arjun tidak mengamalkan konsep SMART untuk mencapai matlamat tersebut. Therefore, Arjun did not apply the SMART concept to achieve the goal. Encik Farhan ingin membeli sebuah telefon bimbit berharga RM2 700 dalam tempoh 18 bulan. Beliau perlu menyimpan RM150 daripada gaji bulanannya sebanyak RM3 000 untuk mencapai matlamat kewangan beliau. Mr Farhan wants to buy a smart phone at a price of RM2 700 within 18 months. He needs to save RM150 from his monthly salary of RM3 000 to achieve his financial goal. • Khusus: Encik Farhan ingin membeli sebuah telefon bimbit dalam tempoh 18 bulan. Specific: Mr Farhan wants to buy a smart phone within 18 months. • Boleh diukur: Encik Farhan perlu menyimpan RM150 sebulan daripada gaji bulanannya. Measurable: Mr Farhan needs to save RM150 every month from his monthly salary. • Boleh dicapai: Encik Farhan mampu menyimpan kerana RM150 merupakan 5% daripada jumlah gaji bulanannya. Attainable: Mr Farhan is able to save because RM150 is 5% of his monthly salary. • Realistik: Matlamat Encik Farhan untuk membeli telefon bimbit dalam tempoh 18 bulan adalah bersifat realistik. Beliau dapat mencapai matlamat tersebut dengan menyimpan RM150 sebulan. Realistic: Mr Farhan’s goal to buy a smart phone within 18 months is realistic. He can achieve his goal by saving RM150 every month. • Tempoh masa: 18 bulan Time-bound: 18 months Oleh itu, Encik Farhan mengamalkan konsep SMART untuk mencapai matlamat tersebut. Therefore, Mr Farhan applied the SMART concept to achieve the goal. Contoh Matematik Pengguna: Pengurusan Kewangan Consumer Mathematics: Financial Management 10 Bab Info Digital 10.1 CONTOH

Matematik Tingkatan 4 Bab 10 141 2 Hitung aliran tunai bulanan dalam setiap situasi berikut. Kemudian, jelaskan jawapan anda. SP: 10.1.1 TP2 Mudah Calculate the monthly cash flow in each of the following situations. Then, explain your answer. (a) Rachel ialah seorang guru sekolah swasta. Pendapatan bulanannya ialah RM3 500. Rachel juga mengajar kelas tuisyen pada masa lapang dan memperoleh pendapatan sampingan sebanyak RM500 sebulan. Rachel mempunyai perbelanjaan tetap sebanyak RM1 200 sebulan dan perbelanjaan tidak tetap sebanyak RM900 sebulan. Rachel is a private school teacher. Her monthly income is RM3 500. She also gives tuition classes during her free time and earns extra RM500 every month. Rachel has monthly fixed expenses of RM1 200 and monthly variable expenses of RM900. RM3 500 + RM500 – RM1 200 – RM900 = RM1 900 (Aliran tunai positif/ Positive cash flow) Aliran tunai positif sebanyak RM1 900 adalah baik kerana Rachel boleh menyimpan wang tersebut di dalam bank dan menikmati faedah. Positive cash flow of RM1 900 is good as Rachel can save the money in a bank and receive interest. (b) Encik Muthu menerima pendapatan aktif sebanyak RM1 900 dan pendapatan pasif sebanyak RM500 sebulan. Encik Muthu mempunyai perbelanjaan tetap sebanyak RM1 700 sebulan dan perbelanjaan tidak tetap sebanyak RM850 sebulan. Mr Muthu earns an active income of RM1 900 and a passive income of RM500 in a month. He has monthly fixed expenses of RM1 700 and monthly variable expenses of RM850. RM1 900 + RM500 – RM1 700 – RM850 = –RM150 (Aliran tunai negatif/ Negative cash flow) Aliran tunai negatif sebanyak RM150 akan membebankan Encik Muthu dan mungkin menyebabkannya menggunakan kad kredit bagi mengatasi masalah kewangan tersebut. Negative cash flow of RM150 will burden Mr Muthu and he will probably use the credit card to facilitate his financial problems. (c) Yuan Xin mempunyai pendapatan aktif sebanyak RM2 800 dan pendapatan pasif sebanyak RM750 sebulan. Yuan Xin membelanjakan sebanyak RM2 250 dan RM1 050 sebulan masing-masing untuk perbelanjaan tetap dan perbelanjaan tidak tetap. Yuan Xin earns an active income of RM2 800 and a passive income of RM750 in a month. She spends RM2 250 and RM1 050 every month on fixed expenses and variable expenses respectively. RM2 800 + RM750 – RM2 250 – RM1 050 = RM250 (Aliran tunai positif/ Positive cash flow) Aliran tunai positif sebanyak RM250 adalah baik kerana Yuan Xin boleh menggunakan lebihan pendapatan tersebut untuk melabur atau menghadapi situasi kecemasan jika perlu. Positive cash flow of RM250 is good as Yuan Xin can use the surplus of income for investment or spend it for emergency situations if necessary. (d) Encik Lim telah bersara. Beliau menerima wang pencen sebanyak RM1 750 sebulan dan tiada pendapatan pasif. Encik Lim mempunyai perbelanjaan tetap sebanyak RM1 100 sebulan dan perbelanjaan tidak tetap sebanyak RM850 sebulan. Mr Lim has retired. He receives a monthly pension of RM1 170 and has no passive income. Mr Lim has monthly fixed expenses of RM1 100 and monthly on variable expenses of RM850. RM1 750 – RM1 100 – RM850 = –RM200 (Aliran tunai negatif/ Negative cash flow) Aliran tunai negatif sebanyak RM200 akan membebankan Encik Lim kerana dia mungkin menggunakan kad kredit untuk mengatasi masalah kewangan tersebut. Negative cash flow of RM200 will burden Mr Lim as he will probably use the credit card to facilitate his financial problems. Dania ialah seorang arkitek dengan pendapatan bulanan sebanyak RM4 300. Dia memperoleh pendapatan pasif sebanyak RM650 sebulan melalui sewaan rumah. Dania mempunyai perbelanjaan tetap sebanyak RM3 200 sebulan dan perbelanjaan tidak tetap sebanyak RM800 sebulan. Dania is an architect with a monthly salary of RM4 300. She earns a passive income of RM650 every month from house rental. Dania spends RM3 200 on fixed expenses and RM800 on variable expenses in a month. RM4 300 + RM650 – RM3 200 – RM800 = RM950 (Aliran tunai positif/ Positive cash flow) Aliran tunai positif sebanyak RM950 adalah baik kerana Dania boleh menyimpan wang tersebut di dalam bank dan menikmati faedah. Positive cash flow of RM950 is good as Dania can save the money in a bank and receive interest. Contoh CONTOH

Matematik Tingkatan 4 Bab 10 142 3 Selesaikan setiap masalah yang berikut. SP: 10.1.2 TP3 TP4 TP5 Sederhana Sukar Solve each of the following problems. (a) Encik Philip dan isterinya ingin membeli sebuah rumah enam bulan selepas mereka berkahwin. Jumlah pendapatan mereka ialah RM8 500 dan jumlah perbelanjaan mereka ialah RM3 000 sebulan. Mereka bercadang untuk membeli sebuah rumah teres dua tingkat yang berharga RM900 000 dengan wang pendahuluan sebanyak RM100 000. Adakah Encik Philip dan isterinya berjaya membeli rumah teres tersebut? Berikan justifikasi anda. Mr Philip and his wife intend to buy a house in six months after getting married. Their total monthly income is RM8 500 and their total monthly expenses is RM3 000. They plan to buy a double-storey terrace house which costs RM900 000 with a down payment of RM100 000. Do Mr Philip and his wife succeed in buying the terrace house? Justify your answer. Pendapatan lebihan/ Surplus of income: RM8 500 – RM3 000 = RM5 500 Jumlah simpanan dalam tempoh enam bulan: Total savings in six months: RM5 500 × 6 = RM33 000 Jumlah bulan yang diperlukan untuk membayar wang pendahuluan: Total number of months needed to make the down payment: RM100 000 RM5 500 = 18.18 bulan/ months ∴ Encik Philip dan isterinya tidak mungkin dapat membeli rumah teres tersebut dalam jangka masa enam bulan kerana jumlah wang yang berjaya dikumpulkan adalah kurang daripada wang pendahuluan. Mereka memerlukan 18.18 bulan untuk mencapai matlamat kewangan mereka. It is impossible for Mr Philip and his wife to buy the terrace house within six months as the total money saved is less than the down payment. They need to save for 18.18 months to achieve their financial goal. (b) Encik Ghafor dan isterinya ingin membeli sebuah rumah dalam masa tujuh tahun selepas mereka berkahwin. Jumlah pendapatan bulanan mereka ialah RM6 000 dan jumlah perbelanjaan bulanan mereka ialah RM4 650. Mereka bercadang untuk membeli sebuah kondominium yang berharga RM630 000 dengan wang pendahuluan sebanyak RM63 000. Mr Ghafor and his wife intend to buy a house in seven years after getting married. Their total monthly income is RM6 000 and their total monthly expenses is RM4 650. They plan to buy a condominium which is priced at RM630 000 with a down payment of RM63 000. (i) Berapakah simpanan bulanan yang harus disimpan oleh Encik Ghafor dan isterinya bagi mencapai matlamat kewangan tersebut? How much monthly saving needs to be saved by Mr Ghafor and his wife to achieve their financial goal? (ii) Pada pendapat anda, adakah Encik Ghafor seorang yang bijaksana dari segi pembelian rumah yang berharga RM630 000? Berikan justifikasi anda. In your opinion, is Mr Ghafor a smart person in buying the house which costs RM630 000? Justify your answer. (i) Simpanan bulanan yang diperlukan oleh Encik Ghafor: Monthly saving needed by Mr Ghafor: RM63 000 7 × 12 = RM63 000 84 = RM750 (ii) Tidak. Hal ini kerana walaupun Encik Ghafor mampu membayar wang pendahuluan sebanyak RM63 000, namun bayaran ansuran bulanan rumah akan membebankan beliau sekiranya jumlah perbelanjaannya adalah tinggi. No. It is because even though Mr Ghafor affords to pay the down payment of RM63 000, but his monthly housing loan instalments will burden him CONTOH if his total expenses are high.

Matematik Tingkatan 4 Bab 10 143 (c) Puan Mary bercadang membeli sebuah kereta baharu yang berharga RM49 000 dalam masa setahun. Beliau ingin membayar wang pendahuluan sebanyak RM6 000 daripada simpanan beliau. Setelah meneliti kedudukan kewangan, Puan Mary memilih tempoh selama sembilan tahun untuk melangsaikan baki pinjaman kereta dari bank Z. Jadual berikut menunjukkan pelan kewangan keluarga Puan Mary. Mrs Mary plans to buy a new car worth RM49 000 within a year. She wishes to make a down payment of RM6 000 from her savings. After evaluating her financial status, Mrs Mary chooses nine years to settle the remaining of her car loan from bank Z. The following table shows the financial plan of Mrs Mary’s family. Pendapatan dan perbelanjaan Income and expenditure (RM) Gaji suami Puan Mary Salary of Mrs Mary’s husband Gaji Puan Mary Mrs Mary’s salary 3 000 2 500 Pendapatan pasif Passive income 0 Jumlah pendapatan bulanan Total monthly income 5 500 Tolak simpanan tetap bulanan Minus fixed monthly savings Tolak simpanan untuk dana kecemasan Minus emergency fund savings 550 100 Baki/ Balance 4 850 Tolak perbelanjaan tetap bulanan Minus monthly fixed expenses Ansuran kereta suami Husband’s car instalment Pinjaman rumah Housing loan Insurans Life insurance 600 900 300 Jumlah perbelanjaan tetap bulanan Total monthly fixed expenses 1 800 Pendapatan dan perbelanjaan Income and expenditure (RM) Tolak perbelanjaan tidak tetap bulanan Minus monthly variable expenses Taska Nursery Keperluan anak-anak Children’s needs Utiliti Utilities Petrol dan tol Petrol and toll Perbelanjaan rumah Household expenses Pemberian kepada ibu bapa Allowances for parents 350 500 400 400 800 400 Jumlah perbelanjaan tidak tetap bulanan Total monthly variable expenses 2 850 Pendapatan lebihan Surplus of income 200 (i) Berdasarkan perancangan kewangan di atas, bolehkah Puan Mary mencapai matlamat kewangannya? Based on the financial plan above, can Mrs Mary achieve her financial goal? (ii) Adakah Puan Mary seorang yang bijak mengurus kewangan? Berikan justifikasi anda. Is Mrs Mary wise in financial management? Justify your answer. (iii) Adakah Puan Mary menggunakan pendekatan SMART untuk mencapai matlamat kewangan beliau? Berikan justifikasi anda. Does Mrs Mary use the SMART approach to achieve her financial goals? Justify your answer. (i) Boleh. Simpanan setahun = RM550 × 12 = RM6 600. Puan Mary masih mempunyai lebihan RM600 selepas wang pendahuluan dibayar. Beliau juga mempunyai pendapatan lebihan sebanyak RM200 sebulan. Can. Annual savings = RM550 × 12 = RM6 600. Mrs Mary still has a surplus of RM600 after paying the down payment. She also has a surplus of income of RM200 every month. (ii) Ya. Hal ini kerana Puan Mary hanya berbelanja untuk keperluan asas, mempunyai simpanan bulanan dan insurans serta tidak memiliki hutang kad kredit. Yes. It is because Mrs Mary only spends her income on basic needs, has monthly savings and life insurance, and does not have any credit card debt. (iii) Ya/ Yes S Membeli sebuah kereta baharu yang berharga RM49 000. To buy a new car worth RM49 000. M Membayar wang pendahuluan sebanyakRM6 000. To make a down payment of RM6 000. A Wang pendahuluan sebanyak RM6 000 boleh dicapai dalam tempoh setahun. The down payment of RM6 000 can be achieved within a year. R Simpanan bulanan sebanyak RM550 adalah 10% daripada jumlah pendapatan bulanan, iaitu RM5 500. The monthly savings of RM550 is 10% of the total monthly income, which is RM5 500. T Dalam tempoh setahun. Within a year. CONTOH

Matematik Tingkatan 4 Bab 10 144 (d) Jadual berikut menunjukkan pelan kewangan Encik Zafran. The following table shows the financial plan of Mr Zafran. Pendapatan bersih Net income (RM) Gaji bersih Encik Zafran Mr Zafran’s net salary 5 800 Bajet perbelanjaan Expenses budget Pinjaman rumah termasuk penyelenggaraan Housing loan including maintenance 700 Ansuran kereta Car instalment 700 Bil utiliti Utility bills 250 Bayaran tol Toll payments 120 Bayaran petrol Petrol payments 200 Perbelanjaan dapur Household expenses 900 Elaun kepada ibu bapa Allowances for parents 300 Perbelanjaan isteri Wife expenses 500 Taska Nursery 400 Keperluan anak-anak Children’s needs 600 Simpanan Savings 100 Insurans Life insurance 150 Bil telefon Mobile bills 150 Melancong Travel 500 Jumlah perbelanjaan Total expenses 5 570 Baki pendapatan Income balance 230 (i) Adakah Encik Zafran mengamalkan perbelanjaan secara berhemat? Does Mr Zafran spend his money wisely? (ii) Encik Zafran ingin membeli harta tanah yang bernilai RM300 000 dalam tempoh lapan tahun. Adakah beliau dapat mencapai matlamat kewangan tersebut dengan tabiat perbelanjaan di atas? Mr Zafran wishes to buy a property worth RM300 000 within eight years. Can he achieve the financial goal with the spending behaviour above? (i) Tidak kerana jumlah simpanan bulanan beliau ialah RM230, iaitu kurang 10% daripada pendapatan bulanan beliau. No because his monthly savings is RM230, which is less than 10% of his monthly income. (ii) RM230 × 12 × 8 = RM22 080 (<RM300 000) Beliau tidak boleh mencapai matlamat pelaburan RM300 000 dengan simpanan bulanan RM230. He cannot achieve his investment goal of RM300 000 with monthly savings of RM230. CONTOH

Matematik Tingkatan 4 Bab 10 145 (e) Jessica ialah seorang akauntan dengan pendapatan bulanan sebanyak RM5 000 yang ingin membeli sebuah kereta baharu berharga RM90 000. Beliau berkemampuan membayar 20% daripada harga kereta itu secara tunai. Bakinya akan dijelaskan melalui pinjaman bank. Jessica is an accountant with monthly income of RM5 000, who wishes to buy a new car worth RM90 000. She is able to pay 20% of the car price by cash. The remaining will be settled through a bank loan. (i) Hitung jumlah pinjaman yang dimohon oleh Jessica daripada bank. Calculate the amount of loan applied by Jessica from the bank. (ii) Jika kadar faedah bagi pinjaman kereta ialah 3% setahun, hitung jumlah pinjaman untuk tempoh bayaran balik selama If the interest rate for the car loan is 3% per annum, calculate the loan amount that she has to pay back for (a) lima tahun, (b) tujuh tahun, (c) sembilan tahun. five years, seven years, nine years. (iii) Hitung bayaran ansuran bulanan (betul kepada ringgit yang terdekat) jika tempoh pinjaman ialah Calculate the payable monthly instalment (correct to the nearest ringgit) if the loan period is (a) lima tahun, (b) tujuh tahun, (c) sembilan tahun. five years, seven years, nine years. (iv) Huraikan pembelian kereta baharu oleh Jessica dengan bayaran ansuran bulanan dalam tempoh masa tujuh tahun dengan menggunakan konsep SMART. Describe the purchase of new car by Jessica with monthly instalment payable within seven years using the SMART concept. (v) Adakah pelan kewangan itu untuk jangka masa pendek atau jangka masa panjang? Is this considered as a short-term financial plan or a long-term financial plan? (i) Deposit kereta/ Car’s deposit: 20 100 × RM90 000 = RM18 000 Jumlah pinjaman dari bank: RM90 000 – RM18 000 = RM72 000 Total loan from the bank: (ii) (a) Jumlah pinjaman/ Total loan amount: RM72 000 + (RM72 000 × 0.03 × 5) = RM72 000 + RM10 800 = RM82 800 (b) Jumlah pinjaman/ Total loan amount: RM72 000 + (RM72 000 × 0.03 × 7) = RM72 000 + RM15 120 = RM87 120 (c) Jumlah pinjaman/ Total loan amount: RM72 000 + (RM72 000 × 0.03 × 9) = RM72 000 + RM19 440 = RM91 440 (iii) (a) Bayaran ansuran bulanan/ Payable monthly instalment: RM82 800 5 × 12 = RM1 380 (b) Bayaran ansuran bulanan/ Payable monthly instalment: RM87 120 7 × 12 = RM1 037 (c) Bayaran ansuran bulanan/ Payable monthly instalment: RM91 440 9 × 12 = RM847 (iv) S Membeli sebuah kereta dengan pinjaman bank sebanyak RM72 000. To buy a car with a loan of RM72 000 from the bank. M Jessica perlu membayar RM1 037 sebulan selama tujuh tahun. Jessica needs to pay RM1 037 every month for seven years. A RM1 037 merupakan 20.74% daripada jumlah pendapatan bulanan Jessica. Dia mampu membayarnya. RM1037 is 20.74% from Jessica’s monthly income. She affords to pay. R Jika perbelanjaan bulanan Jessica kurang daripada 79.26% sebulan, dia boleh membayarnya. If Jessica’s monthly expenses are less than 79.26%, she is able to pay. T Menjelaskan pinjaman dalam masa tujuh tahun. To settle the loan within seven years. (v) Pelan kewangan jangka panjang/ Long-term financial plan CONTOH

Matematik Tingkatan 4 Bab 10 146 (f) Encik Wafiy bekerja sebagai pegawai penyelidik di Agensi Nuklear Malaysia dengan gaji bersih bulanan sebanyak RM3 600. Beliau melakukan kerja sampingan dengan menjadi ejen jualan produk Y. Komisen bulanan yang diperoleh dianggarkan sebanyak RM650. Kutipan sewa rumah keduanya ialah RM700 sebulan. Jadual berikut menunjukkan anggaran perbelanjaan bulanan beliau. Mr Wafiy works as a researcher in Malaysian Nuclear Agency with a monthly net salary of RM3 600. He works part-time as a sales agent of product Y. His monthly commission earned is estimated to be RM650. The rent of his second house is RM700 per month. The following table shows his estimated monthly expenses. Perbelanjaan bulanan Monthly expenses RM Ansuran pinjaman rumah 1 Housing loan instalment 1 750 Ansuran pinjaman rumah 2 Housing loan instalment 2 550 Ansuran kereta Car instalment 560 Perbelanjaan rumah Household expenses 800 Perbelanjaan isteri Wife expenses 300 Utiliti Utility 150 Keperluan anak Children’s needs 400 Perbelanjaan tol dan petrol Toll and petrol expenses 200 Bil telefon Telephone bill 100 Makan luar Eating out 300 Percutian Holiday 400 Takaful suami dan isteri Life insurance for husband and wife 200 Encik Wafiy memperuntukkan 10% daripada gajinya untuk simpanan tetap bulanan bagi mencapai matlamat kewangannya. Mr Wafiy allocates 10% of his salary as fixed monthly savings to achieve his financial goal. (i) Anda dikehendaki menyediakan satu pelan kewangan peribadi bulanan bagi Encik Wafiy. You are required to prepare a monthly personal financial plan for Mr Wafiy. (ii) Kemukakan komen tentang lebihan atau kurangan yang akan dialami oleh Encik Wafiy berdasarkan pelan kewangan ini. Comment on the surplus or deficit that will be experienced by Mr Wafiy based on this financial plan. CONTOH

Matematik Tingkatan 4 Bab 10 147 (i) Pendapatan dan perbelanjaan Income and expenditure RM Gaji bersih/ Net salary Komisen/ Commission 3 600 650 Pendapatan pasif: Sewa rumah Passive income: House rental 700 Jumlah pendapatan bulanan Total monthly income 4 950 Tolak simpanan tetap bulanan Minus fixed monthly savings 360 Baki pendapatan Income balance 4 590 Tolak perbelanjaan tetap bulanan: Minus monthly fixed expenses Ansuran pinjaman rumah (1)/ Housing loan instalment (1) Ansuran pinjaman rumah (2)/ Housing loan instalment (2) Ansuran kereta/ Car instalment Takaful suami dan isteri/ Life insurance for husband and wife 750 550 560 200 Jumlah perbelanjaan tetap bulanan Total monthly fixed expenses 2 060 Tolak perbelanjaan tidak tetap bulanan: Minus monthly variable expenses: Perbelanjaan rumah/ Household expenses Perbelanjaan isteri/ Wife expenses Utiliti/ Utility Keperluan anak/ Children’s needs Perbelanjaan tol dan petrol/ Toll and petrol expenses Bil telefon/ Telephone bill Makan luar/ Eating out Percutian/ Holiday 800 300 150 400 200 100 300 400 Jumlah perbelanjaan tidak tetap bulanan Total monthly variable expenses 2 650 Lebihan/ Kurangan Surplus/ Deficit –120 (ii) Terdapat kurangan dalam pelan kewangan Encik Wafiy, iaitu aliran tunai negatif apabila jumlah perbelanjaan melebihi jumlah pendapatan. Bagi mencapai matlamat kewangannya, Encik Wafiy perlu mengurangkan perbelanjaan tidak tetap seperti percutian dan makan di luar di samping meningkatkan penjualan produk Y. There is a deficit in Mr Wafiy's financial plan, which is negative cash flow as the total expenses is greater than the total income. To achieve his financial goal, Mr Wafiy needs to reduce his variable expenses such as holiday and eating out while increasing the sales of product Y. CONTOH

Matematik Tingkatan 4 Bab 10 148 Kertas 1 Jawab semua soalan./ Answer all questions. 1 Antara berikut, yang manakah merupakan tempoh bagi matlamat kewangan jangka pendek? Which of the following is the duration of a short-term financial goal? A 5 tahun C 2 tahun 5 years 2 years B 3 tahun D Kurang daripada 1 tahun 3 years Less than 1 year 2 Antara berikut, yang manakah bukan tempoh bagi matlamat kewangan jangka panjang? Which of the following is not the duration of a long-term financial goal? A 4 tahun C 9 tahun 4 years 9 years B 7 tahun D 10 tahun 7 years 10 years 3 Antara berikut, yang manakah merupakan keperluan asas yang perlu diutamakan dalam merancang kewangan? Which of the following are the basic needs that should be considered in financial planning? I Makanan III Pakaian Food Clothes II Tempat tinggal IV Telefon pintar Shelter Smart phones A I, II dan IV C I, III dan IV I, II and IV I, III and IV B II, III dan IV D I, II dan III II, III and IV I, II and III 4 Farish ialah seorang guru dengan pendapatan bulanan sebanyak RM3 500. Dia memperoleh pendapatan pasif sebanyak RM550 sebulan melalui sewaan rumah. Farish mempunyai perbelanjaan tetap sebanyak RM2 200 sebulan dan perbelanjaan tidak tetap sebanyak RM700 sebulan. Hitung aliran tunai bulanan Farish. Farish is a teacher with a monthly income of RM3 500. He earns a passive income of RM550 every month with house rental. He spends RM2 200 on fixed expenses and RM700 on variable expenses in a month. Calculate Farish's monthly cash flow. A RM50 C RM1 150 B RM600 D RM1 850 5 Antara berikut, yang manakah bukan pendapatan aktif? Which of the following is not an active income? A Gaji C Elaun Salary Allowance B Komisen D Kad kredit Commission Credit card 6 Antara aspek berikut, yang manakah tidak harus diambil kira untuk membina pelan kewangan jangka panjang? Which of the following aspects should not be included to create a long-term financial plan? A Kadar inflasi C Keinginan Inflation rate Willingness B Kadar faedah D Kesihatan diri Interest rate Personal health 7 Encik Aditya dan isterinya ingin membeli sebuah rumah dalam masa enam tahun selepas mereka berkahwin. Jumlah pendapatan mereka ialah RM4 500 dan jumlah perbelanjaan mereka ialah RM3 450 sebulan. Mereka bercadang membeli sebuah pangsapuri yang berharga RM302 400 dengan wang pendahuluan sebanyak RM30 240. Berapakah simpanan bulanan yang harus disimpan oleh Encik Aditya dan isterinya bagi mencapai matlamat tersebut? Mr Aditya and his wife intend to buy a house within six years after getting married. Their total monthly income is RM4 500 and their total expenses are RM3 450. They plan to buy an apartment which costs RM302 400 with a down payment of RM30 240. How much monthly savings needed to be saved by Mr Aditya and his wife to achieve their goal? A RM420 C RM1 050 B RM630 D RM3 780 8 Daniel memperoleh gaji bersih sebanyak RM3 680 sebulan. Dia menerima komisen sebanyak RM350 dan memperoleh pendapatan pasif sebanyak RM500. Sekiranya Daniel mengamalkan pelan kewangan yang cekap, berapakah simpanan tetap bulanannya? Daniel earns a net salary of RM3 680 a month. He receives a commission of RM350 and a passive income of RM500. If Daniel practices a smart financial plan, how much is his fixed saving? A RM368 C RM416 B RM403 D RM453 Kertas 2 Jawab semua soalan./ Answer all questions. Bahagian A/ Section A 1 David dan rakan sekelasnya ingin bercuti tiga bulan kemudian dan mereka memerlukan RM600 seorang. Dengan menggunakan pendekatan SMART, jelaskan bagaimana David perlu merancang percutian ini. David and his classmates plan to go for vacation three months later and they each need RM600. Using SMART approach, explain how David should plan for this trip. [5 markah/ marks] Praktis Kendiri CONTOH

Matematik Tingkatan 4 Bab 10 149 • Khusus: Menyimpan RM600 dalam masa tiga bulan. Specific: Saves RM600 in three months. • Boleh diukur: Menyimpan RM200 dalam masa sebulan. Measurable: Saves RM200 a month. • Boleh dicapai: David mampu mengumpul RM600 dengan menyimpan sebanyak RM200 sebulan. Attainable: David is able to collect RM600 by saving RM200 every month. • Realistik: David mampu menyimpan RM200 sebulan jika perbezaan pendapatan dengan perbelanjaan lebih daripada RM200 sebulan. Realistic: David is able to save RM200 a month if the difference between his income and expenses is more than RM200 a month. • Tempoh masa: Dalam tempoh tiga bulan. Time-bound: Within three months. Bahagian B/ Section B 2 Edwin ialah seorang doktor dengan pendapatan bulanan sebanyak RM6 000. Beliau ingin membeli sebuah kereta utiliti sukan yang berharga RM130 000. Dia berkemampuan membayar 10% daripada harga kereta itu secara tunai. Bakinya akan dijelaskan melalui pinjaman bank. Edwin is a doctor with a monthly income of RM6 000. He wants to buy a sport utility vehicle worth RM130 000. He is able to pay 10% of the car price by cash. The remaining will be settled through a bank loan. (a) Hitung jumlah pinjaman bank Edwin. Calculate Edwin's total bank loan. [2 markah/ marks] (b) Jika kadar faedah pinjaman kereta ialah 3.3% setahun, hitung jumlah pinjaman bagi tempoh bayaran balik selama If the interest rate of the car loan is 3.3% per annum, calculate the amount of loan for the repayment period of (i) lima tahun, (ii) tujuh tahun, (iii) sembilan tahun. five years, seven years, nine years. [3 markah/ marks] (c) Hitung bayaran ansuran bulanan (betul kepada ringgit yang terdekat) jika tempoh pinjaman ialah Calculate the payable monthly instalment (correct to the nearest ringgit) if the loan period is (i) lima tahun, (ii) tujuh tahun, (iii) sembilan tahun. five years, seven years, nine years. [3 markah/ marks] (d) Dengan menggunakan konsep SMART, huraikan pembelian kereta utiliti sukan oleh Edwin dengan bayaran ansuran bulanan dalam tempoh masa tujuh tahun. By using the SMART concept, describe the purchase of the sport utility vehicle by Edwin with monthly instalment payable within seven years. [2 markah/ marks] (a)Deposit kereta/ Car’s deposit = 10 100 × RM130 000 = RM13 000 Jumlah pinjaman dari bank = RM130 000 – RM13 000 Total loan amount from the bank = RM117 000 (b) (i) Jumlah pinjaman/ Total loan amount: RM117 000 + (RM117 000 × 0.033 × 5) = RM136 305 (ii) Jumlah pinjaman/ Total loan amount: RM117 000 + (RM117 000 × 0.033 × 7) = RM144 027 (iii) Jumlah pinjaman/ Total loan amount: RM117 000 + (RM117 000 × 0.033 × 9) = RM151 749 (c) (i) Bayaran ansuran bulanan/ Payable monthly instalment: RM136 305 5 × 12 = RM2 272 (ii) Bayaran ansuran bulanan/ Payable monthly instalment: RM144 027 7 × 12 = RM1 715 (iii) Bayaran ansuran bulanan/ Payable monthly instalment: RM151 749 9 × 12 = RM1 405 (d) S – Membeli sebuah kereta utiliti sukan dengan pinjaman bank sebanyak RM117 000. To buy a sport utility vehicle with a loan of RM117 000 from the bank. M– Edwin perlu membayar RM1 715 sebulan selama tujuh tahun. Edwin needs to pay RM1 715 every month for seven years. A– RM1 715 merupakan 28.58% daripada jumlah pendapatan bulanan Edwin. Dia mampu membayarnya. RM1 715 is 28.58% from Edwin’s monthly income. He affords to pay. R– Jika perbelanjaan bulanan Edwin kurang daripada 71.42% sebulan, dia boleh membayarnya. If Edwin’s monthly expenses are less than 71.42%, he is able to pay. T– Menjelaskan pinjaman dalam masa tujuh tahun. To settle the loan within seven years. CONTOH

150 Tulis jawapan anda pada ruang jawapan yang disediakan. Anda boleh menggunakan kalkulator saintifik yang tidak boleh diprogramkan. Write your answers in the answer space provided. You may use a non-programmable scientific calculator. 1 Rajah 1 menunjukkan sebiji dadu biasa dan sebiji dadu dengan 12 permukaan. Diagram 1 shows a normal dice and a 12-sided dice. Dadu P/ Dice P Dadu Q/ Dice Q Rajah 1/ Diagram 1 (a) Chew Wai melambung kedua-dua dadu secara serentak manakala Lee Fang hanya melambung dadu Q sekiranya dia mendapat nombor 6 daripada lambungan dadu P. Chew Wai tossed both dice simultaneously whereas Lee Fang only tosses dice Q if she gets a number 6 from tossing dice P. (i) Nyatakan jenis peristiwa yang berlaku daripada lambungan dadu Chew Wai dan lambungan dadu Lee Fang. State the type of events occurred from Chew Wai’s dice-tossing and Lee Fang’s dice-tossing. [2 markah/ marks] (ii) Cari kebarangkalian bagi Chew Wai dan Lee Fang masing-masing mendapat nombor 6 pada dadu Q. Find the probability for Chew Wai and Lee Fang respectively of getting number 6 on dice Q. [4 markah/ marks] (b) Diberi isi padu dadu P ialah 3.375 cm3 dan nisbah jumlah luas permukaan dadu P kepada jumlah luas permukaan dadu Q ialah 2 : 3. Hitung luas, dalam cm2 , bagi satu permukaan dadu Q. Given the volume of dice P is 3.375 cm3 and the ratio of the total surface area of dice P to the total surface area of dice Q is 2 : 3. Calculate the area, in cm2 , of one surface of dice Q. [3 markah/ marks] (c) Chew Wai dan Lee Fang ditugaskan untuk membeli kedua-dua jenis dadu bagi suatu pertandingan mainan berpapan. Chew Wai telah membayar RM7.60 untuk tujuh biji dadu P dan tiga biji dadu Q manakala Lee Fang telah membeli sembilan biji dadu P dan lima biji dadu Q dengan RM10.80. Hitung harga bagi dadu P dan dadu Q. Chew Wai and Lee Fang are assigned to buy both types of dice for a board game competition. Chew Wai had paid RM7.60 for seven dice P and three dice Q whereas Lee Fang had bought nine dice P and five dice Q with RM10.80. Calculate the prices of dice P and dice Q. [4 markah/ marks] (d) Bilangan maksimum peserta bagi pertandingan mainan berpapan itu ialah 35 orang. Jika bilangan peserta tidak mencapai 35 orang, Chew Wai dan Lee Fang akan turut menyertai pertandingan itu. Selepas tarikh akhir pendaftaran, terdapat 30 orang telah mendaftar untuk menyertai pertandingan itu. Nyatakan sama ada Chew Wai dan Lee Fang perlu menyertai pertandingan itu dan berikan satu alasan yang sah dan munasabah. The maximum number of participants for the board game competition is 35. If the number of participants does not reach 35, Chew Wai and Lee Fang will have to join the competition. After the deadline of registration, 30 people have registered to join the competition. State whether Chew Wai and Lee Fang need to join the competition and give a valid and sound reason. [2 markah/ marks] Jawapan/ Answer: (a) (i) Lambungan dadu Chew Wai ialah suatu peristiwa tidak bersandar manakala lambungan dadu Lee Fang ialah suatu peristiwa bersandar. Chew Wai’s dice-tossing is an independent event whereas Lee Fang’s dice-tossing is a dependent event. Soalan SPM Kertas 2 Bahagian C CONTOH

Matematik Tingkatan 4 Soalan SPM Kertas 2 Bahagian C 151 (ii) Kebarangkalian Chew Wai mendapat nombor 6 pada dadu Q: 1 12 The probability of Chew Wai getting a number 6 on dice Q: 1 12 Kebarangkalian Lee Fang mendapat nombor 6 pada dadu Q: The probability of Lee Fang getting a number 6 on dice Q: 1 6 × 1 12 = 1 72 (b) Panjang sisi dadu P/ Side length of dice P: √3.375 3 = 1.5 cm Katakan P sebagai jumlah luas permukaan dadu P dan Q sebagai jumlah luas permukaan dadu Q. Let P be the total surface area of dice P and Q be the total surface area of dice Q. P Q = 2 3 Q = 3P 2 = 3(6 × 1.52 ) 2 = 20.25 cm2 Luas bagi satu permukaan dadu Q/ Area for one surface of dice Q: 20.25 12 = 1.6875 cm2 (c) Katakan P sebagai bilangan dadu P yang dibeli dan Q sebagai bilangan dadu Q yang dibeli. Let P be the number of dice P bought and Q be the number of dice Q bought. 7P + 3Q = 7.6, × 5 35P + 15Q = 38 ……➀ 9P + 5Q = 10.8, × 3 27P + 15Q = 32.4 ……➁ ➀ – ➁: 35P – 27P = 38 – 32.4 8P = 5.6 P = 0.7 P = 0.7 ↷ 7P + 3Q = 7.6, 7(0.7) + 3Q = 7.6 4.9 + 3Q = 7.6 3Q = 2.7 Q = 0.9 \ Harga bagi sebiji dadu P ialah RM0.70 dan sebiji dadu Q ialah RM0.90. The price of a dice P is RM0.70 and a dice Q is RM0.90. (d) Chew Wai dan Lee Fang perlu menyertai pertandingan itu. Ini kerana, jika bilangan peserta tidak mencapai 35 orang, maka Chew Wai dan Lee Fang akan turut menyertai pertandingan itu, dan bilangan peserta tidak mencapai 35 orang, iaitu hanya 30 orang telah mendaftar. Chew Wai and Lee Fang need to join the competition. Because if the number of participants does not reach 35, then Chew Wai and Lee Fang will have to join the competition, and the number of participants does not reach 35, that is only 30 people has registered. CONTOH

Matematik Tingkatan 4 Soalan SPM Kertas 2 Bahagian C 152 2 Rajah 2 menunjukkan jualan bulanan nasi lemak, dalam RM, oleh Makcik Zuraini dalam masa 12 bulan. Diagram 2 shows the monthly sales of nasi lemak, in RM, by Auntie Zuraini in 12 months. Rajah 2/ Diagram 2 (a) Cari min dan sisihan piawai jualan bulanan Makcik Zuraini. Seterusnya, komen tentang serakan jualan dalam 12 bulan ini. Find the mean and standard deviation of Auntie Zuraini’s monthly sales. Hence, comment on the sale distribution in these 12 months. [4 markah/ marks] (b) (i) Margin untung jualan nasi lemak ialah 75%. Dengan menggunakan min jualan, hitung keuntungan bulanan Makcik Zuraini. The profit margin of the sales of nasi lemak is 75%. By using the mean of sales, calculate Auntie Zuraini’s monthly profit. [2 markah/ marks] (ii) Jadual 1 menunjukkan perbelanjaan bulanan Makcik Zuraini. Table 1 shows the monthly expenses of Auntie Zuraini. Perbelanjaan Expense Expense Amaun (RM) Amount (RM) Sewa rumah Rent 500 Premium insurans Insurance premium 250 Bil elektrik dan air Electric and water bills 120 Barangan dapur Groceries 300 Jadual 1/ Table 1 Keuntungan nasi lemak merupakan pendapatan aktif Makcik Zuraini. Demi keluarganya, dia juga bekerja secara sambilan sebagai seorang pencuci pinggan mangkuk di sebuah restoran. Pendapatan daripada kerja sambilannya ialah RM500 setiap bulan. Bina satu pelan kewangan yang ringkas untuk menunjukkan aliran tunai bulanan Makcik Zuraini. The profit of nasi lemak is Auntie Zuraini’s active income. For the sake of her family, she also works part time as a dish washer at a restaurant. The income from her part-time job is RM500 per month. Construct a simple financial plan to show Auntie Zuraini’s monthly cash flow. [5 markah/ marks] (iii) Adakah munasabah bagi Makcik Zuraini untuk membeli satu baju baharu set keluarga yang berharga RM85 jika dia menyimpan 10% daripada pendapatan lebihannya selama enam bulan? Jika tidak, berikan cadangan kepada Makcik Zuraini supaya dia dapat membeli set baju baharu itu dalam masa enam bulan. Is it reasonable for Auntie Zuraini to buy a family set of new clothes which costs RM85 if she saves 10% from her surplus of income for six months? If not, give a suggestion to Auntie Zuraini so that she can buy the new set of clothes in six months. [4 markah/ marks] Jawapan/ Answer: (a) x�: 980 + 990 + 1 000 + 1 050 + 1 020 + 920 + 850 + 1 020 + 1 000 + 1 010 + 880 + 1 040 12 = 11 760 12 = 980 x (x – x̅)2 980 0 990 100 1 000 400 1 050 4 900 x (x – x̅)2 1 020 1 600 920 3 600 850 16 900 1 020 1 600 x (x – x̅)2 1 000 400 1 010 900 880 10 000 1 040 3 600 Σ(x – x̅)2 = 44 000 980 990 1 000 1 050 1 020 920 850 1 020 1 000 1 010 880 1 040 CONTOH

Matematik Tingkatan 4 Soalan SPM Kertas 2 Bahagian C 153 Sisihan piawai/ Standard deviation: Σ(x ‒ x̅)2 N = 44 000 12 = √3666.6667 = ±60.55 ∴ Min jualan bulanan ialah RM980 dan sisihan piawainya ialah RM60.55. Berdasarkan sukatan serakan yang diperoleh, jualan nasi lemak Makcik Zuraini berserak antara julat RM980 ± RM60.55. The mean of monthly sales is RM980 and its standard deviation is RM60.55. Based on the measure of dispersion obtained, Auntie Zuraini’s nasi lemak sales dispersed between the range of RM980 ± RM60.55. (b) (i) Margin untung/ Profit margin = Jualan/ Sales ‒ Kos/ Cost Jualan/ Sales × 100% 75% = RM980 ‒ Kos/ Cost RM980 × 100% RM980 – Kos/ Cost = 0.75 × RM980 Kos/ Cost = RM980 – RM735 = RM245 Keuntungan bulanan/ Monthly profit: RM980 – RM245 = RM735 (ii) Pendapatan dan perbelanjaan Income and expenditure Amaun (RM) Amount (RM) Pendapatan aktif/ Active income Pendapatan pasif/ Passive income 735 500 Jumlah pendapatan/ Total income 1 235 Tolak simpanan/ Minus saving 0 Baki pendapatan/ Income balance 1 235 (‒) Perbelanjaan tetap bulanan/ Monthly fixed saving Sewa rumah/ Rent Premium insurans/ Insurance premium 500 250 Jumlah perbelanjaan tetap bulanan Total monthly fixed expenses 750 (‒) Perbelanjaan tidak tetap bulanan/ Monthly variable expenses Bil elektrik dan air/ Electric and water bills Barangan dapur/ Groceries 120 300 Jumlah perbelanjaan tidak tetap bulanan Total monthly variable expenses 420 Pendapatan lebihan/ Surplus of income 65 (iii) Jumlah simpanan dalam masa 6 bulan: 10% × RM65 × 6 = RM39 Total saving in 6 months: x 100 × RM65 × 6 > RM85 x > RM85 × 100 RM65 × 6 x > 21.79% ∴ Tidak munasabah, kerana RM39 < RM85, simpanan Makcik Zuraini adalah tidak cukup membeli set baju baharu itu. Makcik Zuraini perlu menyimpan sekurang-kurangnya 22% daripada pendapatan lebihannya untuk membeli set baju itu. Not reasonable, because RM39 < RM85, Auntie Zuraini’s saving is not enough to buy the new set of clothes. Auntie Zuraini needs to save at least 22% of her surplus of income to buy the set of clothes. CONTOH

Matematik Tingkatan 4 Soalan SPM Kertas 2 Bahagian C 154 3 Maklumat di Jadual 2 menunjukkan sebuah kereta dan sebuah motosikal yang bergerak antara bandar P dan bandar Q. The information in Table 2 shows a car and a motorcycle which travelled between town P and town Q. Kenderaan Vehicle Perjalanan Journey Kereta Car Mengambil 15 minit untuk bergerak dari bandar P ke bandar Q dan 18 minit untuk pulang ke bandar P Took 15 minutes to move from town P to town Q and 18 minutes to return to town P Motosikal Motorcycle Mengambil 0.4 jam untuk tiba di bandar Q dan tidak kembali ke bandar P Took 0.4 hours to reach town Q and did not return to town P Jadual 2/ Table 2 Diberi jarak antara bandar P dan bandar Q ialah 24 km. Given the distance between town P and town Q is 24 km. (a) (i) Lakar satu graf jarak-masa untuk menunjukkan pergerakan kereta dan motosikal itu dan labelkan paksipaksi dengan d sebagai jarak, dalam km, dan t sebagai masa, dalam minit. Sketch a distance-time graph to show the movement of the car and motorcycle and label the axes with d as the distance, in km, and t as the time, in minute. [4 markah/ marks] (ii) Hitung laju, dalam km/j, bagi kereta semasa kereta bergerak ke bandar Q. Calculate the speed, in km/h, of the car when the car travelled to town Q. [2 markah/ marks] (b) (i) Tulis dua persamaan linear untuk mewakili pergerakan kereta dalam perjalanan pulang dari bandar Q dan pergerakan motosikal ke bandar Q dengan titik persilangan antara pergerakan kereta dan motosikal ialah (t, d). Write two linear equations to represent the movement of the car in its return trip from town P and the movement of the motorcycle to town Q with the intersection point between the movements of the car and motorcycle being (t, d). [4 markah/ marks] (ii) Seterusnya, cari masa dan jarak dari bandar P seketika kedua-dua kenderaan bertemu. Hence, find the time and distance from town P at the moment when both vehicle meets. [2 markah/ marks] (c) Penunggang motosikal tersebut akan pulang ke bandar P jika dan hanya jika dia dapat sampai bandar P sebelum pukul 1 petang. Diberi kerjanya di bandar Q akan mengambil masa 20 minit dan masa dia bertolak dari bandar P ialah pukul 11.50 pagi. Tentukan sama ada penunggang motosikal itu akan pulang ke bandar P selepas menyiapkan kerjanya atau tidak jika laju pemanduannya kekal sama. The motorcyclist will return to town P if and only if he can reach town P before 1 p.m. Given his work in town Q will take 20 minutes and the time he departed from town P is 11.50 a.m. Determine whether the motorcyclist will return to town P after completing his work or not if the speed of his driving remains the same. [3 markah/ marks] Jawapan/ Answer: (a) (i) Jarak, d (km) Distance, d (km) Masa, t (minit) Time, t (minute) (t, d) 0 15 24 24 33 Motosikal Motocycle Kereta Car (ii) Laju/ Speed: 24 km (15 ÷ 60) j/ h = 24 0.25 CONTOH = 96 km/j/ km/h

Matematik Tingkatan 4 Soalan SPM Kertas 2 Bahagian C 155 (b) (i) Kereta/ Car: (15, 24), (33, 0) 0 – 24 33 – 15 = d ‒ 24 t ‒ 15 t – 15 = 18(d ‒ 24) –24 = ‒ 3 4 d + 18 t = ‒ 3 4 d + 33 Motosikal/ Motorcycle: (0, 0), (24, 24) 24 – 0 24 – 0 = t ‒ 0 d ‒ 0 t = d (ii) t = ‒ 3 4 d + 33 …… ➀ t = 3 4 d …… ➁ ➁ ↷ ➀: d = ‒ 3 4 d + 33 d + 3 4 d = 33 7 4 d= 33 d = 33 × 4 7 = 18.857 km t = 18.857 min (c) Jumlah masa yang diambil/ Total time taken: 24 + 20 + 24 = 68 minit/ minutes = 1 jam/ hour 8 minit/ minutes Masa tiba di bandar P/ Arrival time at town P: 11 jam/ hours 50 minit/ minutes + 1 jam/ hour 8 minit/ minutes = 12 jam/ hours 58 minit/ minutes = 12.58 p.m. ∴ Penunggang motosikal itu akan pulang ke bandar P selepas menyiapkan kerjanya. The motorcyclist will return to town P after completing his work. CONTOH

Matematik Tingkatan 4 Soalan SPM Kertas 2 Bahagian C 156 4 Kelas 4 Cempaka mempunyai 48 orang murid. Jadual 3 menunjukkan bilangan murid dengan subjek kegemarannya. Class 4 Cempaka has 48 students. Table 3 shows the number of students with their favourite subjects. Subjek kegemaran Favourite subject Bilangan murid Number of students Matematik Mathematics 29 Sains sahaja Science only x + 2 Bahasa sahaja Language only y Matematik sahaja Mathematics only x + y Matematik dan sains Mathematics and science 14 Matematik dan bahasa Mathematics and language 11 Sains dan bahasa Science and language 10 Semua subjek All subjects 8 Tiada None 3 Jadual 3/ Table 3 (a) Dengan menggunakan M sebagai matematik, S sebagai sains dan B sebagai bahasa, lukis satu gambar rajah Venn untuk mewakili maklumat di Jadual 3. By using M as mathematics, S as science and B as language, draw a Venn diagram to represent the information in Table 3. [3 markah/ marks] (b) Diberi bilangan murid yang suka sains sahaja dan bilangan murid yang suka bahasa sahaja adalah sama. Given the number of students who likes science only and the number of the students who likes language only are the same. (i) Nyatakan dua persamaan linear dalam dua pemboleh ubah yang dapat diperoleh daripada maklumat di atas. State two linear equations in two variables which can be obtained from the information above. [2 markah/ marks] (ii) Seterusnya, cari bilangan murid yang suka bahasa dan bilangan murid yang suka sains. Hence, find the number of students who likes language and the number of students who likes science. [3 markah/ marks] (c) Didapati murid di dalam set (M ∪ S) ∩ B′ adalah ahli persatuan matematik dan sains. It is found that the students in set (M ∪ S) ∩ B′ are members of the mathematics and science society. (i) Tentukan bilangan ahli persatuan matematik dan sains dalam Kelas 4 Cempaka. Determine the number of mathematics and science society members in Class 4 Cempaka. [2 markah/ marks] (ii) Cari kebarangkalian bahawa dua orang murid yang dipilih secara rawak daripada Kelas 4 Cempaka ialah ahli persatuan matematik dan sains. Find the probability that two students chosen randomly from Class 4 Cempaka are members of the mathematics and science society. [2 markah/ marks] (d) Diberi nisbah bilangan ahli persatuan matematik dan sains di Kelas 4 Cempaka kepada bilangan ahli daripada kelas lain ialah 5 : 19. Cari jumlah bilangan ahli persatuan matematik dan sains dan nyatakan jawapan dalam asas 4. Given the ratio of the number of mathematics and science society members in Class 4 Cempaka to the number of members from other classes is 5 : 19. Find the total number of mathematics and science society members and state the answer in base 4. [3 markah/ marks] CONTOH

Matematik Tingkatan 4 Soalan SPM Kertas 2 Bahagian C 157 Jawapan/ Answer: (a) ξ 3 M S B y x + y = 12 x + 2 6 8 3 2 (b) (i) x + y = 12 ……➀ x + 2 = y ……..➁ (ii) ➁ ↷ ➀, x + (x + 2) = 12 2x = 10 x = 5 x = 5 ↷ ➁, 5 + 2 = y y = 7 Bilangan murid yang suka bahasa: The number of students who likes language: 3 + 8 + 2 + 7 = 20 (c) (i) n[(M ∪ S) ∩ B′] = 12 + 6 + 7 = 25 ξ 3 M S B 7 12 7 3 2 8 6 (ii) Kebarangkalian/ Probability: 25 48 × 24 47 = 25 94 (d) Jumlah bilangan ahli persatuan matematik dan sains, p: Total number of mathematics and science society members, p: 5 5 + 19 = 25 p p = 25 × 24 5 = 120 4 120 4 30 … 0 4 7 … 2 4 1 … 3 0 … 1 13204 ∴ p = 13204 Bilangan murid yang suka sains: The number of students who likes science: 6 + 8 + 2 + (5 + 2) = 23 CONTOH

CONTOH

J1 Bab 1 1.1 Fungsi dan Persamaan Kuadratik Quadratic Functions and Equations 1 (f) (h) 2 (a) a = 1, b = ‒5, c = 0 (b) a = 4, b = 0, c = ‒1 (c) a = 1, b = ‒3, c = 2 (d) a = 2, b = 3, c = 0 (e) a = 1 3 , b = 0, c = 2 3 –10 –10 –8 –6 –4 –2 O 2 4 6 f(x) f(x) = 2x2 + x + 7 x 10 30 50 20 40 60 70 Garis mengufuk y = 40 melalui dua titik pada fungsi f(x) = 2x2 + x + 7. Maka, hubungan fungsi itu ialah banyak kepada satu. The horizontal line y = 40 passes through two points on the function f(x) = 2x2 + x + 7. Thus, the relation of the function is many-to-one. 4 (a) a = ‒1, a < 0 Maka, bentuk fungsi ialah ∩. Hence, the shape of the function is ∩. (b) a = ‒3, a < 0 Maka, bentuk fungsi ialah ∩. Hence, the shape of the function is ∩. (c) a = 5, a > 0 Maka, bentuk fungsi ialah ∪. Hence, the shape of the function is ∪. 5 (a) Titik maksimum/ Maximum point (b) Titik minimum/ Minimum point (c) Titik minimum/ Minimum point (d) Titik maksimum/ Maximum point (e) Titik maksimum/ Maximum point 6 (a) x = 2 (b) x = –3 7 (a) Paksi simetri/ Axis of symmetry: x = ‒ 8 2(2) = ‒2 (b) Paksi simetri/ Axis of symmetry: x = ‒ 0 2(3) = 0 8 (a) (i) x –3 –2 1 4 5 f(x) 7 0 –9 0 7 x f(x) –2 –6 1 3 4 6 –4 –8 –4 –3 –2 –1 O 2 5 4 2 6 8 (ii) x –3 –2 1 4 5 f(x) 9 2 –7 2 9 x f(x) –2 –6 1 3 4 6 –4 –8 –4 –3 –2 –1 O 2 5 4 2 6 8 Apabila nilai c berubah, pintasan-y graf turut berubah. When the value of c changes, the y-intercept of the graph also changes. Jawapan CONTOH

Matematik Tingkatan 4 Jawapan J2 (b) (i) x –5 –4 –2 0 1 f(x) –6 –1 3 –1 –6 x f(x) –1 –3 –5 –2 –4 –5 –4 –3 –2 –1 O 1 1 2 3 –6 (ii)  x –3 –2 –1 0 1 f(x) –7 –1 1 –1 –7 x f(x) –1 –3 –5 –7 –8 –2 –4 –6 –5 –4 –3 –2 –1 O 1 1 Apabila nilai a berubah, julat graf turut berubah. When the value of a changes, the range of graph also changes. 9 (a) 4 = c – 2(3)2 + 3(3) c = 4 + 9 = 13 (b) 0 = 4(2)2 – 6(2) + c c = –4 (c) c = 3 10 (a) Luas/ Area: 1 2 (x – 7)(x + 9) = 5 x2 + 2x – 63 = 10 x2 + 2x – 73 = 0 (b) Luas/ Area: 1 2 × 22 7 × (x + 2)2 = 22 x2 + 4x + 4 = 14 x2 + 4x – 10 = 0 (c) Luas/ Area: 1 2 (x)[(2x – 1) + (3x + 4)] = 33 (x)(5x + 3) = 66 5x2 + 3x – 66 = 0 (d) x2 + (x – 3)2 = 202 x2 + x2 – 6x + 9 = 400 2x2 – 6x – 391= 0 (e) Katakan salah satu nombor = x Let one of the number = x x + 60 x = 30 x2 + 60 = 30x x – 30x + 60 = 0 11 (a) Kiri/ Left: 2(–2)2 + (–2) – 6 = 2(4) – 2 – 6 = 0 (= Kanan/ Right) ∴ x = –2 ialah punca bagi 2x2 + x – 6 = 0. x = –2 is a root of 2x2 + x – 6 = 0. (b) Kiri/ Left: 6(2)2 +2 = 6(4) + 2 = 26 (≠ Kanan/ Right) ∴ x = 2 bukan punca bagi 6x2 + x = 2. x = 2 is not a root of 6x2 + x = 2. (c) Kiri/ Left: 2(3)2 – 3(3) – 9 = 2(9) – 9 – 9 = 0 (= Kanan/ Right) ∴ x = 3 ialah punca bagi 2x2 – 3x – 9 = 0. x = 3 is a root of 2x2 – 3x – 9 = 0. 12 (a) 3x2 – 27x = 0 3x(x – 9) = 0 3x = 0 x – 9 = 0 x = 0 x = 9 ∴ x = 0 atau/ or 9 (b) x2 – 6x + 8 = 0 (x – 2)(x – 4) = 0 x – 2 = 0 x– 4 = 0 x = 2 x = 4 ∴ x = 2 atau/ or 4 (c) x2 + 7x – 18 = 0 (x + 9)(x – 2) = 0 x + 9 = 0 x – 2 = 0 x = –9 x = 2 CONTOH ∴ x = –9 atau/ or 2

Matematik Tingkatan 4 Jawapan J3 (d) 6w2 + 17w – 14 = 0 (3w – 2)(2w + 7) = 0 3w – 2 = 0 2w + 7 = 0 w = 2 3 w = – 7 2 ∴ w = – 7 2 atau/ or 2 3 (e) 12x2 – 15 8 = –x 12x2 – 15 = –8x 12x2 + 8x – 15 = 0 (6x – 5)(2x + 3) = 0 6x – 5 = 0 2x + 3 = 0 x = 5 6 x = – 3 2 ∴ x = – 3 2 atau/ or 5 6 (f) (x – 2)(x + 6) = 15 – 2x x2 + 4x – 12 + 2x – 15 = 0 x2 + 6x – 27 = 0 (x + 9)(x – 3) = 0 x + 9 = 0 x – 3 = 0 x = –9 x = 3 ∴ x = –9 atau/ or 3 (g) (x – 4)(x + 5) – x – 16 = 0 x2 + x – 20 – x – 16 = 0 x2 – 36 = 0 (x – 6)(x + 6) = 0 x – 6 = 0 x + 6 = 0 x = 6 x = –6 ∴ x = –6 atau/ or 6 13 (a) –3 3 –9 f(x) x O f(x) = x2 – 9 (b) –5 3 15 f(x) x O f(x) = –x2 – 2x + 15 (–1, 16) (c) –45 f(x) x O f(x) = 4x2 – 8x – 45 (1, –49) – 5 2 9 2 (d) f(x) x O f(x) = –8x2 – 32x + 18 (–2, 50) 18 – 9 2 1 2 (e) f(x) x O f(x) = –4x2 27 + 12x + 27 – 3 2 9 2 ( 3 2 , 36) 14 (a) 625(x + 8)(x + 4) = 120 000 x2 + 12x + 32 = 192 x2 + 12x – 160 = 0 (x + 20)(x – 8) = 0 x = –20 atau/ or 8 (Panjang mesti bernilai positif) (Length must be a positive value) ∴ Nilai x ialah 8./ The value of x is 8. (b) 4( 1 2 )(3x + 2)(4x + 2) + (3x + 2)2 = 224 2(12x2 + 14x + 4) + 9x2 +12x + 4 – 224 = 0 24x2 + 28x + 8 + 9x2 + 12x + 4 – 224 = 0 33x2 + 40x – 212 = 0 (33x + 106)(x – 2) = 0 x = – 106 33 atau/ or 2 (Panjang mesti bernilai positif) (Length must be a positive value) ∴ Nilai x ialah 2./ The value of x is 2. CONTOH

Matematik Tingkatan 4 Jawapan J4 (c) Menggunakan Teorem Pythagoras, Using Pythagoras’ Theorem, x2 + (x + 7)2 = (x + 8)2 x2 + x2 + 14x + 49 = x2 + 16x + 64 x2 – 2x – 15 = 0 (x + 3)(x – 5) = 0 x = –3 atau/ or 5 x mesti bernilai positif bagi panjang, maka x = 5. x must be a positive value for length, thus x = 5. Luas/ Area: 1 2 × 5 × 12 = 30 cm2 Praktis Kendiri Kertas 1 1 –2x2 + 11x = 12 2x2 – 11x + 12 = 0 (x – 4)(2x – 3) = 0 x – 4 = 0 2x – 3 = 0 x = 4 x = 3 2 Jawapan/ Answer: C 2 8x2 + 10x – 3 = 0 (2x + 3)(4x – 1) = 0 2x + 3 = 0 4x – 1 = 0 x = – 3 2 x = 1 4 Jawapan/ Answer: B 3 (x + 2)2 = 2x + 7 x2 + 4x + 4 – 2x – 7 = 0 x2 + 2x – 3 = 0 (x + 3)(x – 1) = 0 x + 3 = 0 x – 1 = 0 x = –3 x = 1 Jawapan/ Answer: D 4 (x + 3)(2x – 4) = x2 – 4 2x2 + 2x – 12 – x2 + 4 = 0 x2 + 2x – 8 = 0 (x + 4)(x – 2) = 0 x + 4 = 0 x – 2 = 0 x = –4 x = 2 Jawapan/ Answer: D 5 (x + 3)(x – 6) = 0 x2 – 3x – 18 = 0 Gantikan x = 3 2 ke dalam persamaan, Substitute x = 3 2 into the equation, ( 3 2 ) 2 – 3( 3 2 ) – 18 = 9 4 – 9 2 – 18 = – 81 4 – 81 4 : 81 2 –1 : 2 x2 – 3x – 18 : –2x2 + 6x + 36 Jawapan/ Answer: C 6 5x + 21 x = 2(x – 3) 5x + 21 = 2x2 – 6x 2x2 – 11x – 21 = 0 (2x + 3)(x – 7) = 0 2x + 3 = 0 x – 7 = 0 x = – 3 2 x = 7 Jawapan/ Answer: A 7 x – 20 3 – 4x = 3 x x2 – 20x = 9 – 12x x2 – 8x – 9 = 0 (x + 1)(x – 9) = 0 x + 1 = 0 x – 9 = 0 x = –1 x = 9 Jawapan/ Answer: A 8 Gantikan (3, 0) ke dalam f(x) = 2x2 – 5x + c, Substitute (3, 0) into f(x) = 2x2 – 5x + c, 0 = 2(3)2 – 5(3) + c c = –2(9) + 15 = –3 Jawapan/ Answer: C 9 y = –2x2 + 8x + 10, a = –2, maka bentuk graf ialah ∩ a = –2, thus the shape of the graph is ∩ c = 10, maka pintasan-y bagi graf ialah 10 c = 10, thus the y-intercept of the graph is 10 Apabila/ When y = 0, –2x2 + 8x + 10 = 0 ÷ –2 x2 – 4x – 5 = 0 (x – 5)(x + 1) = 0 x – 5 = 0 x + 1 = 0 x = 5 x = –1 Jawapan/ Answer: D 10 3 ialah punca, maka 32 – 3k + 15 = 0 3 is a root, thus 32 – 3k + 15 = 0 32 – 3k + 15 = 0 3k = 9 + 15 k = 24 3 k = 8 Jawapan/ Answer: D CONTOH

Matematik Tingkatan 4 Jawapan J5 11 Titik maksimum terletak pada paksi simetri, maka koordinat-x bagi titik maksimum ialah 1. Maximum point is placed at the axis of symmetry, thus the x-coordinate of the maximum point is 1. y = 5 + 4(1) – 2(1)2 = 5 + 4 – 2 = 7 Koordinat titik maksimum = (1, 7) Coordinates of maximum point = (1, 7) Jawapan/ Answer: A 12 1 2 (x)[(2x + 3) + (3 – x)] = 4x + 4 (x)(x + 6) = 8x + 8 x2 + 6x – 8x – 8 = 0 x2 – 2x – 8 = 0 (x + 2)(x – 4) = 0 x + 2 = 0 x – 4 = 0 x = –2 x = 4 Oleh sebab panjang tidak boleh bernilai negatif, maka x = 4. Since length cannot be a negative value, thus x = 4. Jawapan/ Answer: B Kertas 2 Bahagian A/ Section A 1 (a) 1 3 × 22 7 × 14 × (3x + 3)2 = 44 3 (9x2 + 18x + 9) = 132x2 + 264x + 132 (Terbukti/ Proven) (b) 132x2 + 264x + 132 = 1 188 ÷ 132 x2 + 2x + 1 = 9 x2 + 2x – 8 = 0 (x + 4)(x – 2) = 0 x + 4 = 0 x – 2 = 0 x = –4 x = 2 ∴ Nilai x ialah 2./ The value of x is 2. 2 Panjang/ Length: 20 + x + x = 20 + 2x Tinggi/ Height: 15 + 2(x – 1) = 15 + 2x – 2 = 13 + 2x Luas/ Area: (20 + 2x)(13 + 2x) = 494 260 + 66x + 4x2 = 494 4x2 + 66x – 234 = 0 ÷ 2 2x2 + 33x – 117 = 0 (2x + 39)(x – 3) = 0 2x + 39 = 0 x – 3 = 0 x = – 39 2 x = 3 ∴ Nilai x ialah 3./ The value of x is 3. 3 (a) Apabila x = 0, f(x) = 15. Maka, ketinggian Muthu dari atas tanah ialah 15 m. When x = 0, f(x) = 15. Thus, the height of Muthu from above ground is 15 m. (b) Apabila f(x) = 0/ When f(x) = 0, –5x2 + 10x + 15 = 0 ÷ –5 x2 – 2x – 3 = 0 (x – 3)(x + 1) = 0 x = –1 atau/ or 3 ∴ Oleh sebab masa tidak boleh bernilai negatif, x = 3 saat. Since time cannot be a negative value, x = 3 seconds. 4 (a) Katakan umur Syafiah = x./ Let Syafiah’s age = x. x(x – 5) = 126 x2 – 5x – 126 = 0 ATAU/ OR (x – 3)(x – 3 – 5) = 126 (x – 3)(x – 8) = 126 x2 – 11x + 24 – 126 = 0 x2 – 11x – 102 = 0 (b) x2 – 5x – 126 = 0 (x + 9)(x – 14) = 0 x = –9 atau/ or 14 Oleh sebab umur tidak boleh bernilai negatif, maka x = 14. Because age cannot be a negative value, hence x = 14. ∴ Umur Syafiah sekarang ialah 14 + 3 = 17 tahun. Syafiah’s current age is 14 + 3 = 17 years old. ATAU/ OR x2 – 11x – 102 = 0 (x + 6)(x – 17) = 0 x = ‒6 atau/ or 17 Oleh sebab umur tidak boleh bernilai negatif, maka x = 17. Because age cannot be a negative value, hence x = 17. Bahagian B/ Section B 5 (a) (i) x(x + 4) = 140 x2 + 4x – 140 = 0 (ii) x2 + 4 – 140 = 0 (x – 10)(x + 14) = 0 x = 10 atau/ or –14 ∴ x = –14 dan/ and 10 (b) Dengan menggunakan Teorem Pythagoras, By using Pythagoras’ Theorem, x2 + (x + 1)2 = (x + 9)2 x2 + x2 + 2x + 1 = x2 + 18x + 81 2x2 + 2x + 1 = x2 + 18x + 81 x2 – 16x – 80 = 0 (x + 4)(x – 20) = 0 x = –4 atau/ or 20 Oleh sebab panjang tidak boleh bernilai negatif, maka x = 20. Since length cannot be a negative value, thus x = 20. CONTOH

Matematik Tingkatan 4 Jawapan J6 Luas/ Area: 20 × (20 + 1) = 20 cm × 21 cm = 420 cm2 (Ditunjukkan/ Shown) 6 (a) x = ‒3 x = 3 5 x + 3 = 0 5x = 3 5x – 3 = 0 f(x) = (x + 3)(5x – 3) = 5x2 – 3x + 15x – 9 = 5x2 + 12x – 9 Banding dengan: Compare with: f(x) = ax2 + bx + c, a = 5, b = 12 dan c = ‒9 (b) Paksi simetri/ Axis of symmetry: x = ‒ b 2a = ‒ 12 2(5) = ‒ 6 5 Apabila/ When x = ‒ 6 5 , f(‒ 6 5 ) = 5(‒ 6 5 ) 2 + 12(‒ 6 5 ) – 9 = 7 1 5 ‒ 14 2 5 ‒ 9 = ‒16 1 5 ∴ Koordinat titik minimum ialah (‒ 6 5 , ‒16 1 5 ). The coordinates of the minimum point are (– 6 5 , –16 1 5 ). (c) f(x) = 8 5x2 + 12x – 9 = 8 5x2 + 12x – 17 = 0 (5x + 17)(x – 1) = 0 x = ‒ 17 5 atau/ or 1 7 (a) f(4) = 0 16a + 4b + c = 0 c = ‒16a – 4b……➀ f(1) = 9 a + b + c = 9 c = 9 – a – b……➁ ➀ = ➁, ‒16a ‒ 4b = 9 – a – b ‒15a – 3b = 9……➂ Paksi simetri/ Axis of symmetry: ‒ b 2a = 1 2a = ‒b b = ‒2a b = ‒2a ↷ ➂, ‒15a – 3(‒2a) = 9 ‒15a + 6a = 9 ‒9a = 9 a = ‒1 b = ‒2(‒1) = 2 ∴ a = ‒1, b = 2 (b) f(x) = ‒x2 + 2x + c f(1) = 9 ‒1 + 2 + c = 9 c = 8 ∴ f(x) = ‒x2 + 2x + 8 Apabila/ When f(x) = 0, ‒x2 + 2x + 8 = 0 (x – 4)(‒x ‒ 2) = 0 x = 4 atau/ or ‒2 ∴ Satu lagi punca ialah ‒2. The other root is –2. Bab 2 2.1 Asas Nombor Number Bases 1 (a) 84 (b) 61 (c) 106 (d) 50 (e) 94 2 (a) 1 × 22 = 1 × 4 (b) 1 × 26 = 1 × 64 = 4 = 64 (c) 1 × 24 = 1 × 16 (d) 5 × 84 = 5 × 4 096 = 16 = 20 480 (e) 6 × 80 = 6 × 1 (f) 4 × 82 = 4 × 64 = 6 = 256 (g) 4 × 55 = 4 × 3 125 (h) 4 × 53 = 4 × 125 = 12 500 = 500 (i) 2 × 50 = 2 × 1 = 2 3 (a) (1 × 25 ) + (1 × 24 ) + (0 × 23 ) + (1 × 22 )+ (0 × 21 ) + (1 × 20 ) (b) (1 × 23 ) + (0 × 22 ) + (1 × 21 ) + (1 × 20 ) (c) (5 × 84 ) + (2 × 83 ) + (6 × 82 ) + (3 × 81 ) + (4 × 80 ) (d) (1 × 86 ) + (5 × 85 ) + (3 × 84 ) + (4 × 83 ) + (7 × 82 ) + (4 × 81 ) + (2 × 80 ) (e) (3 × 55 ) + (2 × 54 ) + (1 × 53 ) + (3 × 52 ) + (3 × 51 ) + (4 × 50 ) (f) (4 × 54 ) + (3 × 53 ) + (2 × 52 ) + (1 × 51 ) + (4 × 50 ) 4 (a) 1011012 = (1 × 25 ) + (0 × 24 ) + (1 × 23 ) + (1 × 22 ) + (0 × 21 ) + (1 × 20 ) = 32 + 0 + 8 + 4 + 0 + 1 = 4510 CONTOH

Matematik Tingkatan 4 Jawapan J7 (b) 110111002 = (1 × 27 ) + (1 × 26 ) + (0 × 25 ) + (1 × 24 ) + (1 × 23 ) + (1 × 22 ) + (0 × 21 ) + (0 × 20 ) = 128 + 64 + 16 + 8 + 4 = 22010 (c) 748 = (7 × 81 ) + (4 × 80 ) = 56 + 4 = 6010 (d) 53778 = (5 × 83 ) + (3 × 82 ) + (7 × 81 ) + (7 × 80 ) = 2560 + 192 + 56 + 7 = 281510 (e) 3245 = (3 × 52 ) + (2 × 51 ) + (4 × 50 ) = 75 + 10 + 4 = 8910 (f) 24435 = (2 × 53 ) + (4 × 52 ) + (4 × 51 ) + (3 × 50 ) = 250 + 100 + 20 + 3 = 37310 5 (a) Nilai tempat Place value 625 125 25 5 1 Langkah Step 2347 – 1875 472 625 3 472 – 375 97 125 3 97 – 75 22 25 3 22 – 20 2 5 4 2 – 2 0 1 2 Asas 5 Base 5 3 3 3 4 2 ∴ 333425 (b) Nilai tempat Place value 2187 729 243 81 27 9 3 1 Langkah Step 2347 – 2187 160 2187 1 160 – 0 160 729 0 160 – 0 160 243 0 160 – 81 79 81 1 79 – 54 25 27 2 25 – 18 7 9 2 7 – 6 1 3 2 1 – 1 0 1 1 Asas 3 Base 3 1 0 0 1 2 2 2 1 ∴ 100122213 (c) Nilai tempat Place value 1296 216 36 6 1 Langkah Step 2347 – 1296 1051 1296 1 1051 – 864 187 216 4 187 – 180 7 36 5 7 – 6 1 6 1 1 – 1 0 1 1 Asas 6 Base 6 1 4 5 1 1 ∴ 145116 6 (a) 2 198 2 99 ... 0 2 49 … 1 2 24 … 1 2 12 … 0 2 6 … 0 2 3 … 0 2 1 … 1 0 … 1 ∴ 19810 = 110001102 (b) 5 48 5 9 ... 3 5 1 … 4 0 … 1 ∴ 4810 = 1435 (c) 5 692 5 138 … 2 5 27 … 3 5 5 … 2 5 1 … 0 0 … 1 ∴ 69210 = 102325 (d) 8 326 8 40 … 6 8 5 … 0 0 … 5 ∴ 32610 = 5068 (e) 8 5149 8 643 … 5 8 80 … 3 8 10 … 0 8 1 … 2 0 … 1 ∴ 514910 = 120358 (f) 6 2907 6 484 … 3 6 80 … 4 6 13 … 2 6 2 … 1 0 … 2 ∴ 290710 = 21243 CONTOH 6

Matematik Tingkatan 4 Jawapan J8 (g) 9 963 9 107 … 0 9 11 … 8 9 1 … 2 0 … 1 ∴ 96310 = 12809 (h) 3 10250 3 3416 … 2 3 1138 … 2 3 379 … 1 3 126 … 1 3 42 … 0 3 14 … 0 3 4 … 2 3 1 … 1 0 … 1 ∴ 1025010 = 1120011223 7 (a) 101010102 = (1 × 27 ) + (1 × 25 ) + (1 × 23 ) + (1 × 21 ) = 128 + 32 + 8 + 2 = 17010 5 170 5 34 … 0 5 6 … 4 5 1 … 1 0 … 1 ∴ 101010102 = 17010 = 11405 (b) 2435 = (2 × 52 ) + (4 × 51 ) + (3 × 50 ) = 50 + 20 + 3 = 7310 2 73 2 36 … 1 2 18 … 0 2 9 … 0 2 4 … 1 2 2 … 0 2 1 … 0 0 … 1 ∴ 2435 = 7310 = 10010012 (c) 654327 = (6 × 74 ) + (5 × 73 ) + (4 × 72 ) + (3 × 71 ) + (2 × 70 ) = 14406 + 1715 + 196 + 21 + 2 = 1634010 9 16340 9 1815 … 5 9 201 … 6 9 22 … 3 9 2 … 4 0 … 2 ∴ 654327 = 1634010 = 243659 (d) 368 = (3 × 81 ) + (6 × 80 ) = 24 + 6 = 3010 5 30 5 6 … 0 5 1 … 1 0 … 1 ∴ 368 = 3010 = 1105 (e) 45236 = (4 × 63 ) + (5 × 62 ) + (2 × 61 ) + (3 × 60 ) = 864 + 180 + 12 + 3 = 105910 8 1059 8 132 … 3 8 16 … 4 8 2 … 0 0 … 2 ∴ 45236 = 105910 = 20438 8 (a) Asas 2 Base 2 1 0 1 1 0 1 Nilai tempat Place value 22 21 20 22 21 20 Nilai digit Digit value 4 2 1 4 2 1 Asas 8 Base 8 4 + 1 = 5 4 + 1 = 5 ∴ 1011012 = 558 (b) Asas 2 Base 2 1 0 1 0 1 1 Nilai tempat Place value 22 21 20 22 21 20 Nilai digit Digit value 4 2 1 4 2 1 Asas 8 Base 8 4 + 1 = 5 2 + 1 = 3 ∴ 1010112 = 538 (c) Asas 2 Base 2 1 0 1 1 1 0 1 0 Nilai tempat Place value 22 21 20 22 21 20 22 21 20 Nilai digit Digit value 4 2 1 4 2 1 4 2 1 Asas 8 Base 8 2 4 + 2 + 1 = 7 2 ∴ 101110102 = 2728 9 (a) Asas 8 Base 8 6 = 4 + 2 3 = 2 + 1 5 = 4 + 1 Nilai tempat Place value 22 21 20 22 21 20 22 21 20 Asas 2 Base 2 1 1 0 0 1 1 1 0 1 ∴ 6358 = 1100111012 CONTOH

Matematik Tingkatan 4 Jawapan J9 (b) Asas 8 Base 8 4 5 = 4 + 1 1 2 Nilai tempat Place value 22 21 20 22 21 20 22 21 20 22 21 20 Asas 2 Base 2 1 0 0 1 0 1 0 0 1 0 1 0 ∴ 45128 = 1001010010102 (c) Asas 8 Base 8 6 = 4 + 2 2 0 4 Nilai tempat Place value 22 21 20 22 21 20 22 21 20 22 21 20 Asas 2 Base 2 1 1 0 0 1 0 0 0 0 1 0 0 ∴ 62048 = 1100100001002 10 (a) 32245 = (3 × 53 ) + (2 × 52 ) + (2 × 51 ) + (4 × 50 ) = 375 + 50 + 10 + 4 = 43910 8 439 8 54 … 7 8 6 … 6 0 … 6 6678 ∴ A = 6 (b) 8789 = (8 × 92 ) + (7 × 91 ) + (8 × 90 ) = 648 + 63 + 8 = 71910 4 719 4 179 … 3 4 44 … 3 4 11 … 0 4 2 … 3 0 … 2 230334 ∴ A = 0 (c) 54016 = (5 × 63 ) + (4 × 62 ) + (1 × 60 ) = 1080 + 144 + 1 = 122510 8 1225 8 153 … 1 8 19 … 1 8 2 … 3 0 … 2 23118 ∴ A = 2 11 (a) 1 0 12 + 1 1 1 12 1 0 1 0 02 1 1 1 ∴ 101002 (b) 3 2 4 25 + 2 2 3 4 15 3 1 1 3 35 1 1 1 ∴ 311335 (c) 6 3 5 27 + 3 3 3 5 47 4 3 0 3 67 1 1 1 ∴ 430367 12 (a) 1101012 ➝ 5310 1101102 ➝ + 5410 10710 2 107 2 53 … 1 2 26 … 1 2 13 … 0 2 6 … 1 2 3 … 0 2 1 … 1 0 … 1 ∴ 11010112 (b) 76278 ➝ 399110 12318 ➝ + 66510 465610 8 4656 8 582 … 0 8 72 … 6 8 9 … 0 8 1 … 1 0 … 1 ∴ 110608 (c) 324526 ➝ 449610 551236 ➝ + 761110 1210710 6 12107 6 2017 … 5 6 336 … 1 6 56 … 0 6 9 … 2 6 1 … 3 0 … 1 ∴ 1320156 13 (a) 1 1 0 0 12 – 1 0 1 12 1 1 1 02 2 0 1 2 2 ∴ 11102 (b) 7 5 6 6 38 – 3 4 2 1 78 4 1 4 4 48 5 8 ∴ 414448 (c) 4 1 2 3 1 25 – 3 1 2 2 3 15 1 0 0 0 3 15 2 5 ∴ 1000315 CONTOH

Matematik Tingkatan 4 Jawapan J10 14 (a) 887259 ➝ 5891010 224549 ➝ –1495310 4395710 9 43957 9 4884 … 1 9 542 … 6 9 60 … 2 9 6 … 6 0 … 6 ∴ 662619 (b) 412315 ➝ 269110 23235 ➝ – 33810 235310 5 2353 5 470 … 3 5 94 … 0 5 18 … 4 5 3 … 3 0 … 3 ∴ 334035 (c) 5367 ➝ 27210 4537 ➝ – 23410 3810 7 38 7 5 … 3 0 … 5 ∴ 537 15 (a) Markah Fauzi/ Fauzi’s marks: 1000112 = 1(25 ) + 1(21 ) + 1 = 32 + 2 + 1 = 35 Markah Maria/ Maria’s marks: 3105 = 3(52 ) + 1(51 ) = 75 + 5 = 80 ∴ Maria mendapat markah yang lebih tinggi. Maria scored higher marks. (b) 102 = 1 × 21 = 2 110002 = 24 + 23 = 24 Katakan salah satu nombor = x Let one of the numbers = x x + (x + 2) = 24 2x = 22 x = 11 ∴ 11 dan/ and 13 (c) 2 365 2 182 … 1 2 91 … 0 2 45 … 1 2 22 … 1 2 11 … 0 2 5 … 1 2 2 … 1 2 1 … 0 0 … 1 ∴ 1011011012 (d) 1101011102 = 28 + 27 + 25 + 23 + 22 + 21 = 256 + 128 + 32 + 8 + 4 + 2 = 43010 5 430 5 86 … 0 5 17 … 1 5 3 … 2 0 … 3 ∴ 32105 (e) Asas 2 Base 2 1 1 0 0 1 0 0 1 1 1 0 1 Nilai tempat Place value 22 21 20 22 21 20 22 21 20 22 21 20 Nilai digit Digit value 4 2 1 4 2 1 4 2 1 4 2 1 Asas 8 Base 8 4 + 2 = 6 2 2 + 1 = 3 4 + 1 = 5 ∴ 62358 Praktis Kendiri Kertas 1 1 1 × 23 = 1 × 8 = 8 Jawapan/ Answer: D 2 6 × 82 = 6 × 64 = 384 Jawapan/ Answer: A 3 54208 = 5(83 ) + 4(82 ) + 2(8) = 2560 + 256 + 16 = 283210 5 2832 5 566 … 2 5 113 … 1 5 22 … 3 5 4 … 2 0 … 4 423125 Jawapan/ Answer: B 4 Asas 2 Base 2 1 1 0 1 0 1 1 1 Nilai tempat Place value 22 21 20 22 21 20 22 21 20 Nilai digit Digit value 4 2 1 4 2 1 4 2 1 Asas 8 Base 8 2 + 1 = 3 2 4 + 2 + 1 = 7 ∴ 3278 Jawapan/ Answer: A 5 2(54 ) + 4(52 ) + 14 = 1250 + 100 + 14 = 136410 CONTOH

Matematik Tingkatan 4 Jawapan J11 8 1364 8 170 … 4 8 21 … 2 8 2 … 5 0 … 2 25248 Jawapan/ Answer: D 6 14335 = 1(53 ) + 4(52 ) + 3(5) + 3 = 125 + 100 + 15 + 3 = 24310 2 243 2 121 … 1 2 60 … 1 2 30 … 0 2 15 … 0 2 7 … 1 2 3 … 1 2 1 … 1 0 … 1 111100112 Jawapan/ Answer: A 7 111011012 = 27 + 26 + 25 + 23 + 22 + 1 = 128 + 64 + 32 + 8 + 4 + 1 = 23710 5 237 5 47 … 2 5 9 … 2 5 1 … 4 0 … 1 14225 Jawapan/ Answer: C 8 1 0 0 1 1 12 + 1 0 1 1 02 1 1 1 1 0 12 1 1 Jawapan/ Answer: D 9 1 1 1 0 1 1 0 12 – 1 0 1 1 0 02 1 1 0 0 0 0 0 12 Jawapan/ Answer: A 10 4 3 5 67 – 3 3 2 17 1 0 3 57 Jawapan/ Answer: A 11 2 3 4 45 + 3 2 1 15 1 1 1 1 05 1 1 1 1 Jawapan/ Answer: D 12 8 3883 8 485 … 3 8 60 … 5 8 7 … 4 0 … 7 74538 ∴ m = 5 Jawapan/ Answer: A 13 Asas 2 Base 2 1 0 1 1 0 1 0 Nilai tempat Place value 22 21 20 22 21 20 22 21 20 Nilai digit Digit value 4 2 1 4 2 1 4 2 1 Asas 8 Base 8 1 2 + 1 = 3 2 ∴ 1328 Jawapan/ Answer: A 14 1 1 0 0 1 12 – 1 0 0 1 0 02 1 1 1 12 1 0 2 2 23 + 22 + 2 + 1 = 8 + 4 + 2 + 1 = 1510 ∴ A = 5 Jawapan/ Answer: C 15 2 3 0 45 + 3 0 1 15 1 0 3 2 05 1 1 1(54 ) + 3(52 ) + 2(5) = 625 + 75 + 10 = 71010 4 710 4 177 … 2 4 44 … 1 4 11 … 0 4 2 … 3 0 … 2 230124 Jawapan/ Answer: C 16 1 0 1 1 0 12 + 1 1 1 1 12 1 0 0 1 1 0 02 1 1 1 1 1 26 + 23 + 22 = 64 + 8 + 4 = 7610 8 76 8 9 … 4 8 1 … 1 0 … 1 1148 ∴ p = 1 Jawapan/ Answer: A CONTOH

Matematik Tingkatan 4 Jawapan J12 Kertas 2 Bahagian A/ Section A 1 13258 = 1(83 ) + 3(82 ) + 2(81 ) + 5 = 512 + 192 + 16 + 5 = 72510 5 725 5 145 … 0 5 29 … 0 5 5 … 4 5 1 … 0 0 … 1 104005 = 54 + 4(52 ) ∴ m = 2 2 453268 = 4(84 ) + 5(83 ) + 3(82 ) + 2(81 ) + 6(80 ) Maka/ Thus, p = 4, q = 3, r = 1, s = 0 ∴ p + q + r + s = 4 + 3 + 1 + 0 = 8 3 Asas 2 Base 2 1 0 1 1 0 1 1 1 0 Nilai tempat Place value 22 21 20 22 21 20 22 21 20 Nilai digit Digit value 4 2 1 4 2 1 4 2 1 Asas 8 Base 8 4 + 1 = 5 4 + 1 = 5 4 + 2 = 6 1011011102 = 5568 ∴ X = 556 4 (a) 2y314 = 51y6 2(43 ) + y(42 ) + 3(4) + 1 = 5(62 ) + 1(6) + y 128 + 16y + 12 + 1 = 180 + 6 + y 16y – y = 186 – 141 15y = 45 y = 45 15 = 3 (b) Nilai digit Digit value 2y314 51y6 3(42 ) = 48 3 5 (a) 3 × 54 = 3 × 625 = 187510 (b) 3(54 ) + 4(53 ) + 1(52 ) + 2(51 ) + 2 = 1875 + 500 + 25 + 10 + 2 = 241210 9 2412 9 268 … 0 9 29 … 7 9 3 … 2 0 … 3 ∴ 32709 Bab 3 3.1 Pernyataan Statements 1 (a) Ya/ Yes (b) Ya/ Yes (c) Bukan/ No (d) Bukan/ No (e) Ya/ Yes (f) Ya/ Yes (g) Bukan/ No 2 (a) Palsu/ False (b) Benar/ True (c) Benar/ True (d) Palsu/ False (e) Benar/ True (f) Benar/ True (g) Palsu/ False (h) Palsu/ False (i) Benar/ True 3 (a) Sebilangan/ Some (b) Sebilangan/ Some (c) Semua/ All 4 (a) 6 bukan satu nombor gandaan 3. 6 is not a multiple of 3. Palsu False (b) Sudut tirus tidak besar daripada 90°. Acute angles are not bigger than 90°. Benar True (c) Segi tiga tidak mempunyai pepenjuru. Triangles have no diagonals. Benar True 5 (a) 24 ialah satu nombor gandaan 6 dan boleh dibahagi tepat dengan 2. 24 is a multiple of 6 and is divisible by 2. (b) 15 atau 17 ialah satu nombor gandaan 3. 15 or 17 is a multiple of 3. (c) 30 boleh dibahagi tepat dengan 3 atau 5. 30 is divisible by 3 or 5. 6 (a) p: Sebuah segi tiga bersudut tegak mempunyai satu sudut tegak. A right-angled triangle has one right angle. q: Sebuah segi tiga bersudut tegak mempunyai dua sudut tirus. A right-angled triangle has two acute angles. (b) p: Sebuah segi empat sama mempunyai empat sisi. A square has four sides. q: Sebuah segi empat tepat mempunyai empat sisi. A rectangle has four sides. CONTOH

Matematik Tingkatan 4 Jawapan J13 (c) p: Mina telah membeli sehelai skirt. Mina had bought a skirt. q: Mina telah membeli sehelai seluar. Mina had bought a pair of pants. 7 (a) Palsu/ False (b) Benar/ True (c) Palsu/ False (d) Benar/ True 8 (a) Antejadian: Satu nombor tamat dengan digit 5. Antecedent: A number ends with the digit 5. Akibat: Nombor itu boleh dibahagi tepat dengan 5. Consequent: The number is divisible by 5. (b) Antejadian: 4y – 15 = 1 Antecedent: 4y – 15 = 1 Akibat: y = 4 Consequent: y = 4 (c) Jika 42 = 16, maka 42 + 4 = 20. If 42 = 16, then 42 + 4 = 20. (d) Jika x boleh dibahagi tepat dengan 6, maka x boleh dibahagi tepat dengan 2. If x is divisible by 6, then x is divisible by 2. 9 (a) Implikasi 1/ Implication 1: Jika sebuah segi empat selari ialah sebuah rombus, maka pepenjurunya adalah berserenjang. If a parallelogram is a rhombus, then its diagonals are perpendicular. Implikasi 2/ Implication 2: Jika pepenjuru sebuah segi empat selari adalah berserenjang, maka segi empat selari itu ialah sebuah rombus. If the diagonals of a parallelogram is perpendicular, then the parallelogram is a rhombus. (b) Implikasi 1/ Implication 1: Jika B ⊂ A, maka unsur set A ialah unsur set B. If B ⊂ A, then the elements of set A are elements of set B. Implikasi 2/ Implication 2: Jika unsur set A ialah unsur set B, maka B ⊂ A. If the elements of set A are elements of set B, then B ⊂ A. (c) θ ialah sudut tirus jika dan hanya jika 0° N θ N 90°. θ is an acute angle if and only if 0° N θ N 90°. (d) Sisi sebuah segi tiga adalah sama panjang jika dan hanya jika sudut pedalamannya adalah sama. The sides of a triangle are equal if and only if its interior angles are equal. 10 (a) Akas/ Converse: Jika 10 – 7 = 3, maka 7 + 3 = 10. If 10 – 7 = 3, then 7 + 3 = 10. Songsangan/ Inverse: Jika 7 + 3 ≠ 10, maka 10 – 7 ≠ 3. If 7 + 3 ‡ 10, then 10 – 7 ‡ 3. Kontrapositif/ Contrapositive: Jika 10 – 7 ≠ 3, maka 7 + 3 ≠ 10. If 10 – 7 ‡ 3, then 7 + 3 ‡ 10. (b) Akas/ Converse: Jika pintasan-y mempunyai nilai positif, maka c bernilai positif. If the y-intercept has a positive value, then c is a positive value. Songsangan/ Inverse: Jika c bukan bernilai positif, maka pintasan-y tidak mempunyai nilai positif. If c is not a positive value, then the y-intercept does not have a positive value. Kontrapositif/ Contrapositive: Jika pintasan-y tidak mempunyai nilai positif, maka c bukan bernilai positif. If the y-intercept does not have a positive value, then c is not a positive value. (c) Akas/ Converse: Jika sebuah poligon ialah sebuah segi tiga, maka poligon itu mempunyai tiga sisi. If a polygon is a triangle, then the polygon has three sides. Songsangan/ Inverse: Jika sebuah poligon tidak mempunyai tiga sisi, maka poligon itu bukan sebuah segi tiga. If a polygon does not have three sides, then the polygon is not a triangle. Kontrapositif/ Contrapositive: Jika sebuah poligon bukan sebuah segi tiga, maka poligon itu tidak mempunyai tiga sisi. If a polygon is not a triangle, then the polygon does not have three sides. 11 , , (a) Jika 102 ≠ 2, maka 1002 ≠ 20. If 102 ‡ 2, then 1002 ‡ 20. (i) (ii) (iii) (b) Jika 1002 ≠ 20, maka 102 ≠ 2. If 1002 ‡ 20, then 102 ‡ 2. (i) (ii) (iii) (c) Jika 1002 = 20, maka 102 = 2. If 1002 = 20, then 102 = 2. (i) (ii) (iii) 12 (a) Burung unta tidak boleh terbang. Ostrich cannot fly. (b) Buaya mempunyai empat kaki. Crocodiles have four legs. (c) 6 tidak boleh dibahagi tepat dengan 4. 6 is not divisible by 4. 3.2 Hujah Arguments 1 (a) Pernyataan umum/ General statement (b) Pernyataan umum/ General statement CONTOH

Matematik Tingkatan 4 Jawapan J14 (c) Pernyataan khusus/ Specific statement (d) Pernyataan khusus/ Specific statement (e) Pernyataan umum/ General statement 2 (a) Hujah deduktif/ Deductive argument (b) Hujah induktif/ Inductive argument (c) Hujah induktif/ Inductive argument 3 (a) Tidak sah kerana tidak mematuhi bentuk hujah deduktif yang sah. Tidak munasabah kerana kesimpulan palsu. Not valid because it does not comply with a valid form of deductive argument. Not sound because the conclusion is false. (b) Sah dan munasabah./ Valid and sound. (c) Tidak sah kerana tidak mematuhi bentuk hujah deduktif yang sah. Maka, hujah ini tidak munasabah. Not valid because it does not comply with a valid form of deductive argument. Thus, this argument is not sound. (d) Tidak sah dan munasabah kerana tidak mematuhi bentuk hujah deduktif yang sah dan kesimpulannya palsu. Not valid and sound because it does not comply with a valid form of deductive argument and its conclusion is false. 4 (a) 2 ialah satu faktor bagi 18. 2 is a factor of 18. (b) y ≠ 0. (c) 82 = 64. (d) Televisyen tidak terdiri daripada sel-sel. Television is not made up of cells. 5 (a) Jika hari itu hujan, maka Susie akan membawa payung. If the day rains, then Susie will bring an umbrella. (b) Pentagon bukan sebuah segi tiga. Pentagon is not a triangle. (c) Semua kucing mempunyai misai. All cats have whiskers. (d) Tina pergi melancong. Tina goes on a holiday. 6 (a) Hujah ini lemah dan tidak meyakinkan kerana Premis 1 dan kesimpulan adalah palsu. This argument is weak and not cogent because Premise 1 and conclusion are false. (b) Hujah ini kuat dan meyakinkan kerana semua premis dan kesimpulan adalah benar. This argument is strong and cogent because all premises and conclusion are true. (c) Hujah ini lemah dan tidak meyakinkan kerana premis adalah benar tetapi kesimpulan mungkin palsu. This argument is weak and not cogent because the premises are true but the conclusion maybe false. (d) Hujah ini kuat dan meyakinkan kerana semua premis dan kesimpulan adalah benar. This argument is strong and cogent because all premises and conclusion are true. 7 (a) 3n + 8, n = 1, 2, 3, 4, … (b) 100 – n2 , n = 1, 2, 3, 4, … (c) n2 – n, n = 1, 2, 3, 4, … (d) Poligon dibina daripada garis lurus. Polygons are made of straight lines. (e) Unit laju am ialah jarak per unit masa. The general unit of speed is distance per unit time. 8 (a) 3n + 50, n = 1, 2, 3, 4 (b) f(x) = 2x2 – 2x + 3 (c) (i) Ah Chong (ii) Maka, Siva lebih tinggi daripada Mariam dan Ah Chong. Then, Siva is taller than Mariam and Ah Chong. (d) (i) 4n, n = 1, 2, 3, 4, … (ii) 8 (iii) Nilai pH meningkat 1 pada setiap penambahan 4 titis alkali. The value of pH increases 1 at every addition of 4 drops of alkali. (e) (i) Jika Nadia tidak mempunyai RM80, maka dia tidak akan membeli sebuah beg tangan. If Nadia does not have RM80, then she will not buy a handbag. Kesimpulan deduktif/ Deductive conclusion: Nadia mempunyai lebih daripada RM50 tetapi kurang daripada RM80. Nadia has more than RM50 but less than RM80. (ii) RM50+ RM20 = RM70 Praktis Kendiri Kertas 2 Bahagian A/ Section A 1 (a) Pernyataan/ Statement (b) Jika (A ∩ B) ⊂ B, maka (A ∩ B) ⊂ A. If (A ∩ B) ⊂ B, then (A ∩ B) ⊂ A. Benar/ True (c) Kubus mempunyai 6 permukaan. Cube has 6 surfaces. 2 (a) Palsu/ False (b) Implikasi 1/ Implication 1: Jika x < y, maka x + 4 < y + 4. If x < y, then x + 4 < y < 4. Implikasi 2/ Implication 2: Jika x + 4 < y + 4, maka x < y. If x + 4 < y + 4, then x < y. (c) 50 – 5n, n = 0, 1, 2, 3, … 3 (a) Palsu/ False (b) Abu harus mematuhi peraturan sekolah. Abu must follow the school rules. CONTOH

Matematik Tingkatan 4 Jawapan J15 (c) Hujah ini lemah dan tidak meyakinkan kerana semua premis adalah benar tetapi kesimpulan adalah palsu. This argument is weak and not cogent because all premises are true but the conclusion is false. Bahagian B/ Section B 4 (a) (i) Palsu/ False (ii) Palsu/ False (iii) Benar/ True (b) Implikasi 1/ Implication 1: Jika x M 0, maka x ialah satu nombor asli. If x M 0, then x is a natural number. Implikasi 2/ Implication 2: Jika x ialah satu nombor asli, maka x M 0. If x is a natural number, then x M 0. (c) Akas/ Converse: Jika x = 0 atau ‒2, maka x2 + 2x = 0. If x = 0 or –2, then x2 + 2x = 0. Songsangan/ Inverse: Jika x2 + 2x ≠ 0, maka x ≠ 0 atau ‒2. If x2 + 2x ‡ 0, then x ‡ 0 or –2. (d) Isi padu silinder/ Volume of cylinder: π(72 )(14) = 686π cm3 Bab 4 4.1 Persilangan Set Intersection of Sets 1 (a) (i) P = { 5 , 7, 8, 9, 11, 12 , 13, 15} R = {5, 12 , 17} ∴ P ∩ R = {5, 12} (ii) n(P ∩ R) = 2 (b) (i) Q = {7, 9, 11, 12 , 14} R = {5, 12 , 17} ∴ (Q ∩ R) = {12} (ii) n(Q ∩ R) = 1 (c) (i) P = {5, 7, 8, 9, 11, 12 , 13, 15} Q = {7, 9, 11, 12 , 14} R = {5, 12 , 17} ∴ (P ∩ Q ∩ R) = {12} (ii) n(P ∩ Q ∩ R) = 1 2 (a) P = {14 , 15, 16 , 17, 19, 20 } Q = {14 , 16 , 18, 20 } ∴ (P ∩ Q) = {14, 16, 20} (b) P = {14, 15 , 16 , 17 , 19, 20 } R = {15 , 16 , 17 , 20 } ∴ (P ∩ R) = {15, 16, 17, 20} (c) Q = {14, 16 , 18, 20 } R = {15, 16 , 17, 20 } ∴ (Q ∩ R) = {16, 20} (d) P = {14, 15, 16 , 17, 19, 20 } Q = {14, 16 , 18, 20 } R = {15, 16 , 17, 20 } ∴ (P ∩ Q ∩ R) = (16, 20) 3 (a) ξ = {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30} A = {10, 15, 20, 25, 30} B = {10, 15, 30} C = {10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30} (i) A ∩ B = {10, 15, 30} (ii) A ∩ C = {10, 20, 30} (iii) B ∩ C = {10, 30} (iv) A ∩ B ∩ C = {10, 30} (b) •25 •15 •20 A B C •10 •12 •22 •14 •24 •16 •26 •18 •28 •30 4 (a) (P ∩ Q) = {12, 18} (P ∩ Q)� = {10, 11, 13, 14, 15, 16, 17, 19, 20} n(P ∩ Q)� = 9 (b) (P ∩ R) = {12} (P ∩ R)� = {10, 11, 13, 14, 15, 16, 17, 18, 19, 20} n(P ∩ R)� = 10 (c) (Q ∩ R) = {12, 14, 20} (Q R)� = {10, 11, 13, 15, 16, 17, 18, 19} n(Q R)� = 8 (d) (P ∩ Q ∩ R) = {12} (P ∩ Q ∩ R)� = {10, 11, 13, 14, 15, 16, 17, 18, 19, 20} n(P ∩ Q ∩ R)� = 10 5 (a) Katakan ξ = {jumlah murid}, M = {murid suka merah jambu}, U = {murid suka ungu} Let ξ = {total number of students}, M = {students who love pink}, U = {students who love purple} 15 7 5 3 M U ξ (i) Bilangan murid yang menyukai merah jambu atau ungu: Number of students who love pink or purple: 7 + 5 + 3 = 15 orang murid/ students CONTOH

Matematik Tingkatan 4 Jawapan J16 (ii) Bilangan murid yang tidak menyukai keduadua warna tersebut: Number of students who do not love both colours: 30 – 15 = 15 orang murid/ students (b) (i) ξ 96 Bas Bus Kereta Car Basikal Bicycle 16 64 20 12 92 92 8 (ii) (a) 140 – 96 – 16 – 20 = 8 orang murid/ students (b) 400 – 96 – 16 – 64 – 20 – 8 – 12 – 92 = 92 orang murid/ students (c) (i) Katakan K sebagai murid yang suka kucing, D sebagai murid yang suka anjing dan A sebagai murid yang suka arnab. Let K be students who like cats, D be students who like dogs and A be students who like rabbits. ξ K D 27% 28% 2% 20% 2% 3% 1% A 17% (ii) Bilangan murid yang suka satu binatang sahaja: The number of students who likes only one animal: 28 + 20 + 27 100 × 4 500 = 75 100 × 4 500 = 3 375 orang murid/ students (iii) Bilangan murid yang tidak suka ketiga-tiga jenis binatang itu: The number of students who does not like those three animals: 17 100 × 4 500 = 765 orang murid/ students 4.2 Kesatuan Set Union of Sets 1 (a) (i) P ∪ R = {10, 11, 13, 14, 15, 17, 18, 19, 20} (ii) P Q R •13 •10 •11 •14 •15 •12 •17 •19 •16 •20 •18 (b) (i) Q ∪ R = {12, 13, 15, 16, 18, 19} (ii) P Q R •13 •10 •11 •14 •15 •12 •17 •19 •16 •20 •18 (c) (i) P ∪ Q ∪ R = {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} (ii) P Q R •13 •10 •11 •14 •12 •17 •16 •20 •18 •19 •15 2 (a) (P ∪ R) = {1, 2, 3, 4, 5, 6, 7, 9} (P ∪ R)� = {8, 10} n(P ∪ R)� = 2 (b) (Q ∪ R) = {1, 3, 5, 6, 7} (Q ∪ R)� = {2, 4, 8, 9, 10} n(Q ∪ R)� = 5 (c) (P ∪ Q ∪ R) = {1, 2, 3, 4, 5, 6, 7, 9} (P ∪ Q ∪ R)� = {8, 10} n(P ∪ Q ∪ R)� = 2 3 ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} M = {1, 2, 4, 5, 10, 20} N = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} P = {4, 8, 12, 16, 20} (M ∪ N) = {1, 2, 4, 5, 6, 8, 10, 12, 14, 16, 18, 20} (M ∪ N)� = {3, 7, 9, 11, 13, 15, 17, 19} (M ∪ P) = {1, 2, 4, 5, 8, 10, 12, 16, 20} (M ∪ P)� = {3, 6, 7, 9, 11, 13, 14, 15, 17, 18, 19} (N ∪ P) = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} (N ∪ P)� = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} (M ∪ N ∪ P) = {1, 2, 4, 5, 6, 8, 10, 12, 14, 16, 18, 20} CONTOH (M ∪ N ∪ P)� = {3, 7, 9, 11, 13, 15, 17, 19}

Matematik Tingkatan 4 Jawapan J17 ξ M •5 •1 •19 •13 •15 •17 •11 •3 •7 •2 •9 •4 •10 •20 •8 •12 N P •14 •6 •18 •16 4 ξ = {jumlah murid/ total students} P = {pingpong/ ping-pong} S = {bola sepak/ soccer} B = {badminton/ badminton} ξ 32 8 35 20 3 2 P S B Bilangan murid yang tidak suka mana-mana aktiviti: The number of students who does not like any activities: 100 – 32 – 8 – 35 – 2 – 20 = 3 orang murid/ students 4.3 Gabungan Operasi Set Combined Operations on Sets 1 (a) (P ∩ Q) ∪ R = {1, 3, 7, 10, 12, 14, 16, 17, 19, 20, 24} (b) P ∩ (Q ∪ R) = {10, 12, 20, 24} (c) (P ∩ R) ∪ Q = {4, 10, 12, 13, 16, 17, 18, 20, 24, 25, 29} 2 (a) A C ξ B (b) ξ H F G (c) ξ X Y Z 3 (a) ξ P Q R (b) ξ Q R P (c) ξ P R Q 4 (a) (i) ξ X Z Y (ii) ξ X Z CONTOH Y

Matematik Tingkatan 4 Jawapan J18 (b) (i) Jumlah bilangan murid: Total number of students: 5 + 6 + 3 + 4 + 8 + 7 + 7 = 40 orang murid/ students (ii) Bilangan murid yang suka dua jenis sukan: The number of students who likes two types of sports: 6 + 4 + 7 = 17 orang murid/ students (iii) Bilangan murid yang suka sejenis sukan sahaja: The number of students who likes only a type of sport: 5 + 8 + 7 = 20 orang murid/ students (iv) Bilangan murid yang suka bermain badminton atau bola sepak: The number of students who likes to play badminton or football: 5 + 6 + 3 + 4 + 8 + 7 = 33 orang murid/ students (c) (i) 22 + 27 – x = 50 – 7 49 – x = 43 x = 49 – 43 = 6 (ii) M S ξ 7 16 6 21 (iii) 16 + 21 = 37 orang murid/ students (d) (i) x + 3 + 2 + 2 = 2(y + 1 + 2 + 2) x + 7 = 2(y + 5) x + 7 = 2y + 10 x = 2y + 3 …… 12 + 3 + 2 + 1 + x + 2 + y = 32 x + y + 20 = 32 x = 12 – y …… ➁ ➀ = ➁: 2y + 3 = 12 – y 3y = 9 y = 3 x = 12 – 3 = 9 (ii) 12 + 9 + 3 = 24 orang/ people Praktis Kendiri Kertas 1 Bahagian A/ Section A 1 A = {2, 3, 5, 10, 15} B = {5} C = {2, 6, 8, 11, 15} (A ∩ B�) = {2, 3, 10, 15} (A ∩ B�) ∪ C = {2, 3, 6, 8, 10, 11, 15} Jawapan/ Answer: D 2 6 + x + 5 + 8 + 4 = 30 x = 30 – 23 = 7 n[X ∩ (Y ∩ Z)�] = 6 + 7 + 4 = 17 Jawapan/ Answer: C 3 ξ = {1, 2, 3, 5, 6, 7, 8, 9, 10, 11} (A ∩ B) = {2, 3, 6, 7, 9} (A ∩ B)� = {1, 5, 8, 10, 11} (A ∩ B)� ∪ C = {1, 5, 6, 7, 8, 9, 10, 11} Jawapan/ Answer: C 4 C A B 3x 3x x x – 1 2 5 – x 9 – x 3x + (9 – x) + (5 – x) = 17 x + 14 = 17 x = 3 n[B ∩ (A ∪ C)�] = 5 – x = 5 – 3 = 2 Jawapan/ Answer: A 5 n(X ∩ Y�) = 6 + x, n[(Z ∪ Y) X�] = 7 6 + x = 7 x = 1 n(ξ) = x + x + 6 + 2x + 7 = 4x + 13 = 4(1) + 13 = 4 + 13 = 17 Jawapan/ Answer: D 6 (P ∩ Q)� : P Q ξ A : P� Q� P� ∩ Q� ∩ = P P Q P CONTOH Q Q

Matematik Tingkatan 4 Jawapan J19 B : ∩ = P P Q Q� P Q P ∩ Q� P Q C : ∪ = P P Q Q� P Q P ∪ Q� P Q D : ∪ = P� P Q Q� P Q P� ∪ Q� P Q Jawapan/ Answer: D 7 G = {1, 2, 3, 6, 9, 18} H = {1, 2, 3, 4, 6, 12} G ∪ H = {1, 2, 3, 4, 6, 9, 12, 18} ∴ n(ξ) = 8 Jawapan/ Answer: C 8 Q R P P� ∩ R Q R P (P� ∩ R) ∪ Q Q R P Q ∪ = Jawapan/ Answer: A 9 A: A C B B: A C B C: A C B Jawapan/ Answer: D 10 X ∩ Z = Z Z di dalam/ inside X Y = (X ∪ Z)� Y tidak berkaitan dengan set X dan set Y/ Y is not related to set X and set Y Jawapan/ Answer: C Kertas 2 Bahagian A/ Section A 1 (a) P = { 10 , 12 , 14, 16 , 18, 20} Q = { 10 , 11, 12 , 13, 15, 16 } P ∩ Q = {10, 12, 16} (b) ξ P Q •19 •17 •14 •18 •20 •11 •13 •15 •10 •12 •16 2 (a) P Q R ξ (b) (i) B C A (ii) B A C 3 (a) C = {32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52} D = {5, 10, 15, 20, 25, 30, 35, 40, 45} E = {1, 2, 4, 5, 10, 20, 25, 50, 100} F = {1, 4, 9, 16, 25, 36, 49} (b) (i) D = {5, 10, 15, 20, 25 , 30, 35, 40, 45} E = {1, 2, 4, 5, 10, 20, 25 , 50, 100} F = {1, 4, 9, 16, 25 , 36, 49} D ∩ E ∩ F = {25} (ii) A = {31, 32, 33, 34, 35 , 36, 37, 38, 39, 40 , 41, 42, 43, 44, 45 , 46, 47, 48, 49, 50} D = {5, 10, 15, 20, 25, 30, 35 , 40 , 45 } A ∩ D = {35, 40, 45} (A ∩ D) ∪ E = {1, 2, 4, 5, 10, 20, 25, 35, 40, 45, 50,100} CONTOH

Matematik Tingkatan 4 Jawapan J20 (iii) (B ∩ E)� = {1, 2, 4, 5 , 9, 10, 15, 16, 20, 25 , 30, 31, 32, 33, 34 , 35 , 36 , 37, 38 , 39, 40 , 41, 42, 43, 44 , 45 , 46 , 47 , 48 , 49, 52 , 100} (C ∪ D) = {5 , 10 , 15 , 20 , 25 , 30, 32, 34, 35, 36, 38, 40 , 42 , 44 , 45 , 46 , 48 , 50, 52} (B ∩ E)� ∩ (C ∪ D) = {5, 10, 15, 20, 25, 30, 32, 34, 35, 36, 38, 40, 42, 44, 45, 46, 48, 52} (iv) (A ∩ B ∩ D) = {40, 45} (D ∩ E ∩ F) = {25} (A ∩ B ∩ D) ∪ (D ∩ E ∩ F) = {25, 40, 45} Bahagian B/ Section B 4 (a) Q P R x n(Q ∩ R) = n(P ∩ Q ∩ R) = x n(R) + n(Q) – n(Q ∩ R) = n(Q ∪ R) (27 – 1 – 14) + (27 – 1) – x = 30 12 + 26 – x = 30 x = 38 – 30 = 8 (b) Q 10 4 8 5 13 P R (c) 10 + 13 = 23 orang murid/ students (d) (i) P Q R 10 + 8 + 5 + 13 = 36 orang murid/ students (ii) P Q R 10 + 4 + 8 = 22 orang murid/ students 5 (a) K I B 6 ξ 11 – x 9 – x 7 – x x + 5 x + 3 2 + x x n(K) = 23 – (11 – x) – x – (9 – x) = 23 – 11 + x – x – 9 + x = 3 + x n(I) = (23 – 3) – (11 – x) – x – (7 – x) = 20 – 11 + x – x – 7 + x = 2 + x n(B) = (20 + 1) – (9 – x) – x – (7 – x) = 21 – 9 + x – x – 7 + x = 5 + x (b) (x + 3) + (11 – x) + x + (9 – x) + (2 + x) + (7 – x) + (x + 5) + 6 = 48 43 + x = 48 x = 5 (c) 6 + (5 + 3) + (2 + 5) + (5 + 5) = 31 Bab 5 5.1 Rangkaian Network 1 (a) Darjah bucu/ Degree of vertex (b) Bintik/ Dot (c) Graf/ Graph (d) Graf mudah/ Simple graph 2 (a) (b) (c) 3 (a) V = {1, 2, 3, 4} E = {(1, 2), (1, 4), (2, 3), (3, 4)} n(V) = 4 n(E) = 4 Bilangan darjah/ Sum of degrees: 2(4) = 8 (b) V = {J, K, L, M, N} E = {(J, K), (J, L), (K, L), (L, M), (L, N), (M, N)} n(V) = 5 n(E) = 6 Bilangan darjah/ Sum of degrees: 2(6) = 12 (c) V = {1, 2, 3, 4, 5, 6, 7} E = {(1, 2), (1, 7), (2, 3), (2, 7), (3, 4), (3, 6), (4, 5), (5, 6), (6, 7)} n(V) = 7 n(E) = 9 Bilangan darjah/ Sum of degrees: 2(9) = 18 CONTOH