1 © Penerbitan Pelangi Sdn. Bhd. JAWAPAN Nombor Nisbah Rational Numbers BAB 1 1. Positif / Positives: 8, 9, 20, 5 051 Negatif / Negatives: –5, –1, –54, –125 2. (a) (i) Pergerakan kereta / Movement of the car: 87 (ii) Pergerakan van / Movement of the van: –46 (b) (i) Pergerakan lif A / Movement of the lift A: –5 (ii) Pergerakan lif B / Movement of the lift B: 3 (c) (i) Suhu di Oslo / Temperature in Oslo: –3 (ii) Suhu di Kuala Lumpur / Temperature in Kuala Lumpur: 35 3. 8 ü 4.2 –3 ü 1 4 7 ü 4. (a) –9 –6 –3 0 3 6 9 12 –3 –3 –3 +3 +3 +3 +3 (b) –15 –10 –5 0 5 10 15 20 –5 –5 –5 +5 +5 +5 +5 (c) –6 –4 –2 0 2 4 6 8 –2 –2 –2 +2 +2 +2 +2 5. (a) –7 –4 –2 0 5 6 9 Tertib menaik: –7, –4, –2, 0, 5, 6, 9 Ascending order: (b) –6 –4 –1 0 1 2 5 Tertib menurun: 5, 2, 1, 0, –1, –4, –6 Descending order: (c) –6 –3 0 1 4 6 8 Tertib menaik: –6, –3, 0, 1, 4, 6, 8 Ascending order: (d) –7 –5 –4 –3 –1 2 5 Tertib menurun: 5, 2, –1, –3, –4, –5, –7 Descending order: Mahir Diri –3°C –1°C 0°C 2°C 4°C Jawapan / Answer: D 6. (a) +6 + (+4) = 6 + 4 = 10 (b) −7 + 9 = 2 (c) 12 + (+5) = 12 + 5 = 17 (d) − 11 + 4 = −7 (e) − 15 + 9 = −6 7. (a) 4 + (–8) = 4 – 8 = –4 (b) −2 + (−9) = −2 − 9 = −11 (c) −6 + (−6) = −6 − 6 = − 12 (d) 14 − (+7) = 14 − 7 = 7 (e) 1 − (+5) = 1 − 5 = −4
© Penerbitan Pelangi Sdn. Bhd. 2 Matematik Tingkatan 1 Jawapan © Penerbitan Pelangi Sdn. Bhd. 2 8. (a) (i) 6 – (–11) = 6 + 11 = 17 (ii) 9 − (−3) = 9 + 3 = 12 (b) (i) −2 − (−10) = −2 + 10 = 8 (ii) −15 − (−4) = −15 + 4 = −11 9. (a) (i) 8 × (–7) = – (8 × 7 ) = –56 (ii) 11 × (−5) = −(11 × 5) = −55 (b) (i) (−9) × 8 = −(9 × 8) = −72 (ii) (−18) × 10 = −(18 × 10) = −180 (c) (i) (−4) × (−5) = +(4 × 5) = 20 (ii) (−6) × (−12) = +(6 × 12) = 72 10. (a) (i) 64 ÷ (–8) = – (64 ÷ 8 ) = –8 (ii) 132 ÷ (–12) = –(132 ÷ 12) = –11 (b) (i) –88 ÷ 8 = –(88 ÷ 8) = –11 (ii) –42 ÷ 6 = –(42 ÷ 6) = –7 (c) (i) –170 ÷ (–17) = +(170 ÷ 17) = 10 (ii) –54 ÷ (–6) = +(54 ÷ 6) = 9 11. (a) (i) 5 × [2 + (–7)] = 5 × (–5) = – (5 × 5 ) = –25 (ii) 12 × (– 4 – 3) = 12 × (–7) = –(12 × 7) = –84 (iii) 9 × [6 – (–1)] = 9 × (6 + 1) = 9 × 7 = 63 (iv) –6 × [–3 + (–7)] = –6 × (–3 – 7) = –6 × (–10) = +(6 × 10) = 60 (b) (i) –8 ÷ (–10 + 8) = –8 ÷ (–2) = +(8 ÷ 2) = 4 (ii) –16 ÷ (– 4 + 6) = –16 ÷ 2 = –(16 ÷ 2) = –8 (iii) 20 ÷ (–7 + 2) = 20 ÷ (–5) = –(20 ÷ 5) = –4 (iv) 18 ÷ [4 + (–10)] = 18 ÷ (4 – 10) = 18 ÷ (–6) = –(18 ÷ 6) = –3
3 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan 3 © Penerbitan Pelangi Sdn. Bhd. 12. (a) (i) (–7 + 3) × 9 = –4 × 9 = – ( 4 × 9) = –36 (ii) (–2 – 4) × (–7) = –6 × (–7) = +(6 × 7) = 42 (b) (i) –12 + (–36) ÷ 12 = –12 + (–3) = –12 – 3 = –15 (ii) 20 – 48 ÷ (–16) = 20 – (–3) = 20 + 3 = 23 13. (a) (i) 6 × (–2) + 1 = –12 + 1 = –11 (ii) 7 × 4 + (–8) = 28 + (–8) = 28 – 8 = 20 (iii) (–6) × (–2) – 15 = 12 – 15 = –3 (iv) –9 × 4 – 10 = –36 – 10 = –46 (b) (i) –42 ÷ 2 + 8 = –(42 ÷ 2) + 8 = –21 + 8 = –13 (ii) –70 ÷ (–10) – 4 = +(70 ÷ 10) – 4 = 7 – 4 = 3 (iii) 32 ÷ (–4) – (–1) = –(32 ÷ 4) + 1 = –8 + 1 = –7 (iv) 108 ÷ 9 + (–4) = 12 – 4 = 8 14. (a) (i) (–15 + 3) ÷ 4 – (–6) = –12 ÷ 4 – (–6) = –3 + 6 = 3 (ii) –5 + 4 × (–2 + 7) = –5 + 4 × 5 = –5 + 20 = 15 (b) (i) 5 × (–2) + 14 ÷ 2 = –10 + 14 ÷ 2 = –10 + 7 = –3 (ii) –20 ÷ (–5) – 6 × 2 = 4 – 6 × 2 = 4 – 12 = –8 15. (a) – 4 – 5 1 – (–2) = –9 1 + 2 = –93 = –3 (b) –7 + (–3) –2 – 3 = –7 – 3 –2 – 3 = –10 –5 = 2 (c) 3 + (–15) 9 + (–3) = 3 – 15 9 – 3 = –126 = –2
© Penerbitan Pelangi Sdn. Bhd. 4 Matematik Tingkatan 1 Jawapan 16. (a) Beza suhu / Difference in temperature = +58°C – (–6°C) = 58°C + 6°C = 64°C (b) Kedudukan lif / Position of the lift = +9 + (+4) + (–10) Tanda +: pergerakan ke atas. Tanda ‒: pergerakan ke bawah. + sign: movement upwards – sign: movement downwards = 9 + 4 – 10 = 13 – 10 = 3 Maka, lif itu berada di tingkat 3. Therefore, the lift is on the 3rd floor. (c) +RM12 500 + (–RM18 850) Tanda +: keuntungan Tanda ‒: kerugian + sign: profit = RM12 500 – RM18 850 – sign: loss = –RM6 350 Maka, syarikat itu mengalami kerugian RM6 350 dalam masa dua bulan itu. Therefore, the company suffered a loss of RM6 350 in the two months. 17. (a) 7 9 (b) 1 1 2 (c) 3 4 (d) –2 5 6 (e) – 12 7 (f) – 3 8 18. (a) – 1 3 , 5 6 , 1 1 6 , – 8 3 , –1 2 3 ↓ ↓ ↓ ↓ ↓ – 2 6 , 5 6 , 7 6 , – 16 6 , – 10 6 Tanda +: suhu di atas takat beku air. Tanda ‒: suhu di bawah takat beku air. + sign: temperature is above freezing point of water – sign: temperature is below freezing point of water Menyamakan penyebut bagi semua pecahan. Equalise the denominators of all fractions – – 16 6 – – 10 6 – – 2 6 – 5 6 – 7 6 – – 8 3 –1– 2 3 – – 1 3 – 5 6 1– 1 6 Pecahan asal Original fractions Tertib menaik: – 8 3 , –1 2 3 , – 1 3 , 5 6 , 1 1 6 Ascending order (b) 13 2 , –1 1 6 , – 5 6 , 2 1 2 , – 9 2 ↓ ↓ ↓ ↓ ↓ 39 6 , – 7 6 , – 5 6 , 15 6 , – 27 6 – – 27 6 – – 7 6 – – 5 6 – 15 6 – 39 6 – 13 2 – – 9 2 –1– 1 6 2– 1 2 – – 5 6 Pecahan asal Original fractions Tertib menaik: – 9 2 , –1 1 6 , – 5 6 , 2 1 2 , 13 2 Ascending order 19. (a) – 9 4 , 1 1 2 , –1 3 4 , 5 2 , – 7 2 ↓ ↓ ↓ ↓ ↓ – 9 4 , 6 4 , – 7 4 , 10 4 , – 14 4 – – 14 4 – – 9 4 – – 7 4 – 6 4 – 10 4 – 5 2 7 2 – – 9 4 1– 1 2 – – –1– 3 4 Pecahan asal Original fractions Tertib menurun: 5 2 , 1 1 2 , –1 3 4 , – 9 4 , – 7 2 Descending order (b) 7 9 , – 11 3 , 1 1 9 , –1 1 3 , – 2 3 ↓ ↓ ↓ ↓ ↓ 7 9 , – 33 9 , 10 9 , – 12 9 , – 6 9 – – 33 9 – – 12 9 – – 6 9 – 7 9 – 10 9 – 7 9 11 3 –1– 1 3 1– 1 9 – – – – 2 3 Pecahan asal Original fractions Tertib menurun: 1 1 9 , 7 9 , – 2 3 , –1 1 3 , – 11 3 Descending order Menyamakan penyebut bagi semua pecahan. Equalise the denominators of all fractions Menyamakan penyebut bagi semua pecahan. Equalise the denominators of all fractions
5 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan 20. (a) (i) 2 35 + –1 1 10 – – 45 = 135 – 11 10 + 45 = 13 × 2 5 × 2 – 11 10 + 4 × 2 5 × 2 = 26 10 – 11 10 + 8 10 = 26 – 11 + 8 10 = 23 10 = 2 3 10 (ii) –1 89 + – 23 – – 49 = – 179 – 23 + 49 = – 179 – 2 × 3 3 × 3 + 49 = – 179 – 69 + 49 = –17 – 6 + 4 9 = – 199 = –2 19 (b) (i) –2 12 × 43 ÷ – 85 = – 52 × 43 ÷ – 85 = – 52 × 43 × – 58 = 52 × 43 × 58 = 25 12 = 2 1 12 (ii) 2 25 ÷ – 13 × 1 16 = – 125 ÷ 13 × 76 = – 125 × 31 × 76 = – 125 × 31 × 76 = – 425 = –8 25 21. (a) (i) – 34 × –1 18 + 12 = – 34 × – 98 + 12 = – 34 × – 98 + 48 = – 34 × – 58 = + 34 × 58 = 15 32 (ii) 23 – 56 × 1 12 = 46 – 56 × 32 = – 16 × 32 = – 16 × 32 = – 14 2 1 2 1 2 1
© Penerbitan Pelangi Sdn. Bhd. 6 Matematik Tingkatan 1 Jawapan (b) (i) 58 ÷ – 3 10 – 1 15 = 58 ÷ – 3 10 – 65 = 58 ÷ – 3 10 – 12 10 = 58 ÷ – 15 10 = 58 ÷ – 32 = 58 × – 23 = 54 × – 13 = – 5 12 (ii) 23 – – 49 ÷ –2 12 = 23 + 49 ÷ – 52 = 69 + 49 ÷ – 52 = 109 × – 25 = – 49 22. (a) (i) – 58 + 35 × 1 14 – – 12 = – 58 + 35 × 1 14 + 12 = – 58 + 35 × 54 + 12 = – 58 + 34 + 12 = – 58 + 68 + 48 = –5 + 6 + 4 8 = 58 1 4 1 2 (ii) 34 + – 12 – 1 15 ÷ 1 12 = 34 – 12 – 1 15 ÷ 1 12 = 34 – 12 – 65 ÷ 32 = 34 – 12 – 65 × 23 = 34 – 12 – 45 = 15 20 – 10 20 – 16 20 = – 11 20 (b) (i) 1 78 ÷ 59 + – 14 × 1 12 = 1 78 ÷ 59 – 14 × 1 12 = 158 ÷ 59 – 14 × 32 = 158 × 95 – 14 × 32 = 278 – 38 = 248 = 3 (ii) 2 12 × 15 – – 38 ÷ 1 27 = 2 12 × 15 – – 38 ÷ 1 27 = 52 × 15 – – 38 ÷ 97 = 12 + 38 × 79 = 12 + 7 24 = 12 24 + 7 24 = 19 24 1 3 1 1 3 1
7 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan 23. (a) 1 5 Selepas 9 minit After 9 minutes Selepas 4 minit After 4 minutes Aras air asal Initial water level Turun / Dropped 3– cm 1 2 Turun / Dropped 4– cm Perubahan aras air kolam renang Change of water level = –3 1 5 cm – 4 1 2 cm Tanda negatif menunjukkan aras air menurun. Negative sign shows the dropped in water level = – 16 5 cm – 9 2 cm = – 32 10 cm – 45 10 cm = – 77 10 cm = –7 7 10 cm Maka, aras air kolam renang telah turun 7 7 10 cm selepas 9 minit. Therefore, the water level of the swimming pool has dropped 7 7 10 cm after 9 minutes (b) Beza suhu / Difference in temperature = +120 1 5 °C – –10 7 10 °C = 120 1 5 °C + 10 7 10 °C = 120 2 10 °C + 10 7 10 °C = 130 9 10 °C Maka, beza suhu rod itu ialah 130 9 10 °C. Therefore, the difference of the temperature of the rod is 130 9 10°C. Tanda +: suhu di atas takat beku air. Tanda ‒: suhu di bawah takat beku air. + sign: temperature above the freezing point of water. – sign: temperature below the freezing point of water. (c) +1 3 8 juta / million + (–1 3 4 juta / million) = 11 8 juta / million – 7 4 juta / million = 11 8 juta / million – 14 8 juta / million = – 3 8 juta / million Maka, Syarikat Berjaya mengalami kerugian RM 3 8 juta dalam masa dua tahun itu. Therefore, Syarikat Berjaya sufferd a loss of RM 3 8 milllion within those two years. 24. (a) –4.85 –2.004 –0.6 –0.09 1.3 3.07 6.253 Tertib menaik / Ascending order: –4.85, –2.004, –0.6, –0.09, 1.3, 3.07, 6.253 (b) –3.7 –1.59 –0.03 0.789 1.88 2.0 3.09 Tertib menaik / Ascending order: –3.7, –1.59, –0.03, 0.789, 1.88, 2.0, 3.09 25. (a) –1.09 –1.02 –0.7 –0.007 2.9 3.4 3.48 Tertib menurun / Descending order: 3.48, 3.4, 2.9, –0.007, –0.7, –1.02, –1.09 (b) –2.44 –2.09 –1.1 –0.002 0.8 1.28 2.796 Tertib menurun / Descending order: 2.796, 1.28, 0.8, –0.002, –1.1, –2.09, –2.44 Tanda +: keuntungan Tanda ‒: kerugian + sign: profit – sign: loss
© Penerbitan Pelangi Sdn. Bhd. 8 Matematik Tingkatan 1 Jawapan 26. (a) (i) 4.8 + (–5.165) – (–0.08) = 4.8 – 5.165 + 0.08 = –0.365 + 0.08 = –0.285 (ii) –2.30 – 1.752 – (–5.4) = –2.30 – 1.752 + 5.4 = –4.052 + 5.4 = 1.348 (b) (i) –1.46 × (–0.15) ÷ 0.24 = +(1.46 × 0.15) ÷ 0.24 = 0.219 ÷ 0.24 = 0.9125 (ii) 1.84 ÷ (–1.6) × (–2.11) = –(1.84 ÷ 1.6) × (–2.11) = –1.15 × (–2.11) = +(1.15 × 2.11) = 2.4265 27. (a) (i) 1.74 × (–2.274 – 3.176) = 1.74 × (–5.45) = –(1.74 × 5.45 ) = –9.483 (ii) [3.1 – (–1.28)] × 4.5 = (3.1 + 1.28) × 4.5 = 4.38 × 4.5 = 19.71 (b) (i) –5.661 ÷ (–1.24 + 3.04) = –5.661 ÷ 1.8 = –(5.661 ÷ 1.8) = –3.145 (ii) [–1.2 – (–2.24)] ÷ (–0.25) = (–1.2 + 2.24) ÷ (–0.25) = 1.04 ÷ (–0.25) = –(1.04 ÷ 0.25) = –4.16 (c) (i) –2.98 – 0.52 × 1.375 + (–4.88) = –2.98 – 0.52 × 1.375 – 4.88 = –2.98 – 0.715 – 4.88 = –3.695 – 4.88 = –8.575 (ii) 5.43 – (–2.084) – 4.85 × (–2.34) = 5.43 + 2.084 – [4.85 × (–2.34)] = 5.43 + 2.084 + (4.85 × 2.34) = 5.43 + 2.084 + 11.349 = 7.514 + 11.349 = 18.863 (d) (i) 3.78 ÷ 0.3 + 6.54 × (–2.5) = 12.6 + 6.54 × (–2.5) = 12.6 + (–16.35) = 12.6 – 16.35 = –3.75 (ii) 3.32 × (–0.08) + 4.82 × 4.5 = –0.2656 + 4.82 × 4.5 = –0.2656 + 21.69 = 21.4244 28. (a) Perubahan harga saham / Change in share price = RM0.14 + (–RM0.85) + RM0.50 = RM0.14 – RM0.85 + RM0.50 = –RM0.71 + RM0.50 = –RM0.21 Maka, harga saham Syarikat Aman Damai telah turun RM0.21 dalam tiga hari itu. Therefore, the share price of Syarikat Aman Damai decreased by RM0.21 within those three days. (b) Beza jarak antara penyu dengan ikan Difference in distance between the turtle and the fish = –4.27 m – (–7.112 m) = –4.27 m + 7.112 m = 2.842 m Maka, jarak antara penyu dengan ikan ialah 2.842 m. Therefore, the distance between the turtle and the fish is 2.842 m. Tanda +: harga saham naik + sign: share price increase Tanda ‒: harga saham turun – sign: share price decrease Tanda ‒: kedudukan di bawah aras laut – sign: the position is below the sea level
9 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan (c) Panjang kain Ainin – Jumlah panjang kain yang diperlukan Length of Ainin’s cloth – Total length of required cloth = 19 m – 3.82 m × 5 = 19 m – 19.1 m = –0.1 m Maka, kain Ainin tidak mencukupi. Ainin masih kekurangan 0.1 m kain untuk menjahit 5 helai baju.Therefore, Ainin’s cloth is not enough. Ainin still lack 0.1 m of cloth to make 5 shirts. 29. (a) (i) 3 2 3 ü = 11 3 (ii) –2 1 2 ü = – 5 2 (b) (i) 0.7 ü = 7 10 (ii) –0.05 ü = – 5 100 = – 1 20 (c) (i) 1.08 ü = 108 100 = 27 25 (ii) –2.005 ü = – 2 005 1 000 = – 401 200 Tanda ‒ menunjukkan kain itu tidak mencukupi. Masih kekurangan 0.1 m kain. - sign shows that the cloth is not enough. Still lack 0.1 m of cloth 30. (a) (i) – 4.8 + 9 10 – (–2) = –4.8 + 0.9 + 2 = –3.9 + 2 = –1.9 (ii) 1.527 – 3 8 + – 1 2 = 1.527 – 0.375 – 0.5 = 1.152 – 0.5 = 0.652 (b) (i) – 1 6 × 1 1 5 ÷ (–0.7) = – 1 6 × 6 5 ÷ – 7 10 = – 1 5 ÷ – 7 10 = 1 5 × 10 7 = 2 7 (ii) 1.2 × –1 1 2 ÷ 1 2 7 = 12 10 × – 3 2 ÷ 9 7 = – 9 5 × 7 9 = – 7 5 = – 1 2 5 (c) (i) 1 3 × –2 1 2 + 0.25 = 1 3 × – 5 2 + 1 4 = 1 3 × – 10 4 + 1 4 = 1 3 × – 9 4 = – 3 4 6 3 5 1 1 3
© Penerbitan Pelangi Sdn. Bhd. 10 Matematik Tingkatan 1 Jawapan (ii) – 1 10 + 0.75 ÷ 4 5 = – 1 10 + 75 100 ÷ 4 5 = – 10 100 + 75 100 ÷ 4 5 = 65 100 ÷ 4 5 = 65 100 × 5 4 = 13 16 13 1 4 20 (d) (i) 5 8 × (2.3 – 0.06) + –1 3 4 = 0.625 × (2.3 – 0.06) – 1.75 = 0.625 × 2.24 – 1.75 = 1.4 – 1.75 = –0.35 (ii) 3.6 × 1 2 + (–5.4) ÷ 1 1 4 = 3.6 × 0.5 – 5.4 ÷ 1.25 = 1.8 – 5.4 ÷ 1.25 = 1.8 – 4.32 = –2.52 Mahir Diri 1 3 4 × (2.45 – 0.25) – –2 1 2 = 1.75 × (2.45 – 0.25) + 2.5 = 1.75 × 2.20 + 2.5 = 3.85 + 2.5 = 6.35 Jawapan / Answer: C Hitung operasi darab / Calculate the multiplication operation Hitung operasi dalam kurungan / Calculate the operation in the bracket Tukar kepada perpuluhan / Change to decimals Langkah ini sama dengan pilihan jawapan C This step is same as the answer in C. 31. (a) 1.24 juta / million + (–0.228 juta / million) + 1 2 5 juta / million = 1.24 juta / million – 0.228 juta / million + 1.4 juta / million = 1.012 juta / million + 1.4 juta / million = 2.412 juta / million Maka, syarikat itu memperoleh keuntungan RM2.412 juta dalam tiga tahun. Therefore, the company gain a profit of RM2.412 million in three years. (b) 3.4 m – 6 × 5 8 m = 3.4 m – 6 × 0.625 m = 3.4 m – 3.75 m = –0.35 m Maka, reben Lily adalah tidak mencukupi. Lily masih kekurangan 0.35 m reben. Therefore, Lily's ribbon is not enough. She is still lacking 0.35 m of ribbon. 32. Dasar sungai The river bed Selepas 9 minit After 9 minutes Kedudukan baru ikan Fish's new position Aras baru sungai The new water level 5.9 cm 14.6 cm 162 cm Aras asal sungai The initial water level Peningkatan aras sungai / Increase in the river's water level = 2.4 cm + 3 1 2 cm = 2.4 cm + 3.5 cm = 5.9 cm Tinggi asal aras sungai / Initial height of the river's water level = 162 cm + 14.6 cm – 5.9 cm = 176.6 cm – 5.9 cm = 170.7 cm
11 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan Praktis Masteri 1 Bahagian A 1. P = –9 Q = 6 P + Q = –9 + 6 = –3 Jawapan / Answer: B 2. –4 000 –400 –40 –4 4 Jawapan / Answer: C 3. 3 – – 4 5 ÷ 2 2 3 = 3 – – 4 5 – 8 3 = 3 + 4 5 ÷ 8 3 = 3 + 4 5 × 3 8 Jawapan / Answer: D 4. Jika suhu pertama lebih tinggi dari suhu kedua, maka suhu pertama If the first temperature is higher than the second temperature, thus the first temperature = 25°C + 35°C = 60°C (tiada dalam pilihan jawapan) (not in the answers optional) Jika suhu pertama lebih rendah dari suhu kedua, maka suhu pertama If the first temperature is lower than the second temperature, thus the first temperature = 25°C – 35°C = –10°C Jawapan / Answer: C Bahagian B 1. (a) 8.3 –8 0 1.2 4 2 1 2 1 3 14 (b) 92 × (34 – 12) = 92 × 12 – 92 × 34 (18 + 57) + 29 = 18 + (29 + 57) ü 92 × (34 – 12) = 92 × 12 – 92 × 34 () Selesaikan operasi di dalam kurungan terlebih dahulu. (18 + 57) + 29 = 18 + (29 + 57) (ü) Penukaran tanda kurungan tidak menjejaskan operasi kerana hanya melibatkan operasi +. 2. (a) 11 (i) –7 + 18 (ii) 1 2 + 6.5 + 4 1 2 + 6.5 + = 11 = 11 – 1 2 – 6.5 = 4 + 18 = 11 = 11 – 18 = –7 (b) –15, –10 , –5, 0 , 5, 10 –15 + 5 = –10 –5 + 5 = 0 +5 Bahagian C 1. (a) –16 – (–14) –2 = –16 + 14 –2 = –2 –2 = 1 (b) Integer negatif • • P = –2 Negative integer Pecahan positif • • Q = –0.5 Positive fraction Perpuluhan negatif • • S = 1 2 Negative decimals (c) 1 —5 , mempunyai jenis darah A+/ have blood type A+ 1 —3 1 — 3 , kanak-kanak lelaki atau 2 orang / boys or 2 person Bilangan kanak-kanak yang mempunyai jenis darah A+ The number of children that have blood type A+ = 2 × 3 = 6 Bilangan kanak-kanak yang mempunyai jenis darah O The number of children that have blood type O = 6 × 4 = 24
© Penerbitan Pelangi Sdn. Bhd. 12 Matematik Tingkatan 1 Jawapan Faktor dan Gandaan Factors and Multiples BAB 2 1. (a) 17 ü (b) 21 (c) 31 ü (d) 37 ü (e) 49 Mahir Diri Nombor ganjil antara 37 dengan 43 ialah 39 dan 41. 39 boleh dibahagi tepat dengan 3. Maka, nombor perdana antara 37 dengan 43 ialah 41. Nilai y dan nilai z boleh diperoleh dengan cara yang sama. The odd number between 37 and 43 is 39 and 41. 39 is divisible by 3. Thus, the prime number between 37 and 43 is 41. The values of y and z can be determined by using the same method. 31, 37, x, 43, 47, y, 59, z, 67 41 53 61 x = 41, y = 53, z = 61 2. (a) (i) 12 [8] 12 ÷ 8 = 1 baki 4, maka 8 bukan faktor bagi 12. 12 ÷ 8 = 1 remainder 4, therefore 8 is not the factor of 12. (ii) 27 [9] 27 ÷ 9 = 3, maka 9 ialah faktor bagi 27. 27 ÷ 9 = 3, therefore 9 is the factor of 27. (iii) 49 [7] 49 ÷ 7 = 7, maka 7 ialah faktor bagi 49. 49 ÷ 7 = 7, therefore 7 is the factor of 49. (iv) 35 [2] 35 ÷ 2 = 17 baki 1, maka 2 bukan faktor bagi 35. 35 ÷ 2 = 17 remainder 1, therefore 2 is not the factor of 35. Boleh dibahagi tepat dengan 3, bukan nombor perdana Divisible by 3, not a prime number Boleh dibahagi tepat dengan 7, bukan nombor perdana Divisible by 7, not a prime number (v) 90 [5] 90 ÷ 5 = 18, maka 5 ialah faktor bagi 90. 90 ÷ 5 = 18, therefore 5 is the factor of 90. (vi) 26 [6] 26 ÷ 6 = 4 baki 2, maka 6 bukan faktor bagi 26. 26 ÷ 6 = 4 remainder 2, therefore 6 is not the factor of 26. 3. (a) 16 1 16 1 2 8 16 1 2 4 4 8 16 Maka, faktor bagi 16 ialah 1, 2, 4, 8 dan 16. Therefore, the factors of 16 are 1, 2, 4, 8 and 16. (b) 20 1 20 1 2 10 20 1 2 4 5 10 20 Maka, faktor bagi 20 ialah 1, 2, 4, 5, 10 dan 20. Therefore, the factors of 20 are 1, 2, 4, 5, 10 and 20. Mahir Diri 14 7 28 28 4 2 1 Faktor-faktor bagi 28 ialah / Factors of 28 are 1 , 2 , 4 , 7 , 14 , 28 ↑ ↑ ↑ Fakor-faktor yang tidak ada pada rajah. Factors which is not in the diagram 16 = 1 × 16 16 = 2 × 8 16 = 4 × 4 20 = 1 × 20 20 = 2 × 10 20 = 4 × 5
13 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan 4. (a) 15 2 15 ÷ 2 = 7 baki 1, maka 2 bukan faktor perdana bagi 15. 15 ÷ 2 = 7 remainder 1, therefore 2 is not the prime factor of 15. 3 15 ÷ 3 = 5 dan 3 ialah nombor perdana, maka 3 ialah faktor perdana bagi 15. 15 ÷ 3 = 5 and 3 is a prime number, therefore 3 is the prime factor of 5. 7 15 ÷ 7 = 2 baki 1 maka 7 bukan faktor perdana bagi 15. 15 ÷ 7 = 2 remainder 1, therefore 7 is not the prime factor of 15. (b) 18 2 18 ÷ 2 = 9 dan 2 ialah nombor perdana, maka 2 ialah faktor perdana bagi 18. 18 ÷ 2 = 9 and 2 is a prime number, therefore 2 is the prime factor of 18. 7 18 ÷ 7 = 2 baki 4 , maka 7 bukan faktor perdana bagi 18. 18 ÷ 7 = 2 remainder 4, therefore 7 is not the prime factor of 18. 9 9 bukan nombor perdana, maka 9 bukan faktor perdana bagi 18. 9 is not a prime number, therefore 9 is not the prime factor of 18. ü ü 5. (a) 14 Faktor-faktor bagi 14 ialah / The factors of 14 are 14 = 2 × 7 14 = 1 × 14 1 , 2 , 7 , 14 Maka, faktor perdana bagi 14 ialah 2 dan 7. Therefore, the prime factors of 14 are 2 and 7. (b) 45 (c) 30 Faktor-faktor bagi 45 ialah / The factors of 45 are 45 = 5 × 9 45 = 3 × 15 45 = 1 × 45 1 , 3 , 5 , 9 , 15 , 45 Maka, faktor perdana bagi 45 ialah 3 dan 5. Therefore, the prime factors of 45 are 3 and 5. Faktor-faktor bagi 30 ialah / The factors of 30 are 30 = 3 × 10 30 = 5 × 6 30 = 2 × 15 30 = 1 × 30 1 , 2 , 3 , 5 , 6 , 10 , 15 , 30 Maka, faktor perdana bagi 30 ialah 2, 3 dan 5. Therefore, the prime factors of 30 are 2, 3 and 5. 6. (a) 28 2 28 2 14 7 7 1 Maka / Therefore, 28 = 2 × 2 × 7 (b) 30 2 30 3 15 5 5 1 Maka / Therefore, 30 = 2 × 3 × 5 (c) 36 2 36 2 18 3 9 3 3 1 Maka / Therefore, 36 = 2 × 2 × 3 × 3 (d) 60 Maka / Therefo re, 60 = 2 × 2 × 3 × 5 2 60 2 30 3 15 5 5 1
© Penerbitan Pelangi Sdn. Bhd. 14 Matematik Tingkatan 1 Jawapan (e) 84 Maka / Therefore, 84 = 2 × 2 × 3 × 7 7. (a) 9 dan / and 15 Faktor-faktor bagi / The factors of 9: 1 , 3 , 9 15: 1 , 3 , 5 , 15 Maka, faktor sepunya bagi 9 dan 15 ialah 1 dan 3 . Therefore, common factors of 9 and 15 are 1 and 3 . (b) 18, 21 dan / and 27 Faktor-faktor bagi / The factors of 18: 1 , 2 , 3 , 6 , 9 , 18 21: 1 , 3 , 7 , 21 27: 1 , 3 , 9 , 27 Maka, faktor sepunya bagi 18, 21 dan 27 ialah 1 dan 3. Therefore, common factors of 18, 21 and 27 are 1 and 3. (c) 18 dan / and 24 Faktor-faktor bagi / The factors of 18: 1 , 2 , 3 , 6 , 9 , 18 24: 1 , 2 , 3 , 4 , 6 , 8 , 12 , 24 Maka, faktor sepunya bagi 18 dan 24 ialah 1, 2, 3 dan 6. Therefore, common factors of 18 and 24 are 1, 2, 3 and 6. (d) 14, 28 dan / and 42 Faktor-faktor bagi / The factors of 14: 1 , 2 , 7 , 14 28: 1 , 2 , 4 , 7 , 14 , 28 42: 1 , 2 , 3 , 6 , 7 , 14 , 21 , 42 Maka, faktor sepunya bagi 14, 28 dan 42 ialah 1, 2, 7 dan 14. Therefore, common factors of 14, 28 and 42 are 1, 2, 7 and 14. 2 84 2 42 3 21 7 7 1 8. (a) (i) 15 dan / and 30 3 15 30 5 5 10 1 2 FSTB / HCF = 3 × 5 = 15 (ii) 8 dan / and 32 2 8 32 2 4 16 2 2 8 1 4 FSTB / HCF = 2 × 2 × 2 = 8 (iii) 18 dan / and 36 2 18 36 3 9 18 3 3 6 1 2 FSTB / HCF = 2 × 3 × 3 = 18 (b) (i) 12, 18 dan / and 30 2 12 18 30 3 6 9 15 2 3 5 FSTB / HCF = 2 × 3 = 6 (ii) 16, 24 dan / and 32 2 16 24 32 2 8 12 16 2 4 6 8 2 3 4 FSTB / HCF = 2 × 2 × 2 = 8 (iii) 30, 60 dan / and 90 2 30 60 90 3 15 30 45 5 5 10 15 1 2 3 FSTB / HCF = 2 × 3 × 5 = 30
15 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan 9. (a) 2 84 96 24 2 42 48 12 3 21 24 6 7 8 2 Bilangan maksimum anak yatim / Maximum number of orphans = 2 × 2 × 3 = 12 Setiap orang anak yatim mendapat 7 kotak susu, 8 bungkus biskut dan 2 bungkus gula-gula. Every orphan gets 7 boxes of milk, 8 packets of biscuit and 2 packets of sweets. 10. (a) 2 2 × 1, 2 × 2, 2 × 3, 2 × 4 = 2, 4, 6, 8 (b) 5 5 × 1, 5 × 2, 5 × 3, 5 × 4 = 5, 10, 15, 20 (c) 9 9 × 1, 9 × 2, 9 × 3, 9 × 4 = 9, 18, 27, 36 Mahir Diri (i) 2 dan / and 3 Jawapan / Answer: 8 12 ü 18 ü (ii) 3, 4 dan / and 6 Jawapan / Answer: 15 16 24 ü 8 ÷ 3 = 2 baki / remainder 2 Maka, 8 bukan gandaan sepunya bagi 2 dan 3. Therefore, 8 is not a common multiple of 2 and 3. 15 ÷ 4 = 3 baki / remainder 3 Maka, 15 bukan gandaan sepunya bagi 3, 4, dan 6. Therefore, 15 is not a common multiple of 3, 4 and 6. 16 ÷ 3 = 5 baki / remainder 1 Maka, 16 bukan gandaan sepunya bagi 3, 4, dan 6. Therefore, 16 is not a common multiple for 3, 4, and 6. 24 ÷ 3 = 8 24 ÷ 4 = 6 24 ÷ 6 = 4 Maka, 24 ialah gandaan sepunya bagi 3, 4, dan 6. Therefore, 24 is a common multiple of 3, 4 and 6. 12 ÷ 2 = 6 12 ÷ 3 = 4 Maka, 12 ialah gandaan sepunya bagi 2 dan 3. Therefore, 12 is a common multiple for 2 and 3. 18 ÷ 2 = 9 18 ÷ 3 = 6 Maka, 18 ialah gandaan sepunya bagi 2 dan 3. Therefore, 18 is a common multiple for 2 and 3. 11. (a) (i) 6 dan / and 9 Nombor / Number Gandaan / Multiples 6 6, 12 , 18 , 24, …… 9 9, 18 , 27, …… Gandaan sepunya yang pertama ialah 18 . Maka, tiga gandaan sepunya yang pertama ialah 18 × 1, 18 × 2 dan 18 × 3, iaitu 18 , 36 dan 54 . The first common multiple is 18 . Therefore, the first three common multiple are 18 × 1, 18 × 2 and 18 × 3, which are 18 , 36 and 54 .
© Penerbitan Pelangi Sdn. Bhd. 16 Matematik Tingkatan 1 Jawapan (b) (i) 2, 3 dan / and 6 Nombor / Number Gandaan Nombor / Multiples 2 2 , 4 , 6 , 8 , 10 , 12 , 14, …… 3 3 , 6 , 9 , 12 , 15, …… 6 6 , 12 , 18 , …… Gandaan sepunya yang pertama ialah 6. The first common multiple is 6. Maka, tiga gandaan sepunya yang pertama ialah 6 × 1, 6 × 2 dan 6 × 3, iaitu 6, 12 dan 18. Therefore, the first three common multiples are 6 × 1, 6 × 2 and 6 × 3, which are 6, 12 and 18. 12. (a) 6 dan / and 15 3 6 15 2 2 5 5 1 5 1 1 GSTK / LCM = 3 × 2 × 5 = 30 (b) 12 dan / and 18 2 12 18 3 6 9 2 2 3 3 1 3 1 1 GSTK / LCM = 2 × 3 × 2 × 3 = 36 13. (a) 6, 9 dan / and 15 3 6 9 15 2 2 3 5 3 1 3 5 5 1 1 5 1 1 1 GSTK / LCM = 3 × 2 × 3 × 5 = 90 (b) 6, 12 dan / and 18 2 6 12 18 3 3 6 9 2 1 2 3 3 1 1 3 1 1 1 GSTK / LCM = 2 × 3 × 2 × 3 = 36 14. (a) 2 6 8 10 2 3 4 5 2 3 2 5 3 3 1 5 5 1 1 5 1 1 1 Bilangan gula-gula setiap jenis perisa The number of sweets for each flavour = 2 × 2 × 2 × 3 × 5 = 120 Bilangan peket gula-gula berperisa oren yang paling kurang Minimum number of packets of orange flavoured sweet = 120 ÷ 6 = 20 Bilangan peket gula-gula berperisa epal yang paling kurang Minimum number of packets of apple flavoured sweet = 120 ÷ 8 = 15 Bilangan peket gula-gula berperisa durian yang paling kurang Minimum number of packets of durian flavoured sweet = 120 ÷ 10 = 12
17 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan 15. 2 8 12 18 2 4 6 9 2 2 3 9 3 1 3 9 3 1 1 3 1 1 1 Bilangan setiap jenis cenderamata The number of each souvenir = 2 × 2 × 2 × 3 × 3 = 72 Bilangan setiap jenis cenderamata yang diterima oleh seorang rakan kerja The number of each souvenir received by one colleague = 72 ÷ 18 = 4 Jumlah bilangan rantai kunci dan magnet peti sejuk The total number of keychains and fridge magnets = 4 + 4 = 8 Praktis Masteri 2 Bahagian A 1. Faktor-faktor bagi 100 ialah / The factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, 100 Maka, faktor perdana bagi 100 ialah 2 dan 5 Therefore, prime factors of 100 are 2 and 5 Jawapan / Answer: B 2. Faktor-faktor bagi 12: 1, 2, 3, 4, 6, 12 The factors of 12: Faktor-faktor bagi 18: 1, 2, 3, 6, 9, 18 The factors of 18: Maka, faktor sepunya bagi 12 dan 18 ialah 1, 2, 3 dan 6. Therefore, common factors of 12 and 18 are 1, 2, 3 and 6. Jawapan / Answer: D 3. 2 8 24 2 4 12 2 2 6 1 3 ∴ Faktor sepunya terbesar bagi 8 dan 24 ialah The highest common factor of 8 and 24 is 2 × 2 × 2 = 8 Jawapan / Answer: D 4. 36 ÷ 12 = 3 144 ÷ 12 = 12 Maka, 12 ialah faktor sepunya bagi 36 dan 144. Therefore, 12 is the common factor of 36 and 144. Jawapan / Answer: D Bahagian B 1. (a) 12 2 3 6 7 11 ü ü 6 2 12 3 2 Faktor perdana bagi 12 ialah 2 dan 3. (b) 20 1 2 5 10 17 ü ü 10 2 20 5 2 Faktor perdana bagi 20 ialah 2 dan 5. 2. (a) 30 2 15 3 5 (b) 2 4 6 2 2 3 3 1 3 1 1 Gandaan sepunya terkecil bagi 4 dan 6 Lowest common multiple of 4 and 6 = 2 × 2 × 3 = 12
© Penerbitan Pelangi Sdn. Bhd. 18 Matematik Tingkatan 1 Jawapan 3. 2 4 5 8 9 11 21 25 33 41 52 63 Bahagian C 1. (a) Nombor ganjil antara 13 dan 37 ialah Odd number between 13 and 37 is 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35 15, 21, 25, 27, 33 dan 35 boleh dibahagi tepat dengan satu nombor ganjil yang lain. 15, 21, 25, 27, 33 and 35 are divisible by other odd numbers. [15, 21, 27 dan 33 boleh dibahagi tepat dengan 3.] [15, 21, 27 and 33 are divisible by 3.] [25 dan 35 boleh dibahagi tepat dengan 5.] [25 and 35 are divisible by 5.] Maka, nombor perdana yang tinggal ialah 17, 19, 23, 29 dan 31. Thus, the remaining prime numbers are 17, 19, 23, 29 and 31. Maka / Therefore, x = 17, y = 23 dan / and z = 31. x = 17; y = 23; z = 31 (b) (i) 2 18 24 36 3 9 12 18 3 4 6 FSTB bagi 18, 24 dan 36 HCF of 18, 24 and 36 = 2 × 3 = 6 (ii) 3 30 45 5 10 15 3 2 3 2 2 1 1 1 GSTK bagi 30 dan 45 / LCM of 30 and 45 = 3 × 5 × 3 × 2 = 90 (c) 2 36 48 60 2 18 24 30 3 9 12 15 3 4 5 Bilangan kotak hadiah paling banyak The maximum number of gift boxes = 2 × 2 × 3 = 12 Setiap kotak hadiah mengandungi 3 magnet berwarna merah, 4 magnet berwarna kuning dan 5 magnet berwarna hitam. Every gift box consists of 3 red coloured magnets, 4 yellow coloured magnets and 5 black coloured magnets.
19 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan Kuasa Dua, Punca Kuasa Dua, Kuasa Tiga dan Punca Kuasa Tiga Squares, Square Roots, Cubes and Cube Roots BAB 3 1. (a) 49 ü 7 49 7 7 1 49 = 7 × 7 (b) 100 ü 2 100 2 50 5 25 5 5 1 100 = 2 × 2 × 5 × 5 = 2 × 5 × 2 × 5 = 10 × 10 (c) 126 2 126 3 63 3 21 7 7 1 126 = 2 × 3 × 3 × 7 2. (a) (i) 42 = 4 × 4 = 16 (ii) (–9)2 = (–9) × (–9) = 81 (b) (i) 4.12 = 4.1 × 4.1 = 16.81 (ii) (–0.3)2 = (–0.3) × (–0.3) = 0.09 (c) (i) 2 3 2 = 2 3 × 2 3 = 4 9 (ii) –1 1 4 2 = –1 1 4 × –1 1 4 = – 5 4 × – 5 4 = 25 16 3. (a) 64 = 8 × 8 = 8 (b) 1.21 = 1.1 × 1.1 = 1.1 (c) 1 7 9 = 16 9 = 4 3 × 4 3 = 4 3 = 1 1 3 4. (a) (i) 2.24 (ii) 3.32 (iii) 4.80 (iv) 6.63 (v) 10.25 (vi) 11.31 (b) (i) 0.18 (ii) 1.14 (iii) 3.24 (iv) 4.35 (v) 13.51 (vi) 14.53
© Penerbitan Pelangi Sdn. Bhd. 20 Matematik Tingkatan 1 Jawapan (c) (i) 0.76 (ii) 0.66 (iii) 1.34 (iv) 1.25 (v) 1.65 (vi) 1.80 5. (a) (i) 21.82 21.8 adalah antara 21 dan 22 . 21.8 is between 21 and 22 . 21.82 adalah antara 21 2 dan 22 2 . 21.82 is between 21 2 and 22 2 . Maka / Therefore, 21.82 22 2 = 484 (ii) 34.12 34.1 adalah antara 34 dan 35. 34.1 is between 34 and 35. 34.12 adalah antara 342 dan 352 . 34.12 is between 342 and 352 . Maka / Therefore, 34.12 342 = 1 156 (b) (i) 80 80 adalah antara 64 dan 81 . 80 is between 64 and 81 . Maka / Therefore, 80 81 = 9 (ii) 38 38 adalah antara 36 dan 49 . 38 is between 36 and 49 . Maka / Therefore, 38 36 = 6 6. (a) Panjang sisi rangka foto The side length of the photo frame = Panjang kayu / The length of the wood ÷ 4 = 52 ÷ 4 = 13 cm Luas segi empat sama / The area of the square = 13 2 = 13 × 13 = 169 cm2 (b) Panjang segi empat sama / The length of the square = 25 = 5 cm Panjang segi empat tepat The length of the rectangle = 4 × 5 cm = 20 cm Lebar segi empat tepat The width of the rectangle = 2 × 5 cm = 10 cm (atau panjang = 40 cm dan lebar = 5 cm) (or length = 40 cm and wide = 5 cm) 7. (a) 63 3 63 3 21 7 7 1 63 = 3 × 3 × 7 (b) 125 ü 5 125 5 25 5 5 1 125 = 5 × 5 × 5 (c) 180 2 180 2 90 3 45 3 15 5 5 1 180 = 2 × 2 × 3 × 3 × 5 8. (a) 23 = 2 × 2 × 2 = 8 (b) (– 0.4)3 = (– 0.4) × (–0.4) × (–0.4) = –0.064
21 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan (c) 1 1 5 3 = 6 5 3 = 6 5 × 6 5 × 6 5 = 216 125 = 1 91 125 9. (a) 3 –64 = 3 (–4) × (– 4) × (–4) = –4 (b) 3 0.125 = 3 0.5 × 0.5 × 0.5 = 0.5 (c) – 343 1 000 3 = – 7 10 × – 7 10 × – 7 10 3 = – 7 10 10. (a) 3.61 (b) –2.38 (c) –1.08 11. (a) (i) 6.93 6.9 adalah antara 6 dengan 7 . 6.9 is between 6 and 7 . 6.93 adalah antara 6 3 dengan 7 3 . 6.93 is between 6 3 and 7 3 . Maka / Therefore, 6.93 7 3 = 343 (ii) 7.23 7.2 adalah antara 7 dengan 8. 7.2 is between 7 and 8 7.23 adalah antara 73 dengan 83 . 7.23 is between 73 and 83 . Maka / Therefore, 7.23 73 = 343 (b) (i) 218 3 218 3 adalah antara 216 3 dan 343 3 . 218 3 is between 216 3 and 343 3 . Maka / Therefore, 218 3 216 3 = 6 (ii) 510 3 510 3 adalah antara 343 3 dan 512 3 . 510 3 is between 343 3 and 512 3 . Maka / Therefore, 510 3 512 3 = 8 12. (a) Tinggi bekas = Panjang tepi bekas Height of container = Length of the side of the container = 3 729 = 9 cm (b) Panjang tepi / Length of the sides = 84 cm ÷ 12 = 7 cm Isi padu kubus / Volume of the cube = 73 = 343 cm3 (c) 6 cm Panjang segi empat sama / Length of the square = 384 6 = 64 = 8 cm Isi padu tangki air / Volume of the water tank = 83 = 512 cm3
© Penerbitan Pelangi Sdn. Bhd. 22 Matematik Tingkatan 1 Jawapan 13. (a) 52 – 3 –512 = 5 × 5 – 3 (–8) × (–8) × (–8) = 25 – (–8) = 25 + 8 = 33 (b) 93 + 6.25 = 9 × 9 × 9 + 2.5 × 2.5 = 729 + 2.5 = 731.5 (c) 121 – 1 3 3 = 11 × 11 – 1 3 × 1 3 × 1 3 = 11 – 1 27 = 10 26 27 Mahir Diri 2 10 27 3 + 1 9 16 = 64 27 3 + 25 16 = 4 3 + 5 4 = 2 7 12 Praktis Masteri 3 Bahagian A 1. 101 = 10.04 Maka, 101 bukan kuasa dua sempurna. Therefore, 101 is not perfect square. Jawapan / Answer: B 2. 6 × 24 = 6 × 24 = 144 = 12 Jawapan / Answer: C 3. 3 1 – 37 63 = 3 27 64 = 3 4 Jawapan / Answer: A 4. 1 − (0.5)3 = 1 − 0.125 = 0.875 Jawapan / Answer: D 5. 43 − 52 − 3−1253 = 64 − 25 − (−5) = 44 Jawapan / Answer: B 6. Panjang kawasan kebun The length of the orchard = 81 = 9 m Perimeter kebun The perimeter of the orchard = 9 m × 4 = 36 m Jawapan / Answer: B 7. Panjang satu sisi kubus The length of a side of the cube = 3 61 64 = 1 1 4 cm Jumlah luas permukaan kubus The total surface area of a cube = 1 1 4 × 1 1 4 × 6 = 1 9 16 × 6 = 9 3 8 cm2 Jawapan / Answer: C
23 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan Bahagian B 1. 2 36 72 81 99 110 121 150 182 225 2. 64 ü 150 81 343 ü 3. (–2)2 24 1 27 – 2 5 6 × 4 (–2) × (–2) 1 3 3 3 – 8 125 4. (a) (i) 3 64 = 2 × 2 (ii) 32 + 84 = 5 2 (b) (i) 83 = 8 × 8 × 8 (ii) 36 = 3 + 3 Bahagian C 1. (a) Nombor / Number Nilai / Value as as 12 1.4 144 1.96 144 = 12 × 12 = 12 1.96 = 1.4 × 1.4 = 1.4 as – 64 125 3 3 3 8 3 – 4 5 1 1 2 – 64 125 3 = – 4 5 × – 4 5 × – 4 5 3 = – 4 5 3 3 8 3 = 27 8 3 = 3 2 = 1 1 2 (b) (i) 22 – 3 –64 = 22 – –4 = 8 (ii) 1 9 ÷ 1 3 3 = 1 9 × 27 1 = 9 (c) Panjang tapak / Length of the base = 3 512 = 8 cm Luas tapak / Area of the base = 8 cm × 8 cm = 64 cm2
© Penerbitan Pelangi Sdn. Bhd. 24 Matematik Tingkatan 1 Jawapan Nisbah, Kadar dan Kadaran Ratios, Rates and Proportions BAB 4 1. (a) (i) 3 : 5 (ii) 10 : 11 (b) (i) 2 hari : 1 minggu 2 days : 1 week =2 hari / days : (1 × 7 ) / days = 2 : 7 (ii) 9 cm : 40 mm = 9 cm : (40 ÷ 10) cm = 9 : 4 2. (a) (i) 5 : 9 : 12 (ii) 7 : 16 : 2 (b) (i) 1.2 cm : 8 mm :7 mm = (1.2 × 10 ) mm: 8 mm :7 mm = 12 : 8 : 7 (ii) 1 min : 1 jam / hour : 1 4 jam / hour = 1 min : 60 min : 1 4 × 60 min = 1 : 60 : 15 3. (a) (i) 5 : 1, 20 : 4 ü 5 : 1 = 5 × 4 : 1 × 4 = 20 : 4 (ii) 7 : 4, 14 : 16 7 : 4 = 7 × 2 : 4 × 2 = 14 : 8 (tidak sama dengan 14 : 16) (Not the same as 14 : 16) (b) (i) 15 : 10 : 6, 5 : 4 : 2 5 : 4 : 2 = 5 × 3 : 4 × 3 : 2 × 3 = 15 : 12 : 6 (tidak sama dengan 15: 10 : 6) (Not the same as 15 : 10 : 6) (ii) 20 : 15 : 15, 4 : 3 : 3 ü 4 : 3 : 3 = 4 × 5 : 3 × 5 : 3 × 5 = 20 : 15 : 15 (c) (i) 2 5 , 6 15 ü 2 5 = 2 × 3 5 × 3 = 6 15 (ii) 9 7 , 18 16 9 7 = 9 × 2 7 × 2 = 18 14 tidak sama dengan 18 16 not the same as 18 16 4. (a) (i) 5 : 15 = 5 5 : 15 5 = 1 : 3 (ii) 10 : 4 = 10 2 : 4 2 = 5 : 2 (b) (i) 2 3 : 5 9 = 2 3 × 9 : 5 9 × 9 = 6 : 5 (ii) 5 6 : 2 9 = 5 6 × 18 : 2 9 × 18 = 15 : 4
25 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan (c) (i) 400 ml : 1.2 l : 0.8 l = (400 ÷ 1 000) l : 1.2 l : 0.8 l = 0.4 : 1.2 : 0.8 = 4 : 12 : 8 = 4 4 : 12 4 : 8 4 = 1 : 3 : 2 (ii) 1 jam : 120 minit : 1 5 jam 1 hour : 120 minutes : 1 5 hour = 1 jam : 120 60 jam : 1 5 jam = 1 : 2 : 1 5 = 1 × 5 : 2 × 5 : 1 5 × 5 = 5 : 10 : 1 Mahir Diri 4 : 6 = 4 2 : 6 2 = 2 : 3 20 : 16 = 20 4 : 16 4 = 5 : 4 12 : 18 = 12 6 : 18 6 = 2 : 3 20 : 24 = 20 4 : 24 4 = 5 : 6 1 3 : 1 2 = 1 3 × 6 : 1 2 × 6 = 2 : 3 1 4 : 1 6 = 1 4 × 12 : 1 6 × 12 = 3 : 2 5. (a) (i) Kadar = 3 kg 2 biji / cakes Rates (ii) Jisim, kg / Mass, kg Bilangan kek, biji Number of cakes, cake(s) (b) (i) Kadar = 15 km Rates 30 minit / minutes (ii) Jarak, km / Distance, km Masa, minit / Time, minutes hour hour hour 6. (a) 90 km/j / km/h [m/s] = 90 km 1 j / h = (90 × 1 000 ) m (60 × 60) s = 90 000 m 3 600 s = 25 m/s (b) RM38/kg [sen/g] = RM38 1 kg = (38 × 100) sen (1 × 1 000) g = 3 800 sen 1 000 g = 3.8 sen/g 7. (a) 5 m 2 helai / pairs = 15 m 6 helai / pairs (b) RM6.40 8 m3 = RM5.60 7 m3 8. (a) Kain untuk menjahit sehelai seluar Cloth needed to make a pant = 5 m 2 = 2.5 m Kain untuk menjahit 9 helai seluar Cloth needed to make 9 pairs of pants = 9 × 2.5 m = 22.5 m (b) Kos bagi 1 m3 air = RM6.40 Cost for 1 m3 of water 8 = RM0.80 Bil air pada bulan Mac Water bill in March = 15 × RM0.80 = RM12
© Penerbitan Pelangi Sdn. Bhd. 26 Matematik Tingkatan 1 Jawapan 9. (a) (i) p : q = 3 : 4 (ii) p : r = 3 : 12 = 1 : 4 (iii) q : r = 4 : 12 = 1 : 3 (b) (i) p : q = 3 : 6 = 1 : 2 (ii) p : r = 3 : 2 (iii) q : r = 6 : 2 = 3 : 1 10. (a) (i) p : q = 6 : 5 Nilai q adalah sama. Same values of q. q : r = 5 : 4 p : q : r = 6 : 5 : 4 (ii) p : q = 8 : 3 p : r = 8 : 5 p : q : r = 8 : 3 : 5 (b) (i) p : q = 3 : 4 q : r = 2 : 1 Menyamakan nilai q: Equate the values of q: p : q = 3 : 4 q : r = 2 × 2 : 1 × 2 = 4 : 2 p : q : r = 3 : 4 : 2 (ii) p : q = 1 : 6 q : r = 4 : 3 Menyamakan nilai q Equate the values of q: p : q = 1 × 2 : 6 × 2 = 2 : 12 q : r = 4 × 3 : 3 × 3 = 12 : 9 p : q : r = 2 : 12 : 9 11. (a) Guli Akmal : Guli Ben : Guli Chan Akmal’s marbles : Ben’s marbles : Chan’s marbles = 2 : 4 : 5 = 2 4 : 1 : 5 4 Nilai p adalah sama. Same values of p. Nilai q adalah tidak sama. Different values of q Nilai q adalah tidak sama. Different values of q. Bilangan guli Ben diberikan The number of Ben’s marbles is given. Jumlah bilangan guli Akmal dan Chan The total number of Akmal and Chan’s marbles = 2 4 + 5 4 × 180 = 315 (b) Katakan wang simpanan Daniel, Elin dan Farah diwakili oleh d, e dan f. Let Daniel, Elin and Farah’s saving is represented by d, e and f. d : e = 2 : 3 e : f = 6 : 5 d : e = 4 : 6 Maka / Therefore, d : e : f = 4 : 6 : 5 = 4 5 : 6 5 : 1 Wang simpanan Daniel = 4 5 × 150 = RM120 Daniel’s saving 12. Kordial oren : Air = 2 : 5 Orange cordial : Water Isi padu kordial oren yang digunakan = 2 7 × 1 400 The volume of orange cordial used = 400 cm3 Isi padu air yang digunakan = 1 400 – 400 The volume of water used = 1 000 cm3 Nisbah baru / New ratio Kordial oren : Air / Orange cordial : Water = 400 : 1 000 + 120 = 1 : 3 13. (a) (i) Nisbah bilangan buku rujukan kepada jumlah buku The ratio of the number of reference books to the total number of books. = 2 : 2 + 3 = 2 : 5 2 5 = 2 × 20 5 × 20 = 40 100 Maka, peratusan buku rujukan di dalam rak ialah 40%. Therefore, the percentage of reference books on the shelf is 40%. d : e = 2 × 2 : 3 × 2
27 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan (b) (i) Peratusan wang yang digunakan untuk membayar ansuran rumah Percentage of money used to pay the house instalment = 25% = 25 100 Nisbah wang yang digunakan untuk membayar ansuran rumah kepada jumlah pendapatan bulanan The ratio of money used to pay the house instalment to the total monthly income = 25 : 100 = 1 : 4 14. (a) (i) Isi padu kordial oren : Isi padu minuman oren Volume of orange cordial : Volume of orange drink = 250 : 250 + 750 = 250 : 1 000 = 25 : 100 = 25% (ii) Isi padu kordial oren yang diperlukan The volume of orange cordial needed = 25% × 3 000 = 25 100 × 3 000 = 750 ml (b) (i) Bilangan guli kuning : Bilangan guli bukan berwarna kuning The number of yellow marbles : The number of non-yellow marbles = 12 : 72 – 12 = 12 : 60 = 1 5 = 20% (ii) n = 20% × 10 = 20 100 × 10 = 2 Praktis Masteri 4 Bahagian A 1. p : q = 1 : 4 q : r = 4 : 3 p : q : r = 1 : 4 : 3 Jawapan / Answer: C 2. x : y : z = 6 : 9 : 5 Diberi / Given x + y + z = 60 Hasil tambah nisbah / The sum of ratio 6 + 9 + 5 = 20 Maka, 1 bahagian nisbah / Therefore, 1 part of the ratio 60 ÷ 20 = 3 ∴ y + z = (9 + 5) × 3 = 14 × 3 = 42 Jawapan / Answer: C 3. Satu bahagian nisbah murid perempuan One part of the ratio of girls = 15 ÷ 3 = 5 Beza nisbah antara murid lelaki dan perempuan The difference in ratio between boys and girls = 4−3 = 1 Maka, beza bilangan antara murid lelaki dan perempuan Therefore, the difference in number between boys and girls = 1 × 5 = 5 Jawapan / Answer: C 4. Beza nisbah antara bola hijau dan merah The difference in ratio between green and red balls = 12 − 7 = 5 Satu bahagian nisbah bagi beza bilangan antara bola hijau dan merah One part of ratio for the difference in number between green and red balls = 15 ÷ 5 = 3 Jumlah bilangan bola / The total number of balls = (7 + 9 + 12) × 3 = 28 × 3 = 84 Jawapan / Answer: A 5. Pecahan bilangan kamus Bahasa Cina kepada jumlah kamus The fraction of the number of Chinese dictionaries to the total number of dictionaries = 40 100
© Penerbitan Pelangi Sdn. Bhd. 28 Matematik Tingkatan 1 Jawapan Maka, nisbah bilangan kamus Bahasa Cina kepada jumlah kamus Therefore, the ratio of the number of Chinese dictionaries to the total dictionaries = 40 : 100 = 2 : 5 Jawapan / Answer: B 6. Kadar / Rate = RM11.90 300 kg Jawapan / Answer: B Bahagian B 1. (a) (i) 2 : 5 : 7 (ii) 3 : 4 : 12 (b) (i) 1 : 4 dan / and 4 : 16 Ya Tidak Yes No (ii) 5 kg : 9 kg dan / and 10 kg : 16 kg Ya Tidak Yes No 2. (a) (i) m : n : p = 2 : 5 : 7 (ii) p : q : r = 20 : 12 : 3 (b) (i) x = 26 x = 3 Bahagian C 1. (a) (i) 3 : 4 ü (i) 5 : 3 : 4 ü 4 : 16 = 4 ÷ 4 : 16 ÷ 4 = 1 : 4 10 : 16 = 10 ÷ 2 : 16 ÷ 2 = 5 : 8 (b) Satu bahagian nisbah / One part of the ratio = 750 ÷ (5 + 9 + 11) = 30 orang / customers Bilangan pengunjung restoran K Number of customers of restaurant K = 5 × 30 = 150 orang / customers Bilangan pengunjung restoran L Number of customers of restaurant L = 9 × 30 = 270 orang / customers Bilangan pengunjung restoran M Number of customers of restaurant M = 11 × 30 = 330 orang / customers (c) Jumlah bayaran yang dibuat oleh Danish Total payment made by Danish = RM12 × [(1 × 60) minit ÷ 15 minit] = RM48 Peratusan wang saku yang diguna oleh Danish The percentage of pocket money used by Danish = RM48 RM200 × 100% = 24% Jumlah bayaran yang dibuat oleh Selvam Total payment made by Selvam = RM18 × 1 1 2 × 60 minit ÷ 12 minit = RM135 Peratusan wang saku yang diguna oleh Selvam The percentage of pocket money used by Selvam = RM135 RM270 × 100% = 50%
29 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan Ungkapan Algebra Algebraic Expressions BAB 5 1. (a) b mewakili bayaran bil air Encik Kamal pada setiap bulan. b represents Encik Kamal’s monthly water bill payment. b ialah pemboleh ubah yang mempunyai nilai yang berubah sebab bayaran bil bergantung kepada penggunaan air. b is a variable that has a varied value because bill payment is dependent on the usage of water. (b) k mewakili kadar faedah. k represents the interest rate. k ialah pemboleh ubah yang mempunyai nilai yang tetap sebab kadar faedah adalah tetap seperti yang dinyatakan dalam perjanjian. k is a variable with a fixed value because interest rate is fixed as stated in the agreement. 2. (a) Hasil darab dua nombor itu The product of the two numbers = p × q = pq (b) Beza antara nombor r dengan 5 The difference between r and 5 = 5 – (–r) 3. (a) 2x + 5y – 18 = 2( 6 ) + 5( 7 ) – 18 = 12 + 35 – 18 = 29 (b) 6p – 2q + 5 = 6(3) – 2(2) + 5 = 18 – 4 + 5 = 19 (c) 2p – 3q – 21 = 2(8) – 3(–1) – 21 = 16 + 3 – 21 = –2 (d) 1 4 u – 4v – 4 = 1 4 (8) – 4(3) – 4 = 2 – 12 – 4 = –14 (e) 12u + 5v + 2 = 12 1 3 + 5(2) + 2 = 4 + 10 + 2 = 16 4. (a) (i) Jumlah jisim / Total mass = x + y (ii) Jumlah jisim / Total mass = 4.2 + 3.1 = 7.3 kg (b) (i) Katakan bas itu membawa x orang lelaki dan y orang perempuan. let the bus carries x boys and y girls Bilangan murid yang memakai cermin mata The number of students wearing spectacles = 1 5 × 1 3 × x + 1 2 × 1 1 – 1 3 × y = 1 15 x + 1 3 y (ii) Bilangan murid yang memakai cermin mata The number of students wearing spectacles = 1 15 × 30 + 1 3 × 36 = 2 + 12 = 14 (c) (i) Jumlah harga / Total price = 8 × RMp + 12 × RMq = RM8p + RM12q = RM(8p + 12q) (ii) Jumlah harga yang dibayar jika Wong membeli di Kedai A Total price paid if Wong shops at Shop A = 8 × 1.60 + 12 × 1.80 = 12.80 + 21.60 = RM34.40 Jumlah harga yang dibayar jika Wong membeli di Kedai B Total price paid if Wong shops at Shop B = 8 × 1.65 + 12 × 1.75 = 13.20 + 21 = RM34.20 Jumlah bayaran Wong di Kedai B adalah lebih murah. Total amount paid by Wong at Shop B is cheaper.
© Penerbitan Pelangi Sdn. Bhd. 30 Matematik Tingkatan 1 Jawapan 5. (a) (i) 2 (ii) 2x, 7y (b) (i) 3 (ii) 3m2 , 2n, 5mn (c) (i) 3 (ii) 4 p , 2q, 3q (d) (i) 4 (ii) h2 , h 4 , 4h3 , 3h2 (e) (i) 4 (ii) a, 2b, 3abc, 3 5 6. (a) (i) pq : 5 (ii) p : 5q (iii) q : 5p (b) (i) st : –4 (ii) s : –4t (iii) t : –4s (c) (i) pqr : 1 (ii) qr : p (iii) p : qr (d) (i) mn : –2m (ii) m : –2mn (iii) n : –2m2 (e) (i) v2 : u 4 (ii) uv : v 4 (iii) uv2 : 1 4 7. 5pq, 4p2 q 3k, –2k 2p, 2q 0.5ut, –2vt 10abc, 0.2cab mn, nm 5 Bukan sebutan serupa kerana kuasa bagi p tidak sama. Unlike terms because the power of p is not the same. Bukan sebutan serupa kerana pemboleh ubah tidak sama. Unlike terms because the variables are not the same. Sebutan tak serupa / Unlike terms Sebutan serupa / Like terms 5pq, 4p2 q 2p, 2q 0.5ut, –2vt 3k, –2k 10abc, 0.2cab mn, nm 5 Sebutan algebra / Algebraic terms Mahir Diri –3pqr, 1 2 qrp ü hjk 5 , 2jhk ü 1 3 m2 n, 1 2 mn2 0.3wuv, –3uv –6ust, 2vst 5mn2 , n2 m ü Bukan sebutan serupa kerana kuasa bagi pemboleh ubah m dan n tidak sama. Unlike terms because the power of variables m and n is not the same Bukan sebutan serupa kerana pemboleh ubah tidak sama. Unlike terms because the variables are not the same. Bukan sebutan serupa kerana pemboleh ubah tidak sama. Unlike terms because the variables are not the same
31 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan 8. (a) 2x + 1 + 3x = 2x + 3x + 1 = 5x + 1 (b) 5x + 7 – 4x = 5x – 4x + 7 = x + 7 (c) –xy + 3 – 7xy = –xy – 7xy + 3 = –8xy + 3 (d) 5xy + 12 – 2xy + 6y = 5xy – 2xy + 6y + 12 = 3xy + 6y + 12 (e) 1 + 6x – 2xy + 3x – 5xy = 6x + 3x – 2xy – 5xy + 1 = 9x – 7xy + 1 9. (a) (6x + 5y) + (3x – 2y + 8) = 6x + 5y + 3x – 2y + 8 = 6x + 3x + 5y – 2y + 8 = 9x + 3y + 8 (b) (pq + 3qr – 1) – (4qr – 2pq) = pq + 3qr – 1 – 4qr + 2pq = pq + 2pq + 3qr – 4qr – 1 = 3pq – qr – 1 (c) 12 pq + 3p + pqr + 12 pq – 4p = 12 pq + 3p + pqr + 12 pq – 4p = 3p – 4p + 12 pq + 12 pq + pqr = –p + pq + pqr (d) 13 uv – (2u + uv) – v = 13 uv – 2u – uv – v = 13 uv – uv – 2u – v = – 23 uv – 2u – v (e) 0.4wx – 3 – 15 wx – wxy = 25 wx – 3 + 15 wx + wxy = 25 wx + 15 wx + wxy – 3 = 35 wx + wxy – 3 10. (a) p × p = p 2 (b) q × q × q = q3 (c) (p + q) × (p + q) = (p + q)2 (d) (pq – rs)(pq – rs) = (pq – rs)2 (e) (rs + st)(rs + st)(rs + st) = (rs + st)3 11. (a) a2 = a × a (b) b3 = b × b × b (c) (h – k)2 = (h – k) × (h – k) (d) (p + 2q)3 = (p + 2q) × (p + 2q) × (p + 2q) (e) (w – xy)3 = (w – xy) × (w – xy) × (w – xy) 6. (a) (i) 6a × 3ab = 6 × a × 3 × a × b = 6 × 3 × a × a × b = (6 × 3 ) × (a × a ) × b = 18a2b (ii) 3pq × 4p2q = 3 × p × q × 4 × p × p × q = 3 × 4 × p × p × p × q × q = (3 × 4) × (p × p × p) × (q × q) = 12p3q2
© Penerbitan Pelangi Sdn. Bhd. 32 Matematik Tingkatan 1 Jawapan (b) (i) 8r3 s ÷ 4r2 s2 = 8r3 s 4r2 s2 = 8 × r × r × r × s 4 × r × r × s × s = 2r s (ii) 3mn2 ÷ 9m3 n3 = 3mn2 9m3 n3 = 3 × m × n × n 9 × m × m × m × n × n × n = 1 3m2 n 13. (a) 15p2 × 2q3 ÷ 5pq2 = 15 × p × p × 2 × q × q × q 5 × p × q × q = 6pq (b) 12u2 v2 × uv ÷ 18uv2 = 12 × u × u × v × v × u × v 18 × u × v × v = 2u2 v 3 (c) 18mn2 ÷ 6m2 n3 × 2m2 = 18 × m × n × n × 2 × m × m 6 × m × m × n × n × n = 6m n Praktis Masteri 5 Bahagian A 1. 32a dan 23a mempunyai pemboleh ubah yang sama dengan kuasa yang sama. 32a and 23a have same variable with same power. Jawapan / Answer: B 2. 2 5 gk2 m = (gk) 2km 5 Jawapan / Answer: B 3. Jika / If h = –3, 9h2 = 9(–3)2 = 9(9) = 81 Jawapan / Answer: D 2 1 1 3 3 1 2 3 3 1 4. Diberi / Given x = −1, y = 2, z = −3, 3y2 z × 4x2 2x2 yz = 3(2)2 (–3) × 4(–1)2 2(–1)2 (2)(–3) = 3(4)(–3) × 4(1) –12 = –144 –12 = 12 Jawapan / Answer: D 5. (5fg − 7mn + 2) − (6fg − mn −8) = 5fg − 7mn + 2 − 6fg + mn + 8 = −fg − 6mn + 10 Jawapan / Answer: D 6. 10xy – 15w 5 – 2(9 – w + xy) = 2xy – 3w – 18 + 2w – 2xy = –w – 18 Jawapan / Answer: A Bahagian B 1. (a) 3k 4p2 8p2 k 3k2 p (b) 9cd – 3(2cd – 7) = 9cd – 6cd + 21 = 3cd + 21 2. (a) y 3 – 8k y 3 , 8k (b) 1 2 m – 9h + 4 1 2 m, 9h, 4 (c) d – 9d2 + d3 d, 9d2 , d3 (d) 4 5 r – 9t – t 4 5 r, 9t, t
33 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan Bahagian C 1. (a) (i) –6pq × 5pk 18q = –6 × p × q × 5 × p × k 18 × q = –1 × p × 5 × p × k 3 = –5p2 k 3 (ii) –2(d – 5ef) – (d + 6ef) = –2d + 10ef – d – 6ef = –3d + 4ef (b) (i) RM25p + RM12 1 2 p + RM6(3p) = RM25p + RM6p + RM18p (ii) Bilangan pengunjung dewasa = p Number of adult visitors Jumlah jualan harga tiket, RM2 695 Total ticket price, RM2 695 = RM25p + RM6p + RM18p Maka / Therefore, 25p + 6p + 18p = 2 695 49p = 2 695 p = 55 –1 3 (c) Ungkapan algebra bagi isi padu kubus, K Algebraic expression of volume of cube K = 6p cm × 6p cm × 6p cm = (6p × 6p × 6p) cm3 Ungkapan algebra bagi isi padu kuboid, L Algebraic expression of volume of cuboid L = 3p cm × 3p cm × 9p cm = (3p × 3p × 9p) cm3
© Penerbitan Pelangi Sdn. Bhd. 34 Matematik Tingkatan 1 Jawapan Persamaan Linear Linear Equations BAB 6 1. (a) a – 2 = 0 ü Ya. Terdapat satu pemboleh ubah, a, dan kuasa a ialah 1. Yes. There is one variable, a, and the power of a is 1. (b) u2 + 1 = 4 Bukan. Sebab kuasa u ialah 2. No. because the power of u is 2. (c) m + 4n = 0 Bukan. Sebab ada dua pemboleh ubah, m dan n. No. because there are two variables, m and n. (d) 2k + 3 Bukan. Sebab tiada simbol “=”. No because there is no “=” symbol. (e) 5w – 2 = w ü Ya. Terdapat satu pemboleh ubah, w, dan kuasa w ialah 1. Yes. There is one variable, w, and the power of w is 1. (f) t 2 + t = 0 Bukan. Sebab kuasa t ialah 2. No. because the power of t is 2. 2. (a) 4m = 8 (b) Katakan nombor itu ialah n Let the number be n. Maka / therefore, n 8 = 32 (c) Katakan umur adik Yoke Lan ialah p tahun. Let the age of Yoke Lan’s sister is p years. Maka / therefore, 18 – p = 5 3. (a) (i) x – 6 = 0 x – 6 + 6 = 0 + 6 x = 6 (ii) x + 4 = 0 x + 4 – 4 = 0 – 4 x = – 4 (b) (i) x + 2 = 1 x + 2 – 2 = 1 – 2 x = –1 (ii) x – 7 = 3 x – 7 + 7 = 3 + 7 x = 10 (c) (i) 3x = 18 3x 3 = 18 3 x = 6 (ii) 5x = –30 5x 5 = –30 5 x = –6 (d) (i) x 4 = –1 x 4 × 4 = –1 × 4 x = –4 (ii) – x 3 = 3 – x 3 × (–3) = 3 × (–3) x = –9 (e) (i) 3x 5 = 9 3x 5 × 5 3 = 9 × 5 3 x = 15 (ii) 5x 8 = –10 5x 8 × 8 5 = –10 × 8 5 x = –16
35 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan 4. (a) (i) 3x + 4 – 4 = 10 – 4 3x = 6 3x3 = 63 x = 2 (ii) 4x + 5 – 5 = 3 – 5 4x = –2 4x4 = –24 x = – 12 (b) (i) 2x – 3 + 3 = 5 + 3 2x = 8 2x2 = 82 x = 4 (ii) 3x – 2 + 2 = 2 + 2 3x = 4 3x3 = 43 x = 43 (c) (i) 56 x – 4 + 4 = 11 + 4 56 x = 15 56 x × 65 = 15 × 65 x = 18 (ii) 1 + 25 x – 1 = 3 – 1 25 x = 2 25 x × 52 = 2 × 52 x = 5 (d) (i) 4 × x + 4 × 2 = x + 7 4x + 8 = x + 7 4x + 8 – x = x + 7 – x 3x + 8 = 7 3x + 8 – 8 = 7 – 8 3x = –1 x = – 13 – 4 ÷ 3 + 3 ÷ 2 ÷ 3 + 2 + 4× 65 – 1× 52 Kembangkan persamaan. Expand the equation ÷ 3 – x – 8 – 5 ÷ 4 (ii) 3 × 4x + 3 × (–3) = 2x + 1 12x – 9 = 2x + 1 12x – 9 + 9 = 2x + 1 + 9 12x = 2x + 10 12x – 2x = 2x + 10 – 2x 10x = 10 x = 1 5. (a) Katakan PR = x cm dan PQ = 2x cm Let PR = x cm and PQ = 2x cm Maka / Therefore, 9 + x + 2x = 21 9 + 3x = 21 9 + 3x – 9 = 21 – 9 3x = 12 x = 4 P Q R 9 cm 2x cm x cm PQ = 8 cm, PR = 4 cm dan / and QR = 9 cm. Jadi, QR ialah sisi yang paling panjang. Thus, QR is the longest side. (b) Katakan harga sebatang pen ialah RMx. Let the price of a pen is RMx. Maka / Therefore, 25 + 7x = 46 25 + 7x – 25 = 46 – 25 7x = 21 x = 3 Harga sebatang pen ialah RM3. The price of a pen is RM3. (c) Katakan x mewakili nombor yang tidak diketahui itu. Let x represents the unknown number. Maka / Therefore, 16 + 4x = 48 16 + 4x – 16 = 48 – 16 4x = 32 x = 8 Nombor itu ialah 8. / The number is 8. 6. (a) u – 5 = 0 + 9 – 2x ÷ 10 Bukan. Sebab ada satu pemboleh ubah sahaja. No. Because there is only one variable
© Penerbitan Pelangi Sdn. Bhd. 36 Matematik Tingkatan 1 Jawapan (b) m + 3 = 2n Ya. Terdapat dua pemboleh ubah, m dan n, yang kuasanya 1. Yes. There are two variables, m and n, and the power of each variable is 1. (c) e + 4f = g (d) 2u + 5v (e) 5k – 2 = j (f) x + y = y2 Bukan. Sebab kuasa y ialah 2. No. Because the power of y is 2. Mahir Diri 2x + 3y = 0 2a + b = 4b 4u = v 7. (a) Katakan dua nombor itu ialah x dan y. Let the two numbers are x and y. Maka / Therefore, x + y = 120. (b) Katakan harga sebiji epal dan sebiji oren masingmasing ialah RMx dan RMy. Let the price of an apple and an orange are RMx and RMy respectively. Maka / Therefore, 8x + 9y = 25 Bukan. Sebab ada tiga pemboleh ubah. No. Because there are three variables Bukan. Sebab simbol “=” tidak wujud. No. Because the ‘=’ symbol does not exist. Ya. Terdapat dua pemboleh ubah, k dan j, yang kuasanya 1. Yes. There are two variables, k and j, and the power of each variable is 1. (c) Katakan umur Hasnah dan umur adik Hasnah masing-masing ialah x tahun dan y tahun. Let Hasnah’s age and her sister’s age are x years old and y years old respectively. Maka / Therefore, x = 3y 8. (a) x –3 5 y 2y = 3( –3 ) – 1 = –9 – 1 = –10 y = –5 2y = 3( 5 ) – 1 = 15 – 1 = 14 y = 7 (b) x –2 2 y 3(–2) + 2y = 12 –6 + 2y = 12 –6 + 2y + 6 = 12 + 6 2y = 18 y = 9 3(2) + 2y = 12 6 + 2y = 12 6 + 2y – 6 = 12 – 6 2y = 6 y = 3 (c) x –5 0 y 2(–5) – 5y = 20 –10 – 5y = 20 –10 – 20 = 5y 5y = –30 y = –6 2(0) – 5y = 20 0 – 5y = 20 –5y = 20 5y = –20 y = –4 9. (a) Apabila / When x = 0, Apabila / When x = 1, y + 3(0) = 7 y + 3(1) = 7 y + 0 = 7 y + 3 = 7 y = 7 y = 4 Maka, dua pasangan penyelesaian yang mungkin ialah x = 0, y = 7 dan x = 1, y = 4. Therefore, the two pairs of possible solution are x = 0, y = 7 and x = 1, y = 4. (b) Apabila / When x = 0, Apabila / When x = 4, 4y + 0 = –8 4y + 4 = –8 4y = –8 4y = –12 y = –2 y = –3 Maka, dua pasangan penyelesaian yang mungkin ialah x = 0, y = –2 dan x = 4, y = –3. Therefore, the two pairs of possible solution are x = 0, y = –2 and x = 4, y = –3.
37 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan (c) Apabila / When x = –1, Apabila / When x = 4, 5y = 3(–1) – 2 5y = 3(4) – 2 5y = –5 5y = 10 y = –1 y = 2 Maka, dua pasangan penyelesaian yang mungkin ialah x = –1, y = –1 dan x = 4, y = 2. Therefore, the two pairs of possible solution are x = –1, y = –1 and x = 4, y = 2. Mahir Diri 12g + 3k = –4 Gantikan / Substitute k = 2 3 , 12g + 3 2 3 = –4 12g + 2 = –4 12g = –6 g = –6 12 = – 1 2 Jawapan / Answer: A 10. (a) Katakan x mewakili bilangan pen biru dan y mewakili bilangan pen hitam. Let x represents the number of blue pens and y represents the number of black pens. Maka / Therefore, x + y = 5 y = 5 – x Bilangan pen biru Number of blue pens Bilangan pen hitam Number of black pens x = 0 y = 5 – 0 = 5 x = 1 y = 5 – 1 = 4 x = 2 y = 5 – 2 = 3 x = 3 y = 5 – 3 = 2 x = 4 y = 5 – 4 = 1 x = 5 y = 5 – 5 = 0 11. (a) Diberi / Given 2x + y = 5 ..........................a dan / and y + 3x = 12 ........................b Daripada / From a, y = 5 – 2x ..................c Gantikan c dalam b: Substitut into (5 – 2x) + 3x = 12 5 – 2x + 3x = 12 x = 12 – 5 x = 7 Maka, penyelesaian ialah x = 7 dan y = –9 . Therefore, the solution is x = 7 and y = –9 . (b) Diberi / Given y + 2x = 5 ..........................a dan / and 2x = 7 – 3y ..................b Daripada a / From , y = 5 – 2x ..................c Gantikan c dalam b: Substitute into 2x = 7 – 3(5 – 2x) 2x = 7 – 15 + 6x 0 = –8 + 4x 4x = 8 x = 2 Maka, penyelesaian ialah x = 2 dan y = 1. Therefore, the solution is x = 2 and y = 1. (c) Diberi / Given 4y – 3x + 12 = 0...................a dan / and 3y – x = 1...................b Daripada b / From , x = 3y – 1...........c Gantikan c dalam a: Substitute into 4y – 3(3y – 1) + 12 = 0 4y – 9y + 3 + 12 = 0 –5y + 15 = 0 –5y = –15 y = 3 Gantikan y = 3 dalam c: Substitute y = 3 into x = 3(3) – 1 = 9 – 1 = 8 Maka, penyelesaian ialah x = 8 dan y = 3. Therefore, the solution is x = 8 and y = 3. Gantikan x = 2 dalam c: Substitute x = 2 into , y = 5 – 2(2) = 5 – 4 = 1
© Penerbitan Pelangi Sdn. Bhd. 38 Matematik Tingkatan 1 Jawapan 12. (a) Diberi / Given 3x + 2y = 2 ............a dan / and y + 3x = 7 .............b Daripada / From a, 2y + 3x = 2 ............c c – b: 2y – y = 2 – 7 y = –5 Gantikan y = –5 dalam b: Substitute y = –5 into (–5) + 3x = 7 3x = 7 + 5 = 12 x = 4 Maka, penyelesaian ialah x = 4 dan y = –5. Therefore the solution is x = 4 and y = –5. (b) Diberi / Given 3x + 4y = 2 .....................a dan / and 3y + x = –1 ...................b b × 3: 9y + 3x = –3 ...................c c – a: 9y – 4y = –3 – 2 5y = –5 y = –1 Gantikan y = –1 dalam b: Substitute y = –1 into : 3(–1) + x = –1 –3 + x = –1 x = 3 – 1 x = 2 Maka, penyelesaian ialah x = 2 dan y = –1. Therefore, the solution is x = 2 and y = –1 (c) Diberi / Given 3y + 5x = 6 .......................a dan / and 2y + 3x = 5 ........................b a × 2: 6y + 10x = 12 ......................c b × 3: 6y + 9x = 15 ......................d c – d: 10x – 9x = 12 – 15 x = –3 Gantikan x = –3 dalam a: Substitute x = –3 into : 3y + 5(–3) = 6 3y – 15 = 6 3y = 21 y = 7 Maka, penyelesaian ialah x = –3 dan y = 7. Therefore, the solution is x = –3 and y = 7. 13. Katakan harga sekeping tiket orang dewasa ialah RMx dan harga sekeping tiket kanak-kanak ialah RMy. Let the price of an adult ticket is RMx and the price of a child ticket is RMy. Puan Wani membayar RM221 untuk 2 keping tiket orang dewasa dan 5 keping tiket kanakkanak. Puan Wani paid RM221 for 2 adult tickets and 5 child tickets Maka / Therefore, 2x + 5y = 221 ............... a Puan Nadia pula membayar RM194 untuk 3 keping tiket orang dewasa dan 2 keping tiket kanak-kanak. While Puan Nadia paid RM194 for 3 adult tickets and 2 child tickets. Maka / Therefore, 3x + 2y = 194 ............... b a × 3: 6x + 15y = 663 ............................. c b × 2: 6x + 4y = 388 ............................. d c – d: 11y = 275 y = 25 Gantikan y = 25 dalam a: 2x + 5(25) = 221 Substitute y = 25 into 2x + 125 = 221 2x = 96 x = 48 Maka, harga sekeping tiket orang dewasa ialah RM48 dan harga sekeping tiket kanak-kanak ialah RM25. Therefore, the price of an adult ticket is RM48 and the price of a child ticket is RM25. 14. (a) Katakan panjang dan lebar rangka segi empat tepat diwakili oleh x cm dan y cm. Let the length and width of the rectangular frame represented by x cm and y cm Perimeter = 80 cm. Maka / Therefore, 2x + 2y= 80 x + y = 40 ................. a Panjang rangka segi empat tepat adalah 8 cm melebihi lebarnya. The length of the rectangle is 8 cm more than the width. Maka / Therefore, x – y= 8 ...................... b a + b: 2x = 48 x = 24 y cm x cm
39 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan Gantikan x = 24 dalam a: 24 + y = 40 Substitute x = 24 into y = 16 Luas segi empat tepat = 24 × 16 Area of the rectangle, = 384 cm2 (b) Leong membayar RM19 dan mendapat 5 batang pen jenama P dan 4 batang pen jenama Q bermaksud untuk pen jenama P, dia membeli 4 batang pen dan mendapat sebatang pen secara percuma untuk pen jenama Q, dia membeli 3 batang pen dan mendapat sebatang pen secara percuma Leong paid RM19 and got 5 brand P pens and 4 brand Q pens meaning that he bought 4 brand P pens and got one pen for free and he bought 3 brand Q pens and got one pen for free. Maka / Therefore, 4x + 3y = 19 ................ a Ng membayar RM11 untuk 2 batang pen jenama P dan 2 batang pen jenama Q. Ng paid RM11 for 2 brand P pens and 2 brand Q pens. Maka / Therefore, 2x + 2y = 11 ................... b b × 2: 4x + 4y = 22 .................... c c – a: 4y – 3y = 22 – 19 y = 3 Gantikan y = 3 dalam b: 2x + 2(3) = 11 Substitute y = 3 into 2x + 6 = 11 2x = 5 x = 2.50 Maka, x = 2.50 dan y = 3. / Therefore, x = 2.50 and y = 3 Praktis Masteri 6 Bahagian A 1. k + 3h = 2 mempunyai dua pemboleh dengan setiap satu pemboleh ubah kuasanya 1. k + 3h = 2 has two variables with power of 1. Jawapan / Answer: B 2. 8p + r = 14 Gantikan / Substitute p = 3 2 8 3 2 + r = 14 r = 14 – 12 = 2 Jawapan / Answer: A 3. –5p – q = 15 ……..a –4p + q = 3 ……....b a + b: –5p + (–4p) = 15 + 3 –9p = 18 p = –2 Gantikan p = –2 dalam a Substitute p = –2 into a –5(–2) – q = 15 10 – q = 15 q = 10 – 15 = –5 Jawapan / Answer: C Bahagian B 1. (a) 1 3 k + 1 4 h = 9 x2 + y2 = 17 fg – 4g = 4 1 p k – 3q = 11 5t – 8 = w d – 1 d = 8 (b) x 0 1 2 y –3 4 11 2. (a) Ya / Yes (b) Ya / Yes (c) Tidak / No (d) Tidak / No 3. –4(k + 9) = 5 2d = 7 x = y + 6 m + 7 = 2n Persamaan linear dalam satu pemboleh ubah Linear equation in one variable Persamaan linear dalam dua pemboleh ubah Linear equation in two variables Bahagian C 1. (a) (i) (a) 57 (b) + (ii) (a) 45k + 35h = 480 450k + 350h = 48 (b) 188 = 12m – 8n 188 – 12m = 8n
© Penerbitan Pelangi Sdn. Bhd. 40 Matematik Tingkatan 1 Jawapan (b) 2x + 1 2 y = 7 ……..…..........…a 3x – y = 14…..............………b‚ a × 2: 4x + y = 14 …………c b + c: 3x + 4x = 14 + 14 7x = 28 x = 4 Gantikan x = 4 dalam c: Substitute x = 4 into : 4(4) + y = 14 16 + y = 14 y = 14 – 16 = –2 (c) Katakan harga bagi sekilogram mangga ialah RMx dan harga bagi sekilogram epal ialah RMy. Let the price of a kilogram of mangoes is RMx and the price of a kilogram of apples is RMy. Pada hari Rabu, 48 kg mangga dan 30 kg epal dijual dengan jumlah jualan RM375. On Wednesday, 48 kg of mangoes and 30 kg of apples were sold with total sales of RM375 Maka / Therefore, 48x + 30y = 375 .............a Pada hari Khamis, 36 kg mangga dan 24 kg epal dijual dengan jumlah jualan RM288 . On Thursday, 36 kg of mangoes and 24 kg of apples were sold with total sales of RM288 Maka / Therefore, 36x + 24y = 288 ............. b b ÷ 4: 9x + 6y = 72 ............... c c × 5: 45x + 30y = 360 ............. d a – d: 48x – 45x = 375 – 360 3x = 15 x = 5 Gantikan x = 5 dalam c: / Substitute x = 5 into : 9(5) + 6y = 72 45 + 6y = 72 6y = 27 y = 4.50 Maka, harga sekilogram mangga ialah RM5 dan harga sekilogram epal ialah RM4.50. Therefore, the price of a kilogram of mangoes is RM5 and the price of a kilogram of apples is RM4.50.
41 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan Ketaksamaan Linear Linear Inequalities BAB 7 1. (a) (b) < (c) . (d) > 2. (a) (b) . Mahir Diri –2 . –9 –2 lebih besar daripada –9, maka simbol ‘.’ digunakan. –2 is greater than –9, therefore symbol '.' is used. 3. (a) kurang daripada / less than (b) lebih banyak daripada / more than (c) kurang daripada atau sama dengan less than or equal to Sudut Info Emak boleh berbelanja paling banyak RM100 di pasar. Maka, wang perbelanjaan emak adalah termasuk RM100. Mother can spend at most RM100 in the market. Therefore, mother's expenses include RM100. (d) lebih tinggi daripada atau sama dengan higher than or equal to Sudut Info Jika markah ujian Aini ialah 40 markah, Aini lulus dalam ujiannya. Maka, markah Aini boleh 40 markah atau ke atas. If Aini's scores 40 marks, Aini pass her test. Therefore, Aini's mark can be 40 marks or above. (e) kurang daripada atau sama dengan less than or equal to Sudut Info Asmidar dibenarkan membawa kereta dengan 80 km sejam. Maka, laju kereta Asmidar boleh 80 km sejam atau ke bawah. Asmidar is allowed to drive her car with a speed of 80 km per hour. Therefore, the speed of Asmidar's car can be 80 km per hour or below. 4. (a) (i) x kurang daripada / less than 4; x , 4 (ii) x kurang daripada –1; x less than –1; x –1 (b) (i) x lebih besar daripada 2; x is greater than 2; x . 2 (ii) x lebih besar daripada –7; x is greater than –7; x . –7 5. (a) (i) x > 8 8 (ii) x > –7 –7 (b) (i) x < 3 3 (ii) x < –2 –2 (c) (i) x fi 0 0 (ii) x fi –4 –4 (d) (i) x fi 3 3 (ii) x fi –9 –9 6. (a) (i) x > 5 5 x . 5 (ii) x > –1 –1 x . –1 (b) (i) x < 8 8 x 8 (ii) x < –3 –3 x –3
© Penerbitan Pelangi Sdn. Bhd. 42 Matematik Tingkatan 1 Jawapan (c) (i) x fi 10 10 x > 10 (ii) x fi –9 –9 x > –9 Mahir Diri P: x > 11 11 Q: x 7 7 R: x > –9 –9 7. (a) (i) 8 . 4 (ii) 0 . –5 (iii) –0.001 . –0.501 (iv) 23 . 12 (b) (i) 1 , 10 (ii) –4 –1 (iii) –0.2 –0.02 (iv) 13 56 Mahir Diri –2.43 3; 3 . –2.43 8. (a) 1 9 (b) –4 1 (c) –5 –2 (d) –7 5 9. (a) (i) –5 . –11 (ii) 1 . –2 (iii) 3 9 (iv) –6 5 (b) (i) 17 12 (ii) 16 1 (iii) 13 . 15 (iv) 17 . 18 10. (a) (i) (ii) (iii) (iv) (b) (i) . (ii) . (iii) . (iv) . (c) (i) (ii) (iii) (iv) (d) (i) . (ii) . (iii) . (iv) . 11. (a) (i) (ii) (iii) (iv) (b) (i) . (ii) . (iii) . (iv) . (c) (i) . (ii) . (iii) . (iv) . (d) (i) (ii) (iii) (iv)
43 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan 12. (a) r > 25 Sudut Info Sekurang-kurangnya 25 orang bermaksud 25 orang atau ke atas. At least 25 students means 25 students or more. (b) x . 1 800 Sudut Info Jumlah pendapatan bulanan Kassim adalah melebihi RM1 800. Kassim's total monthly income is over RM1 800. (c) y 80 Sudut Info Markah untuk mendapat gred A ialah 80 atau ke atas. Maka, markah Farah adalah kurang daripada 80. The mark to obtain grade A is 80 or above. So, Farah's mark is less than 80. 13. (a) Tinggi murid kelas Tingkatan 1A kurang daripada 2 m. The height of Form 1A students are less than 2 m. (b) Masa yang diambil oleh Encik Ali dari rumah ke pejabatnya adalah melebihi 10 minit. The time taken by Encik Ali from his house to his office is more than 10 minutes. (c) Bilangan penumpang sebuah bas tidak boleh melebihi 40 orang. The number of passengers of a bus must not exceed 40 persons. (d) Gaji satu jam Encik Akmal adalah melebihi atau sama dengan RM25. Encik Akmal's wage for one hour is more than or equal to RM25. (e) Bilangan perkataan bagi sebuah karangan tidak boleh melebihi 120. The number of words in an essay should not exceed 120 words. 14. (a) (i) x = ……, –1, 0, 1 , 2 (semua integer yang kurang daripada 3) (all integers that are less than 3) (ii) x = ……, –9, –8, –7, –6 (semua integer yang kurang daripada –5) (all integers that are less than –5) (b) (i) x = ……, –1, 0, 1, 2, 3 (3 dan semua integer yang kurang daripada 3) (3 and all integers that are less than 3) (ii) x = ……, –9, –8, –7, –6, –5 (–5 dan semua integer yang kurang daripada –5) (–5 and all integers that are less than –5) 15. (a) (i) x = 8 , 9 , 10, 11, …… (semua integer yang lebih besar daripada 7) (all integers that are greater than 7) (ii) x = –3, –2, –1, 0, …… (semua integer yang lebih besar daripada –4) (all integers that are greater than –4) (b) (i) x = 7, 8, 9, 10, 11, …… (7 dan semua integer yang lebih besar daripada 7) (7 and all integers that are greater than 7) (ii) x = –4, –3, –2, –1, 0, …… (–4 dan semua integer yang lebih besar daripada –4) (–4 and all integers that are greater than –4) 16. (a) x + 2 – 2 > 0 – 2 x > –2 (b) x + 3 – 3 0 – 3 x –3 (c) x + 3 4 – 3 4 < 3 4 – 3 4 x < 0 17. (a) x – 4 7 + 4 7 . 0 + 4 7 x . 4 7 (b) x – 6 + 6 0 + 6 x 6 (c) x – 3 4 + 3 4 > 1 4 + 3 4 x > 4 4 x > 1 18. (a) (i) x 5 × 5 < 10 × 5 x < 50 (ii) 2x 3 × 3 2 . 6 × 3 2 x . 9
© Penerbitan Pelangi Sdn. Bhd. 44 Matematik Tingkatan 1 Jawapan (b) (i) – x 8 × (–8) < 2 × (–8) x < –16 (ii) – 3x 2 × – 2 3 > 1 × – 2 3 x > – 2 3 19. (a) (i) 5x 5 < – 20 5 x < –4 (ii) 3x 3 . 12 3 x . 4 (b) (i) –3x –3 . 27 –3 x . –9 (ii) –11x –11 < – 33 –11 x < 3 20. (a) (i) 1 3 x – 1 > 2 1 3 x – 1 + 1 > 2 + 1 1 3 x > 3 1 3 x × 3 > 3 × 3 x > 9 (ii) 3(x + 3) < 12 3(x + 3) 3 < 12 3 x + 3 < 4 x + 3 – 3 < 4 – 3 x < 1 (b) (i) x 2 – 2 + 2 > x + 1 + 2 x 2 > x + 3 x 2 – x > x + 3 – x – x 2 > 3 – x 2 × (–2) < 3 × (–2) x < –6 (ii) 5 – 3x 4 × 4 5 × 4 5 – 3x 20 5 – 3x – 5 20 – 5 –3x 15 –3x –3 . 15 –3 Bahagi dengan –3, disongsangkan menjadi .. Divide by –3, , is reversed to .. x . –5 21. (a) (i) x – 6 + 9 < 60 (ii) x – 6 + 9 < 60 x + 3 < 60 x + 3 – 3 < 60 – 3 x < 57 Bilangan penumpang yang mungkin ialah 57 atau ke bawah. The possible number of passengers is 57 or less. Maka, bilangan maksimum penumpang yang dibawa dari stesen P ialah 57 orang. Therefore, the maximum number of passengers carried from station P is 57. (b) (i) 135 + x + 3x < 240 (ii) 135 + 4x < 240 135 + 4x – 135 < 240 – 135 4x < 105 x < 26.25 Bilangan maksimum guli Ali ialah 26. The maximum number of Ali’s marble is 26. Maka, bilangan maksimum guli Akmal ialah Therefore, the maximum number of Akmal’s marble 26 × 3 = 78. Darab dengan –2, > disongsangkan menjadi <. Multiplied by –2, > is reversed to <.
45 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan (c) (i) Bayaran di P Bayaran di Q Payment in P < Payment in Q 4 + 1.5(x – 2) 1.8x (ii) 4 + 1.5x – 3 1.8x 1 + 1.5x 1.8x 1 – 0.3x 0 –0.3x –1 x . 3.33 Nilai minimum x ialah 4. The minimum value of x is 4. 22. (a) (i) x . 4 dan / and x > 9 4 9 Penyelesaian sepunya Common solution x fi 9 x > 4 Penyelesaian sepunya ialah The common solution is x > 9. (ii) x > –1 dan / and x . 6 –1 6 Penyelesaian sepunya Common solution x > 6 x fi –1 Penyelesaian sepunya ialah The common solution is x . 6. (b) (i) x 10 dan / and x –5 –5 10 Penyelesaian sepunya Common solution x < 10 x < –5 Penyelesaian sepunya ialah The common solution is x –5. (ii) x < 7 dan / and x 8 7 8 x < 8 x fi 7 Penyelesaian sepunya Common solution Penyelesaian sepunya ialah The common solution is x < 7 (c) (i) x . –3 dan / and x < 2 –3 2 x > –3 x fi 2 Penyelesaian sepunya Common solution Penyelesaian sepunya ialah The common solution is –3 x < 2. (ii) x < –4 dan / and x > –7 –7 –4 x fi –7 x ff –4 Penyelesaian sepunya Common solution Penyelesaian sepunya ialah The common solution is –7 < x < –4. Mahir Diri Penyelesaian sepunya Common solution x –6 x –8 –8 –6 fi ff Jawapan / Answer: C
© Penerbitan Pelangi Sdn. Bhd. 46 Matematik Tingkatan 1 Jawapan 23. (a) (i) 2x + 5 19 dan / and 9 – 3x < 6 2x + 5 19 2x + 5 – 5 19 – 5 2x 14 2x 2 14 2 x 7...............a 9 – 3x < 6 9 – 3x – 9 < 6 – 9 –3x < –3 –3x –3 > –3 –3 x > 1...............b 1 7 x fi 1 x < 7 Penyelesaian sepunya Common solution fi ff Penyelesaian sepunya ialah The common solution is 1 < x 7. (ii) 3x 4 + 2 > 11 dan / and x – 3 . 7 3x 4 + 2 > 11 3x 4 + 2 – 2 > 11 – 2 3x 4 > 9 3x 4 × 4 3 > 9 × 4 3 x > 12...............a x – 3 . 7 x – 3 + 3 . 7 + 3 x . 10...............b 10 12 x > 10 x fi 12 Penyelesaian sepunya Common solution fi ff Penyelesaian sepunya ialah The common solution is x > 12. (b) (i) – 2x 3 < 4 dan / and x + 1 2, x ialah integer / x is an integer – 2x 3 < 4 – 2x 3 × – 3 2 > 4 × – 3 2 x > –6...............a x + 1 2 x + 1 – 1 2 – 1 x 1...............b –6 1 x < 1 x fi –6 Penyelesaian sepunya Common solution fi ff –6 < x 1, x ialah integer x is an integer Maka / Therefore, x = –6, –5, –4, –3, –2, –1, 0 (ii) 3x . 2x – 2 dan / and x 2 2, x ialah integer / x is an integer 3x . 2x – 2 3x – 2x . 2x – 2 – 2x x . –2....................a x 2 2 x 2 × 2 2 × 2 x 4......................b –2 4 x < 4 x > –2 Penyelesaian sepunya Common solution fi ff –2 x 4, x ialah integer x is an integer Maka, / Therefore x = –1, 0, 1, 2, 3
47 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan Praktis Masteri 7 Bahagian A 1. 78 cm = 0.78 m 0.78 m . 0.7 m Jawapan / Answer: D 2. Masa yang diambil sekurang-kurang 30 minit. The time spent is at least 30 minutes. ∴ p > 30 minit 30 minutes Jawapan / Answer: C 3. 8 < 2 + 3w , 11 8 − 2 < 2 − 2 + 3w , 11−2 6 < 3w , 9 6 3 < 3w 3 , 9 3 2 < w , 3 Jawapan / Answer: B 4. Jawapan / Answer: C 5. Jawapan / Answer: B Bahagian B 1. (a) (i) 4 (ii) −1 (b) (i) –3 (ii) 7 2. (a) (i) x > −1 (ii) x 8 (b) (ii) –8 –4 (iii) –7 0 Bahagian C 1. (a) (i) . (ii) (iii) . (b) (i) h . 4 (ii) p < 2 (iii) d > 500 (c) (i) 3e > 12 e > 12 ÷ 3 e > 4 (ii) 4t – 14 6 t + 2 > 0 4t 6 + 14 t > –2 t 20 ÷ 4 t 5 ∴ –2 < t 5 Maka / Therefore, t = –2, –1, 0, 1, 2, 3, 4 2. (a) (i) Q (ii) R (iii) P (b) 5 – 4k < –15 –4k < –20 –k < –5 k > 5 (c) (i) 2 – w > 6 – 3w –w + 3w > 6 – 2 2w > 4 ∴ w > 2 3 + h –2 –6 3 + h . 12 ∴ h . 9
© Penerbitan Pelangi Sdn. Bhd. 48 Matematik Tingkatan 1 Jawapan Garis dan Sudut Lines and Angles BAB 8 Mahir Diri (a) 3.2 cm 3.2 cm (b) 35° 35° 1. (a) Sudut tirus / Acute angle (b) Sudut tegak / Right-angled (c) Sudut cakah / Obtuse angle (d) Sudut refleks / Reflex angle 2. (a) A C D B (b) L N M K (c) F K I H J E G 3. (a) Sudut tirus DEF / Acute angle DEF (b) Sudut cakah KLM / Obtuse angle KLM (c) Sudut refleks PQR / Reflex angle PQR Mahir Diri q p B q A p C B q A p C p dan q ialah sudut penggenap p and q are supplementary angles p dan q ialah sudut konjugat p and q are conjugate angles p dan q ialah sudut pelengkap p and q are complementary angles p + q = 360° → Sudut konjugat Conjugate angle p + q = 90° → Sudut pelengkap Complimentary angle p + q = 180° → Sudut penggenap Supplementary angle 4. (a) 38° Sudut pelengkap Complimentary angle = 90° – 38° = 52° (b) 57° Sudut pelengkap Complimentary angle = 90° – 57° = 33° (c) 80° Sudut pelengkap Complimentary angle = 90° – 80° = 10° 5. (a) Sudut penggenap Supplementary angle = 180° – 84° = 96° (b) Sudut penggenap Supplementary angle = 180° – 111° = 69° (c) Sudut penggenap Supplementary angle = 180° – 136° = 44°
49 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan 6. (a) Sudut konjugat Conjugate angle = 360° – 63° = 297° (b) Sudut konjugat Conjugate angle = 360° – 168° = 192° (c) Sudut konjugat Conjugate angle = 360° – 261° = 99° 7. (a) x° + 40° = 90° x° = 90° – 40° = 50° x = 50 (b) 2x° + x° = 90° 3x° = 90° x° = 30° x = 30 8. (a) x° + 88° + 45° = 180° x° + 133° = 180° x° = 180° – 133° = 47° x = 47 (b) 30° + 102° + 3x° = 180° 3x° + 132° = 180° 3x° = 180° – 132° = 48° x° = 16° x = 16 9. (a) x° + 145° + 132° = 360° x° + 277° = 360° x° = 360° – 277° = 83° x = 83 (b) 3x° + 2x° + 65° + 90° = 360° 5x° + 155° = 360° 5x° = 360° – 155° = 205° x° = 41° x = 41 10. (a) A B (b) A B 11. (a) A B K (b) A B K
© Penerbitan Pelangi Sdn. Bhd. 50 Matematik Tingkatan 1 Jawapan 12. (a) A B K 13. (a) A B K (b) A B K 14. (a) R 60° P Q (b) Y Z X 120° Mahir Diri Q 60° R S P RS = 6.3 cm 15. (a) B A O (b) D C O (c) E F O