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REVIEW EXERCISE 15Objective Questions1. Which of the following algebraic terms has two or more unknowns? A 8mn2p B –4k3 C 7q 3t D 102. Which of the following algebraic terms does not have more than two unknowns? A 2h2k3n2 B 5prs z4 C 4 5x D 7 yw3. Unknowns: k and m 1 Coefficient of k3 : – m 6 Based on the above information, which of the following is the algebraic term? m3k A – 6 k 3m B – 6 m C – 3 6k k3 D – 6m4. The coefficient of qr in the algebraic term 7pq2r 3 is A 7p B 7pqr C 7qr2 D 7pqr2© Pearson Malaysia Sdn. Bhd. 2008 Essential Mathematics PMR 1
2 x 2 yz 25. In the algebraic term , the coefficient of x 2 z 2 is 5 y A 5 2y B 5 C 2y 2yz 2 D 56. Which of the following statements is correct? A The coefficient of bc in the algebraic term 5abc is 5. pt 2 p B The coefficient of t 2 in the algebraic term – is . 2 2 C The coefficient of wvz in the algebraic term wvz is 0. D The coefficient of pq2 in the algebraic term –7kpq2 is –7k.7. Which of the following algebraic terms is a like term for 6ap2? A –8p2a B 4bp2 C 12aq2 D 14p2qa8. Which of the following algebraic terms are not like terms? A txy and 7ytx B 9yz and 2yz C 5mk and –10mc D –4pqr and 24pqr 1 29. Which of the following are like terms for xr w? 2 A 6xr2w, wxr2and 9xrw2 B 4xwr2, –r2xw and 2xr2w C 3r2wx, 5wr2x and –x2rw D 7xw2r, 11xr2w and 8r2xw10. The product of 6xy2z and 3x2yz2 is A 6x3y2z3 B 9x3y3z2 C 18x3y3z3 D 8x3yz3© Pearson Malaysia Sdn. Bhd. 2008 Essential Mathematics PMR 2
11. 3rst × 4tr = A 3r2st2 B 3rt2s C 12rst2 D 12r2st2 18 xpy12. = 6 xy A –3x B –6y C 3p D 3y 9 prt ( 8rs)13. = 12 pt A –6r2s B –6rs2 C 6rs D 4r2s 35ab 2 c 16a 3 bc 314. = 56a 2 bc 2 A 5abc B 5a2bc2 C 10a2bc2 D 10a2b2c215. State the number of terms in the expression 4abc + 10kx –6y2z – 2bx2 + 14. A 5 B 6 C 8 D 916. A sack of rice has a mass of x kg. A worker wants to repackage the sack of rice into n smaller packets of mass y kg each. The mass, in kg, of the remaining rice in the sack is A ny – x B n(x – y) C x – ny x y D n© Pearson Malaysia Sdn. Bhd. 2008 Essential Mathematics PMR 3
17. Akmal bought 8nx durians. He found out that ym durians are unripe and cannot be sold. If he sold the remaining durians for a price of RM6 per fruit, then the amount, in RM, of income received is A 8nx – ym B 8nx – 6ym C 48nx – ym D 48nx – 6ym18. The diagram shows a rectangle TUVW. M is the midpoint of TU. Given that PV = QW = 3mzcm and TS = SR = RW. The area, in cm2, of the shaded region is A 40hkxy – 6hmxy B 44hkxy – 12hmxy C 45hkxy – 10hmxy D 48hkxy – 15hmxy 119. Given that p = 3 and q = – . The value of 9pq – 5p = 3 A –24 B –6 C 22 D 32 4x 2 3 yz20. When x = 3, y = 1 and z =–7, then the value of = 3 A 5 B 19 C 41 D 5521. If m = 4, n = 5 and r = –2, then the value of 2mn2 – 7r3 = A 144 B 158 C 242 D 256© Pearson Malaysia Sdn. Bhd. 2008 Essential Mathematics PMR 4
22. Calculate the value of 19 – 2(3a – b – 4c)2 when a = 6, b = –9 and c = 8. A –81 B –31 C 69 D 11923. –3(–ab + 2bc – 3cd) = A –3ab – 6bc + 9cd B –3ab + 6bc – 9cd C 3ab – 6bc – 9cd D 3ab – 6bc + 9cd24. 9hx – 5ky – 7hx + 6ky = A 2hx – ky B 2hx + ky C 4hx – ky D 15hx – 12ky25. (2 + 5xyz – 6tv2c) – (11 – 5yxz – 2ctv2) = A 9 – 10xyz + 4ctv2 B 10xyz + 8ctv2 – 13 C 10xyz – 4ctv2 – 9 D 10xyz – 8ctv2 – 926. 9(ab + 2cd) – 7(ba + 3dc) = A 2ab – cd B 2ab – 3cd C 2ab + 9cd D 2ab + 39cd 12uy 4vx 16 yw 827. = 4 A –3uy – vx + 4yw –2 B –3uy + vx – 4yw + 2 C –3uy + vx – 4yw – 8 D –3uy + 4vx – 16yw + 8 4 fmp 6r 328. – 4(3fmp – 8r3) = 2 A –14fmp + 29r3 B –10fmp + 29r3 C –10fmp – 38r3 D –8fmp + 26r3© Pearson Malaysia Sdn. Bhd. 2008 Essential Mathematics PMR 5
18 p 2 q 15r29. 3(2p2q + 5r) + = 3 A 10r B 12p2q C 12p2q + 10r D 24p2q + 10r30. In the diagram, PMST is a rhombus. M is the midpoint of PQ. The perimeter, in cm, of the whole figure is A 22x + 3y B 22x + 11y C 26x + 3y D 26x + 11y© Pearson Malaysia Sdn. Bhd. 2008 Essential Mathematics PMR 6