Algebraic practices

1. REVIEW EXERCISE 15
Objective Questions
1. Which of the following algebraic terms has two or more unknowns?
A 8mn2p
B –4k3
C 7q
3t
D
10
2. Which of the following algebraic terms does not have more than two unknowns?
A 2h2k3n2
B 5prs
z4
C
4
5x
D
7 yw
3. 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 –
6m
4. The coefficient of qr in the algebraic term 7pq2r 3 is
A 7p
B 7pqr
C 7qr2
D 7pqr2
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2. 2 x 2 yz 2
5. In the algebraic term , the coefficient of x 2 z 2 is
5
y
A
5
2y
B
5
C 2y
2yz 2
D
5
6. 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 14p2qa
8. 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 2
9. 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 8r2xw
10. The product of 6xy2z and 3x2yz2 is
A 6x3y2z3
B 9x3y3z2
C 18x3y3z3
D 8x3yz3
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3. 11. 3rst × 4tr =
A 3r2st2
B 3rt2s
C 12rst2
D 12r2st2
18 xpy
12. =
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 3
14. =
56a 2 bc 2
A 5abc
B 5a2bc2
C 10a2bc2
D 10a2b2c2
15. State the number of terms in the expression 4abc + 10kx –6y2z – 2bx2 + 14.
A 5
B 6
C 8
D 9
16. 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
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4. 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 – 6ym
18.
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
1
19. Given that p = 3 and q = – . The value of 9pq – 5p =
3
A –24
B –6
C 22
D 32
4x 2 3 yz
20. When x = 3, y = 1 and z =–7, then the value of =
3
A 5
B 19
C 41
D 55
21. If m = 4, n = 5 and r = –2, then the value of 2mn2 – 7r3 =
A 144
B 158
C 242
D 256
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5. 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 119
23. –3(–ab + 2bc – 3cd) =
A –3ab – 6bc + 9cd
B –3ab + 6bc – 9cd
C 3ab – 6bc – 9cd
D 3ab – 6bc + 9cd
24. 9hx – 5ky – 7hx + 6ky =
A 2hx – ky
B 2hx + ky
C 4hx – ky
D 15hx – 12ky
25. (2 + 5xyz – 6tv2c) – (11 – 5yxz – 2ctv2) =
A 9 – 10xyz + 4ctv2
B 10xyz + 8ctv2 – 13
C 10xyz – 4ctv2 – 9
D 10xyz – 8ctv2 – 9
26. 9(ab + 2cd) – 7(ba + 3dc) =
A 2ab – cd
B 2ab – 3cd
C 2ab + 9cd
D 2ab + 39cd
12uy 4vx 16 yw 8
27. =
4
A –3uy – vx + 4yw –2
B –3uy + vx – 4yw + 2
C –3uy + vx – 4yw – 8
D –3uy + 4vx – 16yw + 8
4 fmp 6r 3
28. – 4(3fmp – 8r3) =
2
A –14fmp + 29r3
B –10fmp + 29r3
C –10fmp – 38r3
D –8fmp + 26r3
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6. 18 p 2 q 15r
29. 3(2p2q + 5r) + =
3
A 10r
B 12p2q
C 12p2q + 10r
D 24p2q + 10r
30.
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
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