 WinMaths Choi Wing Lam

S4 Form Test 1 
done: 17 / 20
score: 8 / 20

Question 1: 
(8y 8 )4
=
8y 2
A. 64y 8 .
B. 512y 16 .
C. 64y 32 .
D. 512y 30 .

A B C D

Question 2: 
(6x4 )2
=
6x3
A. 6x11 .
B. 36x24 .
C. 36x5 .
D. 6x5 .

A B C D

Question 3: 
(2x9 )4
=
8x4
A. 2x32 .
B. 6x32 .
C. 6x40 .
D. 2x40 .

A B C D

Question 4: 
87n+8
= Answer: B
48n+4
A. 224n+16 . Solution:
B. 25n+16 . 87n+8 23(7n+8)
=
C. 2n+16 . 48n+4 22(8n+4)
221n+24
= 16n+8
D. 221n+8 . 2
= 2(21n+24)−(16n+8)
= 25n+16

A B C D
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Question 5: 
(27 ⋅ 9n+2 )8 = Answer: B
A. 316n+72 . Solution:
B. 316n+56 .
(27 ⋅ 9n+2 )8 = (33 ⋅ 32(n+2) )8
C. 324n+40 . = (33+2n+4 )8
D. 316n+40 . = (32n+7 )8
= 316n+56

A B C D
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Question 6: 
1
( ) 874 = Answer: A
64 176

A. 1 Solution:
.
8278 1 874
B. 0. ( 176 ) 8 = 2 176
74
64 (8 )
C. 1
. 874
8324 = 352
1 8
D. . 1
8102 = 352−74
8
1
= 278
8

A B C D
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Question 7: 
(x + 4)(x2 + 6x + 16) = Answer: C
A. x3 + 16. Solution:
B. (x + 4)3 . (x + 4)(x2 + 6x + 16)
C. x3 + 10x2 + 40x + 64. = (x)(x2 + 6x + 16) + (4)(x2 + 6x + 16)
D. x3 + 8x2 + 32x + 64. = x3 + 6x2 + 16x + 4x2 + 24x + 64
= x3 + 10x2 + 40x + 64

A B C D
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Question 8: 
a b
If + = 4 , then x = Answer: A
7x 9y
A. 9ay Solution:
.
252y − 7b a b
7ay + =4
B. . 7x 9y
9b − 252y a ⋅ 9y b ⋅ 7x
7ay + =4
C. . 7x ⋅ 9y 9y ⋅ 7x
252y − 9b 9ay 7bx
9ay + =4
D. . 63xy 63xy
7y − 252b 9ay + 7bx
=4
63xy
9ay + 7bx = 4 ⋅ 63xy
9ay + 7bx = 252xy
9ay = 252xy − 7bx
9ay = (252y − 7b)x
9ay
x=
252y − 7b

A B C D
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Question 9: 
If 7 − 3m = 8n , then m = Answer: C
A. 8n − 7
. Solution:
3
B. n. 7 − 3m = 8n
C. 7 − 8n 7 − 8n = 3m
.
3 7 − 8n
7 + 8n m=
D. . 3
3

A B C D
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Question 10: 
If a(a − b) = −5(b + a), then b = Answer: D
A. −a2 + 5a
. Solution:
5+a
B. −a2 − 6a a(a − b) = −5(b + a)
.
6−a a2 − ab = −5b − 5a
C. −a2 + 6a 5b − ab = −a2 − 5a
.
6+a
(5 − a)b = −a2 − 5a
D. −a2 − 5a
. −a2 − 5a
5−a b=
5−a

A B C D
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Question 11: 
1 1
− = Answer: A
4x + 7 4x − 7
A. 14 Solution:
.
49 − 16x2
8 1 1
B. . −
49 − 16x2 4x + 7 4x − 7
14 4x − 7 4x + 7
C. . = −
16x2 − 49 (4x + 7)(4x − 7) (4x + 7)(4x − 7)
8 4x − 7 4x + 7
D. . = −
16x2 − 49 16x2 − 49 16x2 − 49
(4x − 7) − (4x + 7)
=
16x2 − 49
4x − 7 − 4x − 7
=
16x2 − 49
−14
=
16x2 − 49
14
=
49 − 16x2

A B C D
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Question 12: 
1 1
− = Answer: D
x2 + 4x + 4 x2 + 8x + 12
A. 1 Solution:
.
(x + 2)(x + 6)
1 1 1 1
B. 2x + 8 − = −
. x2 + 4x + 4 x2 + 8x + 12 (x + 2)2 (x + 2)(x + 6)
(x + 2)2 (x + 6) x+6 x+2
8 = −
C. . (x + 2)2 (x + 6) (x + 2)2 (x + 6)
(x + 2)2 (x + 6)
(x + 6) − (x + 2)
D. 4 =
. (x + 2)2 (x + 6)
(x + 2)2 (x + 6) x+6−x−2
=
(x + 2)2 (x + 6)
4
=
(x + 2)2 (x + 6)

A B C D
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Question 13: 
hl + kl − hm − km + hn + kn = Answer: B
A. (h − k)(l + m − n). Solution:
B. (h + k)(l − m + n).
hl + kl − hm − km + hn + kn
C. (h + k)(l + m + n). = l(h + k) − m(h + k) + n(h + k)
D. (h − k)(l − m − n). = (h + k)(l − m + n)

A B C D
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Question 14: 
a2 − b2 − a − b = Answer: A
A. (a + b)(a − b − 1). Solution:
B. (a − b)(a + b + 1).
a2 − b2 − a − b
C. (a − b)(a + b − 1). = (a + b)(a − b) − (a + b)
D. (a + b)(a − b + 1). = (a + b)(a − b − 1)

A B C D
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Question 15: 
α2 + α − β 2 + β Answer: D
A. (α − β ) (α + β − 1) Solution:
B. (α + β ) (α + β + 1)
α2 + α − β 2 + β
C. (α − β ) (α − β − 1) = α2 − β 2 + α + β
D. (α + β ) (α − β + 1) = (α + β)(α − β) + (α + β)
= (α + β)(α − β + 1)

A B C D
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Question 16: 
4m2 − 6m − 25n2 − 15n = Answer: C
A. (2m − 5n)(2m + 5n − 3). Solution:
B. (2m + 5n)(2m − 5n + 3).
4m2 − 6m − 25n2 − 15n
C. (2m + 5n)(2m − 5n − 3). = 4m2 − 25n2 − 6m − 15n
D. (2m − 5n)(2m + 5n + 3). = (2m + 5n)(2m − 5n) − 3(2m + 5n)
= (2m + 5n)(2m − 5n − 3)

A B C D
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Question 17: 
2m2 + 7mn − 9n2 − 4m + 4n Answer: B
A. (m − n)(2m − 9n − 4). Solution:
B. (m − n)(2m + 9n − 4).
2m2 + 7mn − 9n2 − 4m + 4n
C. (m + n)(2m − 9n + 4). = (m − n)(2m + 9n) − 4(m − n)
D. (m + n)(2m + 9n + 4). = (m − n)(2m + 9n − 4)

A B C D
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Question 18: 
Let p and q be constants. If x2 + p(x + 2) + q ≡ (x − Answer: C
8)(x + 2), then q =
Solution:
A. −6.
Method 1: Substitution
B. −7.
We can put any value of x into an identity.
C. −4.
D. −1. Put x = −2 to kill p:
(−2)2 + p(−2 + 2) + q = (−2 − 8)(−2 + 2)
q = −4

Method 2: Compare Coefficients

Expand:
x2 + p(x + 2) + q = (x − 8)(x + 2)
x2 + px + (2p + q) = x2 − 6x − 16

Compare coeff.:
p = −6
{
2p + q = −16

∴ p = −6 , q = −4

A B C D
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Question 19: 
Let m and n be constants. If m(x + 6)2 + n(x − 5)2 ≡ Answer: B
x2 − 98x − 19 , then m =
Solution:
A. 6.
Method 1: Substitution
B. −4.
We can put any value of x into an identity.
C. 5.
D. −1. Put x = 5 to kill n:
m(5 + 6)2 + n(5 − 5)2 = (5)2 − 98(5) − 19
m = −4

Method 2: Compare Coefficients

Expand:
m(x + 6)2 + n(x − 5)2 ≡ x2 − 98x − 19
(m + n)x2 + (12m − 10n)x + (36m + 25n) ≡ 1x2 − 98x − 19

Compare coeff.:
⎧ m+n=1
⎨12m − 10n = −98
⎩36m + 25n = −19

Solve any two of these equations:

m = −4 , n = 5

A B C D
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Question 20: 
If m and n are constants such that mx2 + m(x + 1) + Answer: C
15 ≡ mx(x + 4) + n(x − 2), then n =
Solution:
A. −7.
Method 1: Substitution
B. 3.
We can put any value of x into an identity.
C. −9.
D. 4. It is impossible to kill m first, so let's kill n and find
m first.

Put x = 2 to kill n :
m(2)2 + m(2 + 1) + 15 = m(2)(2 + 4) + n(2 − 2)
7m + 15 = 12m
m=3

Put x = 0 and m = 3:
0 + 3(0 + 1) + 15 = 0(0 + 4) + n(0 − 2)
n = −9

Method 2: Compare Coefficients

Expand:
mx2 + m(x + 1) + 15 ≡ mx(x + 4) + n(x − 2)
mx2 + mx + (m + 15) ≡ mx2 + (4m + n)x − 2n

Compare coeff.:
m = 4m + n
{
m + 15 = −2n

∴ m = 3 , n = −9

A B C D
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