Integrated exercise a_(book_2_B)_Ans

1. Integrated Exercise A
Integrated Exercise A
Part A
1. 2 x − 3 y − 9 = 0 ...... (1)

3 x − 2 y − 11 = 0 ...... (2)
From (1), we have
2x = 3y + 9
3 9
x= y + ...... (3)
2 2
By substituting (3) into (2), we have
3 9
3 y +  − 2 y − 11 = 0 ∵ The two straight lines shown above intersect at
 2 2
(−2, 3).
9 27
y+ − 2 y − 11 = 0 ∴ The solution is x = −2, y = 3.
2 2
5 5 4. Let $x and $y be the cost of each bottle of milk and each
y=−
2 2 egg respectively.
y = −1 From the question, we have
By substituting y = −1 into (3), we have 3 x + 12 y = 33 ...... (1)

3
x = (−1) +
9 4 x + 6 y = 36 ...... (2)
2 2 (2) × 2: 8 x + 12 y = 72 ...... (3)
=3 (3) − (1) :
∴ The solution is x = 3, y = −1.
8 x + 12 y = 72
−) 3 x + 12 y = 33
2.  4x 3y
 − = 1 ...... (1)
5 5 5 x = 39
5 x − 2 y = 15 ...... (2)
 x = 7.8
From (1), we have By substituting x = 7.8 into (2), we have
4x − 3y = 5 4(7.8) + 6 y = 36
4x = 3y + 5 31.2 + 6 y = 36
3 5 6 y = 4.8
x = y + ...... (3)
4 4 y = 0.8
By substituting (3) into (2), we have Cost of 1 bottle of milk and 2 eggs = $(7.8 × 1 + 0.8 × 2)
3 5
5 y +  − 2 y = 15 = $9.4
 4 4 ∴ The man should pay $9.4.
15 25
y+ − 2 y = 15
4 4 5. (a) 7 + 7 −4 × 7 0 × 7 4 = 7 + 7 −4 + 0 + 4
7 35 = 7 + 70
y=
4 4
= 7 +1
y=5
=8
By substituting y = 5 into (3), we have
3 5
x = (5) + = 5
4 4 (b) (2 p −1 )3 × (3 p 4 ) −3 = 23 p −1×3 × 3−3 p 4× ( −3)
∴ The solution is x = 5, y = 5. 8 1
= ×
p 3 27 p12
3. 2 x + y + 1 = 0 8
 =
x − 2 y + 8 = 0 27 p 3 +12
x − 2y + 8 = 0 8
=
x –4 0 2 27 p15
y 2 4 5
By adding the line x − 2 y + 8 = 0 on the graph, we have (5 × 1014 ) × (5.4 ×10 −9 ) 5 × 5.4
6. = ×1014 + ( −9) − ( −6)
3.6 × 10− 6 3.6
= 7.5 × 1011
139

2. Math in Action (2nd Edition) 2B Full Solutions
7. C5E216 = 12 ×163 + 5 × 162 + 14 × 161 + 2 × 160 By substituting C = 40 into (3), we have
B = 9000 − 100(40)
8. Consider ABE and ACD. = 5000
AB = AC sides opp. equal ∠s ∴ The solution is B = 5000, C = 40.
∠B = ∠C given
(b) The salesman’s salary in that month
∠BAE = ∠BAD + ∠DAE
= $(5000 + 40 × 250)
= ∠CAE + ∠DAE given
= $15 000
= ∠CAD
∴ ∠BAE = ∠CAD
13. (a) Original number = 10y + x
∴ ABE ≅ ACD ASA
New number = 10x + y
9. In PAD,
(b) (i) From the question, we have
90° + ∠PDA = ∠DAQ ext. ∠ of
x + y = 8
= 90° + ∠QAB 
∴ ∠PDA = ∠QAB (10 y + x) − (10 x + y ) = 18
After simplification, we have
Consider DPA and AQB.
∠DPA = ∠AQB = 90° given x + y = 8

∠PDA = ∠QAB proved − x + y = 2
∴ DPA ~ AQB AAA
(ii)  x + y = 8 ...... (1)

− x + y = 2 ...... (2)
10. (a) In ABC,
∠CAB = ∠CBA base ∠s, isos.
(1) + (2) :
∴ ∠CAD = ∠CBE
Consider ACD and BCE. x+ y =8
AD = BE given +) − x + y = 2
AC = BC given 2 y = 10
∠CAD = ∠CBE proved y =5
∴ ACD ≅ BCE SAS
By substituting y = 5 into (1), we have
∴ DC = EC corr. sides, ≅ s
x+5=8
(b) ∵ ACD ≅ BCE proved in (a) x=3
∴ ∠ACD = ∠BCE corr. ∠s, ≅ s ∴ The original number is 53.
∠DCB = ∠ACD + ∠ACB
14. (a) The place value of 0 in 110111112 = 25
= ∠BCE + ∠ACB corr. ∠s, ≅ s
= ∠ECA The place value of 0 in 1B016 = 160
11. (a) In ABC, (b) 110111112
∵ AB = AC given
= 1× 2 7 + 1× 2 6 + 0 × 2 5 + 1 × 2 4 + 1× 2 3
∴ ∠B = ∠C base ∠s, isos.
Consider LBM and HCK. + 1× 2 2 + 1× 21 + 1× 2 0
BM = CK given = 128 + 64 + 0 + 16 + 8 + 4 + 2 + 1
∠B = ∠C proved = 22310
∠BML = ∠CKH = 90° given
∴ LBM ≅ HCK ASA
1B016 = 1×16 2 + 11×161 + 0 × 16 0
(b) Consider LMK and HKM. = 256 + 176 + 0
MK = KM common side = 43210
∠LMK = ∠HKM = 90° given
LM = HK corr. sides, ≅ s
∴ LMK ≅ HKM SAS (c) Arranging 110111112, 1B016 and 30010 in descending
order, we have
1B016 > 30010 > 110111112
Part B
12. (a) From the question, we have
15. (a) 6 x −1 6 1
B + 100C = 9000 ...... (1) (3 xy 3 ) −1 = ×
 x0 y xy 3 xy 3
B + 400C = 21 000 ...... ( 2) 2
From (1), we have = 2 4
x y
B = 9000 − 100C ...... (3)
By substituting (3) into (2), we have
(9000 − 100C ) + 400C = 21 000 (b) (i) 4 2 x −1 = 2 2( 2 x −1)
300C = 12 000
C = 40
140

3. Integrated Exercise A
8−2 = ( 23 ) −2 (c) ∵ ACD ~ ABC proved in (a)
CBD ~ ACD proved in (b)
= 2− 6 ∴ ABC ~ CBD
1 1 AB BC
= ∴ = corr. sides, ~ △s
16 2 4 CB BD
= 2− 4 BC 2 = BD × AB
18. (a)
4 2 x −1 (8−2 )
(ii) = 2 2( 2 x −1) × 2− 6 × 2 − 4
16
= 24 x − 2− 6− 4
= 2 4 x −12
= 2 4 ( x − 3)
16. (a) ∠BAD + ∠ABC = 180° int. ∠s, AD // BC
∠ADC + ∠DCB = 180° int. ∠s, AD // BC
∵ ∠ABC = ∠DCB given
∴ ∠BAD = ∠CDA
Consider ABD and DCA. ∵ ∠COB = 90°
AB = DC given ∠EOB = 45°
AD = DA common side ∠BOF = 60°
∠BAD = ∠CDA proved ∴ ∠EOF = ∠EOB + ∠BOF
∴ ABD ≅ DCA SAS = 45° + 60°
= 105°
(b) ∵ ABD ≅ DCA proved in (a)
1
∴ ∠ABD = ∠DCA corr. ∠s, ≅ s ∠GOF = × 105°
2
∴ ∠CBD = ∠BCA
= 52.5°
∴ OC = OB sides opp. equal ∠s
∴ OBC is an isosceles triangle.
(b)
17. (a) (i) Consider ACD and ABC.
∠A = ∠A common angle
∠ADC = ∠ACB = 90° given
∴ ACD ~ ABC AAA
(ii) ∵ ACD ~ ABC proved in (a)(i)
AC AD
∴ = corr. sides, ~ △s
AB AC
AC 2 = AD × AB
(b) (i) Consider BCD and CAD.
Multiple Choice Questions
1. Answer: B
∵ b = 180° − a − 90° ∠ sum of
= 90° − a 3 x − 2 y = 7 ...... (1)

c1 = 180° − a − 90° ∠ sum of 5 x + 2 y = 1 ...... (2)
= 90° − a (1) + (2) :
∴ b = c1 3x − 2 y = 7
∠ADC = ∠CDB = 90° given +) 5 x + 2 y = 1
∴ BCD ~ CAD AAA
8x = 8
(ii) ∵ △BCD ~ △CAD provided in (b)(i) x =1
CD DB By substituting x = 1 into (2), we have
∴ = corr. sides, ~ △s 5(1) + 2 y = 1
AD DC
CD 2 = AD × BD 2 y = −4
y = −2
∴ The solution is x = 1, y = −2.
141

4. Math in Action (2nd Edition) 2B Full Solutions
2. Answer: C
Point R is the point of intersection of the graphs.
3. Answer: D
2 x + 3 y = 6 ...... (1)

bx + y = 7 ...... (2)
By substituting x = a and y = 4 into (1), we have
2a + 3(4) = 6
2a = −6
a = −3
By substituting x = −3 and y = 4 into (2), we have
−3b + 4 = 7
− 3b = 3
b = −1
∴ The solution is a = −3, b = −1.
4. Answer: D
5. Answer: B
5 4 + 54 + 54 + 5 4 + 5 4 = 5 × 54
= 51+ 4
= 55
6. Answer: C
7. Answer: B
8. Answer: A
9. Answer: B
10. Answer: C
AAS
142