1980 P2

01.
2ab - a² - b²
= -(-2ab + a² + b²)
= -(a² -2ab + b²)
= -(a - b)²...... (E)


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10. B
The hour hand turns (360/12)° in 60 minutes.
The minutes hand turns 360° in 60 minutes.

(360/12)° : 360° = x° : (Angle that the minute hand has turned)
Angle that the minutes hand has turned = 360° * x° ÷ (360/12)° = 12x°
...... (B)


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18.

A. θ = 270° ...... The answer is (A)
B. θ = 180°+30° or 360°-30°
C. θ = 0° or θ = 180°
D. θ = 30° or θ = 180°-30°
E. no root


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31.
First term, a = 8
Common ratio, r = -4/8 = -1/2

The (2n)th term
= 8 × (-1/2)^(2n - 1)
= -2^3 × (1/2)^(2n - 1)
= -(1/2)^(-3) × (1/2)^(2n - 1)
= -(1/2)^(-3 + 2n - 1)
= -(1/2)^(2n - 4)
= -1 / 2^(2n - 4) ...... (E)


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33.
Amount = Principal + Principal ×Rate × Years
Q = P + P × Rate × n
P × Rate × n = Q - P
Rate = (Q - P) / nP
Rate = [(Q - P) / nP] × 100%
Rate = [100(Q - P) / nP]% ...... (C)


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35. C
8 is the factor of a and b.
Let a = 8h and b = 8k where h and k are positive integers.

I. is true
a + b = 8h + 8k = 8(h + k)

II. is false
a + b = 8(h + k) which is divisible by 8, but may not be divisible by 16.

III. is true
ab = (8h)(8k) = 64 hk, which is divisible by 64.

...... The answer is (C).


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39.
Let a cm, b cm and c cm be length, width and height of the cuboid.

ab = 14 ...... [1]
ac = 35 ...... [2]
bc = 10 ...... [3]

[1] × [2] × [3] :
(abc)² = 4900
abc = 70

Volume of the cuboid
= abc cm³
= 70cm³ ...... (B)


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40.
x² + 2x + 7 is even.
Since 2x is even, then x² + 7 must be even. (even + even = even)
Since x is odd, then x² must be odd. (odd + odd = even)
Since x² is odd, then x must be odd. (odd × odd = odd)
Since x is positive, then x must be a positive odd number. ...... (C)



2015-02-05 20:16:19 補充：
Other answers in 意見。

2015-02-05 20:18:58 補充：
49.
In the diagram, the positions of C and D have been interchanged, i.e. the arc is BDC instead of BCD.

Join AD, BD and DC.
Arc BD = Arc DC (given)
BD = DC (equal arcs subtend equal chords)
AB = AC (given)
AD = AD (common side)
ΔADB ≅ ΔADC (SSS)

2015-02-05 20:19:20 補充：
∠BAD = ∠CAD (corr. ∠s of congruent Δs)
...... I is true.

∠ABD = ∠ACD (corr. ∠s of congruent Δs)
But ∠ABD + ∠ACD = 180° (opp. ∠s of concyclic quad.)
Hence ∠ABD = ∠ACD = 90°
AD is a diameter of the circle (∠ in semi-circle is 90°)
...... III is true.

2015-02-05 20:19:36 補充：
AB = AC and BD = DC
ABDC is a kite
BC⊥AD (diagonals of kite)
...... II is true.

All of I, II and III are true.
None of the 5 options is correct.

2015-02-05 20:19:55 補充：
53.
In the 5 Δs, the sum of the 10 base ∠s is twice the sum of exterior angles of the pentagon.

∠ sum of the 5 Δs :
(360° - x) + a + b + c + d + 2 × 360° = 5 × 180°
360° - x + a + b + c + d = 180°
a + b + c + d = 180° - 360° + x
a + b + c + d = x - 180° ...... (A)

2015-02-05 20:20:48 補充：
54.
Let C bevertex of the quadrilateral other than A, B and O.
∠BAC = 180° - θ (adj. ∠s on a st. line)
reflex ∠BOC = 2∠BAC = 360° - 2θ (∠ at centre twice ∠ in segment)
∠BOC = 360° - (360° - 2θ) = 2θ (∠s at a pt.)

2015-02-05 20:20:55 補充：
∠BAC + x + ∠BOC + y = 360° (∠ sum of quad.)
(180° - θ) + x + 2θ + y = 360°
180° + θ + x + y = 360°
x + y = 180° - θ ...... (C)

對，土扁師兄好野！！！

1. E
2ab - a² - b² = -(a² - 2ab + b²) = -(a - b)²

10. B
When the hour hand moves (1/12) 360°, then the minute hand moves 360°.
So, option B is the answer.

18. A
When sinθ = -1, θ = 270°
When sinθ = -1/2, θ = 210° or 330°
When sinθ = 0, θ = 0° or 180°
When sinθ = 1/2, θ = 30° or 150°
When sinθ = 2, no solution.

31. E
a = 8, r = -1/2
T(2n) = ar^(2n-1) = 8(-1/2)^(2n-1) = -1/2^(2n-4)

33. C
Let the rate p.a. be r, Q = P + n*P*r
∴ r = 100(Q-P)/(nP) %

35. C
a = 8m, b = 8n
I. a - b = 8m - 8n = 8(m - n) ...... True
II. a + b = 8m + 8n = 8(m + n) ... Not true if (m+n) = 3.
III. ab = 8m*8n = 64mn ............. True

39. B
bc = 10, ac = 14, ab = 35
Volume = abc = √(10*14*35) = 70

40. C
Let f(x) = x² + 2x + 7 = (x + 1)² + 6
If x is even, then f(x) is odd, if x is odd, then f(x) is even.
So option C is correct.

49. E
(1) are BD = arc DC, so ∠BAD = ∠CAD, thus AD bisects ∠BAC
(2) BH = CH, so AH丄BC
(3) △BDH～△ADB (AAA), ∠BHD = ∠ABD = 90°, so, AD is a diameter.

53. A
360° - x + a + b + c + d = 180°
∴a + b + c + d = x - 180°

54. C
Extend BO to D (on the circle), and ∠ACO = x,
∠CDO = θ (ext.∠= int. opp.∠, cyclic quad.)
∠DCO = θ (base ∠, isos △)
(180° - θ) + (x + θ) + θ + y = 360°
x + y = 180° - θ

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