DSE14 · More about Trigonometry

三角學題庫
Trigonometry Question Bank

120 questions · EN + 中文 bilingual · MC options (A–D) · step-by-step solutions

Questions
120
Sub-topics
18
Templates
37
With Diagrams
53
Basic
59
Intermediate
37
Advanced
24
Languages
2
EN + 中文
Language
Sub-topic
Difficulty
#1 — gen-rectangular-box-1-f654d835
advanced rectangular-box
EN
ABCDEFGH is a rectangular box with AB = 8cm, BC = 13cm and AE = 10cm. Find the angle between diagonal AG and the base ABCD, correct to the nearest degree.
中文
ABCDEFGH 是一個長方體，其中 AB = 8cm，BC = 13cm，AE = 10cm。求 the angle between diagonal AG and the base ABCD（準確至最接近的度）。
A. 38°
B. 33° ✓
C. 23°
D. 57°
Trap Analysis
  • A (38°): Rounding error in inverse trig
  • C (23°): Used wrong triangle for calculation
  • D (57°): Complementary angle confusion
generated81310ABCDEFGH
Step-by-step Solution
1 Base diagonal AC = √(8² + 13²) = 15.2643.
2 AG makes angle θ with base. tan θ = AE/AC = 10/15.2643.
3 θ = arctan(0.6551) ≈ 33°.
1 底面對角線 AC = √(8² + 13²) = 15.2643.
2 AG makes angle θ with base. tan θ = AE/AC = 10/15.2643.
3 θ = arctan(0.6551) ≈ 33°.
Verification
Space diagonal AG = √(8² + 13² + 10²) = 18.2483. sin θ = 10/18.2483. ✓
3d-cosine-formula3d-pythagoras3d-figure-decomposition
#2 — gen-p6-circle-radius-1-a3a3a051
basic p6-circle-radius
EN
In the diagram, the diameter of a circle is 12 cm. Find the radius.
中文
在圖中，圓的直徑是 12 cm。求半徑。
A. 7
B. 12
C. 6 ✓
D. 3
Trap Analysis
  • A (7): Arithmetic slip
  • B (12): Confused radius with diameter
  • D (3): Divided by 4 instead of 2
generated O d = 12
Step-by-step Solution
1 Radius = diameter ÷ 2.
2 r = 12 ÷ 2 = 6.
1 半徑 = 直徑 ÷ 2。
2 r = 12 ÷ 2 = 6.
Verification
2 × 6 = 12. ✓
p6-circle-radius-from-diameter
#3 — gen-right-pyramid-square-base-1-904cd8f0
advanced right-pyramid-square-base
EN
VABCD is a right pyramid with square base ABCD. V is directly above the centre of the base. AB : VA = 6 : 5. Let θ be the angle between △ABV and △CBV. Find cos θ.
中文
VABCD 是一個以 ABCD 為正方形底面的正四角錐。V 在底面中心的正上方。AB : VA = 6 : 5。設 θ 為 △ABV 和 △CBV 之間的角。求 cos θ。
A. 36/5
B. 54/5
C. -9/16 ✓
D. -6/11
Trap Analysis
  • A (36/5): Wrong sign or magnitude in dihedral cosine formula
  • B (54/5): Arithmetic error in pyramid height computation
  • D (-6/11): Used edge ratio directly instead of computing dihedral
generatedABCDV
Step-by-step Solution
1 Square base side = 6k, lateral edge = 5k (ratio 6:5).
2 Half-diagonal of base = 6k√2/2. Height = √((5k)² − (6k√2/2)²) = k√(25 − 36/2).
3 Dihedral angle along edge BV between faces △ABV and △CBV.
4 cos θ = -9/16.
1 正方形底邊 = 6k, 側棱 = 5k (比例 6:5).
2 底面半對角線 = 6k√2/2. 高度 = √((5k)² − (6k√2/2)²) = k√(25 − 36/2).
3 沿棱的二面角 BV 在兩面之間 △ABV and △CBV.
4 cos θ = -9/16.
Verification
Height = 2.6458 (per unit). Angle reasonable for pyramid with ratio 6:5. ✓
3d-dihedral-angle3d-figure-decomposition3d-cosine-formula
#4 — gen-p6-equation-axb-c-1-e21af310
basic p6-equation-axb-c
EN
Solve the equation: 4x + 10 = 62.
中文
解方程：4x + 10 = 62。
A. 13 ✓
B. 18
C. 14
D. 52
Trap Analysis
  • B (18): Moved constant with wrong sign
  • C (14): Arithmetic slip after rearranging
  • D (52): Forgot to divide by coefficient of x
Step-by-step Solution
1 4x + 10 = 62
2 4x = 62 − 10 = 52
3 x = 52 ÷ 4 = 13.
1 4x + 10 = 62
2 4x = 62 − 10 = 52
3 x = 52 ÷ 4 = 13.
Verification
4(13) + 10 = 62. ✓
p6-equation-axb-c
#5 — gen-p6-equation-word-1-a0067918
intermediate p6-equation-word
EN
Twice a number plus 9 is 41. Find the number.
中文
某數的 2 倍加上 9 等於 41。求此數。
A. 16 ✓
B. 32
C. 17
D. 25
Trap Analysis
  • B (32): Forgot to divide by 2
  • C (17): Arithmetic slip in final division
  • D (25): Moved + constant to wrong side
Step-by-step Solution
1 Let the number be x. Then 2x + 9 = 41.
2 2x = 41 − 9 = 32.
3 x = 32 ÷ 2 = 16.
1 設 the number be x. Then 2x + 9 = 41.
2 2x = 41 − 9 = 32.
3 x = 32 ÷ 2 = 16.
Verification
2(16) + 9 = 41. ✓
p6-equation-word
#6 — gen-bearing-1-d4358a1a
advanced bearing
EN
A ship sails from A on a bearing of 135° for 119m to B, then turns to a bearing of 255° and sails 118m to C. Find the distance AC, correct to the nearest metre.
中文
一艘船從 A 以方位角 135° 航行 119m 到 B，然後轉向方位角 255° 航行 118m 到 C。求 AC 的距離（準確至最接近的米）。
A. 119m ✓
B. 237m
C. 61m
D. 132m
Trap Analysis
  • B (237m): Added distances directly (assumed straight line)
  • C (61m): Used sin instead of cos in cosine formula
  • D (132m): Distance from incorrect cosine formula setup
generated119118NN135°255°ABC
Step-by-step Solution
1 Bearing from A: 135°. Bearing from B: 255°.
2 Interior angle at B = 180° − (255 − 135)° = 60° (from parallel North lines).
3 AC² = 119² + 118² − 2(119)(118)cos 60° = 14043.
4 AC = √(14043) ≈ 119m.
1 從 A 的方位角：135°. 從 B 的方位角：255°.
2 B 的內角 = 180° − (255 − 135)° = 60° （由平行北線推得）.
3 AC² = 119² + 118² − 2(119)(118)cos 60° = 14043.
4 AC = √(14043) ≈ 119m.
Verification
118.5032 rounds to 119. ✓
bearingcosine-formula
#7 — gen-right-triangular-prism-1-0ec7cbcf
advanced right-triangular-prism
EN
ABCDEF is a right prism where △ABC and △DEF are congruent triangles with AB = 6cm and AC = 11cm. AD = 15cm. P is on AD with AP = 7cm. Find ∠BPD, correct to the nearest degree.
中文
ABCDEF 是一個直三角柱體，其中 △ABC 和 △DEF 是全等三角形，AB = 6cm，AC = 11cm。AD = 15cm。P 在 AD 上，AP = 7cm。求 ∠BPD（準確至最接近的度）。
A. 24°
B. 9°
C. 71°
D. 19° ✓
Trap Analysis
  • A (24°): Used wrong distance in angle calculation
  • B (9°): Forgot to include point P offset
  • C (71°): Complementary angle confusion
Step-by-step Solution
1 P on AD with AP = 7. So PD = 15 − 7 = 8.
2 BP = √(AB² + AP²) = √(6² + 7²) = 9.2195.
3 BD = √(AB² + AD²) = √(6² + 15²) = 16.1555.
4 cos∠BPD by cosine formula on △BPD.
5 ∠BPD ≈ 19°.
1 P on AD with AP = 7. So PD = 15 − 7 = 8.
2 BP = √(AB² + AP²) = √(6² + 7²) = 9.2195.
3 BD = √(AB² + AD²) = √(6² + 15²) = 16.1555.
4 cos∠BPD by cosine formula on △BPD.
5 ∠BPD ≈ 19°.
Verification
All distances computed from Pythagoras. ✓
3d-cosine-formula3d-pythagoras3d-figure-decomposition
#8 — gen-translation-scaling-1-055ad4f5
intermediate translation-scaling
EN
The variance of x₁, x₂, ..., x₇ is 16. Find the standard deviation of 6x₁ − 5, 6x₂ − 5, ..., 6x₇ − 5.
中文
x₁, x₂, ..., x₇ is 16. Find the standard deviation of 6x₁ − 5, 6x₂ − 5, ..., 6x₇ − 5。
A. 576
B. 24 ✓
C. 96
D. 29
Trap Analysis
  • A (576): Gave new variance instead of new SD
  • C (96): Multiplied variance by |k| (mixed up variance and SD rules)
  • D (29): Added constant to SD
Step-by-step Solution
1 Var(X) = 16 ⟹ SD(X) = 4.
2 Transform: 6xᵢ − 5.
3 SD(6X − 5) = |6| × SD(X) = 6 × 4 = 24.
1 Var(X) = 16 ⟹ SD(X) = 4.
2 變換： 6xᵢ − 5.
3 SD(6X − 5) = |6| × SD(X) = 6 × 4 = 24.
Verification
Translation does not affect SD; scaling by k multiplies SD by |k|. ✓
scaling-effecttranslation-invariance
#9 — gen-translation-scaling-2-934a322e
intermediate translation-scaling
EN
The variance of a set of numbers is 49. Each number is multiplied by -2 and then 1 is added. Find the new variance.
中文
一組數的方差為 49。每個數乘以 -2 再加 1。 求新的方差。
A. 14
B. 98
C. 196 ✓
D. 197
Trap Analysis
  • A (14): Gave SD instead of variance
  • B (98): Multiplied variance by |k| instead of k²
  • D (197): Added c to variance (translation changes variance)
Step-by-step Solution
1 Original variance = 49.
2 Transform: multiply by -2, then add 1.
3 Adding/subtracting a constant does not change variance (translation invariance).
4 Multiplying by -2 scales variance by -2² = 4.
5 New variance = 4 × 49 = 196.
1 原方差 = 49.
2 變換： 乘以 -2, 再加 1.
3 加減常數不改變方差（平移不變性）。
4 乘以 -2 使方差乘以 -2² = 4.
5 新方差 = 4 × 49 = 196.
Verification
Var(kX + c) = k²Var(X). -2² × 49 = 196. ✓
scaling-effecttranslation-invariance
#10 — gen-translation-scaling-3-2c6a4248
intermediate translation-scaling
EN
The variance of a set of numbers is 49. Each number is multiplied by -3 and then 10 is added. Find the new variance.
中文
一組數的方差為 49。每個數乘以 -3 再加 10。 求新的方差。
A. 451
B. 21
C. 147
D. 441 ✓
Trap Analysis
  • A (451): Added c to variance (translation changes variance)
  • B (21): Gave SD instead of variance
  • C (147): Multiplied variance by |k| instead of k²
Step-by-step Solution
1 Original variance = 49.
2 Transform: multiply by -3, then add 10.
3 Adding/subtracting a constant does not change variance (translation invariance).
4 Multiplying by -3 scales variance by -3² = 9.
5 New variance = 9 × 49 = 441.
1 原方差 = 49.
2 變換： 乘以 -3, 再加 10.
3 加減常數不改變方差（平移不變性）。
4 乘以 -3 使方差乘以 -3² = 9.
5 新方差 = 9 × 49 = 441.
Verification
Var(kX + c) = k²Var(X). -3² × 49 = 441. ✓
scaling-effecttranslation-invariance
#11 — gen-central-tendency-1-f9ffc717
intermediate central-tendency
EN
Mean of 3, 5, 8, 8, 10, 12, m, n is 7. If m + n = 10, which MUST be true? I. Mode is 8. II. Median ≤ 8. III. Range ≥ 9.
中文
3, 5, 8, 8, 10, 12，m，n 的平均值為 7。 若 m + n = 10，以下哪些必定成立？ I. 眾數為 8。 II. 中位數 ≤ 8。 III. 全距 ≥ 9。
A. I, II and III
B. I and II
C. II only ✓
D. III only
Trap Analysis
  • A (I, II and III): Did not verify all constraints across all valid pairs
  • B (I and II): Assumed mode is always 8 without checking all valid (m,n)
  • D (III only): Only checked range constraint
Step-by-step Solution
1 m + n = 10. Enumerate valid (m, n) pairs.
2 For each valid pair: check mode, median, range.
3 Only statements true for ALL pairs are "must be true". Answer: II only.
1 m + n = 10. 列舉所有有效的 (m, n) 對.
2 For each valid pair: check mode, median, range.
3 只有對所有組合都成立的陳述才是「必定成立」。 Answer: II only.
Verification
Exhaustive enumeration confirms. ✓
constrained-dataset-reasoningcentral-tendency-combined
#12 — gen-trig-identity-1-24b4b405
basic trig-identity
EN
Simplify (1 − cos²θ) / sin θ.
中文
化簡 (1 − cos²θ) / sin θ。
A. sin θ ✓
B. tan θ
C. cos θ
D. -sin θ
Trap Analysis
  • B (tan θ): Mixed up sin and tan
  • C (cos θ): Mixed up sin and cos
  • D (-sin θ): Sign error in quadrant/identity
Step-by-step Solution
1 Apply standard trig identities:
2 sin(180° + θ) = −sin θ, cos(90° − θ) = sin θ, cos(360° − θ) = cos θ, cos(180° + θ) = −cos θ.
3 sin²θ + cos²θ = 1. tan θ = sin θ / cos θ.
4 Result: sin θ.
1 運用標準三角恒等式：
2 sin(180° + θ) = −sin θ, cos(90° − θ) = sin θ, cos(360° − θ) = cos θ, cos(180° + θ) = −cos θ.
3 sin²θ + cos²θ = 1. tan θ = sin θ / cos θ.
4 得出：sin θ.
Verification
Substitute θ = 30° to verify numerically. ✓
trig-identity-pythagorean
#13 — gen-point-above-ground-1-11db416b
advanced point-above-ground
EN
T is a point vertically above O on a horizontal ground at height h. A and B are two points on the ground with ∠AOB = 90°. The angle of elevation of T from A is 50° and from B is 55°. Express AB in terms of h.
中文
T 是水平地面上 O 點正上方高度為 h 的一點。A 和 B 是地面上兩點，∠AOB = 90°。從 A 觀察 T 的仰角為 50°，從 B 觀察 T 的仰角為 55°。以 h 表示 AB。
A. h × 1.3115
B. h × 1.0929 ✓
C. h × 0.8391
D. h × 0.7002
Trap Analysis
  • A (h × 1.3115): Arithmetic error in cosine formula application
  • C (h × 0.8391): Used OA distance directly (1/tan 50°)
  • D (h × 0.7002): Used OB distance directly (1/tan 55°)
generatedTOAB50°55°90°
Step-by-step Solution
1 Let TO = h. OA = h/tan 50° = 0.8391h. OB = h/tan 55° = 0.7002h.
2 ∠AOB = 90°. AB² = 0.8391²h² + 0.7002²h² − 2(0.8391)(0.7002)h²cos 90°.
3 AB = h√(1.1944) = h × 1.0929.
1 設 TO = h. OA = h/tan 50° = 0.8391h. OB = h/tan 55° = 0.7002h.
2 ∠AOB = 90°. AB² = 0.8391²h² + 0.7002²h² − 2(0.8391)(0.7002)h²cos 90°.
3 AB = h√(1.1944) = h × 1.0929.
Verification
Numeric check with h = 1: AB = 1.0929. ✓
3d-cosine-formula3d-perpendicular-projectiontrig-ratio-basic
#14 — gen-trig-graph-1-7ae9dc6c
basic trig-graph
EN
The figure shows the graph of y = -2 − cos x. What is the value of the amplitude?
中文
圖中所示的是 y = -2 − cos x 的圖像。振幅的值是多少？
A. 0
B. -2
C. 2
D. 1 ✓
Trap Analysis
  • A (0): Off-by-one in root counting
  • B (-2): Read vertical shift instead of amplitude
  • C (2): Off by 1 in reading amplitude from graph
generated 0° 90° 180° 270° 360° -3 -2 -1 x y
Step-by-step Solution
1 y = -2 + -1 cos 1x has amplitude |-1| = 1.
2 Amplitude = 1.
1 y = -2 + -1 cos 1x 的振幅為 |-1| = 1.
2 振幅 = 1.
Verification
Max − Min = -1 − -3 = 2. Amplitude = 1. ✓
trig-graph-reading
#15 — gen-trig-equation-1-9eabf3ec
intermediate trig-equation
EN
Find the number of roots of sin x = -√3/2 in 0° ≤ x < 360°.
中文
求 sin x = -√3/2 在 0° ≤ x < 360° 中的根的數目。
A. 3
B. 4
C. 1
D. 2 ✓
Trap Analysis
  • A (3): Counted a repeated root as distinct
  • B (4): Counted roots from both sin and cos (double-counting)
  • C (1): Missed a root outside primary range
Step-by-step Solution
1 Rewrite equation in terms of a single trig function.
2 Solve the resulting equation: find values of the trig function.
3 Check which solutions lie in 0° ≤ x < 360° — each valid value gives 1 root (if ±1) or 2 roots.
4 Total roots: 2.
1 將方程化為單一三角函數的形式。
2 解所得方程，求出三角函數的值。
3 檢查哪些解落在 0° ≤ x < 360° — 每個有效值對應 1 個根（若為 ±1）或 2 個根。
4 根的總數：2.
Verification
Graph method: count intersections in [0°, 360°). ✓
trig-equation-root-counting
#16 — gen-box-whisker-1-1d76a1a4
basic box-whisker
EN
Which of the following can be obtained from a box-and-whisker diagram? I. Range II. Mean III. Inter-quartile range
中文
以下哪些可以從箱形圖中得知？ I. 全距 II. 平均值 III. 四分位距
A. I, II and III
B. III only
C. I and III only ✓
D. I and II only
Trap Analysis
  • A (I, II and III): Thought mean can be read from box-and-whisker
  • B (III only): Forgot range can also be obtained
  • D (I and II only): Included mean
Step-by-step Solution
1 Box-and-whisker shows: min, Q₁, median, Q₃, max.
2 Range = max − min ✓ (I). Mean cannot be read ✗ (II). IQR = Q₃ − Q₁ ✓ (III).
3 Answer: I and III only.
1 箱形圖顯示： min, Q₁, median, Q₃, max.
2 全距 = 最大值 − 最小值 ✓ (I). 無法讀出平均值 ✗ (II). IQR = Q₃ − Q₁ ✓ (III).
3 Answer: I and III only.
Verification
Mean requires all data points, not just 5-number summary. ✓
box-whisker-readingiqr-definition
#17 — gen-box-whisker-2-67f685c1
basic box-whisker
EN
A box-and-whisker diagram shows: minimum = 10, Q₁ = 24, median = 36, maximum = 56. If the range is 2 times the IQR, find Q₃.
中文
箱形圖顯示： 最小值 = 10, Q₁ = 24, 中位數 = 36, 最大值 = 56. If the range is 2 times the IQR, 求 Q₃。
A. 44
B. 39 ✓
C. 36
D. 41
Trap Analysis
  • A (44): Arithmetic error in solving range = k × IQR
  • C (36): Gave the median instead of Q₃
  • D (41): Subtracted IQR from max instead of solving equation
generated 10 24 36 39 56
Step-by-step Solution
1 Range = 56 − 10 = 46. IQR = Q₃ − Q₁.
2 Given range = k × IQR. Solve for Q₃: 39.
1 Range = 56 − 10 = 46. IQR = Q₃ − Q₁.
2 Given range = k × IQR. Solve for Q₃: 39.
Verification
Check: range / IQR is consistent. ✓
box-whisker-readingiqr-definition
#18 — gen-perpendicular-to-plane-1-4048db88
advanced perpendicular-to-plane
EN
In the figure, △CBD lies on a horizontal plane with CB = 10m, BD = 24m and CD = 26m. AB = 7m is perpendicular to the plane CBD. Find the angle between AC and the plane CBD, correct to the nearest degree.
中文
在圖中，△CBD 位於水平面上，其中 CB = 10m，BD = 24m，CD = 26m。AB = 7m 垂直於平面 CBD。求 the angle between AC and the plane CBD（準確至最接近的度）。
A. 55°
B. 145°
C. 40°
D. 35° ✓
Trap Analysis
  • A (55°): Complementary angle confusion
  • B (145°): Supplementary angle confusion
  • C (40°): Rounding error in inverse trig
generated1024267ABCDE
Step-by-step Solution
1 △CBD has sides 10, 24, 26. Since 10² + 24² = 676 = 676 = 26², it is right-angled at B.
2 The angle between AC and the plane is ∠ACB.
3 tan(angle) = AB/CB = 7/10.
4 angle = arctan(7/10) ≈ 35°.
1 △CBD 的邊長為 10, 24, 26。 由於 10² + 24² = 676 = 676 = 26²，在 B 成直角.
2 AC 與平面的夾角為 ∠ACB。
3 tan(角) = AB/CB = 7/10.
4 角 = arctan(7/10) ≈ 35°.
Verification
tan⁻¹(0.7) = 34.992°. ✓
3d-perpendicular-projection3d-pythagoras
#19 — gen-p6-percent-increase-1-43c061d8
basic p6-percent-increase
EN
A notebook costs $302. The price is increased by 5%. What is the new price (in dollars)?
中文
一本筆記簿原價 302 元。價格上升 5%。求新價格（元）。
A. 287
B. 15
C. 310
D. 317 ✓
Trap Analysis
  • A (287): Percentage direction error (decrease instead of increase)
  • B (15): Gave increase amount instead of new total
  • C (310): Used half of the percentage change
Step-by-step Solution
1 Increase amount = 302 × 5% = 15.
2 New price = 302 + 15 = 317.
1 增加量 = 302 × 5% = 15.
2 新價格 = 302 + 15 = 317.
Verification
302 × 1.05 = 317. ✓
p6-percent-increase
#20 — gen-quadrilateral-1-bd415bed
intermediate quadrilateral
EN
In quadrilateral ABCD, ∠ABC = 30°, AB = 12 and BC = 15. Find AC.
中文
在四邊形 ABCD 中，∠ABC = 30°，AB = 12，BC = 15。求 AC。
A. 27
B. 3
C. 8.5651
D. 7.5651 ✓
Trap Analysis
  • A (27): Added sides instead of using cosine formula
  • B (3): Subtracted sides instead of using cosine formula
  • C (8.5651): Arithmetic error in cosine formula
generatedABCD121530°
Step-by-step Solution
1 By cosine formula: AC² = 12² + 15² − 2(12)(15)cos 30°.
2 = 144 + 225 − 360cos 30° = 57.2309.
3 AC = 7.5651.
1 根據餘弦公式： AC² = 12² + 15² − 2(12)(15)cos 30°.
2 = 144 + 225 − 360cos 30° = 57.2309.
3 AC = 7.5651.
Verification
√(57.2309) = 7.5651. ✓
trig-ratio-basiccosine-formula
#21 — gen-trig-identity-2-16fb6813
basic trig-identity
EN
Simplify sin(180° + θ) + cos(90° − θ).
中文
化簡 sin(180° + θ) + cos(90° − θ)。
A. 0 ✓
B. 2 sin θ
C. -1
D. 1
Trap Analysis
  • B (2 sin θ): Did not recognise sin(180°+θ) = −sin θ
  • C (-1): Sign error
  • D (1): Incorrect identity — sin²+cos²=1 misapplied
Step-by-step Solution
1 Apply standard trig identities:
2 sin(180° + θ) = −sin θ, cos(90° − θ) = sin θ, cos(360° − θ) = cos θ, cos(180° + θ) = −cos θ.
3 sin²θ + cos²θ = 1. tan θ = sin θ / cos θ.
4 Result: 0.
1 運用標準三角恒等式：
2 sin(180° + θ) = −sin θ, cos(90° − θ) = sin θ, cos(360° − θ) = cos θ, cos(180° + θ) = −cos θ.
3 sin²θ + cos²θ = 1. tan θ = sin θ / cos θ.
4 得出：0.
Verification
Substitute θ = 30° to verify numerically. ✓
trig-identity-supplementary
#22 — gen-central-tendency-2-9447abf8
advanced central-tendency
EN
Positive integers: 2, 5, 6, 6, 9, 9, 11, 13, m, n. Mean = 7. Find the range of values of the median.
中文
正整數：2, 5, 6, 6, 9, 9, 11, 13，m，n。平均值 = 7。求中位數的取值範圍。
A. 6 ≤ median ≤ 8
B. 6 ≤ median ≤ 9
C. 5 ≤ median ≤ 7 ✓
D. 7
Trap Analysis
  • A (6 ≤ median ≤ 8): Wrong range — did not enumerate all valid (m,n) pairs
  • B (6 ≤ median ≤ 9): Included invalid pairs outside positive integer constraint
  • D (7): Gave mean instead of analyzing median range
Step-by-step Solution
1 m + n = 9. Both positive integers.
2 Enumerate valid (m, n) pairs. For each, sort and find median.
3 Median range: 5 ≤ median ≤ 7.
1 m + n = 9. 均為正整數.
2 列舉所有有效的 (m, n) 對. For each, sort and find median.
3 中位數範圍： 5 ≤ median ≤ 7.
Verification
All valid pairs checked. ✓
constrained-dataset-reasoningcentral-tendency-combined
#23 — gen-right-triangular-prism-2-0c1dcdf3
advanced right-triangular-prism
EN
ABCDEF is a right prism where △ABC and △DEF are congruent triangles with AB = 12cm and AC = 12cm. AD = 12cm. P is on AD with AP = 6cm. Find ∠BPD, correct to the nearest degree.
中文
ABCDEF 是一個直三角柱體，其中 △ABC 和 △DEF 是全等三角形，AB = 12cm，AC = 12cm。AD = 12cm。P 在 AD 上，AP = 6cm。求 ∠BPD（準確至最接近的度）。
A. 72°
B. 23°
C. 8°
D. 18° ✓
Trap Analysis
  • A (72°): Complementary angle confusion
  • B (23°): Used wrong distance in angle calculation
  • C (8°): Forgot to include point P offset
Step-by-step Solution
1 P on AD with AP = 6. So PD = 12 − 6 = 6.
2 BP = √(AB² + AP²) = √(12² + 6²) = 13.4164.
3 BD = √(AB² + AD²) = √(12² + 12²) = 16.9706.
4 cos∠BPD by cosine formula on △BPD.
5 ∠BPD ≈ 18°.
1 P on AD with AP = 6. So PD = 12 − 6 = 6.
2 BP = √(AB² + AP²) = √(12² + 6²) = 13.4164.
3 BD = √(AB² + AD²) = √(12² + 12²) = 16.9706.
4 cos∠BPD by cosine formula on △BPD.
5 ∠BPD ≈ 18°.
Verification
All distances computed from Pythagoras. ✓
3d-cosine-formula3d-pythagoras3d-figure-decomposition
#24 — gen-standard-score-1-f0258af5
basic standard-score
EN
The standard score of a student who gets 98 marks is 3. If the standard deviation is 6, find the mean score.
中文
一名得 98 分的學生的標準分為 3。 若標準差為 6，求平均分。
A. 86
B. 80 ✓
C. 98
D. 81
Trap Analysis
  • A (86): Added σ to μ instead of computing correctly
  • C (98): Gave the score instead of the mean
  • D (81): Off-by-one in root counting
Step-by-step Solution
1 (98 − μ) / 6 = 3.
2 98 − μ = 3 × 6 = 18.
3 μ = 98 − 18 = 80.
1 (98 − μ) / 6 = 3.
2 98 − μ = 3 × 6 = 18.
3 μ = 98 − 18 = 80.
Verification
z = (98 − 80) / 6 = 3. ✓
standard-score
#25 — gen-bearing-2-e5360ac6
advanced bearing
EN
A ship sails from A on a bearing of 000° for 68m to B, then turns to a bearing of 090° and sails 77m to C. Find the distance AC, correct to the nearest metre.
中文
一艘船從 A 以方位角 000° 航行 68m 到 B，然後轉向方位角 090° 航行 77m 到 C。求 AC 的距離（準確至最接近的米）。
A. 145m
B. 9m
C. 116m
D. 103m ✓
Trap Analysis
  • A (145m): Added distances directly (assumed straight line)
  • B (9m): Used sin instead of cos in cosine formula
  • C (116m): Distance from incorrect cosine formula setup
generated6877NN000°090°ABC
Step-by-step Solution
1 Bearing from A: 000°. Bearing from B: 090°.
2 Interior angle at B = 180° − (90 − 0)° = 90° (from parallel North lines).
3 AC² = 68² + 77² − 2(68)(77)cos 90° = 10553.
4 AC = √(10553) ≈ 103m.
1 從 A 的方位角：000°. 從 B 的方位角：090°.
2 B 的內角 = 180° − (90 − 0)° = 90° （由平行北線推得）.
3 AC² = 68² + 77² − 2(68)(77)cos 90° = 10553.
4 AC = √(10553) ≈ 103m.
Verification
102.7278 rounds to 103. ✓
bearingcosine-formula
#26 — gen-2d-cosine-formula-1-e23e729d
intermediate 2d-cosine-formula
EN
In △ABC, AB = 5, BC = 3 and ∠ABC = 120°. Find cos∠ACB.
中文
在 △ABC 中，AB = 5，BC = 3，∠ABC = 120°。求 cos∠ACB。
A. 13/14 ✓
B. -13
C. 11/14
D. 19/30
Trap Analysis
  • B (-13): Sign error — wrong quadrant
  • C (11/14): Used wrong pair of sides in cosine formula
  • D (19/30): Wrong sign in cosine formula — added instead of subtracted
generatedABC53120°
Step-by-step Solution
1 First find AC = √(5² + 3² − 2(5)(3)cos 120°) = 7.
2 cos∠ACB = (7² + 3² − 5²) / (2 × 7 × 3) = 13/14.
1 首先求 AC = √(5² + 3² − 2(5)(3)cos 120°) = 7.
2 cos∠ACB = (7² + 3² − 5²) / (2 × 7 × 3) = 13/14.
Verification
Angles sum to 180° in triangle. ✓
cosine-formula
#27 — gen-translation-scaling-4-070847c1
intermediate translation-scaling
EN
The variance of a set of numbers is 16. Each number is multiplied by -2 and then 2 is added. Find the new variance.
中文
一組數的方差為 16。每個數乘以 -2 再加 2。 求新的方差。
A. 8
B. 64 ✓
C. 32
D. 66
Trap Analysis
  • A (8): Gave SD instead of variance
  • C (32): Multiplied variance by |k| instead of k²
  • D (66): Added c to variance (translation changes variance)
Step-by-step Solution
1 Original variance = 16.
2 Transform: multiply by -2, then add 2.
3 Adding/subtracting a constant does not change variance (translation invariance).
4 Multiplying by -2 scales variance by -2² = 4.
5 New variance = 4 × 16 = 64.
1 原方差 = 16.
2 變換： 乘以 -2, 再加 2.
3 加減常數不改變方差（平移不變性）。
4 乘以 -2 使方差乘以 -2² = 4.
5 新方差 = 4 × 16 = 64.
Verification
Var(kX + c) = k²Var(X). -2² × 16 = 64. ✓
scaling-effecttranslation-invariance
#28 — gen-p6-circle-circumference-1-4a38de2d
basic p6-circle-circumference
EN
In the circle shown, the diameter is 14 cm. Estimate the circumference (take π as 3.14).
中文
在圖中的圓，直徑為 14 cm。估算圓周（取 π = 3.14）。
A. 22
B. 44 ✓
C. 88
D. 14
Trap Analysis
  • A (22): Used radius in C=pi*d as if it were diameter
  • C (88): Doubled circumference formula by mistake
  • D (14): Used diameter value directly
generated O d = 14
Step-by-step Solution
1 Circumference C = πd.
2 C = 3.14 × 14 = 44 (approx).
1 圓周 C = πd。
2 C = 3.14 × 14 = 44 (approx).
Verification
3.14 × 14 = 44. ✓
p6-circle-circumference
#29 — gen-p6-broken-line-ratio-1-95566f91
basic p6-broken-line-ratio
EN
The broken line graph shows daily steps. Tuesday's steps are how many times Sunday's steps?
中文
折線圖顯示每日步數。星期二的步數是星期日的多少倍？
A. 4
B. 1
C. 2 ✓
D. 3
Trap Analysis
  • A (4): Factor of 2 error
  • B (1): Used difference idea instead of ratio
  • D (3): Ratio computation slip
generated 02000400060008000 SunMonTueWedThuFriSat
Step-by-step Solution
1 From the graph: Sunday = 4000, Tuesday = 8000.
2 Times = Tuesday ÷ Sunday = 8000 ÷ 4000 = 2.
1 由圖可得：星期日 = 4000, 星期二 = 8000.
2 倍數 = 星期二 ÷ 星期日 = 8000 ÷ 4000 = 2.
Verification
4000 × 2 = 8000. ✓
p6-broken-line-ratio
#30 — gen-weighted-mean-stats-1-b4265e6c
basic weighted-mean-stats
EN
Alice types for 5 minutes at 70 words/min and 3 minutes at 50 words/min. Find her average typing speed, correct to the nearest integer.
中文
Alice 以每分鐘 70 字的速度打字 5 分鐘，然後以每分鐘 50 字的速度打字 3 分鐘。求她的平均打字速度（準確至最接近的整數）。
A. 64
B. 62
C. 63 ✓
D. 60
Trap Analysis
  • A (64): Rounding error in weighted average
  • B (62): Arithmetic error in multiplication
  • D (60): Used simple average instead of weighted
Step-by-step Solution
1 Weighted mean = (70 × 5 + 50 × 3) / (5 + 3).
2 = (350 + 150) / 8 = 63.
1 加權平均 = (70 × 5 + 50 × 3) / (5 + 3).
2 = (350 + 150) / 8 = 63.
Verification
500 / 8 ≈ 63. ✓
weighted-mean
#31 — gen-quadrilateral-2-28441eb6
intermediate quadrilateral
EN
In quadrilateral ABCD, AB = 15, BC = 12 and AC = 10.7. Find cos∠ABC.
中文
在四邊形 ABCD 中，AB = 15，BC = 12，AC = 10.7。求 cos∠ABC。
A. √3/2
B. √5/2
C. √2/2 ✓
D. -√2/2
Trap Analysis
  • A (√3/2): Wrong surd — substituted wrong intermediate value
  • B (√5/2): Wrong surd — substituted wrong intermediate value
  • D (-√2/2): Used supplementary angle cos(135°) — sign error
generatedABCD151210.7
Step-by-step Solution
1 Given sides 15, 12 and diagonal 10.6977.
2 cos∠ABC = (15² + 12² − 10.6977²) / (2 × 15 × 12).
3 = √2/2.
1 Given sides 15, 12 and diagonal 10.6977.
2 cos∠ABC = (15² + 12² − 10.6977²) / (2 × 15 × 12).
3 = √2/2.
Verification
arccos(√2/2) ≈ 45°. ✓
trig-ratio-basiccosine-formula
#32 — gen-2d-cosine-formula-2-3b1cad59
intermediate 2d-cosine-formula
EN
In △ABC, AB = 7, BC = 15 and ∠ABC = 60°. Find AC.
中文
在 △ABC 中，AB = 7，BC = 15，∠ABC = 60°。求 AC。
A. 169
B. 13 ✓
C. 9.5987
D. 19.4679
Trap Analysis
  • A (169): Gave AC² instead of AC — forgot to take square root
  • C (9.5987): Used sin instead of cos in cosine formula
  • D (19.4679): Forgot negative sign — used cos(120°) instead of cos(60°)
generatedABC71560°
Step-by-step Solution
1 AC² = 7² + 15² − 2(7)(15)cos 60°.
2 = 49 + 225 − 210 × 0.5 = 169.
3 AC = √(169) = 13.
1 AC² = 7² + 15² − 2(7)(15)cos 60°.
2 = 49 + 225 − 210 × 0.5 = 169.
3 AC = √(169) = 13.
Verification
Check: 13² = 169. ✓
cosine-formula
#33 — gen-trig-identity-3-16fb6813
basic trig-identity
EN
Simplify sin(180° + θ) + cos(90° − θ).
中文
化簡 sin(180° + θ) + cos(90° − θ)。
A. 1
B. 2 sin θ
C. 0 ✓
D. -1
Trap Analysis
  • A (1): Incorrect identity — sin²+cos²=1 misapplied
  • B (2 sin θ): Did not recognise sin(180°+θ) = −sin θ
  • D (-1): Sign error
Step-by-step Solution
1 Apply standard trig identities:
2 sin(180° + θ) = −sin θ, cos(90° − θ) = sin θ, cos(360° − θ) = cos θ, cos(180° + θ) = −cos θ.
3 sin²θ + cos²θ = 1. tan θ = sin θ / cos θ.
4 Result: 0.
1 運用標準三角恒等式：
2 sin(180° + θ) = −sin θ, cos(90° − θ) = sin θ, cos(360° − θ) = cos θ, cos(180° + θ) = −cos θ.
3 sin²θ + cos²θ = 1. tan θ = sin θ / cos θ.
4 得出：0.
Verification
Substitute θ = 30° to verify numerically. ✓
trig-identity-supplementary
#34 — gen-cube-1-d1ade261
advanced cube
EN
ABCDEFGH is a cube where ABCD is the bottom face and EFGH is the top face. Let α be the angle between △AFG and △AFH. Find α, correct to the nearest degree.
中文
ABCDEFGH 是一個正方體，其中 ABCD 為底面，EFGH 為頂面。設 α 為 △AFG 和 △AFH 之間的角。求 α（準確至最接近的度）。
A. 30°
B. 40°
C. 55°
D. 35° ✓
Trap Analysis
  • A (30°): Close angle — arithmetic slip
  • B (40°): Close angle — rounding or formula error
  • C (55°): Complementary angle confusion
generatedABCDEFGH
Step-by-step Solution
1 Let side = a. Set up coordinates: A(0,0,0), B(a,0,0), ..., H(0,a,a).
2 Planes △AFG and △AFH share edge AF.
3 Find normals to each plane, or use perpendicular components along the shared edge.
4 Dihedral angle cos α = 0.8165.
5 α = arccos(0.8165) ≈ 35°.
1 設邊長為 a。 建立坐標系： A(0,0,0), B(a,0,0), ..., H(0,a,a).
2 平面 △AFG 與 △AFH 共用棱 AF。
3 先求各平面的法向量，或利用沿共用棱的垂直分量計算。
4 二面角的餘弦值 cos α = 0.8165.
5 α = arccos(0.8165) ≈ 35°.
Verification
Side length cancels. Result independent of a. ✓
3d-dihedral-angle3d-figure-decomposition
#35 — gen-p6-percent-increase-2-870eb7b2
basic p6-percent-increase
EN
A notebook costs $266. The price is increased by 5%. What is the new price (in dollars)?
中文
一本筆記簿原價 266 元。價格上升 5%。求新價格（元）。
A. 273
B. 13
C. 253
D. 279 ✓
Trap Analysis
  • A (273): Used half of the percentage change
  • B (13): Gave increase amount instead of new total
  • C (253): Percentage direction error (decrease instead of increase)
Step-by-step Solution
1 Increase amount = 266 × 5% = 13.
2 New price = 266 + 13 = 279.
1 增加量 = 266 × 5% = 13.
2 新價格 = 266 + 13 = 279.
Verification
266 × 1.05 = 279. ✓
p6-percent-increase
#36 — gen-p6-broken-line-ratio-2-79577be3
basic p6-broken-line-ratio
EN
The broken line graph shows daily steps. Tuesday's steps are how many times Sunday's steps?
中文
折線圖顯示每日步數。星期二的步數是星期日的多少倍？
A. 1
B. 4
C. 3
D. 2 ✓
Trap Analysis
  • A (1): Used difference idea instead of ratio
  • B (4): Factor of 2 error
  • C (3): Ratio computation slip
generated 0200040006000 SunMonTueWedThuFriSat
Step-by-step Solution
1 From the graph: Sunday = 3000, Tuesday = 6000.
2 Times = Tuesday ÷ Sunday = 6000 ÷ 3000 = 2.
1 由圖可得：星期日 = 3000, 星期二 = 6000.
2 倍數 = 星期二 ÷ 星期日 = 6000 ÷ 3000 = 2.
Verification
3000 × 2 = 6000. ✓
p6-broken-line-ratio
#37 — gen-regular-tetrahedron-1-b8c7fdea
advanced regular-tetrahedron
EN
ABCD is a regular tetrahedron with all edges of length 6. Find its volume.
中文
ABCD 是一個正四面體，所有棱長為 6。求其體積。
A. 19
B. 36
C. 18√2 ✓
D. 9
Trap Analysis
  • A (19): Arithmetic error in height computation
  • B (36): Factor of 2 error
  • D (9): Missing factor in formula
generatedBCDA6
Step-by-step Solution
1 Base area = (√3/4) × 6² = 15.5885.
2 Height of regular tetrahedron = 6√6/3 = 4.899.
3 Volume = ⅓ × base area × height = ⅓ × 15.5885 × 4.899 = 18√2.
1 底面積 = (√3/4) × 6² = 15.5885.
2 正四面體的高 = 6√6/3 = 4.899.
3 體積 = ⅓ × 底面積 × 高 = ⅓ × 15.5885 × 4.899 = 18√2.
Verification
V = 6³/(6√2) = 25.4558. ✓
3d-dihedral-angle3d-figure-decomposition3d-pythagoras
#38 — gen-box-whisker-3-a1e516da
basic box-whisker
EN
The box-and-whisker diagram below shows the distribution of scores. The minimum is 5, Q₁ = 17, median = 25, Q₃ = 31, and maximum = 47. Find the inter-quartile range.
中文
下面的箱形圖顯示了分數的分佈。 最小值為 5, Q₁ = 17, 中位數 = 25, Q₃ = 31, and 最大值 = 47. 求四分位距。
A. 42
B. 30
C. 26
D. 14 ✓
Trap Analysis
  • A (42): Gave range instead of IQR
  • B (30): Used max − Q₁ instead of Q₃ − Q₁
  • C (26): Used Q₃ − min instead of Q₃ − Q₁
generated 5 17 25 31 47
Step-by-step Solution
1 IQR = Q₃ − Q₁ = 31 − 17 = 14.
1 IQR = Q₃ − Q₁ = 31 − 17 = 14.
Verification
Q₃ = 31, Q₁ = 17. Difference = 14. ✓
box-whisker-readingiqr-definition
#39 — gen-p6-broken-line-ratio-3-4515bcde
basic p6-broken-line-ratio
EN
The broken line graph shows daily steps. Tuesday's steps are how many times Sunday's steps?
中文
折線圖顯示每日步數。星期二的步數是星期日的多少倍？
A. 3
B. 0
C. 5
D. 4 ✓
Trap Analysis
  • A (3): Used difference idea instead of ratio
  • B (0): Inverted ratio
  • C (5): Ratio computation slip
generated 0200040006000800010000120001400016000 SunMonTueWedThuFriSat
Step-by-step Solution
1 From the graph: Sunday = 4000, Tuesday = 16000.
2 Times = Tuesday ÷ Sunday = 16000 ÷ 4000 = 4.
1 由圖可得：星期日 = 4000, 星期二 = 16000.
2 倍數 = 星期二 ÷ 星期日 = 16000 ÷ 4000 = 4.
Verification
4000 × 4 = 16000. ✓
p6-broken-line-ratio
#40 — gen-central-tendency-3-c087475c
intermediate central-tendency
EN
Data: 68, 68, 79, 88, 86, 98, 68, 98, 32, c. Mode = 68, mean = 75. Find the median.
中文
數據：68, 68, 79, 88, 86, 98, 68, 98, 32，c。眾數 = 68，平均值 = 75。求中位數。
A. 68
B. 75.5
C. 75
D. 73.5 ✓
Trap Analysis
  • A (68): Gave mode instead of median
  • B (75.5): Did not sort after finding c
  • C (75): Gave mean instead of median
Step-by-step Solution
1 Mode = 68. Mean = 75. Sum of known = 685.
2 c = 75 × 10 − sum = 65.
3 Sort all values, find median of 10 values.
4 Median = 73.5.
1 眾數 = 68. Mean = 75. 已知數的和 = 685.
2 c = 75 × 10 − sum = 65.
3 將所有值排序，求中位數 of 10 values.
4 Median = 73.5.
Verification
Sorted array confirmed. Middle value = 73.5. ✓
central-tendency-combinedmode-mean-median-unknowns
#41 — gen-standard-score-2-37501151
basic standard-score
EN
In an examination, the mean score is 55 and the standard deviation is 8. A student gets 71 marks. Find the standard score of the student.
中文
在一次考試中，平均分為 55，標準差為 8。 一名學生得 71 分。求該學生的標準分。
A. 2 ✓
B. -2
C. 16
D. 1.290909090909091
Trap Analysis
  • B (-2): Sign error — reversed (x-μ) to (μ-x)
  • C (16): Forgot to divide by σ
  • D (1.290909090909091): Divided x by mean instead of using z-score formula
Step-by-step Solution
1 z = (x − μ) / σ = (71 − 55) / 8 = 16 / 8 = 2.
1 z = (x − μ) / σ = (71 − 55) / 8 = 16 / 8 = 2.
Verification
2 × 8 + 55 = 71. ✓
standard-score
#42 — gen-quadrilateral-3-5a608349
intermediate quadrilateral
EN
In quadrilateral ABCD, ∠ABC = 30°, AB = 11 and BC = 11. Find AC.
中文
在四邊形 ABCD 中，∠ABC = 30°，AB = 11，BC = 11。求 AC。
A. 6.694
B. 0
C. 22
D. 5.694 ✓
Trap Analysis
  • A (6.694): Arithmetic error in cosine formula
  • B (0): Subtracted sides instead of using cosine formula
  • C (22): Added sides instead of using cosine formula
generatedABCD111130°
Step-by-step Solution
1 By cosine formula: AC² = 11² + 11² − 2(11)(11)cos 30°.
2 = 121 + 121 − 242cos 30° = 32.4219.
3 AC = 5.694.
1 根據餘弦公式： AC² = 11² + 11² − 2(11)(11)cos 30°.
2 = 121 + 121 − 242cos 30° = 32.4219.
3 AC = 5.694.
Verification
√(32.4219) = 5.694. ✓
trig-ratio-basiccosine-formula
#43 — gen-p6-percent-decrease-1-be5e6827
basic p6-percent-decrease
EN
A jacket is sold at $306. During a sale, the price is reduced by 25%. Find the sale price (in dollars).
中文
一件外套原價 306 元。減價期間打折 25%。求減價後價格（元）。
A. 229 ✓
B. 77
C. 383
D. 267
Trap Analysis
  • B (77): Gave decrease amount instead of sale price
  • C (383): Percentage direction error (increase instead of decrease)
  • D (267): Used half of the percentage change
Step-by-step Solution
1 Decrease amount = 306 × 25% = 77.
2 Sale price = 306 − 77 = 229.
1 減少量 = 306 × 25% = 77.
2 減價後價格 = 306 − 77 = 229.
Verification
306 × 0.75 = 229. ✓
p6-percent-decrease
#44 — gen-trig-equation-2-34f0ccec
intermediate trig-equation
EN
Find the number of roots of cos x = 1/2 in 0° ≤ x < 360°.
中文
求 cos x = 1/2 在 0° ≤ x < 360° 中的根的數目。
A. 4
B. 1
C. 3
D. 2 ✓
Trap Analysis
  • A (4): Counted roots from both sin and cos (double-counting)
  • B (1): Missed a root outside primary range
  • C (3): Counted a repeated root as distinct
Step-by-step Solution
1 Rewrite equation in terms of a single trig function.
2 Solve the resulting equation: find values of the trig function.
3 Check which solutions lie in 0° ≤ x < 360° — each valid value gives 1 root (if ±1) or 2 roots.
4 Total roots: 2.
1 將方程化為單一三角函數的形式。
2 解所得方程，求出三角函數的值。
3 檢查哪些解落在 0° ≤ x < 360° — 每個有效值對應 1 個根（若為 ±1）或 2 個根。
4 根的總數：2.
Verification
Graph method: count intersections in [0°, 360°). ✓
trig-equation-root-counting
#45 — gen-translation-scaling-5-8ea4565b
intermediate translation-scaling
EN
The variance of a set of numbers is 36. Each number is multiplied by 3 and then 5 is added. Find the new variance.
中文
一組數的方差為 36。每個數乘以 3 再加 5。 求新的方差。
A. 324 ✓
B. 18
C. 108
D. 329
Trap Analysis
  • B (18): Gave SD instead of variance
  • C (108): Multiplied variance by |k| instead of k²
  • D (329): Added c to variance (translation changes variance)
Step-by-step Solution
1 Original variance = 36.
2 Transform: multiply by 3, then add 5.
3 Adding/subtracting a constant does not change variance (translation invariance).
4 Multiplying by 3 scales variance by 3² = 9.
5 New variance = 9 × 36 = 324.
1 原方差 = 36.
2 變換： 乘以 3, 再加 5.
3 加減常數不改變方差（平移不變性）。
4 乘以 3 使方差乘以 3² = 9.
5 新方差 = 9 × 36 = 324.
Verification
Var(kX + c) = k²Var(X). 3² × 36 = 324. ✓
scaling-effecttranslation-invariance
#46 — gen-central-tendency-4-f9ffc717
intermediate central-tendency
EN
Mean of 3, 5, 8, 8, 10, 12, m, n is 7. If m + n = 10, which MUST be true? I. Mode is 8. II. Median ≤ 8. III. Range ≥ 9.
中文
3, 5, 8, 8, 10, 12，m，n 的平均值為 7。 若 m + n = 10，以下哪些必定成立？ I. 眾數為 8。 II. 中位數 ≤ 8。 III. 全距 ≥ 9。
A. II only ✓
B. I, II and III
C. III only
D. I and II
Trap Analysis
  • B (I, II and III): Did not verify all constraints across all valid pairs
  • C (III only): Only checked range constraint
  • D (I and II): Assumed mode is always 8 without checking all valid (m,n)
Step-by-step Solution
1 m + n = 10. Enumerate valid (m, n) pairs.
2 For each valid pair: check mode, median, range.
3 Only statements true for ALL pairs are "must be true". Answer: II only.
1 m + n = 10. 列舉所有有效的 (m, n) 對.
2 For each valid pair: check mode, median, range.
3 只有對所有組合都成立的陳述才是「必定成立」。 Answer: II only.
Verification
Exhaustive enumeration confirms. ✓
constrained-dataset-reasoningcentral-tendency-combined
#47 — gen-p6-circle-area-1-d3c2a2fb
basic p6-circle-area
EN
In the circle shown, the radius is 3 cm. Find the area (take π as 3.14).
中文
在圖中的圓，半徑為 3 cm。求面積（取 π = 3.14）。
A. 9
B. 28 ✓
C. 19
D. 50
Trap Analysis
  • A (9): Used pi*r instead of pi*r^2
  • C (19): Used circumference formula instead of area
  • D (50): Squared wrong radius value
generated O r = 3
Step-by-step Solution
1 Area A = πr².
2 A = 3.14 × 3² = 28 (approx).
1 面積 A = πr²。
2 A = 3.14 × 3² = 28 (approx).
Verification
3.14 × 9 = 28. ✓
p6-circle-area
#48 — gen-connected-triangles-1-1b7f3e6f
intermediate connected-triangles
EN
In the figure, AC ⊥ BD. AC = 21 and AB = 75. Find sin∠ABC.
中文
在圖中，AC ⊥ BD。AC = 21，AB = 75。求 sin∠ABC。
A. 25/7
B. 7/25 ✓
C. 24/25
D. 8/25
Trap Analysis
  • A (25/7): Inverted fraction
  • C (24/25): Used adjacent/hypotenuse (cos instead of sin)
  • D (8/25): Wrong numerator — likely arithmetic slip
generatedABCD2175
Step-by-step Solution
1 AC ⊥ BD. AB = 75.
2 sin∠ABC = AC/AB = 21/75 = 7/25.
1 AC ⊥ BD. AB = 75.
2 sin∠ABC = AC/AB = 21/75 = 7/25.
Verification
sin = opp/hyp in right △ACB. ✓
trig-ratio-chaintrig-ratio-basic
#49 — gen-p6-average-missing-1-ed9f2526
intermediate p6-average-missing
EN
The average of five numbers is 19.2. Four of them are 21, 12, 19, 23. Find the fifth number.
中文
五個數的平均數是 19.2，其中四個數是 21, 12, 19, 23。求第五個數。
A. 96
B. 75
C. 19
D. 21 ✓
Trap Analysis
  • A (96): Gave total sum, not missing value
  • B (75): Gave sum of known numbers only
  • C (19): Subtraction slip in missing term
Step-by-step Solution
1 Total sum = average × 5 = 19.2 × 5 = 96.
2 Sum of known numbers = 21 + 12 + 19 + 23 = 75.
3 Missing number = 96 − 75 = 21.
1 總和 = average × 5 = 19.2 × 5 = 96.
2 已知數之和 = 21 + 12 + 19 + 23 = 75.
3 缺少的數 = 96 − 75 = 21.
Verification
(75 + 21) ÷ 5 = 19.2. ✓
p6-average-missing
#50 — gen-p6-equation-axb-c-2-d319b7ac
basic p6-equation-axb-c
EN
Solve the equation: 5x + 4 = 74.
中文
解方程：5x + 4 = 74。
A. 15
B. 70
C. 14 ✓
D. 13
Trap Analysis
  • A (15): Moved constant with wrong sign
  • B (70): Forgot to divide by coefficient of x
  • D (13): Off-by-one in root counting
Step-by-step Solution
1 5x + 4 = 74
2 5x = 74 − 4 = 70
3 x = 70 ÷ 5 = 14.
1 5x + 4 = 74
2 5x = 74 − 4 = 70
3 x = 70 ÷ 5 = 14.
Verification
5(14) + 4 = 74. ✓
p6-equation-axb-c
#51 — gen-right-pyramid-square-base-2-62df4d75
advanced right-pyramid-square-base
EN
VABCD is a right pyramid with square base ABCD. V is directly above the centre of the base. AB : VA = 4 : 3. Let θ be the angle between △ABV and △CBV. Find cos θ.
中文
VABCD 是一個以 ABCD 為正方形底面的正四角錐。V 在底面中心的正上方。AB : VA = 4 : 3。設 θ 為 △ABV 和 △CBV 之間的角。求 cos θ。
A. -4/5 ✓
B. 24/5
C. -4/7
D. 16/5
Trap Analysis
  • B (24/5): Arithmetic error in pyramid height computation
  • C (-4/7): Used edge ratio directly instead of computing dihedral
  • D (16/5): Wrong sign or magnitude in dihedral cosine formula
generatedABCDV
Step-by-step Solution
1 Square base side = 4k, lateral edge = 3k (ratio 4:3).
2 Half-diagonal of base = 4k√2/2. Height = √((3k)² − (4k√2/2)²) = k√(9 − 16/2).
3 Dihedral angle along edge BV between faces △ABV and △CBV.
4 cos θ = -4/5.
1 正方形底邊 = 4k, 側棱 = 3k (比例 4:3).
2 底面半對角線 = 4k√2/2. 高度 = √((3k)² − (4k√2/2)²) = k√(9 − 16/2).
3 沿棱的二面角 BV 在兩面之間 △ABV and △CBV.
4 cos θ = -4/5.
Verification
Height = 1 (per unit). Angle reasonable for pyramid with ratio 4:3. ✓
3d-dihedral-angle3d-figure-decomposition3d-cosine-formula
#52 — gen-box-whisker-4-8a15bc45
basic box-whisker
EN
The box-and-whisker diagram below shows the distribution of scores. The minimum is 26, Q₁ = 36, median = 48, Q₃ = 60, and maximum = 71. Find the inter-quartile range.
中文
下面的箱形圖顯示了分數的分佈。 最小值為 26, Q₁ = 36, 中位數 = 48, Q₃ = 60, and 最大值 = 71. 求四分位距。
A. 24 ✓
B. 45
C. 34
D. 35
Trap Analysis
  • B (45): Gave range instead of IQR
  • C (34): Used Q₃ − min instead of Q₃ − Q₁
  • D (35): Used max − Q₁ instead of Q₃ − Q₁
generated 26 36 48 60 71
Step-by-step Solution
1 IQR = Q₃ − Q₁ = 60 − 36 = 24.
1 IQR = Q₃ − Q₁ = 60 − 36 = 24.
Verification
Q₃ = 60, Q₁ = 36. Difference = 24. ✓
box-whisker-readingiqr-definition
#53 — gen-p6-average-missing-2-36ec4881
intermediate p6-average-missing
EN
The average of five numbers is 20.2. Four of them are 23, 20, 23, 19. Find the fifth number.
中文
五個數的平均數是 20.2，其中四個數是 23, 20, 23, 19。求第五個數。
A. 101
B. 14
C. 16 ✓
D. 85
Trap Analysis
  • A (101): Gave total sum, not missing value
  • B (14): Subtraction slip in missing term
  • D (85): Gave sum of known numbers only
Step-by-step Solution
1 Total sum = average × 5 = 20.2 × 5 = 101.
2 Sum of known numbers = 23 + 20 + 23 + 19 = 85.
3 Missing number = 101 − 85 = 16.
1 總和 = average × 5 = 20.2 × 5 = 101.
2 已知數之和 = 23 + 20 + 23 + 19 = 85.
3 缺少的數 = 101 − 85 = 16.
Verification
(85 + 16) ÷ 5 = 20.2. ✓
p6-average-missing
#54 — gen-p6-equation-a-x+b-c-1-5c940081
intermediate p6-equation-a-x+b-c
EN
Solve the equation: 8(x + 1) = 64.
中文
解方程：8(x + 1) = 64。
A. 6
B. 8
C. 9
D. 7 ✓
Trap Analysis
  • A (6): Arithmetic slip in final step
  • B (8): Stopped after dividing by coefficient (forgot final subtraction)
  • C (9): Off-by-one in root counting
Step-by-step Solution
1 8(x + 1) = 64
2 x + 1 = 64 ÷ 8 = 8
3 x = 8 − 1 = 7.
1 8(x + 1) = 64
2 x + 1 = 64 ÷ 8 = 8
3 x = 8 − 1 = 7.
Verification
8(7 + 1) = 64. ✓
p6-equation-a-x+b-c
#55 — gen-box-whisker-5-6602ad90
basic box-whisker
EN
From a box-and-whisker diagram: minimum = 23, Q₁ = 30, median = 36, Q₃ = 40, maximum = 54. Find the range.
中文
根據箱形圖： 最小值 = 23, Q₁ = 30, 中位數 = 36, Q₃ = 40, 最大值 = 54. 求全距。
A. 10
B. 17
C. 24
D. 31 ✓
Trap Analysis
  • A (10): Gave IQR instead of range
  • B (17): Used Q₃ − min
  • C (24): Used max − Q₁
generated 23 30 36 40 54
Step-by-step Solution
1 Range = max − min = 54 − 23 = 31.
1 全距 = 最大值 − 最小值 = 54 − 23 = 31.
Verification
Direct subtraction. ✓
box-whisker-readingiqr-definition
#56 — gen-standard-score-simult-1-1e8c16de
advanced standard-score-simult
EN
In a test, a student who scores 52.5 has standard score -1.5, and another student who scores 70 has standard score 2. Find the mean.
中文
在一次測驗中，一名得 52.5 分的學生的標準分為 -1.5，另一名得 70 分的學生的標準分為 2。求平均分。
A. 55
B. 60 ✓
C. 61.25
D. 65
Trap Analysis
  • A (55): Subtracted σ from μ — sign error in simultaneous equations
  • C (61.25): Averaged the two scores instead of solving z-score equations
  • D (65): Added σ to μ
Step-by-step Solution
1 (52.5 − μ) / σ = -1.5 … ①
2 (70 − μ) / σ = 2 … ②
3 ② − ①: (70 − 52.5) / σ = 2 − (-1.5) = 3.5.
4 σ = 17.5 / 3.5 = 5.
5 From ②: μ = 70 − 2 × 5 = 60.
1 (52.5 − μ) / σ = -1.5 … ①
2 (70 − μ) / σ = 2 … ②
3 ② − ①: (70 − 52.5) / σ = 2 − (-1.5) = 3.5.
4 σ = 17.5 / 3.5 = 5.
5 From ②: μ = 70 − 2 × 5 = 60.
Verification
Check ①: (52.5 − 60) / 5 = -1.5. ✓
standard-score-simultaneous
#57 — gen-perpendicular-to-plane-2-2853aaa0
advanced perpendicular-to-plane
EN
In the figure, △CBD lies on a horizontal plane with CB = 16m, BD = 30m and CD = 34m. AB = 14m is perpendicular to the plane CBD. Find the angle between AC and the plane CBD, correct to the nearest degree.
中文
在圖中，△CBD 位於水平面上，其中 CB = 16m，BD = 30m，CD = 34m。AB = 14m 垂直於平面 CBD。求 the angle between AC and the plane CBD（準確至最接近的度）。
A. 41° ✓
B. 46°
C. 49°
D. 139°
Trap Analysis
  • B (46°): Rounding error in inverse trig
  • C (49°): Complementary angle confusion
  • D (139°): Supplementary angle confusion
generated16303414ABCDE
Step-by-step Solution
1 △CBD has sides 16, 30, 34. Since 16² + 30² = 1156 = 1156 = 34², it is right-angled at B.
2 The angle between AC and the plane is ∠ACB.
3 tan(angle) = AB/CB = 14/16.
4 angle = arctan(14/16) ≈ 41°.
1 △CBD 的邊長為 16, 30, 34。 由於 16² + 30² = 1156 = 1156 = 34²，在 B 成直角.
2 AC 與平面的夾角為 ∠ACB。
3 tan(角) = AB/CB = 14/16.
4 角 = arctan(14/16) ≈ 41°.
Verification
tan⁻¹(0.875) = 41.1859°. ✓
3d-perpendicular-projection3d-pythagoras
#58 — gen-p6-pie-percent-count-1-95e166bb
basic p6-pie-percent-count
EN
A pie chart represents 160 students. The Apple sector is 25% of the circle. How many students chose Apple?
中文
某圓形圖代表 160 名學生。Apple 扇形佔全圓 25%。有多少名學生選擇 Apple？
A. 25
B. 4000
C. 120
D. 40 ✓
Trap Analysis
  • A (25): Gave percentage value instead of count
  • B (4000): Forgot to divide by 100
  • C (120): Used complement percentage
generated Apple (25)Banana (30)Orange (20)Grape (25)
Step-by-step Solution
1 Apple count = 25% of 160.
2 = 160 × 0.25 = 40.
1 Apple 人數 = 25% 佔 160。
2 = 160 × 0.25 = 40.
Verification
40 ÷ 160 = 25%. ✓
p6-pie-percent-to-count
#59 — gen-central-tendency-5-0c53596a
intermediate central-tendency
EN
Data: 79, 68, 68, 32, 68, 86, 88, 98, 98, c. Mode = 68, mean = 77. Find the median.
中文
數據：79, 68, 68, 32, 68, 86, 88, 98, 98，c。眾數 = 68，平均值 = 77。求中位數。
A. 68
B. 77
C. 84
D. 82 ✓
Trap Analysis
  • A (68): Gave mode instead of median
  • B (77): Gave mean instead of median
  • C (84): Did not sort after finding c
Step-by-step Solution
1 Mode = 68. Mean = 77. Sum of known = 685.
2 c = 77 × 10 − sum = 85.
3 Sort all values, find median of 10 values.
4 Median = 82.
1 眾數 = 68. Mean = 77. 已知數的和 = 685.
2 c = 77 × 10 − sum = 85.
3 將所有值排序，求中位數 of 10 values.
4 Median = 82.
Verification
Sorted array confirmed. Middle value = 82. ✓
central-tendency-combinedmode-mean-median-unknowns
#60 — gen-p6-percent-decrease-2-bda2b096
basic p6-percent-decrease
EN
A jacket is sold at $138. During a sale, the price is reduced by 25%. Find the sale price (in dollars).
中文
一件外套原價 138 元。減價期間打折 25%。求減價後價格（元）。
A. 103 ✓
B. 35
C. 173
D. 120
Trap Analysis
  • B (35): Gave decrease amount instead of sale price
  • C (173): Percentage direction error (increase instead of decrease)
  • D (120): Used half of the percentage change
Step-by-step Solution
1 Decrease amount = 138 × 25% = 35.
2 Sale price = 138 − 35 = 103.
1 減少量 = 138 × 25% = 35.
2 減價後價格 = 138 − 35 = 103.
Verification
138 × 0.75 = 103. ✓
p6-percent-decrease
#61 — gen-p6-broken-line-readmax-1-c45bdce4
basic p6-broken-line-readmax
EN
The broken line graph shows the number of steps walked in a week. What is the greatest number of steps shown in the graph?
中文
折線圖顯示一星期步數。圖中最大的步數是多少？
A. 7000 ✓
B. 6000
C. 7001
D. 8000
Trap Analysis
  • B (6000): Axis reading error by one tick
  • C (7001): Off-by-one in root counting
  • D (8000): Axis reading error by one tick
generated 02000400060008000 SunMonTueWedThuFriSat
Step-by-step Solution
1 Read each data point from the graph.
2 The largest value shown is 7000.
1 從圖中讀取各數據點。
2 圖中最大值為 7000.
Verification
max(5000, 3000, 7000, 4000, 5000, 5000, 7000) = 7000. ✓
p6-broken-line-read
#62 — gen-translation-scaling-6-57003401
intermediate translation-scaling
EN
The standard deviation of a set of numbers is 7. If each number is multiplied by 6 and then 9 is added, find the new standard deviation.
中文
一組數的標準差為 7。 若每個數乘以 6 再加 9，求新的標準差。
A. 252
B. 51
C. 7
D. 42 ✓
Trap Analysis
  • A (252): Multiplied SD by k² instead of |k|
  • B (51): Added c to SD (translation changes SD)
  • C (7): Thought scaling does not change SD
Step-by-step Solution
1 Original SD = 7.
2 Adding 9 does not change SD (translation invariance).
3 Multiplying by 6 scales SD by |6| = 6.
4 New SD = 6 × 7 = 42.
1 原標準差 = 7.
2 加常數不改變標準差（平移不變性）。
3 乘以 6 使標準差乘以 |6| = 6.
4 新標準差 = 6 × 7 = 42.
Verification
SD(kX + c) = |k| × SD(X). ✓
scaling-effecttranslation-invariance
#63 — gen-p6-pie-angle-1-705320b9
basic p6-pie-angle
EN
In a class survey, 30 out of 72 students chose option A. In a pie chart, what is the angle of sector A?
中文
班內調查中，72 名學生中有 30 名選 A。若用圓形圖表示，A 扇形的角度是多少？
A. 150 ✓
B. 132
C. 42
D. 210
Trap Analysis
  • B (132): Arithmetic slip in scaling to 360°
  • C (42): Gave percentage instead of angle
  • D (210): Used complement sector angle
generated A (30)B (30)C (12)
Step-by-step Solution
1 Sector angle = (part ÷ total) × 360°.
2 = (30 ÷ 72) × 360° = 150°.
1 扇形角 =（部分 ÷ 總數）× 360°。
2 = (30 ÷ 72) × 360° = 150°.
Verification
30/72 = 150/360. ✓
p6-pie-angle
#64 — gen-p6-broken-line-readmax-2-06dd2eda
basic p6-broken-line-readmax
EN
The broken line graph shows the number of steps walked in a week. What is the greatest number of steps shown in the graph?
中文
折線圖顯示一星期步數。圖中最大的步數是多少？
A. 5000
B. 4000
C. 6000 ✓
D. 7000
Trap Analysis
  • A (5000): Axis reading error by one tick
  • B (4000): Read second-highest point as maximum
  • D (7000): Axis reading error by one tick
generated 0200040006000 SunMonTueWedThuFriSat
Step-by-step Solution
1 Read each data point from the graph.
2 The largest value shown is 6000.
1 從圖中讀取各數據點。
2 圖中最大值為 6000.
Verification
max(4000, 2000, 3000, 4000, 4000, 4000, 6000) = 6000. ✓
p6-broken-line-read
#65 — gen-standard-score-simult-2-088be167
advanced standard-score-simult
EN
Scores: 61, x, 77. Standard scores: -1, 1, 3. Find x.
中文
分數：61, x, 77。標準分：-1, 1, 3。求 x。
A. 61
B. 70
C. 69 ✓
D. 65
Trap Analysis
  • A (61): Gave x₁ instead of computing from z = 1
  • B (70): Arithmetic error in final substitution
  • D (65): Gave the mean instead of the target score
Step-by-step Solution
1 From the two known z-scores, solve for μ and σ.
2 (77 − 61) / σ = 4 ⟹ σ = 4.
3 μ = 65. x = σ(1) + μ = 4 + 65 = 69.
1 From the two known z-scores, solve for μ and σ.
2 (77 − 61) / σ = 4 ⟹ σ = 4.
3 μ = 65. x = σ(1) + μ = 4 + 65 = 69.
Verification
z = (69 − 65) / 4 = 1. ✓
standard-score-simultaneous
#66 — gen-connected-triangles-2-20522413
intermediate connected-triangles
EN
In the figure, AC ⊥ BD. AC = 15, BC = 36 and CD = 26. Find tan∠ABC.
中文
在圖中，AC ⊥ BD。AC = 15，BC = 36，CD = 26。求 tan∠ABC。
A. 12/5
B. 15/62 ✓
C. 5/13
D. 5/12
Trap Analysis
  • A (12/5): Inverted ratio (adjacent/opposite)
  • C (5/13): Used AB as denominator instead of BD
  • D (5/12): Used BC instead of BD for denominator
generatedABCD153626
Step-by-step Solution
1 AC ⊥ BD, so △ACB is right-angled at C.
2 AB = √(15² + 36²) = 39. BD = 36 + 26 = 62.
3 tan∠ABC = AC/BC = 15/62 = 15/62.
1 AC ⊥ BD, 因此 △ACB 在 C 成直角.
2 AB = √(15² + 36²) = 39. BD = 36 + 26 = 62.
3 tan∠ABC = AC/BC = 15/62 = 15/62.
Verification
Direct from right triangle. ✓
trig-ratio-chaintrig-ratio-basic
#67 — gen-single-right-triangle-1-43818dec
basic single-right-triangle
EN
In △ABC, ∠B = 90°. AB = 6 and BC = 8. Find tan∠C.
中文
在 △ABC 中，∠B = 90°。AB = 6，BC = 8。求 tan∠C。
A. 4/5
B. 4/3
C. 3/5
D. 3/4 ✓
Trap Analysis
  • A (4/5): Used cos instead of tan
  • B (4/3): Inverted ratio (adjacent/opposite)
  • C (3/5): Used sin instead of tan
generated68ABC
Step-by-step Solution
1 ∠B = 90°. In right △ABC:
2 tan∠C = opp/adj = 6/8 = 3/4.
1 ∠B = 90°. 在直角 △ABC:
2 tan∠C = 對邊/鄰邊 = 6/8 = 3/4.
Verification
6² + 8² = 100 = 100 = 10². Pythagorean triple confirmed. ✓
trig-ratio-basic
#68 — gen-p6-broken-line-readmax-3-50451e7e
basic p6-broken-line-readmax
EN
The broken line graph shows the number of steps walked in a week. What is the greatest number of steps shown in the graph?
中文
折線圖顯示一星期步數。圖中最大的步數是多少？
A. 6000
B. 8000
C. 7001
D. 7000 ✓
Trap Analysis
  • A (6000): Axis reading error by one tick
  • B (8000): Axis reading error by one tick
  • C (7001): Off-by-one in root counting
generated 02000400060008000 SunMonTueWedThuFriSat
Step-by-step Solution
1 Read each data point from the graph.
2 The largest value shown is 7000.
1 從圖中讀取各數據點。
2 圖中最大值為 7000.
Verification
max(7000, 3000, 3000, 5000, 6000, 7000, 6000) = 7000. ✓
p6-broken-line-read
#69 — gen-weighted-mean-stats-2-05512ac5
basic weighted-mean-stats
EN
14 teachers and 100 students have an overall mean height of 38 cm. The students' mean height is 108 cm. Find the teachers' mean height.
中文
14 名老師和 100 名學生的整體平均身高為 38 cm。學生的平均身高為 108 cm。求老師的平均身高。
A. -448
B. 467
C. -462 ✓
D. -177
Trap Analysis
  • A (-448): Miscounted total number of items
  • B (467): Arithmetic error in weighted mean equation
  • D (-177): Used simple average instead of weighted mean
Step-by-step Solution
1 Total sum = 114 × 38 = 4332.
2 Sub-group sum = 100 × 108 = 10800.
3 Remaining mean = (4332 − 10800) / 14 = -462.
1 總和 = 114 × 38 = 4332.
2 子組總和 = 100 × 108 = 10800.
3 其餘平均值 = (4332 − 10800) / 14 = -462.
Verification
14 × -462 + 100 × 108 = 114 × 38. ✓
weighted-mean
#70 — gen-p6-pie-angle-2-3f1a91e6
basic p6-pie-angle
EN
In a class survey, 30 out of 120 students chose option A. In a pie chart, what is the angle of sector A?
中文
班內調查中，120 名學生中有 30 名選 A。若用圓形圖表示，A 扇形的角度是多少？
A. 90 ✓
B. 25
C. 72
D. 270
Trap Analysis
  • B (25): Gave percentage instead of angle
  • C (72): Arithmetic slip in scaling to 360°
  • D (270): Used complement sector angle
generated A (30)B (78)C (12)
Step-by-step Solution
1 Sector angle = (part ÷ total) × 360°.
2 = (30 ÷ 120) × 360° = 90°.
1 扇形角 =（部分 ÷ 總數）× 360°。
2 = (30 ÷ 120) × 360° = 90°.
Verification
30/120 = 90/360. ✓
p6-pie-angle
#71 — gen-regular-tetrahedron-2-15c1cce5
advanced regular-tetrahedron
EN
ABCD is a regular tetrahedron with all edges of length 6. E is the foot of the perpendicular from A to BC. Let θ be the dihedral angle ∠AED between faces ABC and DBC. Find cos θ.
中文
ABCD 是一個正四面體，所有棱長為 6。E 是 A 到 BC 的垂足。設 θ 為面 ABC 和面 DBC 之間的二面角 ∠AED。求 cos θ。
A. 1/3 ✓
B. 1/2
C. √3/3
D. 2/3
Trap Analysis
  • B (1/2): Used face angle (60°) cosine instead of computing 3D angle
  • C (√3/3): Confusion with direction cosine
  • D (2/3): Wrong denominator in cos formula — used 2·edge² not 2·AO·BO
generatedBCDA6
Step-by-step Solution
1 All edges = 6. E is foot of perpendicular from B to BC.
2 In equilateral △BCD, the perpendicular from any vertex to the opposite side has length 6√3/2.
3 By symmetry, the dihedral angle between any two faces of a regular tetrahedron satisfies cos θ = 1/3.
1 所有邊長 = 6. E is 垂足 from B to BC.
2 在正三角形 △BCD 中，任一顶點到對邊的垂線長度為 6√3/2.
3 由對稱性，正四面體任意兩面間的二面角滿足 cos θ = 1/3.
Verification
arccos(1/3) ≈ 70.5°. Well-known result for regular tetrahedra. ✓
3d-dihedral-angle3d-figure-decomposition3d-pythagoras
#72 — gen-standard-score-simult-3-596cba70
advanced standard-score-simult
EN
In a test, a student who scores 47 has standard score -3, and another student who scores 77 has standard score 2. Find the mean.
中文
在一次測驗中，一名得 47 分的學生的標準分為 -3，另一名得 77 分的學生的標準分為 2。求平均分。
A. 59
B. 71
C. 65 ✓
D. 62
Trap Analysis
  • A (59): Subtracted σ from μ — sign error in simultaneous equations
  • B (71): Added σ to μ
  • D (62): Averaged the two scores instead of solving z-score equations
Step-by-step Solution
1 (47 − μ) / σ = -3 … ①
2 (77 − μ) / σ = 2 … ②
3 ② − ①: (77 − 47) / σ = 2 − (-3) = 5.
4 σ = 30 / 5 = 6.
5 From ②: μ = 77 − 2 × 6 = 65.
1 (47 − μ) / σ = -3 … ①
2 (77 − μ) / σ = 2 … ②
3 ② − ①: (77 − 47) / σ = 2 − (-3) = 5.
4 σ = 30 / 5 = 6.
5 From ②: μ = 77 − 2 × 6 = 65.
Verification
Check ①: (47 − 65) / 6 = -3. ✓
standard-score-simultaneous
#73 — gen-central-tendency-6-f9ffc717
intermediate central-tendency
EN
Mean of 3, 5, 8, 8, 10, 12, m, n is 7. If m + n = 10, which MUST be true? I. Mode is 8. II. Median ≤ 8. III. Range ≥ 9.
中文
3, 5, 8, 8, 10, 12，m，n 的平均值為 7。 若 m + n = 10，以下哪些必定成立？ I. 眾數為 8。 II. 中位數 ≤ 8。 III. 全距 ≥ 9。
A. I, II and III
B. I and II
C. III only
D. II only ✓
Trap Analysis
  • A (I, II and III): Did not verify all constraints across all valid pairs
  • B (I and II): Assumed mode is always 8 without checking all valid (m,n)
  • C (III only): Only checked range constraint
Step-by-step Solution
1 m + n = 10. Enumerate valid (m, n) pairs.
2 For each valid pair: check mode, median, range.
3 Only statements true for ALL pairs are "must be true". Answer: II only.
1 m + n = 10. 列舉所有有效的 (m, n) 對.
2 For each valid pair: check mode, median, range.
3 只有對所有組合都成立的陳述才是「必定成立」。 Answer: II only.
Verification
Exhaustive enumeration confirms. ✓
constrained-dataset-reasoningcentral-tendency-combined
#74 — gen-p6-percent-whole-1-c216b72f
intermediate p6-percent-whole
EN
In a class, 30 students are girls. This is 60% of the class. How many students are there in total?
中文
某班有 30 名女生，佔全班 60%。求全班總人數。
A. 50 ✓
B. 90
C. 1
D. 18
Trap Analysis
  • B (90): Added part and percent directly
  • C (1): Divided by percent value instead of decimal form
  • D (18): Found the part of part, not the whole
Step-by-step Solution
1 30 is 60% of the total.
2 Total = 30 ÷ 60% = 30 ÷ 0.6.
3 Total = 50.
1 30 是總數的 60%。
2 總數 = 30 ÷ 60% = 30 ÷ 0.6.
3 總數 = 50.
Verification
50 × 60% = 30. ✓
p6-percent-whole
#75 — gen-p6-circle-circumference-2-161728c4
basic p6-circle-circumference
EN
In the circle shown, the diameter is 12 cm. Estimate the circumference (take π as 3.14).
中文
在圖中的圓，直徑為 12 cm。估算圓周（取 π = 3.14）。
A. 75
B. 19
C. 38 ✓
D. 12
Trap Analysis
  • A (75): Doubled circumference formula by mistake
  • B (19): Used radius in C=pi*d as if it were diameter
  • D (12): Used diameter value directly
generated O d = 12
Step-by-step Solution
1 Circumference C = πd.
2 C = 3.14 × 12 = 38 (approx).
1 圓周 C = πd。
2 C = 3.14 × 12 = 38 (approx).
Verification
3.14 × 12 = 38. ✓
p6-circle-circumference
#76 — gen-translation-scaling-7-4a053ae4
intermediate translation-scaling
EN
The mean and variance of a set of numbers {x₁, x₂, ..., xₙ} are 10 and 36 respectively. Which of the following must be true? I. The mean of {6x + 10} is 70. II. The standard deviation of {6x + 10} is 36. III. The variance of {6x + 10} is 216.
中文
The mean and variance of a set of numbers {x₁, x₂, ..., xₙ} are 10 and 36 respectively. 以下哪些必定成立？ I. The mean of {6x + 10} is 70. II. The standard deviation of {6x + 10} is 36. III. The variance of {6x + 10} is 216。
A. I only
B. II only
C. I and II only ✓
D. I, II and III
Trap Analysis
  • A (I only): Only checked the mean transformation
  • B (II only): Only checked the SD transformation
  • D (I, II and III): Thought variance scales by k instead of k²
Step-by-step Solution
1 Mean of kX + c = k × mean + c. SD of kX + c = |k| × SD.
2 Var of kX + c = k² × Var, NOT k × Var.
3 Check each statement against these rules. Answer: I and II only.
1 Mean of kX + c = k × mean + c. SD of kX + c = |k| × SD.
2 Var of kX + c = k² × Var, NOT k × Var.
3 Check each statement against these rules. Answer: I and II only.
Verification
Standard transformation rules for mean, SD, variance. ✓
scaling-effecttranslation-invariance
#77 — gen-cube-2-4b272399
advanced cube
EN
ABCDEFGH is a cube where ABCD is the bottom face and EFGH is the top face. Let α be the angle between △DFG and △DFH. Find α, correct to the nearest degree.
中文
ABCDEFGH 是一個正方體，其中 ABCD 為底面，EFGH 為頂面。設 α 為 △DFG 和 △DFH 之間的角。求 α（準確至最接近的度）。
A. 65°
B. 55°
C. 60° ✓
D. 30°
Trap Analysis
  • A (65°): Close angle — rounding or formula error
  • B (55°): Close angle — arithmetic slip
  • D (30°): Complementary angle confusion
generatedABCDEFGH
Step-by-step Solution
1 Let side = a. Set up coordinates: A(0,0,0), B(a,0,0), ..., H(0,a,a).
2 Planes △DFG and △DFH share edge DF.
3 Find normals to each plane, or use perpendicular components along the shared edge.
4 Dihedral angle cos α = 0.5.
5 α = arccos(0.5) ≈ 60°.
1 設邊長為 a。 建立坐標系： A(0,0,0), B(a,0,0), ..., H(0,a,a).
2 平面 △DFG 與 △DFH 共用棱 DF。
3 先求各平面的法向量，或利用沿共用棱的垂直分量計算。
4 二面角的餘弦值 cos α = 0.5.
5 α = arccos(0.5) ≈ 60°.
Verification
Side length cancels. Result independent of a. ✓
3d-dihedral-angle3d-figure-decomposition
#78 — gen-p6-percent-decrease-3-fcefcfd7
basic p6-percent-decrease
EN
A jacket is sold at $373. During a sale, the price is reduced by 15%. Find the sale price (in dollars).
中文
一件外套原價 373 元。減價期間打折 15%。求減價後價格（元）。
A. 345
B. 429
C. 317 ✓
D. 56
Trap Analysis
  • A (345): Used half of the percentage change
  • B (429): Percentage direction error (increase instead of decrease)
  • D (56): Gave decrease amount instead of sale price
Step-by-step Solution
1 Decrease amount = 373 × 15% = 56.
2 Sale price = 373 − 56 = 317.
1 減少量 = 373 × 15% = 56.
2 減價後價格 = 373 − 56 = 317.
Verification
373 × 0.85 = 317. ✓
p6-percent-decrease
#79 — gen-tetrahedron-on-ground-1-50c69e57
advanced tetrahedron-on-ground
EN
In the figure, PQRS is a tetrahedron. The base △PQR lies on a horizontal plane. Q is vertically below S. ∠PQR = 120°, ∠QPS = 40° and ∠QRS = 50°. Find cos∠RPS.
中文
在圖中，PQRS 是一個四面體。底面 △PQR 位於水平面上。Q 在 S 的正下方。∠PQR = 120°，∠QPS = 40°，∠QRS = 50°。求 cos∠RPS。
A. -1/2
B. 0.766
C. -0.6983
D. 0.6983 ✓
Trap Analysis
  • A (-1/2): Used ground angle cos(120°) directly
  • B (0.766): Used elevation angle cos(40°) instead of computing 3D angle
  • C (-0.6983): Sign error in cosine formula
generatedQPRS40°50°120°
Step-by-step Solution
1 Let QS = h (vertical). tan 40° = h/PQ ⟹ PQ = h/tan 40° = 1.1918h.
2 tan 50° = h/RQ ⟹ RQ = h/tan 50° = 0.8391h.
3 ∠PQR = 120°. By cosine formula: PR² = 1.1918²h² + 0.8391²h² − 2(1.1918)(0.8391)h²cos 120° = 3.1244h².
4 PS = h/sin 40° = 1.5557h. RS = h/sin 50° = 1.3054h.
5 Apply cosine formula to △PRS: 0.6983.
1 設 QS = h (vertical). tan 40° = h/PQ ⟹ PQ = h/tan 40° = 1.1918h.
2 tan 50° = h/RQ ⟹ RQ = h/tan 50° = 0.8391h.
3 ∠PQR = 120°. 根據餘弦公式： PR² = 1.1918²h² + 0.8391²h² − 2(1.1918)(0.8391)h²cos 120° = 3.1244h².
4 PS = h/sin 40° = 1.5557h. RS = h/sin 50° = 1.3054h.
5 對 △PRS 運用餘弦公式： 0.6983.
Verification
All lengths proportional to h (cancels). Angle ≈ 45.7092°. ✓
3d-cosine-formula3d-perpendicular-projection3d-figure-decomposition
#80 — gen-central-tendency-7-9c41533e
intermediate central-tendency
EN
Data: 86, 98, 79, 88, 98, 68, 68, 32, 68, c. Mode = 68, mean = 80. Find the median.
中文
數據：86, 98, 79, 88, 98, 68, 68, 32, 68，c。眾數 = 68，平均值 = 80。求中位數。
A. 84.5
B. 80
C. 68
D. 82.5 ✓
Trap Analysis
  • A (84.5): Did not sort after finding c
  • B (80): Gave mean instead of median
  • C (68): Gave mode instead of median
Step-by-step Solution
1 Mode = 68. Mean = 80. Sum of known = 685.
2 c = 80 × 10 − sum = 115.
3 Sort all values, find median of 10 values.
4 Median = 82.5.
1 眾數 = 68. Mean = 80. 已知數的和 = 685.
2 c = 80 × 10 − sum = 115.
3 將所有值排序，求中位數 of 10 values.
4 Median = 82.5.
Verification
Sorted array confirmed. Middle value = 82.5. ✓
central-tendency-combinedmode-mean-median-unknowns
#81 — gen-p6-equation-axb-c-3-82e3e541
basic p6-equation-axb-c
EN
Solve the equation: 8x + 2 = 50.
中文
解方程：8x + 2 = 50。
A. 48
B. 6 ✓
C. 5
D. 7
Trap Analysis
  • A (48): Forgot to divide by coefficient of x
  • C (5): Off-by-one in root counting
  • D (7): Arithmetic slip after rearranging
Step-by-step Solution
1 8x + 2 = 50
2 8x = 50 − 2 = 48
3 x = 48 ÷ 8 = 6.
1 8x + 2 = 50
2 8x = 50 − 2 = 48
3 x = 48 ÷ 8 = 6.
Verification
8(6) + 2 = 50. ✓
p6-equation-axb-c
#82 — gen-standard-score-simult-4-52e38b3f
advanced standard-score-simult
EN
In a test, a student who scores 58 has standard score -1.5, and another student who scores 78 has standard score 1. Find the mean.
中文
在一次測驗中，一名得 58 分的學生的標準分為 -1.5，另一名得 78 分的學生的標準分為 1。求平均分。
A. 70 ✓
B. 78
C. 68
D. 62
Trap Analysis
  • B (78): Added σ to μ
  • C (68): Averaged the two scores instead of solving z-score equations
  • D (62): Subtracted σ from μ — sign error in simultaneous equations
Step-by-step Solution
1 (58 − μ) / σ = -1.5 … ①
2 (78 − μ) / σ = 1 … ②
3 ② − ①: (78 − 58) / σ = 1 − (-1.5) = 2.5.
4 σ = 20 / 2.5 = 8.
5 From ②: μ = 78 − 1 × 8 = 70.
1 (58 − μ) / σ = -1.5 … ①
2 (78 − μ) / σ = 1 … ②
3 ② − ①: (78 − 58) / σ = 1 − (-1.5) = 2.5.
4 σ = 20 / 2.5 = 8.
5 From ②: μ = 78 − 1 × 8 = 70.
Verification
Check ①: (58 − 70) / 8 = -1.5. ✓
standard-score-simultaneous
#83 — gen-trig-graph-2-e46deba5
basic trig-graph
EN
The figure shows the graph of y = 1 + 2cos x. What is the value of the amplitude?
中文
圖中所示的是 y = 1 + 2cos x 的圖像。振幅的值是多少？
A. 1
B. 4
C. 2 ✓
D. 3
Trap Analysis
  • A (1): Read vertical shift instead of amplitude
  • B (4): Factor of 2 error
  • D (3): Off by 1 in reading amplitude from graph
generated 0° 90° 180° 270° 360° -1 0 1 2 3 x y
Step-by-step Solution
1 y = 1 + 2 cos 1x has amplitude |2| = 2.
2 Amplitude = 2.
1 y = 1 + 2 cos 1x 的振幅為 |2| = 2.
2 振幅 = 2.
Verification
Max − Min = 3 − -1 = 4. Amplitude = 2. ✓
trig-graph-reading
#84 — gen-single-right-triangle-2-244dab4e
basic single-right-triangle
EN
In △ABC, ∠B = 90°. AB = 24, BC = 45 and AC = 51. Find cos∠C.
中文
在 △ABC 中，∠B = 90°。AB = 24，BC = 45，AC = 51。求 cos∠C。
A. 17/15
B. 15/17 ✓
C. 15/8
D. 8/17
Trap Analysis
  • A (17/15): Inverted fraction
  • C (15/8): Used cot instead of cos
  • D (8/17): Used sin instead of cos (opposite/hypotenuse)
Step-by-step Solution
1 ∠B = 90°. In right △ABC:
2 cos∠C = adj/hyp = 45/51 = 15/17.
1 ∠B = 90°. 在直角 △ABC:
2 cos∠C = 鄰邊/斜邊 = 45/51 = 15/17.
Verification
24² + 45² = 2601 = 2601 = 51². Pythagorean triple confirmed. ✓
trig-ratio-basic
#85 — gen-p6-circle-area-2-d430a82a
basic p6-circle-area
EN
In the circle shown, the radius is 6 cm. Find the area (take π as 3.14).
中文
在圖中的圓，半徑為 6 cm。求面積（取 π = 3.14）。
A. 19
B. 154
C. 113 ✓
D. 38
Trap Analysis
  • A (19): Used pi*r instead of pi*r^2
  • B (154): Squared wrong radius value
  • D (38): Used circumference formula instead of area
generated O r = 6
Step-by-step Solution
1 Area A = πr².
2 A = 3.14 × 6² = 113 (approx).
1 面積 A = πr²。
2 A = 3.14 × 6² = 113 (approx).
Verification
3.14 × 36 = 113. ✓
p6-circle-area
#86 — gen-p6-circle-circumference-3-405cdeb2
basic p6-circle-circumference
EN
In the circle shown, the diameter is 20 cm. Estimate the circumference (take π as 3.14).
中文
在圖中的圓，直徑為 20 cm。估算圓周（取 π = 3.14）。
A. 126
B. 20
C. 63 ✓
D. 31
Trap Analysis
  • A (126): Doubled circumference formula by mistake
  • B (20): Used diameter value directly
  • D (31): Used radius in C=pi*d as if it were diameter
generated O d = 20
Step-by-step Solution
1 Circumference C = πd.
2 C = 3.14 × 20 = 63 (approx).
1 圓周 C = πd。
2 C = 3.14 × 20 = 63 (approx).
Verification
3.14 × 20 = 63. ✓
p6-circle-circumference
#87 — gen-p6-average-basic-1-33e3d0ae
basic p6-average-basic
EN
Find the average of these five numbers: 21, 28, 24, 11, 10.
中文
求以下五個數的平均數：21, 28, 24, 11, 10。
A. 18.8 ✓
B. 94
C. 20
D. 24
Trap Analysis
  • B (94): Gave total sum instead of average
  • C (20): Rounding/arithmetic slip
  • D (24): Divided by wrong count
Step-by-step Solution
1 Sum = 21 + 28 + 24 + 11 + 10 = 94.
2 Average = 94 ÷ 5 = 18.8.
1 Sum = 21 + 28 + 24 + 11 + 10 = 94.
2 Average = 94 ÷ 5 = 18.8.
Verification
Average × 5 = 18.8 × 5 = 94. ✓
p6-average-basic
#88 — gen-p6-circle-radius-2-a3a3a051
basic p6-circle-radius
EN
In the diagram, the diameter of a circle is 12 cm. Find the radius.
中文
在圖中，圓的直徑是 12 cm。求半徑。
A. 12
B. 3
C. 7
D. 6 ✓
Trap Analysis
  • A (12): Confused radius with diameter
  • B (3): Divided by 4 instead of 2
  • C (7): Arithmetic slip
generated O d = 12
Step-by-step Solution
1 Radius = diameter ÷ 2.
2 r = 12 ÷ 2 = 6.
1 半徑 = 直徑 ÷ 2。
2 r = 12 ÷ 2 = 6.
Verification
2 × 6 = 12. ✓
p6-circle-radius-from-diameter
#89 — gen-p6-equation-a-x+b-c-2-7f51bf9f
intermediate p6-equation-a-x+b-c
EN
Solve the equation: 7(x + 8) = 84.
中文
解方程：7(x + 8) = 84。
A. 3
B. 12
C. 5
D. 4 ✓
Trap Analysis
  • A (3): Arithmetic slip in final step
  • B (12): Stopped after dividing by coefficient (forgot final subtraction)
  • C (5): Off-by-one in root counting
Step-by-step Solution
1 7(x + 8) = 84
2 x + 8 = 84 ÷ 7 = 12
3 x = 12 − 8 = 4.
1 7(x + 8) = 84
2 x + 8 = 84 ÷ 7 = 12
3 x = 12 − 8 = 4.
Verification
7(4 + 8) = 84. ✓
p6-equation-a-x+b-c
#90 — gen-translation-scaling-8-75c5b776
intermediate translation-scaling
EN
The standard deviation of a set of numbers is 2. If each number is multiplied by 4 and then 5 is added, find the new standard deviation.
中文
一組數的標準差為 2。 若每個數乘以 4 再加 5，求新的標準差。
A. 8 ✓
B. 32
C. 13
D. 2
Trap Analysis
  • B (32): Multiplied SD by k² instead of |k|
  • C (13): Added c to SD (translation changes SD)
  • D (2): Thought scaling does not change SD
Step-by-step Solution
1 Original SD = 2.
2 Adding 5 does not change SD (translation invariance).
3 Multiplying by 4 scales SD by |-4| = 4.
4 New SD = 4 × 2 = 8.
1 原標準差 = 2.
2 加常數不改變標準差（平移不變性）。
3 乘以 4 使標準差乘以 |-4| = 4.
4 新標準差 = 4 × 2 = 8.
Verification
SD(kX + c) = |k| × SD(X). ✓
scaling-effecttranslation-invariance
#91 — gen-p6-average-basic-2-85c80c16
basic p6-average-basic
EN
Find the average of these five numbers: 22, 12, 13, 27, 15.
中文
求以下五個數的平均數：22, 12, 13, 27, 15。
A. 22
B. 89
C. 17.8 ✓
D. 19
Trap Analysis
  • A (22): Divided by wrong count
  • B (89): Gave total sum instead of average
  • D (19): Rounding/arithmetic slip
Step-by-step Solution
1 Sum = 22 + 12 + 13 + 27 + 15 = 89.
2 Average = 89 ÷ 5 = 17.8.
1 Sum = 22 + 12 + 13 + 27 + 15 = 89.
2 Average = 89 ÷ 5 = 17.8.
Verification
Average × 5 = 17.8 × 5 = 89. ✓
p6-average-basic
#92 — gen-p6-average-basic-3-9ed996e9
basic p6-average-basic
EN
Find the average of these five numbers: 22, 25, 17, 24, 8.
中文
求以下五個數的平均數：22, 25, 17, 24, 8。
A. 24
B. 20
C. 19.2 ✓
D. 96
Trap Analysis
  • A (24): Divided by wrong count
  • B (20): Rounding/arithmetic slip
  • D (96): Gave total sum instead of average
Step-by-step Solution
1 Sum = 22 + 25 + 17 + 24 + 8 = 96.
2 Average = 96 ÷ 5 = 19.2.
1 Sum = 22 + 25 + 17 + 24 + 8 = 96.
2 Average = 96 ÷ 5 = 19.2.
Verification
Average × 5 = 19.2 × 5 = 96. ✓
p6-average-basic
#93 — gen-single-right-triangle-3-0e3321ae
basic single-right-triangle
EN
In △ABC, ∠B = 90°. AB = 10, BC = 24 and AC = 26. Find sin∠C.
中文
在 △ABC 中，∠B = 90°。AB = 10，BC = 24，AC = 26。求 sin∠C。
A. 5/12
B. 13/5
C. 12/13
D. 5/13 ✓
Trap Analysis
  • A (5/12): Used tan instead of sin
  • B (13/5): Inverted fraction
  • C (12/13): Used cos instead of sin (adjacent/hypotenuse)
Step-by-step Solution
1 ∠B = 90°. In right △ABC:
2 sin∠C = opp/hyp = 10/26 = 5/13.
1 ∠B = 90°. 在直角 △ABC:
2 sin∠C = 對邊/斜邊 = 10/26 = 5/13.
Verification
10² + 24² = 676 = 676 = 26². Pythagorean triple confirmed. ✓
trig-ratio-basic
#94 — gen-weighted-mean-stats-3-3c29c95b
basic weighted-mean-stats
EN
20 teachers and 40 students have an overall mean height of 150 cm. The students' mean height is 45 cm. Find the teachers' mean height.
中文
20 名老師和 40 名學生的整體平均身高為 150 cm。學生的平均身高為 45 cm。求老師的平均身高。
A. 360 ✓
B. 355
C. 380
D. 203
Trap Analysis
  • B (355): Arithmetic error in weighted mean equation
  • C (380): Miscounted total number of items
  • D (203): Used simple average instead of weighted mean
Step-by-step Solution
1 Total sum = 60 × 150 = 9000.
2 Sub-group sum = 40 × 45 = 1800.
3 Remaining mean = (9000 − 1800) / 20 = 360.
1 總和 = 60 × 150 = 9000.
2 子組總和 = 40 × 45 = 1800.
3 其餘平均值 = (9000 − 1800) / 20 = 360.
Verification
20 × 360 + 40 × 45 = 60 × 150. ✓
weighted-mean
#95 — gen-box-whisker-6-ad6c448d
basic box-whisker
EN
The box-and-whisker diagram below shows the distribution of scores. The minimum is 6, Q₁ = 12, median = 23, Q₃ = 32, and maximum = 40. Find the inter-quartile range.
中文
下面的箱形圖顯示了分數的分佈。 最小值為 6, Q₁ = 12, 中位數 = 23, Q₃ = 32, and 最大值 = 40. 求四分位距。
A. 28
B. 26
C. 34
D. 20 ✓
Trap Analysis
  • A (28): Used max − Q₁ instead of Q₃ − Q₁
  • B (26): Used Q₃ − min instead of Q₃ − Q₁
  • C (34): Gave range instead of IQR
generated 6 12 23 32 40
Step-by-step Solution
1 IQR = Q₃ − Q₁ = 32 − 12 = 20.
1 IQR = Q₃ − Q₁ = 32 − 12 = 20.
Verification
Q₃ = 32, Q₁ = 12. Difference = 20. ✓
box-whisker-readingiqr-definition
#96 — gen-weighted-mean-stats-4-c1999746
basic weighted-mean-stats
EN
Alice types for 4 minutes at 56 words/min and 4 minutes at 50 words/min. Find her average typing speed, correct to the nearest integer.
中文
Alice 以每分鐘 56 字的速度打字 4 分鐘，然後以每分鐘 50 字的速度打字 4 分鐘。求她的平均打字速度（準確至最接近的整數）。
A. 52
B. 54
C. 53 ✓
D. 55
Trap Analysis
  • A (52): Arithmetic error in multiplication
  • B (54): Rounding error in weighted average
  • D (55): Off-by-one in root counting
Step-by-step Solution
1 Weighted mean = (56 × 4 + 50 × 4) / (4 + 4).
2 = (224 + 200) / 8 = 53.
1 加權平均 = (56 × 4 + 50 × 4) / (4 + 4).
2 = (224 + 200) / 8 = 53.
Verification
424 / 8 ≈ 53. ✓
weighted-mean
#97 — gen-weighted-mean-stats-5-364e8cbc
basic weighted-mean-stats
EN
14 teachers and 56 students have an overall mean height of 57 cm. The students' mean height is 30 cm. Find the teachers' mean height.
中文
14 名老師和 56 名學生的整體平均身高為 57 cm。學生的平均身高為 30 cm。求老師的平均身高。
A. 98
B. 179
C. 160
D. 165 ✓
Trap Analysis
  • A (98): Used simple average instead of weighted mean
  • B (179): Miscounted total number of items
  • C (160): Arithmetic error in weighted mean equation
Step-by-step Solution
1 Total sum = 70 × 57 = 3990.
2 Sub-group sum = 56 × 30 = 1680.
3 Remaining mean = (3990 − 1680) / 14 = 165.
1 總和 = 70 × 57 = 3990.
2 子組總和 = 56 × 30 = 1680.
3 其餘平均值 = (3990 − 1680) / 14 = 165.
Verification
14 × 165 + 56 × 30 = 70 × 57. ✓
weighted-mean
#98 — gen-p6-circle-area-3-d430a82a
basic p6-circle-area
EN
In the circle shown, the radius is 6 cm. Find the area (take π as 3.14).
中文
在圖中的圓，半徑為 6 cm。求面積（取 π = 3.14）。
A. 154
B. 19
C. 113 ✓
D. 38
Trap Analysis
  • A (154): Squared wrong radius value
  • B (19): Used pi*r instead of pi*r^2
  • D (38): Used circumference formula instead of area
generated O r = 6
Step-by-step Solution
1 Area A = πr².
2 A = 3.14 × 6² = 113 (approx).
1 面積 A = πr²。
2 A = 3.14 × 6² = 113 (approx).
Verification
3.14 × 36 = 113. ✓
p6-circle-area
#99 — gen-point-above-ground-2-97164259
advanced point-above-ground
EN
T is a point vertically above O on a horizontal ground. A and B are two points on the ground with ∠AOB = 120°. The angle of elevation of T from A is 50° and from B is 40°. Find ∠ATB, correct to the nearest degree.
中文
T 是水平地面上 O 點正上方的一點。A 和 B 是地面上兩點，∠AOB = 120°。從 A 觀察 T 的仰角為 50°，從 B 觀察 T 的仰角為 40°。求 ∠ATB（準確至最接近的度）。
A. 90°
B. 14°
C. 10°
D. 76° ✓
Trap Analysis
  • A (90°): Added elevation angles directly
  • B (14°): Complementary angle confusion
  • C (10°): Subtracted elevation angles directly
generatedTOAB50°40°120°
Step-by-step Solution
1 Let TO = h. OA = h/tan 50° = 0.8391h. OB = h/tan 40° = 1.1918h.
2 ∠AOB = 120°. AB² = 0.7041h² + 1.4203h² − 2(0.8391)(1.1918)h²cos 120° = 3.1244h².
3 TA = h/sin 50° = 1.3054h. TB = h/sin 40° = 1.5557h.
4 cos∠ATB = (TA² + TB² − AB²) / (2·TA·TB).
5 h cancels. Result ≈ 76°.
1 設 TO = h. OA = h/tan 50° = 0.8391h. OB = h/tan 40° = 1.1918h.
2 ∠AOB = 120°. AB² = 0.7041h² + 1.4203h² − 2(0.8391)(1.1918)h²cos 120° = 3.1244h².
3 TA = h/sin 50° = 1.3054h. TB = h/sin 40° = 1.5557h.
4 cos∠ATB = (TA² + TB² − AB²) / (2·TA·TB).
5 h 會被消去。 Result ≈ 76°.
Verification
Angle independent of h. ✓
3d-cosine-formula3d-perpendicular-projectiontrig-ratio-basic
#100 — gen-translation-scaling-9-ac1f31d4
intermediate translation-scaling
EN
The variance of a set of numbers is 9. Each number is multiplied by 3 and then 3 is subtracted. Find the new variance.
中文
一組數的方差為 9。每個數乘以 3 再減 3。 求新的方差。
A. 9
B. 81 ✓
C. 78
D. 27
Trap Analysis
  • A (9): Gave SD instead of variance
  • C (78): Added c to variance (translation changes variance)
  • D (27): Multiplied variance by |k| instead of k²
Step-by-step Solution
1 Original variance = 9.
2 Transform: multiply by 3, then subtract 3.
3 Adding/subtracting a constant does not change variance (translation invariance).
4 Multiplying by 3 scales variance by 3² = 9.
5 New variance = 9 × 9 = 81.
1 原方差 = 9.
2 變換： 乘以 3, 再減 3.
3 加減常數不改變方差（平移不變性）。
4 乘以 3 使方差乘以 3² = 9.
5 新方差 = 9 × 9 = 81.
Verification
Var(kX + c) = k²Var(X). 3² × 9 = 81. ✓
scaling-effecttranslation-invariance
#101 — gen-standard-score-3-d4ff6665
basic standard-score
EN
The standard score of a student who gets 70 marks is 1.5. If the mean is 55, find the standard deviation.
中文
一名得 70 分的學生的標準分為 1.5。 若平均分為 55，求標準差。
A. 100
B. 15
C. 11
D. 10 ✓
Trap Analysis
  • A (100): Gave variance instead of SD
  • B (15): Gave |x-μ| instead of dividing by z
  • C (11): Arithmetic error when solving (x-μ)/σ = z
Step-by-step Solution
1 (70 − 55) / σ = 1.5.
2 15 / σ = 1.5.
3 σ = 15 / 1.5 = 10.
1 (70 − 55) / σ = 1.5.
2 15 / σ = 1.5.
3 σ = 15 / 1.5 = 10.
Verification
z = (70 − 55) / 10 = 1.5. ✓
standard-score
#102 — gen-standard-score-4-a7df5692
basic standard-score
EN
The standard score of a student who gets 90 marks is 3. If the mean is 72, find the standard deviation.
中文
一名得 90 分的學生的標準分為 3。 若平均分為 72，求標準差。
A. 18
B. 6 ✓
C. 36
D. 7
Trap Analysis
  • A (18): Gave |x-μ| instead of dividing by z
  • C (36): Gave variance instead of SD
  • D (7): Arithmetic error when solving (x-μ)/σ = z
Step-by-step Solution
1 (90 − 72) / σ = 3.
2 18 / σ = 3.
3 σ = 18 / 3 = 6.
1 (90 − 72) / σ = 3.
2 18 / σ = 3.
3 σ = 18 / 3 = 6.
Verification
z = (90 − 72) / 6 = 3. ✓
standard-score
#103 — gen-2d-cosine-formula-3-80369205
intermediate 2d-cosine-formula
EN
In △ABC, AB = 3, BC = 8 and ∠ABC = 60°. Find AC.
中文
在 △ABC 中，AB = 3，BC = 8，∠ABC = 60°。求 AC。
A. 5.6063
B. 49
C. 9.8489
D. 7 ✓
Trap Analysis
  • A (5.6063): Used sin instead of cos in cosine formula
  • B (49): Gave AC² instead of AC — forgot to take square root
  • C (9.8489): Forgot negative sign — used cos(120°) instead of cos(60°)
generatedABC3860°
Step-by-step Solution
1 AC² = 3² + 8² − 2(3)(8)cos 60°.
2 = 9 + 64 − 48 × 0.5 = 49.
3 AC = √(49) = 7.
1 AC² = 3² + 8² − 2(3)(8)cos 60°.
2 = 9 + 64 − 48 × 0.5 = 49.
3 AC = √(49) = 7.
Verification
Check: 7² = 49. ✓
cosine-formula
#104 — gen-single-right-triangle-4-88ad38bc
basic single-right-triangle
EN
In △ABC, ∠B = 90°. AB = 14, BC = 48 and AC = 50. Find cos∠C.
中文
在 △ABC 中，∠B = 90°。AB = 14，BC = 48，AC = 50。求 cos∠C。
A. 24/25 ✓
B. 24/7
C. 25/24
D. 7/25
Trap Analysis
  • B (24/7): Used cot instead of cos
  • C (25/24): Inverted fraction
  • D (7/25): Used sin instead of cos (opposite/hypotenuse)
Step-by-step Solution
1 ∠B = 90°. In right △ABC:
2 cos∠C = adj/hyp = 48/50 = 24/25.
1 ∠B = 90°. 在直角 △ABC:
2 cos∠C = 鄰邊/斜邊 = 48/50 = 24/25.
Verification
14² + 48² = 2500 = 2500 = 50². Pythagorean triple confirmed. ✓
trig-ratio-basic
#105 — gen-connected-triangles-3-1da2197e
intermediate connected-triangles
EN
In the figure, AC ⊥ BD. AC = 36 and AB = 60. Find sin∠ABC.
中文
在圖中，AC ⊥ BD。AC = 36，AB = 60。求 sin∠ABC。
A. 2/5
B. 5/3
C. 3/5 ✓
D. 4/5
Trap Analysis
  • A (2/5): Wrong numerator — likely arithmetic slip
  • B (5/3): Inverted fraction
  • D (4/5): Used adjacent/hypotenuse (cos instead of sin)
generatedABCD3660
Step-by-step Solution
1 AC ⊥ BD. AB = 60.
2 sin∠ABC = AC/AB = 36/60 = 3/5.
1 AC ⊥ BD. AB = 60.
2 sin∠ABC = AC/AB = 36/60 = 3/5.
Verification
sin = opp/hyp in right △ACB. ✓
trig-ratio-chaintrig-ratio-basic
#106 — gen-weighted-mean-stats-6-96d2eac0
basic weighted-mean-stats
EN
15 teachers and 50 students have an overall mean height of 132 cm. The students' mean height is 30 cm. Find the teachers' mean height.
中文
15 名老師和 50 名學生的整體平均身高為 132 cm。學生的平均身高為 30 cm。求老師的平均身高。
A. 487
B. 467
C. 251
D. 472 ✓
Trap Analysis
  • A (487): Miscounted total number of items
  • B (467): Arithmetic error in weighted mean equation
  • C (251): Used simple average instead of weighted mean
Step-by-step Solution
1 Total sum = 65 × 132 = 8580.
2 Sub-group sum = 50 × 30 = 1500.
3 Remaining mean = (8580 − 1500) / 15 = 472.
1 總和 = 65 × 132 = 8580.
2 子組總和 = 50 × 30 = 1500.
3 其餘平均值 = (8580 − 1500) / 15 = 472.
Verification
15 × 472 + 50 × 30 = 65 × 132. ✓
weighted-mean
#107 — gen-trig-identity-4-a018003b
basic trig-identity
EN
Simplify (sin²θ + cos²θ)² − 2sin²θ cos²θ.
中文
化簡 (sin²θ + cos²θ)² − 2sin²θ cos²θ。
A. sin⁴θ + cos⁴θ ✓
B. 1 - 2sin²θ cos²θ
C. (sin²θ - cos²θ)²
D. 1
Trap Analysis
  • B (1 - 2sin²θ cos²θ): Correct expansion but did not simplify to required form
  • C ((sin²θ - cos²θ)²): Wrong algebraic identity
  • D (1): Confused (sin²+cos²)² with sin⁴+cos⁴
Step-by-step Solution
1 Apply standard trig identities:
2 sin(180° + θ) = −sin θ, cos(90° − θ) = sin θ, cos(360° − θ) = cos θ, cos(180° + θ) = −cos θ.
3 sin²θ + cos²θ = 1. tan θ = sin θ / cos θ.
4 Result: sin⁴θ + cos⁴θ.
1 運用標準三角恒等式：
2 sin(180° + θ) = −sin θ, cos(90° − θ) = sin θ, cos(360° − θ) = cos θ, cos(180° + θ) = −cos θ.
3 sin²θ + cos²θ = 1. tan θ = sin θ / cos θ.
4 得出：sin⁴θ + cos⁴θ.
Verification
Substitute θ = 30° to verify numerically. ✓
trig-identity-algebraic
#108 — gen-central-tendency-8-f9ffc717
intermediate central-tendency
EN
Mean of 3, 5, 8, 8, 10, 12, m, n is 7. If m + n = 10, which MUST be true? I. Mode is 8. II. Median ≤ 8. III. Range ≥ 9.
中文
3, 5, 8, 8, 10, 12，m，n 的平均值為 7。 若 m + n = 10，以下哪些必定成立？ I. 眾數為 8。 II. 中位數 ≤ 8。 III. 全距 ≥ 9。
A. III only
B. II only ✓
C. I and II
D. I, II and III
Trap Analysis
  • A (III only): Only checked range constraint
  • C (I and II): Assumed mode is always 8 without checking all valid (m,n)
  • D (I, II and III): Did not verify all constraints across all valid pairs
Step-by-step Solution
1 m + n = 10. Enumerate valid (m, n) pairs.
2 For each valid pair: check mode, median, range.
3 Only statements true for ALL pairs are "must be true". Answer: II only.
1 m + n = 10. 列舉所有有效的 (m, n) 對.
2 For each valid pair: check mode, median, range.
3 只有對所有組合都成立的陳述才是「必定成立」。 Answer: II only.
Verification
Exhaustive enumeration confirms. ✓
constrained-dataset-reasoningcentral-tendency-combined
#109 — gen-p6-percent-increase-3-6e66ca66
basic p6-percent-increase
EN
A notebook costs $255. The price is increased by 25%. What is the new price (in dollars)?
中文
一本筆記簿原價 255 元。價格上升 25%。求新價格（元）。
A. 191
B. 287
C. 64
D. 319 ✓
Trap Analysis
  • A (191): Percentage direction error (decrease instead of increase)
  • B (287): Used half of the percentage change
  • C (64): Gave increase amount instead of new total
Step-by-step Solution
1 Increase amount = 255 × 25% = 64.
2 New price = 255 + 64 = 319.
1 增加量 = 255 × 25% = 64.
2 新價格 = 255 + 64 = 319.
Verification
255 × 1.25 = 319. ✓
p6-percent-increase
#110 — gen-rectangular-box-2-0592c3b0
advanced rectangular-box
EN
ABCDEFGH is a rectangular box with AB = 20cm, BC = 13cm and AE = 17cm. Find the angle between diagonal AG and the base ABCD, correct to the nearest degree.
中文
ABCDEFGH 是一個長方體，其中 AB = 20cm，BC = 13cm，AE = 17cm。求 the angle between diagonal AG and the base ABCD（準確至最接近的度）。
A. 55°
B. 40°
C. 25°
D. 35° ✓
Trap Analysis
  • A (55°): Complementary angle confusion
  • B (40°): Rounding error in inverse trig
  • C (25°): Used wrong triangle for calculation
generated201317ABCDEFGH
Step-by-step Solution
1 Base diagonal AC = √(20² + 13²) = 23.8537.
2 AG makes angle θ with base. tan θ = AE/AC = 17/23.8537.
3 θ = arctan(0.7127) ≈ 35°.
1 底面對角線 AC = √(20² + 13²) = 23.8537.
2 AG makes angle θ with base. tan θ = AE/AC = 17/23.8537.
3 θ = arctan(0.7127) ≈ 35°.
Verification
Space diagonal AG = √(20² + 13² + 17²) = 29.2916. sin θ = 17/29.2916. ✓
3d-cosine-formula3d-pythagoras3d-figure-decomposition
#111 — gen-standard-score-simult-5-812d92c6
advanced standard-score-simult
EN
Scores: 48, x, 72. Standard scores: -1.5, 1, 1.5. Find x.
中文
分數：48, x, 72。標準分：-1.5, 1, 1.5。求 x。
A. 68 ✓
B. 69
C. 60
D. 48
Trap Analysis
  • B (69): Arithmetic error in final substitution
  • C (60): Gave the mean instead of the target score
  • D (48): Gave x₁ instead of computing from z = 1
Step-by-step Solution
1 From the two known z-scores, solve for μ and σ.
2 (72 − 48) / σ = 3 ⟹ σ = 8.
3 μ = 60. x = σ(1) + μ = 8 + 60 = 68.
1 From the two known z-scores, solve for μ and σ.
2 (72 − 48) / σ = 3 ⟹ σ = 8.
3 μ = 60. x = σ(1) + μ = 8 + 60 = 68.
Verification
z = (68 − 60) / 8 = 1. ✓
standard-score-simultaneous
#112 — gen-translation-scaling-10-55ecb964
intermediate translation-scaling
EN
The mean and variance of a set of numbers {x₁, x₂, ..., xₙ} are 10 and 25 respectively. Which of the following must be true? I. The mean of {7x + 4} is 74. II. The standard deviation of {7x + 4} is 35. III. The variance of {7x + 4} is 175.
中文
The mean and variance of a set of numbers {x₁, x₂, ..., xₙ} are 10 and 25 respectively. 以下哪些必定成立？ I. The mean of {7x + 4} is 74. II. The standard deviation of {7x + 4} is 35. III. The variance of {7x + 4} is 175。
A. I and II only ✓
B. I, II and III
C. II only
D. I only
Trap Analysis
  • B (I, II and III): Thought variance scales by k instead of k²
  • C (II only): Only checked the SD transformation
  • D (I only): Only checked the mean transformation
Step-by-step Solution
1 Mean of kX + c = k × mean + c. SD of kX + c = |k| × SD.
2 Var of kX + c = k² × Var, NOT k × Var.
3 Check each statement against these rules. Answer: I and II only.
1 Mean of kX + c = k × mean + c. SD of kX + c = |k| × SD.
2 Var of kX + c = k² × Var, NOT k × Var.
3 Check each statement against these rules. Answer: I and II only.
Verification
Standard transformation rules for mean, SD, variance. ✓
scaling-effecttranslation-invariance
#113 — gen-trig-equation-3-9eabf3ec
intermediate trig-equation
EN
Find the number of roots of sin x = -√3/2 in 0° ≤ x < 360°.
中文
求 sin x = -√3/2 在 0° ≤ x < 360° 中的根的數目。
A. 3
B. 2 ✓
C. 1
D. 4
Trap Analysis
  • A (3): Counted a repeated root as distinct
  • C (1): Missed a root outside primary range
  • D (4): Counted roots from both sin and cos (double-counting)
Step-by-step Solution
1 Rewrite equation in terms of a single trig function.
2 Solve the resulting equation: find values of the trig function.
3 Check which solutions lie in 0° ≤ x < 360° — each valid value gives 1 root (if ±1) or 2 roots.
4 Total roots: 2.
1 將方程化為單一三角函數的形式。
2 解所得方程，求出三角函數的值。
3 檢查哪些解落在 0° ≤ x < 360° — 每個有效值對應 1 個根（若為 ±1）或 2 個根。
4 根的總數：2.
Verification
Graph method: count intersections in [0°, 360°). ✓
trig-equation-root-counting
#114 — gen-p6-pie-percent-count-2-0bf6be81
basic p6-pie-percent-count
EN
A pie chart represents 100 students. The Apple sector is 30% of the circle. How many students chose Apple?
中文
某圓形圖代表 100 名學生。Apple 扇形佔全圓 30%。有多少名學生選擇 Apple？
A. 70
B. 30 ✓
C. 31
D. 3000
Trap Analysis
  • A (70): Used complement percentage
  • C (31): Off-by-one in root counting
  • D (3000): Forgot to divide by 100
generated Apple (30)Banana (30)Orange (20)Grape (20)
Step-by-step Solution
1 Apple count = 30% of 100.
2 = 100 × 0.3 = 30.
1 Apple 人數 = 30% 佔 100。
2 = 100 × 0.3 = 30.
Verification
30 ÷ 100 = 30%. ✓
p6-pie-percent-to-count
#115 — gen-p6-equation-word-2-838e1eaf
intermediate p6-equation-word
EN
Twice a number plus 6 is 30. Find the number.
中文
某數的 2 倍加上 6 等於 30。求此數。
A. 12 ✓
B. 13
C. 24
D. 18
Trap Analysis
  • B (13): Arithmetic slip in final division
  • C (24): Forgot to divide by 2
  • D (18): Moved + constant to wrong side
Step-by-step Solution
1 Let the number be x. Then 2x + 6 = 30.
2 2x = 30 − 6 = 24.
3 x = 24 ÷ 2 = 12.
1 設 the number be x. Then 2x + 6 = 30.
2 2x = 30 − 6 = 24.
3 x = 24 ÷ 2 = 12.
Verification
2(12) + 6 = 30. ✓
p6-equation-word
#116 — gen-p6-pie-percent-count-3-73ff8fb4
basic p6-pie-percent-count
EN
A pie chart represents 120 students. The Apple sector is 20% of the circle. How many students chose Apple?
中文
某圓形圖代表 120 名學生。Apple 扇形佔全圓 20%。有多少名學生選擇 Apple？
A. 96
B. 20
C. 24 ✓
D. 2400
Trap Analysis
  • A (96): Used complement percentage
  • B (20): Gave percentage value instead of count
  • D (2400): Forgot to divide by 100
generated Apple (20)Banana (30)Orange (20)Grape (30)
Step-by-step Solution
1 Apple count = 20% of 120.
2 = 120 × 0.2 = 24.
1 Apple 人數 = 20% 佔 120。
2 = 120 × 0.2 = 24.
Verification
24 ÷ 120 = 20%. ✓
p6-pie-percent-to-count
#117 — gen-p6-percent-whole-2-b86c15d8
intermediate p6-percent-whole
EN
In a class, 12 students are girls. This is 30% of the class. How many students are there in total?
中文
某班有 12 名女生，佔全班 30%。求全班總人數。
A. 4
B. 42
C. 40 ✓
D. 0
Trap Analysis
  • A (4): Found the part of part, not the whole
  • B (42): Added part and percent directly
  • D (0): Divided by percent value instead of decimal form
Step-by-step Solution
1 12 is 30% of the total.
2 Total = 12 ÷ 30% = 12 ÷ 0.3.
3 Total = 40.
1 12 是總數的 30%。
2 總數 = 12 ÷ 30% = 12 ÷ 0.3.
3 總數 = 40.
Verification
40 × 30% = 12. ✓
p6-percent-whole
#118 — gen-p6-pie-angle-3-18d52f46
basic p6-pie-angle
EN
In a class survey, 24 out of 90 students chose option A. In a pie chart, what is the angle of sector A?
中文
班內調查中，90 名學生中有 24 名選 A。若用圓形圖表示，A 扇形的角度是多少？
A. 264
B. 78
C. 27
D. 96 ✓
Trap Analysis
  • A (264): Used complement sector angle
  • B (78): Arithmetic slip in scaling to 360°
  • C (27): Gave percentage instead of angle
generated A (24)B (54)C (12)
Step-by-step Solution
1 Sector angle = (part ÷ total) × 360°.
2 = (24 ÷ 90) × 360° = 96°.
1 扇形角 =（部分 ÷ 總數）× 360°。
2 = (24 ÷ 90) × 360° = 96°.
Verification
24/90 = 96/360. ✓
p6-pie-angle
#119 — gen-tetrahedron-on-ground-2-65b71dd2
advanced tetrahedron-on-ground
EN
In the figure, PQRS is a tetrahedron. The base △PQR lies on a horizontal plane. Q is vertically below S. ∠PQR = 60°, ∠QPS = 45° and ∠QRS = 60°. Find cos∠RPS.
中文
在圖中，PQRS 是一個四面體。底面 △PQR 位於水平面上。Q 在 S 的正下方。∠PQR = 60°，∠QPS = 45°，∠QRS = 60°。求 cos∠RPS。
A. √2/2
B. 0.5785 ✓
C. -0.5785
D. 1/2
Trap Analysis
  • A (√2/2): Used elevation angle cos(45°) instead of computing 3D angle
  • C (-0.5785): Sign error in cosine formula
  • D (1/2): Used ground angle cos(60°) directly
generatedQPRS45°60°60°
Step-by-step Solution
1 Let QS = h (vertical). tan 45° = h/PQ ⟹ PQ = h/tan 45° = 1h.
2 tan 60° = h/RQ ⟹ RQ = h/tan 60° = 0.5774h.
3 ∠PQR = 60°. By cosine formula: PR² = 1²h² + 0.5774²h² − 2(1)(0.5774)h²cos 60° = 0.756h².
4 PS = h/sin 45° = 1.4142h. RS = h/sin 60° = 1.1547h.
5 Apply cosine formula to △PRS: 0.5785.
1 設 QS = h (vertical). tan 45° = h/PQ ⟹ PQ = h/tan 45° = 1h.
2 tan 60° = h/RQ ⟹ RQ = h/tan 60° = 0.5774h.
3 ∠PQR = 60°. 根據餘弦公式： PR² = 1²h² + 0.5774²h² − 2(1)(0.5774)h²cos 60° = 0.756h².
4 PS = h/sin 45° = 1.4142h. RS = h/sin 60° = 1.1547h.
5 對 △PRS 運用餘弦公式： 0.5785.
Verification
All lengths proportional to h (cancels). Angle ≈ 54.6549°. ✓
3d-cosine-formula3d-perpendicular-projection3d-figure-decomposition
#120 — gen-standard-score-5-f916967d
basic standard-score
EN
In an examination, the mean score is 70 and the standard deviation is 10. A student gets 95 marks. Find the standard score of the student.
中文
在一次考試中，平均分為 70，標準差為 10。 一名學生得 95 分。求該學生的標準分。
A. 2.5 ✓
B. 25
C. -2.5
D. 1.3571428571428572
Trap Analysis
  • B (25): Forgot to divide by σ
  • C (-2.5): Sign error — reversed (x-μ) to (μ-x)
  • D (1.3571428571428572): Divided x by mean instead of using z-score formula
Step-by-step Solution
1 z = (x − μ) / σ = (95 − 70) / 10 = 25 / 10 = 2.5.
1 z = (x − μ) / σ = (95 − 70) / 10 = 25 / 10 = 2.5.
Verification
2.5 × 10 + 70 = 95. ✓
standard-score