Secondary 4 Elementary Mathematics Statistics Probability Quiz

Secondary 4 Elementary Mathematics Quiz - Statistics Probability: Answer Key

Total Marks: 40

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Section A: Data Analysis and Measures of Central Tendency

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1. [4 marks]

Given data: 52, 48, 65, 72, 48, 55, 61, 48, 78, 64, 55, 80, 55, 67, 55

(a) Mode and Median [2 marks]

Method: The mode is the most frequent value. The median is the middle value when data is ordered.

Ordered data: 48, 48, 48, 52, 55, 55, 55, 55, 61, 64, 65, 67, 72, 78, 80

• Mode = 55 (appears 4 times, most frequent) [1 mark]
• Median = 55 (8th value out of 15, i.e. the middle position $\frac{15+1}{2} = 8$) [1 mark]

Common mistake: Forgetting to order data before finding median, or confusing mode with mean.

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(b) Mean mark [1 mark]

$\text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}} = \frac{48+48+48+52+55+55+55+55+61+64+65+67+72+78+80}{15}$

$= \frac{903}{15} = \mathbf{60.2}$ [1 mark]

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(c) Total marks of other 25 students [1 mark]

Total marks of 40 students = $40 \times 62.5 = 2500$

Total marks of 15 students = 903 (from part b)

Total marks of other 25 students = $2500 - 903 = \mathbf{1597}$ [1 mark]

Teaching note: The mean of combined groups is NOT the average of the two means unless the groups are equal in size. Always work with totals.

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2. [3 marks]

Expected visual features: Smooth cumulative frequency curve passing through (40,2), (45,8), (50,22), (55,45), (60,62), (65,74), (70,80). Median at cumulative frequency 40 (half of 80), lower quartile at 20, upper quartile at 60.

(a) Median mass [1 mark]

At cumulative frequency 40, read across to curve then down to x-axis: approximately 54-55 kg (accept ~54.5 kg)

(b) Interquartile range [1 mark]

• Lower quartile $Q_1$: at cumulative frequency 20, approximately 50 kg
• Upper quartile $Q_3$: at cumulative frequency 60, approximately 58 kg

$\text{IQR} = Q_3 - Q_1 \approx 58 - 50 = \mathbf{8 \text{ kg}}$ (accept 7-9 kg)

(c) Percentage with mass > 58 kg [1 mark]

At 58 kg, cumulative frequency ≈ 62. Number above 58 kg = $80 - 62 = 18$

$\text{Percentage} = \frac{18}{80} \times 100\% = \mathbf{22.5\%}$ (accept 20-25% depending on graph reading)

Marking note: Allow ±1 kg for readings from curve; method marks awarded for correct procedure even with slight reading errors.

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3. [4 marks]

(a) Modal class [1 mark]

The modal class is the class with the highest frequency: $800 \leq x < 1000$ [1 mark]

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(b) Estimated mean [2 marks]

Class	Mid-value ($x$)	Frequency ($f$)	$fx$
400 ≤ x < 600	500	8	4000
600 ≤ x < 800	700	14	9800
800 ≤ x < 1000	900	22	19800
1000 ≤ x < 1200	1100	12	13200
1200 ≤ x < 1400	1300	4	5200
Total		60	52000

$\text{Estimated mean} = \frac{\sum fx}{\sum f} = \frac{52000}{60} = \mathbf{866.67} \approx \mathbf{\$867} \text{ (3 sf)}$ [2 marks]

Method mark: Correct mid-values and attempt at $\frac{\sum fx}{\sum f}$. Accuracy mark: Correct calculation.

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(c) Why the mean is an estimate [1 mark]

The exact values of individual data within each class are not known; we only know they fall within intervals. Using mid-values assumes values are evenly distributed within each class, which may not be true. [1 mark]

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4. [2 marks]

(a) Finding $k$ [1 mark]

$\frac{3+7+8+12+k}{5} = 8$

$30 + k = 40$

$k = \mathbf{10}$ [1 mark]

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(b) New mean with 16 added [1 mark]

New sum = $40 + 16 = 56$; new count = 6

$\text{New mean} = \frac{56}{6} = \mathbf{9.33} \text{ (3 sf)}$ or exactly $\frac{28}{3}$ [1 mark]

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5. [3 marks]

(a) Mean [1 mark]

$\bar{x} = \frac{\sum x}{n} = \frac{240}{10} = \mathbf{24}$ [1 mark]

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(b) Standard deviation [1 mark]

$\sigma = \sqrt{\frac{\sum x^2}{n} - \bar{x}^2} = \sqrt{\frac{7250}{10} - 24^2} = \sqrt{725 - 576} = \sqrt{149} = \mathbf{12.2} \text{ (3 sf)}$ [1 mark]

Formula note: This is the population standard deviation. For sample standard deviation, divide by $n-1$, but for Sec 4 E-Math, population SD is standard unless specified.

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(c) New mean and new SD after transformation [1 mark]

Transformation: $y = 2x + 3$

• New mean = $2 \times 24 + 3 = \mathbf{51}$ [0.5 mark]
• New SD = $2 \times 12.2 = \mathbf{24.4}$ (SD is multiplied by 2 only; adding 3 does not affect spread) [0.5 mark]

Teaching note: For $y = ax + b$: new mean = $a\bar{x} + b$, new SD = $|a| \times$ old SD (adding constant doesn't change spread).

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Section B: Probability

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6. [3 marks]

Sample space: There are $6 \times 2 = 12$ equally likely outcomes.

All outcomes: (1,H), (1,T), (2,H), (2,T), (3,H), (3,T), (4,H), (4,T), (5,H), (5,T), (6,H), (6,T) [1 mark for complete list]

(b) P(even and head) [1 mark]

Even numbers: 2, 4, 6. Favourable: (2,H), (4,H), (6,H)

$P = \frac{3}{12} = \mathbf{\frac{1}{4}}$ [1 mark]

(c) P(>4 or tail or both) [1 mark]

Method 1 — Direct: Numbers > 4: 5, 6. Tail on coin: any number with T.

• (5,H), (6,H), (1,T), (2,T), (3,T), (4,T), (5,T), (6,T): 8 outcomes... wait, let's be careful.

Better — Complement or direct count:

• 4: 5, 6 → (5,H), (5,T), (6,H), (6,T)

• Tail: (1,T), (2,T), (3,T), (4,T), (5,T), (6,T)

Union: (5,H), (6,H), (1,T), (2,T), (3,T), (4,T), (5,T), (6,T) = 8 outcomes

No — let's use formula: $P(>4 \text{ or } T) = P(>4) + P(T) - P(>4 \text{ and } T) = \frac{4}{12} + \frac{6}{12} - \frac{2}{12} = \frac{8}{12} = \mathbf{\frac{2}{3}}$ [1 mark]

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7. [4 marks]

(a) Tree diagram [1 mark]

Expected structure:

• First branch: R (5/10), B (3/10), G (2/10)
• From R: R (4/9), B (3/9), G (2/9)
• From B: R (5/9), B (2/9), G (2/9)
• From G: R (5/9), B (3/9), G (1/9)

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(b) P(same colour) [2 marks]

$P(\text{RR}) = \frac{5}{10} \times \frac{4}{9} = \frac{20}{90}$

$P(\text{BB}) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90}$

$P(\text{GG}) = \frac{2}{10} \times \frac{1}{9} = \frac{2}{90}$

$P(\text{same colour}) = \frac{20+6+2}{90} = \frac{28}{90} = \mathbf{\frac{14}{45}}$ [2 marks]

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(c) P(at least one red) [1 mark]

Method — Complement: $1 - P(\text{no red})$

$P(\text{no red}) = P(\text{BB}) + P(\text{BG}) + P(\text{GB}) + P(\text{GG})$... or simpler: second draw from non-reds.

First draw not red: 5/10, then second not red: 4/9. Or:

$P(\text{at least one R}) = 1 - \frac{5}{10} \times \frac{4}{9} = 1 - \frac{20}{90} = \frac{70}{90} = \mathbf{\frac{7}{9}}$ [1 mark]

Alternatively: $P(\text{RR}) + P(\text{RB}) + P(\text{RG}) + P(\text{BR}) + P(\text{GR}) = \frac{20+15+10+15+10}{90} = \frac{70}{90} = \frac{7}{9}$

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8. [3 marks]

Venn diagram approach: Let $x$ = both basketball and tennis.

Total = Basketball only + Tennis only + Both + Neither $30 = (18-x) + (15-x) + x + 7$ $30 = 40 - x$ $x = 10$

(a) Both sports [1 mark]

10 students play both basketball and tennis. [1 mark]

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(b) P(basketball only) [1 mark]

Basketball only = $18 - 10 = 8$

$P = \frac{8}{30} = \mathbf{\frac{4}{15}}$ [1 mark]

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(c) P(basketball | tennis) [1 mark]

$P(\text{B}|\text{T}) = \frac{P(\text{B and T})}{P(\text{T})} = \frac{\frac{10}{30}}{\frac{15}{30}} = \frac{10}{15} = \mathbf{\frac{2}{3}}$

Or directly: of 15 tennis players, 10 also play basketball: $\frac{10}{15} = \frac{2}{3}$ [1 mark]

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9. [2 marks]

(a) $P(A \cap B)$ [1 mark]

Using: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$0.85 = 0.6 + 0.5 - P(A \cap B)$

$P(A \cap B) = 1.1 - 0.85 = \mathbf{0.25}$ [1 mark]

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(b) Independent? [1 mark]

For independence: $P(A) \times P(B) = P(A \cap B)$

$P(A) \times P(B) = 0.6 \times 0.5 = 0.3$

$P(A \cap B) = 0.25 \neq 0.3$

Not independent because $P(A) \times P(B) \neq P(A \cap B)$ [1 mark]

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10. [3 marks]

(a) Tree diagram [1 mark]

Expected structure:

• First branch: Rain (2/5), No rain (3/5)
• From Rain: Umbrella (4/5), No umbrella (1/5)
• From No rain: Umbrella (1/10), No umbrella (9/10)

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(b) P(carry umbrella) [1 mark]

$P(\text{Umbrella}) = \frac{2}{5} \times \frac{4}{5} + \frac{3}{5} \times \frac{1}{10} = \frac{8}{25} + \frac{3}{50} = \frac{16+3}{50} = \frac{19}{50} = \mathbf{0.38}$ [1 mark]

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(c) P(no rain | umbrella) [1 mark]

$P(\text{No rain}|\text{Umbrella}) = \frac{P(\text{No rain and Umbrella})}{P(\text{Umbrella})} = \frac{\frac{3}{5} \times \frac{1}{10}}{\frac{19}{50}} = \frac{\frac{3}{50}}{\frac{19}{50}} = \mathbf{\frac{3}{19}}$ [1 mark]

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11. [2 marks]

(a) All 8 outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT [not separately marked but required]

(b) P(exactly two heads) [1 mark]

Favourable: HHT, HTH, THH — 3 outcomes

$P = \mathbf{\frac{3}{8}}$ [1 mark]

(c) P(at least one head) [1 mark]

Complement: $1 - P(\text{TTT}) = 1 - \frac{1}{8} = \mathbf{\frac{7}{8}}$ [1 mark]

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12. [2 marks]

Total outcomes: $4 \times 4 = 16$

(a) P(sum = 6) [1 mark]

Favourable: (2,4), (3,3), (4,2) — 3 outcomes

$P = \mathbf{\frac{3}{16}}$ [1 mark]

(b) P(spinner > die) [1 mark]

Favourable: (2,1), (3,1), (3,2), (4,1), (4,2), (4,3) — 6 outcomes

$P = \frac{6}{16} = \mathbf{\frac{3}{8}}$ [1 mark]

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Section C: Statistical Diagrams and Interpretation

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13. [3 marks]

Expected visual: Pie chart with angles: Walk 90°, Bus 120°, MRT 100°, Car 50° (total 360°)

(a) Number by bus [1 mark]

$\text{Bus} = \frac{120°}{360°} \times 120 = \mathbf{40 \text{ students}}$ [1 mark]

(b) Angle for walk [1 mark]

Already given as 90° in diagram (or verify: walk = 120 - 40 - ($\frac{100}{360} \times 120$ = 33.3) - ($\frac{50}{360} \times 120$ = 16.7)... actually from given: walk students = $\frac{90}{360} \times 120 = 30$). Angle is 90° [1 mark]

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(c) Height of bar for Car [1 mark]

Walk: 30 students → 6 cm, so scale is 5 students per cm.

Car: $\frac{50°}{360°} \times 120 = \frac{50}{3} \approx 16.67$ students... wait let me check: total should be 120.

Actually: walk = 90/360 × 120 = 30, bus = 120/360 × 120 = 40, MRT = 100/360 × 120 = 33.33... this doesn't give whole numbers. The problem states angles, so we work with given info.

Height proportional to frequency: $\frac{\text{Car}}{\text{Walk}} = \frac{50°}{90°} = \frac{5}{9}$

Height for Car = $\frac{5}{9} \times 6 = \mathbf{\frac{10}{3}} = 3.33$ cm [1 mark]

Or if using actual count assuming rounded: Car ≈ 16.7 students, Walk = 30, ratio ≈ 5:9, height ≈ 3.33 cm.

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14. [3 marks]

Expected visual: Back-to-back stem-and-leaf with Class A: 42,43,45, 50,54,57,59, 62,63,66,68, 71,75,78 (14 students) Class B: 41,46,48, 52,53,55,55,58, 60,64,65,67,69, 73,74,76,78 (17 students)

(a) Students in Class A [1 mark]

Count leaves: 14 students [1 mark]

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(b) Median for Class B [1 mark]

17 values: median is 9th value. Ordered: 41, 46, 48, 52, 53, 55, 55, 58, 60, 64, 65, 67, 69, 73, 74, 76, 78

Median = 60 marks [1 mark]

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(c) Two comparisons [1 mark]

• Central tendency: Class B has higher median (60 vs Class A median of 63... let me recheck Class A: 14 values, median is average of 7th and 8th = (59+62)/2 = 60.5. Class B median = 60. So medians are similar, OR Class A slightly higher.
• Spread: Class B has wider range (78-41 = 37) vs Class A (78-42 = 36), similar. Class B IQR: Q1 = 53, Q3 = 67, IQR = 14. Class A Q1 = 50, Q3 = 66, IQR = 16. So Class A slightly more spread out.
• Performance: Generally similar, but Class B has more students (17 vs 14).

Any two valid, supported comparisons: [1 mark]

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15. [2 marks]

Expected visual: Histogram with frequency densities: 10-15 (width 5, fd 2.4), 15-20 (width 5, fd 4.0), 20-25 (width 5, fd 3.2), 25-35 (width 10, fd 1.2), 35-50 (width 15, fd 0.4)

(a) Frequency 15-20 cm [1 mark]

Frequency = frequency density × class width = $4.0 \times 5 = \mathbf{20 \text{ fish}}$ [1 mark]

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(b) Number > 25 cm [1 mark]

25-35 cm: $1.2 \times 10 = 12$ 35-50 cm: $0.4 \times 15 = 6$

Total = $12 + 6 = \mathbf{18 \text{ fish}}$ [1 mark]

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(c) Why modal class needs calculation [1 mark]

The modal class is the class with the highest frequency, not highest frequency density. Since class widths vary, a class with high frequency density but narrow width may have lower actual frequency than a wider class with lower density. [1 mark]

Example: 15-20 cm has fd 4.0, width 5, frequency = 20. But 25-35 cm has fd 1.2, width 10, frequency = 12. Need to calculate frequencies to compare.

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16. [2 marks]

Expected visual: Box plots with Bank P: min=2, Q1=5, med=8, Q3=12, max=18; Bank Q: min=3, Q1=4, med=6, Q3=10, max=15

(a) Range and IQR for Bank P [1 mark]

• Range = $18 - 2 = \mathbf{16}$ min [0.5 mark]
• IQR = $12 - 5 = \mathbf{7}$ min [0.5 mark]

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(b) Two comparisons [1 mark]

• Median waiting time: Bank P's median (8 min) > Bank Q's median (6 min), so Bank P typically has longer waits.
• Spread: Bank P has larger range (16 vs 12) and larger IQR (7 vs 6), so waiting times at Bank P are more variable.
• Consistency: Bank Q is more consistent with generally shorter waits.

Any two valid comparisons: [1 mark]

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17. [1 mark]

Expected visual: Dual line graph; temperature peaks in May at 30°C.

(a) Highest temperature [0 mark embedded in total]

May [no separate mark]

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(b) Relationship [0.5 mark]

As temperature increases from January to May, rainfall decreases (negative correlation). Rainfall increases slightly in June as temperature drops slightly. [0.5 mark]

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(c) Mean rainfall [0.5 mark]

$\text{Mean} = \frac{150+120+80+60+40+70}{6} = \frac{520}{6} = \mathbf{86.7 \text{ mm}} \text{ (3 sf)}$ [0.5 mark]

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Section D: Combined and Conditional Probability

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18. [2 marks]

Table of information:

	Smartphone	No smartphone	Total
Female	75	45	120
Male	55	25	80
Total	130	70	200

(a) P(female and no smartphone) [1 mark]

$P = \frac{45}{200} = \mathbf{\frac{9}{40}} = 0.225$ [1 mark]

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(b) P(male | smartphone) [1 mark]

$P = \frac{55}{130} = \mathbf{\frac{11}{26}}$ or ≈ 0.423 [1 mark]

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19. [2 marks]

(a) Showing $n^2 + 15n - 36 = 0$ [1 mark]

Total counters: $8 + n$

$P(\text{WW}) = \frac{8}{8+n} \times \frac{7}{7+n} = \frac{14}{45}$

$\frac{56}{(8+n)(7+n)} = \frac{14}{45}$

Cross multiply: $56 \times 45 = 14 \times (8+n)(7+n)$

$2520 = 14(56 + 15n + n^2)$

$180 = 56 + 15n + n^2$

$n^2 + 15n + 56 - 180 = 0$

$n^2 + 15n - 36 = 0$ [1 mark] ✓

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(b) Solving for $n$ [1 mark]

$(n+18)(n-3) = 0$ (or quadratic formula)

$n = -18$ (reject, since $n \geq 0$) or $n = \mathbf{3}$ [1 mark]

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20. [2 marks]

Expected visual: 6×6 table with cell (i,j) showing max(i,j)

(a) Table completed with max values [not separately marked]

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(b) P(score = 4) [1 mark]

Score = 4 when: max = 4. This occurs when both ≤ 4 and at least one = 4.

• Both = 4: (4,4)
• One = 4, other < 4: (4,1), (4,2), (4,3), (1,4), (2,4), (3,4)

Total favourable: 7 cells

Wait — let me recount using the table structure:

Cells where max = 4:

• Row 4: max(4,1)=4, max(4,2)=4, max(4,3)=4, max(4,4)=4, max(4,5)=5, max(4,6)=6 → 4 cells
• Rows 1,2,3 with column 4: max(1,4)=4, max(2,4)=4, max(3,4)=4 → 3 cells (excluding row 4 already counted)

Total: 7 cells? But (4,4) counted once. So 4 + 3 = 7.

Actually let's be systematic. |Die 2 values across, Die 1 values down|:

For max(i,j) = 4: need all cells where largest entry is exactly 4.

• (4,1), (4,2), (4,3), (4,4), (1,4), (2,4), (3,4)

That's 7 cells out of 36.

$P(\text{score} = 4) = \mathbf{\frac{7}{36}}$ [1 mark]

Wait — let me verify: (4,5), (4,6), (5,4), (6,4) all give max > 4. Yes.

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(c) P(score ≤ 3) [1 mark]

Score ≤ 3 means max ≤ 3, so both dice ≤ 3. Favourable cells: 3 × 3 = 9 cells: (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)

$P = \frac{9}{36} = \mathbf{\frac{1}{4}}$ [1 mark]

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END OF ANSWER KEY