(a) Coordinates of A:

From the description, OA = 10 km on Delhi-Agra highway. Since Delhi-Agra highway is along x-axis (as O is at origin and highways cross at O), point A lies on x-axis at distance 10 from O.

Therefore, coordinates of A = (10, 0).

✅ Answer: (i) (10,0)

(b) Equation of line AB:

Point A = (10, 0), Point B = (0, 12) (since OB = 12 km on YY’ which is y-axis).

Equation of line through two points: \(\frac{y – y_1}{y_2 – y_1} = \frac{x – x_1}{x_2 – x_1}\)

\(\frac{y – 0}{12 – 0} = \frac{x – 10}{0 – 10}\)

\(\frac{y}{12} = \frac{x – 10}{-10}\)

Cross multiply: \(-10y = 12(x – 10)\)

\(-10y = 12x – 120\)

\(12x + 10y = 120\)

Divide by 2: \(6x + 5y = 60\)

✅ Answer: (ii) 6x + 5y = 60

(c) Distance of AB from O(0,0):

Distance from point to line formula: \(d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\)

Line AB: \(6x + 5y – 60 = 0\) (from part b)

Point O(0,0): \(d = \frac{|6(0) + 5(0) – 60|}{\sqrt{6^2 + 5^2}} = \frac{|-60|}{\sqrt{36 + 25}} = \frac{60}{\sqrt{61}}\)

✅ Answer: (ii) \(\frac{60}{\sqrt{61}}\) km

(d) Slope of line AB:

Slope \(m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{12 – 0}{0 – 10} = \frac{12}{-10} = -\frac{6}{5}\)

✅ Answer: (iii) \(-\frac{6}{5}\)

(a) P(Ashima does not qualify):

P(S’) = 1 – P(S) = 1 – 0.10 = 0.9

✅ Answer: (i) 0.9

(b) P(at least one qualifies) = P(A ∪ S):

P(A ∪ S) = P(A) + P(S) – P(A ∩ S)

= 0.05 + 0.10 – 0.02 = 0.13

✅ Answer: (ii) 0.13

(c) P(at least one does not qualify):

This is complement of P(both qualify):

P(at least one does not qualify) = 1 – P(both qualify)

= 1 – 0.02 = 0.98

✅ Answer: (iv) 0.98

(d) P(both do not qualify):

P(both do not qualify) = P(A’ ∩ S’) = 1 – P(A ∪ S) [De Morgan’s Law]

= 1 – 0.13 = 0.87

Alternatively: P(A’ ∩ S’) = 1 – P(A ∪ S) = 1 – 0.13 = 0.87

✅ Answer: (iii) 0.87

(a) All 4 cards from hearts:

Number of heart cards = 13,

Ways to choose 4 from 13

= C(13,4) \(= \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = \frac{17160}{24} = 715\)

✅ Answer: (iii) 715

(b) All 4 cards from different suits:

We need one card from each of the 4 suits.

Number of ways = \(C(13,1) \times C(13,1) \times C(13,1)\)\( \times C(13,1) = 13^4 = 28561\)

✅ Answer: (i) 28561

(c) All face cards:

Total face cards = 4 suits × 3 face cards (J, Q, K) = 12

Ways to choose 4 from 12 = \(C(12,4) = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \frac{11880}{24} = 495\)

✅ Answer: (iii) 495

(d) 2 red and 2 black cards:

Choose 2 red from 26 red cards: \(C(26,2) = \frac{26 \times 25}{2} = 325\)

Choose 2 black from 26 black cards: \(C(26,2) = 325\)

Total ways = \(325 \times 325 = 105625\)

Check given options: 106525, 105525, 105265, none of these.

105625 is not listed, so correct is “None of these”.

✅ Answer: (iv) None of these