a - b = c - a :∴ 2a = b + c ∴ 3a = a + b + c

Required mean =\(\frac{a + b + c}{3}=\frac{3a}{3}\)= a

Let radius of inscribed circle be 1

∴ Area = π1² which is area of unshaded region.

∴ Radius of semicircle = 2

∴ Area of semicircle = \(\frac{\pi 2^{2}}{2}\) = 2π

∴ Area of shaded region = 2π - π = π

Let there be 5 seats on each bus.

∴ Total seats = 15

Occupied 4 seats = ⁴⁄₅ ^{× 15 = 12}

After ¹⁄₄ of them get off, 9 passengers will travel.

Two busses will have total of 10 seats.

∴ Required fraction = \(\frac{9}{10}\)

Average is 2 ⇒ sum of 3 digits must be 6.

There are 6 nos. between 100 and 200.

105,114,123,132,141,150

200 to 300 → 204, 213, 222, 231, 240

300 to 400 → 303, 312, 321, 330

400 to 500 → 402,411,420

500 to 600 → 501,510

600 to 700 → 600

∴ Total numbers 6 + 5 + 4 + 3 + 2 + 1 = 21

Area of unshaded region = πr²

Area of shaded region = πR² - πr² = π(R² - r²)

∴ π(R² - r²) = πr²

∴ R² = 2r² ∴ ^R⁄_r= √2 = 1.414

Average of a and b = \(\frac{a+b}{2}\)

∴ Addition of one more b will raise the average as b > a

The problem can also be attempted by plugging in nos. such as a = 2, b = 3.

Col. A-

When q is divided by p, quotient is zero and remainder is q.

Col. B-

When p is divided by q, the remainder r is always less than the divisor q.

If x = 90°, then PQ = PR = 5 and QR = 5√2 = (5 × 1.414) > 7

Since X > 90°, QR > 5 √2 > 7

Since we do not know measurement of p and q, it is not possible to find out the answer.

Adding the two equations

2a + 2b + 2c= 18d

∴ a + b + C = 9d

∴ a + b + C + d = 10d

∴ \(\frac{a + b + c + d}{4}\) = 2. 5d ⇒ Col. A

If d is negative answer will be B.

∴ The actual answer should be D.