(¹⁄₅)^P = √7 ∴ 5^{- P} = 7^1⁄₂

∴ 7 = 5^{- 2p} -I

Also (¹⁄₇)^q = √5 ∴ 7^-q = 5^1⁄₂

∴ 7 = 5^-1⁄₂q -II

From I and II

5^-2p = 5^-1⁄₂q ∴ -2p = -\(\frac{1}{2q}\)

∴ 4pq = 1 ∴ ^pq⁄₂ = ¹⁄₈

Col. A

Let initial edge be a ∴ Volume = a³

Let new edge be 2a ∴ Volume = 8a³

∴ Change in volume = \(\frac{8a^{3}-a^{3}}{a^{3}}\times 100\)

= 700% increase.

Col. B

Let initial radius be r

∴ Volume = ⁴⁄₃πr³

Let the new radius be 2r

∴ New volume = ⁴⁄₃π8r²

= 700% increase.

Let p and q be positive numbers.

∴ If p > q (e..g., 4 > 3), then |p| > |q|

Let p and q be negative numbers.

(e.g. -3 > -4)

∴ |P| is positive p and |q| is positive q.

∴ |p| < |q|

8 × 9 × 10 = 720

∴ K = 7 K + 2 = 9

Col. A-

\(\frac{\frac{12m-n+5}{6}}{\frac{n-5-12m}{3}}=\frac{12m-n+5}{2n-10-24m}\)

=-\(\frac{12m-n+5}{24m-2n+10}=\frac{1(12m-n+5)}{2(12m-n+5)}\)

= -¹⁄₂ = -0.5

Col. B - -0.5

Col. A-

Events = HH, HT, TH, TT

(H = Head, T = Tail)

∴ Probability = ³⁄₄ (At least one head)

Redraw the figure in following manner

Clearly d > c

Note that all numbers are positive.

Let w = 4, x = 3, Y = 2, Z = 1

∴ Col. A - Col. B -

^w⁄_y = 2 ^x⁄_z = 3

∴ Col. B > Col. A

Consider a second set of values

w = 40, x = 14, Y = 2, Z = 1

∴ ^w⁄_y=20 x=\(\frac{14}{1}\)=14

∴ Col. A> Col. B

Since triangles are equilateral the central angles measure 60° and their sum is 120° So unshaded area ¹⁄₃ of circle and shaded are ²⁄₃ of circle.

∴ Area of shaded region = ²⁄₃ × π × (3)³ = 6π

(Shaded region is semicircle + sector of 60 = ²⁄₃ of the circle.)

2m = 11 - 3n

Col. A-

4m + 11 = 22 - 6n + 11 = 33 - 6n

Col. B-

Canceling -6n from both sides