ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120° and ∠BAC = 30°, then ∠BCD is

Solution
∠COD = 120°
∠BAC= 30°
∠CAD = ^{1}⁄_{2} × 120° = 60°
∴ ∠BAD = 90°
∴ ∠BCA = 180°90° = 90°
The value of \(\frac{cot30^{\circ}cot75^{\circ}}{tan15^{\circ}tan60^{\circ}}\) is

Solution
cot 30° = cot (90°  60°) = tan 60°
cot 75° = cot (90°  15°) = tan 15°
\(frac{cot30^{circ}cot75^{circ}}{tan15^{circ}tan60^{circ}}=frac{tan60^{circ}tan15^{circ}}{tan15^{circ}tan60^{circ}}=1\)
The radius of the base of a right circular cone is doubled keeping its height fixed. The volume of the cone will be

Solution
Original volume of cone =\(frac{1}{3}pi r^{2}h\)
New volume of cone =\(frac{1}{3}pi (2r^{2})h=frac{4}{3}pi r^{2}h\)
An equation whose graph passes through the origin, out of the given equations 2x + 3y = 2,2x – 3y = 3, 2x+ 3y = 5 and 2x + 3y = 0 is ,

Solution
Putting x= 0 in equation 2x+ 3y= 0, we get y= 0 Hence, this straight line passes through the origin.
The greatest among 1.5, 1.05, \(1.0\overline{5}\), \(1.\overline{5}\) is

Solution
1.05 < \(1.0overline{5}\) < 1.5 < \(1.overline{5}\) Remember: \(1.0overline{5}\) = 1.0555 .... \(1.overline{5}\) = 1.555 ....
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