Which of the following is the major constituents of LPG?
In addition to hydrogen, the other abundant element present on Sun’s surface is
Natural rubber is the polymer of
Which of the following elements is used in the manufacture of fertilizers?
When a group of computers is connected together in a small area without the help of telephone lines it is called
Velocity of sound in air does not depend on
‘Comose’ seeds are seeds with
The technique of calculating the age of fossil organisms is
Haemoglobin contains the metal
A matured mammalian cell without nucleus is
Ravi’s father has a son Rohit who has an aunt Laxmi who has a husband Rao whose fatherinlaw is Mohan. What is the relation of Mohan to Ravi ?
Directions : A series is given with one term missing. Choose the correct alternative from the given ones that will complete the series
2,5,8, 11, 14, ?
Directions : A series is given with one term missing. Choose the correct alternative from the given ones that will complete the series.
BDF, KMO, TVX, ?
Which one set of letters when sequentially placed at the gaps in the given letter series shall complete it ?
abbabba
Directions : Select the one which is different from the other three responses.
In a ΔABC, \(\overline{AB^{2}} + \overline{AC^{2}} = \overline{BC^{2}}\) and \(\overline{BC} = \sqrt{2AB}\),then ∠ABC is

Solution
AB^{2}+ AC^{2}= BC^{2}⇒∠BAC = 90°
⇒AB^{2}+ AC^{2}= 2AB^{2}
⇒AB^{2}= AC^{2}⇔AB = AC
∴ ∠ABC = ∠ACB = 45°
The third proportional to 0.8 and 0.2 is

Solution
If the third proportional be x, then
0.8 : 0.2 : : 0.2 : x
⇒0.8 × x= 0.2 x 0.2
⇒x= 0.8 = 80 = \(\frac{0.2\times O.2}{0.8}=\frac{4}{80}= 0.05\)
A train covers a distance of 10 km in 12 minutes. If its speed is decreased by 5 km/hr, the time taken by it to cover the same distance will be

Solution
Speed of train =\(=\frac{Distance}{Time}=\frac{10}{\frac{12}{60}}kmps\)
\(=\left ( \frac{10\times 60}{12} \right )=50 kmph\)
New speed = 45 kmph
∴ Required time = ^{10}⁄_{45} hour = ^{2}⁄_{9} x 60
minutes = ^{40}⁄_{3} minutes = 13 minutes 20 seconds
If x^{3}+ y^{3} = 35 and x + y = 5, then the value of ^{1}⁄_{x}+^{1}⁄_{y} will be

Solution
(x+ y)^{3} = x^{3}+ y^{3} + 3 (xy) (x+ y)
⇒125 = 35 + 3(5) xy⇒15xy= 125  35 = 90
\(⇒xy=\frac{90}{15}=6⇔\frac{x+y}{y}=\frac{1}{y}+\frac{1}{x}=\frac{5}{6}\)
If A denotes the volume of a right circular cylinder of same height as its diameter and B is the volume of a sphere of same radius, then ^{A}⁄_{B} is

Solution
Volume of cylinder = πr^{2}h.
⇒A = πr^{2}(2r)= 2πr^{2}
[∵ h = 2r]Volume of sphere =\(\frac{4}{3}\pi r^{3}\)
\(B=\frac{4}{3}\pi r^{3}\Rightarrow \frac{A}{B}=\frac{2\pi r^{3}}{\frac{4}{3}\pi r^{3}}=\frac{6}{4}=\frac{3}{2}\)