Five students are sitting in a row. ‘T’ is on the right of ‘Z’. ‘M’ is on the left of ‘Z’ but is on the right of ‘L’. ‘T’ is on the left of ‘Q’. Who is sitting first from the left ?
Shyam was facing East. He walked 5 km forward and then after turning to his right walked 3 km. Again he turned to his right and walked 4 km. After this he turned back. Which direction was he facing at that time ?
If A = 1, CAT = 24. Then POLICE = ?
If you are 9th person in a queue starting from one end and 11th from another end, what is the number of persons in the queue?
Directions : Which answer figure will complete the question figure?
Van Mahotsav is associated with
Shivpuri National Park of Madhya Pradesh is important for
The first women to swim across seven important seas is
The famous painting ‘Monalisa’ was the creation of
Who discovered America?
The term ‘genetics’ was coined by
Insectivorous plants grow in soil deficient in
Vegetables and fruits should be a part of our diet because they stimulate
Surface water is a better source of drinking water because it is poor in
An insect is an orgainism having
The greatest value among the fractions ^{2}⁄_{7},^{1}⁄_{3},^{5}⁄_{6},^{3}⁄_{4} is

Solution
^{2}⁄_{7} = 0.286;^{1}⁄_{3} = 0.33
^{5}⁄_{6} = 0.833;^{3}⁄_{4} = 0.75
Clearly, the greatest fraction is ^{5}⁄_{6}
A can do a work in 20 days and B in 40 days. If they work on it together for 5 days, then the fraction of the work that is left is

Solution
(A+ B)'s 5 days' work
\(=5\left ( \frac{1}{20}+\frac{1}{40} \right )=5\left ( \frac{2+1}{40} \right )=\frac{15}{10}=\frac{3}{8}\)
∴ Remaining work = 1 ^{3}⁄_{8}=^{5}⁄_{8}
If the mean of 4 observations is 20, when a constant C is added to each observation, the mean becomes 22. The value of C is

Solution
4C = 22 × 4  20 × 4
= 88  80 = 8
⇒C=^{8}⁄_{4}=2
If 2x + y:= 6 and x = 2 are two linear equations, then graph of two equations meet at a point

Solution
Putting x = 2 in the equation 2x + y = 6, we
have
2×2+y=6
⇒y=64=2
∴ Required point = (2, 2)
The digit in unit’s place of the number (1570)^{2}+ (1571)^{2}+ (1572)^{2}+ (1573)^{2} is

Solution
Unit's digit in (1570)^{2} = 0
Unit's digit in (1571)^{2} = 1
Unit's digit in (1572)^{2} = 4
Unit's digit in (1573)^{2} = 9
∴ Required unit's digit = Unit's digit in
(0 + 1 + 4 + 9) = 4