The sum of all the natural numbers between 301 and 501 (including 301 and 501) which are divisible by 7 is

Solution
The smallest number = 301
The largest number = 497,
Common difference = d = 7
t_{n} = a+ (n 1) d
⇒ 497 = 301 + (n 1) × 7
⇒ (nl) × 7 = 497  301 = 196
⇒ n 1 = 196 + 7 = 28
⇒ n= 28 + 1 = 29
∴ Required sum
=^{n}⁄_{2} (first term + last term)
=^{29}⁄_{2}(301 + 497)\(=\frac{29\times 798}{2}= 11571\)
Natural: Artificial: : Spontaneous:?

Solution
Natural is the antonym of Arttficial. Similarly, Spontaneous is the antonym of Calculated.
Find the odd one out.

Solution
Campaign is used as both Verb and Noun. All other words are Nouns.
If X is a player then X must be stout. This statement can be deduced from

Solution
The statement implies that all players are stout. If all players are stout, them some players must be stout.
In a certain code language ‘278’ means ‘run very fast’.’853′ means ‘come back fast’ and ‘376’ means ‘run and come’ then ‘back’ may be represented by the digit

Solution
back ⇒ 5
If ‘A + B’ stands for ‘A is the father of B’, ‘A – B’ stands for ‘A is the brother of B’. ‘A × B’ stands for ‘A is the wife of B’ and ‘A + B’ stands for ‘A is the sister of B’ then ‘P + Q + R’ means

Solution
P + Q → P is father of Q.
Q + R → Q is the sister of R.
Therefore, P is the father of Q and R.
If 1st July. 1977 was a Friday then 1st July 1970 was a

Solution
Number of odd days upto 1st July 1977:
Number of odd days in January 1977 =3
Number of odd days in February 1977 =0
Number of odd days in March 1977 =3
Number of odd days in April 1977 =2
Number of odd days in May 1977 =3
Number of odd days in June 1977 =2
13 odd days1976 was a leap year, so number of odd days = 2
In 1975 number of odd day = 1
In 1974 number of odd day = 1
In 1973 number of odd day = 1
In 1972 was a leap year, so number of odd days = 2
In 1971, number of odd day = 1
Number of odd days in 1970
from December to 1st July
= 3 + 2 + 3 + 2 + 3 + 3 = 16
Total number of odd days
= 13 + 8 + 16 = 37, i.e. 2 odd days
Therefore, 1st July, 1970 was
two days before Friday, i.e., Wednesday
If A stands for ‘+’, B stands for ‘‘, C stands for ‘x’ and D for ‘÷’ then ^{1}⁄_{2}A^{1}⁄_{3}B^{1}⁄_{4}C^{1}⁄_{5}D^{1}⁄_{6}=

Solution
^{1}⁄_{2}A^{1}⁄_{3}B^{1}⁄_{4}C^{1}⁄_{5}D^{1}⁄_{6}
=^{1}⁄_{2}+^{1}⁄_{3}^{1}⁄_{4}×^{1}⁄_{5}÷^{1}⁄_{6}
=^{1}⁄_{2}+^{1}⁄_{3}^{3}⁄_{10}
\(=\frac{15+109}{30}=\frac{16}{30}=\frac{8}{15}\)
In the following sequence of alphabets
a a b a b a a b a b a a a b b a b a b b a a a a
the number of a’s in between 7th a from left and 7th a from right is

Solution
22 boys are standing in front of Kunal in a queue. 12 boys are standing to the back of Rohit in the same queue. If total number of boys is 30 then the number of boys standing in between Kunal and Rohit is

Solution
There are four boys in between Rohit and Kunal.