2013 RRB Allahabad

A>B>C and C<D<E and A<E,

So, C is the youngest and E is the oldest.

The total distance of his friend's house from his house is 10 kilometers north-east. This can be obtained by using Pythagoras theorem where the distance is the sum of the squared distance of 3 and 4 which is 5 kilometers.

There are 32 boys ahead of X and 26 ahead of Y. Since Y is also ahead of X, there are 31 other boys ahead of X. ie there are 5 boys between X and Y.

As per the given coding, it equals 4 /4 + 2x2 + 8 = 13

$$ 1 ^ {3} + 2 ^ {3} $$ = 9

$$ 3 ^ {3} + 4 ^ {3} $$ = 91

$$ 5 ^ {3} + 6 ^ {3} $$ = 341

Since A, B and C are statements such that if both A and B are true then C is false, if A is always true then B is true when C is false. hence (d) is the correct option.

"imp" is common in all the 4 words. In the next 3 letters, "rom" comes before "rov" which comes before "rud" and finally "ude" comes in the dictionary. The reverse order would be as mentioned in option (d). In a dictionary, impudent,imprudent,improvise,impromptu come in the reverse alphabetical order.

$$ a ^ {3} + b ^ {3} = (a+b) ( a^{2} + b^{2} - ab) $$

Substituting b = 1/a, we get the answer as 0

The roots of a quadratic equation are given by $$ \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Hence, the roots are = $$ \frac{6\pm\sqrt{36-4}}{2}$$ = $$3 \pm \sqrt{8}$$
From the eqn, product of roots = c/a =1/1 = 1
Hence, the roots are reciprocals.
Thus, 1/(3+√8) = 3-√8
$$x^{2}+ \frac{1}{x^{2}}$$ = $$(3+\sqrt{8})^2 + (3-\sqrt{8})^2$$ = 9 + 8 +6√8 + 9 + 8 - 6√8 = 34

When 5 is added to the numerator, it becomes one. So, the numerator is 5 less than the denominator. When six is added to the denominator, it becomes double the numerator. 

So, let the numerator be x, denominator = x + 5

x + 5 + 6 = 2x

So, x = 11

So, the fraction is 11/16