**Answer : N=11 , x=2 **

**Solution :**

#### let N(N-1) = t;

#### (N+x)(N+x-1) = t+46;

#### trail and error method x=2 and N=11

**Answer : A**

**Solution :**

#### Total number of outcomes in the simultaneous throw of 3 dice is 6 x 6 x 6 = 216.

That is, n(S) = 216.

Then E = {(1,1,1),(1,1,3),(1,1,5),(2,2,1),(2,2,3),(2,2,5),(3,3,1),(3,3,3),(3,3,5),(4,4,1),(4,4,3),(4,4,5),(5,5,1),(5,5,3),(5,5,5),(6,6,1),(6,6,3),(6,6,5)}

Therefore, n(E) = 18

Required p(E) = 18 / 216 = 1 / 12.

**Answer : 30**

**Solution :**

#### Total no of balls = 89 and (89-29) /2 = 60/2 = 30 and also 14 + 16 = 5 + 7 + 18 = 30

**Answer : C**

**Solution :**

#### below. (We recommend you to try with other options, to prove that they are wrong.)

On day 1: Medals awarded = (1 + 24/6) = 5 : Remaining 20 medals

On day 2: Medals awarded = (2 + 18/6) = 5 : Remaining 15 medals

On day 3: Medals awarded = (3 + 12/6) = 5 : Remaining 10 medals

On day 4: Medals awarded = (4 + 6/6) = 5 : Remaining 5 medals

On day 5: Medals awarded = 5

**Answer : B**

**Solution :**

#### Jeya Celebrated her 18^{th} birthday on 28 February 2009, Saturday.

#### Then her 19^{th} birthday will be on 28 February 2010, Sunday. (This is because 2010 is not a leap year and hence contains 365 days. 365 divided by 7 gives remainder 1 which means any date on 2009 corresponds to the same date on 2010 advanced by one day.)

#### Her 20^{th} birthday will be on 28 February 2011, Monday.

#### Her 21^{st} birthday will be on 28 February 2012, Tuesday.

#### Her 22^{nd} birthday will be on 28 February 2013, Thursday. (Since, 2012 is a leap year hence contains 366 days. 366 divided by 7 gives remainder 2 which means any date on 2013 corresponds to the same date on 2012 advanced by two days.)

#### Her 23^{rd} birthday will be on 28 February 2014, Friday.

#### Her 24^{th} birthday will be on 28 February 2015, Saturday.

#### Her 25^{th} birthday will be on 28 February 2016, Sunday.

#### So, her 25^{th} birthday will be on Sunday.

**Answer : D**

**Solution :**

#### Inspecting how RELATED was written as EFUBKDQ, we can easily find that,

RELATED – is first written as DETALER

DETALER becomes EFUBKDQ **(+1,+1,+1, +1,-1,-1,-1 – alphabetically)**

Applying the same logic to RETAINS we get,

RETAINS is first written as SNIATER

SNIATER becomes TOJBSDQ (applying the above rule)

**Answer : B**

**Solution :**

#### Though all animals are reptiles from the two statements, there is no indication that all reptiles are monkeys (and in turn animals). Hence option I cannot be true always. But second conclusion is always true which is straightforward to interpret.

**Answer : A**

**Solution :**

#### Let the two unequal numbers be a and b.

We have to find the product of numbers = ab

Sum of numbers = a + b

Difference of numbers = a – b.

#### The squares of sum and difference of the two unequal numbers are (a + b)^{2} and (a – b)^{2}

Then the difference between the squares of sum and difference of two unequal numbers = (a + b)^{2} – (a – b)^{2}

#### Given that, (a + b)^{2} – (a – b)^{2}= 4928.

Remember the formulae (x – y)^{2} = x^{2} + y^{2} – 2xy and (x + y)^{2} = x^{2} + y^{2} + 2xy

Therefore, we have

(a + b)^{2} – (a – b)^{2} = 4928 = a^{2} + b^{2} + 2ab – (a^{2} + b^{2} – 2ab)

4928 = a^{2} + b^{2} + 2ab – a^{2} – b^{2} + 2ab

4928 = 4ab

ab = 1232.

Hence the required product is 1232.

**Answer : A**

**Solution :**

#### Let PQ be the height of the lighthouse.

And Given that, angle(PAQ) = a, angle(PBQ) = 2a and angle(PCQ) = 3a

#### We know that, “The size of an exterior angle at a vertex of a triangle equals the sum of the sizes of the interior angles at the other two vertices of the triangle”.

Here, angle(PBQ) = angle(A) + angle(APB)

2a = a + angle(APB)

angle(APB) = a

#### Since the opposite sides with equal angles are equal,

AB = PB

Therefore, PB = x

#### From triangle PQB,

PQ / PB = sin(2a)

PQ = xsin2a.

Hence the height of the lighthouse is xsin2a.

**Answer : A**

**Solution :**

Given is an isosceles triangle. (two sides are equal in length)