Aptitude and Reasoning Questions

The answer is 'mare' because a "mare’s nest" doesn't exist just like "hen's teeth" don't exist.

The given figure can be labelled as :

Straight lines :

The number of straight lines are 19

i.e. BC, CD, BD, AF, FE, AE, AB, GH, IJ, KL, DE, AG, BH, HI, GJ, IL, JK, KE and DL.

Ans. 
540 = 3240/6
108 = 540/5
27 = 108/4
? = 27/3, i.e. ? = 9

On carefully inspecting this question, one can understand that there are two days which are important and these are:

A. My Birthday.

B. The day when I am making this statement.

If you think for a while, you will understand that such statements can be made only around the year’s end. So, if my birthday is on 31 December, then I will be making this statement on 1 January.

I will further explain using the following example:

1. Consider that today is 01 January 2017.

2. Then, the day before yesterday was 30 December 2016 and according to the question I was 25 then.

3. Yesterday was 31 December 2016, which happens to be my birthday too (Woohoo!), and my age increases by one to become 26.

4. I will turn 27 on my birthday this year (31 December 2017).

5. I will turn 28 on my birthday next year (31 December 2018).

Now, if you read the question again, it will make more sense:

The day before yesterday(30 December 2016), I was 25 years old and next year(31 December 2018) I will be 28.

NOTE :

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2:  Add +6 to the Given Year.

Solution : 

Given Year is 2024, Which is a leap year.

So, Add +28 to the given year (i.e 2024 + 28) = 2052

Therfore, The calendar of the year 2024 can be used again in the year 2052.

Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then 

 Required probability = PA∩B∩C 

= PA PBA PCA∩B

 Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9

When  a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white

 ∴PBA=317 

Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.

 ∴PCA∩B =216=18

 Hence the required probability = 29×317×18=1204

Let the sum be Rs. x. Then,

 C.I=x*50300*1003-x=343x216-x=127x216 

⇔127x216=1270 

x=2160 

Thus, the sum is Rs. 2160

 S.I=Rs.2160*503*3*1100=Rs 1080