Aptitude and Reasoning Questions

1. How to find Speed(s) if distance(d) & time(t) is given:

 Speed (S) = Distance (D)Time (T) 

Ex: Find speed if a person travels 4 kms in 2 hrs?

Speed = D/T = 4/2 = 2 kmph.

2.  Similarly, we can find distance (d)  if speed (s) & time (t) is given by

Distance (D) = Speed (S) x Time (T)

Ex : Find distance if a person with a speed of 2 kmph in 2 hrs?

Distance D = S X T = 2 x 2 = 4 kms. 

3.  Similarly, we can find time (t) if speed (s) &  distance (d) is given by

Time (T) = Distance (D)Speed (S)

Ex : Find in what time a person travels 4 kms with a speed of 2 kmph?

Time T = D/S = 4/2 = 2 hrs.

4. How to convert  km/hr into m/sec :

P km/hr = P x 518 m/sec

Ex : Convert 36 kmph into m/sec?

36 kmph = 36 x 5/18 = 10 m/sec 

5. How to convert  m/sec into km/hr :

P m/sec = P x 185 kmph.

Ex : Convert 10 m/sec into km/hr?

10 m/sec = 10 x 18/5 = 36 kmph.

6.  If the ratio of the speeds of A and B is a:b, then the ratio of the times taken by them to cover the same distance is b : a.

7.  Suppose a man covers a certain distance at x km/ hr and an equal distance at y km/hr . Then, the average speed during the whole journey is 2xyx+y km/hr.

Let original length = x and original breadth = y.

Original area = xy.

New length = x/2 and New breadth = 3y.

New area = (x/2 * 3y) = (3/2 )xy

Therefore,  Increase % = [(1/2)xy * (1/xy) * 100] % = 50%

Let total amount = Rs. P

Total expenditure = (23 + 33 + 19 + 16)% of P = 91%ofP

Remaining money = (100 - 91)% of P = 9% of P

But given in the question that,

9% of P = 504

9p/100 = 504

=> P = 50400/9

=> P = Rs. 5600

Required money spent on Food & Insurance = 23 + 33 %

= 56% of 5600

= 56 x 5600/100 = 56 x 56 = Rs. 3136

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.

Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.

Year : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017
Odd day : 1 2 1 1 1 2 1 1 1 2 1

Sum = 14 odd days => 0 odd days.
Calendar for the year 2018 will be the same as for the year 2007.

Almond Selling price = CP(1 + P%/100)

=> CP = 20000, SP = 23000

=> 23000 = 20000 (1 + P/100)

=> P% = 100(23000/20000 - 1) = 15

Then, Olive's P% = 30%

SP for Olive = CP for Sarah = 23000(1 + 30/100) = 29900

Now Sarah wants to earn profit of 8%

=> Sp for Sarah = 29900(1 + 8/100)

= Rs. 32,292.

Let the initial speed of Akhil be '4p' kmph

Then speed after decrease in speed = '3p' kmph

We know that, 

Change in speed == Change in time

According to the given data,

$$\begin{gathered} 243p - 244p = 35 - 2560 \\ => p = 12 \end{gathered}$$

Hence, the initial speed of Akhil = 4p = 4 x 12 = 48 kmph.