Aptitude and Reasoning Questions

Let the distance be 'd' kms.

According to the given data,

$$\begin{gathered} d30 - d50 = 4060 hrs \\ => 6d = 300 \\ => d = 50 kms. \\ \end{gathered}$$

Let the speed of Kaushik be Sk and speed of Piyush be Sp and let the time taken by Piyush be t second then,

 SkSp = 800t+20600t = 23since given ratio

t = 20 s

Speed of Piyush = Sp = 600/20 = 30 m/s

and    Speed of Kaushik = 800/40 = 20 m/s

Here I am very happy to start this Program is correct statement.

"Too" is not used with pleasant adjectives. It means 'more than required' and is usually used with unpleasant adjectives.

So, 'too healthy', 'too happy', 'too glad' etc are wrong.

Let the total number of pages in the book be 'x'.

From given data,

2x/5 + 34x/100 + 52 + 52 = x

 => 40x + 34x + 10400 = 100x

 => 26x = 10400

 => x = 400.

 Hence, total pages in the book are 400.

Let the 6 digits of the required 6 digit number be abcdef

Then, the number to be odd number the last digit must be odd digit i.e 3, 5, 7 or 9

The first digit cannot be ‘0’ => possible digits = 3, 5, 7, 2, 6, 8

Remaining 4 places can be of 6 x 5 x 4 x 3 ways

This can be easily understood by

Therefore, required number of ways = 6 x 6 x 5 x 4 x 3 x 4 = 36 x 20 x 12 = 720 x 12

= 8640 ways.

From above diagram
AC represents the hill and DE represents the tower

Given that AC = 100 m

angleXAD = angleADB = 30° (∵ AX || BD )
angleXAE = angleAEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE

tan 60°=AC/CE => √3 = 100/CE =>CE = 100/√3 ----- (1)

tan 30° = AB/BD => 1/√3 = 100−h/BD => BD = 100−h(√3)
∵ BD = CE and Substitute the value of CE from equation 1
100/√3 = 100−h(√3) => h = 66.66 mts

The height of the tower = 66.66 mts.

Let the sum be Rs.x. Then, 

 x1+101002-x=1155

⇒x11102-1=1155

=> x =5500

sum = Rs. 5500.

So, S.I = Rs. 5500×5×4100 = 1100