Aptitude and Reasoning Questions

OKG
Remaining all are not in series
only OKG is in series of(15,11,7)

Here it follows the random difference of '4'.

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.

Putting A=1, B=2, C=3,..., M=13,....,X=24,Y=25, Z=26, we have :

 JR = J + R2 + 2 = 10 + 18 2 + 2 = 16

CK = C + K2 + 2 = 3 + 112 + 2 = 9

TX = T + X2+ 2 = 20 + 242 + 2 = 24

OS = O + S2 + 2 = 15 + 192 + 2 = 19

 So, the missing numbers are :

21 and 14

Hence, the answer is (b).

Let O's share = Rs. P

=> N's share = 75100 x P = 3P4

M's share = 54 x 3P4 = 15P16

$$\begin{gathered} 15P16 + 3P4 + P = 4300 \\ => P = 4300 x 1643 = 1600 \end{gathered}$$ 

Ankit's daughter's mother---Ankit's wife; Ankit's wife's father---Ankit's father-in-law;
Father-in-law's daughter--- Ankit's sister-in-law. So, Sanvi is sister-in-law of Ankit.

Relative speed = 120 + 80 = 200 km/hr.
= 200 x 5/18 = 500/9 m/sec.

Let the length of the other train be L mts.
Then, (L + 260)/9 = 500/9 => L = 240 mts.

Each corresponding letter moves five places forward in the alphabet.
Here as BE is related to GJ then PS will relate to UX.

Profit received by Chinna as working partner = 14.5% of Rs. 19600

= 14.5x19600/100 = Rs. 2842

Balance in profit = 19600-2842 = Rs. 16758

Ratio of investment of Chinna & Munna = 80,000 : 1,40,000 = 4 : 7

Hence share of Chinna in investment = 4x16758/100 = Rs. 6093.85

Therefore, Share of Munna = 19600 - 2842 - 6093.85 = Rs. 10664.15