Aptitude and Reasoning Questions

The horizontal lines are IK, AB, HG and DC i.e. 4 in number.

The vertical lines are AD, EH, JM, FG and BC i.e. 5 in number.

The slanting lines are IE, JE, JF, KF, DE, DH, FC and GC i.e. 8 is number.

Thus, there are 4 + 5 + 8 = 17 straight lines in the figure.

T = 4 years

R = 5%

Banker's Gain, BG = Rs.200

TD= BG×100TR=200×1004×5=Rs.1000

 TD=PW×BG

⇒1000=PW×200

=>PW = Rs.5000

Let us assume A is the faster and B is slower, A takes 10s to finish a lap of 50 m, B takes 17s(approx) to finish a lap of 50 m.
So at 11th sec A starts his 2nd lap by the time B will be in between so that is the first time they will meet each other, at 21st sec A starts his 3rd lap by the time B will be coming back to finish his 2nd lap. So that is the second time they meet in opposite. Then at 31st sec A will start his 4th lap by the time B would not even finish 2nd lap as he take 34(2*17) sec to finish 2 laps. So at 35th sec B will start his 3rd lap by the time A will be coming back to finish his 4th lap in the half way so they will meet 3rd time. At 41st sec A will start its fifth lap by the time B will be going to finish 3rd lap in same direction. At 50th second both will complete A's 5th lap and B's 3rd lap. So they will start next lap(A's 6th and B's 4th) from the same point in same direction. And A will finish the race after completing 300m(6*50) total distance and then they wont meet opposite to each other.

So totally they met 3 times in opposite direction.

Given 0.875 and 0.375

0.875 ::

After decimal point there are 3 digits 875

Now, mutiply and divide 0.875 with 1000

$$\begin{gathered} 0.875 x 10001000 = 8751000 \\ Now, this can be reduced to fraction 78 \\ Therefore, 0.875 as fraction = 78 \end{gathered}$$

Similarly,

0.375 :: 

After decimal point there are 3 digits 375

Now, mutiply and divide 0.375 with 1000

$$\begin{gathered} 0.375 x 10001000 = 3751000 \\ Now, this can be reduced to fraction 38 \\ Therefore, 0.375 as fraction = 38 \end{gathered}$$

Let S be the sample space.
Here n(S)= 23 = 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.