Aptitude and Reasoning Questions

Here if you observe, add and subtract are opposite in nature. Similarly, mutiply and divide are also opposite in nature.

Hence, Add : Multiply :: Subtract : Divide.

Q, the Doctor, is the father of T. S, the Housewife, is the grandmother of T and hence the mother of Q. Since there are only two married couples one being that of Q, the grandfather of R. i.e, U must be married to S. Thus, R and T will be both children of Q and these must be the students So, P, who remains, shall be the wife of Q and she alone can be the nurse. Thus, U must be the contractor 

P is the nurse. 

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.

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 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.

Draw the two chords as shown in the figure. Let O be the center of the circle. Draw OC
perpendicular to both chords. That divides the two chords in half.
So CD = 10 and AB = 14. Draw radii OA and OD, both equal to radius r.
We are given that BC = 5, the distance between the two chords. Let
OB = x.

We use the Pythagorean theorem on right triangle ABO

AO² = AB² + OB²
r² = 14² + x²

We use the Pythagorean theorem on right triangle DCO

DO² = CD² + OC²

We see that OC = OB+BC = x+5, so

r² = 10² + (x+5)²

So we have a system of two equations:

r² = 14² + x²
r² = 10² + (x+5)²

Since both left sides equal r², set the right sides
equal to each other.

14² + x² = 10² + (x+5)²
196 + x² = 100 + x² + 10x + 25
196 = 125 + 10x
71 = 10x
7.1 = x

r² = 14² + x²
r² = 196 + (7.1)²
r² = 196 + 50.41
r² = 246.41
r = √246.41
r = 15.69745202 cm

We know that, M1 × D1W1 = M2 × D2W2

Here M1 = 1, D1 = 6 min, W1 = 1 and M2 = M, D2 = 90 min, W2 = 1845

1 × 61 = M × 901845 

=> M = 123