GATE Questions

---

Let the distance be 'd' kms.

According to the given data,

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

Given 

1 = 1 => 1 x 11 = 11

2 = 22 => 2 x 11 = 22

3 = 33 => 3 x 11 = 33

...

...

Similarly,

11 => 11 x 11 = 121.

The main component of bones and teeth is Calcium Phosphate.

I = (600x11x5)/100 = 330 Paise = Rs. 3.30

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 'd' be the distance between A and B
K -time = d/10 + d/9 = 19d/90 hours
L -time = 2d/12 = d/6 hours
We know that, 10 min = 1/6 hours
Thus, time difference between K and L is given as 10 minutes.
=> 19d/90 - d/6 = 1/6
=> (19d-15d)/90 = 1/6
=> 4d/90 = 1/6
Thus,
d= 15/4 km = 3.75 km.

hence the distance between A and B is 3.75 km.

At 1 pm ... Distance between A and B = 2x3.5 = 7 kms
at 2 pm .....Distance between A and B = 7 + 3.5 - 1 = 9.5 kms
At 3 pm ....Distance between A and B = 9.5 + 3.5 - 2 = 11 kms.
..
..
..
At 7 pm distance between A and B = 3 kms.
From 7 pm to 8 pm, speed of A = 3.5 kmph
speed of B = 8 kmph
Relative speed = 4.5 kmph

Time taken by B to meet A = 3/4.5 hrs = 30/45 hrs = 120/3 mins = 40 mins.

So meeting time is 7 : 40 PM.