**Please read Part-1 before reading this post. In this article I have covered questions related to Boat and Streams, Gun-shots and some other left over topics.**

Average speed for the complete journey = 2XY/(X + Y) = 2*20*30/50 = 24 km/hr

**Answer : (D)**

Let the total distance be 100 km

**Average Speed = Total Distance/Total Time**

Total time = 70/20 + 10/25 + 20/8 = 3.5 + 0.4 + 2.5 = 6.4

Average Speed = 100/6.4 = 15.625 m.p.h

This is again a very frequently asked question. Let the distance of his school be X km.

(Time taken to reach the school at 3 km/hr) - (Time taken to reach the school at 4 km/hr) = (10 + 10) minutes or 1/3 hours

X/3 - X/4 = 1/3

Hence X = 4 km

**Answer : (B)**

__Direct Formula__
Distance = S1*S2/(S1 - S2) * Time difference

S1 = 4 km/hr, S2 = 3 km/hr, Time Difference = 10 - (-10) = 20 minutes or 1/3 hours

Distance = 4*3/(4-3) * 1/3 = 4 km

**Note:**In the above formula, while calculating the time difference, "late" time is written with negative sign.

__Method 1__
Let time taken by second runner = t. So time taken by first runner = t + 32/60 = t + 8/15

Since distance is constant, hence speed and time are inversely proportional

S2/S1 = T1/T2

16/15 = (t + 8/15)/t

16/15 = 1 + 8/15t

1/15 = 8/15t

t = 8 hours

So second runner takes 8 hours to cover the distance with a speed of 16 km/hr

Hence distance = 8*16 = 128 km

__Method 2__
Let the distance be X km. Then,

X/15 - X/16 = 32/60

Solve for X

X = 128 km

**Answer: (A)**

__Method 3 (Direct formula)__-
In such questions you can use the same formula you used for Q. (3)

Distance = S1*S2/(S1 - S2) * Time difference

Distance = 16*15/(16 - 15) * (32/60) = 16 * 15 * 32/60 = 128 km

A man rows down a river 15 km in 3 hrs.

Hence, Downstream Speed(v) = 15/3 = 5 km/hr

Similarly, Upstream Speed(u) = 15/7.5 = 2 km/hr

v = Rate in still water + Rate of stream

u = Rate in still water - Rate of stream

Add the above 2 equations-

Rate in still water = (v + u)/2 = (5 + 2)/2 = 3.5 km/hr

**Answer: (C)**

We have in the above question-

Speed of the current = (v - u)/2

u = 36/6 = 6 km/hr

v = 48/6 = 8 km/hr

Speed of the current = (8 - 6)/2 = 1 km/hr

**Answer: (D)**

Let the distance be X km. Let he takes 't' time downstream, then he will take '2t' time upstream.

Downstream speed(v) = X/t

Upstream speed(u) = X/2t

Speed of the boat in still water/Speed of the current = (v + u)/(v - u) = (X/t + X/2t)/(X/t - X/2t)

= 3/2 : 1/2

= 3 : 1

**Answer : (B)**

**Direct formula-**

**So,**Speed of the boat in still water/Speed of the current = (2t + t)/(2t - t) = 3 : 1

Given,

24/u + 28/v = 6 or 12/u + 14/v = 3 ... (1)

30/u + 21/v = 6.5 ... (2)

The best way to solve (1) and (2) is by eliminating a variable.

Multiply equation (1) by 3

36/u + 42/v = 9 ... (3)

Multiply equation (2) by 2

60/u + 42/v = 13 ... (4)

Subtract equation (3) from (4)

24/u = 4

u = 6 km/hr

Put u = 6 in equation (1)

v = 14 km/hr

Speed of the boat in still water = (u + v)/2 = (6 + 14)/2 = 10 km/hr

**Answer: (D)**

**Q. 9)**

**Two guns were fired from the same place at an interval of 13 minutes but a person in a train approaching the place hears the second shot 12 mins 30 seconds after the first. Find the speed of the train(approx) supposing that sound travels at 330 m/s.**

**A. 40 B. 47 C. 55 D. 60**

Distance travelled by sound in 30 sec = Distance travelled by train in 12 min 30 sec

Let the speed of the train be X m/sec

Distance travelled by sound in 30 sec = 330*30 metres

Distance travelled by train in 12 min 30 sec (750 sec) = X*750

330*30 = X*750

X = 13.2 m/sec = 13.2 * 18/5 km/hr = 47.52 km/hr

**Answer: 47 km/hr**

__Explanation__

When you hear the gun shot, that means the sound has travelled to your ears.

First consider a simple scenario when the train is not moving. When the two shots are fired from A, a person sitting in the train will hear them at an interval of 13 minutes only. The sound travels the distance from A to B.

Now let us consider the scenario when the train is moving from B to A. When the first shot is fired, the sound will travel from A to B and the person sitting inside the train will hear it instantly. Now when the second shot is fired after 13 minutes, the sound would not have to travel from A to B, because the person sitting inside the train is not at B any more. He has moved from position B to X. Hence the sound only needs to travel from A to X.

Hence in this case, the person is hearing the shot after 12 minutes 30 seconds. Instead of travelling for 13 minutes (from A to B), now the sound is travelling only for 12 min 30 sec (from A to X). Hence we can say,

AB = Distance travelled by sound in 13 minutes

AX = Distance travelled by sound in 12 minutes 30 seconds

XB = Distance travelled by sound in 30 seconds ... (1)

After 12 minutes 30 seconds, the sound moves from A to X and also the train moves from B to X.

BX = Distance travelled by train in 12 minutes 30 seconds ... (2)

Hence from (1) and (2) we can say-

**Distance travelled by sound in 30 sec =**

**Distance travelled by train in 12 min 30 sec**

**Q. 10) Two guns were fired from the same place at an interval of 10 minutes and 30 seconds, but a person in a train approaching the place hears second shot 10 minutes after the first. The speed of train (in km/hr), supposing that sound travels at 330m/s is:**

**A. 19.8 B. 58.6 C. 59.4 D. 111.8**

Distance travelled by sound in 30 sec = Distance travelled by train in 10 minutes (600 sec)

330*30 = X*600

X = 16.5 m/sec or 59.4 km/hr

**Answer: (C)**

**I have covered almost all the type of questions that are asked by SSC from this topic. If you have any doubt in STD, please drop comment.**

**Keep Reading :)**

**To buy the super-hit SSC Hack-book, follow the below link-**

last 2 ques were brilliant !!

ReplyDeletethanks !!

Thnks so much sir.. itna achha to kbhi teacher ne nhi smjhaya...

ReplyDeleteIs question 8 correct?

ReplyDeleteIn 6 hrs, he goes 24km Up 28km

But in 6.5 hrs, he goes 30km up 21km down!? :)

Going up faster than going down by defying the laws of physics? or am I missing something?

This comment has been removed by the author.

ReplyDelete@ankita : Please refrain from any kind of marketing/promotions here...

ReplyDeleteKiller Blog !! Thnx a ton fr sparing your precious time ...

ReplyDeleteCan u please help me with this problem (CGL 2014 T-I)

A car travels from P to Q at a constant speed. If its speed is increased by kmph, it would've taken 1 hour less to cover the distance. It wouldve taken further 45 minutes lesser if the speed was further increased by 10kmph. The distance between P and Q is ?

Answer: 540km

st = (s+10)(t-1) = st - s + 10t - 10

Delete10t = s+10 ... (1)

again,

st = (s+20)(t-1.75) = st - 1.75s + 20t - 35

20t = 1.75s + 35 ... (2)

solve (1) and (2)

0.25s = 15

s = 60

t = 7

answer = 60*7 = 420

420 is the correct answer

*10 kmph

ReplyDeleteThank u so much (:

ReplyDelete