# SSC Quantitative Aptitude tips & tricks: Boats & Streams

In SSC exams including CGL & CHSL, there are always 1-2 questions are asked from Boats and Streams concepts. In this article, we are going to cover all criteria, tricks and shortcut methods of solving all type of questions based on this concept. Let us go through it-

*SSC Aptitude tips*

In SSC exams including CGL & CHSL, there are always 1-2 questions are asked from Boats and Streams concepts. The type of questions is very typical and sometimes it is not easy to comprehend the question as well. In this article, we are going to cover all criteria, tricks and shortcut methods of solving all type of questions based on this concept. Let us go through it-

**Boats & Streams**

The concept of questions based on boats are upstream and downstream parameters on the basis of speed, distance and time. Following types of questions are generally inquired in the exams-

- Calculate upstream/downstream speed of boat/person.
- Calculate the boat speed when speed of stream is given.
- Calculate the speed of stream when upstream/downstream speeds are given.

Before heading to the questions, A few terms must be discussed.

**The Speed of Boat:- **Speed of any floating body means body’s speed in still water. This body may include boat, swimmer and etc.

**Downstream Motion:-** If the motion of a boat is along the direction of the stream, then such motion is called downstream motion.

**Upstream Motion:- **Motion of boat is against the direction of the stream, then this motion should be identified as Upstream Motion.

Suppose, the speed of boat in still water is x and speed of the stream is y, then

- Speed of downstream= (x + y)
- Speed of upstream= (x – y)
- Speed of boat in still water(x)= ½( Upstream Speed + Downstream Speed)
- Speed of stream= ½(Downstream speed – Upstream speed)

**SSC Quantitative Aptitude tricks: Simple & Compound Interest**

** Points to remember**

* If the speed of boat/swimmer is **x km/h** in still water and it takes **n times** as long time to row up and row down the river, then the speed of boat in still water

* If A person’s swimming speed is **x km/h** in still water and stream is flowing at a speed of **y km/h**; it takes **time t** to row to a place and back, then the distance between two places will be

* A person row in downstream distance in **t hours** and return to the same distance in **T hours**; if the speed of stream is **x km/h**; then, speed of the person will be

* If speed of boat in still water is **x km/h** and speed of water is **y km/h**, then average speed in reaching and returning from a place will be

* When boat’s speed in still water is x km/h and speed of river is y km/h; the time taken to cover a certain distance upstream is t more than the time taken to cover the same distance downstream, the distance will be

**SSC Quantitative Aptitude tricks: Algebraic formulae & their applications**

**Examples**

**1. ****Rajesh can swim 24 km/h in still water. It takes him 3 times to row up as to row down the river. Find the rate of the stream?**

**Ans.:-** The speed of swimmer x= 12 and n=3;

Stream’s speed;

= 12*(3+1)/(3-1)= 12*4/2=**24 km/h.**

**2. ****A boat’s speed in still water is 20 km/h while the river is flowing with a speed of 2 km/h and time taken to cover a distance upstream is 5 h more than the time taken to cover the same distance downstream. Find this distance?**

Ans.:- The speed of boat x= 20;

The speed of river= 2 and time t= 5h;

Hence, the distance will be ;

= 5(20*20-2*2)/2*20

=49.5 km.

**3. ****Ravi can swim in still water with a speed of 5 km/h to reach a certain place and to come back. Find his average speed for the whole journey, if the speed of river is 2 km/h.**

Ans.:- Given x= 5 km/h and y= 2 km/h.

Then as per the formulae applied in this situation will be;

= (5+2)*(5-2)/5

=6 km/h.

**SSC Quantitative Aptitude Tips and Tricks: Partnership**

**4. ****A man can row 21 km/h in still water and he takes 2 hours time to row from one place & to come back in a river flowing at a rate of 4 km/h. How far is the place?**

Ans.:- Given x= 21 km/h and y= 4 km/h;

The time t= 2 hours;

The distance in this case will be;

= 2*(21^{2}-4^{2})/2*21

**=20.23 km.**

Above type of questions are generally asked from this section. If you have gone through these and understood the concept clearly, then we can assume that there will be no question in the exam asked from this topic can remain unanswered.

So, keep on looking for further tips and tricks in question solving at www.jagranjosh.com .

**All the best!**

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