How to Solve Boats and Stream Maths Quickly

Extending from the basic knowledge rules of boats and stream math problems, here we have given some examples on how to solve boat and stream math problems quickly and easily using quick tricks and basic methods too. The aim is to give you knowledge of basic rules and math shortcut tricks to prepare you for all kind of competitive exams.

• What time will be taken by a boat to cover a distance of 128 km along the stream, if speed of boat in still water is 24 km/h and speed of stream is 8 km/h?

Answer: Here we have

B = 24 and S = 8.

So, D = 24 + 8 = 32

Since the boat is moving along the stream, it means the boat is moving downstream.

Therefore, 128/32 = 4 is the required time.

• A boat goes 48 km downstream in 20 hour. It takes 4 hour more to cover the same distance against the stream. What is the speed of the boat in still water?

Answer: From the question we have

Downstream speed of the boat = 48/20 = 2.4

And Upstream speed of the boat is 48/24 = 2.

Therefore, the speed of the boat in still water = (2 + 2.4) / 2 = 2.2 km/h.

• Ashutosh can row 24 km/h in still water. It takes him twice as long to row up as to row down the river. Find the rate of Stream.

Answer: Let’s consider that, Ashotosh takes 1 unit of time to row downstream and 2 units of time to row upstream.

Here we see that Boat’s speed is 3/2. But in question we have been given that the boat’s speed is 24 km/h.

Therefore we can write 24 * 1/3/2 = 24 * 2/3 = 16 unit.

So from the figure we can write

Downstream speed = 16 * 2 = 32 km/h.

Again,

 D = B + S

32 = 24 + S

S = 8 km/h.

• A motor boat can travel at 10 km/h in still water. It travelled 91 km downstream in a river and then returned to the same place, taking altogether 20 hour. The rate of flow of river is how much?

Answer: We know that upstream speed is

U = boat’s speed (B) – speed of the current (S)

And downstream speed is

D = boat’s speed (B) + speed of the current (S)

Therefore,

Placing B = 10, i.e. motorboat’s speed, we get

10² – S² = 91

S (rate of river flow) = 3 km/h.

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Boats and Streams Tricks for Bank Exams

Boats and Stream math is one of the important topic in bank exam questions as well as in SSC exams. Timing plays an important role in competitive exams both bank exams and SSC exams. Boats and Stream math questions are one of the easiest problems that we usually miss in Competitive exams. 

Here i will provide you all type of shortcut tricks on boats and stream questions come in competitive exams like bank exams. Not just stopped at shortcut tricks on boats and stream, here you will be provided with all basic math formulas on boats and stream and also basic concepts. 

To the readers of this blog, please go through all the boat and stream formulas, basic concepts and short tricks on boat and stream problems. Based on banking exams questions some difficult problems with solutions for bank exams have also been given here.

Important things to keep in mind:

Basic Concepts:

Downstream: Downstream (D) is calculated by adding up the boat's speed running in the same direction with that of the current.

∴ D = Boat's Speed (B) + Speed of the Current (S)

Upstream: Upstream (U) is calculated by subtracting the speed of the current from the speed of the Boat running in the opposite direction of the current.

∴ U = Boat's Speed (B) - Speed of the Current (S)

Boat's speed in still water: (Downstream + Upstream)/2

Speed of Stream: (Downstream - Upstream)/2

Quick Formulas: Click here

The formulas are always important for faster calculations and find out answers quickly. But here, using basic concepts problems or questions on boats and stream are being tried to solve without remembering formulas. 

Example:

• Rajesh can row 12 km/h in still water. It takes him twice as long to row up as to row down the river. Find the rate of stream?

Answer: Let's consider Rajesh as a boat (B) who's speed is 12 km/h.

We know that there is an inverse relation between time and speed, which is 

Time ∝ 1/Speed

So, let's consider that to go downstream Rajesh takes 1 unit of time and to go upstream he takes 2 unit of times, since Rajesh takes twice as long to row up as to row down the river.

We know that speed of Boat in still water = (D+U)/2

From above we get Downstream speed is 2 and upstream speed is 1.

So, (2 + 1)/2 = 3/2

But it is given that boat's speed or Rajesh's speed in still water is 12 km/h.

∴ we can write 12 * 1/3/2 = 12 * 2/3 = 8.

So the downstream speed is 8 * 2 = 16

                                                                                                        

We know that 

Downstream = Boat's speed + Speed of Current

∴ 16 = 12 + Speed of Current

Speed of Current = 4 km/h.

For more examples visit: Examples with Solutions

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Work and Time Problems with Solutions

Q1. Ajay can do a piece of work in 25 days and Sanjay can finish it in 20 days. They work together for 5 days and then Ajay goes away. In how many days will Sanjay finish the remaining work?

Answer: As we have done in our Previous Example, we consider that Ajay does 25 units of work in 25 days and Sanjay does 20 units of work in 20 days.

Since efficiency of Sanjay is 5, 
so the remaining work will done in 55/5 = 11 days.

Q2. It is given that 16 men working 18 hours a days can build a wall 36 m long, 4 m broad and 24 m high in 20 days. How many men will be required to build a wall 64 m long, 6 m broad and 18 m high working 12 hour a day in 16 days?

Answer: There is a technique to solve this type of work and time problems quickly. The short trick to solve this problem is

M₁D₁T₁W₂ = M₂D₂T₂W₁

By applying this rule we get

16*18*20*(64*6*18) = M₂*12*16*(36*4*24)

M₂ = 60

Q3. P can do a piece of work in 12 days, while Q alone can finish it in 8 days. With the help of R they can finish the work in 4 days. How long will R take to finish the work alone?

Answer: As we have done previously, here also we consider that P does 12 units of work, Q does 8 units of work and P,Q,R all together do 4 units of work. (1 unit/days)

So taking LCM of 12, 8, 4 we get 24 as total units of work. Now,

So we have seen in the picture that R’s efficiency is 1, i.e. R can do 1 unit of work in one day. As we found that the total work is 24 units,

so to do the whole work alone 

R will take 24/1 = 24 days.

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Time and Work Tricks

In time and work mathematics problems we are going to learn techniques and tricks solve work and time problems. To solve time and work problems we have to remember certain basic rules which are

Rule 1:

If a person can do a piece of work in n days (hours), then that person’s

1 day work = 1/n

Rule 2:

If a person is n times efficient than the second person, then work done by

First person : Second person = n : 1

And time to complete a work by

First person : Second person = 1 : n

Rule 3:

If ratio of numbers of men required to complete a work is m : n, then the ratio of time taken by them will be n : m.

To do the solutions quickly here are some quick tricks to solve time and work problems and formulas to do time and work easily.

Let’s take a look

Q1. A can do a piece of work in 10 days, while B can do the same work in 6 days. B worked for 4 days. How long will A take to complete the remaining work?

Answer: Take a careful note on this type of work and time problem. To do this type of problems here we shall consider total times which are given in days as total works that a person can do, i.e. here A does 10 units of work in 10 days (1 unit/day) and B does 6 units of work in 6 days (1 unit/day).

Now we take L.H.S of 10 and 6, which will give us the total work that can be done together. Here L.H.S of 10 and 6 is 30.

So according to the question the remaining work is done by A alone. We know that A’s efficiency is 3.

Therefore, A will complete the remaining work in 10/3 days.

For more examples ClickHere.

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Problems on Ages

Problems on ages come both in bank exams and SSC exams, but the frequency that is the number questions on problems on ages is not very high. There are so many formulas on age problems which definitely reduce your time in examination hall but what if you don’t remember the formula at the time of exam or you can not recognize the formula? Well, this is why I am sharing some tricks to solve problems on ages with question and answers.

Q1.  A father is twice as old as his son. 20 years beck he was twelve times as old as his son. What is the present age of father?

Answer: Here I have tried to explain the whole process graphically.

Figure 1

Then,

Figure 2

Now, here we have calculated that the time period is 10, but we have been given that the time period is 20 years. 

So

(20/10) * 22 

[father’s present age as derived in the figure above]

= 2 * 22 = 44 years.

Q2. 5 years ago, age of Sachin was 4 times the age of Vineet and after 10 years, Sachin will be twice as old as Vineet. Find the present ages of Vineet.

Answer: Graphical Presentation,

Figure 3

After that,

Figure 4

Now, here we have found out the total time period, from 5 years ago to 10 years after, which is 2. But according to the questions the total time period is 15 years.

So, the present age of Vineet is

(15/2)  * 1 + 5 

= 7.5 + 5 = 12.5 years.

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How to find Square Root

How to find square root of 5 digit number, square root of 4 digit number or square root of any number without calculator has now become very easy. To find square roots you need to remember certain things that will help you find square roots quickly.

Let’s take a look,

The tricks to find square roots in seconds you have to remember this chart.

Number	Square	Last Digits
0	0	0
1	1	1
2	4	4
3	9	9
4	16	6
5	25	5
6	36	6
7	49	9
8	64	4
9	81	1

Now Carefully see the the following chart to find how square roots are derived.

Note that in 5^th column, if (d) that is the value of (c)*(c+1) is greater than the first two digits of the example [28 in the example 2809 or 50 in the example 5041] that is 30>28 or less than the first two digits of the example that is 72<73 in column 6^th, then the last digit of the answer will be accordingly, derived from the 7^th column (f) of the chart.

So this is how the square of any number is derived quickly even without calculator. One suggestion, the more you practice the more quickly you can calculate square roots.

In the similar way you can also find the Cube Roots. 

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How to find Cube Root

To find the cube root of a number you have to keep in mind certain rules and also squares & cubes of numbers, at least 2 digit number, which are necessary. Click here. Here I will show you how to find Cube roots of numbers in seconds.

Step 1
Keep in mind cubes of numbers up to 9 from 0.

   Chart

Number	Cube	Last digit
0	0	0
1	1	1
2	8	8
3	27	7
4	64	4
5	125	5
6	216	6
7	343	3
8	512	2
9	729	9

Step 2: 

Now let’s take examples to figure out how to find cube root of any number.

1728 and 21952 are the examples. To find the cube root first of all you have to ignore last three digits. To find cube root you always have to ignore the last three digits.

Number	Ignore Last 3 digits
1728	728
21952	952

Step 3:
Now considering the number of remaining digits that is the first 1, 2 or 3 digits, we have remember the lower cube to this number. Like, 1³ = 1 for the number 1728 after ignoring 728, 2³ = 8 which is lower than 21 of the number 21952 after ignoring 952. So we select 1 and 2 for the numbers respectively.

Now notice that the last digit of 1728 or 21952 is 8 and 2 respectively which are the last digits of cubes of 2 and 8 respectively. So we take 2 and 8 respectively.

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How to solve Calendar Maths

How to solve calendar mathematics is a great fun. Here you will get the idea about how to solve calendar math problems and can help yourself  find the day of a week for both Ordinary Year and Leap Year and even find the day of the week without knowing any particular given dates.

Days Gain or Loss

• Ordinary Year

When we move forward or backward by 1 year, then 1 day is gained or lost.
Example: 8th July 2014 was Friday, then 8th July 2015 would be = Friday + 1 = Saturday.
If 4th February 2014 was Monday, then 4th February 2013 would be = Monday - 1 = Sunday.

• Leap Year

When we move forward or backward by 1 leap year, then 2 days are gained or lost.

Example: 24th December 2011 was Tuesday, then 24th December 2012 would be

Tuesday + 2 = Thursday.

If  17th December 2012 was Tuesday, then 17th December 2011 would be

Tuesday - 2 = Sunday.

[Note: For finding the days in leap years the day must have crossed 29th of February, or else 1 should be added instead of 2.]

• Find a Particular Day on the basis of Given Day and Date

If 5th March 1999 was Friday, what day of the week was it on 9th May 2000?

Ans: The number of days between 5th March 1999 and 9th May 2000,

 Since 5th March 1999 was Friday then 5th March 2000 would be

Friday + 2 = Sunday 

[2000 is leap year and 5th March has crossed 29th February.]

After 5th March 2000 we have 26 days of March + 30 days of April + 9 days of May = 65 days.

In 65 days we have 9 weeks and 2 odd or extra days.

So, 5th March 2000 is

Sunday + 9 Weeks 

= Sunday + 2 odd or extra days 

= Tuesday.

• Find a Particular Day without Given Date and Day

Mahatma Gandhi was born on 2nd October 1869. What was the day of the week?

Ans: The nearest Century to 1869 is 1600.

To come to 1869 we have 1600 + 200 + 68

1600 = 0 odd days

200 = 3 odd days

68 = (17 leap year + 51 ordinary year)

So, 1600 + 200 + 68 = Sum of odd days

      = 0 + 3 + (17*2 + 51*1)

      = 0 + 3 + 85 = 88

      = 12 weeks + 4 odd day (88/7= 12 wk + 4 days)

Now total odd days from January to 2 October of 69 or 1869

(odd days are derived by dividing total days of a month by 7, and the remaining are odd days)

January = 3     February = 0
March = 3       April = 2
May = 3         June = 2
July = 3         August = 2
September = 2   October = 2 (up to birth day date)

Total days = 23 = 3 weeks + 2 odd days

So, 4 odd days + 2 odd days = 6 days.

Therefore, 2nd October 1869 is Saturday.

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