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Quantitative Aptitude – Percentage

A very important chapter for competitive exam is Percentage. Every competitive exam question paper must have at least one question from this topic. And so many other questions which are related to this chapter. So, you should learn this chapter very carefully.

Percentage is related to fraction. So if you do not know what fraction is or if you need to brush up your previous knowledge of fraction, then first learn “Fraction“.

But, if you are comfortable with fraction then we can start our lesson.

What is Percentage ?

Definition: Percentage is the number per hundred.

Let’s explain this,

• 50% means 50/100 or 50 per 100.
• 25% means 25/100 or 25 per 100.
• 10% means 10/100 or 10 per 100.

Examples

Suppose a hotel has 100 rooms in it. Out of those rooms 50 rooms are vacant and other rooms are booked. So the percentage of vacant room in that hotel is 50/100. That means 50% room of that hotel is vacant.

Similarly if 25 rooms were vacant in that hotel, then we can say that 25% rooms are vacant.

Now if the hotel has 80 rooms in it and out of those 80 rooms 60 rooms are vacant, then what will be the percentage of vacant rooms?

To calculate this we need formula. And formula for calculating percent is given below.

What are the Formulas ?

Suppose we want to find 50% of 100. The formula to find this is,

Percent
—————  X Number = Percentage
100

Therefore, 50% of 100 is,

50
——– X 100 = 50
100

So, 50% of 100 is 50.

Using this formula we can also find out “50 is how much percent of 100?“.

Percent
—————  X Number = Percentage
100

putting values,

Percent
—————  X 100 = 50
100

Therefore, Percent = 50

So, 50 is 50% percent of 100.

And also we can also find out “If 50% of a number is 50 then what is the number?“.

Percent
—————  X Number = Percentage
100

putting values,

50
———  X Number = 50
100

After calculating, Number = 100

If 50% of a number is 50 then the number is 100.

Now back to the hotel room problem. We asked that if a hotel has 80 rooms in it and out of those 80 rooms 60 rooms are vacant. Then what will be the percentage of vacant room.

Formula for this is,

Percent
—————  X Number = Percentage
100

putting values,

Percent
—————  X 80 = 60
100

Therefore, Percent = 75%

So, If 60 rooms out of 80 rooms are vacant in the hotel, it means 75% rooms are vacant in that hotel.

So, we hope you can now do any kind of problem using the above rule. If you need any farther help on this chapter, then please let us know. We will discuss on those problems here in www.AptitudeTricks.com.

Quantitative Aptitude – HCF and LCM

HCF and LCM is one of the most important chapter in Quantitative Aptitude. We all learn HCF and LCM in our school days. These are the basics of math calculations. In competitive exam preparation also we need HCF and LCM very much. So now we will recall what we have learned in our school days.

HCF (Highest Common Factor)

Definition: The largest whole number which is a Factor of two other whole numbers is called HCF.

Let’s explain this,

First of all we need to know What is Factor ?

Factors are Numbers which we multiply to get another number. For example,

2 and 5 are the Factors of 10. Because, 2 X 5 = 10.
Similarly 3 and 4 are Factors of 12 (3 X 4 = 12).

But we can also do this, 2 X 6 = 12. So, 2 and 6 are also factors of 12.
So, 1, 2, 3, 4, 6 and 12 itself are factors of 12.

Now we will find out HCF of two numbers using above rule.

Let’s take 12 and 18 and find out the HCF of these two numbers. So, First of all we need to find out the factors of these two numbers.

Factors of 12 are : 1, 2, 3, 4, 6 and 12.
Factors of 18 are : 1, 2, 3, 6, 9 and 18.

So, common factors between these two numbers are: 1, 2, 3 and 6. And the largest common factor between these two number is 6. So, HCF of 12 and 18 is 6.

LCM (Least Common Multiple)

Difinition: The smallest whole number which is divisible by two other whole number is called LCM.

Let’s explain this,

First of all we need to know What is Multiple of a Number ?

It’s very easy, Multiple of a number is the product of that number with any counting numbers. For example,

Multiples of 5 are : (5 X 1) = 5, (5 X 2) = 10, (5 X 3) = 15, (5 X 4) = 20, (5 X 5) = 25, (5 X 6) = 30, …

Now we will find out LCM of two numbers using above rule.

Let’s take 12 and 18 again and find out the LCM of these two numbers. So, First of all we need to find out the multiples of these two numbers.

Multiples of 12 are : 12, 24, 36, 48, 60, 72, 84, 96, …
Multiples of 18 are : 18, 36, 54, 72, 90, 108, 126, …

So, common multiples between these two numbers are: 36, 72 and many more. But we need only the smallest common multiples between these two numbers. And the smallest common multiple between these two number is 36. So, LCM of 12 and 18 is 36.

If you need any farther help on HCF and LCM, then please let us know. We will discuss on those problems here in www.AptitudeTricks.com.

Quantitative Aptitude – Cube and Cube Root

Cube and Cube root, these are the basics of any mathematical calculations. In quantitative aptitude you will need cube and cube root in many calculations. So you must do this chapter very carefully.

If you remember, you do this thing in your school days. Its not so hard to learn. You just need to know the rules of How to find out Cube and Cube roots. In exam you may get questions directly from cube and cube root chapter, but you will surely get a question which is related with cube and cube root.

Now we will discuss Cube and Cube root in details.

What is Cube ?

A cube is a nothing but multiply a number with the same number by three times. Say, we need to find the cube of number 4. So we have to multiply the number 4 with 4 by three times.

• 4 x 4 x 4 = 64

We denote a cube as X³. A small 3 is written at the upper right corner of that number. Few other cubes are :

• 2 x 2 x 2 =8
• 3 x 3 x 3 =27
• 4 x 4 x 4 =64
• 5 x 5 x 5 =125 …

What is Cube Root ?

Finding Cube root is the opposite of finding Cube. Its the exact reverse process of finding Cube. Cube root is denoted as . Here X is any  number.

For example, Cube root of 64 is 4.  because we know that Cube of 4 is 64.

Students should memorize cubes upto 10. This would help you to do your cube root problems quickly. Sometimes you may get a question where the cube root be in fractional number. Those are also done in the similar way as you do in normal cube root problems.

For example, What is the Cube root of 66. The answer is 4.04124. The smaller but nearest exact cube root of 66 is 64. cube root of 64 is 4. Now the rest 2 will be calculated as same way and you will get an answer of 04124.

If you need any farther help on fractional cube root, then let us know. We will discuss on those problems here in www.AptitudeTricks.com.

Quantitative Aptitude – Decimal Fraction

As we already discussed “What is Fraction“. A Decimal Fraction is one of the type of Fraction. It’s a basic and very common term in mathematics. A Decimal Fraction or popularly known as Decimal Numbers.

When we do math, most of the mathematical calculation is having decimal fraction in it. As we already says that it’s a basic thing in mathematics so you need to learn this chapter very carefully.

Now, let’s dig into the details of this topic.

What is Decimal Fraction ?

Basically, a Decimal Fraction is a combination of two numbers, one is Nominator and other is Denominator. The denominator of a decimal fraction is always a power of ten. And, the nominator represents the number of parts of denominator.

But, we don’t use decimal fraction in this format in our calculations (this rule is not applicable every time). We represent it in a more easier way by using a decimal point.

Few examples of Decimal Fractions:

• Suppose, 5/10 is a Fraction. So we represent it as 0.5 .

• Similarly, 5/100 can be represent as 0.05 .

Basically, the number of zeros in denominator is same as the number of digits in decimal places.

• So, in the first example 5/10, here 10 is a denominator which has only one zero. So, the decimal number must have minimum one digit in decimal places. Nominator of this fraction is 5, so, 5 will be added after the decimal point. So 5/10 is equal to 0.5 .

• And, in the next example, i.e, 5/100. Here denominator has two zeros. So the decimal number must have at least two digits in the decimal places. But, we have only one digit in nominator (5). So, here we have to add one extra digit after decimal point. And, we will add Zero to fill the gap. We add zero before the nominator. So, 5/100 is equal to 0.05.

So, you can add as much zeros as you want before a nominator, unless the number of digit in nominator will equals the number of zeros present in denominator.

Few Examples to Test your learnings:

Example #1

Example #2

If 3x = 0.08y, then find the value of

A.

B.

C.

D.

Correct Answer is: [Option A].

How to Solve:

So, to find the value of the given equation, we need to simplify it.

Firstly, as we know that, 3x = 0.08y.
So,


or,

or,

Now, to find the value of the given equation we will apply the value of
So,


or,

or, putting the value of x/y

so,

or,

finally,

Therefore, the value of the given equation is .

Example #3

Example #4

Example #5

Example #6

Example #7

Example #8

How to Solve:
10/13, 15/19, 19/22, 23/26
So, to find out the difference between the biggest and smallest fraction, we need to convert every vulgar fraction into decimal.

Therefore, by converting every fraction into decimal we get,
10/13 = 0.769
15/19 = 0.789
19/22 = 0.863
23/26 = 0.884

So, now we can clearly say that,
0.769 < 0.789 < 0.863 < 0.884
or, 10/13 < 15/19 < 19/22 < 23/26

Therefore, the biggest fraction here is 23/26. And, the smallest fraction is 10/13.

Now, the difference between these two is,


or,

or,

So, the required answer of the given question is, 3/26.

Example #9

How to Solve:
So, from the question we can form the fraction as:



As we know, 1 hour is equal to 60 minute.
And, 1 minute is equal to 60 seconds.

So, we can arrange the equation as,


or,

or,

So, the decimal fraction of the following question is, .00027.

Example #10

So, if you need any farther help on this chapter, then please let us know. And, surely We will discuss on those problems here in www.AptitudeTricks.com.

Quantitative Aptitude – Average

Average is one of the most important topic in quantitative aptitude. In so many aptitude questions in the exams, you will need to use the formulas of this chapter. So it’s very important to clear all your doubts of this chapter. And those who already know this topic they need to just brush up their knowledge. This will help you in your further chapters.

Here we will discuss all the basics of this topic, and we also show you that how you can use average in your exam. So let’s starts the topic.

What is an Average ?

Definition: An Average is a calculated Central value of a set of Numbers.

Let’s explain this,
Suppose we have a set of numbers, Like,

1, 5, 7, 12, 15, 20

So, if we need to find the central value or mid value of this set of numbers, then we need to add all numbers present in the number set first.

1 + 5 + 7 + 12 + 15 + 20 = 60

So, the sum of all numbers is 60. Now divide 60 with the number of numbers present in the number set, i.e, 6.

Sum of Numbers / Total Number of Numbers = Average

Therefore,
60 / 6 = 10

So, the central value or mid value of 1, 5, 7, 12, 15, 20 is 10.

Arithmetic Mean

In mathematics and statistics an average is called Arithmetic Mean. We all seen average form our school days. After final examination when our teacher showed us our final marks. He or she would get that marks by averaging all the term marks and that marks would be our final marks. Like wise when we calculating our car or bike’s millage we always averaging total run divided by how many liter we put. So, this topic is vary handy in our daily life also.

So, we hope you can now do any kind of problem using the above rule. If you need any farther help on this chapter, then please let us know here in www.AptitudeTricks.com. We will discuss on those problems here later.

Quantitative Aptitude – Square and Square Root

Square and Square root, these are the basics of any mathematical calculations. In quantitative aptitude you will need Square and Square root in many calculations. So you must do this chapter very carefully.

If you remember, you do this thing in your school days. It’s not so hard to learn. You just need to know the rules of How to find out Square and Square roots. In exam you may get questions directly from this chapter, but you will surely get a question which is related with Square and Square root.

Now we will discuss Square and Square root in details.

What is Square ?

A Square is a nothing but multiplication a number with the same number. Say, we need to find the Square of number 4. So we have to multiply the number 4 with 4.

• 4 x 4 = 16

We denote a Square as X². A small 2 is written at the upper right corner of that number. Few other Squares are :

• 2 x 2 = 4
• 3 x 3 = 9
• 4 x 4 = 16
• 5 x 5 = 25 …

What is Square Root ?

Finding Square root is the opposite of finding Square. Its the exact reverse process of finding Square. It is denoted as √9. Here X is any number.

For example, √16 is 4. because we know that Square of 4 is 16.

Students should memorize Squares upto 20. This would help you to do your Square root problems quickly. Sometimes you may get a question where the Square root be in fractional number. Those are also done in the similar way as you do in normal problems.

For example, What is the Square root of 20. The answer is 4.4721. The smaller but nearest exact Square root of 20 is 16. Square root of 16 is 4. Now the rest 4 will be calculated as same way and you will get an answer of 0.4727.

Few Examples to Test your learnings:

Example #1

How to Solve:
So, to find the required number, we need to arrange it into an equation.

Let, the number be X.
Therefore, we can say,


or,

or,

so,

or,

finally,

Therefore, the number is 231.

Example #2

is how much bigger than ?

A.

B.

C.

D.

Correct Answer is: [Option A].

How to Solve:
So, to find the answer, we need to find the difference between these two numbers.

Therefore,


or,

or,

so,

finally,

Therefore, the is bigger than .

Example #3

Find the value of:

A. 10

B. 12

C. 14

D. 15

How to Solve:

So, to find the value of the given expression, we need to simplify it.

Therefore,


or,

or,

so,

or,

or,

finally,

Therefore, the value of the given expression is 10.

Example #4

Find the value of:

A. 15

B. 18

C. 21

D. 23

How to Solve:

So, to find the value of the given expression, we need basic square root techniques.

Therefore, the value would be,


or,

or,

so,

or,

finally,

So, the value of the given expression is, 18.

Example #5

Find the value of:

A. 190/15

B. 195/17

C. 200/19

D. 210/21

How to Solve:

So, to find out the value of this equation, we need to simplify it.

Therefore,


or,

or,

finally,

So, the value of the given equation is 190/15.

Example #6

For what value of X, the statement is true?

A. 50

B. 54

C. 60

D. 62

How to Solve:

So, to find the value of X, we need to simplify the statement.

Now,


or,

or,

so,

or,

finally,

Therefore, the statement will be true if the value of X is 62.

Example #7

If, , then find the value of n.

A. 5

B. 6

C. 7

D. 8

How to Solve:

So, to find the value of n in this statement, we need to simplify the statement.

Therefore,


or, As 5⁴ is equal to 625

or, Squaring both sides

so,

or,

finally,

Therefore, the value on n in the given statement will be 8.

Example #8

Find the square root of:

A. 2

B. 3

C. 4

D. 5

How to Solve:
So, to find the square root of the given equation, we need to arrange it properly.

Therefore,


or, applying the formula of a² - b²

or,

so,

finally,

Therefore, the square root of is 3.

Example #9

If , then find the value of 8 x 32.

A. 42

B. 47

C. 52

D. 56

How to Solve:
So, from the question itself we know that, .

Now, we need to find the value of 8 x 32.
Therefore, putting the value in the expression we get,


or,

or,

so,

finally,

So, the value of 8 x 32 is, 56.

Example #10

If you need any farther help on fractional Square root, then let us know. We will discuss on those problems here in www.AptitudeTricks.com.

Quantitative Aptitude – Fractions

Fractions represents as the part of a whole number. It shows, how many equal parts are there in that whole number. We use two numbers to write a fraction. One line separate these two numbers, like ¹/₄. The upper number or top number represents the number of parts and lower number or bottom number represents the total number of parts.

So, in examination you will get two to three questions from fraction chapter. Therefore, prepare this chapter very well. Now, we will discuss few types of fractions –

Types of Fractions

• Simple Fraction:

  The most simplest form of fraction is Simple Fraction. And, it’s written as ^x/_y , where x and y both are integers. x is called numerator and y is called denominator.

• Decimal Fraction:

  In Decimal Fraction, numbers are written like 0.50, 0.25, 0.70 etc. These numbers are not separated by any separator, but you can convert these fraction as simple fraction. A decimal number 0.50 can be written in decimal fraction as ¹/₂.

• Proper Fraction:

  A fraction is said to be a Proper Fraction if numerator is less than denominator. For example, ¹/₂, ⁵/₇, ¹³/₂₀ etc.

• Improper Fraction:

  A fraction is said to be a Improper Fraction if numerator is bigger than denominator. For example, ⁴/₃, ⁹/₇, ³¹/₂₃ etc.

• Mixed Fraction:

  A mixed fraction is a combination of two numbers. One of them is an integer number (non zero) and another one is a proper fraction. And, we write a mixed fraction as 3¹/₂ . It’s actually the sum of two different numbers. When we write mixed fraction numbers, we do not add ‘+’ operator. So, 3¹/₂ is actually 3 + ¹/₂.

• Complex Fractions:

  Complex Fraction is a combination of Proper Improper fraction or Mixed fraction or both.

Few Examples to Test your learnings:

Example #1

How to Solve:
X + Y = 2
X - Y = 3/5
So, in this example we need to find out the value of each fraction by evaluating the expression.

So, by adding two expression we get,


or,

or,

so,

or,

Similarly, by putting the value of X in expression (i) we get,


or,

or,

so,

or,

So, the value of X is 13/10 and value of Y is 7/10.

Example #2

How to Solve:
So, in this example we need to find out the fraction value of 35%.

And, this is so simple to find out. For instance, we need to write 35% as fraction.
Like,
35%

so,

or, 5 is the HCF of 35 and 100

or,

So, the fraction value of 35% is 7/20.

Example #3

Example #4

Example #5

How to Solve:
So, in this example we need to find out the fraction multiple value of 13/37.

So, for this we need to divide each option with the question fraction. And, the option which gave us the result 1, that would be our choice.
Like,


or,

or, This is not the correct option

Then,


or,

or, This is also not the correct option

Then,


so,

or,

or, This is the correct option

So, the fraction equivalent to 13/37 is 91/259.

Example #6

Evaluate the expression:

A. 1.34

B. 1.83

C. 3.17

D. 5

How to Solve:

So, in this example we need to evaluate the expression and find the value.

And, here we need to apply some basic formula to get the value. You can also use the conventional way of evaluating the value. But, it will be very quick if you use basic math formula.

Let, is a and is b.
So,


or,

or,

so,

or,

or,

so,

or, 5

Example #7

In a math class, teacher asked every students to multiply with a fraction. But instead of multiplying Jay divided the fraction with . As a result he gets more than the expected result. Find out the fraction value to whom teacher asked to multiply .

A. 1/2

B. 1/4

C. 1/8

D. 1

How to Solve:
So, in this example we need to form the equation and find out the value of that equation.

Let, the fraction be X.
So, according to the question we get,


so,

or,

or,

So, the fraction value here is 1/2 to whom teacher asked to multiply .

Example #8

How to Solve:
So, in this example we need to form the equation and find out the value of that equation.

Let, the integer number be X.
So, according to the question we get,


or,

or,

so,

or,

or,

So, the integer number is 60.

Example #9

Between and which equation is greater?

A. First Equation

B. Second Equation

C. Both are Equal

D. Can't Determine

How to Solve:
In this example we need to find out value of both the equations.

So, from the first equation we get,


or,

or,

And, from the first equation we get,


or,

or,

So, no one is greater. Both equations are equal.

Example #10

Evaluate the expression.

A. 49/113

B. 53/136

C. 68/157

D. 77/169

How to Solve:

In this example we need to simplify the expression to get the result.

So, by evaluating the expression we get,


or,

or,

so,

or,

or,

so,

or,

or,

so,

Therefore, the answer is, 68/157.

So, if you need any farther help on Fractions, then please let us know. And, definitely we will discuss on those problems here in www.AptitudeTricks.com.

Quantitative Aptitude – Simplification

Simplification is the process of finding the most simplest form of a given expression. In competitive exam simplification is another important topic. At least two or three questions are from simplification. So, now we will discuss the details of simplification here in this page.

An equation may consist few numbers and few arithmetic operators in it. And, we can simplify the equation on the basis of priority of the operators. So, we need to know which operators has most priority and which has less. And, here comes a rule known as BODMAS rule.

As we know that BODMAS rule tells us about the priority of the operators. So, what does BODMAS stands for-

• B = Bracket
• O = Of
• D = Division
• M = Multiplication
• A = Addition
• S = Subtraction

This means, first of all we need to remove brackets from the expression, if any. Brackets should be removed in order of, (), {} and then []. Then we do “Of” operations and so on in the following order.

And, one more important thing is Virnaculum or Bar. If an expression has virnaculam then apply the virnaculam first on the expression. BODMAS rule will be applied on the expression after virnaculam.

Few examples of Simplification

Simplification: Example 1

Simplification of: 16200 / 270 / 30

Answer: If your answer is 1800 then you are wrong. If your answer is 2 then your are right. And, do division from left to right. So, first of all (16200 / 270) = 60. Then divide 60 with 30 and the result will be 2.

Simplification: Example 2

Simplification of: 2540 + (340 / 17)

Answer: According to BODMAS rule, first of all we need to simplify the expression which are within brackets. So, 340 / 17 = 20. Now, add this result with the remaining portion of the expression. 2540 + 20 = 2560 will be the answer.

Simplification: Example 3

A man has some hens and cows. Now, total number of heads be 100 and total number of feet equals 316. So, find the number of hens and cows that man have.

Answer: Let the number of hens be x and the number of cows be y.
x + y = 100 …….. (i)
As hens has two legs and cows has four legs, so, 2x + 4y = 316 or x + 2y = 158 …….. (ii)
Solving (i) and (ii) we get:

x + 2y = 158
x + y = 100
(-)
———————-
y = 58

So, x = (100 – y) = 42
Therefore, the number of hens be 42 and number of cows be 58.

Few Examples to Test your learnings:

Example #1

Example #2

Example #3

Example #4

Example #5

Example #6

Example #7

Example #8

Example #9

Example #10

So, if you need any farther help on Simplification, then let us know. And, We will discuss on those problems here in www.AptitudeTricks.com. Feel free to ask your questions.

Quantitative Aptitude – Calendar

Calendar Aptitude is a very important chapter for any competitive exams. Most of the competitive exams like SSC, PSC, Bank has this topic in their syllabus. So it’s very important to learn this topic carefully.

Few questions from Calendar chapter will come in exams. To solve those questions, you need to learn the basic formulas of Calendar aptitude. Now here we will discuss those formulas. Learn these formulas and practice.

What is Calendar ?

Calendar is a series of pages which shows days, weeks, months and specific year. A calendar is also showing a seasonal information of a year.

Basic Ideas of Calendar

• Ordinary Years: Years which has 365 days and is not a leap year is called an ordinary year.

• Leap Year: A leap year has 366 days. In a leap year the month of February has 29 days. Except centuries, years which is divisible by 4 is called a leap year. And centuries which is divisible by 400 is a leap year.
  

• Odd Days: 7 days makes a week. So in a time period of few days the left over days more than the complete weeks are called odd days. Suppose, we need to find the number of odd days in 40 days. So after dividing 40 by 7, we get 5 complete weeks and 5 left over days. These 5 days are called odd days.

• Counting of Odd Days:

1. One Ordinary Year = 365 Days. This means, it has 52 complete weeks and 1 odd day. So, One ordinary year has 1 odd day.

2. One Leap Year = 366 Days. This means, it has 52 complete weeks and 2 odd days. So, a leap year has 2 odd days.

3. In 100 years the odd days will be,

100 years = 76 ordinary years + 24 leap years.

∴ (76 x 1 + 24 x 2) = 124 odd days.

∴ 17 complete weeks and 5 odd days.

So, Number of odd days in 100 years is 5.

4. In 200 years the odd days will be,

(5 x 2) = 10 odd days.

∴ 1 complete week and 3 odd days.

So, Number of odd days in 200 years is 3.

5. In 300 years the odd days will be,

(5 x 3) = 15 odd days = 1 Complete week and 1 odd day.

6. In 400 years the odd days will be,

(5 x 4 + 1) = 21 odd days = 3 Complete week and 0 odd day.

So, this is basically the idea of aptitude calendar chapter. Now if you have any query regarding this topic then please feel free to ask us here. Write your query in the comment section down below. Our team of www.AptitudeTricks.com will answer your question and try to clear your doubts.

Quantitative Aptitude – Boats and Streams

“Boats and Streams” is very important topic in competitive exams. Banking aptitude exams, Rail exams, PSC’s, SSC’s and almost all competitive exams have aptitude paper and all of them must have at least one question from boats and streams chapter. So this is very important chapter to learn for the exam.

So, now we will discuss the basic concepts and formulas of Boats and Streams.

What is Boats and Streams ?

Boats and Streams is a very important chapter of Quantitative Aptitude. Questions of this topic is related to find out the speed of boats in the streams of water.

Basically stream is the force of water. Water flows from up to down. And here comes two main thing of this chapter, one is Downstream and another is Upstream.

What is Downstream ?

• The direction, which way the water flows is called Downstream.

What is Upstream ?

• The opposite direction of flowing water is called Upstream.

Basic formulas of Boats and Streams

Suppose, the speed of a stream of a flowing river is x km/hr and the speed of a boat in still water is y km/hr, then

• Downstream Speed = (x + y) km/hr.

So, If the boat is floating in the water with the stream then the speed of the boat will be – Sum of downstream speed and boat speed.

• Upstream Speed = (y – x) km/hr.

And, If the boat is floating in the water against the stream then the speed of the boat will be – Upstream speed minus the boat speed.

Suppose, the speed of downstream is x km/hr and the speed upstream is y km/hr, then

• Speed in still water is =     1/2 (x + y) km/hr.

• Rate of stream =     1/2 (x – y) km/hr.

So, this is all basic concepts of Boats and Streams chapter. Learn this formulas and concept very carefully. If you have face any problem in this chapter of you have anything to say regarding this topic, then feel free to ask us. We will answer your query here in www.AptitudeTricks.com.