# 7.1: Partial Fractions - Mathematics

7.1: Partial Fractions - Mathematics

Integrate the following function with respect to x :

Decompose the given rational function into partial fractions.

1/(x - 1) (x + 2) 2   =  A/(x - 1) + B/(x + 2) + C/(x + 2) 2

=  (1/9) ∫(1 /(x-1))dx-(1/9) (1/(x+2)-(1/3) (1/(x+2) 2 dx

=  (1/9) [log(x - 1) - log(x + 2)] - (1/3)(1/(x+2)) + c

=  (1/9) [log(x - 1)/(x - 2)] - (1/3(x + 2)) + c

Integrate the following function with respect to x :

Decompose the given rational function into partial fractions.

3x - 9  =  A(x+2) (x 2 +1) +B(x-1) (x 2 +1)+(Cx+D)(x-1)(x+2)

3x - 9  =  A(x+2) (x 2 +1) +B(x-1) (x 2 +1)+(Cx+D)(x-1)(x+2)

∫[ (3x - 9)/(x - 1)(x + 2)(x 2  + 1)] dx

=   - ∫1 /(x - 1) dx +  1/(x + 2) dx +  ∫ 3 / (x 2  + 1) dx

=  - log (x - 1) + log (x + 2) + 3 tan -1 (x)

Integrate the following function with respect to x :

In the given rational fraction, the highest exponent of x in numerator is greater than the highest exponent of x in denominator.

So, we can use long division to decompose the given rational function. From the above long division, we have

x 3 /(x - 1)(x - 2)  =  (x + 3) + (7x - 6)/(x 2 - 3x + 2)

x 3 /(x - 1)(x - 2)  =  (x + 3) + (7x - 6)/(x - 1)(x - 2) ----(1)

Decompose (7x - 6)/(x - 1)(x - 2) into partial fractions.

(7x - 6)/(x - 1)(x - 2)  =  A/(x - 1) + B(x - 2)

=   (x + 3) dx - ∫1 /(x - 1) dx + 8 ∫1 /(x - 2) dx

=  x 2 /2 + 3x -  log(x - 1) + 8log(x -  2) + C After having gone through the stuff given above, we hope that the students would have understood, how to integrate rational functions u sing partial fractions.

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## 7.4 Partial Fractions

Introduction: In this lesson we will learn to use partial fraction decomposition to integrate rational functions. The technique called partial fraction decomposition shows us how to break a rational function down into a sum of simpler rational functions. These simpler rational functions can then be integrated routinely.

Objectives: After this lesson you should be able to:

• Understand the concept of partial fraction decomposition.
• Use partial fraction decomposition with linear and quadratic factors (some of which may be repeated) to integration rational functions.

Video & Notes: Fill out the note sheet for this lesson (7-4-Partial-Fractions) as you watch the video. If you prefer, you could read Section 7.4 of your textbook and work out the problems on the notes on your own as practice. Remember, notes must be uploaded to Blackboard weekly for a grade! If for some reason the video below does not load you can access it on YouTube here.

Homework: Go to WebAssign and complete the 𔄟.4 Partial Fractions” assignment.

## Fraction Calculator

The values can be simple fractions, mixed fractions or improper fractions.

### Overview of Fractions:

A fraction names part of a region or part of a group. A fraction is the number of shaded parts divided by the number of equal parts. The numerator is the number above the fraction bar, and the denominator is the number below the fraction bar.

A proper fraction is a fraction in which the numerator is less than the denominator. An improper fraction is a fraction in which the numerator is greater than or equal to the denominator. A number can be classified as a proper fraction, an improper fraction, or as a mixed number. Any number divided by itself is equal to one. A mixed number consists of a whole-number part and a fractional part.

Equivalent fractions are different fractions that name the same number. Equivalent fractions are different fractions that name the same number. The numerator and the denominator of a fraction must be multiplied by the same nonzero whole number in order to have equivalent fractions.

To simplify a fraction (reduce it to lowest terms), the numerator and the denominator must be divided by the same nonzero whole number. A fraction is in lowest terms when the greatest common factor (GCF) of its numerator and denominator is one.

When comparing two fractions with like denominators, the larger fraction is the one with the greater numerator. To compare fractions with unlike denominators, use the LCD to write equivalent fractions with a common denominator then compare the numerators.

You can convert an improper fraction greater than one to a mixed number through long division of its numerator and denominator. Comparison of numerator and denominator: If the numerator < denominator, then the fraction < 1.

To order fractions with like denominators, look at the numerators and compare them two at a time. To order fractions with unlike denominators, use the LCD to write them as equivalent fractions with like denominators. Then compare two fractions at a time. It is helpful to write a number in a circle next to each fraction to compare them more easily.

### Introduction to Fractions

Introduction, Classify Fractions, Equivalent Fractions, Simplify, Compare and Order. Convert Fractions to Mixed Numbers. Convert Mixed Numbers to Fractions. Math instruction is visual and conceptual.

### Add and Subtract Fractions and Mixed Numbers

Add and subtract fractions with like and unlike denominators, LCD, add and subtract mixed numbers, solve real-world problems. These lessons use both visual and conceptual approaches.

### Multiply and Divide Fractions and Mixed Numbers

Multiply fractions with and without cancelling, multiply mixed numbers, reciprocals, divide fractions, divide mixed numbers, solving real-world problems. Instruction is visual and conceptual.

## Questions in Exercise 7.1

(i) frac <1>

(ii) frac <1>

(iii) frac <1>

(iv) frac <3>

(v) frac <4>

Q1) Write the fraction representing the shaded portion.

(i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) ## Introduction to the Laplace Transform

### 8.3 Solving Initial-Value Problems with the Laplace Transform

In this section, we show how the Laplace transform is used to solve initial-value problems. To do this, we first need to understand how the Laplace transform of the derivatives of a function relates to the function itself. We begin with the first derivative.

Theorem 45 (Laplace Transform of the First Derivative). Suppose that f(t) is continuous for all t ≥ 0 and is of exponential order b for t > T. Also, suppose that f'(t) is piecewise continuous on any closed subinterval of [0,∞). Then, for s > b

Proof. Using integration by parts with u = e − st and dv = f'(t) dt, we have

Proof of Theorem 45 assumes that f' is a continuous function. If we use the assumption that f' is continuous on 0 < t1 < t2 < ⋯tn < ∞, we complete the proof by using

This is the same integration by parts formula shown in the proof of Theorem 45 for each integral. Now we make the same assumptions of f' and f" as we did of f and f', respectively, in the statement of Theorem 45 and use Theorem 45 to develop an expression for ℒ<f"(t)>:

Continuing this process, we can construct similar expressions for the Laplace transform of higher order derivatives, which leads to the following theorem.

Theorem 46 (Laplace Transform of Higher Derivatives). More generally, if f (i) (t) is a continuous function of exponential order b on [0, ∞) for i = 0,1,…,n − 1 and f (n) (t) is piecewise continuous on any closed subinterval of [0,∞), then for s > b

We will use this theorem and corollary in solving initial-value problems. However, we can also use them to find the Laplace transform of a function when we know the Laplace transform of the derivative of the function.

#### Solution

We now show how the Laplace transform can be used to solve initial-value problems. Typically, when we solve an initial-value problem that involves y(t), we use the following steps:

compute the Laplace transform of each term in the differential equation

solve the resulting equation for ℒ<y(t)> = Υ(s) and

determine y(t) by computing the inverse Laplace transform of Υ(s).

The advantage of this method is that through the use of the property

we change the differential equation to an algebraic equation that can be solved for ℒ<f(t)>.

Solve the initial-value problem y' − 4y = e 4t , y(0) = 0.

How does the solution change if y(0) = 1?

#### Solution

We begin by taking the Laplace transform of both sides of the differential equation. Because ℒ<y'> = sΥ(s) − y(0) = sΥ(s), we have

In many cases, we must determine a partial fraction decomposition of Υ( s) to obtain terms for which the inverse Laplace transform can be found.

#### Solution

Let Υ(s) = ℒ<y(t)>. Then, applying the Laplace transform to the equation gives us ℒ<y" − 4y'> = ℒ<0>. Because

A partial fraction decomposition involving a repeated linear factor is illustrated in the following example.

#### Solution

Use Laplace transforms to solve y' − y = 0.

In some cases, F(s) involves irreducible quadratic factors as we see in the next example.

#### Solution

Let ℒ<y(t)> = Υ(s). Taking the Laplace transform of the equation and solving for Υ(s) gives us,

## Partial fractions

Every fractional number , i. e. such a rational number m n that the integer m is not divisible by the integer n , can be decomposed to a sum of partial fractions as follows:

 m n = m 1 p 1 ν 1 + m 2 p 2 ν 2 + ⋯ + m t p t ν t

Here, the p i ’s are distinct positive prime numbers , the ν i ’s positive integers and the m i ’s some integers. Cf. the partial fractions of expressions.

 6 289 = 6 17 2
 - 1 24 = - 3 2 3 + 1 3 1
 1 504 = - 1 2 3 + 32 3 2 - 24 7 1

How to get the numerators m i for decomposing a fractional number 1 n to partial fractions? First one can take the highest power p ν of a prime p which divides the denominator n . Then n = p ν ⁢ u , where gcd ⁡ ( u , p ν ) = 1 . Euclid’s algorithm gives some integers x and y such that

Dividing this equation by p ν ⁢ u gives the

 1 n = 1 p ν ⁢ u = x p ν + y u .

If u has more than one distinct prime factors, a similar procedure can be made for the fraction y u , and so on.

Note. The numerators m 1 , m 2 , …, m t in the decomposition are not unique. E. g., we have also

## Contents

If the starting number is rational, then this process exactly parallels the Euclidean algorithm applied to the numerator and denominator of the number. In particular, it must terminate and produce a finite continued fraction representation of the number. The sequence of integers that occur in this representation is the sequence of successive quotients computed by the Euclidean algorithm. If the starting number is irrational, then the process continues indefinitely. This produces a sequence of approximations, all of which are rational numbers, and these converge to the starting number as a limit. This is the (infinite) continued fraction representation of the number. Examples of continued fraction representations of irrational numbers are:

• √ 19 = [42,1,3,1,2,8,2,1,3,1,2,8. ] (sequence A010124 in the OEIS). The pattern repeats indefinitely with a period of 6.
• e = [21,2,1,1,4,1,1,6,1,1,8. ] (sequence A003417 in the OEIS). The pattern repeats indefinitely with a period of 3 except that 2 is added to one of the terms in each cycle.
• π = [37,15,1,292,1,1,1,2,1,3,1. ] (sequence A001203 in the OEIS). No pattern has ever been found in this representation.
• ϕ = [11,1,1,1,1,1,1,1,1,1,1. ] (sequence A000012 in the OEIS). The golden ratio, the irrational number that is the "most difficult" to approximate rationally. See: A property of the golden ratio φ.

Continued fractions are, in some ways, more "mathematically natural" representations of a real number than other representations such as decimal representations, and they have several desirable properties:

• The continued fraction representation for a rational number is finite and only rational numbers have finite representations. In contrast, the decimal representation of a rational number may be finite, for example
• 137 / 1600 = 0.085625 , or infinite with a repeating cycle, for example
• 4 / 27 = 0.148148148148.
• Every rational number has an essentially unique continued fraction representation. Each rational can be represented in exactly two ways, since [a0a1. an−1,an] = [a0a1. an−1,(an−1),1] . Usually the first, shorter one is chosen as the canonical representation.
• The continued fraction representation of an irrational number is unique.
• The real numbers whose continued fraction eventually repeats are precisely the quadratic irrationals.  For example, the repeating continued fraction [11,1,1. ] is the golden ratio, and the repeating continued fraction [12,2,2. ] is the square root of 2. In contrast, the decimal representations of quadratic irrationals are apparently random. The square roots of all (positive) integers, that are not perfect squares, are quadratic irrationals, hence are unique periodic continued fractions.
• The successive approximations generated in finding the continued fraction representation of a number, that is, by truncating the continued fraction representation, are in a certain sense (described below) the "best possible".

A continued fraction is an expression of the form

where ai and bi can be any complex numbers. Usually they are required to be integers. If bi = 1 for all i the expression is called a simple continued fraction. If the expression contains finitely many terms, it is called a finite continued fraction. If the expression contains infinitely many terms, it is called an infinite continued fraction. 

Thus, all of the following illustrate valid finite simple continued fractions:

Examples of finite simple continued fractions
Formula Numeric Remarks
a 0 > 2 All integers are a degenerate case
a 0 + 1 a 1 +>>> 2 + 1 3 <3>>> Simplest possible fractional form
a 0 + 1 a 1 + 1 a 2 ++>>>>> − 3 + 1 2 + 1 18 <2+<18>>>>> First integer may be negative
a 0 + 1 a 1 + 1 a 2 + 1 a 3 +++>>>>>>> 1 15 + 1 1 + 1 102 <15+<1+<102>>>>>>> First integer may be zero

For simple continued fractions of the form

To calculate a continued fraction representation of a number r , write down the integer part (technically the floor) of r . Subtract this integer part from r . If the difference is 0, stop otherwise find the reciprocal of the difference and repeat. The procedure will halt if and only if r is rational. This process can be efficiently implemented using the Euclidean algorithm when the number is rational. The table below shows an implementation of this procedure for the number 3.245, resulting in the continued fraction expansion [3 4,12,4].

or in the notation of Pringsheim as

or in another related notation as

Sometimes angle brackets are used, like this:

The semicolon in the square and angle bracket notations is sometimes replaced by a comma.  

One may also define infinite simple continued fractions as limits:

Every finite continued fraction represents a rational number, and every rational number can be represented in precisely two different ways as a finite continued fraction, with the conditions that the first coefficient is an integer and the other coefficients are positive integers. These two representations agree except in their final terms. In the longer representation the final term in the continued fraction is 1 the shorter representation drops the final 1, but increases the new final term by 1. The final element in the short representation is therefore always greater than 1, if present. In symbols:

The continued fraction representations of a positive rational number and its reciprocal are identical except for a shift one place left or right depending on whether the number is less than or greater than one respectively. In other words, the numbers represented by [ a 0 a 1 , a 2 , … , a n ] a_<1>,a_<2>,ldots ,a_]> and [ 0 a 0 , a 1 , … , a n ] ,a_<1>,ldots ,a_]> are reciprocals.

The last number that generates the remainder of the continued fraction is the same for both x and its reciprocal.

Every infinite continued fraction is irrational, and every irrational number can be represented in precisely one way as an infinite continued fraction.

An infinite continued fraction representation for an irrational number is useful because its initial segments provide rational approximations to the number. These rational numbers are called the convergents of the continued fraction.   The larger a term is in the continued fraction, the closer the corresponding convergent is to the irrational number being approximated. Numbers like π have occasional large terms in their continued fraction, which makes them easy to approximate with rational numbers. Other numbers like e have only small terms early in their continued fraction, which makes them more difficult to approximate rationally. The golden ratio ϕ has terms equal to 1 everywhere—the smallest values possible—which makes ϕ the most difficult number to approximate rationally. In this sense, therefore, it is the "most irrational" of all irrational numbers. Even-numbered convergents are smaller than the original number, while odd-numbered ones are larger.

For a continued fraction [a0 a1, a2, . ] , the first four convergents (numbered 0 through 3) are

The numerator of the third convergent is formed by multiplying the numerator of the second convergent by the third coefficient, and adding the numerator of the first convergent. The denominators are formed similarly. Therefore, each convergent can be expressed explicitly in terms of the continued fraction as the ratio of certain multivariate polynomials called continuants.

If successive convergents are found, with numerators h 1, h 2, . and denominators k 1, k 2, . then the relevant recursive relation is:

hn = anhn − 1 + hn − 2 , kn = ankn − 1 + kn − 2 .

The successive convergents are given by the formula

Thus to incorporate a new term into a rational approximation, only the two previous convergents are necessary. The initial "convergents" (required for the first two terms) are 0 ⁄1 and 1 ⁄0. For example, here are the convergents for [01,5,2,2].

 n an hn kn −2 −1 0 1 2 3 4 0 1 5 2 2 0 1 0 1 5 11 27 1 0 1 1 6 13 32

When using the Babylonian method to generate successive approximations to the square root of an integer, if one starts with the lowest integer as first approximant, the rationals generated all appear in the list of convergents for the continued fraction. Specifically, the approximants will appear on the convergents list in positions 0, 1, 3, 7, 15, . , 2 k −1 , . For example, the continued fraction expansion for √ 3 is [11,2,1,2,1,2,1,2. ]. Comparing the convergents with the approximants derived from the Babylonian method:

### Properties Edit

A Baire space is a topological space on infinite sequences of natural numbers. The infinite continued fraction provides a homeomorphism from the Baire space to the space of irrational real numbers (with the subspace topology inherited from the usual topology on the reals). The infinite continued fraction also provides a map between the quadratic irrationals and the dyadic rationals, and from other irrationals to the set of infinite strings of binary numbers (i.e. the Cantor set) this map is called the Minkowski question mark function. The mapping has interesting self-similar fractal properties these are given by the modular group, which is the subgroup of Möbius transformations having integer values in the transform. Roughly speaking, continued fraction convergents can be taken to be Möbius transformations acting on the (hyperbolic) upper half-plane this is what leads to the fractal self-symmetry.

The limit probability distribution of the coefficients in the continued fraction expansion of a random variable uniformly distributed in (0, 1) is the Gauss–Kuzmin distribution.

### Some useful theorems Edit

Corollary 2: The difference between successive convergents is a fraction whose numerator is unity:

Corollary 3: The continued fraction is equivalent to a series of alternating terms:

Corollary 4: The matrix

Corollary 1: A convergent is nearer to the limit of the continued fraction than any fraction whose denominator is less than that of the convergent.

Corollary 2: A convergent obtained by terminating the continued fraction just before a large term is a close approximation to the limit of the continued fraction.

are consecutive convergents, then any fractions of the form

where m is an integer such that 0 ≤ m ≤ a n + 1 > , are called semiconvergents, secondary convergents, or intermediate fractions. The ( m + 1 ) -st semiconvergent equals the mediant of the m -th one and the convergent h n k n ><>>>> . Sometimes the term is taken to mean that being a semiconvergent excludes the possibility of being a convergent (i.e., 0 < m < a n + 1 > ), rather than that a convergent is a kind of semiconvergent.

It follows that semiconvergents represent a monotonic sequence of fractions between the convergents h n − 1 k n − 1 ><>>>> (corresponding to m = 0 ) and h n + 1 k n + 1 ><>>>> (corresponding to m = a n + 1 > ). The consecutive semiconvergents a b >> and c d >> satisfy the property a d − b c = ± 1 .

1. Truncate the continued fraction, and reduce its last term by a chosen amount (possibly zero).
2. The reduced term cannot have less than half its original value.
3. If the final term is even, half its value is admissible only if the corresponding semiconvergent is better than the previous convergent. (See below.)

For example, 0.84375 has continued fraction [01,5,2,2]. Here are all of its best rational approximations.

The strictly monotonic increase in the denominators as additional terms are included permits an algorithm to impose a limit, either on size of denominator or closeness of approximation.

The "half rule" mentioned above requires that when a k is even, the halved term a k /2 is admissible if and only if |x − [a0 a1, . ak − 1]| > |x − [a0 a1, . ak − 1, ak/2]|  This is equivalent  to: 

The convergents to x are "best approximations" in a much stronger sense than the one defined above. Namely, n / d is a convergent for x if and only if |dxn| has the smallest value among the analogous expressions for all rational approximations m / c with cd that is, we have |dxn| < |cxm| so long as c < d . (Note also that |dkxnk| → 0 as k → ∞ .)

### Best rational within an interval Edit

A rational that falls within the interval (x, y) , for 0 < x < y , can be found with the continued fractions for x and y . When both x and y are irrational and

x = [a0 a1, a2, . ak − 1, ak, ak + 1, . ] y = [a0 a1, a2, . ak − 1, bk, bk + 1, . ]

where x and y have identical continued fraction expansions up through ak−1 , a rational that falls within the interval (x, y) is given by the finite continued fraction,

z(x,y) = [a0 a1, a2, . ak − 1, min(ak, bk) + 1]

This rational will be best in the sense that no other rational in (x, y) will have a smaller numerator or a smaller denominator. [ citation needed ]

If x is rational, it will have two continued fraction representations that are finite, x1 and x2 , and similarly a rational y will have two representations, y1 and y2 . The coefficients beyond the last in any of these representations should be interpreted as +∞ and the best rational will be one of z(x1, y1) , z(x1, y2) , z(x2, y1) , or z(x2, y2) .

For example, the decimal representation 3.1416 could be rounded from any number in the interval [3.14155, 3.14165) . The continued fraction representations of 3.14155 and 3.14165 are

3.14155 = [3 7, 15, 2, 7, 1, 4, 1, 1] = [3 7, 15, 2, 7, 1, 4, 2] 3.14165 = [3 7, 16, 1, 3, 4, 2, 3, 1] = [3 7, 16, 1, 3, 4, 2, 4]

and the best rational between these two is

### Interval for a convergent Edit

A rational number, which can be expressed as finite continued fraction in two ways,

z = [a0 a1, . ak − 1, ak, 1] = [a0 a1, . ak − 1, ak + 1]

will be one of the convergents for the continued fraction expansion of a number, if and only if the number is strictly between

x = [a0 a1, . ak − 1, ak, 2] and y = [a0 a1, . ak − 1, ak + 2]

The numbers x and y are formed by incrementing the last coefficient in the two representations for z . It is the case that x < y when k is even, and x > y when k is odd.

Consider x = [a0 a1, . ] and y = [b0 b1, . ] . If k is the smallest index for which ak is unequal to bk then x < y if (−1) k (akbk) < 0 and y < x otherwise.

If there is no such k , but one expansion is shorter than the other, say x = [a0 a1, . an] and y = [b0 b1, . bn, bn + 1, . ] with ai = bi for 0 ≤ in , then x < y if n is even and y < x if n is odd.

[37,15,1,292,1,1. ] (sequence A001203 in the OEIS).

Let us suppose that the quotients found are, as above, [37,15,1]. The following is a rule by which we can write down at once the convergent fractions which result from these quotients without developing the continued fraction.

## CBSE NCERT Solutions for Class 6 Maths Chapter 7 Fractions

All the Class 6 NCERT Solutions provided on this page are solved by experts of Embibe based on the CBSE guidelines. With the help of these NCERT Solutions for Class 6 Maths, students can solve their assignments and home works easily on time.

Before getting into the details of Class 6 NCERT Solutions for Maths Chapter 7, let’s have a look at the sections along with downloadable PDF links:

 Exercise Topics 7.1 Introduction 7.2 A Fraction 7.3 Fraction on the Number Line 7.4 Proper Fractions 7.5 Improper and Mixed Fractions 7.6 Equivalent Fractions 7.7 Simplest Form of a Fraction 7.8 Like Fractions 7.9 Comparing Fractions 7.10 Addition and Subtraction of Fractions

### NCERT Solutions for Class 6 Maths Chapter 7: Fractions PDF Download

Here we have provided the solutions for chapter 7 of NCERT Books for Class 6 Maths.

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Chapter – 1: Knowing Our Numbers
• Chapter – 2: Whole Numbers
• Chapter – 3: Playing with Numbers
• Chapter – 4: Basic Geometrical Ideas
• Chapter – 5: Understanding Elementary Shapes
• Chapter – 6: Integers
• Chapter – 8: Decimals
• Chapter – 9: Data Handling
• Chapter – 10: Mensuration
• Chapter – 11: Algebra
• Chapter – 12: Ratio and Proportion
• Chapter – 13: Symmetry
• Chapter – 14: Practical Geometry

### CBSE Class 6 Maths Chapter 7: Fractions – Chapter Summary

A fraction is a number representing a part of a whole. Every fraction has a point associated with it on the number line. Students will come across the different types of fractions in this chapter – Proper Fractions, Improper Fractions, Mixed Fractions, Like Fractions and Unlike Fractions.

A fraction is the simplest form if its numerator and denominator have no common factor except 1. In this chapter, students will be learning about how to compare fractions. It is easy to compare fractions, but the comparison of the unlike fractions requires special attention where common denominators and LCM are required.

 CBSE Class 6 Maths Syllabus CBSE Class 6 Syllabus (All Subjects) NCERT Books for Class 6 Maths NCERT Books For Class 6

### FAQs Related to CBSE Class 6 Maths Chapter 7 Solutions

Here we have provided some of the most frequently asked questions from CBSE Class 6 Maths Chapter 7:

A: There are 24 hours in a day
We have 8 hours
Hence, the required fraction is 8/24 or 1/3

A: There are 60 minutes in 1 hour
∴ 1 hour = 60 minutes
Hence, the required fraction is 40/60 or 2/3

A: Arya has divided the sandwich into 3 equal parts. So each person will get one part.
(b) Each boy receive 1/3 part
∴ Required Fraction is 1/3

A: Total number of dresses Kanchan has to dye = 30 dresses
Number of dresses she has finished = 20 dresses
∴ Required Fraction = 20/30 or 2/3

A: Natural numbers from 102 to 113 are
102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113
Total number of natural numbers given = 12
Number of prime numbers = 4 [103, 107, 109, 113]
∴ Required Fraction = 4/12 or 1/3

A. Fraction on the Number Line is the exercise that is properly explained in NCERT Solutions for Class 6 Maths Chapter 7 Fractions Exercise 7.3.

A. Proper Fractions is the exercise that is properly explained in NCERT Solutions for Class 6 Maths Chapter 7 Fractions Exercise 7.4.

Now you are provided with the detailed CBSE NCERT Solutions For Class 6 Maths Chapter 7. We hope this detailed article helps you in your preparation.

If you have any doubt regarding NCERT Solutions, you can drop your comments below. We will get back to you at the earliest.