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Page No 82:

(i) The shaded portion is 3 parts of the whole figure
$\therefore$ $\frac{3}{4}$
(ii) The shaded portion is 1 parts of the whole figure
$\therefore$ $\frac{1}{4}$
(iii) The shaded portion is 2 parts of the whole figure.
$\therefore$ $\frac{2}{3}$
(iv) The shaded portion is 3 parts of the whole figure.
$\therefore$$\frac{3}{10}$
(v)The shaded portion is 4 parts of the whole figure.
$\therefore$$\frac{4}{9}$
(vi) The shaded portion is 3 parts of the whole figure.
$\therefore$ $\frac{3}{8}$

Page No 82: Page No 82:

The given rectangle is not divided into four equal parts.

Thus, the shaded region is not equal to $\frac{1}{4}$ of the whole.

Page No 82:

(i) $\frac{3}{4}$        (ii) $\frac{4}{7}$             (iii) $\frac{2}{5}$           (iv) $\frac{3}{10}$           (v) $\frac{1}{8}$
(vi) $\frac{5}{6}$             (vii)$\frac{8}{9}$              (viii) $\frac{7}{12}$

Page No 83:

Numerator        Denominator
(i) 4                         9
(ii) 6                       11
(iii) 8                      15
(iv) 12                     17
(v) 5                        1

Page No 83:

(i)$\frac{3}{8}$        (ii) $\frac{5}{12}$          (iii)$\frac{7}{16}$             (iv) $\frac{8}{15}$

Page No 83:

(i) two-thirds
(ii) four$-$ninths
(iii) two$-$fifths
(iv) seven$-$tenths
(v) one$-$thirds
(vi) three$-$fourths
(vii) three$-$eighths
(viii) nine$-$fourteenths
(ix) five$-$elevenths
(x) six$-$fifteenths

Page No 83:

We know: 1 hour = 60 minutes
∴ The required fraction = $\frac{24}{60}=\frac{2}{5}$

Page No 83:

There are total 9 natural numbers from 2 to 10. They are 2, 3, 4, 5, 6, 7, 8, 9, 10.
Out of these natural numbers, 2, 3, 5, 7 are the prime numbers.
∴ The required fraction = $\frac{4}{9}$.

Page No 83:

(i) $\frac{2}{3}$ of 15 pens =
(ii) $\frac{2}{3}$ of 27 balls =
(iii) $\frac{2}{3}$ of 36 balloons = ​

Page No 83:

(i) $\frac{3}{4}$ of 16 cups =
(ii) $\frac{3}{4}$ of 28 rackets =
(iii) $\frac{3}{4}$ of 32 books =

Page No 83:

Neelam gives $\frac{4}{5}$ of 25 pencils to Meena.

Thus, Meena gets 20 pencils.
∴ Number of pencils left with Neelam = 25 $-$ 20 = 5 pencils
Thus, 5 pencils are left with Neelam.

Page No 83:

Draw a 0 to 1 on a number line. Label point 1 as A and mark the starting point as 0.

(i) Divide the number line from 0 to 1 into 8 equal parts and take out 3 parts from it to reach point P. (ii) Divide the number line from 0 to 1 into 9 equal parts and take out 5 parts from it to reach point P

. (iii) Divide the number line from 0 to 1 into 7 equal parts and take out 4 parts from it to reach point P. (Iv) Divide the number line from 0 to 1 into 5 equal parts and take out 2 parts from it to reach point P. (v) Divide the number line from 0 to 1 into 4 equal parts and take out 1 part from it to reach point P. Page No 85:

A fraction whose numerator is greater than or equal to its denominator is called an improper fraction. Hence,  are improper fractions.

Page No 85:

Clearly,  are improper fractions, each with 5 as the denominator.

Page No 85:

Clearly, are improper fractions, each with 13 as the numerator.

We have:
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

Page No 85:

(i) On dividing 17 by 5, we get:
Quotient = 3
Remainder = 2
∴

(ii) On dividing 62 by 7, we get:
Quotient = 8
Remainder = 6
∴

(iii) On dividing 101 by 8, we get:
Quotient = 12
Remainder = 5
∴

(iv) On dividing 95 by 13, we get:
Quotient = 7
Remainder = 4
∴

(v) On dividing 81 by 11, we get:
Quotient = 7
Remainder = 4
∴

(vi) On dividing 87 by 16, we get:
Quotient = 5
Remainder = 7
∴

(vii) On dividing 103 by 12, we get:
Quotient = 8
Remainder = 7
∴

(viii) On dividing 117 by 20, we get:
Quotient = 5
Remainder = 17
∴

Page No 85:

An improper fraction is greater than 1. Hence, it is always greater than a proper fraction, which is less than 1.
(i)

(ii)

(iii)

(iv)

(v)

(vi)

Page No 86:

(i) Draw a number line. Mark 0 as the starting point and 1 as the ending point.
Then, divide 0 to 1 in four equal parts, where each part is equal to 1/4.
Show the consecutive parts as 1/4, 1/2, 3/4 and at 1 show 4/4 = 1. (ii) Draw 0 to 1 on a number line. Divide the segment into 8 equal parts, each part corresponds to 1/8. Show the consecutive parts as 1/8, 2/8, 3/8, 4/8, 5/8, 6/8, 7/8 and 8/8. Highlight the required ones only. (iii) Draw 0 to 2 on a number line. Divide the segment between 0 and 1 into 5 equal parts, where each part is equal to 1/5.
Show 2/5, 3/5, 4/5 and 8/5 3 parts away from 1 towards 2. (1 < 8/5 < 2) Page No 89:

(i)

∴

Hence, the five fractions equivalent to $\frac{2}{3}$ are .

(ii) ​

∴

Hence, the five fractions equivalent to $\frac{4}{5}$ are .

(iii) ​

∴

Hence, the five fractions equivalent to $\frac{5}{8}$ are .

(iv) ​

∴

Hence, the five fractions equivalent to $\frac{7}{10}$ are .

(v) ​​

∴

Hence, the five fractions equivalent to $\frac{3}{7}$ are .

(vi)  ​

∴

Hence, the five fractions equivalent to $\frac{6}{11}$ are .

(vii)

∴

Hence, the five fractions equivalent to $\frac{7}{9}$ are .

(viii)

∴

Hence, the five fractions equivalent to $\frac{5}{12}$ are .

Page No 89:

The pairs of equivalent fractions are as follows:
(i)
(ii)
(iv)

Page No 89:

(i) Let
Clearly, 30 = 5 $×$ 6
So, we multiply the numerator by 6.

∴ ​
Hence, the required fraction is $\frac{18}{30}$.
(ii)  ​Let
Clearly, 24 = 3 $×$ 8
So, we multiply the denominator by 8.

∴ ​
Hence, the required fraction is $\frac{24}{40}$.

Page No 89:

(i) Let
Clearly, 54 = 9 $×$ 6
So, we multiply the numerator by 6.
∴ ​
Hence, the required fraction is $\frac{30}{54}$.
(ii)  ​Let
Clearly, 35 = 5 $×$ 7
So, we multiply the denominator by 7.
∴ ​
Hence, the required fraction is $\frac{35}{63}$.

Page No 89:

(i) Let
Clearly, 77 = 11 $×$ 7
So, we multiply the numerator by 7.

∴ ​
Hence, the required fraction is $\frac{42}{77}$.
(ii)  ​Let
Clearly, 60 = 6 $×$ 10
So, we multiply the denominator by 10.

∴ ​
Hence, the required fraction is $\frac{60}{110}$.

Page No 89:

Let
Clearly, 4 = 24 $÷$ 6
So, we divide the denominator by 6.
∴ ​
Hence, the required fraction is $\frac{4}{5}$.

Page No 89:

(i) Let
Clearly, 9 = 36 $÷$ 4
So, we divide the denominator by 4.
∴ ​
Hence, the required fraction is $\frac{9}{12}$.
(ii)  ​Let
Clearly, 4 = 48 $÷$ 12
So, we divide the numerator by 12.
∴ ​
Hence, the required fraction is $\frac{3}{4}$.

Page No 89:

(i) Let
Clearly, 4 = 56 $÷$ 14
So, we divide the denominator by 14.
∴ ​
Hence, the required fraction is $\frac{4}{5}$.
(ii)  ​Let
Clearly, 10 = 70 $÷$ 7
So, we divide the numerator by 7.
∴ ​
Hence, the required fraction is $\frac{8}{10}$.

Page No 89:

(i) Here, numerator = 9 and denominator = 15
Factors of 9 are 1, 3 and 9.
Factors of 15 are 1, 3, 5 and 15.
Common factors of 9 and 15 are 1 and 3.
H.C.F. of 9 and 15 is 3.
∴
Hence, the simplest form of .

(ii) Here, numerator = 48 and denominator = 60
Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
Common factors of 48 and 60 are 1, 2, 3, 4, 6 and 12.
H.C.F. of 48 and 60 is 12.
∴
Hence, the simplest form of .

(iii) Here, numerator = 84 and denominator = 98
Factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 42 and 84.
Factors of 98 are 1, 2, 7, 14, 49 and 98.
Common factors of 84 and 98 are 1, 2, 7 and 14.
H.C.F. of 84 and 98 is 14.
∴
Hence, the simplest form of .

(iv) Here, numerator = 150 and denominator = 60
Factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 75 and 150.
Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
Common factors of 150 and 60 are 1, 2, 3, 5, 6, 10, 15 and 30.
H.C.F. of 150 and 60 is 30.
∴
Hence, the simplest form of .

(v) ​Here, numerator = 72 and denominator = 90
Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.
Factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45 and 90.
Common factors of 72 and 90 are 1, 2, 3, 6, 9 and 18.
H.C.F. of 72 and 90 is 18.
∴
Hence, the simplest form of .

Page No 89:

(i) Here, numerator = 8 and denominator = 11
Factors of 8 are 1, 2, 4 and 8.
Factors of 11 are 1 and 11.

Common factor of 8 and 11 is 1.
Thus, H.C.F. of 8 and 11 is 1.
Hence, $\frac{8}{11}$ is the simplest form.

(ii) Here, numerator = 9 and denominator = 14
Factors of 9 are 1, 3 and 9.
Factors of 14 are 1, 2, 7 and 14.
Common factor of 9 and 14 is 1.
Thus, H.C.F. of 9 and 14 is 1.
Hence, $\frac{9}{14}$ is the simplest form.

(iii) Here, numerator = 25 and denominator = 36
Factors of 25 are 1, 5 and 25.
Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.
Common factor of 25 and 36 is 1.
Thus, H.C.F. of 25 and 36 is 1.
Hence, $\frac{25}{36}$ is the simplest form.

(iv) Here, numerator = 8 and denominator = 15
Factors of 8 are 1, 2, 4 and 8.
Factors of 15 are 1, 3, 5 and 15.
Common factor of 8 and 15 is 1.
Thus, H.C.F. of 8 and 15 is 1.
Hence, $\frac{8}{15}$ is the simplest form.
(v) Here, numerator = 21 and denominator = 10
Factors of 21 are 1, 3, 7 and 21.
Factors of 10 are 1, 2, 5 and 10.
Common factor of 21 and 10 is 1.
Thus, H.C.F. of 21 and 10 is 1.
Hence, $\frac{21}{10}$ is the simplest form.

Page No 90:

(i) 28
(ii) 21
(iii) 32
(iv) 12
(v) 5
(vi) 9

Page No 93:

Like fractions:
Fractions having the same denominator are called like fractions.
Examples:

Unlike fractions:
Fractions having different denominators are called unlike fractions.
Examples:

Page No 93:

The given fractions are L.C.M. of 5, 10, 15 and 30 = (5 $×$ 2 $×$ 3) = 30
So, we convert the given fractions into equivalent fractions with 30 as the denominator.
(But, one of the fractions already has 30 as its denominator. So, there is no need to convert it into an equivalent fraction.)
Thus, we have:

Hence, the required like fractions are

Page No 93:

The given fractions are
L.C.M. of 4, 8, 12 and 24 = (4 $×$ 2 $×$ 3) = 24
So, we convert the given fractions into equivalent fractions with 24 as the denominator.
(But one of the fractions already has 24 as the denominator. So, there is no need to convert it into an equivalent fraction.)
Thus, we have:

Hence, the required like fractions are

Page No 93:

Between two fractions with the same denominator, the one with the greater numerator is the greater of the two.

(i) >
(ii) >
(iii) <
(iv) >
(v) >
(vi) <

Page No 93:

Between two fractions with the same numerator, the one with the smaller denominator is the greater of the two.

(i) >
(ii) >
(iii)<
(iv) >
(v) <
(vi) >

Page No 93:

By cross multiplying:
5 $×$ 5 = 25 and 4 $×$ 7 = 28
Clearly, 28 > 25
$\therefore$

Page No 93:

By cross multiplying:
3 $×$ 6 = 18 and 5 $×$ 8 = 40
Clearly, 18 < 40
$\therefore$

Page No 93:

By cross multiplying:
7 $×$ 7 = 49 and 11 $×$ 6 = 66
Clearly, 49 < 66
$\therefore$

Page No 93:

By cross multiplying:
5 $×$ 11 = 55 and 9 $×$ 6 = 54
Clearly, 55 > 54
$\therefore$

Page No 93:

By cross multiplying:
2 $×$ 9 = 18 and 4 $×$ 3 = 12
Clearly, 18 > 12
$\therefore$

Page No 93:

By cross multiplying:
6 $×$ 4 = 24 and 13 $×$ 3 = 39
Clearly, 24 < 39
$\therefore$

Page No 93:

By cross multiplying:
3 $×$ 6 = 18 and 4 $×$ 5 = 20
Clearly, 18 < 20
$\therefore$

Page No 93:

By cross multiplying:
5 $×$ 12 = 60 and 8 $×$ 7 = 56
Clearly, 60 > 56
$\therefore$

Page No 93:

L.C.M. of 9 and 6 = (3 $×$ 3 $×$ 2) = 18
Now, we convert  into equivalent fractions having 18 as the denominator.
∴​ 4949

Clearly,
$\therefore$

Page No 93:

L.C.M. of 5 and 10 = (5 $×$ 2) = 10
Now, we convert  into an equivalent fraction having 10 as the denominator as the other fraction has already 10 as its denominator.
∴​ 4949

Clearly,
$\therefore$

Page No 93:

L.C.M. of 8 and 10 = (2 $×$ 5 $×$ 2 $×$ 2) = 40
Now, we convert  into equivalent fractions having 40 as the denominator.
∴​ 4949

Clearly,
$\therefore$

Page No 93:

L.C.M. of 12 and 15 = (2 $×$ 2 $×$ 3 $×$ 5) = 60
Now, we convert  into equivalent fractions having 60 as the denominator.
∴​ 4949

Clearly,
$\therefore$

Page No 93: The given fractions are .
L.C.M. of 2, 4, 6 and 8 = (2 $×$ 2 $×$ 2 $×$ 3) = 24
We convert each of the given fractions into an equivalent fraction with denominator 24.
Now, we have:

Clearly,

∴ ​
Hence, the given fractions can be arranged in the ascending order as follows:

Page No 93:

The given fractions are L.C.M. of 3, 6, 9 and 18 = (3 $×$ 2  $×$ 3) = 18
So, we convert each of the fractions whose denominator is not equal to 18 into an equivalent fraction with denominator 18.
Now, we have:

Clearly,
∴ ​

Hence, the given fractions can be arranged in the ascending order as follows:

Page No 93:

The given fractions are
L.C.M. of 5, 10, 15 and 30 = (2 $×$ 5 $×$ 3) = 30 So, we convert each of the fractions whose denominator is not equal to 30 into an equivalent fraction with denominator 30.
Now, we have:

Clearly,
∴ ​

Hence, the given fractions can be arranged in the ascending order as follows:

Page No 93:

The given fractions are
L.C.M. of 4, 8, 16 and 32 = (2 ⨯ 2 ⨯ 2 ⨯ 2 ⨯ 2) = 32 So, we convert each of the fractions whose denominator is not equal to 32 into an equivalent fraction with denominator 32.
Now, we have:

Clearly,
∴ ​

Hence, the given fractions can be arranged in the ascending order as follows:

Page No 93:

The given fractions are
L.C.M. of 4, 8, 12 and 24 = (2 ⨯ 2 ⨯ 2 ⨯ 3) = 24 So, we convert each of the fractions whose denominator is not equal to 24 into an equivalent fraction with denominator 24.
Thus, we have;

Clearly,

∴ ​

Hence, the given fractions can be arranged in the descending order as follows:

Page No 93:

The given fractions are
L.C.M. of 9, 12, 18 and 36 = (3 ⨯ 3 ⨯ 2 ⨯ 2) = 36 We convert each of the fractions whose denominator is not equal to 36 into an equivalent fraction with denominator 36.
Thus, we have:

Clearly,

∴ ​

Hence, the given fractions can be arranged in the descending order as follows:

Page No 93:

The given fractions are
L.C.M. of 3, 5,10 and 15 = (2 ⨯ 3 ⨯ 5) = 30 So, we convert each of the fractions into an equivalent fraction with denominator 30.
Thus, we have:

Clearly,
∴ ​

Hence, the given fractions can be arranged in the descending order as follows:

Page No 93:

The given fractions are
L.C.M. of 7, 14, 21 and 42 = (2 ⨯ 3 ⨯ 7) = 42 We convert each one of the fractions whose denominator is not equal to 42 into an equivalent fraction with denominator 42.
Thus, we have:

Clearly,
∴ ​
Hence, the given fractions can be arranged in the descending order as follows:

Page No 93:

The given fractions are
As the fractions have the same numerator, we can follow the rule for the comparison of such fractions.
This rule states that when two fractions have the same numerator, the fraction having the smaller denominator is the greater one.
Clearly,
Hence, the given fractions can be arranged in the descending order as follows:

Page No 93:

The given fractions are
As the fractions have the same numerator, so we can follow the rule for the comparison of such fractions.
This rule states that when two fractions have the same numerator, the fraction having the smaller denominator is the greater one.

Clearly,
Hence, the given fractions can be arranged in the descending order as follows:

Page No 94:

Lalita read 30 pages of a book having 100 pages.
Sarita read $\frac{2}{5}$ of the same book.
$\frac{2}{5}$ of 100 pages = ​
Hence, Sarita read more pages than Lalita as 40 is greater than 30.

Page No 94:

To know who exercised for a longer time, we have to compare .
On cross multiplying:
4 $×$ 2 = 8 and 3 $×$ 3 = 9
Clearly, 8 < 9
$\therefore$
Hence, Rohit exercised for a longer time.

Page No 94:

Fraction of students who passed in VI A =

Fraction of students who passed in VI B =
In both the sections, the fraction of students who passed is the same, so both the sections have the same result.

Page No 96:

The given fractions are like fractions.
We know:
Sum of like fractions  =
Thus, we have:

Page No 96:

The given fractions are like fractions.
We know:
Sum of like fractions  =
Thus, we have:

Page No 96:

The given fractions are like fractions.
We know:
Sum of like fractions  =
Thus, we have:

Page No 96:

L.C.M. of 9 and 6 = (2 $×$ 3 $×$ 3) = 18 Now, we have:

Page No 96:

L.C.M. of 12 and 16 = (2 $×$ 2 $×$ 2 $×$ 2 $×$ 3) = 48 Now, we have:

Page No 96:

L.C.M. of 15 and 20 = (3 $×$ 5 $×$ 2 $×$ 2) = 60 Page No 96:

We have: 234+556

Page No 96:

We have: 234+556

Page No 96:

We have: 234+556

Page No 96:

We have: 234+556

Page No 96:

We have: 234+556

Page No 96:

We have: 234+556

Page No 96:

We have: 234+556

Page No 96:

We have: 234+556

Page No 96:

We have: 234+556

Page No 96:

Total cost of both articles = Cost of pencil + Cost of eraser
Thus, we have:

Hence, the total cost of both the articles is .

Page No 96:

Total cloth purchased by Sohini = Cloth for kurta + Cloth for pyjamas
Thus, we have:

$\therefore$ Total length of cloth purchased =

Page No 96:

Distance from Kishan's house to school = Distance covered by him by rickshaw + Distance covered by him on foot
Thus, we have: Hence, the distance from Kishan's house to school is .

Page No 96:

Weight of the cylinder filled with gas = Weight of the empty cylinder + Weight of the gas inside the cylinder
Thus, we have:

Hence, the weight of the cylinder filled with gas is .

Page No 99:

Difference of like fractions = Difference of numerator $÷$ Common denominator

Page No 99:

Difference of like fractions = Difference of numerator $÷$ Common denominator

Page No 99:

Difference of like fractions = Difference of numerator $÷$ Common denominator

Page No 99:

L.C.M. of 6 and 9 = (3 $×$ 2 $×$ 3) = 18
Now, we have:

Page No 99:

L.C.M. of 2 and 8 = (2 $×$ 2 $×$ 2) = 8
Now, we have:

Page No 99:

L.C.M. of 8 and 12 = (2 $×$ 2$×$ 2$×$3) = 24
Now, we have:

We have:

Page No 99:

We have:

234+556

Page No 99:

We have:
234+556

Page No 99:

We have:

234+556

Page No 99:

We have:

Page No 99:

We have:

Page No 99:

We have:

Page No 99:

We have:

Page No 99:

We have:

Page No 99:

Let x be added to $9\frac{2}{3}$ to get 19.

923

Page No 99:

Let x be added to $6\frac{7}{15}$ to get $8\frac{1}{5}$.

Page No 99:

Let us compare .
3 $×$ 7 = 21 and 4 $×$ 5 = 20
Clearly, 21 > 20

Required difference:

Hence, .

Page No 99:

Amount of milk left with Mrs. Soni = Total amount of milk bought by her $-$ Amount of milk consumed
$\therefore$ Amount of milk left with Mrs. Soni

$\therefore$ Milk left with Mrs. Soni =

Page No 99:

Actual duration of the film = Total duration of the show $-$ Time spent on advertisements

Thus, the actual duration of the film was .

Page No 99:

Money left with the rickshaw puller = Money earned by him in a day $-$ Money spent by him on food

Hence, Rs $80\frac{3}{4}$ is left with the rickshaw puller.

Page No 99:

The length of the other piece = (Length of the wire $-$ Length of one piece)

Hence, the other piece is  long.

(c)

(c)

Page No 100:

(a) 15

Explanation:

(a) 4

Explanation:

Page No 100:

(c)

(Fractions having the same denominator are called like fractions.)

Page No 100:

(d) none of these

In a proper fraction, the numerator is less than the denominator.

Page No 100:

(a) $\frac{7}{8}$
In a proper fraction, the numerator is less than the denominator.

Page No 100:

(b)
Between the two fractions with the same numerator, the one with the smaller denominator is the greater.

Page No 100:

(c) $\frac{3}{5}$

L.C.M. of 5, 3, 6 and 10 = (2 $×$ 3 $×$ 5) = 30
Thus, we have:

Page No 100:

( b ) $\frac{4}{5}$
Among the given fractions with the same numerator, the one with the smallest denominator is the greatest.

Page No 100:

(a) $\frac{6}{11}$
Among like fractions, the fraction with the smallest numerator is the smallest.

Page No 100:

(d) $\frac{7}{12}$

Explanation:

​​L.C.M. of 4, 6, 12 and 3 = (2 $×$ 2 $×$ 3) = 12
Thus, we have:

Page No 100:

(b) $\frac{23}{5}$

Page No 100:

(c) $4\frac{6}{7}$
On dividing 34 by 7:
Quotient = 4
Remainder = 6

Page No 101:

(b) $\frac{3}{4}$

Explanation:

Addition of like fractions = Sum of the numerators / Common denominator

Page No 101:

(b) $\frac{1}{2}$
Explanation:

Page No 101:

(d) $1\frac{1}{18}$

Explanation:

Page No 101:

(a) $3\frac{1}{3}$

Explanation:
Let us compare  .
10 ⨯ 10 = 100 and 3 ​⨯ 33 = 99
Clearly, 100 > 99
∴

Page No 103:

A fraction is defined as a number representing a part of a whole, where the whole may be a single object or a group of objects.

Examples:

Page No 103:

An hour has 60 minutes.
$\therefore$ Fraction for 35 minutes =
Hence, $\frac{7}{12}$ part of an hour is equal to 35 minutes.

Page No 103:

56 = 8 ⨯ 7
So, we need to multiply the numerator by 7.

$\therefore$
Hence, the required fraction is $\frac{35}{56}$.

Page No 103:

Let OA = AB = BC = 1 unit
$\therefore$ OB = 2 units and OC = 3 units
Divide BC into 5 equal parts and take 3 parts out to reach point P.
Clearly, point P represents the number $2\frac{3}{5}$.

We have:

Page No 103:

Cost of a pen =

Cost of a pencil =

So, the cost of a pen is more than the cost of a pencil.
Difference between their costs:

Hence, the cost of a pen is Rs $12\frac{1}{2}$ more than the cost of a pencil.

Page No 103:

Let us compare .
By cross multiplying:
3 ⨯ 7 = 21 and ​4 ⨯ 5 = 20
Clearly, 21 > 20
∴​$\frac{3}{4}>\frac{5}{7}$
Their difference:

Hence,

Page No 103:

L.C.M. of 2, 3, 9 and 6 = (2 ⨯ 3 ​⨯ 3) = 18
Now, we have:

Page No 103:

30 = 5 ​⨯ 6
So, we have to multiply the numerator by 6 to get the equivalent fraction having denominator 30.

Thus,

Page No 103:

The factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84.
The factors of 98 are 1, 2, 7, 14, 49, 98.
The common factors of 84 and 98 are 1, 2, 7, 14.
The H.C.F. of 84 and 98 is 14.
Dividing both the numerator and the denominator by the H.C.F.:

Page No 103:

(b) an improper fraction

In an improper fraction, the numerator is greater than the denominator.

Page No 103:

(a) proper fraction

In a proper fraction, the numerator is less than the denominator.

Page No 103:

(b) $\frac{3}{8}<\frac{5}{12}$

Considering :

Page No 103:

(a) $\frac{2}{3}$

Explanation:
L.C.M. of 3, 9, 2 and 12 = ( 2 ⨯ 2 ⨯ 3 ​⨯ 3) = 36
Now, we have:

Page No 103:

(b) $2\frac{1}{4}$
Explanation:

Page No 103:

(c)
Like fractions have same the denominator.

Page No 104:

(d) $\frac{16}{21}$

Page No 104:

(ii)

(iii)

(iv)

(v)