# Chapter 2: Fractions and Decimal Numbers

Exercise 2.1

1) Solve the following

(a) 2-$$\frac{3}{5}$$

Soln:

To solve ,we have to make both numbers in fraction form

$$\Rightarrow \frac{2}{1}-\frac{3}{5}$$

Lets take LCM of 1,5 = 1*5 =5

$$\Rightarrow \frac{2}{1}*\frac{5}{5}=\frac{10}{5}$$

Since both numbers has the denominator as 5, we can solve the equation now,

The solution is

$$\Rightarrow \frac{10}{5}-\frac{3}{5}=\frac{7}{5}$$

(b) $$4+\frac{7}{8}$$

To solve we have to make both numbers in fraction form

$$\Rightarrow \frac{4}{1}+\frac{7}{8}$$

Lets take LCM of 1,8 = 1*8 =8

$$\Rightarrow \frac{4}{1}*\frac{8}{8}=\frac{32}{8}$$

Since both numbers has the denominator as 8, we can solve the equation

now,

The solution is

$$\Rightarrow \frac{32}{8}+\frac{7}{8}=\frac{39}{8}$$

(c) $$\frac{3}{5}+\frac{2}{7}$$

Since, the numbers are in fraction form,

Lets take LCM of 5,7= 5*7=35

For first number,

$$\Rightarrow \frac{3}{5}*\frac{7}{7}=\frac{21}{35}$$

For second number,

$$\Rightarrow \frac{2}{7}*\frac{5}{5}=\frac{10}{35}$$

The solution is

$$\Rightarrow \frac{21}{35}+\frac{10}{35}=\frac{31}{35}$$

(d) $$\frac{9}{11}-\frac{4}{15}$$

Since ,the numbers are in fraction form

Lets take LCM of 11,15= 11*15=165

For first number,

$$\Rightarrow \frac{9}{11}*\frac{15}{15}=\frac{135}{165}$$

For second number,

$$\Rightarrow \frac{4}{15}*\frac{11}{11}=\frac{44}{165}$$

The solution is

$$\Rightarrow \frac{135}{165}-\frac{44}{165}=\frac{91}{165}$$

(e) $$\frac{7}{10}+\frac{2}{5}+\frac{3}{2}$$

Since, the numbers are in fraction form,

Lets take LCM of 10,5,2= 2*5=10

For first number,

$$\Rightarrow \frac{7}{10}*\frac{1}{1}=\frac{7}{10}$$

For second number,

$$\Rightarrow \frac{2}{5}*\frac{2}{2}=\frac{4}{10}$$

For third number,

$$\Rightarrow \frac{3}{2}*\frac{5}{5}=\frac{15}{10}$$

The solution is

$$\Rightarrow \frac{7}{10}+\frac{4}{10} + \frac{15}{10} =\frac{26}{10}$$

(f) $$2\frac{2}{3}+3\frac{1}{2}$$

Since, the numbers are in mixed fraction form ,we have to convert it into fractional form

For first number in mixed fraction,

$$2\frac{2}{3}=\frac{\left ( 3*2 \right )+2}{3}=\frac{8}{3}$$

For second number in mixed fraction,

$$3\frac{1}{2}=\frac{\left ( 2*3 \right )+1}{2}=\frac{7}{2}$$

Lets take LCM of 3,2=3*2=6

For first number,

$$\Rightarrow \frac{8}{3}*\frac{2}{2}=\frac{16}{6}$$

For second number,

$$\Rightarrow \frac{7}{2}*\frac{3}{3}=\frac{21}{6}$$

The solution is

$$\Rightarrow \frac{16}{6}+\frac{21}{6}=\frac{37}{6}$$

(g) $$8\frac{1}{2}-3\frac{5}{8}$$

Since, the numbers are in mixed fraction form we have to convert it into fractional form

For first number in mixed fraction,

$$8\frac{1}{2}=\frac{\left ( 2*8 \right )+1}{}2=\frac{17}{2}$$

For second number in mixed fraction,

$$3\frac{5}{8}=\frac{\left ( 8*3 \right )+5}{}8=\frac{29}{8}$$

Lets take LCM of 2,8=2*8=8

For first number,

$$\Rightarrow \frac{17}{2}*\frac{4}{4}=\frac{68}{8}$$

For second number,

$$\Rightarrow \frac{29}{8}*\frac{1}{1}=\frac{29}{8}$$

The solution is

$$\Rightarrow \frac{68}{8}-\frac{29}{8}=\frac{39}{8}$$

2) Arrange the following numbers in descending order:

1. a)$$\frac{2}{9},\frac{2}{3},\frac{8}{21}$$

Solution:

$$\frac{2}{9},\frac{2}{3},\frac{8}{21}$$

Let’s take LCM of 9,3,21

9 = 3*3

21= 3* 7

3= 3*1

So, we can take two 3’s as a common term since we four 3’s and 7 is coming only once

Therefore ,the required LCM is= 3*3*7=63

For the first number,

$$\Rightarrow \frac{2}{9}*\frac{7}{7}=\frac{14}{63}$$

For the second number,

$$\Rightarrow \frac{2}{3}*\frac{21}{21}=\frac{42}{63}$$

For the third number,

$$\Rightarrow \frac{8}{21}*\frac{3}{3}=\frac{24}{63}$$

Descending order means arranging the numbers from largest to smallest

So,

$$\frac{14}{63}=0.22$$ $$\frac{42}{63}=0.66$$ $$\frac{24}{63}=0.38$$

Therefore, the decreasing order of rational numbers are

$$\frac{42}{63}> \frac{24}{63}> \frac{14}{63}$$

i.e)

$$\frac{2}{3}>\frac{8}{21}> \frac{2}{9}$$

b)$$\frac{1}{5},\frac{3}{7},\frac{7}{10}$$

Solution:

$$\frac{1}{5},\frac{3}{7},\frac{7}{10}$$

Let’s take LCM of 5,7,10

7=7*1

5=5*1

10=5*2

So, we can take one 5 as a common term .Since, we have two 5’s and 7,2 is coming only once

Therefore the required LCM is= 5*7*2= 70

For the first number,

$$\Rightarrow \frac{1}{5}*\frac{14}{14}=\frac{14}{70}$$

For the second number,

$$\Rightarrow \frac{3}{7}*\frac{10}{10}=\frac{30}{70}$$

For the third number,

$$\Rightarrow \frac{7}{10}*\frac{7}{7}=\frac{49}{70}$$

Descending order means arranging the numbers from largest to smallest

So,

$$\frac{14}{70}=0.2$$ $$\frac{30}{70}=0.42$$ $$\frac{49}{70}=0.70$$

Therefore, the decreasing order of rational numbers are

$$\frac{49}{70}> \frac{30}{70}> \frac{14}{70}$$

i.e)

$$\frac{7}{10}>\frac{3}{7}>\frac{1}{5}$$

3) If the sum of the numbers are same along both rows and columns, Will it form a magic square?

 $$\frac{5}{13}$$ $$\frac{7}{13}$$ $$\frac{3}{13}$$ $$\frac{3}{13}$$ $$\frac{5}{13}$$ $$\frac{7}{13}$$ $$\frac{7}{13}$$ $$\frac{3}{13}$$ $$\frac{5}{13}$$

Solution:

• Sum of first row = $$\frac{5}{13}$$+ $$\frac{7}{13}$$+ $$\frac{3}{13}$$= $$\frac{15}{13}$$

• Sum of second row=$$\frac{3}{13}$$+ $$\frac{5}{13}$$+ $$\frac{7}{13}$$= $$\frac{15}{13}$$

• Sum of third row=$$\frac{7}{13}$$+ $$\frac{3}{13}$$+ $$\frac{5}{13}$$= $$\frac{15}{13}$$

• Sum of first column=$$\frac{5}{13}$$+ $$\frac{3}{13}$$+ $$\frac{7}{13}$$= $$\frac{15}{13}$$

• Sum of second column=$$\frac{7}{13}$$+ $$\frac{5}{13}$$+ $$\frac{3}{13}$$= $$\frac{15}{13}$$

• Sum of third column=$$\frac{3}{13}$$+ $$\frac{7}{13}$$+ $$\frac{5}{13}$$= $$\frac{15}{13}$$

• Sum of first diagonal (left to right)= $$\frac{5}{13}$$+ $$\frac{5}{13}$$+ $$\frac{5}{13}$$= $$\frac{15}{13}$$

• Sum of second diagonal (right to left)= $$\frac{3}{13}$$+ $$\frac{5}{13}$$+ $$\frac{7}{13}$$= $$\frac{15}{13}$$

• Yes, it forms a magic square. Since, the sum of fractions in each row , each column and along the diagonals are same.

4) A rectangular block of length $$6\frac{1}{4}$$cm and $$3\frac{2}{3}$$cm of width is noted. Find the perimeter and area of the rectangular block.

Solution:

As the block is rectangular in shape

W.K.T

Perimeter of rectangle= $$2*\left ( length+breadth \right )$$

Since, both the numbers are in mixed fraction, it is first converted to fractional form

For length,

$$6\frac{1}{4}$$= $$\frac{\left ( 4*6 \right )+1}{4}=\frac{25}{4}$$

$$3\frac{2}{3}$$= $$\frac{\left ( 3*3 \right )+2}{3}=\frac{11}{3}$$

LCM of 3,4= 12

$$\Rightarrow \frac{25}{4}*\frac{3}{3}=\frac{75}{12}$$ $$\Rightarrow \frac{11}{3}*\frac{4}{4}=\frac{44}{12}$$

Perimeter of rectangle= $$2*\left ( length+breadth \right )$$

= $$2*\left ( \frac{75}{12}+\frac{44}{12} \right )$$

= $$2*\left ( \frac{119}{12} \right )$$

= $$\frac{119}{6}$$cm

= $$\frac{75}{12}*\frac{44}{12}=\frac{3300}{12}$$

$$\Rightarrow 275 cm^{2}$$

5) Find the perimeter of

(i) DXYZ

(ii) The rectangle YMNZ in this figure given below.

Out of two perimeters, which is greater?

Solution:

(i)

In DXYZ,

XY = $$\frac{5}{2}$$cm,

YZ = $$2\frac{3}{4}$$ cm,

XZ=$$3\frac{3}{5}$$cm

The perimeter of triangle XYZ = XY + YZ + ZX

=($$\frac{5}{2}$$+$$2\frac{3}{4}$$+ $$3\frac{3}{5}$$)

=$$\frac{5}{2}+\frac{11}{4}+\frac{18}{5}$$

LCM of 2,4,5=

2=2*1

4=2*2

5=5*1

We have, 2 as a common term as we have two 2’s and take 2,5 into account because it’s a single term

LCM=2*5*2=20

$$\Rightarrow \frac{5}{2}*\frac{10}{10}=\frac{50}{20}$$ $$\Rightarrow \frac{11}{4}*\frac{5}{5}=\frac{55}{20}$$ $$\Rightarrow \frac{18}{5}*\frac{4}{4}=\frac{72}{20}$$ $$\Rightarrow \left ( \frac{50+55+72}{20} \right )$$

$$\Rightarrow \frac{177}{20}$$cm

(ii)

In rectangle YMNZ,

YZ = $$2\frac{3}{4}$$ cm,

MN= $$\frac{7}{6}$$

W.K.T

Perimeter of rectangle = 2 (length + breadth)

= $$2\left(2\frac{3}{4}+\frac{7}{6}\right )$$

= $$2\left(\frac{11}{4}+\frac{7}{6}\right )$$

LCM of 4, 6

4=2*2

6=2*3

We have ,2 as a common term as we have two 2’s and take 2,3 into account because it’s a single term

LCM=2*3*2=12

$$\Rightarrow \frac{11}{4}*\frac{3}{3}=\frac{33}{12}$$ $$\Rightarrow \frac{7}{6}*\frac{2}{2}=\frac{14}{12}$$ $$2 \times \frac{33+14}{12} = \frac{47}{6}$$

So the greatest perimeter out of this is,

(i)

$$\frac{177}{20}=8.85$$cm

(ii)

$$\frac{47}{6}=7.83$$cm

Comparing the perimeter of rectangle and triangle

$$\frac{177}{20}=8.85$$cm > $$\frac{47}{6}=7.83$$cm

Therefore, the perimeter of triangle XYZ is greater than that of rectangle YMNZ.

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### 1 Comment

1. Haritha
May 5, 2018

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