## Class 6 Maths Chapter -7 Fractions

**Exercise 2.1**

**1) Solve the following**

**(a) 2-⅗**

**Soln:**

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

⇒^{2}⁄_{1}−⅗

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

⇒ ^{2}⁄_{1}∗^{5}⁄_{5 }= ^{10}⁄_{5
}Since both numbers has the denominator as 5, we can solve the equation now,

The solution is

⇒^{10}⁄_{5 }−^{3}⁄_{5 } =^{7}⁄_{5 }

**(b) 4+ ^{7}⁄**

_{8 }To solve we have to make both numbers in fraction form

⇒

^{4}⁄

_{1}+

^{7}⁄

_{8 }Lets take LCM of 1,8 = 1*8 =8

⇒

^{4}⁄

_{1}∗

^{8}⁄

_{8}=

^{32}⁄

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

now,

The solution is

⇒

^{32}⁄

_{8}+

^{7}⁄

_{8}=

^{39}⁄

_{8}

**(c) ^{3}⁄_{5}+^{2}⁄_{7}**

Since, the numbers are in fraction form,

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

For first number,

⇒^{3}⁄_{5}∗^{7}⁄_{7}=^{21}⁄_{35}

For second number,

⇒^{2}⁄_{7}∗^{5}⁄_{5}=^{10}⁄_{35}

The solution is

⇒^{21}⁄_{35}+^{10}⁄_{35}=^{31}⁄_{35}

**d) **^{9}⁄_{11}−^{4}⁄_{15}

^{9}⁄

_{11}−

^{4}⁄

_{15}

Since ,the numbers are in fraction form

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

For first number,

⇒^{9}⁄_{11}∗^{15}⁄_{15}=^{135}⁄_{165}

For second number,

⇒^{4}⁄_{15}∗^{11}⁄_{11}=^{44}⁄_{165}

The solution is

⇒^{135}⁄_{165} − ^{44}⁄_{165} = ^{91}⁄_{165}

**(e) ^{7}⁄_{10}+^{2}⁄_{5}+^{3}⁄_{2}**

Since, the numbers are in fraction form,

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

For first number,

⇒ ^{7}⁄_{10}∗ ^{1}⁄_{1}=^{7}⁄_{10}

For second number,

⇒ ^{2}⁄_{5}∗ ^{2}⁄_{2}= ^{4}⁄_{10}

For third number,

⇒ ^{3}⁄_{2}∗ ^{5}⁄_{5}= ^{15}⁄_{10}

The solution is

⇒ ^{7}⁄_{10} + ^{4}⁄_{10} + ^{15}⁄_{10} = ^{26}⁄_{10}

**(f) 2**^{2}⁄_{3}+3^{1}⁄_{2}

^{2}⁄

_{3}+3

^{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^{2}⁄_{3}=^{(3∗2)+2}⁄_{3}=^{8}⁄_{3}

For second number in mixed fraction,

3^{1}⁄_{2} = ^{(2∗3)+1}⁄_{2}=^{7}⁄_{2}

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

For first number,

⇒^{8}⁄_{3}∗^{2}⁄_{2}=^{16}⁄_{6}

For second number,

⇒^{7}⁄_{2}∗^{3}⁄_{3}=^{21}⁄_{6}

The solution is

⇒^{16}⁄_{6}+^{21}⁄_{6}=^{37}⁄_{6}

**(g) 8 ^{1}⁄_{2}−3^{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^{1}⁄_{2(2∗8)+1⁄2=17⁄2}

For second number in mixed fraction,

3^{5}⁄_{8}=^{(8∗3)+5⁄2=29⁄8}

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

For first number,

⇒^{17}⁄_{2}∗^{4}⁄_{4}=^{68}⁄_{8}

For second number,

⇒^{29}⁄_{8}∗^{1}⁄_{1}=^{68}⁄_{8}

**The solution is**

⇒ ^{68}⁄_{8} – ^{68}⁄_{8} = ^{39}⁄_{8}

**2) Arrange the following numbers in descending order:**

** a)**^{2}⁄_{9},^{2}⁄_{3},^{8}⁄_{21}

^{2}⁄

_{9},

^{2}⁄

_{3},

^{8}⁄

_{21}

**Solution:**

^{2}⁄_{9},^{2}⁄_{3},^{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,

⇒^{2}⁄_{9}∗^{7}⁄_{7}=^{14}⁄_{63}

For the second number,

⇒^{2}⁄_{3}∗^{21}⁄_{21}=^{42}⁄_{63}

For the third number,

⇒^{8}⁄_{21}∗^{3}⁄_{3}=^{24}⁄_{63}

Descending order means arranging the numbers from largest to smallest

So,

^{14}⁄_{63}=0.22 ^{42}⁄_{63}=0.66 ^{24}⁄_{63}=0.38

Therefore, the decreasing order of rational numbers are

^{42}⁄_{63 }> ^{24}⁄_{63 }> ^{14}⁄_{63}

i.e)

^{2}⁄_{3 }> ^{8}⁄_{21 }> ^{2}⁄_{9}

**b) **^{1}⁄_{5 }, ^{3}⁄_{7 }, ^{7}⁄_{10}

^{1}⁄

_{5 },

^{3}⁄

_{7 },

^{7}⁄

_{10}

**Solution:**

^{1}⁄_{5 }, ^{3}⁄_{7 }, ^{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,

⇒^{1}⁄_{5} ∗ ^{14}⁄_{14} = ^{14}⁄_{70}

For the second number,

⇒^{3}⁄_{7}∗^{10}⁄_{10}=^{30}⁄_{70}

For the third number,

⇒^{7}⁄_{10}∗^{7}⁄_{7}=^{49}⁄_{70}

Descending order means arranging the numbers from largest to smallest

So,

^{14}⁄_{70} = 0.2 ^{30}⁄_{70} = 0.42 ^{49}⁄_{70} = 0.70

Therefore, the decreasing order of rational numbers are

^{49}⁄_{70} > ^{30}⁄_{70} > ^{14}⁄_{70}

i.e)

^{7}⁄_{10 } > ^{3}⁄_{7} > ^{1}⁄_{5}

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

^{5}⁄_{13} | ^{7}⁄_{13} | ^{3}⁄_{13} |

^{3}⁄_{13} | ^{5}⁄_{13} | ^{7}⁄_{13} |

^{7}⁄_{13} | ^{3}⁄_{13} | ^{5}⁄_{13} |

**Answers:**

**Sum of third row** = ^{7}⁄_{13} + ^{3}⁄_{13 } + ^{5}⁄_{13} = ^{15}⁄_{13}

**Sum of first column**= ^{5}⁄_{13} + ^{3}⁄_{13} + ^{7}⁄_{13} = ^{15}⁄_{13}

**Sum of second column** = ^{7}⁄_{13} + ^{5}⁄_{13} + ^{3}⁄_{13} = ^{15}⁄_{13}

**Sum of third column**= ^{3}⁄_{13}+ ^{7}⁄_{13} + ^{5}⁄_{13} = ^{15}⁄_{13}

**Sum of first diagonal (left to right) **= ^{5}⁄_{13} + ^{5}⁄_{13} + ^{5}⁄_{13} = ^{15}⁄_{13}

**Sum of second diagonal (right to left) **= ^{3}⁄_{13} + ^{5}⁄_{13} + ^{7}⁄_{13} = ^{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**^{1}⁄_{4}cm and 3^{2}⁄_{3}cm of width is noted. Find the perimeter and area of the rectangular block.

**Solution:**

^{1}⁄

_{4}cm and 3

^{2}⁄

_{3}cm of width is noted. Find the perimeter and area of the rectangular block.

As the block is rectangular in shape

W.K.T

Perimeter of rectangle= 2∗(length+breadth)

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

For length,

6^{1}⁄_{4} = ^{(4∗6)+1}⁄_{4}=^{25}⁄_{4}

For breadth.

33^{2}⁄_{3} = ^{(3∗3)+2}⁄_{3}=^{11}⁄_{3}

LCM of 3,4= 12

⇒^{25}⁄_{4} ∗ ^{3}⁄_{3} = ^{75}⁄_{12} ⇒^{11}⁄_{3} ∗ ^{4}⁄_{4} = ^{44}⁄_{12}

Perimeter of rectangle= 2∗(length+breadth)

= 2∗(^{75}⁄_{12} + ^{44}⁄_{12})

= 2∗(^{119}⁄_{12})

= ^{119}⁄_{6} cm

Area of rectangle = Length*breadth

= ^{75}⁄_{12} ∗ ^{44}⁄_{12} =^{3300}⁄_{12}

⇒275cm2

**5) Find the perimeter of**

(i) DXYZ

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

Out of two perimeters, which is greater?

(i) DXYZ

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

Out of two perimeters, which is greater?

**Solution:
**(i) In DXYZ,

XY =

^{5}⁄

_{2}cm,

YZ = 2

^{3}⁄

_{4}cm,

XZ = 3

^{3}⁄

_{5}cm

The perimeter of triangle XYZ = XY + YZ + ZX

=(

^{5}⁄

_{2}+ 2

^{3}⁄

_{4}+ 3

^{3}⁄

_{5})

=

^{5}⁄

_{2}+

^{11}⁄

_{4}+

^{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

⇒^{5}⁄_{2} ∗ ^{10}⁄_{10} = ^{50}⁄_{20} ⇒ ^{11}⁄_{4} ∗ ^{5}⁄_{5} = ^{55}⁄_{20 } ⇒ ^{18}⁄_{5} ∗ ^{4}⁄_{4 }= ^{72}⁄_{20} ⇒

(^{50+55+72}⁄_{20})

⇒17720cm

(ii) In rectangle YMNZ,

YZ = 2^{3}⁄_{4} cm,

MN= ^{7}⁄_{6}

W.K.T

Perimeter of rectangle = 2 (length + breadth)

= 2(2^{3}⁄_{4} + ^{7}⁄_{6})

= 2(^{11}⁄_{4} + ^{7}⁄_{6})

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

⇒^{11}⁄_{4} ∗ ^{3}⁄_{3}=^{33}⁄_{12} ⇒^{7}⁄_{6} ∗^{2}⁄_{2} = ^{14}⁄_{12} 2×^{33+14}⁄_{12}=^{47}⁄_{6}

So the greatest perimeter out of this is,

(i) ^{177}⁄_{20} =8.85cm

(ii)^{47}⁄_{6} = 7.83cm

Comparing the perimeter of rectangle and triangle

^{177}⁄_{20} = 8.85cm > ^{47}⁄_{6} = 7.83cm

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