# NCERT Solution 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}**

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}**

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}**

**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}**

**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:**

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? 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.