NCERT Solution Class 6 Maths Chapter -7 Fractions

Class 6 Maths Chapter -7 Fractions

Exercise 2.1

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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

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(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

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(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

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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

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(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

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(f) 2²⁄₃+3¹⁄₂

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

For first number in mixed fraction,
2²⁄₃=^(3∗2)+2⁄₃=⁸⁄₃

For second number in mixed fraction,
3¹⁄₂ = ^(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

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(g) 8¹⁄₂−3⁵⁄₈

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

For first number in mixed fraction,
8¹⁄_{2(2∗8)+1⁄2=¹⁷⁄2}

For second number in mixed fraction,
3⁵⁄₈=^{(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

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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

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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

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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.

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4) A rectangular block of length 6¹⁄₄cm and 3²⁄₃cm 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,

6¹⁄₄ = ^(4∗6)+1⁄₄=²⁵⁄₄

For breadth.

33²⁄₃ = ^(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

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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 = ⁵⁄₂cm,
YZ = 2³⁄₄ cm,
XZ = 3³⁄₅cm
The perimeter of triangle XYZ = XY + YZ + ZX
=(⁵⁄₂ + 2³⁄₄ + 3³⁄₅)
=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 = 2³⁄₄ cm,

MN= ⁷⁄₆

W.K.T

Perimeter of rectangle = 2 (length + breadth)

= 2(2³⁄₄ + 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׳³⁺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.

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