Home » Class 6 Maths » NCERT Solution Class 6 Maths Chapter -7 Fractions

# NCERT Solution Class 6 Maths Chapter -7 Fractions

## 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
21−⅗
Lets take LCM of 1,5 = 1*5 =5
215105
Since both numbers has the denominator as 5, we can solve the equation now,
The solution is
105 35 =75

(b) 4+78
To solve we have to make both numbers in fraction form
41+78
Lets take LCM of 1,8 = 1*8 =8
4188=328
Since both numbers has the denominator as 8, we can solve the equation
now,
The solution is
328+78=398

(c) 35+27

Since, the numbers are in fraction form,

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

For first number,
3577=2135

For second number,
2755=1035

The solution is
2135+1035=3135

###### d) 9⁄11−4⁄15

Since ,the numbers are in fraction form

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

For first number,
9111515=135165

For second number,
4151111=44165

The solution is
13516544165 = 91165

(e) 710+25+32

Since, the numbers are in fraction form,

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

For first number,
71011=710

For second number,
2522= 410

For third number,
3255= 1510

The solution is
710 + 410 + 1510 = 2610

###### (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,
223=(3∗2)+23=83

For second number in mixed fraction,
312 = (2∗3)+12=72

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

For first number,
8322=166

For second number,
7233=216

The solution is
166+216=376

(g) 812−358

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

For first number in mixed fraction,
812(2∗8)+1⁄2=172

For second number in mixed fraction,
358=(8∗3)+52=298

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

For first number,
17244=688

For second number,
29811=688

The solution is
688688 = 398

###### 2) Arrange the following numbers in descending order: a)2⁄9,2⁄3,8⁄21

Solution:
29,23,821

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,
2977=1463

For the second number,
232121=4263

For the third number,
82133=2463

Descending order means arranging the numbers from largest to smallest

So,
1463=0.22 4263=0.66 2463=0.38

Therefore, the decreasing order of rational numbers are

4263  2463  1463

i.e)

2821 29

###### b) 1⁄5  ,  3⁄7 ,  7⁄10

Solution:
15 3710

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,
151414 = 1470

For the second number,
371010=3070

For the third number,
71077=4970

Descending order means arranging the numbers from largest to smallest

So,
1470 =  0.2   3070 = 0.42    4970 = 0.70

Therefore, the decreasing order of rational numbers are

4970  >  3070  >  1470

i.e)
710   > 37   >   15

###### 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

Sum of third row713  +  313 513  = 1513

Sum of first column513 + 313 + 713 = 1513

Sum of second column = 713 + 513 + 313 = 1513

Sum of third column=   313+ 713 + 513 = 1513

Sum of first diagonal (left to right)  = 513 + 513 + 513 = 1513

Sum of second diagonal (right to left)  313 + 513 + 713 = 1513

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 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

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

For length,

614 = (4∗6)+14=254

3323 = (3∗3)+23=113

LCM of 3,4= 12
25433 = 751211344 = 4412

= 2∗(7512 + 4412)

= 2∗(11912)

= 1196 cm

= 75124412 =330012
⇒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 = 52cm,
YZ = 234 cm,
XZ = 335cm
The perimeter of triangle XYZ = XY + YZ + ZX
=(52 + 234 + 335)
=52 + 114 + 185
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
52  ∗ 1010   =   5020   ⇒  114  ∗  55  =  5520 18544 = 7220
(50+55+7220)

⇒17720cm

(ii) In rectangle YMNZ,

YZ = 234 cm,

MN= 76

W.K.T

Perimeter of rectangle = 2 (length + breadth)

= 2(234 + 76)

= 2(114 + 76)

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
11433=33127622 = 141233+1412=476

So the greatest perimeter out of this is,

(i) 17720 =8.85cm

(ii)476 = 7.83cm

Comparing the perimeter of rectangle and triangle

17720 = 8.85cm > 476 = 7.83cm

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