Ncert Solutions For Class 7 Maths Ex 2.1

Ncert Solutions For Class 7 Maths Chapter 2 Ex 2.1

 

1) Solve the following

(a) 2-35

Soln:

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

2135

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

2155=105

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

The solution is

10535=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) 911415

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) 223+312

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

For first number in mixed fraction,

223=(32)+23=83

For second number in mixed fraction,

312=(23)+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) 812358

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

For first number in mixed fraction,

812=(28)+12=172

For second number in mixed fraction,

358=(83)+58=298

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

For first number,

17244=688

For second number,

29811=298

The solution is

688298=398

 

2) Arrange the following numbers in descending order:

  1. a)29,23,821

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)

23>821>29

 

b)15,37,710

Solution:

15,37,710

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?

513 713 313
313 513 713
713 313 513

 

Solution:

  • Sum of first row = 513+ 713+ 313= 1513

 

  • Sum of second row=313+ 513+ 713= 1513

 

  • Sum of third row=713+ 313+ 513= 1513

 

  • Sum of first column=513+ 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 614cm and 323cm 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,

614= (46)+14=254

For breadth.

323= (33)+23=113

LCM of 3,4= 12

25433=7512 11344=4412

Perimeter of rectangle= 2(length+breadth)

= 2(7512+4412)

= 2(11912)

= 1196cm

Area of rectangle = Length*breadth

                              = 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

521010=5020 11455=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=3312 7622=1412 2×33+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.

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