# 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$\frac{3}{5}$

Soln:

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

2135$\Rightarrow \frac{2}{1}-\frac{3}{5}$

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

2155=105$\Rightarrow \frac{2}{1}*\frac{5}{5}=\frac{10}{5}$

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

The solution is

10535=75$\Rightarrow \frac{10}{5}-\frac{3}{5}=\frac{7}{5}$

(b) 4+78$4+\frac{7}{8}$

To solve we have to make both numbers in fraction form

41+78$\Rightarrow \frac{4}{1}+\frac{7}{8}$

Lets take LCM of 1,8 = 1*8 =8

4188=328$\Rightarrow \frac{4}{1}*\frac{8}{8}=\frac{32}{8}$

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

now,

The solution is

328+78=398$\Rightarrow \frac{32}{8}+\frac{7}{8}=\frac{39}{8}$

(c) 35+27$\frac{3}{5}+\frac{2}{7}$

Since, the numbers are in fraction form,

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

For first number,

3577=2135$\Rightarrow \frac{3}{5}*\frac{7}{7}=\frac{21}{35}$

For second number,

2755=1035$\Rightarrow \frac{2}{7}*\frac{5}{5}=\frac{10}{35}$

The solution is

2135+1035=3135$\Rightarrow \frac{21}{35}+\frac{10}{35}=\frac{31}{35}$

(d) 911415$\frac{9}{11}-\frac{4}{15}$

Since ,the numbers are in fraction form

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

For first number,

9111515=135165$\Rightarrow \frac{9}{11}*\frac{15}{15}=\frac{135}{165}$

For second number,

4151111=44165$\Rightarrow \frac{4}{15}*\frac{11}{11}=\frac{44}{165}$

The solution is

13516544165=91165$\Rightarrow \frac{135}{165}-\frac{44}{165}=\frac{91}{165}$

(e) 710+25+32$\frac{7}{10}+\frac{2}{5}+\frac{3}{2}$

Since, the numbers are in fraction form,

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

For first number,

71011=710$\Rightarrow \frac{7}{10}*\frac{1}{1}=\frac{7}{10}$

For second number,

2522=410$\Rightarrow \frac{2}{5}*\frac{2}{2}=\frac{4}{10}$

For third number,

3255=1510$\Rightarrow \frac{3}{2}*\frac{5}{5}=\frac{15}{10}$

The solution is

710+410+1510=2610$\Rightarrow \frac{7}{10}+\frac{4}{10} + \frac{15}{10} =\frac{26}{10}$

(f) 223+312$2\frac{2}{3}+3\frac{1}{2}$

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$2\frac{2}{3}=\frac{\left ( 3*2 \right )+2}{3}=\frac{8}{3}$

For second number in mixed fraction,

312=(23)+12=72$3\frac{1}{2}=\frac{\left ( 2*3 \right )+1}{2}=\frac{7}{2}$

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

For first number,

8322=166$\Rightarrow \frac{8}{3}*\frac{2}{2}=\frac{16}{6}$

For second number,

7233=216$\Rightarrow \frac{7}{2}*\frac{3}{3}=\frac{21}{6}$

The solution is

166+216=376$\Rightarrow \frac{16}{6}+\frac{21}{6}=\frac{37}{6}$

(g) 812358$8\frac{1}{2}-3\frac{5}{8}$

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$8\frac{1}{2}=\frac{\left ( 2*8 \right )+1}{}2=\frac{17}{2}$

For second number in mixed fraction,

358=(83)+58=298$3\frac{5}{8}=\frac{\left ( 8*3 \right )+5}{}8=\frac{29}{8}$

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

For first number,

17244=688$\Rightarrow \frac{17}{2}*\frac{4}{4}=\frac{68}{8}$

For second number,

29811=298$\Rightarrow \frac{29}{8}*\frac{1}{1}=\frac{29}{8}$

The solution is

688298=398$\Rightarrow \frac{68}{8}-\frac{29}{8}=\frac{39}{8}$

2) Arrange the following numbers in descending order:

1. a)29,23,821$\frac{2}{9},\frac{2}{3},\frac{8}{21}$

Solution:

29,23,821$\frac{2}{9},\frac{2}{3},\frac{8}{21}$

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,

2977=1463$\Rightarrow \frac{2}{9}*\frac{7}{7}=\frac{14}{63}$

For the second number,

232121=4263$\Rightarrow \frac{2}{3}*\frac{21}{21}=\frac{42}{63}$

For the third number,

82133=2463$\Rightarrow \frac{8}{21}*\frac{3}{3}=\frac{24}{63}$

Descending order means arranging the numbers from largest to smallest

So,

1463=0.22$\frac{14}{63}=0.22$ 4263=0.66$\frac{42}{63}=0.66$ 2463=0.38$\frac{24}{63}=0.38$

Therefore, the decreasing order of rational numbers are

4263>2463>1463$\frac{42}{63}> \frac{24}{63}> \frac{14}{63}$

i.e)

23>821>29$\frac{2}{3}>\frac{8}{21}> \frac{2}{9}$

b)15,37,710$\frac{1}{5},\frac{3}{7},\frac{7}{10}$

Solution:

15,37,710$\frac{1}{5},\frac{3}{7},\frac{7}{10}$

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,

151414=1470$\Rightarrow \frac{1}{5}*\frac{14}{14}=\frac{14}{70}$

For the second number,

371010=3070$\Rightarrow \frac{3}{7}*\frac{10}{10}=\frac{30}{70}$

For the third number,

71077=4970$\Rightarrow \frac{7}{10}*\frac{7}{7}=\frac{49}{70}$

Descending order means arranging the numbers from largest to smallest

So,

1470=0.2$\frac{14}{70}=0.2$ 3070=0.42$\frac{30}{70}=0.42$ 4970=0.70$\frac{49}{70}=0.70$

Therefore, the decreasing order of rational numbers are

4970>3070>1470$\frac{49}{70}> \frac{30}{70}> \frac{14}{70}$

i.e)

710>37>15$\frac{7}{10}>\frac{3}{7}>\frac{1}{5}$

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

 513$\frac{5}{13}$ 713$\frac{7}{13}$ 313$\frac{3}{13}$ 313$\frac{3}{13}$ 513$\frac{5}{13}$ 713$\frac{7}{13}$ 713$\frac{7}{13}$ 313$\frac{3}{13}$ 513$\frac{5}{13}$

Solution:

• Sum of first row = 513$\frac{5}{13}$+ 713$\frac{7}{13}$+ 313$\frac{3}{13}$= 1513$\frac{15}{13}$

• Sum of second row=313$\frac{3}{13}$+ 513$\frac{5}{13}$+ 713$\frac{7}{13}$= 1513$\frac{15}{13}$

• Sum of third row=713$\frac{7}{13}$+ 313$\frac{3}{13}$+ 513$\frac{5}{13}$= 1513$\frac{15}{13}$

• Sum of first column=513$\frac{5}{13}$+ 313$\frac{3}{13}$+ 713$\frac{7}{13}$= 1513$\frac{15}{13}$

• Sum of second column=713$\frac{7}{13}$+ 513$\frac{5}{13}$+ 313$\frac{3}{13}$= 1513$\frac{15}{13}$

• Sum of third column=313$\frac{3}{13}$+ 713$\frac{7}{13}$+ 513$\frac{5}{13}$= 1513$\frac{15}{13}$

• Sum of first diagonal (left to right)= 513$\frac{5}{13}$+ 513$\frac{5}{13}$+ 513$\frac{5}{13}$= 1513$\frac{15}{13}$

• Sum of second diagonal (right to left)= 313$\frac{3}{13}$+ 513$\frac{5}{13}$+ 713$\frac{7}{13}$= 1513$\frac{15}{13}$

• 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 614$6\frac{1}{4}$cm and 323$3\frac{2}{3}$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)$2*\left ( length+breadth \right )$

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

For length,

614$6\frac{1}{4}$= (46)+14=254$\frac{\left ( 4*6 \right )+1}{4}=\frac{25}{4}$

323$3\frac{2}{3}$= (33)+23=113$\frac{\left ( 3*3 \right )+2}{3}=\frac{11}{3}$

LCM of 3,4= 12

25433=7512$\Rightarrow \frac{25}{4}*\frac{3}{3}=\frac{75}{12}$ 11344=4412$\Rightarrow \frac{11}{3}*\frac{4}{4}=\frac{44}{12}$

Perimeter of rectangle= 2(length+breadth)$2*\left ( length+breadth \right )$

= 2(7512+4412)$2*\left ( \frac{75}{12}+\frac{44}{12} \right )$

= 2(11912)$2*\left ( \frac{119}{12} \right )$

= 1196$\frac{119}{6}$cm

= 75124412=330012$\frac{75}{12}*\frac{44}{12}=\frac{3300}{12}$

275cm2$\Rightarrow 275 cm^{2}$

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 = 52$\frac{5}{2}$cm,

YZ = 234$2\frac{3}{4}$ cm,

XZ=335$3\frac{3}{5}$cm

The perimeter of triangle XYZ = XY + YZ + ZX

=(52$\frac{5}{2}$+234$2\frac{3}{4}$+ 335$3\frac{3}{5}$)

=52+114+185$\frac{5}{2}+\frac{11}{4}+\frac{18}{5}$

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

521010=5020$\Rightarrow \frac{5}{2}*\frac{10}{10}=\frac{50}{20}$ 11455=5520$\Rightarrow \frac{11}{4}*\frac{5}{5}=\frac{55}{20}$ 18544=7220$\Rightarrow \frac{18}{5}*\frac{4}{4}=\frac{72}{20}$ (50+55+7220)$\Rightarrow \left ( \frac{50+55+72}{20} \right )$

17720$\Rightarrow \frac{177}{20}$cm

(ii)

In rectangle YMNZ,

YZ = 234$2\frac{3}{4}$ cm,

MN= 76$\frac{7}{6}$

W.K.T

Perimeter of rectangle = 2 (length + breadth)

= 2(234+76)$2\left(2\frac{3}{4}+\frac{7}{6}\right )$

= 2(114+76)$2\left(\frac{11}{4}+\frac{7}{6}\right )$

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$\Rightarrow \frac{11}{4}*\frac{3}{3}=\frac{33}{12}$ 7622=1412$\Rightarrow \frac{7}{6}*\frac{2}{2}=\frac{14}{12}$ 2×33+1412=476$2 \times \frac{33+14}{12} = \frac{47}{6}$

So the greatest perimeter out of this is,

(i)

17720=8.85$\frac{177}{20}=8.85$cm

(ii)

476=7.83$\frac{47}{6}=7.83$cm

Comparing the perimeter of rectangle and triangle

17720=8.85$\frac{177}{20}=8.85$cm > 476=7.83$\frac{47}{6}=7.83$cm

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