1) Solve the following
(a) 2-\(\frac{3}{5}\)
Soln:
To solve ,we have to make both numbers in fraction form
\(\Rightarrow \frac{2}{1}-\frac{3}{5}\)Lets take LCM of 1,5 = 1*5 =5
\(\Rightarrow \frac{2}{1}*\frac{5}{5}=\frac{10}{5}\)Since both numbers has the denominator as 5, we can solve the equation now,
The solution is
\(\Rightarrow \frac{10}{5}-\frac{3}{5}=\frac{7}{5}\)
(b) \(4+\frac{7}{8}\)
To solve we have to make both numbers in fraction form
\(\Rightarrow \frac{4}{1}+\frac{7}{8}\)Lets take LCM of 1,8 = 1*8 =8
\(\Rightarrow \frac{4}{1}*\frac{8}{8}=\frac{32}{8}\)Since both numbers has the denominator as 8, we can solve the equation
now,
The solution is
\(\Rightarrow \frac{32}{8}+\frac{7}{8}=\frac{39}{8}\)
(c) \( \frac{3}{5}+\frac{2}{7}\)
Since, the numbers are in fraction form,
Lets take LCM of 5,7= 5*7=35
For first number,
\(\Rightarrow \frac{3}{5}*\frac{7}{7}=\frac{21}{35}\)For second number,
\(\Rightarrow \frac{2}{7}*\frac{5}{5}=\frac{10}{35}\)The solution is
\(\Rightarrow \frac{21}{35}+\frac{10}{35}=\frac{31}{35}\)
(d) \(\frac{9}{11}-\frac{4}{15}\)
Since ,the numbers are in fraction form
Lets take LCM of 11,15= 11*15=165
For first number,
\(\Rightarrow \frac{9}{11}*\frac{15}{15}=\frac{135}{165}\)For second number,
\(\Rightarrow \frac{4}{15}*\frac{11}{11}=\frac{44}{165}\)The solution is
\(\Rightarrow \frac{135}{165}-\frac{44}{165}=\frac{91}{165}\)
(e) \(\frac{7}{10}+\frac{2}{5}+\frac{3}{2}\)
Since, the numbers are in fraction form,
Lets take LCM of 10,5,2= 2*5=10
For first number,
\(\Rightarrow \frac{7}{10}*\frac{1}{1}=\frac{7}{10}\)For second number,
\(\Rightarrow \frac{2}{5}*\frac{2}{2}=\frac{4}{10}\)For third number,
\(\Rightarrow \frac{3}{2}*\frac{5}{5}=\frac{15}{10}\)The solution is
\(\Rightarrow \frac{7}{10}+\frac{4}{10} + \frac{15}{10} =\frac{26}{10}\)
(f) \(2\frac{2}{3}+3\frac{1}{2}\)
Since, the numbers are in mixed fraction form ,we have to convert it into fractional form
For first number in mixed fraction,
\(2\frac{2}{3}=\frac{\left ( 3*2 \right )+2}{3}=\frac{8}{3}\)For second number in mixed fraction,
\(3\frac{1}{2}=\frac{\left ( 2*3 \right )+1}{2}=\frac{7}{2}\)Lets take LCM of 3,2=3*2=6
For first number,
\(\Rightarrow \frac{8}{3}*\frac{2}{2}=\frac{16}{6}\)For second number,
\(\Rightarrow \frac{7}{2}*\frac{3}{3}=\frac{21}{6}\)The solution is
\(\Rightarrow \frac{16}{6}+\frac{21}{6}=\frac{37}{6}\)
(g) \(8\frac{1}{2}-3\frac{5}{8}\)
Since, the numbers are in mixed fraction form we have to convert it into fractional form
For first number in mixed fraction,
\(8\frac{1}{2}=\frac{\left ( 2*8 \right )+1}{}2=\frac{17}{2}\)For second number in mixed fraction,
\(3\frac{5}{8}=\frac{\left ( 8*3 \right )+5}{}8=\frac{29}{8}\)Lets take LCM of 2,8=2*8=8
For first number,
\(\Rightarrow \frac{17}{2}*\frac{4}{4}=\frac{68}{8}\)For second number,
\(\Rightarrow \frac{29}{8}*\frac{1}{1}=\frac{29}{8}\)The solution is
\(\Rightarrow \frac{68}{8}-\frac{29}{8}=\frac{39}{8}\)
2) Arrange the following numbers in descending order:
a)\(\frac{2}{9},\frac{2}{3},\frac{8}{21}\)
Solution:
\(\frac{2}{9},\frac{2}{3},\frac{8}{21}\)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,
\(\Rightarrow \frac{2}{9}*\frac{7}{7}=\frac{14}{63}\)For the second number,
\(\Rightarrow \frac{2}{3}*\frac{21}{21}=\frac{42}{63}\)For the third number,
\(\Rightarrow \frac{8}{21}*\frac{3}{3}=\frac{24}{63}\)Descending order means arranging the numbers from largest to smallest
So,
\(\frac{14}{63}=0.22\) \(\frac{42}{63}=0.66\) \(\frac{24}{63}=0.38\)Therefore, the decreasing order of rational numbers are
\(\frac{42}{63}> \frac{24}{63}> \frac{14}{63}\)
i.e)
\(\frac{2}{3}>\frac{8}{21}> \frac{2}{9}\)
b) \(\frac{1}{5},\frac{3}{7},\frac{7}{10}\)
Solution:
\(\frac{1}{5},\frac{3}{7},\frac{7}{10}\)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,
\(\Rightarrow \frac{1}{5}*\frac{14}{14}=\frac{14}{70}\)For the second number,
\(\Rightarrow \frac{3}{7}*\frac{10}{10}=\frac{30}{70}\)For the third number,
\(\Rightarrow \frac{7}{10}*\frac{7}{7}=\frac{49}{70}\)Descending order means arranging the numbers from largest to smallest
So,
\(\frac{14}{70}=0.2\) \(\frac{30}{70}=0.42\) \(\frac{49}{70}=0.70\)Therefore, the decreasing order of rational numbers are
\(\frac{49}{70}> \frac{30}{70}> \frac{14}{70}\)
i.e)
\(\frac{7}{10}>\frac{3}{7}>\frac{1}{5}\)
3) If the sum of the numbers are same along both rows and columns, Will it form a magic square?
| \(\frac{5}{13}\) | \(\frac{7}{13}\) | \(\frac{3}{13}\) |
| \(\frac{3}{13}\) | \(\frac{5}{13}\) | \(\frac{7}{13}\) |
| \(\frac{7}{13}\) | \(\frac{3}{13}\) | \(\frac{5}{13}\) |
Solution:
- Sum of first row = \(\frac{5}{13}\)+ \(\frac{7}{13}\)+ \(\frac{3}{13}\)= \(\frac{15}{13}\)
- Sum of second row=\(\frac{3}{13}\)+ \(\frac{5}{13}\)+ \(\frac{7}{13}\)= \(\frac{15}{13}\)
- Sum of third row=\(\frac{7}{13}\)+ \(\frac{3}{13}\)+ \(\frac{5}{13}\)= \(\frac{15}{13}\)
- Sum of first column=\(\frac{5}{13}\)+ \(\frac{3}{13}\)+ \(\frac{7}{13}\)= \(\frac{15}{13}\)
- Sum of second column=\(\frac{7}{13}\)+ \(\frac{5}{13}\)+ \(\frac{3}{13}\)= \(\frac{15}{13}\)
- Sum of third column=\(\frac{3}{13}\)+ \(\frac{7}{13}\)+ \(\frac{5}{13}\)= \(\frac{15}{13}\)
- Sum of first diagonal (left to right)= \(\frac{5}{13}\)+ \(\frac{5}{13}\)+ \(\frac{5}{13}\)= \(\frac{15}{13}\)
- Sum of second diagonal (right to left)= \(\frac{3}{13}\)+ \(\frac{5}{13}\)+ \(\frac{7}{13}\)= \(\frac{15}{13}\)
- 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 \(6\frac{1}{4}\)cm and \(3\frac{2}{3}\)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= \(2*\left ( length+breadth \right )\)
Since, both the numbers are in mixed fraction, it is first converted to fractional form
For length,
\(6\frac{1}{4}\)= \(\frac{\left ( 4*6 \right )+1}{4}=\frac{25}{4}\)
For breadth.
\(3\frac{2}{3}\)= \(\frac{\left ( 3*3 \right )+2}{3}=\frac{11}{3}\)
LCM of 3,4= 12
\(\Rightarrow \frac{25}{4}*\frac{3}{3}=\frac{75}{12}\) \(\Rightarrow \frac{11}{3}*\frac{4}{4}=\frac{44}{12}\)Perimeter of rectangle= \(2*\left ( length+breadth \right )\)
= \(2*\left ( \frac{75}{12}+\frac{44}{12} \right )\)
= \(2*\left ( \frac{119}{12} \right )\)
= \(\frac{119}{6}\)cm
Area of rectangle = Length*breadth
= \(\frac{75}{12}*\frac{44}{12}=\frac{3300}{12}\)
\( \Rightarrow 275 cm^{2}\)
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 = \(\frac{5}{2}\)cm,
YZ = \(2\frac{3}{4}\) cm,
XZ=\(3\frac{3}{5}\)cm
The perimeter of triangle XYZ = XY + YZ + ZX
=(\(\frac{5}{2}\)+\(2\frac{3}{4}\)+ \(3\frac{3}{5}\))
=\(\frac{5}{2}+\frac{11}{4}+\frac{18}{5}\)
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
\(\Rightarrow \frac{5}{2}*\frac{10}{10}=\frac{50}{20}\) \(\Rightarrow \frac{11}{4}*\frac{5}{5}=\frac{55}{20}\) \(\Rightarrow \frac{18}{5}*\frac{4}{4}=\frac{72}{20}\) \(\Rightarrow \left ( \frac{50+55+72}{20} \right )\)\(\Rightarrow \frac{177}{20}\)cm
(ii) In rectangle YMNZ,
YZ = \(2\frac{3}{4}\) cm,
MN= \(\frac{7}{6}\)
W.K.T
Perimeter of rectangle = 2 (length + breadth)
= \(2\left(2\frac{3}{4}+\frac{7}{6}\right )\)
= \(2\left(\frac{11}{4}+\frac{7}{6}\right )\)
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
\(\Rightarrow \frac{11}{4}*\frac{3}{3}=\frac{33}{12}\) \(\Rightarrow \frac{7}{6}*\frac{2}{2}=\frac{14}{12}\) \( 2 \times \frac{33+14}{12} = \frac{47}{6}\)So the greatest perimeter out of this is,
(i) \(\frac{177}{20}=8.85\)cm
(ii) \(\frac{47}{6}=7.83\)cm
Comparing the perimeter of rectangle and triangle
\(\frac{177}{20}=8.85\)cm > \(\frac{47}{6}=7.83\)cm
Therefore, the perimeter of triangle XYZ is greater than that of rectangle YMNZ.
