Questions answered correctly by Jeyanth = \(\frac { 19 }{ 30 } \times 180=19\times 6=114\)
Questions answered incorrectly by Jeyanth = \(\frac { 5 }{ 18 } \times 180=50\)
∴ Number of questions not attended by Jeyanth = 180 − (114 + 50)
= 180 − 164 = 16

\(\left( \frac { 4 }{ 3 } -\left( \frac { -3 }{ 2 } \right) \right) +\left( \frac { -5 }{ 3 } \div \frac { 30 }{ 12 } \right) +\left( \frac { -12 }{ 9 } \times \frac { -27 }{ 16 } \right) =\left( \frac { 4 }{ 3 } +\frac { 3 }{ 2 } \right) +\left( \frac { -5 }{ 3 } \times \frac { 12 }{ 30 } \right) +\left( \frac { -12 }{ 9 } \times \frac { -27 }{ 16 } \right) \)
= \(\left( \frac { 8 }{ 9 } +\frac { 9 }{ 6 } \right) +\left( \frac { -1 }{ 1 } \times \frac { 4 }{ 6 } \right) +\left( \frac { -3 }{ 1 } \times \frac { -3 }{ 4 } \right) \)
= \(\left( \frac { 17 }{ 6 } \right) +\left( \frac { -4 }{ 6 } \right) +\left( \frac { 9 }{ 4 } \right) \)
= \(\left( \frac { 17-4 }{ 6 } \right) +\frac { 9 }{ 4 } =\frac { 13 }{ 6 } +\frac { 9 }{ 4 } \)
= \(\frac { 26+27 }{ 12 } =\frac { 53 }{ 12 } \)

Let the number be x .
The student had to find \(\frac { 8x }{ 9 } \) but, he had found \(\frac { x }{ \left( \frac { 8 }{ 9 } \right) } \) that is \(\frac { 9x }{ 8 } \) 
Now, \(\frac { 9x }{ 8 } -\frac { 8x }{ 9 } =34\)
\(\frac { 81x-64x }{ 72 } =34\Rightarrow \frac { 17x }{ 72 } =34\)
\(x=\frac { 34\times 72 }{ 17 } =144\)

First, make the denominator of the rational number \(\frac { 14 }{ -5 } \) to be positive as \(\frac { -14 }{ 5 } \)
Then, make the denominators the same by finding the LCM of the denominators.
So, \(\frac { -10 }{ 3 } =\frac { -10 }{ 3 } \times \frac { 5 }{ 5 } =\frac { -50 }{ 15 } \) and \(\frac { -14 }{ 5 } =\frac { -14 }{ 5 } \times \frac { 3 }{ 3 } =\frac { -42 }{ 50 } \)
As 50 > 42, we have −50 < -42
Hence, \(\frac { -50 }{ 15 } <\frac { -42 }{ 50 } \) and so \(\frac { -42 }{ 50 } >\frac { -50 }{ 15 } \). Thus, \(\frac { -14 }{ 5 } >\frac { -10 }{ 3 } \)

Now, \(\frac { 5 }{ -4 } =\frac { 5\times (-1) }{ -4\times (-1) } =\frac { -5 }{ 4 } \)
Also, \(\frac { -11 }{ -7 } =\frac { -11\times (-1) }{ -7\times (-1) } =\frac { 11 }{ 7 } \)
Here, \(\frac { 11 }{ 7 } \) is positive and \(\frac { -5 }{ 4 } \) is a negative rational number.
∴ \(\frac { 11 }{ 7 } >\frac { -5 }{ 4 } \), that is \(\frac { -11 }{ -7 } >\frac { 5 }{ -4 } \)

Method 1:
\(\frac { -18 }{ -42 } =\frac { -18\div (-2) }{ -42\div (-2) } =\frac { 9\div 3 }{ 21\div 3 } =\frac { 3 }{ 7 } \) (dividing by –2 and 3 successively)
Method 2:
The HCF of 18 and 42 is 6 (Find it). Thus, we can get its standard form by dividing it by 6.
\(\frac { -18 }{ -42 } =\frac { -18\times (-1) }{ -42\times (-1) } =\frac { 18 }{ 42 } =\frac { 18\div 6 }{ 42\div 6 } =\frac { 3 }{ 7 } \)

Method 1:
\(\frac { 48 }{ -84 } =\frac { 48\div (-2) }{ -84\div (-2) } =\frac { -24\div 2 }{ 42\div 2 } =\frac { -12\div 3 }{ 21\div 3 } =\frac { -4 }{ 7 } \) (dividing by –2, 2 and 3 successively)
Method 2:
The HCF of 48 and 84 is 12 (Find it). Thus, we can get its standard form by dividing it by  -12.
\(=\frac { 48\div (-12) }{ -84\div (-12) } =\frac { -4 }{ 7 } \)  

Capacity of the small water jug = 3\(\frac{4}{5}\)litres.
Capacity of the big jug = 2\(\frac{2}{3}\) times the small one.
\(2\frac { 2 }{ 3 } \times 3\frac { 4 }{ 5 } =\frac { 8 }{ 3 } \times \frac { 19 }{ 5 } =\frac { 152 }{ 15 } \)
= \(10\frac { 2 }{ 5 } \) litres
Capacity of the large jug = \(10\frac { 2 }{ 5 } \) litres

Let the total weight of a box of apple = x kg.
Weight of \(\frac{3}{4}\) of a box apples = 3 kg 225 gm.
= 3.225 kg
\(\frac{3}{4}\times\)x = 3225
x = \(\frac{3.225\times 4}{3}kg\)
= 1..075 x 4 kg
= 4.3 kg
= 4 kg 300 gm
Weight of the box of apples = 4 kg 300 gm

\(x=\frac { -31 }{ 45 } \)
\(-x=-\left( \frac { -31 }{ 45 } \right) \)
\(-x=\frac { 31 }{ 45 } \)
\(-(-x)=\frac { 31 }{ 45 } =x\)
\(\therefore -(-x)=x\)

The rational number for the point marked on the number line is 1\(\frac{3}{4}=\frac{7}{4}\)

The rational number for the point marked on the number line is  -3\(\frac{2}{3}\) = \(\frac{-11}{3}\)

\(\frac { 7 }{ -6 } =\frac { 7\times 2 }{ -6\times 2 } =\frac { 14 }{ -12 } \)
\(\frac { 7 }{ -6 } =\frac { 7\times 3 }{ -6\times 3 } =\frac { 21 }{ -18 } \)
\(\frac { 7 }{ -6 } =\frac { 7\times 4 }{ -6\times 4 } =\frac { 28 }{ -24 } \)
\(\frac { 7 }{ -6 } =\frac { 7\times 5 }{ -6\times 5 } =\frac { 35 }{ -30 } \)
Four equivalent rational numbers of \(\frac { -7 }{ 6 } \) are \(\frac { 14 }{ -12 } ,\frac { 21 }{ -18 } ,\frac { 28 }{ -24 } ,\frac { 35 }{ -30 } \)

\(\frac { -3 }{ 5 } =\frac { -3\times 2 }{ 5\times 2 } =\frac { -6 }{ 10 } \)
\(\frac { -3 }{ 5 } =\frac { -3\times 3 }{ 5\times 3 } =\frac { -9 }{ 15 } \)
\(\frac { -3 }{ 5 } =\frac { -3\times 4 }{ 5\times 4 } =\frac { -12 }{ 20 } \)
\(\frac { -3 }{ 5 } =\frac { -3\times 5 }{ 5\times 5 } =\frac { -15 }{ 25 } \)
Four equivalent rational numbers of \(\frac { -3 }{ 5 } \)are \(\frac { -6 }{ 10 } ,\frac { -9 }{ 15 } ,\frac { -12 }{ 20 } ,\frac { -15 }{ 25 } \)

–1.2 and -2.3
\(-1.2=\frac { -1.2\times 10 }{ 1\times 10 } =\frac { -12 }{ 10 } \)
\(-2.3=\frac { -2.3\times 10 }{ 1\times 10 } =\frac { -23 }{ 10 } \)
∴ Five rational numbers between -1.2\(\\ \\ (=\frac { -12 }{ 10 } )\) and -2.3 \((=\frac { -23 }{ 10 } )\) are \(\frac { -21 }{ 10 } ,\frac { -20 }{ 10 } ,\frac { -15 }{ 10 } ,\frac { -14 }{ 10 } ,\frac { -13 }{ 10 } \)

0.25 and 0.35
\(0.25=\frac { 0.25\times 100 }{ 1\times 100 } =\frac { 25 }{ 100 } \)
\(0.35=\frac { 0.35\times 100 }{ 1\times 100 } =\frac { 35 }{ 100 } \)
∴ Five rational numbers between 0.25\((=\frac { 25 }{ 100 } )\) and 0.35 \((=\frac { 35 }{ 100 } )\) are \(\frac { 26 }{ 100 } ,\frac { 27 }{ 100 } ,\frac { 30 }{ 100 } ,\frac { 32 }{ 100 } ,\frac { 33 }{ 100 } \)

The average of a and b is \(\frac{1}{2}(a+b)\)
The average of \(\frac{14}{5}\) and \(\frac{16}{3}\) is C₁=\(\frac { 1 }{ 2 } \left( \frac { 14 }{ 5 } +\frac { 16 }{ 3 } \right) \)
\({ C }_{ 1 }=\frac { 1 }{ 2 } \left( \frac { 42+80 }{ 15 } \right) \)
\({ C }_{ 1 }=\frac { 122 }{ 30 } \)
\({ C }_{ 1 }=\frac { 61 }{ 15 } \)
\(\\ \frac { 14 }{ 5 } <\frac { 61 }{ 15 } <\frac { 16 }{ 3 } \)   ..(1)
The average of \(\frac { 14 }{ 5 }\) and \(\frac{61}{15}\)is C₂=\(\frac { 1 }{ 2 } \left( \frac { 14 }{ 5 } +\frac { 61 }{ 15 } \right) \)
\(\frac { 1 }{ 2 } \left( \frac { 14 }{ 5 } +\frac { 61 }{ 15 } \right) \)
\({ C }_{ 2 }=\frac { 1 }{ 2 } \times \left( \frac { 42+61 }{ 15 } \right) \)
\({ C }_{ 2 }=\frac { 1 }{ 2 } \times \frac { 103 }{ 15 } \)
\(\frac { 103 }{ 15 } \)
\(\therefore \frac { 14 }{ 5 } <\frac { 103 }{ 30 } <\frac { 61 }{ 15 } \)  ..(2)
The average of \(\frac{103}{30}\) and \(\frac{61}{15}\) is C₃=\(\frac { 1 }{ 2 } \left( \frac { 103 }{ 30 } +\frac { 61 }{ 15 } \right) \)
\({ C }_{ 3 }=\frac { 1 }{ 2 } \times \left( \frac { 103+122 }{ 30 } \right) \)
\({ C }_{ 3 }=\frac { 1 }{ 2 } \times \frac { 225 }{ 30 } \)
\({ C }_{ 3 }=\frac { 1 }{ 2 } \times \frac { 15 }{ 2 } \)
\(\\ { C }_{ 3 }=\frac { 15 }{ 4 } \)
\(\therefore \frac { 103 }{ 30 } <\frac { 15 }{ 4 } <\frac { 61 }{ 15 } \)...(3)
From (1), (2) and (3) we get,       
\(\frac { 14 }{ 5 } <\frac { 103 }{ 30 } <\frac { 15 }{ 4 } <\frac { 61 }{ 15 } <\frac { 16 }{ 3 } \)