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

Now, \(3\frac { 1 }{ 2 } +1\frac { 3 }{ 4 } +2\frac { 3 }{ 8 } \)
= \(\frac { 7 }{ 2 } +\frac { 7 }{ 4 } +\frac { 19 }{ 8 } \)
= \(\frac { 7\times 4+7\times 2+19\times 1 }{ 8 } =\frac { 28+14+19 }{ 8 } \)
= \(\frac { 61 }{ 8 } =7\frac { 5 }{ 8 } \) which lies between the whole numbers 7 and 8.
Now, \(\frac { 64 }{ 8 } =8\) and \(\frac { 56 }{ 8 } =7\)
Therefore, the rational number to be added to \(\frac { 61 }{ 8 } \) and get \(\frac { 64 }{ 8 } \) is \(\frac { 64 }{ 8 } -\frac { 61 }{ 8 } =\frac { 3 }{ 8 } \) and the rational number to be subtracted from \(\frac { 61 }{ 8 } \) to get \(\frac { 56 }{ 8 } \) is \(\frac { 61 }{ 8 } -\frac { 56 }{ 8 } =\frac { 5 }{ 8 } \) it becomes = 7 + 1 = 8

Number of metres of ribbon Sankari has in total
= \(18\frac { 3 }{ 4 } \times 4\frac { 1 }{ 3 } \)
= \(\frac { 75 }{ 4 } \times \frac { 13 }{ 3 } =\frac { 325 }{ 4 } =81\frac { 1 }{ 4 } m\)

\(3.0=\frac { 30 }{ 10 } =\frac { 3 }{ 1 } \)

Method 1:
LCM of 11 and 9 = 11 × 9 = 99
\(\frac { -7 }{ 11 } =\frac { -7 }{ 11 } \times \frac { 9 }{ 9 } =\frac { -63 }{ 99 } \)
\(\frac { 5 }{ -9 } =\frac { 5\times (-1) }{ 9\times (-1) } =\frac { -5 }{ 9 } \times \frac { 11 }{ 11 } =\frac { -55 }{ 99 } \)
Therefore, 6 rational numbers between \(\frac { -7 }{ 11 } \left( =\frac { -63 }{ 99 } \right) \) and \(\frac { 5 }{ -9 } \left( =\frac { -55 }{ 99 } \right) \)
\(\frac { -63 }{ 99 } \frac { -56 }{ 99 } ,\frac { -57 }{ 99 } ,\frac { -59 }{ 99 } ,\frac { 60 }{ 99 } ,\frac { -61 }{ 99 } ,\frac { -62 }{ 99 } ,\frac { -55 }{ 99 } \)
Method 2:
The average of a and b is \(\frac { 1 }{ 2 } \) (a + b)
The average of \(\frac { -7 }{ 11 } \) and \(\frac { 5 }{ -9 } \) is \({ c }_{ 1 }=\frac { 1 }{ 2 } \left( \frac { -7 }{ 11 } +\frac { -5 }{ 9 } \right) \) 
\(=\frac { 1 }{ 2 } \left( \frac { -63-55 }{ 99 } \right) =\frac { 1 }{ 2 } \left( \frac { -118 }{ 99 } \right) \)
\({ c }_{ 1 }=\frac { -59 }{ 99 } \)
\(\therefore \frac { -7 }{ 11 } <\frac { -59 }{ 99 } <\frac { -5 }{ 9 } \)          ......(1)
 The average of \(\frac { -7 }{ 11 } \) and \(\frac { -5 }{ 9 } \) is \({ c }_{ 2 }=\frac { 1 }{ 2 } \left( \frac { -7 }{ 11 } +\frac { -59 }{ 99 } \right) \)
\(=\frac { 1 }{ 2 } \left( \frac { -63-59 }{ 99 } \right) \)
\({ c }_{ 2 }=\frac { 1 }{ 2 } \times \frac { -122 }{ 99 } =\frac { -61 }{ 99 } \)
\(\therefore \frac { -7 }{ 11 } <\frac { -61 }{ 99 } <\frac { -59 }{ 99 } \)         .......(2)
The average of \(\frac { -59 }{ 99 } \) and \(\frac { -5 }{ 9 } \) is \({ c }_{ 3 }=\frac { 1 }{ 2 } \left( \frac { -5 }{ 9 } +\frac { -59 }{ 99 } \right) \)
\(=\frac { 1 }{ 2 } \left( \frac { -55-59 }{ 99 } \right) \)
\({ c }_{ 3 }=\frac { 1 }{ 2 } \left( \frac { -114 }{ 99 } \right) =\frac { -57 }{ 99 } \)
\(\therefore \frac { -59 }{ 99 } <\frac { -57 }{ 99 } <\frac { -5 }{ 9 } \)          .........(3)
Combining (1), (2) and (3), we get \(\frac { -7 }{ 11 } <\frac { -61 }{ 99 } <\frac { -59 }{ 99 } <\frac { -57 }{ 99 } <\frac { -5 }{ 9 } \). Thus we have found 3 rational numbers between \(\frac { -7 }{ 11 } \) and \(\frac { -5 }{ 9 } \) Similarly, try to find 3 more rational numbers in between \(\frac { -7 }{ 11 } \) and \(\frac { -5 }{ 9 } \) in the same way.

That is, \(\frac { -7 }{ 11 } <\frac { -61 }{ 99 } <\frac { -59 }{ 99 } <\frac { -57 }{ 99 } <\frac { -5 }{ 9 } \)