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

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, \(\frac { 9x }{ 8 } -\frac { 8x }{ 9 } =34\)
\(\frac { 81x-64x }{ 72 } =34\Rightarrow \frac { 17x }{ 72 } =34\)
\(x=\frac { 34\times 72 }{ 17 } =144\)

Let the other number be x
Given, \(\frac { 2 }{ 15 } +x=\frac { 4 }{ 5 } \)
\(\Rightarrow x=\frac { 4 }{ 5 } -\frac { 2 }{ 15 } =\frac { 12-2 }{ 15 } =\frac { 10 }{ 15 } \)
\(\Rightarrow x=\frac { 2 }{ 3 } \)

Now, \(\frac { -12 }{ 17 } -\frac { 9 }{ 17 } =\frac { -12-9 }{ 17 } =\frac { -21 }{ 17 } \)

First, make the denominators positive and write the numbers in standard form as \(\frac { -3 }{ 5 } ,\frac { 7 }{ -10 } ,\frac { -15 }{ 20 } ,\frac { 14 }{ -30 } ,\frac { -8 }{ 15 } \) Now, the LCM of 5,10,15,20 and 30 is 60 (How?). Change the given rational numbers to their equivalent form with common denominator 60.
\(\frac { -3 }{ 5 } =\frac { -3 }{ 5 } \times \frac { 12 }{ 12 } =\frac { -36 }{ 60 } \)
\(\frac { 7 }{ -10 } =\frac { -7 }{ 10 } \times \frac { 6 }{ 6 } =\frac { -42 }{ 60 } \)
\(\frac { -15 }{ 20 } =\frac { -15 }{ 20 } \times \frac { 3 }{ 3 } =\frac { -45 }{ 60 } \)
\(\frac { -14 }{ 30 } =\frac { -14 }{ 30 } \times \frac { 2 }{ 2 } =\frac { -28 }{ 60 } \)
\(\frac { -8 }{ 15 } =\frac { -8 }{ 15 } \times \frac { 4 }{ 4 } =\frac { -32 }{ 60 } \) 
Now, comparing the numerators − 36, − 42, − 45, − 28 and – 32 we see that
− 28 > − 32 > − 36 > − 42 > − 45
That is, \(\frac { -28 }{ 60 } >\frac { -32 }{ 60 } >\frac { -36 }{ 60 } >\frac { -42 }{ 60 } >\frac { -45 }{ 60 } \)
and so, \(\frac { -14 }{ 30 } >\frac { -8 }{ 15 } >\frac { -3 }{ 5 } >\frac { 7 }{ -10 } >\frac { -15 }{ 20 } \)
Hence, the descending order of the given rational numbers is \(\frac { -14 }{ 30 } ,\frac { -8 }{ 15 } ,\frac { -3 }{ 5 } ,\frac { 7 }{ -10 } ,and\frac { -15 }{ 20 } \) and its reverse order gives the ascending order. Hence the ascending order of the given rational numbers is \(\frac { -15 }{ 20 } ,\frac { 7 }{ -10 } ,\frac { -3 }{ 5 } ,\frac { -8 }{ 15 } ,and\frac { -14 }{ 30 } \) 

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:
\(\frac { 48 }{ -84 } =\frac { 48\div (-2) }{ -84\div (-2) } =\frac { -24\div 2 }{ 42\div 2 } =\frac { -12\div 3 }{ 21\div 3 } =\frac { -4 }{ 7 } \) (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.
\(=\frac { 48\div (-12) }{ -84\div (-12) } =\frac { -4 }{ 7 } \)  

\(-5.8=\frac { -58 }{ 10 } =\frac { -29 }{ 5 } =-5\frac { 4 }{ 5 } \)

\(0.25=\frac { 25 }{ 100 } =\frac { 1 }{ 4 } \)

We know (x + a)(x + b) = x² + (a + b)x + ab
(5x + 3)(5x + 4) = (5x)² + (3x + 4)(5x) + (3)(4)
= 52x² + (7)(5x) + 12
(5x + 3)(5x + 4) = 25 x ² + 35x + 12

∠TSA = 90°
∠P = ∠A as TA = TP
∴ ∠ATS = ∠PTS
AS = PS (if sides then angles )

Proof:

Here, ΔABC ～ ΔADE (given)
∴ Their corresponding sides are proportional (by RHS similarity).
∴ \(\frac { AC }{ AE } =\frac { BC }{ DE } \)
⇒ \(\frac { 12 }{ 96 } =\frac { 54 }{ h } \)
⇒ h = \(\frac { 5\times 96 }{ 12 } \) = 40 feet
∴ The height of the tree is 40 feet.

Proof

Amount deposited by Ram
= (x × Rs.2000 + y × Rs.500 + z × Rs.100)
= Rs.( 2000x + 500y + 100z )
Amount deposited by Velan = 3xy times × Amount deposited by Ram
= 3xy × (2000x + 500y + 100z)
=(3 × 2000)(x × x × y)+(3 × 500)(x × y × y)+(3 × 100)(x × y × z)
= (6000x²y + 1500xy²+ 300xyz)

= 3x²y(2x³y³) − 3x²y(5x²y) + 3x²y(9xy)
multiplying each term of the polynomial by the monomial
= (3 × 2)(x² × x³)(y × y³)−(3 × 5)(x² × x²)(y × y) + (3 × 9)(x² × x)(y × y)
= 6x⁵y⁴−15x⁴y² + 27x3y²

Area of the rectangular painting, A = (l × b) sq.units
= (4xy³) × (3x²y)
= (4 × 3)(x × x²)(y3 × y)
A = 12x³y³ sq.units.

(i) To elect a leader
In class VIII, a math club has four members namely M, A, T, and H,
Therefore, there are 4 different ways by which they can be elected a leader.
(ii) To elect a leader and an assistant leader
In Figure, the red shaded boxes show that same member comes twice. As one person cannot have two leadership. Therefore, the red shaded boxes cannot be counted. There are only 12 different ways (either shown in yellow boxes or green boxes) to choose a leader and an assistant leader for a math club.

There are 2 different outcomes in tossing a coin namely a Head and a Tail. Both events cannot occur simultaneously.

Now, from Fig PQ = PR
⇒\(\angle\)Q = \(\angle\)R (angles opposite to equal sides are equal)
⇒ \(\angle\)x = \(\angle\)y
⇒\(\angle\)x+ \(\angle\)y + 50^o = 180^o (angle sum property in ΔPQR)
⇒ 2\(\angle\)x = 130^o
⇒\(\angle\)x = 65^o
⇒ \(\angle\)y = 65^o

Diameter, d = 7 cm, therefore r = \(\frac{7}{2}\) cm
Area of the shaded region = Area of the square – 4 x Area of the circular quadrant
\(={ a }^{ 2 }-4\times \frac { 1 }{ 4 } { \pi r }^{ 2 }\)
\(=(7\times 7)-\left( 4\times \frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \right) \)
= 49 – 38.5 = 10.5 sq.cm. (approx.)

(i) The length of the iron frame required = length of the arc + 4r
= πr + 4r
\(=\left( \frac { 22 }{ 7 } \times 49 \right) +(\times 49)\)
= 154 + 196
= 350 cm (approximately)
(ii) Area of each of the mirror sectors
\(=\frac { { \theta }^{ 0 } }{ { 360 }^{ 0 } } \times { \pi r }^{ 2 }\)
\(=\frac { { 60 }^{ 0 } }{ { 360 }^{ 0 } } \times \frac { 22 }{ 7 } \times 49\times 49\)
= 1257.67 sq.cm (approximately)

Radius, r = 7.5 cm and n = 6
Area of each of the sectors, A = \(\frac{1}{n}\times \pi r^2\) sq. units
\(\frac{1}{6}\pi \times 7.5\times 7.5\)
= 9.375 x \(\frac{22}{7}\)

Perimeter of the palm leaf fan = 43 cm
That is, l + 2r = 43
l + 2 x (10, 5) = 43
l = 43 - 21
∴ the length of the arc l = 22 cm.
Length of the arc \(l=\frac { { \theta }^{ 0 } }{ { 360 }^{ 0 } } \times 2\pi { r }\) units

\(=\frac { { 360 }^{ 0 } }{ { 3 } } \times { 120 }^{ 0 }\)
Also, area of the palm leaf fan
A = \(\frac{lr}{2}\) sq.units
\(=\frac { 22\times 10.5 }{ 2 } \)
A = 115.5cm² (approximately)

Area of the sector, \(A=\frac { { \theta }^{ 0 } }{ { 360 }^{ 0 } } \times \pi { r }^{ 2 }\) sq.units
\(=\frac { { 120 }^{ 0 } }{ { 360 }^{ 0 } } \times \pi \times 21\times 21\)
\(=147\times \frac { 22 }{ 7 } \)
A = 462 sq.cm (approximately)