Let the number to be find out = x
Its two third =\(\frac{2x}{3}\)
Given x - \(\frac{2}{3}\) x = \(\frac{1}{5}x+30\)
\(x-\frac { 2 }{ 3 } -\frac { 1 }{ 5 } x=30\)
\(\\ x\left( 1-\frac { 2 }{ 3 } -\frac { 1 }{ 5 } \right) =30\)
\(x\left( \frac { 15-10-3 }{ 15 } \right) =30\)
\(x\times \frac { 2 }{ 15 } =30\)
\(x=30\div \frac { 2 }{ 15 } \)
\(x=\frac { 30 }{ 1 } \times \frac { 15 }{ 2 } \)
x = 225

\(LHS=\left( \frac { \frac { 7 }{ 9 } -5 }{ \frac { 4 }{ 3 } } \right) \div \frac { 3 }{ 2 } +\frac { 4 }{ 9 } -\frac { 1 }{ 3 } =\left( \frac { \frac { 7-(5\times 9) }{ 9 } }{ \frac { 4 }{ 3 } } \right) \div \frac { 3 }{ 2 } +\frac { 4 }{ 9 } -\frac { 1 }{ 3 } \)
\(=\left( \frac { \frac { 7 }{ 9 } -5 }{ \frac { 4 }{ 3 } } \right) \div \frac { 3 }{ 2 } +\frac { 4 }{ 9 } -\frac { 1 }{ 3 } =\left( \frac { \frac { -38 }{ 9 } }{ \frac { 4 }{ 3 } } \right) \div \frac { 3 }{ 2 } +\frac { 4 }{ 9 } -\frac { 1 }{ 3 } \)
\(=\left( \frac { -38 }{ 9 } \times \frac { 3 }{ 4 } \right) \div \frac { 3 }{ 2 } +\frac { 4 }{ 9 } -\frac { 1 }{ 3 } =\frac { -19 }{ 6 } \div \frac { 3 }{ 2 } +\frac { 4 }{ 9 } -\frac { 1 }{ 3 } \)
\(=\frac { -19 }{ 6 } \times \frac { 2 }{ 3 } +\frac { 4 }{ 9 } -\frac { 1 }{ 3 } =\frac { -19 }{ 9 } +\frac { 4 }{ 9 } -\frac { 1 }{ 3 } =\frac { -19+4-(1\times 3) }{ 9 } \)
\(=\frac { -15-3 }{ 9 } =\frac { -18 }{ 9 } =-2=RHS\)

Total spending of subbu = \(\frac{1}{3}+\frac{2}{5}+\frac{1}{10}\)
\(=\frac { (1\times 10)+(2\times 6)+(1\times 3) }{ 30 } =\frac { 10+12+3 }{ 30 } =\frac { 25 }{ 30 } =\frac { 5 }{ 6 } \)
Let the total earning be 1.
Total spending \(=\frac{5}{6}\)
∴ fraction of money left with subbu \(=1-\frac { 5 }{ 6 } =\frac { (1\times 6)-(5\times 1) }{ 6 } =\frac { 6-5 }{ 6 } =\frac { 1 }{ 6 } \)

Since multiplication is commutative.
We have \(=\left( \frac { -3 }{ 8 } \times \frac { 4 }{ 5 } \right) +\left( \frac { -3 }{ 8 } \times \frac { 1 }{ 4 } \right) +\frac { 19 }{ 20 } \)
\(\\ =\left\{ \frac { -3 }{ 8 } \times \left( \frac { 4 }{ 5 } +\frac { 1 }{ 4 } \right) \right\} +\frac { 19 }{ 20 } \)
\(=\left\{ \frac { -3 }{ 8 } \times \left( \frac { (4\times 4)+(1\times 5) }{ 20 } \right) \right\} +\frac { 19 }{ 20 } \)
\(\\ =\left\{ \frac { -3 }{ 8 } \times \left( \frac { 16+5 }{ 20 } \right) \right\} +\frac { 19 }{ 20 } =\left\{ \frac { -3 }{ 8 } \times \frac { 21 }{ 20 } \right\} +\frac { 19 }{ 20 } \)
\(=\frac { -63 }{ 160 } +\frac { 19 }{ 20 } =\frac { (-63\times 1)+(19\times 8) }{ 160 } \)
\(=\frac { -63+152 }{ 160 } =\frac { 89 }{ 160 } \)
\(\therefore \frac { -3 }{ 8 } \times \frac { 4 }{ 5 } -\frac { 3 }{ 8 } \times \frac { 1 }{ 4 } +\frac { 19 }{ 20 } =\frac { 89 }{ 160 } \)

\(\frac{-3}{13} \div-3=\frac{-3}{13} \times \frac{1}{-3}=\frac{1}{13} \)

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

–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 \(\frac { -21 }{ 10 } ,\frac { -20 }{ 10 } ,\frac { -15 }{ 10 } ,\frac { -14 }{ 10 } ,\frac { -13 }{ 10 } \)

-2 and 0 
i,e. \(\frac{-2}{1}\) and \(\frac{0}{1}\)
\(\frac { -2 }{ 1 } =\frac { -2\times 10 }{ 1\times 10 } =\frac { -20 }{ 10 } \)
\(\frac { 0 }{ 1 } =\frac { 0\times 10 }{ 1\times 10 } =\frac { 0 }{ 10 } \)
∴ Five rational numbers between \(\frac{-20}{10}\) ( = -2) and \(\frac{0}{10}\)( = 0)are
\(\frac { -20 }{ 10 } ,\frac { -19 }{ 10 } ,\frac { -18 }{ 10 } ,\frac { -7 }{ 10 } ,\frac { -6 }{ 10 } ,\frac { -5 }{ 10 } ,\frac { 0 }{ 10 } (=0)\)

(4x² + 9)(3x - 2) = 12x³ - 8x² + 27x - 18

(a + b)² = a² + 2ab + b²
(3m+5)² = (3m)² + 2(3m)(5) + 5²
= 3²m² + 30m + 25 = 9m² + 30m + 25

AG = 4.5 cm, GR = 3.8 cm, ∠A = 60°, ∠G = 110° and ∠R = 90°.

steps:
1. Draw a line segment AG = 4.5 cm
2. At G on AG made ∠AGX = 110°
3. With G as centre drawn an arc of radius 3.8 cm let it cut GX at R.
4. At R on GR made ∠GRZ = 90°
5. At A on AG made ∠GAY = 90°
6. AY and RZ meet at I.
7. AGRI is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral AGRI
\( =1 / 2 \times \mathrm{d} \times\left(\mathrm{h}_{1}+\mathrm{h}_{2}\right) \text { sq. units } \)
\(=1 / 2 \times 7 \times(2.4+2.7) \)
\(=1 / 2 \times 7 \times 5.1=17.85 \text { sq. cm. }\)

Triangular Prism

\(\triangle\)BCF ≡ \(\triangle\)DEF
AC = AD (given)
\( \angle \mathrm{CBF} =\angle \mathrm{DEF}(\text { given }) \)
\(\angle \mathrm{BFC} =\angle \mathrm{EFD}(\text { vertically opposite angles }) \)
\(\angle \mathrm{BCF} =\angle \mathrm{EDF}(\text { Third angle }) \ldots . .(2)\)
From (1), (2)
\(\angle \mathrm{FCD}=\angle \mathrm{FDC}\)
CF = DF
\(\Delta \mathrm{BCF} \equiv \triangle \mathrm{EDF}(\mathrm{ASA})\) 
Hence proved.

Given in\( \triangle \mathrm{WHY} and \ \triangle \mathrm{WET}, \)
\(\angle \mathrm{W} =\angle \mathrm{W} \)
\(\angle \mathrm{WYH} =\angle \mathrm{WTE} \)
\(\angle \mathrm{WHY} =\angle \mathrm{WET} \\ \therefore \Delta \mathrm{WHY} \sim \Delta \mathrm{WET} \)
Also \(\triangle \)WHY ~ \(\triangle \)WET
∴ Corresponding sides are proportionated
\(\frac { WH }{ WE } =\frac { HY }{ ET } =\frac { WY }{ WT } \)
\(\\ \frac { 6 }{ 6+HE } =\frac { 4 }{ ET } =\frac { 4 }{ 16 }\)
\( \\ \frac { 6 }{ 6+HE } =\frac { 4 }{ 16 }\)
\( \\ 6+HE=\frac { 6 }{ 4 } \times16\)
\(\\ 6+HE=24\)
\(\\ \therefore HE=24-6\\ HE=18\)
\(\\ Again\quad \frac { 4 }{ ET } =\frac { 4 }{ 16 } \)
\(\\ ET=\frac { 4 }{ 4 } \times 16\)
\(\\ ET=16\)

Area = Area of a rectangle + Area of a triangle + Area of a trapezium
For rectangle length I = 120 - 20 - 20 cm = 80 cm
Breadth b = 30 cm
For the triangle base = 30 cm
Height = 20 cm
For the trapezium height h = 20 cm
Parallel sided a = 50 cm
b = 30cm
∴ Area of the figure = (l x b) + (\(\frac12\) x base x height) + \(\frac12\) x h x (a + b) sq. units
= (80 x 30) + (\(\frac12\) x 30 x 20) + (\(\frac12\)x 20 x (50 + 30) cm²
= 2400 + 300 + 800 cm² = 3500 cm²
Area of the figure = 3500 cm²

The tree diagram for this may be

∴ Number of possible ways to select one questions from each of 3 sections is 3 x 5 = 15 ways

The given figure is a combination of a semicircle and a triangle.
diameter = 6 cm
radius = 3 cm
base = 6 cm
height = 9 cm
The shaded area of the figure = Area of the semicircle + Area of the triangle
\(=\frac{1}{2} \pi \mathrm{r}^{2}+\frac{1}{2} \mathrm{bh} \)
\(=\frac{1}{2} \times 3.14 \times 3 \times 3+\frac{1}{2} \times 6 \times 9 \)
= 14.13 + 27 = 41.13 cm²

From this figure
Perimeter = 2l + 2 x 6 + 2r
\(=2 \times \frac{\theta}{360} \times 2 \pi \mathrm{r}+12+2 \times 3.5 \)
\(=2 \times \frac{90}{360} \times 2 \times \frac{22}{7} \times 3.5+12+7 \)
= 11+ 12 + 7
= 30 cm
The area of shaded part
= Area of two circular quadrants + Area of the rectangle
\(=\left(2 \times \frac{\theta}{360} \times \pi r^{2}\right)+(l \times b) \)
\(=\left(2 \times \frac{90}{360} \times \frac{22}{7} \times 3.5 \times 3.5\right)+(6 \times 3.5) \)
= 19. 25 + 21
= 40.25 cm²

(2x + 3) (2x - 4) 
= 4x² - 8x + 6x - 12 
= 4x² - 2x - 12

(2p²q³)\(\times\)(-9pq)² = (+)(-) \(\times\) (2\(\times\)9) (p²\(\times\) p (q³\(\times\)q²)) = -18p³q⁵

6x \(\times\) 4= (6 \(\times\) 4)(x) = 24x

The teacher is going to select one student from class VI out of 10 students in 10 ways from class VII out of 15 students in 15 ways and from class VIII out of 20 students in 20 ways.
ஃ Number of ways 3 students can be selected = 10 + 15 + 20 = 45 ways

From the figure the tile is in a circular quadrant, Then the central angle is 90^o
Area of the sector = \(\frac{\theta}{360} \times \pi r^{2}\)
\(=\frac{90}{360}\) x 3.14 x 30 x 30
  = 706.5 cm²

Length of the arc of the sector l = 50 mm
Radius r = 14mm
Area of the sector = \(\frac { lr }{ 2 } \) sq. units
= \(\frac { 50\times 14 }{ 2 } \) mm² = 50 x 7 mm² = 350 mm²
Area of the sector = 350mm²

\(\frac { -5 }{ 12 } ,\frac { -11 }{ 8 } ,\frac { -15 }{ 24 } ,\frac { -7 }{ -9 } ,\frac { 12 }{ 36 } \)
LCM of 12, 8, 24, 9, 36 is 4 x 3 x 2 x 3
= 72
\(\frac { -5 }{ 12 } =\frac { -5\times 6 }{ 12\times 6 } =\frac { -30 }{ 72 } \)
\(\frac { -11 }{ 8 } =\frac { -11\times 9 }{ 8\times 9 } =\frac { -99 }{ 72 } \)
\(\frac { -15 }{ 24 } =\frac { -15\times 3 }{ 24\times 3 } =\frac { -45 }{ 72 } \)
\(\\ \frac { -7 }{ -9 } =\frac { 7\times 9 }{ 9\times 8 } =\frac { 56 }{ 72 } \)
\(\frac { 12 }{ 36 } =\frac { 12\times 2 }{ 36\times 2 } =\frac { 24 }{ 72 } \)
Now comparing the numerators -30, -99, -45, 56, 24 we get 56 > 24 > - 30 > -45 > -99
i.e  \(\frac { 56 }{ 72 } >\frac { 24 }{ 72 } >\frac { -30 }{ 72 } >\frac { -45 }{ 72 } >\frac { -99 }{ 70 } \) and so \(\\ \\ \\ \frac { -7 }{ -9 } >\frac { 12 }{ 36 } >\frac { -5 }{ 12 } >\frac { -15 }{ 24 } >\frac { -11 }{ 8 } \)
∴ Descending order: \(\frac { -7 }{ -9 } ,\frac { 12 }{ 36 } ,\frac { -5 }{ 12 } ,\frac { -15 }{ 24 } ,\frac { -11 }{ 8 } \)
Ascending order  \(\frac { -11 }{ 8 } ,\frac { -15 }{ 24 } ,\frac { -5 }{ 12 } ,\frac { 12 }{ 36 } ,\frac { -7 }{ -9 } \)