Let the present age of Smita be x years. Present age of Smita's mother is 34 years.
Age of mother of Smita two years from now
= 34 + 2 = 36 years
Age of Smita two years from now
= (x + 2) years
according to the question,
4(x + 2) = 36 \(\Rightarrow \)x+2=\(\frac { 36 }{ 4 } \)
\(\Rightarrow \) x+2=9 \(\Rightarrow \)x=9-2
\(\Rightarrow \) x=7
Hence, the present age of Smita is 7 years.

Given, \(\left( \frac { 2 }{ 9 } \right) ^{ 3 }\times \left( \frac { 2 }{ 9 } \right) ^{ 6 }=\left( \frac { 2 }{ 9 } \right) ^{ 2m-1 }\)
We know that. a^m x aⁿ = a^m+n
Let a=\(\frac { 2 }{ 9 } \)
So, \(\left( \frac { 2 }{ 9 } \right) ^{ 3 }\times \left( \frac { 2 }{ 9 } \right) ^{ 6 }=\left( \frac { 2 }{ 9 } \right) ^{ 3+6 }=\left( \frac { 2 }{ 9 } \right) ^{ 2m-1 }\)
⇒ \(\left( \frac { 2 }{ 9 } \right) ^{ 9 }=\left( \frac { 2 }{ 9 } \right) ^{ 2m-1 }\)
If am =an, then m = n
So, 9=2m-1 ⇒ 9+1 = 2m
∴ m=\(\frac { 10 }{ 2 } \)=5

Area covered by swimming pool = 30 rn x 20 m = 600 m²
Length of outer rectangle = (30 + 8 + 8) m = 46 m and its breadth = (20 + 5 + 5) m = 30 m
So, the area of outer rectangle = 46 m x 30 m = 1380 m 2
Area of cemented path = Area of outer rectangle - Area of swimming pool = 1380 - 600 m² = 780 m²
Cost of cementing 1 m2 path = Rs 200
So, total cost of cementing the path = Rs 780 x 200 = Rs 156000.

Number of lines of symmetry for the given figures arc as follows:

Given, perimeter of a rectangular sheet = 100 cm and length of a rectangular sheet, I = 35 cm
Let b be the breadth of a rectangular sheet .
We know that, perimeter of a rectangular sheet = 2 (/ + b)
\(\Rightarrow\)  2(/ + b) = 100  \(\Rightarrow\) 2(35 + b) = 100
\(\Rightarrow\) \(35+b=\frac{100}{2}\)    [dividing by 2 on both sides]
\(\Rightarrow\) 35 + b = 50  \(\Rightarrow\) b = 50 - 35  \(\Rightarrow\) b = 15 cm
\(\therefore\) Area of a rectangular sheet = I X b = 35 x 15 = 525 cm²
Hence, the breadth and area of a rectangular sheet are 15 cm and 525 cm², respectively.

\(\frac { 137 }{ 120 } \) 

\(-\frac{1}{3}=\frac{-1\times5}{3\times5}=-\frac{5}{15}\)
\(-\frac{1}{5}=\frac{-1\times3}{5\times3}=\frac{-3}{15}\)
\(\because \frac{-3}{15}\) is to right of \(\frac{-5}{15}\)
\(\therefore \frac{-3}{15}>\frac{-5}{15}\)
\(\Rightarrow-\frac{1}{5}>(-\frac{1}{3})\ or\ -\frac{1}{3}<(-\frac{1}{5})\)

(i) Here, Base (b) = 4 cm and Height (h) = 3 cm
\(\therefore\) Area of triangle ABC=\(\frac{1}{2}\) \(\times\)b\(\times\)h=\(\frac{1}{2}\)\(\times\)4\(\times\)3 cm² = 6 cm²

8 x 10⁴ + 6 x 10³ + 0 x 10² + 4 x 10¹ + 5 x 10⁰
= 8 x 10000 + 6 x 1000 + 0 x100 + 4 x 10 + 5 x 1
= 86045                                                   [\(\because\)10⁰ = 1]

Rs 720.

3² x5⁹ x7^-1

Let the age of father be x and the age of son be y. So, the algebraic expression will be 3y = x

154 cm²

6.25 cm², 10 cm 

\(\frac { 75 }{ 182 } \)

\(\left( \frac { { a }^{ 4 } }{ { a }^{ 2 } } \right) \times { a }^{ 3 }\)=a^4-2 x a³ =a² x a³

Area of \(\triangle ABC\) =\(\frac { 3\times 4 }{ 2 } \)
= \(\frac { 5\times AD }{ 2 } \)⇒ AD = 2.4 cm

Area of one tile (parallelogram shape)
= base x height
= 3 x 4.5
= 13.5 cm²
Area of such 2760 tiles = 2760 x 13.5
= 37,260 cm²
= 3.726 m²
Cost of polishing = 3.726 x 20
= Rs. 74.52.

Given, \(\frac { \left( -\frac { 3 }{ 4 } \right) ^{ 4 }\times \left( \frac { 125 }{ 27 } \right) }{ \left( \frac { 5 }{ 3 } \right) ^{ 2 }\times \left( \frac { 9 }{ 16 } \right) } \)
∵ \(\frac { 125 }{ 27 } =\frac { 5\times 5\times 5 }{ 3\times 3\times 3 } =\frac { 5^{ 3 } }{ 3^{ 3 } } \)
and \(\frac { 9 }{ 16 } =\frac { (-3)\times (-3) }{ 4\times 4 } =\frac { (-3)^{ 2 } }{ { 4 }^{ 2 } } \)
So, \(\frac { \left( -\frac { 3 }{ 4 } \right) ^{ 4 }\times \frac { { 5 }^{ 3 } }{ { 3 }^{ 3 } } }{ \left( \frac { 5 }{ 3 } \right) ^{ 2 }\times \frac { (-3)^{ 2 } }{ { 4 }^{ 2 } } } =\frac { \left( -\frac { 3 }{ 4 } \right) ^{ 4 }\times \left( \frac { 5 }{ 3 } \right) ^{ 3 } }{ \left( \frac { 5 }{ 3 } \right) ^{ 2 }\times \left( -\frac { 3 }{ 4 } \right) ^{ 2 } } \)      \(\left[ \because \frac { { a }^{ n } }{ b^{ n } } =\left( \frac { a }{ b } \right) ^{ n } \right] \)
= \(\left( \frac { -3 }{ 4 } \right) ^{ 4-2 }\times \left( \frac { 5 }{ 3 } \right) ^{ 3-2 }\)     [∵ a^m ÷ aⁿ = a^m-n]
=\(\left( \frac { -3 }{ 4 } \right) ^{ 2 }\times \left( \frac { 5 }{ 3 } \right) ^{ 1 }\)
= \(\frac { (-3)\times (-3) }{ 4\times 4 } \times \frac { 5 }{ 3 } =\frac { 9 }{ 16 } \times \frac { 5 }{ 3 } =\frac { 3\times 5 }{ 16 } =\frac { 15 }{ 16 } \).

96 m²

Length of paper = 324 cm
Breadth of paper = 172 cm
Total, area of paper for envelopes = 324x 172 cm²
If each envelope requires a piece of paper size 18 cm by 12 cm.
So, length of each envelope = 18 cm and breadth of each envelope = 12 cm
Area of paper required for each envelope
= 18 x 12 cm².
\(\therefore\) Number of envelopes = \(\frac{Total\ areas\ of\ paper}{Area\ of\ paper\ required\ for\ one\ envelope}\)
= \(\frac{324\times 172}{18\times 12}\) = 258.

Steps of construction
(i) Draw a line AC= 4 cm.
(ii) Construct \(\angle XAC=75^0\) at A.
(iii) Also, construct \(\angle YCA=60^0\)at C.
(iv) Ray X and ray Y cut each other at point B.

Hence, \(\triangle ABC\) is the required triangle, having \(\angle A=75^0,\angle C=60^0\) and AC = 4 cm.

Steps of construction
(i) Draw a line AB= 5 cm and mark a point C outside it.
(ii) Mark another point D on AB, join CD.
(iii) With point D as centre, draw an arc cutting AB at E and also, CD at F.
(iv) With point C as centre and radius equal to DE, draw another arc cutting CD at M.
(v) With point M as centre and radius equal to EF, mark a point P on the previous arc.
(vi) Now, join PC to draw a line n.
Hence, line n is required line parallel to AB.

Substituting a = 3 and b = 2 in
(a) a + b, we get
a + b = 3 + 2 = 5
(b) 7a - 4b,  we get
7a - 4b = 7 x 3 - 4 x 2
= 21 - 8 = 13
(c) a² + 2ab + b², we get
a² + 2ab + b² = 3² + 2 x 3 x 2 + 2²
= 9 + 2 x 6 + 4
= 9 + 12 + 4 = 25.
(d) a3 - b3, we get
a³ - b³ = 3³ - 2³ = 3 x 3 x 3 - 2 x 2 x 2
= 27 -8 = 19.

(a) \(\frac{12^4\times 9^3\times 4}{6^3\times 8^2\times 27} =\frac{(3\times 2^2)^4\times (3^2)^3\times 2^2}{(2\times 3)^3\times (2^3)^2\times 3^3}\)
\(=\frac{3^4\times 2^8\times 3^6\times 2^2}{2^3\times3^3\times2^6\times3^3}\)
\(=\frac{2^{8+2}\times 3^{6+4}}{2^{6+3}\times 3^{3+3}}\)
\(=\frac{2^{10}\times 3^{10}}{2^9\times 3^6}=2^{10-9}\times 3^{10-6}\)
= 2 x 3⁴ = 2 x 81 = 162.
(b) 2³ x a³ x 5a⁴
= 8 x a³ x 5 x a⁴
= 8 x 5 x a³ x a⁴
= 40 x a³ + 4
= 40 x a⁷
= 40a⁷.