\(\frac{10\times 5^{n+1}+25\times 5^n}{3\times 5^{n+2}+10\times 5^{n+1}}\)
\(=\frac{2\times 5\times 5^n\times 5^1+5\times5\times5^n}{3\times5^n\times5^2+2\times5\times5^1\times5^n}\)  
\(=\frac{2\times5\times5^n\times5+5\times5\times5^n}{3\times5^n\times5\times5+2\times5\times5\times5^n}\)
\(=\frac{5\times5\times5^n\times2+5\times5\times5^n}{5\times5\times5^n\times3+5\times5\times5^n\times2}\)
\(=\frac{5\times5\times5^n(2+1)}{5\times5\times5^n(3+2)}=\frac{(2+1)}{(3+2)}\)
\(=\frac{3}{5}\)

Given, ratio 3 : 2
For 3 : 2 of 350,
Sum of ratio = 3 + 2 = 5
So, \(\frac { 3 }{ 5 } \) of 350 = \(\frac { 3 }{ 5 } \)x 350 = 3 x 70 = 210
and \(\frac { 2 }{ 5 } \) of 350 = \(\frac { 2 }{ 5 } \) x 350 = 2 x 70 = 140
∴  3 : 2 of 350 is 210, 140.

Given, one side of a square is 18x - 20.
Adjacent side is 42 -13x.
We know that, all the sides of a square are always equal. So, 18x - 20 = 42 -13x
              18x + 13x = 42 + 20
                       31x = 62
\(\Rightarrow\)            \(x=\frac{62}{31}\) = 2 units
Hence, side of the square is (18 x 2) - 20 = 36 - 20 = 16 units.

(i) The three pairs of equal parts in ΔCBD and ΔBCE are
ㄥBDC = ㄥBEC = 90° [given]
CB = BC    [common]
BD = CE      [given]
(ii) Yes, In ΔCBD and ΔBCE, we have
ㄥBDC = ㄥBEC = 90° [given]
CB = BC [common]
BD = CE [given]
Therefore, by RHS congruence rule, two triangles are congruent. The correspondence is B H C, E HD,
C ↔️ B.
In symbolic form, ΔCBD ≅ ΔBCE.
(iii) Yes, here ΔCBD ≅ ΔBCE
We know that, the corresponding parts of two congruent triangles are equal.
Therefore, ㄥDCB = ㄥEBC.

Rs.1600

JK=22.9 ft, mㄥJLM =34⁰

3

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]
=\(\left( \frac { -3 }{ 4 } \right) ^{ 2 }\times \left( \frac { 5 }{ 3 } \right) ^{ 1 }\)
= \(\frac { (-3)\times (-3) }{ 4\times 4 } \times \frac { 5 }{ 3 } =\frac { 9 }{ 16 } \times \frac { 5 }{ 3 } =\frac { 3\times 5 }{ 16 } =\frac { 15 }{ 16 } \).

Given, \({ \left( \frac { 1 }{ 5 } \right) }^{ 5 }\times { \left( \frac { 1 }{ 5 } \right) }^{ 19 }={ \left( \frac { 1 }{ 5 } \right) }^{ 8x }\)
\(\Rightarrow ​​{ \left( \frac { 1 }{ 5 } \right) }^{ 5+19 }={ \left( \frac { 1 }{ 5 } \right) }^{ 8x }\)      [\(\because\)a^m x aⁿ = a^m+n]
\(\Rightarrow ​​{ \left( \frac { 1 }{ 5 } \right) }^{ 24 }={ \left( \frac { 1 }{ 5 } \right) }^{ 8x }\)
Since, bases are equal, by equating their exponents, we get
8x = 24
\(\therefore\)  x = 24/8 = 3

As per the given information in question,
cost price of per table = Rs 450
Number of tables = 24
So, cost price of 24 tables = 24 x 450 = Rs 10800
Selling price of per table = Rs 600
Number of tables sold at rate Rs  600 =16
Selling price of 16 tables = 16 x 600 =Rs 9600
\(\therefore\) Remaining tables = 24 -16 = 8
\(\because\) 8 tables sold at Rs 400.
Selling price for 8 tables = 8 x 400 = Rs 3200
Total selling price = 9600 + 3200 = Rs12800
\(\therefore\) Profit or Gain = Rs12800 - Rs10800 = Rs2000
Now, Gain. percentage=\({Gain \over Total \ cost \ price}\times 100\)
\(={2000\over10800}\times100={2000\over108}=18.51\%\)
Hence, his gain is 18.51%.

(i) In ΔABC and ΔDCB, ㄥABC = ㄥDCB [70° each]
BC = BC            [common]
ㄥACB= ㄥDBC     [30° each]
∴ ΔABC ≅ ΔDCB [by ASA congruence rule]
(ii) ΔABC ≅ ΔDCB  ⇒ AB = DC
(iii) ΔABC ≅ ΔDCB ⇒ AC = BD

As per the given information in the question, the total number of students is 1070.
Let x be the number of boys in the school.
So, the number of girls in the school will be x + 50.
Then, x + (x + 50) = 1070
   2x + 50 = 1070 \(\Rightarrow\) 2x = 1070 - 50
           2x =1020 \(\Rightarrow\) x = 510
Hence, the number of boys in the school is 510.
So, the number of girls in the school will be
          = 510 + 50 = 560.