$$\eqalign{ & {\text{A : B}} \cr & \frac{3}{2}:\frac{8}{3} \cr} $$
take L.C.M. of denominator and multiply)
$$ \Rightarrow \frac{3}{2} \times 6:\frac{8}{3} \times 6 = 9x:16x$$
After adding 15 in each we get
$$\eqalign{ & \therefore \frac{{9x + 15}}{{16x + 15}} = \frac{5}{3} \times \frac{2}{5} \cr & \Rightarrow \frac{{9x + 15}}{{16x + 15}} = \frac{2}{3}\left( {{\text{cross multiply}}} \right) \cr & \Rightarrow 27x + 45 = 32x + 30 \cr & \Rightarrow 5x = 15 \cr & \Rightarrow x = 3 \cr & \therefore {\text{Smaller number}} \cr & = {\text{9}} \times {\text{3 = 27}} \cr & {\text{Greater number}} \cr & {\text{ = 16}} \times {\text{3 = 48}} \cr} $$

Let the initial fares of 1st, 2nd and 3rd class be Rs. 8x, Rs. 6x, and Rs. 3x respectively
$$\eqalign{ & {\text{Revised fare of 1st class}} \cr & {\text{ = }}\frac{5}{6}{\text{of Rs}}.8x \cr & = {\text{Rs}}.\left( {\frac{{20x}}{3}} \right) \cr & {\text{Revised fare of }}2{\text{nd class}} \cr & {\text{ = }}\frac{{11}}{{12}}{\text{of Rs}}.6x \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{11x}}{2}} \right) \cr} $$
Let the number of passengers of 1st, 2nd and 3rd class be 9y, 12y and 26y respectively
Then,
$$\eqalign{ & = \frac{{20x}}{3} \times 9y + \frac{{11x}}{2} \times 12y + 3x \times 26y = 1088 \cr & \Rightarrow 60xy + 66xy + 78xy = 1088 \cr & \Rightarrow 204xy = 1088 \cr & \Rightarrow xy = \frac{{1088}}{{204}} = \frac{{16}}{3} \cr} $$
∴ Collection from passengers of 1st class = 60xy
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {60 \times \frac{{16}}{3}} \right) \cr & = {\text{Rs}}.320 \cr} $$

$$\eqalign{ & \frac{{{m_1} \times {h_1}}}{{{w_1}}} = \frac{{{m_2} \times {h_2}}}{{{w_2}}} \cr & \Rightarrow \frac{{5 \times 5}}{5} = \frac{{10 \times 10}}{{{w_2}}} \cr & \Rightarrow {w_2} = 20 \cr} $$

$$\eqalign{ & {\text{Let }}\frac{x}{2} = \frac{y}{3} = \frac{z}{4} = l \cr & {\text{Then,}} \cr & = x = 2l,y = 3l,z = 4l \cr & \therefore \frac{x}{2} = \frac{{2x - 3y + 5z}}{k} \cr & \Rightarrow \frac{{2l}}{2} = \frac{{2 \times 2l - 3 \times 3l + 5 \times 4l}}{k} \cr & \Rightarrow k = 4 - 9 + 20 = 15 \cr} $$

The first mixture contains wine and water in the ratio of 3 : 2
Wine in 3 liters of mixture = $$\frac{3}{5} \times 3 = \frac{9}{5}$$
Water in 3 liters of mixture = $$\frac{2}{5} \times 3 = \frac{6}{5}$$

Let us consider that the first mixture is mixed with the second mixture that has quantity as 9x liters (4x liters of wine and 5x liters of water).

After mixing,
The total quantity of wine = Total quantity of water
$$\eqalign{ & \Rightarrow \frac{9}{5} + 4x = \frac{6}{5} + 5x \cr & \Rightarrow x = \frac{9}{5} - \frac{6}{5} \cr & \Rightarrow x = \frac{3}{5} \cr} $$

Second Mixture required = 9x = $$9 \times \frac{3}{5}$$ = $$5\frac{2}{5}{\text{ litres}}$$

	A	:	B	:	C
	2	:	3	:	5
Let	2x	:	3x	:	5x

C - A = 12000
5x - 2x = 12000
3x = 12000
x = 4000
Monthly salary of B is = 3x = Rs. 12000
∴ Annual salary of B is
= 12 × 1200
= Rs. 144000

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,{\text{A}}:{\text{B}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,5:3 \cr & {\text{Let }}5x:3x = 8x \cr & \Rightarrow {\text{New comers}} \cr & 5y:7y = 12y \cr & \therefore 8x + 12 = 1200 \cr & \Rightarrow 2x + 3y = 300......(i) \cr & {\text{Again,}}\frac{{5x + 5y}}{{3x + 7y}} = \frac{7}{5} \cr & 25x + 25y = 21x + 49y \cr & \Rightarrow 4x - 24y = 0 \cr & \Rightarrow 4x = 24y \cr & \Rightarrow x = 6y...........(ii) \cr & \therefore {\text{From equation }}.......(i) \cr & 12y + 3y = 300 \cr & y = 20 \cr & \therefore x = 6 \times 20 = 120 \cr} $$
The number of students initially
8x = 8 × 120 = 960

$$\eqalign{ & {\text{Let }}\frac{a}{b} = \frac{c}{d} = k \cr & {\text{Then,}} \cr & \Rightarrow a = bk,\,c = dk \cr & \therefore \frac{{ma + nc}}{{mb + nd}} \cr & = \frac{{mbk + ndk}}{{mb + nd}} \cr & = \frac{{k\left( {mb + nd} \right)}}{{\left( {mb + nd} \right)}} \cr & \Rightarrow k = \frac{a}{b} \cr & \Rightarrow a:b \cr} $$

$$\eqalign{ & {4^{3.5}}:{2^5} \cr & = {\left( {{2^2}} \right)^{3.5}}:{2^5} \cr & = {2^{2 \times 3.5}}:{2^5} \cr & = {2^7}:{2^5} \cr & = {2^2}{.2^5}:{2^5} \cr & = {2^2}:1 \cr & = 4:1 \cr} $$

Given,
A : B = 1 : 2 ------(i)
B : C = 3 : 4 ------(ii)
C : D = 6 : 9 ------(iii)
D : E = 12 : 16 ------(iv)

In equation (i) multiply with 3 and In equation (ii) multiply with 2 we get
A : B = 3 : 6 and B : C = 6 : 8
∴ A : B : C = 3 : 6 : 8 -----(v)

In equation (v) multiply with 3 and In equation (iii) multiply with 4 we get
A : B : C = 9 : 18 : 24 and C : D = 24 : 36
∴ A : B : C : D = 9 : 18 : 24 : 36
i.e. A : B : C : D = 3 : 6 : 8 : 12 -----(vi)

Now A : B : C : D = 3 : 6 : 8 : 12 and
D : E = 12 : 16
∴ A : B : C : D = 3 : 6 : 8 : 12 : 16

	A	:	B
	3	:	5
Let	3x	:	5x

L.C.M. of A & B is = 3 × 5 × x = 15x
15x = 225
Given
x = 15
∴ A = 15 × 3 = 45
B = 15 × 3 = 75
Smaller number is = 45

$$\eqalign{ & {\text{Let }}\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k \cr & {\text{Then,}} \cr & a = bk, \cr & b = ck, \cr & c = dk \cr & {\text{Also,}} \cr & \Rightarrow \frac{a}{b} \times \frac{b}{c} \times \frac{c}{d} = {k^3} \cr & \Rightarrow {k^3} = \frac{a}{{d}} \cr & \therefore \frac{{{b^3} + {c^3} + {d^3}}}{{{a^3} + {b^3} + {c^3}}} \cr & = \frac{{{b^3} + {c^3} + {d^3}}}{{{{\left( {bk} \right)}^3} + {{\left( {ck} \right)}^3} + {{\left( {dk} \right)}^3}}} \cr & = \frac{{{b^3} + {c^3} + {d^3}}}{{{k^3}\left( {{b^3} + {c^3} + {d^3}} \right)}} \cr & = \frac{1}{{{k^3}}} \cr & = \frac{d}{a} \cr} $$

$$\eqalign{ & = \frac{p}{q} = \frac{3}{4},\frac{r}{s} = \frac{8}{5},\frac{x}{y} = \frac{{10}}{6} \cr & \Rightarrow p = \frac{{3q}}{4},r = \frac{{8s}}{5},x = \frac{{5y}}{3} \cr & \therefore \frac{{psx}}{{qry}} = \frac{{\frac{{3q}}{4} \times s \times \frac{{5y}}{3}}}{{q \times \frac{{8s}}{5} \times y}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\frac{5}{4}qsy}}{{\frac{8}{5}qsy}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{4} \times \frac{5}{8} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{25}}{{32}} \cr} $$

Given, A : B = 3 : 4
B : C = 8 : 9
A : B = 3 : 4 (multiply with 2)
i.e. A : B = 6 : 8 and B : C = 8 : 9
∴ A : B : C = 6 : 8 : 9

$$\eqalign{ & \,\,\,{\text{A}}\,\,\,\,\,\,\,{\text{:}}\,\,\,\,\,\,{\text{B}}\,\,\,\,\,\,{\text{:}}\,\,\,\,\,{\text{C}} \cr & 3 \times 4:2 \times 4:2 \times 3 \cr & \,\,\,12\,\,\,\,:\,\,\,\,\,{\text{8}}\,\,\,\,\,:\,\,\,\,{\text{6}} \cr & \,\,\,\boxed{\,\,6\,\,\,\,:\,\,\,\,4\,\,\,\,\,\,:\,\,\,\,3\,\,} \cr} $$

$$\eqalign{ & {\text{Let}}\,\frac{x}{y} = \frac{y}{z} = \frac{z}{u} = k. \cr & {\text{Now, }}\frac{x}{u} = \frac{{64}}{{27}} \cr & \Rightarrow \frac{x}{y} \times \frac{y}{z} \times \frac{z}{u} = \frac{{64}}{{27}} \cr & \Rightarrow {k^3} = {\left( {\frac{4}{3}} \right)^3} \cr & \Rightarrow k = \frac{4}{3}. \cr & {\text{So}},\,x:y = y:z = z:u = 4:3. \cr & \therefore \frac{x}{z} = \frac{x}{y} \times \frac{y}{z} = \frac{4}{3} \times \frac{4}{3} = \frac{{16}}{{9.}} \cr} $$

$$\eqalign{ & p = k,\,q = 2k,\,r = 4k \cr & {\text{Then,}} \cr & \sqrt {5{p^2} + {q^2} + {r^2}} \cr & = \sqrt {5{k^2} + {{\left( {2k} \right)}^2} + {{\left( {4k} \right)}^2}} \cr & = \sqrt {5{k^2} + 4{k^2} + 16{k^2}} \cr & = \sqrt {25{k^2}} \cr & = 5k \cr & = 5p. \cr} $$

	1st	:	2nd
Fare	4x	:	x
Passengers	1	:	40

Total fare = 4x : 40x = 44x
⇒ 44x = 1100
⇒ x = $$\frac{1100}{44}$$ = 25
∴ Fare = 1^st class amount received per day
⇒ 4x = 4 × 25
⇒ 4x = 100

$$\eqalign{ & = \frac{a}{{b + c}} = \frac{1}{3} \cr & \Rightarrow a = \frac{{b + c}}{3} \cr & \Rightarrow \frac{c}{{a + b}} = \frac{5}{7} \cr & \Rightarrow 7c = 5a + 5b \cr & \Rightarrow 7c = \frac{{5\left( {b + c} \right)}}{3} + 5b \cr & \Rightarrow 7c - \frac{5}{3}c = 5b + \frac{5}{3}b \cr & \Rightarrow \frac{{16c}}{3} = \frac{{20b}}{3} \cr & \Rightarrow 16c = 20b \cr & \Rightarrow b = \frac{4}{5}c. \cr & a = \frac{{b + c}}{3} = \frac{{\frac{4}{5}c + c}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{9c}}{5} \times \frac{1}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{5}c. \cr & \therefore \frac{b}{{a + c}} = \frac{{\left( {\frac{4}{5}c} \right)}}{{\left( {\frac{3}{5}c + c} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{4c}}{5} \times \frac{5}{{8c}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1:2 \cr} $$

$$\eqalign{ & = A:B = \frac{1}{2}:\frac{1}{3} = 3:2, \cr & = {\text{B}}:{\text{C}} = \frac{1}{2}:\frac{1}{3} = 3:2 \cr & = \left( {3 \times \frac{2}{3}} \right):\left( {2 \times \frac{2}{3}} \right) \cr & = 2:\frac{4}{3}. \cr & \Rightarrow {\text{A}}:{\text{B}}:{\text{C}} = 3:2:\frac{4}{3} \cr & = 9:6:4 \cr} $$

$$\eqalign{ & {\text{let A}} = 2{\text{k}},\,{\text{B}} = 3{\text{k,}}\,{\text{C}} = 4{\text{k}} \cr & {\text{Then,}} \cr & \Rightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{{2{\text{k}}}}{{3{\text{k}}}} = \frac{2}{3}, \cr & \Rightarrow \frac{{\text{B}}}{{\text{C}}} = \frac{{3{\text{k}}}}{{4{\text{k}}}} = \frac{3}{4}, \cr & \Rightarrow \frac{{\text{C}}}{{\text{A}}} = \frac{{4{\text{k}}}}{{2{\text{k}}}} = 2 \cr & \Rightarrow \frac{{\text{A}}}{{\text{B}}}:\frac{{\text{B}}}{{\text{C}}}:\frac{{\text{C}}}{{\text{A}}} = \frac{2}{3}:\frac{3}{4}:2 \cr & = 8:9:24 \cr} $$

$$\eqalign{ & = a:b = 3:4 \cr & = b:c = 4:7 \cr & \Rightarrow a:b:c = 3:4:7 \cr & {\text{Let }}a = 3k, \cr & \,\,\,\,\,\,\,\,\,\,b = 4k, \cr & \,\,\,\,\,\,\,\,\,\,c = 7k. \cr & {\text{Then,}} \cr & \frac{{a + b + c}}{c} \cr & = \frac{{3k + 4k + 7k}}{{7k}} \cr & = \frac{{14k}}{{7k}} \cr & = 2 \cr} $$

Let A = 2k, B = 3k, C = 5k
A = x% of (B + C)
⇒ 2k = x% of (3k + 5k) = x% of 8k
$$\eqalign{ & \Rightarrow \frac{x}{{100}} = \frac{{{\text{2k}}}}{{{\text{8k}}}} = \frac{1}{4} \cr & \Rightarrow x = \frac{{100}}{4} = 25 \cr} $$

$$\eqalign{ & = \frac{x}{y} = \frac{3}{1} \cr & \Rightarrow \frac{{{x^3}}}{{{y^3}}} = \frac{{27}}{1} \cr & \Rightarrow \frac{{{x^3} - {y^3}}}{{{x^3} + {y^3}}} = \frac{{\left( {\frac{{{x^3}}}{{{y^3}}}} \right) - 1}}{{\left( {\frac{{{x^3}}}{{{y^3}}}} \right) + 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{27 - 1}}{{27 + 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{26}}{{28}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{13}}{{14}} \cr & = \left( {{x^3} - {y^3}} \right):\left( {{x^3} + {y^3}} \right) \cr & = 13:14 \cr} $$

$$\eqalign{ & a + b\sqrt 3 = \frac{1}{{2 - \sqrt 3 }} \cr & \Rightarrow \frac{1}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} \cr & \Rightarrow \frac{{2 + \sqrt 3 }}{{4 - 3}} \cr & \Rightarrow 2 + \sqrt 3 \cr} $$
By rationalisation of denominator
⇒ a + b$$\sqrt 3 $$ = 2 + $$\sqrt 3 $$
⇒ Now compare the rational & irrational parts
∴ a = 2
   b = 1
∴ a : b
   2 : 1

$$\eqalign{ & = \frac{{5a + 3b}}{{2a - 3b}} = \frac{{23}}{5} \cr & \Rightarrow 5\left( {5a + 3b} \right) = 23(2a - 3b) \cr & \Rightarrow 25a + 15b = 46a - 69b \cr & \Rightarrow 21a = 84b \cr & \Rightarrow \frac{a}{b} = 4. \cr & {\text{Hence}},\,a:b = 4:1 \cr} $$

$$\eqalign{ & = {\text{8a}} = {\text{9b}} \Rightarrow a = \frac{9}{8}b \cr & \therefore \frac{{\text{a}}}{{\text{9}}}:\frac{{\text{b}}}{{\text{8}}} = \frac{{\left( {\frac{9}{8}b} \right)}}{9}:\frac{{\text{b}}}{{\text{8}}} \cr & = \frac{{\text{b}}}{{\text{8}}}:\frac{{\text{b}}}{{\text{8}}} = 1:1 \cr} $$

$$\eqalign{ & = \frac{a}{b} = \frac{b}{c} \cr & \Rightarrow {b^2} = ac \cr & \Rightarrow \frac{{{a^4}}}{{{b^4}}} = \frac{{{a^4}}}{{{{\left( {ac} \right)}^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{a^4}}}{{{a^2}{c^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{a^2}}}{{{c^2}}} \cr & \Rightarrow {a^4}:{b^4} = {a^2}:{c^2} \cr} $$

$$\eqalign{ & = \frac{x}{y} = \frac{7}{3} \cr & = \frac{{xy + {y^2}}}{{{x^2} - {y^2}}} \cr & = \frac{{\left( {\frac{x}{y}} \right) + 1}}{{\left( {\frac{{{x^2}}}{{{y^2}}}} \right) - 1}} \cr & = \frac{{\frac{7}{3} + 1}}{{{{\left( {\frac{7}{3}} \right)}^2} - 1}} \cr & = \frac{{10}}{3} \times \frac{9}{{40}} \cr & = \frac{3}{4} \cr} $$

	A	:	B	:	C
	1	:	3	:	4
Let	100x	:	300x	:	400x
	↓+5%		↓+10%		↓+15%
Let	105x	:	330x	:	460x
	105	:	330	:	460
	21	:	66	:	92