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<P>Let (4/5)P = (3/2)Q = (2/3)R = k

P = 5k/4, Q = 2k/3, R = 3k/2

P : Q : R = 5k/4 : 2k/3 : 3k/2

Multiplying ratio by 12/k

P : Q : R = 15 : 8 : 18

Let the price of the TICKET be y

And the number of visitors be z

Then if price of ticket is INCREASED in RATIO 3 : 5 the visitors DECREASED in ratio 7 : 2

i.e. when ticket price = 3y/8

then number of visitors = 7z/9

So total collection before = (3y/8) × (7z/9) = 7yz/24

Now after increasing price tickets

Ticket price = 5y/8

⇒ Number of visitors = 2z/9

⇒ Total collection after = (5y/8) × (2z/9) = 5yz/36

So, it is clear that the collection after has decreased.

Ratio = (7yz/24) : (5yz/36) = 21 : 10

Let two number be 5x and 6X

Product = 5x × 6x = 30x²

According to the question,

30x² = 1470

⇒ X² = 49

⇒ x = 7

Now,

Let the present age of two friends be A and B

A + B = 20 and (A – 5) (B – 5) = 24

AB – 5(A + B) + 25 = 24

AB – 75 = 24

AB = 99

(A – B) ² = (A + B) ² – 4AB

A – B = 2

According to the GIVEN information,

LCM of 5, 13 and 7 is 455.

On dividing the given equation by 455, we get,

5A/455 = 13B/455 = 7C/455

⇒ A/91 = B/35 = C/65

⇒ A/B = 91/35

⇒ B/C = 35/65

∴ A : B : C = 91 : 35 : 65

Hence (a) is correct

Rule:

If (a/b) = (c/d), then,

Componendo-

→ (a + b)/b = (c + d)/d

Divinendo-

→ (a - b)/b = (c - d)/d

Now, given that m/n = 9/7, then according to the rule-

⇒ (m - n)/n = (9 - 7)/7

= 2/7

Now, cube of 2/7

= (2 × 2 × 2)/(7 × 7 × 7)

= 8/343 = 0.02

Hence, the REQUIRED answer is 0.02.

 Given 7x - 15y = 4x + y ⇒ 3X = 16y ⇒ $(\frac{{\RM{x}}}{{\rm{y}}} = \frac{{16}}{3})$

Squaring both sides, we get

$(\begin{array}{l} \Rightarrow {\rm{\;}}\frac{{{{\rm{x}}^2}}}{{{{\rm{y}}^2}}} = \frac{{256}}{9}\\ \Rightarrow {\rm{\;}}\frac{{3{{\rm{x}}^2}}}{{2{{\rm{y}}^2}}} = \frac{3}{2} \TIMES \frac{{256}}{9}\\ \Rightarrow {\rm{\;}}\frac{{3{{\rm{x}}^2}}}{{2{{\rm{y}}^2}}} = \frac{{128}}{3} \END{array})$

$(\Rightarrow {\rm{\;}}\frac{{3{{\rm{x}}^2}{\rm{\;}} + {\rm{\;}}2{{\rm{y}}^2}}}{{3{{\rm{x}}^2}\; -\;2{{\rm{y}}^2}}} = \frac{{128{\rm{\;}} + {\rm{\;}}3}}{{128\; -\; 3}})$ (By componendo and dividendo)

$(\Rightarrow {\rm{\;}}\frac{{3{{\rm{x}}^2}{\rm{\;}} + {\rm{\;}}2{{\rm{y}}^2}}}{{3{{\rm{x}}^2}\; - \;2{{\rm{y}}^2}}} = \frac{{131}}{{125}})$

Let present age of Mahesh is X years

Let present age of Ajay is y years

⇒ x : y = 3 : 2

⇒ 3Y = 2x---(i)

After 8 years

⇒ (x + 8) : (y + 8) = 11 : 8

⇒ 8X + 64 = 11y + 88

⇒ 8x = 11y + 24---(ii)

From eq (i) and (ii), we get,

⇒ y = 24

⇒ x = 36

Let the NUMBER of marks scored by SANJAY in ENGLISH be ‘7x’

And the number of marks scored by Sanjay in Hindi be ‘11X’

According to question,

11x - 7x = 20

⇒ x = 5

∴ The number of marks scored by Sanjay in English = 7x = 7 × 5 = 35

The third proportional to P and Q is 24.

⇒ P/Q = Q/24

⇒ Q²/P = 24

the FOURTH proportional to P, Q and R is 48.

⇒ P/Q = R/48

⇒ QR/P = 48

Dividing the two EQUATIONS, we get Q/R = 24/48 = 1 : 2

LET the INCOMES of X and Y be 7A and 8A

Let the expenditures of X and Y be 4b and 5B.

Then the income of both will be,

⇒ 7a = 4b + 960----(1)

⇒ 8a = 5b + 960----(2)

By solving the above equation, we get

a = 320 and b = 320

Suppose total books in two racks = 270

Since the ratio of CAPACITY of racks is 1 ? 2;

∴ Number of books in SMALLER rack = 1/3 × 270 = 90

Number of books in larger rack = 2/3 × 270 = 180

Since the ratio of total Engineering and Finance books in two racks together is 2 ? 3;

∴ Number of Engineering books in two racks together = 2/5 × 270 = 108

Number of Finance books in two racks together = 3/5 × 270 = 162

Since the number of books in the smaller rack is 90 and the ratio of total Engineering and Finance books in smaller rack is 4 ? 5;

∴ Number of Engineering books in smaller rack = 4/9 × 90 = 40

Number of Finance books in smaller rack = 5/9 × 90 = 50

∴ Number of Engineering books in larger rack = 108 - 40 = 68

Number of Finance books in larger rack = 162 - 50 = 112

Let x and y be 3a and 5a

$( \RIGHTARROW \frac{{2x + 5y}}{{2Y - 3x}} = \frac{{2 \times 3a + 5 \times 5a}}{{2 \times 5a - 3 \times 3a}})$

$( \Rightarrow \frac{{2x + 5y}}{{2y - 3x}} = \frac{{6a + 25A}}{{10a - 9a}} = \frac{{31a}}{{1a}} = \;\frac{{31}}{1})$

LET the acid and water CONTENT in the mixture be 2X and 5x respectively

New QUANTITY of acid = 2x + 5

Total quantity of mixture = (2x + 5) + 5x = 7x + 5

(2x + 5)/5x = ½

x = 10

Total final mixture = 7x + 5 = 7 × 10 + 5 = 75 litres

Given, ratio of father’s age to his SON’s age is 8 : 5.

Let their ages be 8x and 5x respectively.

Given, PRODUCT of their age is 1440.

∴ 8x × 5x = 1440

⇒ x² = 36

⇒ x = 6

Father’s present age = 8 × 6 = 48

Son’s present age = 5 × 6 = 30

As the boy has the maximum number of 50 notes POSSIBLE, HENCE, we can write,

⇒ 340 = (50 × 6) + (10 × 4)

⇒ No. of 50 RUPEE notes = 6

⇒ No. of 10 rupee notes = 4

Let US assume NUMBER of boys and girls in the school is 4x and 3x respectively.

Total number of student = 4x + 3x = 7x = 945 (given)

⇒ x = 945/7 = 135

Number of boys = 4x = 4 × 135 = 540

Number of girls = 3x = 3 × 135 = 405

Let us assume fee of 1 BOY is Rs. Y

Total Fees of 540 boys = Y × 540 = Rs. 540Y

Fees of 1 girl = 80% of Y = 0.8Y

Fees of 405 girls = 0.8Y × 405 = Rs. 324Y

Sum of fees of boys and girls = 540Y + 324Y = 864Y = Rs. 535680

⇒ Y = 535680/864 = 620

Given, Supriya’s share = 10 × Sayathri’s share

⇒ (Supriya’s share) ? (Sayathri’s share) = 10 ? 1

Now, if the whole SUM of Rs. 10000 was divided between Supriya and Sayathri, then,

⇒ Sayathri’s share = 1/11 × 10000 = Rs. 909.09

P/Q = 3/4

⇒ P/Q + 1 = 3/4 + 1 (ADDED 1 both side)

⇒ (P + Q)/Q = 7/4 ----(i)

Q/R = 2/5

⇒ R/Q = 5/2

⇒ R/Q + 1 = 5/2 + 1

⇒ (R + Q)/Q = 7/2 ----(II)

Then equation (i) ÷ (ii)

Let the present age of Raman is A and the present age of Mohan is B.

(A – 5) = (B­­­­­­ – 5) × 3

A – 3B = -10----(1)

(A + 10) = (B + 10) × 2

A – 2B = 10----(2)

Solving equation (1) and (2)

A = 50 and B = 20

⇒ Ratio = 1/2 : 1/3 : 1/4

Multiplying the ratios by LCM of (2, 3, 4) = 12

⇒ Ratio = 6 : 4 : 3

Let the Ratio Factor be x

 ⇒ A = 6x, B = 4X, C = 3X

⇒ 6x + 4x + 3x = 3900

⇒ 13x = 3900

⇒ x = 300

⇒ A = 6x = 1800

⇒ B = 4x = 1200

⇒ C = 3x = 900

FRESH fruits are of 300 kg

Fresh FRUIT contains 75% of water

∴ Fresh fruit contains 25% of pulp

⇒ Pulp in 300 kg fresh fruit = 75 kg

Let the QUANTITY of dry fruits obtained be x

⇒ (100 – 20)% of x = 75

⇒ 80/100 × x = 75

⇒ x = 93.75

Given, POPULATION of a TOWN is 10,000. If the males increases by 5% and the females by 6%, the population will be 10540.

Let the NUMBER of males be ‘a’

Number of females = 10000 – a

Now, 5% of a + 6% of (10000 – a) = 10540 – 10000

⇒ 0.05a + 600 – 0.06a = 540

⇒ 0.01a = 60

⇒ a = 6000

Number of females = 10000 – 6000 = 4000

Let the AMOUNTS of silver, gold and platinum in the first ALLOY be 4x, 7x and 3x.

Total amount used = 14x.

Let the amounts of iron, gold and platinum in the second alloy be 8y, 9y and 11y.

Total amount used = 28Y

Since equal amounts are mixed, 14x = 28y ⇒ x = 2y

Total alloy = 28y + 28y = 56y = 1 KG. ⇒ y = 1/56 Kg

Amount of gold = 7x + 9y = 14y + 9y = 23y = 23/56 Kg.

 Let the REQUIRED NUMBER be X.

Let, x be the mean proportional

As we KNOW that, mean proportional = √a × b

Where a and b are OUTER elements of proportion

14 : X = X : 15

X² = 14 × 15

X = 14.5

So, the mean proportional between 14 and 15 = 14.5 Ans.

Let the FRACTION be X.

$(\therefore \frac{{\RM{X}}}{{\frac{1}{{15}}}} = \frac{{\frac{5}{6}}}{{\frac{2}{3}}})$

$(\Rightarrow 15{\rm{X}} = \frac{{15}}{{12}})$

As we know, the COMPOUND ratio of the ratios (a ? b) and (C ? d) is (ac ? bd)

LET a = (p + 1), b = (p – 1), c = (p + 2), d = (p – 2)

Given, compound ratio = p/2 + 1 = (p + 2)/2

⇒ [(p + 1).(p + 2)]/[(p – 1).(p – 2)] = (p + 2)/2

⇒ (p + 1)/(p² – 3p + 2) = 1/2

⇒ 2p + 2 = p² – 3p + 2

⇒ p² – 5p = 0

⇒ p(p – 5) = 0

⇒ p = 0, 5

? p ≠ 0

LET the TWO numbers be x and y

xy = 108 ⇒ x = 108/y

x : y = y : 16

x × 16 = y²

108/y × 16 = y²

y³ = 108 × 16

⇒ y = 12

x = 108/12 = 9

 Given $(\frac{{\rm{x}}}{1} = \frac{{\sqrt {{\rm{a\;}} + {\rm{\;}}3{\rm{b}}} {\rm{\;}} + {\rm{\;}}\sqrt {{\rm{a}} - 3{\rm{b}}} }}{{\sqrt {{\rm{a\;}} + {\rm{\;}}3{\rm{b}}} - \sqrt {{\rm{a}} - 3{\rm{b}}} }})$

$(\Rightarrow {\rm{\;}}\frac{{{\rm{x\;}} + {\rm{\;}}1}}{{{\rm{x}} - 1}} = \frac{{(\sqrt {{\rm{a\;}} + {\rm{\;}}3{\rm{b}}} {\rm{\;}} + {\rm{\;}}\sqrt {{\rm{a}} - 3{\rm{b}})} {\rm{\;}} + {\rm{\;}}(\sqrt {{\rm{a\;}} + {\rm{\;}}3{\rm{b}}} - \sqrt {{\rm{a}} - 3{\rm{b}})} }}{{(\sqrt {{\rm{a\;}} + {\rm{\;}}3{\rm{b}}} {\rm{\;}} + {\rm{\;}}\sqrt {{\rm{a}} - 3{\rm{b}})} - (\sqrt {{\rm{a\;}} + {\rm{\;}}3{\rm{b}}} - \sqrt {{\rm{a}} - 3{\rm{b}})} }})$ (By componendo and dividendo)

$(\Rightarrow {\rm{\;}}\frac{{{\rm{x\;}} + {\rm{\;}}1}}{{{\rm{x}} - 1}} = \frac{{2\sqrt {{\rm{a\;}} + {\rm{\;}}3{\rm{b}}} }}{{2\sqrt {{\rm{a}} - 3{\rm{b}})} }})$

$(\Rightarrow {\rm{\;}}\frac{{{{\left( {{\rm{x\;}} + {\rm{\;}}1} \right)}^2}}}{{{{\left( {{\rm{x}} - 1} \right)}^2}}} = \frac{{{\rm{a\;}} + {\rm{\;}}3{\rm{b}}}}{{{\rm{a}} - 3{\rm{b}}}})$ (On squaring both SIDES)

$(\Rightarrow {\rm{\;}}\frac{{{{\left( {{\rm{x\;}} + {\rm{\;}}1} \right)}^2}{\rm{\;}} + {\rm{\;}}{{\left( {{\rm{x}} - 1} \right)}^2}}}{{{{\left( {{\rm{x\;}} + {\rm{\;}}1} \right)}^2} - {{\left( {{\rm{x}} - 1} \right)}^2}}} = \frac{{\left( {{\rm{a\;}} + {\rm{\;}}3{\rm{b}}} \right){\rm{\;}} + {\rm{\;}}\left( {{\rm{a}} - 3{\rm{b}}} \right)}}{{\left( {{\rm{a\;}} + {\rm{\;}}3{\rm{b}}} \right) - \left( {{\rm{a}} - 3{\rm{b}}} \right)}})$ (By componendo and dividendo)

$(\Rightarrow {\rm{\;}}\frac{{2{{\rm{x}}^2}{\rm{\;}} + {\rm{\;}}2}}{{4{\rm{x}}}} = \frac{{2{\rm{a}}}}{{6{\rm{b}}}}{\rm{\;}})$

$(\Rightarrow {\rm{\;}}\frac{{{{\rm{x}}^2}{\rm{\;}} + {\rm{\;}}1}}{{2{\rm{x}}}} = \frac{{\rm{a}}}{{3{\rm{b}}}})$

$(\Rightarrow {\rm{\;}}3{\rm{b}}{{\rm{x}}^2}{\rm{\;}} + {\rm{\;}}3{\rm{b}} = 2{\rm{ax}})$

$(\Rightarrow {\rm{\;}}3{\rm{b}}{{\rm{x}}^2} - 2{\rm{ax\;}} + {\rm{\;}}3{\rm{b}} = 0.)$

Ratio of selected to unselected applicants = 19 ? 17

Let’s take selected candidates be 19x, then unselected candidates be 17x, so total candidates is 36x

According to given CONDITIONS,

No. of job applicants = 36x - 1200 and selected candidates = 19x - 800,

⇒ Selected candidates/unselected candidates = (19x - 800)/(36x - 1200 - 19x + 800) = 1/1

⇒ 19x - 800 = 17x - 400

⇒ 2x = 400

⇒ X = 200

<P>P/Q = 4

(P - Q)/(P + Q) = (4Q – Q)/(4Q + Q) = 3/5 

$(\Rightarrow {\rm{}}\frac{{2A}}{3} = \frac{{3B}}{{10}} = \frac{C}{2} = k)$ (k is a constant)

$(\Rightarrow A = \frac{{3{\rm{K}}}}{2}{\rm{}},{\rm{B}} = \frac{{10{\rm{K}}}}{3},{\rm{C}} = 2{\rm{K}})$

⇒ A : B : C = $(\frac{{3{\rm{K}}}}{2}:\frac{{10K}}{3}:2K)$

⇒ A : B : C = $(\frac{{3{\rm{K}}}}{2}:\frac{{10K}}{3}:2K)$

Let share of B be x

⇒ A’s share = 3x/4

⇒ C’s share $( = \frac{2}{5} \times \frac{{3x}}{4} = \frac{{3x}}{{10}})$

 $(x + \frac{{3x}}{4} + \frac{{3x}}{{10}} = 3444)$

$(\frac{{20x + 15x + 6x}}{{20}} = 3444)$ 

⇒ $(41x = 68880)$

⇒ x = 68880/41 = RS 1680

∴ B’s share = Rs 1680

0.02% of 150% of 600 = 600 × 1.5 × 0.0002 = 0.18

Let THIRD PROPORTIONAL be a.

Then,

⇒ $(\FRAC{9}{{36}} = \frac{{36}}{a})$

⇒ a = 36²/9

⇒ a = 144

Let the PAY of Raj, Rahul's and Sohan be RS. x, y and Z resp.

⇒ x/y = 3/4

⇒ x = Rs. 1500

⇒ 1500/y = 3/4

⇒ y = 1500 × (4/3) = Rs. 2000

Also, the ratio of Rahul's pay to Sohan's pay is 2 ? 9

⇒ y/z = 2/9

⇒ y = Rs. 2000

⇒ 2000/z = 2/9

⇒ Z = 2000 × (9/2) = Rs. 9000

∴ Sohan is PAID Rs. 9000

Total AMOUNT collected = Rs 5400

The notes are in the ratio = 20 ? 15 ? 8

⇒ Let the no. of notes be 20x, 15X, 8X respectively for 10, 20 and 50 rupees notes.

The no. of notes is MULTIPLIED by the corresponding amount to find the CORRECT amount by each denomination.

⇒ Total amount by 10-rupee notes = 10 × 20x = 200x----(1)

⇒ Total amount by 20-rupee notes = 20 × 15x = 300x----(2)

⇒ Total amount by 50-rupee notes = 50 × 8x = 400x----(3)

⇒ 200x + 300x + 400x = 5400(From (1), (2) and (3))

⇒ 900x = 5400

⇒ x = 5400/900 = 6

The notes are 20 × 6 = 120; 15 × 6 = 90; 8 × 6 = 48

From the GIVEN DATA,

TOTAL decrease = 3000 - 120 = 2880

C’s share $(= \frac{5}{{12}} \times 2880 = 5 \times 240 = 1200)$

Total money = RS. 1050

Share of M = 3/(3 + 5 + 7) = 3/15 = 1/5

Share of M = (1/5) × 1050 = Rs. 210

Share of N = 5/(3 + 5 + 7) = 5/15 = 1/3

Share of N = (1/3) × 1050 = Rs. 350

Difference between the shares of M and N = 350 - 210 = Rs. 140

Time taken to cross the tree = Length of train/SPEED of train

⇒ Speed of train = 180/15 = 12 m/sec.

Let the speed of the second train be ‘x’ m/sec.

? The TRAINS are moving in opposite directions, HENCE, their relative speed is the sum of their individual speeds

⇒ Relative speed = x + 12

Now,

Time taken to cross each other = Sum of lengths of trains/Relative speed

⇒ 20 = (180 + 180)/(x + 12)

⇒ x + 12 = 360/20

⇒ x = 18 - 12 = 6 m/sec.

There are 396 CHILDREN with 5 marbles. If 198 children with 5 marbles move to second hall, there will be 198 children left in first hall with 5 marbles.

⇒ Number of marbles in second hall = (225 × 7 + 198 × 5)

⇒ Number of marbles in first hall = (198 × 5)

∴ Ratio of number of marbles in second hall to that in first hall = (225 × 7 + 198 × 5)/(198 × 5)

U : V = 6 : 7

Let U = 6x and V = 7x

V : W = 21 : 6

Let V = 21Z and W = 6z

3z can be written as x

Let V = 7x and W = 2x

U : V : W = 6x : 7x : 2x = 6 : 7 : 2

Let the Total AMOUNT be X.

Then According to the question,

x/3 + x/9 + 5000 = x

(3x + x + 45000)/9 = x

4x + 45000 = 9x

5x = 45000

x = 9000

Let the three numbers be X, y and z.

Ratio of x and y = 4 ? 7

Ratio of y and z = 2 ? 3

Then, x ? y ? z = 8 ? 14 ?21

Let the three NUMBER be 8A, 14a and 21a.

The product of x and z = 4200

⇒ x × z = 4200

⇒ 8a × 21a = 4200

⇒ a² = 25

⇒ a = 5

First number = x = 8a = 40

Second number = y = 14a = 70

Third number = z = 21a = 105

The SUM of three numbers = 40 + 70 + 105 = 215

Let the price of the ticket be 8x and revised price be 11x (SINCE the ratio given is 8 ? 11)

Number of tickets sold be 23y and revised number be 21Y,

⇒ Revenue before increase in price = 8x × 23y = 184xy

⇒ Revenue after increase in price = 11x × 21y = 231xy

Given, 184xy = 36800

⇒ XY = 200

⇒ INCREASED revenue = 231xy = 231 × 200 = Rs. 46200

Given, Anil’s capital is 2/7^th of the total

∴ The RATIO of their capitals is 2 : 5

Given, Anil invested for 2/3 year

Let Mukesh invested for ‘x’ year

Given, Anil’s SHARE is 4/9^th of total profit

The ratio of their profits is 4 : 5

(2 × (2/3)) : (5 × x) = 4 : 5

⇒ (4/3)/5x = 4/5

⇒ x = 1/3

Given numbers are $(1\frac{1}{2})$ and $(1\frac{1}{3})$

⇒$(1\frac{1}{2})$= 3/2 and $(1\frac{1}{3})$ = 4/3

Reciprocals of the above numbers are 2/3 and 3/4.

x² : 4 = 4 : y²

x²y² = 16

xy = 4

Let the MEAN PROPORTIONAL of x³ and y³ be a

a² = x³y³

⇒ a = (4)^3/2 = 8