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Red ? Green ? Blue = 5 ? 6 ? 9

Thus, let red, green, blue chocolates be 5x, 6x and 9x.

Now, 16 green and 44 blue were added and the new ratio became 6 ? 8 ? 13

Thus, 5x ? 6x + 16 ? 9x + 44 = 6 ? 8 ? 13

∴ 5x ? 6x + 16 = 3 ? 4

∴ 20x = 18x + 48

∴ x = 24

∴ TOTAL amount = 5(24) + 6(24) + 16 + 9(24) + 44 = 540

Let the number of 25p coins, 50P coins and 1 rupee coins be 8X, 4x and 3x respectively.

⇒ (0.25)8x + (0.50)4x + (1)3x = 35

⇒ x = 5

Required difference = 15 – 40 × 0.25 = Rs. 5

Given: A/3 = B/2 = C/5

⇒ B = 2A/3

⇒ C = 5A/3

To find: ratio (C + A)² : (A + B)² : (B + C)²

⇒ (5A/3 + A)² : (A + 2A/3)² : (2A/3 + 5A/3)²

⇒ (8A/3)² : (5A/3)² : (7A/3)²

⇒ (8A)² : (5A)² : (7A)²

∴ 64 : 25 : 49

Let 3X = 5Y = 2Z = k

X = k/3, Y = k/5, Z = k/2

Sum of shares of all of them together = 7750

⇒ (k/3) + (k/5) + (k/2) = 7750

⇒ (10K + 6k + 15k)/30 = 7750

⇒ 31k = 7750 × 30

⇒ k = (7750 × 30/31) = 7500

According to the given information,

NUMBER of total workers in the FARM = 420

Ratio of men and women = 13 :17

Number of men workers = total number of workers $(\times \;\frac{{13}}{{13 + 17}})$

Number of men workers $(= \;420\; \times \frac{{13}}{{13 + 17}}{\rm{\;}} = {\rm{\;}}182)$

Number of FEMALE workers = 420 – 182 = 238

According to the given information,

4 female workers leave the work,

the remaining number of female workers = 238 – 4 = 234

New ratio of MALE to female workers = 182 : 234 = 7 : 9

Let the denominator of rational number be x.

Numerator = (x - 5)

New denominator = (x + 6)

New numerator = (x - 5 + 8) = (x + 3)

Given that, (x + 3)/ (x + 6) = 5/7

⇒ 7x + 21 = 5x + 30

⇒ 2x = 9

⇒ x = 9/2

Original rational number = (x - 5)/x = -(1/2)/(9/2) = -1/9

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) + (5 × 7)

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

Since the RATIO of marks obtained in Physics and Biology is 6 : 5

Let the marks obtained in Physics and biology be 6x and 5x respectively.

According to the given INFORMATION,

⇒ Gopal’s marks in Chemistry = 3x

? Total Marks in Chemistry, Biology and Physics = 210

⇒ 3x + 5x + 6x = 210

⇒ 14X = 210

∴ x = 15

∴ Marks in Chemistry = 15x = 3 × 15 = 45

Let Hari’s salary be ‘x’

⇒ GEETA’s salary = 9x/5

⇒ SHALU’s salary = 9x/2

∴ Ratio of Geeta’s to shalu’s salary = ((9x/5) /(9x/2) ) = 2/5

<P>Let the INCOME of Saurabh and Truptil be 4p and 5p and their expenses be 3q and 4q.

Given, each saves RS 12000

So, 4p – 3q = 12000 and 5p - 4q = 12000

On solving, we GET p = 12000 and q = 12000.

We KNOW,

Profit = investment × time

⇒ Time = profit/investment

Given,

⇒ RATIO of their profit = 4 ? 15 ? 6

⇒ Ratio of their investment = 8 ? 5 ? 3

Let $(E\; = \;\frac{{6{P^2}\; + \;5PQ\; + \;{Q^2}}}{{6{P^2} - 5PQ\; + \;{Q^2}}})$ 

$(\BEGIN{array}{l} \RIGHTARROW E\; = \;\frac{{6{P^2}\; + \;5PQ\; + \;{Q^2}}}{{6{P^2} - 5PQ\; + \;{Q^2}}}\; = \;\frac{{6{P^2}\; + \;3PQ\; + \;2PQ\; + \;{Q^2}}}{{6{P^2} - 3PQ - 2PQ\; + \;{Q^2}}}\\ \Rightarrow E\; = \;\frac{{\left[ {\left( {3P\; + \;Q} \right)\left( {2P\; + \;Q} \right)} \right]}}{{\left[ {\left( {3P - Q} \right)\left( {2P - Q} \right)} \right]}}\; = \;\left[ {\frac{{3P\; + \;Q}}{{3P - Q}}} \right] \times \left[ {\frac{{2P\; + \;Q}}{{2P - Q}}} \right]\\ \Rightarrow E\; = \;\left[ {\frac{{3 \times 3\; + \;2}}{{3 \times 3-2}}} \right] \times \left[ {\frac{{2 \times 3\; + \;2}}{{2 \times 3-2}}} \right]\; = \;\left( {\frac{{11}}{7}} \right) \times \left( {\frac{8}{4}} \right)\; = \;\frac{{22}}{7} \end{array})$

∴ E = 22/7

Initial collection = 17 × 11 = 187

New collection = 19 × 10 = 190

An office OPENS at 8 AM and closes at 5 PM.

So, total PERIOD of office = 9 hours = 540 minutes

And lunch interval = 90 minutes.

a ? b = 1 ? 2, b ? C = 3 ? 1

⇒ a ? b ? c = 3 ? 6 ? 2

c ? d = 3 ? 2

⇒ a ? b ? c ? d = 9 ? 18 ? 6 ? 4

d ? e = 2 ? 1

GIVEN, three parts are in the ratio $(1:\frac{1}{3}:\frac{1}{6})$

Let the first PART be a.

Middle part = a/3

Last part = a/6

Given, 78 is DIVIDED into these three parts

∴ a + a/3 + a/6 = 78

⇒ 9a/6 = 78

⇒ a = 156/3 = 52

Now, Middle part = a/3

? DIAMETER of the vessel = 14 CM

∴ RADIUS = 7 cm;

We need to find out the surface area of the spherical vessel;

Surface area of spherical vessel = 4πr²

⇒ 4 × 22/7 × 7²

⇒ 616 cm²

? The cost of PAINTING the spherical vessel is Rs. 8008;

∴ Cost of painting PER square centimetre = 8008/616 = Rs. 13

Let the average weight of the 69 students be A.

∴ total weight of the 69 of them will be 69A.

As question STATES, the total weight of the class = 69A + 38

When this STUDENT is also included, the average weight decreases by 0.2 KG;

⇒ (69A + 38)/70 = A - 0.2

⇒ 69A + 38 = 70A - 14

⇒ 70A - 69A = 38 + 14

⇒ A = 52

As we know that, Fourth proportional = (B × c)/a 

Where a,b and c are the 1^st, 2^nd and 3^rd elements of proportion respectively

Formula = Fourth proportional = (b × c)/a

a = 7, b = 15 and c = 14

(14 × 15)/7

210/7 = 30

As per the given data

Rise of the BALL for first bounce = 48 × 2/3 = 32 m

Rise of the ball for SECOND bounce = 32 × 2/3 = 64/3 m

Rise of the ball for third bounce = 64/3 × 2/3 = 128/9 m

Rise of the ball for FOURTH bounce = 128/9 × 2/3 = 9.48 m

Let salaries of Rs. A and Rs. B be m and n respectively.

Given,

⇒ m : n = 13 : 15

⇒ m/n = 13/15

⇒ n = 15m/13

Given,

⇒ (m + 3000)/(n + 3000) = 29 / 33

⇒ 33M + 99000 = 29n + 87000

⇒ 29n - 33m = 12000

⇒ 29(15m/13) - 33m = 12000

⇒ 435m/13 - 33m = 12000

⇒ 435m - 429m = 12000 × 13

⇒ 6m = 156000

⇒ m = 26000

NEW salary of A = 26000 + 3000 = 29000

Let the NUMBERS are x and y,

Mean proportional = √xy = 18

⇒ xy = 324---- (1)

THIRD proportional = y²/x = 144

⇒ y² = 144x---- (2)

Putting the value of x from this in 1^st equation,

⇒ y³/144 = 324

⇒ y³ = 324 × 144

⇒ y³ = 18 × 18 × 18 × 8

⇒ y = 18 × 2

⇒ y = 36 and x = 324/36 = 9

GIVEN $(\frac{{3{\rm{a\;}} + {\rm{\;}}4{\rm{B}}}}{{3{\rm{c\;}} + {\rm{\;}}4{\rm{d}}}} = \frac{{3{\rm{a}} - 4{\rm{b}}}}{{3{\rm{c}} - 4{\rm{d}}}})$

$(\Rightarrow {\rm{\;}}\frac{{3{\rm{a\;}} + {\rm{\;}}4{\rm{b}}}}{{3{\rm{a}} - 4{\rm{b}}}} = \frac{{3{\rm{c\;}} + {\rm{\;}}4{\rm{d}}}}{{3{\rm{c}} - 4{\rm{d}}}})$ (By alternendo)

$(\Rightarrow {\rm{\;}}\frac{{\left( {3{\rm{a\;}} + {\rm{\;}}4{\rm{b}}} \RIGHT){\rm{\;}} + {\rm{\;}}\left( {3{\rm{a}} - 4{\rm{b}}} \right)}}{{\left( {3{\rm{a\;}} + {\rm{\;}}4{\rm{b}}} \right) - \left( {3{\rm{a}} - 4{\rm{b}}} \right)}} = \frac{{\left( {3{\rm{c\;}} + {\rm{\;}}4{\rm{d}}} \right){\rm{\;}} + {\rm{\;}}\left( {3{\rm{c}} - 4{\rm{d}}} \right)}}{{\left( {3{\rm{c\;}} + {\rm{\;}}4{\rm{d}}} \right) - \left( {3{\rm{c}} - 4{\rm{d}}} \right)}})$ (By componendo and dividendo)

$(\Rightarrow {\rm{\;}}\frac{{6{\rm{a}}}}{{8{\rm{b}}}} = \frac{{6{\rm{c}}}}{{8{\rm{d}}}})$ 

$( \Rightarrow {\rm{\;}}\frac{{\rm{a}}}{{\rm{b}}} = \frac{{\rm{c}}}{{\rm{d}}})$.

TOTAL Sum = RS. 2489

New sum after diminish the amount = 2489 – 12 – 12 – 5 = 2489 – 29 = 2460

∴ C’s share = 4/12 × 2460 = 820

given that,

2x – 4Y = x + 3y

⇒ 2x – x = 3Y + 4Y

⇒ x = 7Y

$(\Rightarrow \;\FRAC{x}{y} = \;\frac{7}{1})$

$(\Rightarrow \;\frac{{2x}}{{4y}} = \;\frac{{2 \times 7}}{{4 \times 1}})$(MULTIPLYING both SIDES by 2 and dividing both sides by 4)

$(\Rightarrow \;\frac{{2x}}{{4y}} = \frac{{14}}{4})$

$(\Rightarrow \;\frac{{2x + 4y}}{{2x - 4y}} = \;\frac{{14 + 4}}{{14 - 4}})$(by componendo-dividendo)

The ratio of two NATURAL numbers P and Q is 2 : 3. Let the numbers be 2T and 3T.

If 12 is added to both of them, their ratio BECOMES 11 : 16.

⇒ (2T + 12)/(3T + 12) = 11/16

⇒ 32T + 192 = 33T + 132

⇒ T = 60

⇒ P = 2T = 120 and Q = 3T = 180

Total number of 25P coins = (5/10) × 200 = 100

Total number of 50p coins = (3/10) × 200 = 60

Total number of 1 RUPEE coins = (2/10) × 200 = 40

Total MONEY with the person = 0.25 × 100 + 0.50 × 60 + 1 × 40 = Rs. 95

 GIVEN p = $(\FRAC{{4{\rm{xy}}}}{{{\rm{x\;}} + {\rm{\;y}}}})$

$(\Rightarrow {\rm{\;}}\frac{{\rm{p}}}{{2{\rm{x}}}} = \frac{{2{\rm{y}}}}{{{\rm{x\;}} + {\rm{\;y}}}})$ and $(\frac{{\rm{p}}}{{2{\rm{y}}}} = \frac{{2{\rm{x}}}}{{{\rm{x\;}} + {\rm{\;y}}}})$.

Applying COMPONENDO and dividendo on both, we get

$(\Rightarrow {\rm{\;}}\frac{{{\rm{p\;}} + {\rm{\;}}2{\rm{x}}}}{{{\rm{p}} - 2{\rm{x}}}} = \frac{{2{\rm{y\;}} + {\rm{\;}}\left( {{\rm{x\;}} + {\rm{\;y}}} \RIGHT)}}{{2{\rm{y}} - \left( {{\rm{x\;}} + {\rm{\;y}}} \right)}})$ and $(\frac{{{\rm{p\;}} + {\rm{\;}}2{\rm{y}}}}{{{\rm{p}} - 2{\rm{y}}}} = \frac{{2{\rm{x\;}} + {\rm{\;}}\left( {{\rm{x\;}} + {\rm{\;y}}} \right)}}{{2{\rm{x}} - \left( {{\rm{x\;}} + {\rm{\;y}}} \right)}})$

$(\Rightarrow {\rm{\;}}\frac{{{\rm{p\;}} + {\rm{\;}}2{\rm{x}}}}{{{\rm{p}} - 2{\rm{x}}}} = \frac{{{\rm{x\;}} + {\rm{\;}}3{\rm{y}}}}{{{\rm{y}} - {\rm{x}}}})$ and $(\frac{{{\rm{p\;}} + {\rm{\;}}2{\rm{y}}}}{{{\rm{p}} - 2{\rm{y}}}} = \frac{{3{\rm{x\;}} + {\rm{\;y}}}}{{{\rm{x}} - {\rm{y}}}})$

$(\begin{array}{l}\therefore {\rm{\;}}\frac{{{\rm{p\;}} + {\rm{\;}}2{\rm{x}}}}{{{\rm{p}} - 2{\rm{x}}}}{\rm{\;}} + {\rm{\;}}\frac{{{\rm{p\;}} + {\rm{\;}}2{\rm{y}}}}{{{\rm{p}} - 2{\rm{y}}}} = \frac{{{\rm{x\;}} + {\rm{\;}}3{\rm{y}}}}{{{\rm{y}} - {\rm{x}}}}{\rm{\;}} + {\rm{\;}}\frac{{3{\rm{x\;}} + {\rm{\;y}}}}{{{\rm{x}} - {\rm{y}}}}\\ = \frac{{3{\rm{x\;}} + {\rm{\;y}}}}{{{\rm{x}} - {\rm{y}}}} - \frac{{{\rm{x\;}} + {\rm{\;}}3{\rm{y}}}}{{{\rm{x}} - {\rm{y}}}} = \frac{{\left( {3{\rm{x\;}} + {\rm{\;y}}} \right) - \left( {{\rm{x\;}} + {\rm{\;}}3{\rm{y}}} \right)}}{{{\rm{x}} - {\rm{y}}}}\\ = \frac{{2{\rm{x}} - 2{\rm{y}}}}{{{\rm{x}} - {\rm{y}}}} = \frac{{2\left( {{\rm{x}} - {\rm{y}}} \right)}}{{{\rm{x}} - {\rm{y}}}} = 2\end{array})$

As we know, the compound ratio of ratios (a ? c) and (B ? d) is (ab ? cd)

Hence, we have,

⇒ 1 ? (X + y)(x - y) = 1 ? y²

⇒ 1 ? (x² - y²) = 1 ? y²

⇒ y² = x² - y²

⇒ 2y² = x²

⇒ x²/y² = 2

⇒ x/y =√2

Given that, a : b = 4 : 3, b : c = 2 : 7, c : d = 3 : 4

⇒ a : c = 8 : 21 ⇒ a = 8c/21

⇒ b : d = 3 : 14 ⇒ b = 3d/14

⇒ a : d = 2 : 7 ⇒ a = 2d/7

⇒ b = 3a/4, c = 21a/8, d = 7a/2

⇒ a : b : c : d = a : 3a/4 : 21a/8 : 7a/2

∴ a : b : c : d = 8 : 6 : 21 : 28

RATIO of amount of A and B = 5 : 6 = 5/6 = (5 × 2)/(6 × 2) =10/12

Ratio of amount of B and C = 4 : 7 = 4/7 = (4 × 3)/(7 × 3) = 12/21

∴ Ratio of SHARES of A, B and C = 10 : 12 : 21

Given,

Amount to be divided = Rs. 1290

∴ SHARE of A $(= \frac{{10}}{{10 + 12 + 21}} \times 1290 = \frac{{10}}{{43}} \times 1290 = Rs.\ 300)$

∴ Share of B $(= \frac{{12}}{{10 + 12 + 21}} \times 1290 = \frac{{12}}{{43}} \times 1290 = Rs.\ 360)$

We KNOW that a : b : : c : x, then x = bc/a

⇒ Fourth PROPORTIONAL x = (17 × 27)/9 = 51