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For the respective quantity CONSUMED in 1 kg of MIXTURE,

Quantity of type A WHEAT in 1 kg of mixture = (2/5) × 1 = 0.4 kg

Quantity of type B wheat in 1 kg of mixture = 1 – 0.4 = 0.6 kg

For the respective COST price of wheat type,

Cost price for type A = 0.4 × 50 = Rs. 20

Cost price for type B = 0.6 × 30 = Rs. 18

Cost price per kg of mixture = Rs. 38

For the SELLING price,

Selling price – cost price = Profit

⇒ Selling price – 38 = 25% of cost price

⇒ Selling price = (0.25 × 38) + 38

∴ Selling price = Rs. 47.5

Let ‘x’ UNITS of both the mixtures be mixed together

TOTAL quantity of NEW mixture = x + x = 2X

Quantity of honey in first mixture = 10% of x = x/10

Quantity of honey in second mixture = 40% of x = 2x/5

⇒ Total quantity of honey in new mixture = x/10 + 2x/5 = x/2

⇒ Quantity of water in new mixture = 2x – x/2 = 3x/2

∴ Ratio of new mixture = x/2 ? 3x/2 = 1 ? 3

LET the water added be X ML

Total solution = (x+50) ml

% of alcohol = 50%

50/(x + 50) × 100= 50

QUANTITY of water in 300 ml of MIXTURE = [x/(x + 1)] × 300

When 50 ml of water is mixed in the mixture, ratio of acid and water BECOMES 2 ? 5

⇒ Quantity of water in in the new mixture = 5/7 × (300 + 50)

⇒ (x/x + 1) × 300 + 50 = 5/7 × (300 + 50)

⇒ (x/x + 1) × 300 = 200

⇒ (x/x + 1) = 2/3

∴ x = 2

Let the jar initially contains 3x and x LITRES of mixture A and B respectively

∴ Quantity of A in mixture left = (3x – ¾ × 6) = (3x – 18/4) litres

∴ Quantity of B in mixture left = (x – ¼ × 6) = (x – 6/4) litres

Given,

⇒ (3x – 18/4)/((x – 6/4) + 10) = 1/3

⇒ 9X – 54/4 = x – 6/4 + 10

⇒ (36x – 54)/4 = (4x – 6 + 40)/4

⇒ 36x – 4x = 34 + 54

⇒ 32X = 88

⇒ x = 2.75

Quantity of alcohol in 50 litres of first mixture = 17/25 × 50 = 34 litres

Quantity of water in 50 litres of first mixture = 8/25 × 50 = 16 litres

Similarly,

Quantity of alcohol in 60 litres of second mixture = 11/15 × 60 = 44 litres

Quantity of water in 60 litres of second mixture = 4/15 × 60 = 16 litres

⇒ Total quantity of alcohol in new mixture = 34 + 44 = 78 litres

⇒ Total quantity of water in new mixture = 16 + 16 = 32 litres

Let ‘x’ litres of water be added to obtain the final mixture

⇒ Ratio of final mixture = 78 ? (32 + x) = 2 ? 1

⇒ 78 = 64 + 2x

⇒ x = 14/2 = 7 litres

LET initial quantity of sugar = 4a and milk = 5A

(Initial quantity of sugar)/(Initial quantity of milk + Quantity of milk ADDED) = 2/3

⇒ 4a/(5a + 7) = 2/3

⇒ 10a + 14 = 12a

⇒ 2a = 14

⇒ a = 7 liters

∴ Quantity of Sugar in new MIXTURE is same as initial so = 4a = 4 × 7 = 28 liters

Average MONEY per student = 3900/65 = 60 paise

Applying the allegation rule:

⇒ (80 - 60)/(60 - 30) = 2/3

⇒ BOYS : Girls = 3 : 2

⇒ Number of boys = (65 × 3)/(3 + 2)

⇒ Number of boys = 39

⇒ Number of girls = 65 - 39 = 26

Suppose M KG of first ALLOY and N kg of second alloy is taken.

Amount of Iron in first alloy = 7M/12 kg

Amount of Silver in first alloy = 2M/12 kg

Amount of Silver in second alloy = N/3 kg

RATIO of Iron and Silver in mixture is 5 : 3.

⇒ (7M/12)/(2M/12 + N/3) = 5/3

⇒ (7M/(2M + 4N)) = 5/3

⇒ 21M = 10M + 20N

⇒ 11M = 20N

⇒ M/N = 20/11

MILK left = Capacity X (1 – fraction of milk withdrawn)ⁿ, where n is the no of process

⇒ Milk left = 50 × (1 – 5/50)³ = 50 × (45/50)³ = 36.45 L

⇒ Total cost price = 36.45 × 100 = Rs. 3645

⇒ Total selling price = 50 × 100 = Rs. 5000

Total weight of pulses is 1 kg

Let the weight of RS. 11.10 per kg PRICE be x and Rs. 15.20 per kg PULSE be (1 - x)

Let the ratio in which mixture is mixed be x : (1 - x)

Total price of pulses at Rs. 11.10 per kg = 11.10x

Total price of pulses at Rs. 15.20 per kg = 15.2(1 - x)

Total cost of mixture = (15.2(1 - x) + 11.10x)

(15.2(1 - x) + 11.10x) = 13.20

15.2 - 4.1x = 13.2 ⇒ x = 20/41

Ratio in which mixture is mixed = x : (1 - x) = 20/21 = 20 : 21

Let the quantity of A and B in the solution be X and Y.

⇒ After adding 10ltr of water, total solution = X + Y + 10

B is only 10% of the total solution

⇒ Y/ (X + Y + 10) = 0.1----(1)

Initial percentage of A in the solution = (X × 100)/ (X + Y)

New percentage of A in solution = (X × 100)/(X + Y + 10)

This new percentage is 30% less than before

[(X × 100)/(X+Y) - (X × 100)/ (X + Y + 10)]/ [(X × 100)/(X+Y)] = 0.3

 10 /(X+Y+10) = 0.3

X + Y + 10 = 100/3

Putting this in (1) we get

Y = 0.1 × 100/3 = 10/3

X = 70/3 - 10/3 = 20

If 2lts of A is added the amount of A BECOMES 2 + 20 = 22

Trick:

⇒ Difference between the prices of TWO tea (a) = 300 - 200 = 100

⇒ Difference between the net price and cheapest (b) = 225 - 200 = 25

∴ REQUIRED ratio = b : (b - a) = 25 : 75 = 1 : 3

Detailed Solution:

Let amount of RS. 300/kg tea added be x and Rs. 200/kg added be 1

⇒ Total cost of tea = 300 × x + 200 × 1 = 200 + 300x

⇒ Amount of tea = 1 + x

⇒ Cost of per kg of mixed tea = Rs 225

⇒ 200 + 300x = 225 × (1 + x)

⇒ x = 1/3

∴ Required ratio = 1 : 3

We know the formula:

X = A(1 - R/C)^N

Here X = Liquid remaining after replacement

A = Total quantity of liquid before replacement

R = Quantity of replaced liquid

C = Total Capacity of drum

n = No. of times the liquid was replaced

⇒ A = 80, R = 20, C = 80 and n = 2

Putting these VALUES in the formula,

⇒ X = 80 × (1 - 1/4)²

⇒ X = 80 × 9/16

⇒ X = 45 litres

⇒ AMOUNT of ethanol present after replacement = 45 litres

3/5 is the CORRECT ANSWER

? Alloy CONTAINS copper and tin in the ratio 3 : 2

Let the amount of copper in an alloy be 3X

And the amount of tin in an alloy be 2x

According to QUESTIONS,

⇒ 3x + 250 = 2 × 2x

⇒ x = 250 gm

∴ The amount (in gm) of tin in the alloy = 2x = 2 × 250 = 500 gm

Initial ratio = 5 ? x

After ADDING 1 liter of water to the 4 LITERS of blend, the final QUANTITY of the blend BECOMES 5 Liters

Final Ratio = 1 ? 1

⇒ Value of 2 = 5 Liters

⇒ Quantity of juice = Quantity of water in the blend = 2.5 Liters

Since the Quantity of juice is same,

⇒ Quantity of juice in initial mixture = 2.5 Liters

⇒ Quantity of water in initial mixture = 1.5 Liter

⇒ Initial total quantity = 4 Liters

4 Liters in mixed in the ratio = 2.5 ? 1.5 = 5 ? 3

∴ Value of x = 3

INITIAL AMOUNT of wine = 60 LITERS.

After replacing 10 liters of wine with water

⇒ There is 50 liters of wine and 10 liters of water.

⇒ Wine to water RATIO = 5: 1

10 liters of this mixture is removed and REPLACED by water.

⇒ Amount of wine removed from the mixture = (5/6) × 10 = 25/3 liters.

Amount of water removed = (1/6) × 10 = 5/3 liters.

In 1 KG of alloy A, Gold = 7/9 and Copper = 2/9

In 1 kg of alloy Q, Gold = 7/18 and Copper = 11/18

Ratio of Gold and Copper in alloy C

= 7/9 + 7/18 ? 2/9 + 11/18

Initially, in the container amount of milk = 64 litres

Amount of water = 0 litres

First, 8 litres of milk is replaced

∴ Amount of milk = 64 - 8 = 56 litres

Amount of water = 8 litres

RATIO of milk and water = 56 ? 8 = 7 ? 1

Second, 8 litres of MIXTURE is replaced

In this mixture, ratio of milk and water is = 7 ? 1

Thus, amount of milk replaced = $(\frac{7}{8} \times 8 = 7)$ litres

Amount of milk REMAINING = 56 - 7 = 49 litres

Amount of water = 64 - 49 = 15 litres

Now NEW ratio = 49 ? 15

Alternate method :

To find amount of milk

 $({\rm{Original\;amount\;}}{\left( {1 - \;\frac{{Amount\;replaced\;by\;water}}{{Original\;amount}}} \right)^n})$

Where n = number of operations

$(64{\left( {1 - \;\frac{8}{{64}}} \right)^2} = 49)$ litres

Amount of water = 64 - 49 = 15 litres

Now new ratio = 49 ? 15

Amount of water in the MIXTURE = 15L

⇒ Milk = 85 L

On adding the milk to make the concentration of milk 87.5%, the quantity of the milk remains the same,

⇒ 12.5% Water = 15 L

⇒ 100% = 120 L

⇒ Out of 120 L, concentration of water is 15L

⇒ Quantity of milk = 105 L

TOTAL COST of type 1 wheat = 28 × 56 = Rs. 1,568

Total cost of type 2 wheat = 32 × 80 = Rs. 2,560

Total cost of type 3 wheat = 16 × 33 = Rs. 528

Total cost = 1568 + 2560 + 528 = Rs. 4,656

Total kg = 28 + 32 + 16 = 76kg

∴ Average price = $(\frac{{4656}}{{76}} = Rs.\;61\frac{5}{{19}})$ per kg

Let US denote by y gallons the AMOUNT of 4% solution.

⇒ The amount of solution of 11% acid solution = 35 - y gallons

SOLVING 0.07 (35) = 0.04y + 0.11 (35 - y),we get

y = 20 gallons

Let the number of RED and blue BALLS in the box be 5x and 3x respectively.

Given that 16 balls are removed from the box and replaced with 16 other blue balls

⇒ Number of red balls in the box left after this PROCESS = 5x – (5/8 × 16)

⇒ 5x – 10

⇒ Number of blue balls in the box left after this process = 3x – (3/8 × 16)

⇒ 3x – 6

Since 16 additional balls were added to the box,

⇒ Total number of blue balls in the box will be (3x – 6 + 16)

⇒ 3x + 10

Given that the ratio of number of red and blue balls after this process = 3/5

⇒ (5x – 10)/(3x + 10) = 3/5

⇒ (5x – 10) × 5 = (3x + 10) × 3

⇒ 25x – 50 = 9x + 30

⇒ 16x = 80

⇒ x = 5

∴ INITIAL number of blue balls PRESENT in the box initially = 3 × 5 = 15

⇒ Let ‘x’ be the quantity of milk in the original mixture

⇒ $(\frac{7}{3} = \frac{x}{{120}})$

⇒ Quantity of milk, x = (120 × 7)/3 = 280ml

⇒ the mixture contains 280 ml of milk and 120 ml of water

⇒ The expected new RATIO of milk : water is 1 : 1

⇒ Let ‘y’ be the EXTRA water added to make the mixture have milk and water in equal parts

⇒ THUS, 120 + y = 280

⇒ y = 160 ml

∴ Extra water added = 160 ml

MILK ? Water ? Sugar syrup = 9 ? 4 ? 1

Amount = 84 litres

Amount of milk = $(\FRAC{9}{{14}} \TIMES 84 = 54)$ litres

Amount of water = $(\frac{4}{{14}} \times 84 = 24)$ litres

Amount of sugar syrup = $(\frac{1}{{14}} \times 84 = 6)$ litres

Let x be amount of water added and y be amount of sugar syrup, such that RATIO of mixture becomes 9 ? 6 ? 2.

54 ? (24 + x) ? (6 + y) = 9 ? 6 ? 2

Now, 54 ? (24 + x) = 9 ? 6

∴ (54)(6) = (24 + x)(9)

∴ x = 12 litres

Also, 54 ? (6 + y) = 9 ? 2

∴ (54)(2) = (6 + y)(9)

∴ y = 6 litres

Amount of water = 24 + 12 = 36

Amount of sugar syrup = 6 + 6 = 12 litres

Amount of milk and sugar syrup = 54 + 12 = 66 litres

Given, LOTION of 7 ml contains 50% alcohol.

∴ Amount of water = 50% of 7ml = 3.5ml

Amount of alcohol = 50% of 7ml = 3.5 ml

Let the water added be ‘a’ ml.

Now, in order to obtain 20% alcohol SOLUTION.

$(\frac{{3.5}}{{7 + a}} \times 100\%= 20\% \;)$

⇒ 5 × 3.5 = 7 + a

⇒ a = 10.5 ml

Let the milkman buy X litre of milk at 20 per ltr

⇒ Cost Price = 20x

⇒ ADDS 1/3 of water to it so quantity = x + (x/3) = 4x/3

⇒ Selling Price = 24 × (4x/3) = (96x/3)

⇒ Gain % = (SP - CP) × (100/CP)

⇒ Gain % = {(96x/3) - 20x} × (100/20x)

⇒ Gain % = (3600x)/60x

Total price of type 1 rice = 450 × 96 = RS. 43,200

Total price of type 2 rice = 550 × x = Rs. 550x

Total kg = 450 + 550 = 1000kg

Average price = Rs. 66.3

Total price of the mixture = 66.30 × 1000 = Rs. 66,300

Price of type 2 rice per kg = x = $(\frac{{66300\; - \;43200}}{{550}})$ = Rs. 42 per kg

Let the capacity of each bottle be y

For bottle 1, milk be 5A and water be 7a

⇒ 5a + 7a = y

⇒ a = y/12…(i)

For bottle 2, milk be 7b and water be 9b

⇒ 7b + 9b = y

⇒ b = y/16…(ii)

For bottle 3, milk be 2C and water be c

⇒ 2c + c = y

⇒ c = y/3…(iii)

⇒ PERCENTAGE of milk can be given as = (5a + 7b + 2c) × 100/(5a + 7b + 2c + 7a + 9b + c)

⇒ Percentage of milk = (5a + 7b + 2c) × 100/(12a + 16b + 3c) …from (i), (ii) and (iii)

⇒ Percentage of milk = (5 × y/12 + 7 × y/16 + 2 × y/3) × 100/(y + y + y)

⇒ Percentage of milk = 50.7%

∴ the percentage of milk be 50.7%

DIFFERENCE of Pepsi and WATER = 5 unit - 2 unit = 3 UNITS

⇒ 3 units = 15 litres

⇒ 1 unit = 5 litres

Quantity of Pepsi = 5 units = 5 × 5 = 25 litres

Quantity of water = 2 units = 2 × 5 = 10 litres

Now 5 litres of water is added to the mixture

New Quantity of water = 10 + 5 = 15 litres

New ratio of Pepsi and water = 25 : 15

Let ‘X’ quantity of each of three MIXTURES to be mixed

Quantity of milk in the new MIXTURE = (1/4)x + (3/8)x + (11/16)x = 21x/16

Quantity of water in the new mixture = (3/4)x + (5/8)x + (5/16)x = 27x/16

Let y liters of the second SOLUTION must be added.

Then, [15 × 20 + 5 × y]/(20 + y) = 13

Let the mixture is of ‘13x’ litres

⇒ Quantity of milk in the mixture = 11x litres

⇒ Quantity of WATER in the mixture = 2x litres

After adding 35 litres of water, new mixture has (2x + 35) part water

After adding 35 Litres of water, the new mixture will contain twice as much pure milk as water.

⇒ 11x/ (2x + 35) = 2/1

⇒ 4x + 70 = 11x

⇒ 11x - 4x = 70

⇒ 7X = 70

⇒ x = 10 litres

According to QUESTION,

x(50/100) + 3x(40/100) + 8X(60/100) = 26

⇒ 0.5x + 1.2x + 4.8x = 26

⇒ x = 26/6.5 = 4 L

The total volume of SOLUTION A and B = x + 3x + 8x = 12x = 12 × 4 = 48 L

∴ The total volume of solution A and B = 48 L

⇒ Let the quantity of milk and water in 1^st and 2^nd vessel be 3x, 4x and 2X, 3x respectively.

⇒ Milk in the first mixture = 3x/(3x + 4x) = 3/7

⇒ Water in the first mixture = 4x/(3x + 4x) = 4/7

⇒ Milk in the SECOND mixture = 2x/(2x + 3x) = 2/5

⇒ Water in the second mixture = 3x/(2x + 3x) = 3/5

⇒ Ratio of Milk and water in new mixture = (3/7 + 2/5)/(4/7 + 3/5)

⇒ Ratio of Milk and water in new mixture = {(3 × 5 + 2 × 7)/35}/{(4 × 5 + 3 × 7)/35} 

⇒ Ratio of Milk and water in new mixture = (15 + 14)/(20 + 21)

LET the quantity of new mixture is ‘a’ litres

ACID in the mixture = 20% of 80 LITRE = 16 litres

Water in the mixture = 80 – 16 = 64 litres = 80% of the mixture

In new mixture water is only 40% and acid is 60%

Water in new mixture = 40% of a litres = 64 litres

⇒ 2/5 × a = 64 litres

⇒ a = 160 litres

Acid in new mixture = 160 – 64 = 96 litres = 60% of the new mixture

LET the QUANTITY of the wine in the CASK originally be x litres

Then, quantity of wine left in cask after 4 OPERATIONS = {x (1 – 8/x)?} litres

⇒ {X (1 – 8/x)?}/x = 16/81

⇒ (1 – 8/x)? = (2/3)?

⇒ (X – 8)/x = 2/3

⇒ 3x – 24 = 2x

∴ x = 24 litres

TOTAL salt present in 1 kg of 5% solution = 5 × 1000/100 = 50gm.

Saturation LIMIT of salt is 200gm per Litre.

∴ AMOUNT of WATER required to dissolve 50gm of salts = 50 × 1000/200 = 250ml.

Rate of evaporation = 20gm/ hour.

∴ Amount to be evaporated for sedimenting = 1000 - 250 = 750GM.

Time taken to evaporate 750gm = 750/20 = 37.5 hours.

Given, he earns a profit of $(66\frac{2}{3}\% .)$ LET C be the COST price per kg of the rasgulla.

$(\Rightarrow 66\frac{2}{3} = \frac{{15 - C}}{C} \times 100)$ 

⇒ 2C = 45 – 3C

⇒ C = 9

Let the cost price of one kg of sugar be 7x and that of one kg of flour be 3x. Since they are MIXED in the ratio 3 : 5, we have:

$(9 = \frac{3}{8}\left( {7x} \RIGHT) + \frac{5}{8}\left( {3x} \right) = \frac{{36x}}{8})$ 

⇒ x = 2

LET the Initial quantity of LIQUID be 8 litres

∴ After replacing ‘x’ litres with water,

Water in mixture = 3 - 3x/8 + x = 3 + 5x/8

Syrup in mixture = 5 - 5x/8

According to the given CONDITION,

⇒ 3 + 5x/8 = 5 - 5x/8

∴ 5x/4 = 2

∴ x = 8/5

∴ For 8 litres, 8/5 of the mixtures must be TAKEN off

Quantity of liquid B = 70% of 450 = 315 ML

Let x ml of liquid B be added.

315 + x = 80% of (450 + x)

⇒ 315 + x = 0.8 × (450 + x)

⇒ 315 + x = 360 + 0.8x

⇒ 0.2x = 45

Let the ratio of 20% ethanol to 40% ethanol be X ? 1,

According to the QUESTION,

⇒ x × 0.2 + 1 × 0.4 = (x + 1) × 0.28

⇒ 0.08x = 0.12

⇒ x = 3/2

LET y g of WATER and chemical was taken.

Initially 1st process

First test TUBE = (y – 20) g

Second test tube = (y + 20) g

2nd process

First test tube = (y – 20) + (y + 20) × (2/3)

Second test tube = (y + 20) × (1/3)

Now,

(y -20) + (2/3)(y +20) = 4 × (1/3)(y + 20)

The Pomegranate ADDED = (¼) of glass

Price of Pomegranate juice = (1/4) × 40 = RS. 10

CARROTS added = (¼) of glass

Price of carrots juice = (1/4) × 40 = Rs. 10

Orange added = ½ of glass

Price of orange juice = (1/2) × 30 = Rs. 15

Price of MIXTURE juice = 10 + 10 + 15 = Rs. 35 

Juice ? MILK ? Water = 2 ? 9 ? 7

Amount = 108 litres

Amount of juice = $(\frac{2}{{18}} \times 108 = 12)$ litres

Amount of milk = $(\frac{9}{{18}} \times 108 = 54)$ litres

Amount of water = $(\frac{7}{{18}}\; \times 108 = 42)$ litres

Let x be amount of milk added and y be amount of water, such that ratio of mixture BECOMES 1 ? 7 ? 6

12 ? (54 + x) ? (42 + y) = 1 ? 7 ? 6

Now, 12 ? (54 + x) = 1 ? 7

∴ (12)(7) = 54 + x

∴ x = 30 litres

Also, 12 ? (42 + y) = 1 ? 6

∴ (12)(6) = 42 + y

∴ y = 30 litres

Amount of water = 42 + 30 = 72 litres

Amount of juice and water = 12 + 72 = 84 litres

LET the quantity of additional milk is x liters

In the mixture of 800 liters

Quantity of milk = (5/8) × 800 = 500 liters

So the quantity of water is = 300 liters

According to the QUESTION,

⇒ (500 + x)/300 = 2/1

⇒ 500 + x = 600

⇒ x = 600 – 500 = 100 liters

So, the new quantity of the mixture is 800 + 100 = 900 liters

∴ The final quantity of the mixture is 900 liters