Selling price is Rs 15 and Profit is 25%

=> Cost Price = $$\frac{15}{1.25}$$ = Rs 12

=> In every litre of solution, there is Rs 12 worth milk.

But the actual cost of milk is Rs 16.

=> Ratio = 12 : 4 = 3 : 1

Let the ratio in which they should be mixed = x:y

==> $$\dfrac{90x+120y}{x+y} = 100$$

==> 90x + 120y = 100x + 100y

==> 20y = 10x

==> x:y = 2:1

Required ratio = 2 : 1

Total cost of rice = 4 * 5 + 8 * 6 = 20 + 48 = Rs. 68

Total amount of rice = 4 + 8 = 12 kg

Average price per mixture = (Total cost of rice)/(Total amount of rice) =  $$\dfrac{68}{12}\ =\ 5.66$$ Rs. per kg

Hence, the correct answer is option C.

The mixture contains 20% water and 80% wine. The pure water that we are mixing consists of 100% water.
The resultant mixture after mixing these two has 25% water.
By the rule of allegation, Ratio of wine: water = (100 - 25) : (25 - 20) = 75 : 5 = 15 : 1
For 15 L of wine, 1 L of water is to be added.
∴ For 150 L of wine,  $$\dfrac{(150*1)}{15}=10$$ L to be added.

CP of 1 liter of the mixture = (6*10 + 4*25) / (6+4) = 16
Profit = 25%
=> SP = (1 + profit%/100)CP = 125/100 * 16 = 5/4 * 16 = Rs 20
Hence, the correct answer is option B.

Profit = 0.2 CP
Profit = SP - CP
1.2CP = SP
Hence, CP = (SP/1.2) = 75.6/1.2 = 63
Now, let the cost of type 1 rice is T1 and cost of type 2 be T2.
63 = $$\frac{T1 + 3T2}{4}$$
= (75 + 3*x)/4
x = 59
Therefore, cost of type 2 price is 59.

The quantity of salt in 3 litres of the solution is 10%*3 = 0.3 litres.
Let the quantity of water added be X.
So, (X+3)*5% = 0.3
Or, X+3 = 0.3*20 = 6
Hence, X=3 litres

Let the volume of water to be added be x ml
Volume of alcohol in 100 ml of the solution = 50 ml
$$\dfrac{50}{100+x} = 0.25$$

50 = 25+0.25x
x = 100 ml

The volume of the mixture is 100 ml.
Volume of water = 20 ml
Let x ml water is added.
$$\frac{20+x}{100+x} = 0.4$$
20+x = 40+0.4x
x = 200/6 = 33.33ml

Let the required amount be x and y
0.4x+0.25y = 0.3x+0.3y
0.1x = 0.05y
x:y = 1:2

Let x:y be the ratio at which the two varieties of wheat be mixed.
Then, $$\frac{12x}{x + y} + \frac{18y}{x + y} = 16$$
$$\frac{12(x + y) + 6y}{x + y} = 16$$
$$\frac{6y}{x + y} = 4$$
thus, $$\frac{y}{x + y} = \frac{2}{3} = \frac{2}{2 + 1}$$
thus, y = 2 and x = 1.
∴ the ratio in which the two varieties of wheat costing Rs 12 and Rs 18 should be mixed is 1:2

The formula to calculate the amount of wine left after repeating the process n times is $$W\ =\ I\left(1\ +\ \dfrac{x}{I}\right)^n$$ where x is the amount being replaced n times and I is the initial quantity of the mixture.

So, the amount of wine left after repeating the process 4 times = $$35 * (1-3.5/35)^4$$ = 22.96 ~ 23L

So the correct answer is option C - 23L

Rahul drinks 7 litres of water a day. So, he is on the tour for 1239/7 = 177 days.

This can be done using the concept of alligations.
4 10
\ /
8
/ \
2 4

Hence the required ratio is 2:4 or 1:2.

Let the cost price of the mixture be x. Hence,
(100-x)/x = 25%
100 - x = 0.25x
x = 80

Hence, by the alligation formula
75              100
\             /
80
/            \
20                5

i.e. 20: 5 or 4:1 ratio

We use the alligation method
5%             40%
\             /
25%
/           \
15               20

The ratio at which the acids must be mixed is 15:20 or 3:4.

Number of liters of milk in the initial mixture = (120/10)*7 = 84
Number of liters of water = (120/10)*3 = 36
Let x be the amount of water that should be added.
==> 84/(36+x) = 7/5 ==> 84*5 = (36+x)*7 ==> x = 24.
So the correct option to choose is D.

Let the quantities mixed be $$x, y.$$
Based on the given information, the following equation could be derived.
$$\frac{8x+13y}{x+y} = 10$$ ==> $$3y = 2x$$
==> $$x:y = 3 :2$$
Hence, the correct option to choose is E.

The amount of Zinc present in 10kg of Alloy

$$A = \dfrac{2}{5}$$ of $$10 = 4kg$$

The amount of Zinc present in 14kg of Alloy

$$B = \dfrac{3}{7}$$ of $$14 = 6kg$$

Thus, total amount of Zinc present in 24 kg Alloy

$$C = 6 + 4 = 10 kg$$

Thus, the proportion if Zinc present in Alloy

$$C = \dfrac{10}{24} = \dfrac{5}{12}$$

Let's assume the milk seller buys 1 Liter of milk from the farm.

Quantity of milk in this solution = 700ml

He adds water equal to 20% of the total solution = 200ml

Thus, the amount of water in the final quantity = 300ml + 200ml = 500 ml

Thus, the ratio of pure milk to water in the final solution = 700 : 500 = 7 : 5

Volume of hydrochloric acid in solution 1 = 20% of 2 L = 0.4 L
Volume of hydrochloric acid in solution 2 = 50% of 3 L = 1.5 L
Total volume of hydrochloric acid = 0.4+1.5 = 1.9 L
Total volume of resultant solution = 2L + 3L = 5 L
% of acid = (1.9/5)*100 = 38%

Let us take 4 kg and 5 kg of the two varieties of rice.
Price of P = 20*4 = 80
Price of Q = 38*5 = 190
Price per kg of the mixture = (80+190)/9 = 270/9 = Rs. 30

The amount of acid in 2 litres of 70% sulphuric acid = 1400 ml
The amount of acid in 1 litre of 40% sulphuric acid = 400 ml
Total amount of acid = 1400+400 = 1800 ml
Total amount of solution = 2000+1000 = 3000 ml
Concentration of acid = $$\dfrac{1800}{3000}*100=60$$%

Every time Aditya replaces 100 ml of the solution, the concentration of milk gets (100-10) =90% of what it was earlier. (10% because 100 ml is 10% of 1L)
So, initially the concentration of milk in the solution is 100% and he repeats the process 2 times.
So, the final concentration of milk in the solution will be 0.9 X 0.9 X 100% = 81%

Let the initial volume of the mixture be 7V
Amount of milk in the mixture = 4V
Amount of water in the mixture = 3V
After adding 300 ml of water the ratio becomes 1:1. So we get,
4V = 3V + 300
V = 300 ml
Total volume of the initial mixture = 7*300 = 2100 ml
Hence Option B.

We know that,  $$\dfrac{QNR final}{Total} = \dfrac{QNR initial}{Total}(1- \dfrac{quantity replaced}{Total})^n$$

Here, QNR is the quantity which is only removed in the whole process, and hence here it is milk, and 'n' is the number of times the replacement process is repeated

So using the formula

$$\dfrac{QNR final}{Total}$$ = $$\dfrac{81}{81}$$ x $$(1- \dfrac{27}{81})^2$$

$$\dfrac{QNR final}{Total}$$ = 4/9

The value of Milk/(Milk + Water) = 4/9.

So after two replacement cycles, the ratio of Milk : Water = 4:5

The ratio of milk to water in Jar A

= 180 : 36 = 5:1

Now, let 6x litres of the mixture be taken out from Jar A and put in Jar B.

Then, milk in Jar B = 5x

Water in Jar B = x

So, after adding 6L of water, the ratio becomes

5x/(x+6) = 5/2

or, 10x=(5x + 30)

or, 5x=30,

:. x=6

Hence the mixture that was taken out from Jar A = 6x = 6 × 6 = 36 litres

As the Milkman is making 25% profit on selling milk and water solution at Rs 15, the Cost price of water and Milk solution is $$\dfrac{15}{1.25}$$ = Rs 12

Let the amount of pure milk be x litres and the amount of water mixed with it be y litres.

So using , $$\dfrac{16x}{x + y} = 12$$

= $$\dfrac{x}{y} = \dfrac{3}{1}$$

Let wholesaler mixed 7 kg and 2 kg of tea of variety 60 Rs/kg and 105 Rs/kg respectively.
Cost of tea of variety 60 Rs/kg=$$60\times7$$.
=Rs. 420.
Cost of tea of variety 105 Rs/kg =$$105\times2$$.
=Rs. 210.
Total cost=420+210.
=630.
Selling price=$$9\times100$$.
=900.
Profit percentage=$$\frac{900-630}{630}\times100$$.
=$$\frac{300}{7}$$%.
=$$42\frac{6}{7}$$%.
Hence, Option D is correct.

Let the quantities of Milk and water be M and W, respectively.
M : W =10 : 3
3M=10W
In 26 litres of the mixture
M =26(10/13) = 20 litre and
W =26(3/13) = 6 litre
8 litres of water are added.
The resulting ratio of M and W is
M-20 : W-6+8 = 5 : 2
2(M-20) = 5(W+2)
2M - 40 = 5W + 10
Multiply all the terms by 2.
4M - 80 = 10 W + 20
Replacing 10W with 3M.
4M - 80 = 3M + 20
M = 100
Hence, W would be 30.
Total quantity = 100+30 = 130.
Option E is the correct answer.

18 litres of pure water was added to a vessel containing 80 litres of pure milk.

Total mixture = 80 + 18 = 98 litres

Now, 49 litres i.e., $$\frac{1}{2}$$is removed, => Milk left = $$\frac{80}{2}$$ = 40 litres

Water left = $$\frac{18}{2}$$ = 9 litres

Let milk added be $$2x$$ litres and water added is $$x$$ litres

=> $$\frac{40 + 2x}{9 + x} = \frac{4}{1}$$

=> $$40 + 2x = 36 + 4x$$

=> $$2x = 40 - 36 = 4$$

=> $$x = \frac{4}{2} = 2$$

$$\therefore$$ Quantity of milk added = $$2 \times 2 = 4$$ litres

Mixture remaining after selling 10 litres = 40 - 10 = 30 litres

Now, quantity of water in 30 litres of mixture = $$\frac{15}{100} * 30$$ = 4.5 litres

Milk = 30 - 4.5 = 25.5 litres

After adding 12.5 litres of water, the total quantity of water = 12.5 + 4.5 = 17 litres

Required ratio of milk and water = 25.5 : 17 = 1.5 : 1 = 3 : 2

Quantity of milk in 27 litre mixture = $$\frac{4}{4 + 5} \times 27 = 12$$ litre

Quantity of water = $$27 - 12 = 15$$ litre

Ratio of milk and water in the new mixture = $$70 : 30 = 7 : 3$$

Acc to ques,

=> $$\frac{X + 12}{15} = \frac{7}{3}$$

=> $$3X + 36 = 15 \times 7 = 105$$

=> $$3X = 105 - 36 = 69$$

=> $$X = \frac{69}{3} = 23$$ litre

Let mixture in vessel A = $$x$$ ml

and mixture in vessel B = $$y$$ ml

=> Milk in vessel A = $$\frac{4x}{7}$$

Milk in vessel B = $$\frac{2y}{5}$$

Acc to ques,

=> $$\frac{4x}{7} + \frac{2y}{5} = \frac{1}{2} (x + y)$$

=> $$\frac{4x}{7} - \frac{x}{2} = \frac{y}{2} - \frac{2y}{5}$$

=> $$\frac{x}{14} = \frac{y}{10}$$

=> $$\frac{x}{y} = \frac{14}{10} = \frac{7}{5}$$

Quantity of milk in 30 l mixture = $$\frac{7}{3} \times 30 = 21$$ litre

Quantity of water = $$30 - 21 = 9$$ litre

Let the quantity of water to be added = $$x$$ litre

Acc to ques,

=> $$\frac{9 + x}{30 + x} = \frac{40}{100}$$

=> $$\frac{9 + x}{30 + x} = \frac{2}{5}$$

=> $$45 + 5x = 60 + 2x$$

=> $$5x - 2x = 60 - 45 = 15$$

=> $$x = \frac{15}{3} = 5$$ litre

15 ltr of apple juice, 20 ltr of mango juice and 5 litres of litchi juice are mixed.
So percentage of mango juice in the mixture = 20 / (15+20+5) = 50%
So in 5 ltr of this mixture 50% will be mango juice, i.e. 2.5 litres will be mango juice.
So E is the right choice.

If we are taking out 6 litres out of a 60 litre solution and replacing it with water,

=> We are replacing $$\frac{1}{10}$$th of the solution with water

=> Fraction of milk will become $$\frac{9}{10}$$th of original.

If the process is repeated 'n' times, fraction of milk will become $$(\frac{9}{10})^n$$ of the original.

Here, the process is done twice.

=> Final quantity of milk = $$(\frac{9}{10})^2 \times 60$$ = 48.6 litres

and Final quantity of water = 60 - 48.6 = 11.4 litres

$$\therefore$$ Required ratio = $$\frac{48.6}{11.4}$$ = 81 : 19

Jar A has 78 litres of a mixture of milk and water in the respective ratio of 6 : 7

=> Quantity of milk in Jar A = $$\frac{6}{13} \times 78 = 36$$ litres

Quantity of water in Jar A = $$78 - 36 = 42$$ litres

26 litres of the mixture were taken out from Jar A, i.e., $$\frac{26}{78} = (\frac{1}{3})^{rd}$$

=> Milk left = $$36 - \frac{1}{3} \times 36 = 24$$

Water left = $$42 - \frac{1}{3} \times 42 = 28$$

Let milk added to jar A = $$x$$ litres

Acc. to ques, => $$\frac{24 + x}{28} = \frac{60}{40}$$

=> $$\frac{24 + x}{28} = \frac{3}{2}$$

=> $$48 + 2x = 84$$

=> $$2x = 84 - 48 = 36$$

=> $$x = \frac{36}{2} = 18$$ litres

Jar A has 36 litres of mixture of milk and water in the respective ratio of 5 : 4

=> Quantity of milk in Jar A = $$\frac{5}{9} \times 36 = 20$$ litres

Quantity of water in Jar A = $$36 - 20 = 16$$ itres

Let quantity of water in Jar B = $$x$$ litres

=> Quantity of milk in Jar B = $$(20 - x)$$ litres

Acc. to ques, => $$\frac{20 + (20 - x)}{16 + x} = \frac{5}{3}$$

=> $$120 - 3x = 80 + 5x$$

=> $$5x + 3x = 120 - 80$$

=> $$8x = 40$$

=> $$x = \frac{40}{8} = 5$$ litres

Jar A has 60 litres of mixture of milk and water in the respective ratio of 2 : 1

=> Quantity of milk in Jar A = $$\frac{2}{3} \times 60 = 40$$ litres

Quantity of water in Jar A = $$60 - 40 = 20$$ itres

Let quantity of water in Jar B = $$x$$ litres

=> Quantity of milk in Jar B = $$(40 - x)$$ litres

Acc. to ques, => $$\frac{40 + (40 - x)}{20 + x} = \frac{13}{7}$$

=> $$560 - 7x = 260 + 13x$$

=> $$13x + 7x = 560 - 260$$

=> $$20x = 300$$

=> $$x = \frac{300}{20} = 15$$ litres

The shopkeeper wants to earn a profit of 50% by selling the mixture at Rs.37.5
Therefore, 1.5*cost price of the mixture = Rs.37.5
Cost price of the mixture = Rs.25

Let the ratio in which coffee and the chicory must be mixed be x:1

30*x + 10*1 = 25(x+1)
30x+10 = 25x + 25
5x = 15
=> x = 3
Therefore, the ratio in which coffee and chicory must be mixed is 3:1.
Therefore, option B is the right answer.

Ratio of liquid A and B in the mixture X = 2:5
Concentration of liquid A in the mixture X = 2/7.

Let the amount of liquid A in the mixture be 2x and the volume of the mixture be 7x.
Let the amount of liquid A that has to be additionally added be y.
We know that (2x+y)/(7x+y) = 0.5
2x + y = 3.5x + 0.5y
0.5y = 1.5x
y = 3x.

7x units of mixture X must be mixed with 3x units of liquid A. Therefore, mixture X and liquid A must be mixed in the ratio 7:3  for the concentration of liquid A to be 50%. Hence, option C is the right answer.

Let us say that we have 10 litres of mixture initially. So it has 8 litres of milk and 2 litres of water. Now we have to make 5 litres of milk and 5 litres of water. Hence, effectively we need to remove 3 litres of milk. Now for every 4 units of milk, 1 unit of water will also be removed. Hence, total quantity which we need to draw from the mixture will be 3*5/4 = 3.75 litres. Thus, the value of x must be 37.5 %
Hence, option D is the correct answer.

From 200 litres, he sells 15% of the milk, i.e. he sells 30 litres of the milk and replaces it with water. The quantity of pure milk with him is 170 litres.
He sells 20% of the solution, i.e. he sells 40L of the solution.
In this 40L, the amount of pure milk was 85% = 34 litres.
Therefore, the amount of pure milk in the mixture = 170-34 = 136 litres.
Therefore, our answer is option ‘C’.

Let us assume that 30 litres each of P and Q are mixed.
In 30 litres of P, milk is 25 litres and water is 5 litres.
In 30 litres of Q, milk is 18 litres and water is 12 litres.
When P and Q are mixed, quantity of milk = (25 + 18) litres = 43 litres
When P and Q are mixed, quantity of water = (5 + 12) litres = 17 litres
Required ratio is 43 : 17
Hence, option C is the correct answer.

Let 12 litres of the first solution be mixed with 24 litres of the second solution.
Therefore, total volume of the solution = (12 + 24) litres = 36 litres
Quantity of milk in the first solution = $$\frac{3}{4} * 12$$ litres = 9 litres
Quantity of milk in the second solution = $$\frac{2}{3} * 24$$ litres = 16 litres
Total quantity of milk = (9 + 16) litres = 25 litres
% of milk = $$\frac{25}{36} * 100$$% = 69.44%

Let ‘X’ litres of solution A be mixed with 100 litres of solution B.
Concentration of milk in solution B = 100 - 60 = 40%.
The resultant solution is a 50% water-milk solution.
Hence,
$$\dfrac{0.70*X + 0.40*100}{X+100} = \dfrac{50}{100}$$
$$70*X + 4000 = 50*X + 5000$$
$$20X= 1000$$
$$X= 50$$liters.
Hence, option E is the correct answer.

In an 820 litres of mixture, the ratio between the quantities of milk and water is 4:1, respectively.
Initial quantity of milk = $$\frac{4}{5}\times820$$
= 656
Initial quantity of water = $$\frac{1}{5}\times820$$
= 164
If 40 litres of the mixture is taken out and some quantity of water is added to the mixture, then the ratio between the quantities of milk and water will be 3:1, respectively.
Let’s assume the quantity of water added to the mixture is y.
$$\frac{656-32}{164+y-8} = \frac{3}{1}$$

$$\frac{624}{156+y} = \frac{3}{1}$$

156+y = 208
y = 208-156 = 52 litres
Hence, option E is the correct answer.

Statement 1 is True. Asheville had the highest percentage increase in the number of independent bookstores from 2023 to 2024. The number of bookstores grew by 13.33% - from 45 to 51.

Statement 2 is also True. The only cities that saw a decline in the number of independent bookstores between 2023 and 2024 were Austin (-2.4%) and Boston (-3.5%).

Statement 3 is False. The number of bookstores in San Francisco in 2024 was 70 - this made up roughly 10.2% of the total independent bookstores across all listed cities.