CAT Mixtures and Solutions Questions

Let us assume the amount of Milk in the container to be X and the amount of water in the container to be Y. 
We are told that X+Y=300. 

Now, an operation is given where "an amount of this solution was taken out from the container which was twice the volume of water that was earlier poured into it, and water was poured to refill the container again" 
Volume of the water initially is Y. If twice that amount is taken out, the percentage of the solution that is taken out will be, $$\frac{2Y}{X+Y}$$
That means the quantity of milk that will remain in the solution will be, $$X\left(1-\frac{2Y}{X+Y}\right)$$
This value is given to be 72%, 72% of 300 will be 216

$$X\left(1-\frac{2Y}{X+Y}\right)=216$$
$$X\left(\frac{X+Y-2Y}{X+Y}\right)=216$$
Writing $$X=300-Y$$
$$\left(300-Y\right)\left(300-2Y\right)=64800$$

Expanding this we have, 
$$2Y^2-900Y+25200=0$$
Factorising this equation we have, 
$$2\left(Y-30\right)\left(Y-420\right)=0$$

Y is either 30 or 420. 

Given that the capacity of the container itself is 300, Y has to be 30. 

Hence the amount of water initially is 30 Litres. 

Let us say the capacity of the glass is X, and it is completely filled with milk, 
If two-thirds of its content is poured out and replaced with water, the remaining fraction of the milk will be one third. 
And this is said to be done three more times, that means a total of 4 times. 
So the contents of Milk initially being X,
And after 4 times the contents will be, $$X\left(1-\frac{2}{3}\right)^4=\frac{X}{81}$$

Since the total contents is X, and the milk contents is X/81, the water contents will be 80X/81. 

Ratio of milk to water will be $$\frac{X}{81}:\ \frac{80X}{81}$$

Answer is $$1:80$$

Let's start from the step when there was 50% concentration. 
Let's take there to ee 2T solution: T acid and T water.

Adding 15 litres of acid increases the acid concentration to  80%, giving the equation $$\frac{T+15}{2T+15}=\frac{4}{5}$$
Solving this would give us T=5

This means that there were 5 litres of acid and 5 litres of water after mixing 2 litres of water. 

Therefore, there would be 5 litres of acid and 3 litres of water before adding the water. 
We are asked the ratio of water to acid, which would be 3:5

Therefore, Option B is the correct answer. 

Given that in the final mixture, the ratio of coffee and cocoa is 16:9

Let us assume coffee is 16 units and cocoa is 9 units.

=> Initially, there are 25 units of coffee and 0 units of cocoa

Let's say x units of the mixture is removed and replaced with cocoa

=> Now, we have (25-x) coffee and x units of cocoa. => Mixture P

Now, if x units of the mixture is removed:

Amount of coffee present = (25-x) - $$\dfrac{\left(25-x\right)}{25}\times\ x$$

=> $$\left(25-x\right)\left(1-\dfrac{x}{25}\right)=16$$

=> $$\left(25-x\right)^2=16\times\ 25$$

=> 25 - x = 20 => x = 5.

In mixture P, cocoa = x = 5

In mixture Q, cocoa = 9 units.

=> Required ratio = 5:9

Let the volume of mixture A be 200 ml, which implies the quantity of cocoa in the mixture is 120 ml, and the quantity of sugar In the mixture 80 ml.

Similarly, let the volume of the mixture be 300 ml, which implies the quantity of coffee, and sugar in the mixture is 210, and 90 ml, respectively.

Now we combine mixture A, and B in the ratio of 2:3 (if 200 ml mixture A, then 300 ml of mixture B).

Hence, the volume of the mixture C is (200+300) = 500 ml, and the quantity of the sugar is (90+80) = 170 ml. 

Now he mixes C with an equal amount of milk to make a drink, which implies the quantity of the final mixture is (500+500) = 1000 ml.

The quantity of sugar in the final mixture is 170 ml.

Hence, the percentage is 17% 

The correct option is A

Let's assume that after n iteration, the volume of the milk will be less than 50%, which is less than 20 liters.

Initially, the amount of milk is 40 liters, after the first iteration, the volume of milk is$$40\cdot\frac{9}{10}$$

After the second iteration, the volume of milk is $$40\times\left(\frac{9}{10}\right)^{^2}$$

Similarly, after the n iterations, the volume of milk is $$40\times\left(\frac{9}{10}\right)^{^n}$$

Now,

$$40\times\left(\frac{9}{10}\right)^{^n}\ \le\ 20$$

=> $$\left(\frac{9}{10}\right)^{^n}\ \le\ \frac{1}{2}$$

=> $$n\ge\ 7$$

Hence, the correct answer is 7

It is given,

7C = 30P = 9A and Ankita bought 4C, 14P and 6A.

Let 7C = 30P = 9A = 630k

C = 90k, P = 21k, and A = 70k

Cost price of 4C, 14P and 6A = 4(90k)+14(21k)+6(70k) = 1074k

Marked up price = 1074k + 1752

S.P = $$\frac{1}{6}\left(1074k+1752\right)+\left(\frac{4}{5}\right)\left(\frac{5}{6}\right)\left(1074k+1752\right)$$ = $$\frac{5}{6}\left(1074k+1752\right)$$

S.P - C.P = profit

$$1460-\frac{1074k}{6}=744$$

$$\frac{1074k}{6}=716$$

k = 4

Money spent on buying almonds = 420k = 420*4 = Rs 1680

The answer is option A.

Initially: a glass 500cc milk and a cup 500cc water

Step 1: 150 cc of milk is transferred to the cup from glass

After step 1: Glass - 350 cc milk, Cup - 150 cc milk and 500 cc water

Step 2: 150 cc of this mixture is transferred from the cup to the glass

After step 2: 

Glass - 350 cc milk + 150 cc mixture with milk:water ratio 3:10

Cup - 500 cc mixture with milk:water ratio 3:10

water in glass : milk in cup = $$\frac{10}{13}\times150\ :\ \frac{3}{13}\times500=1:1$$

The answer is option A.

Lemon juice : sugar syrup in the mixture is 1:1, i.e. 50% Lemon juice and 50% sugar syrup.

In sugar syrup, 100% is sugar syrup.

These two are mixed in the ratio 1:3.

Lemon juice = $$\ \dfrac{\ 1\left(50\%\right)}{1+3}$$

Sugar syrup = $$\ \dfrac{\ 1\left(50\%\right)+3\left(100\%\right)}{1+3}=\dfrac{350}{4}$$

Required ratio = 50:350 = 1:7

The answer is option D.

Step 1: Half the content of the first container is transferred to the second container

Step 2: Half of the mixture of second container is transferred back to the first container

Step 3:  Half the content of the first container is transferred back to the second container

Sugar syrup : Milk in second container = 62.5 : 75 = 5 : 6

The answer is option D.

Let initial volume be V, final be F for milk.

The formula is given by : $$F\ =\ V\cdot\left(1-\frac{K}{V}\right)^n$$ n is the number of times the milk is drawn and replaced.

so we get $$F=\ V\left(1-\frac{K}{V}\right)^{^2}$$
here K =9
we get
$$\frac{16}{25}V\ =\ V\ \left(1-\frac{9}{V}\right)^{^2}$$
we get $$1-\frac{9}{V}=\ \frac{4}{5}or\ -\frac{4}{5}$$

If considering $$1-\frac{9}{V}=-\frac{4}{5}$$

V =5, but this is not possible because 9 liters is drawn every time.

Hence : $$1-\frac{9}{V}=\frac{4}{5},\ V\ =\ 45\ liters$$

Let the alloy contain x Kg silver and y kg copper 
Now when mixed with 3Kg Pure silver 
we get $$\frac{\left(x+3\right)}{x+y+3}=\frac{9}{10}$$
we get 10x+30 =9x+9y+27
9y-x=3    (1)
Now as per condition 2
silver in 2nd alloy = 2(0.9) =1.8
so we get$$\frac{\left(x+1.8\right)}{x+y+2}=\frac{21}{25}$$
we get 21y-4x =3   (2)
solving (1) and (2) we get y= 0.6 and x =2.4
so x+y = 3

Considering the three kinds of tea are A, B, and C.

The price of kind A = Rs 800 per kg.

The price of kind B = Rs 500 per kg.

The price of kind C = Rs 300 per kg.

They were mixed in the ratio of 2 : 3: 5.

1/6 of the total mixture is sold for Rs 700 per kg.

Assuming the ratio of mixture to A = 12kg, B = 18kg, C =30 kg.

The total cost price is 800*12+500*18+300*30 = Rs 27600.

Selling 1/6 which is 10kg for Rs 700/kg the revenue earned is Rs 7000.

In order to have an overall profit of 50 percent on Rs 27600.

Thes selling price of the 60 kg is Rs 27600*1.5 = Rs 41400.

Hence he must sell the remaining 50 kg mixture for Rs 41400 - Rs 7000 = 34400.

Hence the price per kg is Rs 34400/50 = Rs 688

Let Bottle A have an indigo solution of strength 33% while Bottle B have an indigo solution of strength 17%.

The ratio in which we mix these two solutions to obtain a resultant solution of strength 21% : $$\frac{A}{B}=\frac{21-17}{33-21}=\frac{4}{12}or\ \frac{1}{3}$$

Hence, three parts of the solution from Bottle B is mixed with one part of the solution from Bottle A. For this process to happen, we need to displace 600 cc of solution from Bottle A and replace it with 600 cc of solution from Bottle B {since both bottles have 800 cc, three parts of this volume = 600cc}.As a result, 200 cc of the solution remains in Bottle B.

Hence, the correct answer is 200 cc.  

Initially let's consider A and B as one component

The volume of the mixture is doubled by adding A(60% alcohol) i.e they are mixed in 1:1 ratio and the resultant mixture has 72% alcohol.

Let the percentage of alcohol in component 1 be 'x'.

Using allegations , $$\frac{\left(72-60\right)}{x-72}=\frac{1}{1}$$ => x= 84

Percentage of alcohol in A = 60% => Let's percentage of alcohol in B = x%

 The resultant mixture has 84% alcohol. ratio = 1:3

Using allegations , $$\frac{\left(x-84\right)}{84-60}=\frac{1}{3}$$

=> x= 92%

Initially the amount of Dye and Water are 16,24 respectively.

To make the ratio of Dye to Water to 2:5 the amount of water should be 40l for 16l of Dye=> 16l of water is added.

Now, the Dye and Water arr 16,40 respectively.

After removing 1/4th of solution the amount of Dye and Water will be 12,30l respectively.

To have Dye and Water in the ratio of 2:3, for 30l of water we need 20l of Dye => 8l of Dye should be added.

Hence , 8 is correct answer. 

Let the volume of Metals A,B,C we 3x, 4x, 7x

Ratio weights of given volume be 5y,2y,6y

.'. 15xy+8xy+42xy=130 => 65xy=130 => xy=2.

.'.`The weight, in kg. of the metal C is 42xy=84.

Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively.

The amount of salt in vessels A, B, C = 50 ml, 110 ml, 160 ml respectively.

The amount of water in vessels A, B, C = 450 ml, 390 ml, 340 ml respectively.

In 100 ml solution in vessel A, there will be 10ml of salt and 90 ml of water

Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A

i.e after the first transfer, the amount of salt in vessels A, B, C = 40, 120, 160 ml respectively.

after the second transfer, the amount of salt in vessels A, B, C =40, 100, 180 ml respectively.

After the third transfer, the amount of salt in vessels A, B, C = 70, 100, 150 respectively.

Each transfer can be captured through the following table.

Percentage of salt in vessel A =$$\ \frac{\ 70}{500}\times\ 100$$

=14%

The weight/volume(g/L) for liquid 1 = 1000

The weight/volume(g/L) for liquid 2 = 800

The weight/volume(g/L) of the mixture = 480/(1/2) = 960

Using alligation the ratio of liquid 1 and liquid 2 in the mixture = (960-800)/(1000-960) = 160/40 = 4:1

Hence the percentage of liquid 1 in the mixture = 4*100/(4+1)=80

Let the volume of the first and the second solution be 100 and 300.
When they are mixed, quantity of ethanol in the mixture 
= (20 + 300S)
Let this solution be mixed with equal volume i.e. 400 of third solution in which the strength of ethanol is 20%.
So, the quantity of ethanol in the final solution 
= (20 + 300S + 80) = (300S + 100)
It is given that, 31.25% of 800 = (300S + 100)
or, 300S + 100 = 250
or S = $$\frac{1}{2}$$ = 50%
Hence, 50 is the correct answer.

The selling price of the mixture is Rs.40/kg. 
Let a be the price of 1 kg of tea A in the mixture and b be the price per kg of tea B. 
It has been given that the profit is 10% if the 2 varieties are mixed in the ratio 3:2
Let the cost price of the mixture be x. 
It has been given that 1.1x = 40
x = 40/1.1
Price per kg of the mixture in ratio 3:2 = $$\frac{3a+2b}{5} $$
$$\frac{3a+2b}{5} = \frac{40}{1.1}$$
$$3.3a+2.2b=200$$ --------(1)

The profit is 5% if the 2 varieties are mixed in the ratio 2:3.
Price per kg of the mixture in ratio 2:3 = $$\frac{2a+3b}{5}$$
$$\frac{2a+3b}{5} = \frac{40}{1.05}$$
$$2.1a+3.15b=200$$ ------(2)

Equating (1) and (2), we get, 
$$3.3a+2.2b = 2.1a+3.15b$$
$$1.2a=0.95b$$
$$\frac{a}{b} = \frac{0.95}{1.2}$$
$$\frac{a}{b} = \frac{19}{24}$$

Therefore, option C is the right answer. 

Let the price of peanuts be Rs. 100x per kg
Then, the price of walnuts = Rs. 300x per kg
Cost price of peanuts for the shopkeeper = Rs. 110x per kg
Cost price of walnuts for the shopkeeper = Rs. 360x per kg
Total cost incurred to the shopkeeper while buying = Rs.(8 * 110x + 16 * 360x) = Rs. 6640x
Since, 5kg walnut and 3kg peanuts are lost in transit, the shopkeeper will be remained with (16-5)+(8-3)=16kgs of nuts
Total selling price that the shopkeeper got = Rs. (166 * 16) = Rs. 2656
Profit = 25%
So, cost price = Rs. 2124.80
Therefore, 6640x = 2124.80
On solving, we get x = 0.32
Therefore, price of walnuts = Rs. (300 * 0.32) = Rs. 96 per kg.
Hence, option A is the correct answer

Let the price of paint B be x. 
Price of paint A = x+8

We know that the amount of paint B in the mixture does not exceed the amount of paint A. Therefore, paint B can at the maximum compose 50% of the mixture. 

The seller sells 10 litres of paint at Rs.264 earning a profit of 10%. 
=> The cost price of 10 litres of the paint mixture = Rs. 240

Therefore, the cost of 1 litre of the mixture = Rs.24

We have to find the highest possible cost of paint B. 
When we increase the cost of paint B, the cost of paint A will increase too. If the cost price of the mixture is closer to the cost of paint B, then the amount of paint B present in the mixture should be greater than the amount of paint A present in the mixture.
The highest possible cost of paint B will be obtained when the volumes of paint A and paint B in the mixture are equal. 

=> (x+x+8)/2 = 24
2x = 40
x = Rs. 20

Therefore, option C is the right answer. 

Final quantity of alcohol in the mixture = $$\dfrac{700}{700+175}*(\dfrac{90}{100})^2*[700+175]$$ = 567 ml

Therefore, final quantity of water in the mixture = 875 - 567 = 308 ml

Hence, we can say that the percentage of water in the mixture = $$\dfrac{308}{875}\times 100$$ = 35.2 %

It is given that in drum 1, A and B are in the ratio 18 : 7. 

Let us assume that in drum 2, A and B are in the ratio x : 1.

It is given that drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7.

By equating concentration of A 

$$\Rightarrow$$ $$\dfrac{3*\dfrac{18}{18+7}+4*\dfrac{x}{x+1}}{3+4} = \dfrac{13}{13+7}$$

$$\Rightarrow$$ $$\dfrac{54}{25}+\dfrac{4x}{x+1} = \dfrac{91}{20}$$

$$\Rightarrow$$ $$\dfrac{4x}{x+1} = \dfrac{239}{100}$$

$$\Rightarrow$$ $$x = \dfrac{239}{161}$$

Therefore, we can say that in drum 2, A and B are in the ratio $$\dfrac{239}{161}$$ : 1 or 239 : 161. 

Let 'a', 'b' and 'c' be the concentration of salt in solutions A, B and C respectively. 

It is given that three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%.

$$\Rightarrow$$ $$\dfrac{a+2b+3c}{1+2+3} = 20$$

$$\Rightarrow$$ $$a+2b+3c = 120$$ ... (1)

If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%.

$$\Rightarrow$$ $$\dfrac{3a+2b+c}{1+2+3} = 30$$

$$\Rightarrow$$ $$3a+2b+c = 180$$ ... (2)

From equation (1) and (2), we can say that 

$$\Rightarrow$$ $$b+2c = 45$$

$$\Rightarrow$$ $$b = 45 - 2c$$

Also, on subtracting (1) from (2), we get

$$a - c = 30$$

$$\Rightarrow$$ $$a = 30 + c$$

In solution D, B and C are mixed in the ratio 2 : 7

So, the concentration of salt in D = $$\dfrac{2b + 7c}{9}$$ = $$\dfrac{90 - 4c + 7c}{9}$$ = $$\dfrac{90 + 3c}{9}$$

Required ratio = $$\dfrac{90 + 3c}{9a}$$ = $$\dfrac{90 + 3c}{9 (30 + c)}$$ = $$1 : 3$$

Hence, option B is the correct answer.

The ratio of milk and water in Bottle 1 is 7:2 and the ratio of milk and water in Bottle 2 is 9:4
Therefore, the proportion of milk in Bottle 1 is $$\frac{7}{9}$$ and the proportion of milk in Bottle 2 is $$\frac{9}{13}$$

Let the ratio in which they should be mixed be equal to X:1.

Hence, the total volume of milk is $$\frac{7X}{9}+\frac{9}{13}$$
The total volume of water is $$\frac{2X}{9}+\frac{4}{13}$$
They are in the ratio 3:1

Hence, $$\frac{7X}{9}+\frac{9}{13} = 3*(\frac{2X}{9}+\frac{4}{13})$$
Therefore, $$91X+81=78X+108$$

Therefore $$X = \frac{27}{13}$$

The proportion of water in the first mixture is $$\frac{1}{3}$$
The proportion of Liquid A in the first mixture is $$\frac{2}{3}$$

The proportion of water in the second mixture is $$\frac{1}{4}$$
The proportion of Liquid B in the second mixture is $$\frac{3}{4}$$

The proportion of water in the third mixture is $$\frac{1}{5}$$
The proportion of Liquid C in the third mixture is $$\frac{4}{5}$$

As they are mixed in the ratio 4:3:2, the final amount of water is $$4 \times \frac{1}{3} + 3 \times \frac{1}{4} + 2 \times \frac{1}{5} = \frac{149}{60}$$
The final amount of Liquid A in the mixture is $$4\times\frac{2}{3} = \frac{8}{3}$$
The final amount of Liquid B in the mixture is $$3\times\frac{3}{4} = \frac{9}{4}$$
The final amount of Liquid C in the mixture is $$2\times\frac{4}{5} = \frac{8}{5}$$

Hence, the ratio of Water : A : B : C in the final mixture is $$\frac{149}{60}:\frac{8}{3}:\frac{9}{4}:\frac{8}{5} = 149:160:135:96$$

From the given choices, only option C  is correct.

After selling 1/4th of the mixture, the remaining quantity of water is 15 liters and milk is 60 liters. So the milkman would add 25 liters of water to the mixture. The total amount of water now is 40 liters and milk is 60 liters. Therefore, the required ratio is 2:3.

Fresh grapes contain 90% water so water in 20kg of fresh pulp = (90/100)x20= 18kg. 

In 20kg fresh grapes, the weight of water is 18kg and the weight of pulp is 2kg.

The concept that we apply in this question is that the weight of pulp will remain the same in both dry and fresh grapes. If this grape is dried, the water content will change but pulp content will remain the same.

Suppose the weight of the dry grapes be D.

80% of the weight of dry grapes = weight of the pulp = 2 kg

(80/100) x D =2 kg.

D = 2.5 kg

For the mixture to be 100 times as sweet as glucose, its sweetness relative to the mixture should be at least 74.

1 gm of saccharin = 675

Let the number of grams of sucrose to be added be N. Thus, the total weight of the mixture = N + 1.

So, (675 + N) / (N+1) = 74

=> 675 + N = 74N + 74

=> 601 = 73N => N = 8.23

When N=9, sweetness will be S = (675+9)/10 = 684/10 = 68.4

When N=8, sweetness will be S = (675+8)/9 = 683/9 = 75.8

So, option b) is the correct answer.

The relative sweetness of the mixture is (1*0.74 + 2*1 + 3*1.7) / (1+2+3) = 7.84/6 = 1.30

Option a) is the correct answer.

Let the volume of the cup be V.
Hence, after removing three cups of alcohol from the first container,

Volume of alcohol in the first container is 500-3V
Volume of water in the second container is 500 and volume of alcohol in the second container is 3V.
So, in each cup, the amount of water contained is $$\frac{500}{500+3V}*V$$

Hence, after adding back 3 cups of the mixture, amount of water in the first container is $$0+\frac{1500V}{500+3V} $$
Amount of alcohol contained in the second container is $$3V - \frac{9V^2}{500+3V} = \frac{1500V}{500+3V}$$

So, the required proportion of water in the first container and alcohol in the second container are equal.

Let the total weight of fresh grapes be 100 gm.

=> Fresh grapes have 90 gm of water and 10 gm of fruit.

When these grapes are dried, the amount of fruit does not change.

=> 10 grams will become 80% of the content in dry grapes

=> Weight of dry grapes = $$\frac{10}{0.8}$$ = 12.5 gm

So, the weight of fresh grapes reduces to 1/8th of its original weight.

=> 20 kg of fresh grapes give 2.5 kg of dry grapes.

Fraction of A in contained 1 = $$\frac{5}{6}$$

Fraction of A in contained 2 = $$\frac{1}{4}$$

Let the ratio of liquid required from containers 1 and 2 be x:1-x

x($$\frac{5}{6}$$) + (1-x)($$\frac{1}{4}$$) = $$\frac{1}{2}$$

$$\frac{7x}{12}$$ = $$\frac{1}{4}$$

=> x = $$\frac{3}{7}$$

=> Ratio = 3:4