Let us say the capacity of the glass is X, and it is completely filled with milk,&nbsp;If two-thirds of its content is poured out and replaced with water, the remaining fraction of the milk will be one third.&nbsp;And this is said to be done three more times, that means a total of 4 times.&nbsp;So the contents of Milk initially being X,And after 4 times the contents will be,&nbsp;$X\left(1-\frac{2}{3}\right)^4=\frac{X}{81}$Since the total contents is X, and the milk contents is X/81, the water contents will be 80X/81.&nbsp;Ratio of milk to water will be&nbsp;$\frac{X}{81}:\ \frac{80X}{81}$Answer is&nbsp;$1:80$

Let's start from the step when there was 50% concentration.&nbsp;Let's assume total 2T solution: T acid and T water.Adding 15 litres of acid increases the acid concentration to&nbsp;80%, giving the equation&nbsp;$\frac{T+15}{2T+15}=\frac{4}{5}$Solving this would give us T=5This means that there were 5 litres of acid and 5 litres of water after mixing 2 litres of water.&nbsp;Therefore, there would be 5 litres of acid and 3 litres of water before adding the water.&nbsp;We are asked the ratio of water to acid, which would be 3:5Therefore, Option B is the correct answer.&nbsp;

Let us assume the amount of Milk in the container to be X and the amount of water in the container to be Y.&nbsp;We are told that X+Y=300.&nbsp;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"&nbsp;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,&nbsp;$\frac{2Y}{X+Y}$That means the quantity of milk that will remain in the solution will be,&nbsp;$X\left(1-\frac{2Y}{X+Y}\right)$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,&nbsp;$2Y^2-900Y+25200=0$Factorising this equation we have,&nbsp;$2\left(Y-30\right)\left(Y-420\right)=0$Y is either 30 or 420.&nbsp;Given that the capacity of the container itself is 300, Y has to be 30.&nbsp;Hence the amount of water initially is 30 Litres.&nbsp;

We are told that Rajesh manages 20 hectares and Vimal manages 30 hectaresFor Vimal, we know the distribution of the land between Wheat and Mustard, 5:3So, wheat area will be,&nbsp;$\frac{5}{8}\left(30\right)$Mustard area will be,&nbsp;$\frac{3}{8}\left(30\right)$Similarly, let us assume that the distribution of crops between Wheat and Mustard to be k:1Wheat will be,&nbsp;$\frac{k}{k+1}\left(20\right)$Mustard will be,&nbsp;$\frac{1}{k+1}\left(20\right)$We are told that total area of Wheat and Mustard is in the ratio 11:9Adding them up we get,&nbsp;$\dfrac{\left(\dfrac{150}{8}+\dfrac{20k}{k+1}\right)}{\left(\dfrac{90}{8}+\dfrac{20}{k+1}\right)}=\dfrac{11}{9}$$\dfrac{\left(\dfrac{15}{8}+\dfrac{2k}{k+1}\right)}{\left(\dfrac{9}{8}+\dfrac{2}{k+1}\right)}=\dfrac{11}{9}$$\dfrac{135}{8}+\dfrac{18k}{k+1}=\dfrac{99}{8}+\dfrac{22}{k+1}$$\dfrac{36}{8}=\dfrac{22-18k}{k+1}$$44-36k=9k+9$$45k=35$$k=\dfrac{7}{9}$Hence the ratio of distribution of area between Wheat and Mustard for Rajesh is&nbsp;$\dfrac{7}{9}$

Given that in the final mixture, the ratio of coffee and cocoa is 16:9Let us assume coffee is 16 units and cocoa is 9 units.=&gt; Initially, there are 25 units of coffee and 0 units of cocoaLet's say x units of the mixture is removed and replaced with cocoa=&gt; Now, we have (25-x) coffee and x units of cocoa. =&gt; Mixture PNow, if x units of the mixture is removed:Amount of coffee present = (25-x) - $ \dfrac{\left(25-x\right)}{25}\times\ x $=&gt; $ \left(25-x\right)\left(1-\dfrac{x}{25}\right)=16 $=&gt; $ \left(25-x\right)^2=16\times\ 25 $=&gt; 25 - x = 20 =&gt; x = 5.In mixture P, cocoa = x = 5In mixture Q, cocoa = 9 units.=&gt; Required ratio = 5:9

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 \leq 20 $=&gt; $ \left(\frac{9}{10}\right)^n \leq \frac{1}{2} $=&gt; $ n \geq 7 $Hence, the correct answer is 7

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.&nbsp;Now he mixes&nbsp;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%&nbsp;The correct option is A

Initially: a glass 500cc milk and a cup 500cc waterStep 1: 150 cc of milk is transferred to the cup from glassAfter step 1: Glass - 350 cc milk, Cup - 150 cc milk and 500 cc waterStep 2: 150 cc of this mixture is transferred from the cup to the glassAfter step 2:Glass - 350 cc milk + 150 cc mixture with milk:water ratio 3:10Cup - 500 cc mixture with milk:water ratio 3:10water in glass : milk in cup = $\frac{10}{13} \times 150 : \frac{3}{13} \times 500 = 1 : 1$

CAT Averages Mixtures Alligations

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Free CAT Quant Questions