In what ratio must rice at ₹30/kg be mixed with rice at ₹45/kg to get a mixture worth ₹36/kg?

3 : 2. By alligation the ratio (cheaper : dearer) = (45 - 36) : (36 - 30) = 9 : 6 = 3 : 2.

A 40-litre mixture of milk and water is in the ratio 3 : 1. How much water must be added to make the ratio 3 : 2?

10 L. The 40 L holds 30 L milk and 10 L water. Keeping milk at 30 L, a 3 : 2 ratio needs 20 L water, so 10 L must be added.

How many kg of sugar at ₹9/kg must be mixed with 27 kg of sugar at ₹7/kg so that the mixture is worth ₹8/kg?

27 kg. By alligation the two grades are needed in ratio (9 - 8) : (8 - 7) = 1 : 1, so 27 kg of the ₹9 sugar is required.

From a 60-litre vessel full of milk, 6 litres are drawn out and replaced with water. This is repeated once more. How much milk is now left?

48.6 L. After n replacements of x from V, milk left = V (1 - xV )^n = 60(0.9)^2 = 48.6 L.

A vessel holds milk and water in the ratio 5 : 1. On adding 5 litres of water the ratio becomes 5 : 2. How much milk is in the original mixture?

25 L. Let the mixture be 5x milk and x water. Then 5xx + 5 = 52, so 10x = 5x + 25 and x = 5. Milk = 5x = 25 L.

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Quantitative Aptitude · Averages, Ratios & Percentages

Mixtures & Alligation — Questions & Answers

Alligation is a fast way to find the ratio in which two ingredients of known value must be mixed to reach a target value. Repeated-replacement questions use a separate decay formula.

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Alligation: $\dfrac{\text{quantity cheaper}}{\text{quantity dearer}} = \dfrac{\text{dearer} - \text{mean}}{\text{mean} - \text{cheaper}}$
After removing $x$ units from $V$ and refilling $n$ times, pure quantity left $= V\left(1 - \dfrac{x}{V}\right)^n$

5 solved Mixtures & Alligation questions

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01
In what ratio must rice at ₹30/kg be mixed with rice at ₹45/kg to get a mixture worth ₹36/kg?
5 steps · reveal one at a time
1. Step 1: Understand the Goal
  We need to find the specific ratio in which two types of rice, with different costs, must be mixed to achieve a desired average cost for the final mixture. This is a classic problem for the method of alligation.
2. Step 2: Calculate Price Differences
  First, find the difference between the dearer rice price and the mixture price: $45 - 36 = 9$ . Next, find the difference between the mixture price and the cheaper rice price: $36 - 30 = 6$ .
3. Step 3: Form the Ratio
  The ratio of the cheaper rice quantity to the dearer rice quantity is found by taking the difference of the dearer price from the mean price, to the difference of the mean price from the cheaper price. So, the ratio is $(45 - 36) : (36 - 30)$ .
4. Step 4: Simplify the Ratio
  From our calculations, the ratio is $9 : 6$ . To simplify this, we can divide both sides by their greatest common divisor, which is $3$ . This gives us $9 \div 3 : 6 \div 3 = 3 : 2$ .
5. Step 5: The answer
  So the answer is (B) 3 : 2. This ratio means for every 3 units of cheaper rice, you need to mix 2 units of dearer rice to get the desired average price.
02
A 40-litre mixture of milk and water is in the ratio 3 : 1. How much water must be added to make the ratio 3 : 2?
4 steps · reveal one at a time
1. Step 1: Initial Mixture
  The total volume of the mixture is $40$ L, with milk and water in a ratio of $3 : 1$ . This means milk is $3/4$ of the total and water is $1/4$ . So, initially, there are $30$ L of milk and $10$ L of water.
2. Step 2: Target Ratio
  We want to change the ratio of milk to water to $3 : 2$ by adding only water. The amount of milk will stay the same at $30$ L. In the new ratio, if $3$ parts represent $30$ L of milk, then $1$ part represents $10$ L.
3. Step 3: Water Needed
  Since $1$ part is $10$ L, the $2$ parts for water in the new $3 : 2$ ratio will be $2 \times 10 = 20$ L. This is the total amount of water required in the new mixture to achieve the desired ratio.
4. Step 4: The answer
  So the answer is (D) 10 L. We started with $10$ L of water and need $20$ L, so we must add $20 - 10 = 10$ L of water.
03
How many kg of sugar at ₹9/kg must be mixed with 27 kg of sugar at ₹7/kg so that the mixture is worth ₹8/kg?
5 steps · reveal one at a time
1. Step 1: Understand the Goal
  We need to find out how much sugar at \text{\char₹}9/kg should be mixed with 27 kg of sugar at \text{\char₹}7/kg to get a mixture worth \text{\char`₹}8/kg. This type of problem is efficiently solved using the alligation method.
2. Step 2: Apply Alligation Rule
  The alligation rule helps find the ratio in which two ingredients must be mixed. We compare the difference between the higher price and the mean price to the difference between the mean price and the lower price.
3. Step 3: Calculate Price Ratios
  The difference for the \text{\char₹}9/kg sugar is $9 - 8 = 1$. The difference for the \text{\char₹}7/kg sugar is $8 - 7 = 1$ . So, the ratio of the quantities of \text{\char₹}9/kg sugar to \text{\char₹}7/kg sugar must be $1:1$ .
4. Step 4: Find Unknown Quantity
  Since the ratio of the two types of sugar is $1:1$ , and we have 27 kg of the \text{\char₹}7/kg sugar, we must also have 27 kg of the \text{\char₹}9/kg sugar.
5. Step 5: The answer
  So the answer is (B) 27 kg, because the alligation rule shows that the two types of sugar must be mixed in a $1:1$ ratio.
04
From a 60-litre vessel full of milk, 6 litres are drawn out and replaced with water. This is repeated once more. How much milk is now left?
5 steps · reveal one at a time
1. Step 1: Understand the process
  We start with 60 litres of pure milk. A specific amount, 6 litres, is removed and then replaced with water. This entire process is repeated one more time. We need to find the final quantity of milk.
2. Step 2: Milk after first step
  After the first draw, 6 litres of milk are removed, leaving $60 - 6 = 54$ litres of milk. When 6 litres of water are added, the total volume returns to 60 litres, but it's now a mixture. The milk concentration is $54/60 = 0.9$ .
3. Step 3: Milk after second step
  In the second step, 6 litres of the mixture are drawn out. The amount of milk removed is $6 \times 0.9 = 5.4$ litres. So, the milk remaining is $54 - 5.4 = 48.6$ litres. Adding water doesn't change the milk quantity.
4. Step 4: Using the formula
  Alternatively, we can use the formula for such problems: Milk left $= V(1 - x/V)^n$ , where $V$ is initial volume, $x$ is amount drawn/replaced, and $n$ is number of repetitions. Here, $V=60$ , $x=6$ , $n=2$ .
5. Step 5: The answer
  So the answer is (C) 48.6 L. Plugging the values into the formula gives us $60(1 - 6/60)^2 = 60(1 - 0.1)^2 = 60(0.9)^2 = 60 \times 0.81 = 48.6$ litres. This matches our step-by-step calculation.
05
A vessel holds milk and water in the ratio 5 : 1. On adding 5 litres of water the ratio becomes 5 : 2. How much milk is in the original mixture?
4 steps · reveal one at a time
1. Step 1: Set up variables
  Let the initial quantity of milk in the vessel be $5x$ litres and the quantity of water be $x$ litres. This represents the given ratio of $5:1$ between milk and water.
2. Step 2: Formulate new ratio
  When 5 litres of water are added, the quantity of water becomes $x + 5$ litres. The quantity of milk remains $5x$ . The new ratio of milk to water is given as $5:2$ . So, we can write the equation: $\frac{5x}{x + 5} = \frac{5}{2}$ .
3. Step 3: Solve for x
  Now, we solve the equation for $x$ . Cross-multiplying gives us $5x \times 2 = 5 \times (x + 5)$ . This simplifies to $10x = 5x + 25$ . Subtracting $5x$ from both sides, we get $5x = 25$ , which means $x = 5$ .
4. Step 4: The answer
  So the answer is (C) 25 L. The question asks for the amount of milk in the original mixture. We defined the original amount of milk as $5x$ . Since we found $x = 5$ , the amount of milk is $5 \times 5 = 25$ litres.

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