## Mixture and Alligation Problem with Explanations

For all competitive exams Mixture and Alligation is an important chapter. From Mixture and Alligation chapter at least 2-3 problems are appear in all competitive exams like SSC and bank exams (SBI and IBPS-clerk and PO). This section will teach you to understand Mixture and Alligation problems with explanation and short cut to solve problems quickly

**Mixture and Alligation – Study material for all competitive exams**

Mixture and Alligation is a part of quantitative or numerical aptitude section for all competitive exams. This chapter will give brief explanation to Mixture and Alligation Problems. This will be very helpful to crack all competitive exams with more marks. “Practice makes a man perfect” so please practice more problems with different models. All the best for your preparations and success.

### Important formulas for Mixture and Alligation:

**Alligation:**It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of desired price.**Mean Price:**The cost of a unit quantity of the mixture is called the mean price.**Rule of Alligation:**If two ingredients are mixed, then

Quantity of cheaper/Quantity of dearer = C.P. of dearer/Mean price – C.P. of cheaper

(Cheaper quantity) : (Dearer quantity) = (*d*–*m*) : (*m*–*c*).- Suppose a container contains
*x*of liquid from which*y*units are taken out and replaced by water.

After*n*operations, the quantity of pure liquid = (X(1-Y/X)^{n}) units.

### Mixture and Alligation Problem with Solutions

Before go through the problems of Mixture and Alligation. Please understand and learn basic formulas of Mixture and Alligation.

Complexity of the question is differ for each exams. While learning solve high difficulty questions to boost your knowledge and success your exams.

**1) How much milk (in litres) costing Rs 60 per litres should be mixed with 35 litres of milk costing Rs 84 per litres so that there is a profit of 50% on selling the mixture at Rs 111 per litres?**

** A) 25 l**

** B) 32 l**

** C) 17 l**

** D) 36 l**

** E) 46 l**

Explanation:

CP of mixture = 100/150 * 111 = Rs 74

Let x l of milk to be mixed. So by method of allegation:

(x)……………..…..(35)

60…………………..84

………….74

10……………………..14

So ratio is 10 : 14 = 5 : 7

So x/35 = 5/7

x = 25 l

**2) A container whose capacity is 60 l contains milk and water in the ratio 3 : 2. How much quantity of the mixture should be replaced with pure milk so that in the final mixture, ratio of milk to water is 7 : 3?**

** A) 22 l**

** B) 20 l**

** C) 15 l**

** D) 17 l**

** E) 14 l**

**Explanation: **

In 60 l of mixture, milk = 3/5 * 60 = 36 l, so water = 24 l

Let x litres of mixture is replaced

So

Remaining Milk after replacement is = 36 – (3/5)*x + x = 36 + 2x/5

So (36 + 2x/5)/60 = 7/10

Solve, x = 15 l

**3) 3 containers having capacities in the ratio 2 : 3 : 1 contain mixture of liquids A and B such that the ratio of A to B in them is 2 : 3, 1 : 4 and 3 : 7 respectively. If all the three containers are emptied in a single container, what will be the ratio of A to B in the final mixture?**

** A) 13 : 58**

** B) 11 : 54**

** C) 22: 13 **

** D) 17 : 43**

** E) None of these**

Explanation:

2+3 = 5, 1+4 = 5, 3+ 7 = 10

LCM pf 5, 5, 10 = 10

Capacities are in the ratio 2 : 3 : 1

Suppose the capacities are 20, 30 and 10

So A in final mixture is 2/5 * 20 + 1/5 * 30 + 3/10 * 10 = 17

And B in final mixture is (20+30+10) – 17 = 43

So final ratio = 17 : 43

**4) 12 litres of water is drawn out from a container full of water and replaced by milk. Again 12 litres of mixture are drawn and the container is again filled with milk. The ratio of final quantity of water to milk in the container is 11 : 25. How much did the container hold?**

** A) 60 litres**

** B) 65 litres**

** C) 72 litres**

** D) 39 litrers**

** E) None of these**

**Explanation: **

Let x litres is total capacity of container

Using formula, amount of water left = x [1 – 12/x]<sup>2</sup>

[1 – 12/x]<sup>2</sup>/x = 25/(25+11)

Solving we get, x = 72 l

**5) Two containers A and B contain mixture of milk and water such that A contains 40% milk and B contains 22% milk. Some part of mixture in container A is replaced by equal quantity of mixture from container B. How much quantity of the mixture was replaced if final mixture contains 32% milk?**

** A) 3/7**

** B) 2/5**

** C) 7/10**

** D) 4/7**

** E) 5/9**

**Explanation: **

By the method allegation:

Reaming……………….Replaced

22………………………….…..40

…………….…..32

8……………………..………..10

So ratio is 8 : 10 = 4 : 5

So replaced part is 5/(4+5) = 5/9

**6) 25 litres are drawn from a cask full of wine and is then filled with water. This operation is performed one more time. The ratio of the quantity of wine now left in cask to that of the water is 36 : 85. How much wine the cask hold originally?**

** A) 66 l**

** B) 85 l**

** C) 59 l**

** D) 55 l**

** E) 46 l**

**Explanation: **

Let x l wine was there originally. So

36/(36+85) = (1 – 25/x)^{2}

Solve, x = 55 l

**7) A 24 litres of milk and water mixture contains milk and water in the ratio 3 : 5. What litres of the mixture should be taken out and replaced with pure milk so that the final mixture contains milk and water in equal proportions?**

** A) 22/3 l**

** B) 20/3 l**

** C) 3 l**

** D) 32/5 l**

** E) 24/5 l**

Explanation:

In 24 l of mixture, milk = 3/8 * 24 = 9 l, so water = 15 l

Now since the mixture is to be replaced with pure milk, the amount of mixture will remain same after replacement too.

In 24 l mixture, to have 12 l water and 12 l milk, 3 l of water should be taken out, since we are only adding milk.

Let x l of mixture taken out. So 5/8 * x = 3,

Solve, x = 24/5 l

**8) How much milk (in litres) costing Rs 50 per litres should be mixed with 18 litres of milk costing Rs 56 per litres so that there is a profit of 25% on selling the mixture at Rs 65 per litres?**

** A) 25 l**

** B) 32 l**

** C) 17 l**

** D) 36 l**

** E) 46 l**

**Explanation: **

CP of mixture = 100/125 * 65 = Rs 52

Let x l of milk to be mixed. So by method of allegation:

(x)……………..…..(18)

50…………………..56

………….52

4……………………..2

So x/18 = 4/2

x = 36 l

**9) A mixture of 30 litres contains milk and water in the ratio 7 : 3. 10 litres of the mixture is taken out and replaced with pure milk and the same operation is repeated one more time. Find the final ratio of milk to water in the mixture.**

** A) 12 : 7**

** B) 9 : 4**

** C) 13 : 2**

** D) 15 : 7**

** E) 11 : 5**

Explanation:

In 30 l of mixture, milk = 7/10 * 30 = 21l, so water = 9 l

let x = amount of water after replacement and y = amount of water before replacement, so y = 9

Now

x/y = [1 – 10/30]^{2}

Solve, x = 4 l

Now since mixture is 30 l only after replacement also. So milk in mixture after replacement = 30 – 4 = 26 l

So final ratio = 26 : 4 = 13 : 2

**10) A 56 litre mixture contains milk and water in the ratio of 5 : 2 . How much water should be added to the mixture so as make the resultant mixture containing 40% water in it?**

** A) 35/6 l**

** B) 40/3 l**

** C) 29/3 l**

** D) 27/2 l**

** E) 32/3 l**

Explanation:

In 56 l, milk = 5/(5+2) * 56 = 40 l, so water = 56 – 40 = 16 l

Final ratio of milk to water will be = 60 : 40 = 3 : 2

Let x litres of water to be added. So

40/(16+x) = 3/2

Solve, x = 32/3 l