# Difficult Alligations or Mixtures Solved QuestionAptitude Discussion

 Q. A merchant mixes three varieties of rice costing Rs.20/kg, Rs.24/kg and Rs.30/kg and sells the mixture at a profit of 20% at Rs.30 / kg. How many kgs of the second variety will be in the mixture if 2 kgs of the third variety is there in the mixture?
 ✖ A. 1 kg ✖ B. 3 kgs ✔ C. 5 kgs ✖ D. 6 kgs

Solution:
Option(C) is correct

If the selling price of the mixture is Rs.30/kg and the merchant makes a profit of 20%, then the cost price of the mixture $=\dfrac{30}{1.2}$ = Rs.25/kg.

We need to find out the ratio in which the three varieties are mixed to obtain a mixture costing Rs.25 /kg.

Let variety $A$ cost Rs.20/kg, variety $B$ cost Rs.24 / kg and variety $C$ cost Rs.30/kg.

The mean desired price falls between $B$ and $C$.

Step 1:
Find out the ratio $Q_a : Q_c$ using alligation rule.

$\dfrac{Q_a}{Q_c}=\dfrac{30-25}{25-20}=\dfrac{1}{1}$

Be warned if we do not simplify the ratio and let if remain $\dfrac{Q_a}{Q_c}=\dfrac{5}{5}$, it changes the final answer, which will be the CORRECT answer as well. See at the end of this answer for more details.

Step 2:
Find out the ratio $Q_b:Q_c$ using alligation rule.

$\dfrac{Q_b}{Q_c}=\left(\dfrac{30-25}{25-24}\right)=\dfrac{5}{1}$

Step 3:
$Q_c$ is found by adding the value of $Q_c$ in step 1 and step 2 $= 1 + 1 = 2$

Many people (Rahul KumarMaksudDheerajPragya) are arguing over this step as for why $Q_c$ is being added, find details for the same at the end of the solution.

Therefore, the required ratio $= 1 : 5 : 2$

If there are 2 kgs of the third variety in the mixture, then there will be 5 kgs of the second variety in the mixture.

Note:
When appearing in actual exam problems of this kind are to be skipped, at least in the first go.

If you were able to solve a decent number of other problems, then you should look at this problem. Moreover, such tye of questions may have MORE THAN ONE SOLUTION which may not match the given answer choices with the correct alternate answer of yours. For an alternate (and CORRECT) answer check SundararajC's comment.

Edit1: Thank you VINAY and bimal for mentioning the anomaly, changed the option from choice $B$ to choice $C$

Edit2: As pointed by SundararajC, 1.6666 kg of 2nd variety is ASLO a correct answer (check out how) but since it is not given in the answer choices we have settled with 5 Kgs.

Edit 3: Many people are arguing over the addition of $Q_c$ in step 3.

From step 1, we have,

$\dfrac{Q_a}{Q_c}=\dfrac{1}{1}$

Say we have, 1kg of both $Q_a$ & $Q_c$.

From step 2, we have,

$\dfrac{Q_b}{Q_c}=\dfrac{5}{1}$

Say we have, 5 kgs of $Q_b$ and 1 kg of $Q_c$.

Now if we mix both these mixtures, we have following quantities in the FINAL mixture.

$Q_a=1\text{ kg}$ from mixture 1 only

$Q_b=5\text{ kg}$ from mixture 2 only

$Q_c=1\text{ kg}$ from mixture 1 AND $1\text{ kg}$ from mixture 2

Thus, $Q_c=(1+1)=2\text{2kgs}$.

Thus, Final ratio of $Q_a: Q_b:Q_c=\textbf{1:5:2}$ and NOT $1:5:1$ as some people are saying.

Hope this helps.

Edit 4: For an alternative solution, check comment by Mohit Mehta.

## (18) Comment(s)

Arun K P
()

here

ratio = 1:5:1

given c = 2 kg

hence b= 10 kg

Mohit Mehta
()

easy way:A=20/kg,B=24/kg ,C =30/kg

let $x$ kg be rice A ,$y$ be rice B and 2 is rice C

also final CP = 25

therefore,

$20x+24y+60=25(x+y+2)$

Solving we get,

$y=10-5x$

Since, $y$ cannot be zero or negative

Hence, $x$ can only be 1 giving $y = \textbf{5 kg}$

Pragya
()

cannot we do it like

b:c=5:1

so c=2kg we have to find 2nd variety let x so

5:1=x:2

x=10?

Kiran
()

Can someone explain me why resultant of type 3(Q_C) is coming into picture in this problem and not in

Question no 3:

In what ratio must a person mix three kinds of wheat costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg?

Prince
()

the answer would be 5 (option c) if we assume first quantity is 1 kg.

Kiran- In what ratio must a person mix three kinds of wheat costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg?

Correct answer-Infinitely many solution and the solution given is wrong. 12:7:7 is one possibilty

SundararajC
()

sundar:

I got the answer 1.6666 kg as follows.

Cost price of mixture is 25 R/kg

20 5

24 5

25

30 1

30 5

totally 16 parts in which third variety contains 6 parts,2 nd variety contains 5 parts and 1st variety contains 5 parts.

6 parts is 2 kg.therefore 5 parts is 1.6666kg.

so 1 st and 2 nd varieties both having 1.6666kg each.

Deepak
()

SundararajC
()

profit 20 percent. So (selling price-cost price)/cost price=0.2

So cost price becomes 25r/kg.

now

rs/kg | ave |difference

20 | | 5

|

24 | | 5

| 25 |

30 | | 1

30 | | 5

so totally 16 parts.

totally 16 parts in which third variety contains 6 parts,2 nd variety contains 5 parts and 1st variety contains 5 parts.

6 parts is 2 kg.therefore 5 parts is 1.6666kg.

so 1 st and 2 nd varieties both having 1.6666kg each.

SundararajC
()

Depak i explain simply as follows.

3rd variety (30/kg variety) is 2 kg

Selling price of mixture is 30/kg

My answer is both 1 st and 2nd variety is 1.6666kg.

Let me take this as correct answer.

Now Cost price,

$=(1.66666 \times 20)+(1.66666 \times 24)+(2 \times 30)$

$= \text{ Rs.}133.29$

Total selling amount $=(2 +1.66666+1.66666)\times 30=\text{ Rs.}160$

Now let us calculate the profit percentage,

$=\left(\dfrac{160-133.329}{133.329} \right)\times 100=20\%$

Which complies with the question.

Deepak
()

Thank you SundararajC, for your input. There is no error with your solution, multiple answers to the given question is possible. Modified the solution to incorporate your input.

This might really help others, thank you for your inputs.

Rahul Kumar
()

Answer is neither C nor B. The answer should be 10 Kg for the second variety of rice.

When $Q_a: Q_c = 1:1$ and $Q_b: Q_c = 5:1$, then it is clear that $Q_a: Q_b: Q_c = 1 : 5 : 1$ in the final mixture

Now it is given that $Q_c = 2kg$ in the final mixture.

hence, $Q_b = 5 x 2 = 10 \text{ Kg}$ in the final mixture.

Rakesh
()

I believe the solution makes it clear that the 3rd ratio is to be calculated by ADDING the values received of $Q_c$ from step 1 and 2.

That is $Q_c=1+1=2$, which makes final ratio to be, $1:5:2$ and not $1:5:1$.

So, 5 kgs appear to be right answer to me.

Maksud
()

why are few people suggesting addition of q(c)??

my ansns is 10.

ratio is 1:5:1

why are some of u guys adding 1+1 and making the ratio 1:5:2??

VINAY AHUJA
()

ans should option c ie 5

the required proportion would be 1:5:2 . it clearly indicates that if 3 quantity is 2 kg the 2 quantity must be 5 kg.

Bimal
()

is there any short-cut sir.......another thing the answer will be 'c' not 'b'....

Dheeraj
()

If we calculate the allegations of the 1st and 2nd mixtures(the ratio obtained is 1:5) followed by the 2nd and 3rd mixtures (the ratio is 5:1).

Thus if we proceed our calculations as per the original ratios concept we get 1:5:1 and the answer obtained is eqaul to 10.

How should we analyze beforehand that the question is of such type?

Naresh
()

$a:c = 2:2$

$c:b= 2:5$

So,

$a:c:b = 2:2:5$

or

$a:b:c = 2:5:2$

Naresh
()

Sorry, my mistake... since the mixture contains all the parts... it's 1:5:(1+1)
so 1:5:2 is correct ..