Aptitude Questions and Answers

- 1. Alligation
- 2. Mean Price
- 3. Basic Formula
- 4. Replacement of Part of Solution Formula
- 5. Mixture of More Than Two Elements

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 a desired price.

*A process or rule for the solution of problems concerning the compounding or mixing of ingredients differing in price or quality.*

*Merrian - Webster Medical Dictionary*

The cost price of a unit quantity of the mixture is called the mean price.

If two ingredients A and B of price x and y respectively are mixed and the price of resultant mixture is M (mean price)then the ratio (R) in which ingredients are mixed is given by, **the rule of allegation**

$$ R = \left(\dfrac{M-y}{x-M}\right)$$

The above formula can be represented in diagram below, which in turn is more intuitive to grasp.

Thus the required ratio is,

$$ R = \left(\dfrac{M-y}{x-M}\right) = \left(\dfrac{y - M}{M - x}\right)$$

Suppose a container contains a solution from which some quantity of solution is taken out and replaced with one of the ingredients. This process is repeated n times then,

$\text{Final Amount of ingredient that is not replaced}=$

$$\text{Initial Amount} \times \left(\dfrac{\text{Vol. after removal}}{\text{Vol. after replacing}}\right)^n$$

Above formula is not only true for absolute amounts but for ratios as well. So following formula is also valid:

$\text{Final ratio of ingredient not replaced to total}=$

$$\text{Initial ratio} * \left(\dfrac{\text{Vol. after removal}}{\text{Vol. after replacing}}\right)^n$$

These questions may seem a little tricky at first, but it is similar concept applied repeteadly.

In order to calculate final ratio of ingredients when mixture contains more than two ingredients,

- Take two ingredients such that 1st ingredient is LOWER than the mean value and the other one is HIGHER than the mean value.
- Calculate the ratio of ingredients
- Repeat for all possible pairs
- Final ratio is the ratio obtained from step 2 (if an ingredient is common in the ratios, add values for this particular ingredient)

In what ratio must a person mix three kind of tea each of which has a price of 70, 80 and 120 rupees per kg, in such a way that the mixture costs him 100 rupees per kg?

Here the prices of tea are 70, 80 and 120 And mean price is 100.

Here, prices LOWER than the mean are 70 and 80. And prices HIGHER than the mean is 120.

Thus possible pairs which can give mean value of 100 is: {70, 120} and {80, 120}

Let us denote tea of Rs. 70 with $t_{70}$, tea of Rs. 80 with $t_{80}$ and tea of Rs. 120 with $t_{120}$

We apply the old alligation rule to ALL (two in this case) the pairs

**For the 1st pair ($t_{70}$, $t_{120}$)**

$\Rightarrow t_{70}:t_{120}=20:30$

Similarly, **for the 2st pair ($t_{80}$, $t_{120}$)**

$t_{80}:t_{120}=-20:-20$

$\Rightarrow t_{80}:t_{120}=20:20$

Thus, **Final ratio:**

$t_{70}:t_{80}:t_{120}=20:20:(30+20)$

$\Rightarrow t_{70}:t_{80}:t_{120}=2:2:5$

It's best to simplify the intermediate ratios at the end, else you may get wrong answer.

Here, if we had simplified the ratio to be, $t_{70}:t_{120}=2:3$ after first pair and $t_{80}:t_{120}=1:1$ after second pair.We could have easily gone to calculate ratio to be, $t_{70}:t_{80}:t_{120}=2:1:(3+1)$ or $t_{70}:t_{80}:t_{120}=2:1:4$ which is different from the final ratio $(t_{70}:t_{80}:t_{120}=2:2:5)$

How must a shop owner mix 4 types of rice worth Rs 95, Rs 60, Rs 90 and Rs 50 per kg so that he can make the mixture of these rice worth Rs 80 per kg?

Here the prices of sugars are 50, 60, 90 and 95.

And the mean price is 80.

Here, prices LOWER than the mean are 50 and 60. And prices HIGHER than the mean are 90 and 95.

Thus possible pairs which can give mean value of 80 are: {50, 95} and {60, 90}

Let us denote rice of Rs. 50 with $r_{50}$, rice of Rs. 60 with $r_{60}$, rice of Rs. 90 with $r_{90}$ and rice of Rs. 90 with $r_{95}$

We apply the old alligation rule to ALL (two in this case) the pairs

**For the 1st pair ($r_{50}$, $r_{95}$)**

$r_{50}:r_{95}=-15:-30$

$\Rightarrow r_{50}:r_{95}=15:30$

Similarly, **for the 2st pair ($r_{60}$, $r_{90}$)**

$r_{60}:r_{90}=-10:-20$

$\Rightarrow r_{60}:r_{90}=10:20$

Thus, **Final ratio:**

$r_{50}:r_{60}:r_{90}:r_{95}=15:10:20:30$

$\Rightarrow r_{50}:r_{60}:r_{90}:r_{95}=3:2:4:6$