Numerical Applications

Alligations and Mixtures

Mixtures and Alligation:

When two or more substances are mixed together, they form a Mixture. The method to find out the quantity of each substance in a mixture is called Alligation.

Alligation is an old practice used to find out the quantities of the ingredients. There are two types of allegation:           

(i) Alligation Medial: It is used to find out the quantity of mixture when we are given the quantities of its ingredients.
(ii)Alligation Alternate: It is used to find out the quantity of each ingredient needed to make the mixture.

Here, we will discuss Alligation alternate.

Let us mix two ingredients A and B to form a mixture X. Let the quantities of A and B be a and b and the cost price per unit be C₁ and C₂ respectively. 
The per unit price of the mixture X is C_m. 
Then, the cost price of mixture X = Cost price of ingredient A + Cost price of ingredient B

$$\begin{gathered} \Rightarrow C_m\left(a+b\right)=C_{1}a+C_{2}b ...\left(1\right) \left(\because \text{Total quantity of mixture} X=a+b\right) \\ \Rightarrow C_{m}a-C_{1}a=C_{2}b-C_{m}b \\ \Rightarrow \left(C_m-C_1\right)a=\left(C_2-C_m\right)b \\ \Rightarrow \frac{a}{b}=\frac{\left(C_2-C_m\right)}{\left(C_m-C_1\right)} ...\left(2\right) \end{gathered}$$

From (1), we get

$C_m=\frac{C_{1}a+C_{2}b}{a+b}$

This shows that C_m is the weighted mean of cost price C₁ and C₂ of ingredients A and B respectively.
So, C_m lies between C₁ and C_2. Suppose $C_1<C_2$ then C₁ < C_m < C₂

Also, from (2), we get

$\frac{\text{Quantity of cheaper ingredient} A}{\text{Quantity of dearer ingredient} B}=\frac{ \text{Cost price of dearer ingredient} C_2-\text{Mean cost price} C_m}{\text{Mean cost price} C_m-\text{Cost price of cheaper ingredient} C_1}$


Alligation grid:

The result of (2), can be represented using the alligation grid:


                         $\frac{a}{b}=\frac{\left(C_2-C_m\right)}{\left(C_m-C_1\right)}$


Example: The cost of oil 1 is ₹ 50 per litre and oil 2 is ₹ 70 per litre. We make a mixture by taking oil 1 and oil 2 in the ratio of 2 : 3, find the price per litre of the mixture?

Solution: Let the price per litre of the mixture of oil be P.

Then,
 $$\begin{gathered} P=\frac{\mathrm{Price} \mathrm{of} \mathrm{Oil} 1\times \mathrm{Quantity} \mathrm{of} \mathrm{Oil} 1+\mathrm{Price} \mathrm{of} \mathrm{Oil} 2\times \mathrm{Quantity} \mathrm{of} \mathrm{Oil} 2}{\mathrm{Quantity} \mathrm{of} \mathrm{Oil} 1+\mathrm{Quantity} \mathrm{of} \mathrm{Oil} 2} \\ \Rightarrow P=\frac{50\times 2+70\times 3}{2+3} \\ \Rightarrow P=\frac{100+210}{5} \\ \Rightarrow P=\frac{310}{5} \\ \Rightarrow P=₹62 \end{gathered}$$

Hence, the price of the mixture of oil is ₹ 62 per litre.


Repeated Dilution:

Suppose a container contains x units of  A. From this, y units are taken out and replaced by the same units of B.
Then, after the first replacement
A left in x units of the mixture of  A and B $=\left(x-y\right)=x\left(1-\frac{y}{x}\right)$.
$\therefore$ A left in 1 unit of the mixture of  A and B $=\left(1-\frac{y}{x}\right)$.
$\Rightarrow$A left in y units of the mixture of  A and B $=y\left(1-\frac{y}{x}\right)$.

∴ Quantity of A left after taking out y units of the mixture of A and B
 $$\begin{gathered} =x\left(1-\frac{y}{x}\right)-y\left(1-\frac{y}{x}\right) \\ =\left(x-y\right)\left(1-\frac{y}{x}\right) \\ =x\left(1-\frac{y}{x}\right)\left(1-\frac{y}{x}\right) \\ =x{\left(1-\frac{y}{x}\right)}^2 \end{gathered}$$
Thus, the quantity of A after 2 repetition$=x{\left(1-\frac{y}{x}\right)}^2$
Similarly, the quantity of A after n repetition$=x{\left(1-\frac{y}{x}\right)}^n$
$\therefore \frac{\text{Quantity of A left in the mixture}}{\text{Original quantity of A}}={\left(1-\frac{y}{x}\right)}^n$


The mixture of two mixtures having the same ingredients:

If mixture X contains ingredients A and B in a ratio of a : b and mixture Y contains ingredients A and B in a ratio of n : m.
Now, n units of mixture X and m units of mixture Y are mixed together to form a mixture Z with ingredients A and B in the ratio of q_A : q_B.
Now, 
Quantity of A in n units of mixture X $=\left(\frac{a}{a+b}\right)n$
Quantity of B in n units of mixture X $=\left(\frac{b}{a+b}\right)n$
Quantity of A in m units of mixture Y $=\left(\frac{x}{x+y}\right)m$
Quantity of B in m units of mixture Y $=\left(\frac{x}{x+y}\right)m$
Therefore, the total quantity of A in the mixture Z = $q_{\mathrm{A}}=\left(\frac{a}{a+b}\right)n+\left(\frac{x}{x+y}\right)m$
And, the total quantity of B in the mixture Z = $q_{\mathrm{B}}=\left(\frac{b}{a+b}\right)n+\left(\frac{y}{x+y}\right)m$
 
 $\therefore \frac{q_{\mathrm{A}}}{q_{\mathrm{B}}}=\frac{\left(\frac{a}{a+b}\right)n+\left(\frac{x}{x+y}\right)m}{\left(\frac{b}{a+b}\right)n+\left(\frac{y}{x+y}\right)m}$

Then,
Total quantity of A in the mixture Z  $$\begin{gathered} =q_{\mathrm{A}}=\frac{q_{\mathrm{A}}}{q_{\mathrm{A}}+q_{\mathrm{B}}}\times \left(q_{\mathrm{A}}+q_{\mathrm{B}}\right) \\ =\frac{q_{\mathrm{A}}}{q_{\mathrm{A}}+q_{\mathrm{B}}}\times \left(n+m\right) \left(\because q_{\mathrm{A}}+q_{\mathrm{B}}=n+m\right) \end{gathered}$$

Similarly, the total quantity of B in the mixture Z $=\frac{q_{\mathrm{B}}}{q_{\mathrm{A}}+q_{\mathrm{B}}}\times \left(n+m\right) \left(\because q_{\mathrm{A}}+q_{\mathrm{B}}=n+m\right)$
 Stream:- Flow of water in a river is called the stream and the direction of the flow of water is called the direction of the stream.

Downstream:- Direction along with the flow of water in a river is called downstream and the speed of the boat along the flow of water is called downstream speed.

Upstream: Direction against the flow of water in a river is called upstream and the speed of the boat against the flow of water is called upstream speed.

Let the speed of the boat in still water be u m/sec and the speed of the stream in a river be v m/sec.

The speed of the boat along with the flow of the water in a river has assisted the movement of the boat and increases its speed.

∴ Downstream speed = (u + v) m/s                   .....(1)

The speed of the boat against the flow of the water in a river has resisted the movement of the boat and decreases its speed.

∴ Upstream speed = (u − v)m/s                          .....(2)

From (1) and (2), we get

Upstream speed$<$Downstream speed

Some important results on Boats and streams problems:-

(1) If the downstream speed of the boat is x m/s and the upstream speed of the boat is y m/s, then
Speed of the boat in still water $=\frac{1}{2}\left(x+y\right)$
Speed of the stream $=\frac{1}{2}\left(x-y\right)$

(2) Let the speed of the boat in still water and the speed of the stream be u and v m/sec respectively, then
   
     $\begin{array}{rcl}\mathrm{Average} \mathrm{Speed}& =& \frac{\left(u+v\right)\left(u-v\right)}{u}\\ & =& \frac{\mathrm{Do\dots }}{}\end{array}$