Direct Indirect proportion and relation


Direct and Inverse proportionality Method:
Most problems in arithmetic can be solved by observing the relationship between the given variables.  Especially inverse relationship is omnipresent in most of the problems.  This relationship can be observed in the following chapters. 

Quantity x Concentration = Constant (Mixtures and Allegations Problems)
Price x Quantity = Constant (Profit and Loss)
Time x Speed = Distance (Time Speed and Distance)
Days x Efficiency = Total Work (Time and Work) 

Let us throw some light on the problems based on the above concept.

Example 1:
Due to reduction in the price of mangoes by 30%, A person got  15 mangoes more for the same amount.  What is the number of mangoes originally purchased?
Traditional Method:
We know that Price x quantity = Expenditure.  Assume Price = P; Quantity = Q; Expenditure = E.  Then we can write 
$ \Rightarrow \displaystyle\frac{E}{P} = Q$
But we know that if price of the mangoes got reduced by 30%, new price could be 70% (P).  Since the expenditure is constant, that person got 15 mangoes extra at the reduced price. 
$ \Rightarrow \displaystyle\frac{E}{{70\% (P)}} = Q + 15$
we can substitute $\displaystyle\frac{E}{P} = Q$
$ \Rightarrow \displaystyle\frac{Q}{{70\% }} = Q + 15$  $ \Rightarrow \displaystyle\frac{Q}{{\frac{{70}}{{100}}}} = Q + 15$
$ \Rightarrow \displaystyle\frac{{100(Q)}}{{70}} = Q + 15$  $ \Rightarrow \displaystyle\frac{3}{7}Q = 15$
So Q = 35

Inverse proportionality Method:
We know this relationship  Price x Quantity = Expenditure (or) P x Q = E
If price got changed to 70% (Price) or $\displaystyle\frac{7}{{10}}$ (price), quantity must change to $\displaystyle\frac{{10}}{7}$ of the original quantity to keep Expenditure constant.
$ \Rightarrow \displaystyle\frac{7}{{10}}P \times \frac{{10}}{7}Q = E$
But we know that the quantity difference is 15 mangoes.  
$ \Rightarrow \displaystyle\frac{{10}}{7}Q - Q = 15$
So Q= 35.

Example 2:
A boy reached school 10 min early if he travels at a speed of 4 Kmph.  But If he reduces his speed to 3 Kmph, He is 10 min late.  What is the Distance between his school and House?
Traditional method: 
Assume the distance between the school and the boy's home is D km and actual time is T hours.
If he travels at the speed of  4 kmph he is early by 10 min or $\displaystyle\frac{{10}}{{60}}Hours$.  On the other hand he is 10 min or $\displaystyle\frac{{10}}{{60}}Hours$ late when he travels at 3 kmph. 
$ \Rightarrow \displaystyle\frac{D}{4} = T - \frac{{10}}{{60}}Hours$  ---- (1)
$ \Rightarrow \displaystyle\frac{D}{3} = T + \frac{{10}}{{60}}Hours$  ----(2)
By (2) - (1) we get
$ \Rightarrow \displaystyle\frac{D}{3} - \frac{D}{4} = \frac{{20}}{{60}}$  $ \Rightarrow \displaystyle\frac{{4D - 3D}}{{12}} = \frac{1}{3}$
$ \Rightarrow D = \displaystyle\frac{{12}}{3} = 4$


Inverse proportionality Method:
If we assume his original speed is S kmph, then his speed changed from 4 kmph to 3 kmph.  or we can say the speeds are in the ratio 4:3.  otherwise his speeds are S, $\displaystyle\frac{3}{4}S$. 
But we know that S x T = D
$ \Rightarrow \displaystyle\frac{3}{4}S \times \frac{4}{3}T = D$.
But we know that the total time he is taking extra while reducing his speed from 4 to 3 is 20 min or 1/3 Hr.
$ \Rightarrow \displaystyle\frac{4}{3}T - T = \frac{1}{3}$
or T = 1 hr.
So the boys takes one hour to reach his school while traveling at 4 kmph. so distance is equal to Speed x time = 4 x 1 hr = 4 km.

Two more important relations between variables:
Apart from direct proportion and indirect proportion, there exist another two relations between variables.
1. Direct relation
2. Indirect relation

1. Direct Relation:
Here the relation between variables can be best defined as Y = K + mX
Even though x is zero, y is not zero.  It take a value of K.  But Y varies directly in relation to x.  If x increases, y also increases, If x decreases y also decreases.

Example 3:
The expenses of organizing a garden party increased from Rs.9000 to Rs.12000 when the number of registered candidates increased from 25 to 40. Find the total cost of organizing the party if there are 50 final registrations.
We know that the expenses of party increases not directly proportional but directly relational.
Assume the fixed component of expense is K rupees and Variable component is M rupees.
Then the total cost is given by
K + 25 x M = 9000 ............(1)
When there are 40 registrations the total cost is
 K + 40 x M = 12000 .........(2)
By substracting (1) from (2)
$ \Rightarrow 15M = 3000 \Rightarrow M = 200$
So the variable cost per head is Rs.200
To find the fixed cost we can substitute Rs.200 in either (1) or (2), then K = 4000
If there are 50 registrations then the total cost = 4000 + 50 x 200 = Rs.14000

2. Inverse Relation:
Here Y = K -mX holds good. If x increases, y decreases and vice versa.


Example 4:
The reduction in the speed of an engine is directly proportional to the square of the number of bogies attached to it.  The speed of the train is 100 km/hr when there are 4 bogies and 55 km/hr when there are 5 bogies.  What is the maximum number of bogies that can be attached to the train so that, even with those many numbers of bogies, it can just move?
We know that  $S = {S_{\max }} - K \times {(bogies)^2}$
When there are 4 bogies speed of the train is 100km/hr $100 = {S_{\max }} - K \times {(4)^2}$ $\Rightarrow 100 = {S_{\max }} - K \times 16$ .......(1)
When there are 5 bogies speed of the train is 55km/hr
$55 = {S_{\max }} - K \times {(5)^2}$ $\Rightarrow$ 55 = ${S_{\max }} - K \times 25$ ........(2)
(2) - (1) gives $9K = 45 \Rightarrow K = 5$
By substituting K = 5 in either (1) or (2) we can find the maximum speed of the train = 180 km/hr
Assume, for N number of bogies the train does not move.
$ \Rightarrow 0 = 180 - 5{(N)^2}$$ \Rightarrow N = 6$
If the engine is attached to 6 bogies, it does not move. If we want the train to move we need to attach maximum of 5 bogies