# Gyan Room : Arithmetic - Concepts & Shortcuts

• Rule of Alligation

Eg: A mixture of water and milk is worth 96 Rupees per litre. If there is 18 L of water present in the mixture and Milk is worth 112 rupees. What is the quantity of the Milk in the mixture if we assume Water is free?

Water = 0 .... Milk = 112 Rupees
.... 96 rupees ....
112 - 96 = 16 : 96

Water : Milk = 1:6
So the mixture has 6 x 18 = 108 Litres of Milk.

• Wine Water Formula

If a container has a liters of liquid A; and b liters are withdrawn and replaced by another liquid B of equal quantity and the operation is repeated n times, then:

1. (Liquid A left after nth operation)/(Initial quantity of A in the vessel) = (a-b)^n/(a)^n
2. (Liquid A left after nth operation )/(Liquid B left after nth operation) = ((a-b)^n/(a)^n )/(1- (a-b)^n/(a)^n )

Eg: A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

Quantity of milk after 3 operations = 40 x (40 - 4)^3/40^3 = 29.16

• Application of Rule of Alligation in TSD and Profit/Loss

Eg: A person covers 800km partly at a speed of 50kmph and partly at a speed of 150kmph, in 10 hours over all. What is the distance covered at the speeds of 150 kmph?

Average speed = 800/10 = 80 kmph
Apply Rule of Alligation

50 kmph .... 150 kmph
... 80 kmph ...
70 : 30 ( 150 - 80 = 70 : 80 - 50 = 30 )

So out of 10 hours, 30 x 10 / 100 = 3 hours in 150 kmph.
So distance covered = 150 kmph x 3 = 450 km.

Eg: Shopkeeper sold 45 kg of goods. If he sells some quantity at a loss of 3% and rest at 17% profit, making 5% profit on the whole, find the quantity sold at profit?

17% ... -3% (Loss)
.... 5% ....
8 : 12

At Profit : At loss = 8 : 12
Quantity sold at profit = 45 x 8/20 = 18 kg

1. Where to Use Weighted Average and Where to use Alligation ?
See any Question which you can solve by Weighted average can 100% be solved by Alligation .. Because Alligation , in its core is weighted Average Formula rearranged a bit
Use Alligation When : OVERALL AVERAGE is Given
Use Weighted Average : When OVERALL AVERAGE is not given and only Weights or Ratios of Weights are given

2. For faster calculation, use the Method of DEVIATION while solving Average n Alligation question..This makes the calculations very fast

3. Don't use ACTUAL WEIGHTs rather always USE RATIO OF WEIGHTS while solving Weighted average or Alligations

MOST MOST IMPORTANT POINTS : (SHORTCUTS )

a) When You use Alligation on SPEEDS , you get ratio of TIME (not Distance )

b) When you use Alligation on INVERSE OF SPEED, then you get ratio of DISTANCES

• Circular Motion

When two persons A, and B are running around a circular track of length L mts with speeds of a, b m/s in the same direction

• They meet each other at any point on the track is L/(a−b) seconds
• They meet each other at exactly at the starting point = (L/a, L/b) seconds

When two persons A, and B are running around a circular track of length L mts with speeds of a, b m/s in opposite direction,

• They meet each other at any point on the track is L/(a+b) seconds
• They meet each other at exactly at the starting point = (L/a,L/b) seconds

When three persons A, B and C are running around a circular track of length L mts with speeds of a, b and c m/s in the same direction direction,

• They meet each other at any point on the track is L/(a−b), L/(b−c) seconds

• A neat TSD trick from Deepak Mehta (CAT 100 percentiler)

Ram and Shyam start at same end of a swimming competition.The length of pool is 50m.The race is for completing 1000m. If Ram beats Shyam and meets him 17 times during the race what could be the speed of Shyam if speed of Ram is 5m/s?

Every time Ram and Shyam meet, they (combined) cover a distance of 100m (2x the length of the pool).

To illustrate this, consider the first time they meet. Ram will be on the second leg (back from the other end), and Shyam will be on his first (since he is slower).

You can see that Ram’s and Shyam’s combined trajectory is 2x the length of the pool; i.e. 100m.

Now consider the second time they meet. Depending on Shyam’s speed, he could still be in his 1st lap (in which case Ram would be on his 4th), or he could be in his 2nd (with Ram being in his 3rd).

In either case, we have 4 laps in total; i.e. a distance of 200m.

So, if they meet 17 times, total distance covered is 17∗100=1700m.

Out of which Ram covers 1000m (since he wins the race). So Shyam covers 1700–1000=700m.

Since Ram’s speed is 5m/s, Shyam’s is: 5 ∗ 700/1000=3.5m/s.

• Successive Percentage Change

This is a very important concept in percentages and we can use it when the percentage values are good.

The formula goes like this: If a certain value is first changed by a%, then by b%; then the net percentage change is given by
(a + b + ab/100)%

3 important things to NOTE here:

1. The order of the changes does NOT matter. So, if it was b% change followed by a% change, even then the net change would have been the same.

2. Note that I wrote percentage change, not percentage increase/decrease. It works good with both. Just take the appropriate signs. For increase, use a (+) value and for decrease a (-) value. If the final result is (+) or (-), that will denote whether there is a net increase or decrease

3. Remember that the final result is also in percentage terms, not the value.

Examples:
20% increase followed by 30% decrease
20 - 30 - 20*30/100 = -16%

20% increase, followed by 50% increase, followed by 30% decrease
Use the formula on any two of them first
20+50+20 * 50/100 = 80%
Then use the result with the 3rd value
80 - 30 - 80 * 30/100 = 26%

• Componendo-Dividendo

Now will see another interesting observation. Most of you by now must be knowing that reapplying componendo-dividendo on the componendo-dividendo form will give back the original form
That is if we have,
(x+y)/(x-y) = 5/2
Reapplying comp-div on the left side, we will get x/y
But if we apply on one side, we should be applying on both sides, to get:
x/y = (5+2)/(5-2) = 7/3
Even if you know this, do you know why this happens?
The reason is that the coefficient of 'x' in the numerator and that of 'y' in the denominator adds to 0
+1-1 = 0
So, if we reapply the same process we get back the original form.
This is a very helpful understanding, example:

(2x+3y)/(2x-3y) = 5/4 ... this is a com-div form over 2x and 3y
=> 2x/3y = (5+4)/(5-4) = 9
=>x/y = 27/2

Let us extend this to something a bit different
(2x+3y)/(3x-2y) = 5/4
It's not a comp-div form, but observe that the coefficient of 'x' in the numerator (+2) and that of 'y' in the denominator (-2), adds to 0
thus, if we reapply the whole process we should get back x/y
But what is the process?
(2 * numerator + 3 * denominator)/(3 * numerator - 2 * denominator)
so, x/y = (2 * 5 + 3 * 4)/(3 * 5 - 2 * 4) = 22/7

• Multiplication factor

This is a very important understanding of percentages from calculations point of view. What this basically does is: it converts a percentage question to a ratio question. And dealing with ratios are always easier than to deal with percentages, as ratios are easier to calculate.

So, if a value increases by 20%, which is 1/5th: the multiplication factor must be (1+1/5) = 6/5, as the value is increasing
If the value reduces by 30%, which is 3/10: the multiplication factor must be (1 - 3/10) = 7/10, as the value is reducing

Try to use multiplication factors in the basic questions to ease up your calculations

• If two objects A and B start simultaneously from opposite points and, after meeting, reach their destinations in ‘a’ and ‘b’ hours respectively (i.e. A takes ‘a hrs’ to travel from the meeting point to his destination and B takes ‘b hrs’ to travel from the meeting point to his destination), then the ratio of their speeds is given by:
Sa/Sb = √(b/a)

i.e. Ratio of speeds is given by the square root of the inverse ratio of time taken.

Two trains A and B starting from two points and travelling in opposite directions, reach their destinations 9 hours and 4 hours respectively after meeting each other. If the train A travels at 80kmph, find the rate at which the train B runs.

Sa/Sb = √(4/9) = 2/3
This gives us that the ratio of the speed of A : speed of B as 2 : 3.
Since speed of A is 80 kmph, speed of B must be 80 * (3/2) = 120 kmph

A and B start from Opladen and Cologne respectively at the same time and travel towards each other at constant speeds along the same route. After meeting at a point between Opladen and Cologne, A and B proceed to their destinations of Cologne and Opladen respectively. A reaches Cologne 40 minutes after the two meet and B reaches Opladen 90 minutes after their meeting. How long did A take to cover the distance between Opladen and Cologne?

Sa/Sb = sqrt(90/40) = 3/2

This gives us that the ratio of the speed of A : speed of B as 3:2. We know that time taken is inversely proportional to speed. If ratio of speed of A and B is 3:2, the time taken to travel the same distance will be in the ratio 2:3. Therefore, since B takes 90 mins to travel from the meeting point to Opladen, A must have taken 60 (= 90*2/3) mins to travel from Opladen to the meeting point

So time taken by A to travel from Opladen to Cologne must be 60 + 40 mins = 1 hr 40 mins

[Credits: Veritasprep]

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