# Gyan Room : Arithmetic - Concepts & Shortcuts

• My 0.02\$

Sum of first n natural numbers = n (n + 1)/2
Sum of first n even numbers = n(n + 1)
Sum of first n odd numbers = n^2

Sum of square of first n natural numbers = n(n + 1)(2n + 1) / 6
Sum of square of first n even natural numbers = 2n(n + 1)(2n + 1)/3
Sum of square of first n odd natural numbers = n(4n^2 - 1)/3

Sum of cube of first n natural numbers = [n(n + 1)/2]^2
Sum of cube of first n even natural numbers = 2[ n(n + 1)]^2
Sum of cube of first n odd natural numbers = n^2(2n^2 - 1)

Plug in values for n and you can verify the above results. This results are pretty useful for Averages, Number theory and LR.

• Very useful

If Speeds forms an Arithmetic Progression, then time taken will be in Harmonic Progression (Constant distance)
If Time taken forms an AP, then speed forms an HP. (Again for constant distance)

Ex: CAT 2006 Question - Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30,40 and 60 km per hour respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start?

Speeds are in HP and distance is constant. So time should be AP
t, t + 2 and ? should be in AP. So unknown = t + 4, and answer is Kiranmala started 4 hours after Arun.

• If Cost Price of X things = Selling Price of Y things, then Profit/Loss = [(x - y)/y] x 100

ex: The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, find the value of x
25 = [(20 - x)/x] * 100
x = 16

If X articles are bought for P rupees and Y articles are sold for Q rupees then Profit/Loss = [ (XQ/YP) - 1] x 100

ex: 9 Pens are purchased for Rs. 10 and 15 Pens are sold for Rs. 17. Find profit/loss percent.
P/L = [ (9 x 17)/(15 x 10) - 1] x 100 = + 2% (Positive, so Profit)

If after selling x things P/L is equal to SP of y things then P/L = [y/(x -/+ y)] x 100

ex: By selling 100 bananas, a fruit-seller gains the selling price of 20 bananas. His gain percentage is?
P = [20/(100 - 20)] x 100 = 25% Gain

• False Weights

If an item is claimed to be sold at cost price, by using false weight, then the overall percentage gain/loss = [(Claimed weight/Actual weight) - 1] x 100

Example: A dishonest dealer plans to sell his goods at cost price, but he uses a weight of 960 grams instead of 1 Kg weight. then the percentage of gain or profit is

[ 1000/960 - 1] x 100 = 40/960 * 100 = 25/6 % profit.

Same shop keeper plans to sell his goods at a profit of 10% AND use a weight which is 20% less. find his total percentage gain.

Formula is profit% = [(100 + % profit)(claimed value/ actual value)] - 100

So here, [(100 + 10)(1000/800)] - 100 = 37.5% profit.

PS: I am not a big fan of mugging up formulas, especially if it is hard to remember. But this type of questions are pretty common in our entrance exams, so sharing!
If you feel it is better to go with the theoretical method, I am with you :)

• Product Constancy (VERY IMPORTANT)

Let say, A x B = Constant
x% of increase/decrease in A can be compensated by 100x/(100 + x)% decrease/increase of B.

Applications in cases like

1. If Price is increased by 60% then Consumption should be decreased by (100 x 60)/160 = 37.5% for keeping the expenditure same as before.
2. If Length is decreased by 25% then breadth should be increase by 20% to keep the area same.
3. If speed is increased by 50% then time taken to cover the same distance will decrease by 33.33%

Example - A reduction of 10% in the price of sugar enables a person to buy 6.2kg more for Rs 279. Find reduced price per kilogram.
(a) Rs 5
(b) Rs 4.5
(c) Rs 4.05
(d) None of these

10% reduction -> 11.11% increase in consumption = 6.2 kg
=> original consumption was 55.8 kg
(how we did it ? 11.11% increase -> 1/9 times Original = 6.2 kg, so Original value = 6.2 x 9 = 55.8)
So total the person can buy 62 Kg for 279 -> 4.5 Rs / Kg

It helps to learn some fractions

1/2 = 50%
1/3 = 33%
1/4 = 25%
1/5 = 20%
1/6 = 16.66%
1/7 = 14.28%
1/8 = 12.5%
1/9 = 11.11% (which we used in the above problem)
1/10 = 10%
1/11 = 9.09%
1/12 = 8.33%
1/13 = 7.69%
1/14 = 7.14%
1/15 = 6.66%
1/16 = 6.25%
1/17 = 5.88%
1/18 = 5.55%
1/19 = 5.26%
1/20 = 5%

• Profit and Loss Basic Formula • P & L Concepts:

%Profit = (Profit/Cost Price)*100 = (Selling Price – Cost price)*100/Cost Price

%Loss = (Loss/Cost Price)*100 = (Cost Price – Selling Price)*100/Cost price

%Discount = Discount*100/Marked Price

Selling Price = Marked Price – Discount

If two items are sold, each at Rs.x one at a gain of P% and the other at a loss of P%, there is an overall loss given by (P^2/100) %. The absolute value of the loss is given by (2P^2x/ (100^2 – P^2)).

If cost price of two items is same and %Loss and %Gain on the two items are equal, then net loss or net profit is zero.

If (x+y) articles are sold at cost price of x articles then the percentage discount = Y*100/(x+Y).

In case of successive discounts a% and b% the effective discount is (a+b-(ab/100)) %

By using false weight , If a substance is sold at cost price, the overall gain % is given by (100+Gain % )/100 = True scale Weight/False scale Weight.

• Time & Work Concepts:

If A can do a piece of work in ‘a’ number of days, then in one day (1/a)th of the work is done. Similarly if an inlet pipe fills a cistern in ‘a’ hours, then (1/a)th part is filled in one hour. If an outlet pipe empties a cistern in ‘a’ hours, then (1/a)th part is emptied in 1 hour.

If A is ‘x ‘ times as good a workman as B , then he will take (1/x)th of the time taken by B to do the same work. If Pipe A is ‘x’ times bigger than pipe B, then pipe A will take (1/x)th of the time taken by pipe B to fill the cistern.

If A and B can do a piece of work in ‘x’ and ‘y’ days respectively, then working together they will take (xy/x+y) days to finish the work and one day they will finish (x+y/xy)th part of the work. If A and B fill a cistern in ‘m’ and ‘n’ hours respectively then together they will take (mn/m+n) hours to fill the cistern and in one hour (m+n/mn)th part of the cistern will be filled. Similarly A and B empty a cistern in ‘m’ and ‘n’ hours, respectively then together they will take (m+n/mn) hours to empty the cistern.

If an outlet pipe empties the cistern in ‘n’ hours and an inlet pipe fills a cistern in ‘m’ hours then the net part filled in 1 hour when both the pipes are opened is (1/m – 1/n) and the cistern will get filled in (mn/n-m) hours. For the Cistern to get filled its necessary that mn the cistern will never get filled. In general Net part filled of a cistern = (Sum of work done by inlets) – (Sum of work done by outlets)

If two pipes A and B can fill a cistern in ‘x’ minutes and if A alone can fill it in ‘a’ minutes more than ‘x’ minutes and B alone can fill it in ‘b’ minutes more than x minutes then x= root(ab). If two men A and B together can finish a job in x days and if A working alone takes ‘a’ days more than A and B working together and B working alone takes ‘b’ days more than A and B working together then x = root(ab).

• Partnership Concepts:

Simple Partnership:

Here capitals of partners are invested for same period and profit or loss is divided in the ratio of their investments.

(Investment of A/Investment of B ) = (Profit of A/Profit of B ) or (Loss of A/Loss of B )

(Investment of A: Investment of B: Investment of C) = (Profit of A: Profit of B: Profit of C) or (Loss of A: Loss of B: Loss of C)

Compound Partnership

Here capitals are invested for different periods and profit or loss is divided in the ratio of their monthly equivalent investment.

Monthly equivalent is the product of the capital invested and period for which it is invested.

(Monthly Equivalent investment of A/ Monthly Equivalent Investment of B ) = (Investment of A * Period of Investment of A) / (Investment of B * Period of Investment of B ) = (Profit of A)/(Profit of B ) or (Loss of A)/(Loss of B )

If three partners are there profit or loss is divided in the ratio of their monthly equivalent investment.

• Simple & Compound Interest Concepts :

Simple Interest, SI = PNR/100
Where P is Principal, R is rate of interest and N is the time period in years.

Let P is Principal, R is rate of interest and N is time period in years,
When interest is compounded annually,
Amount = P (1+(R/100))^N

When interest is compounded half yearly,
Amount = P (1+(R/2)/100)^2N

When interest is compounded quarterly,
Amount = P (1+(R/4)/100) ^4N

When interest is compounded annually, but time period in fraction say a b/c years
Amount = P*(1+R/100)^a * (1+bR/100c)

When rates are different for different years say R1 % , R2 % and R3 % for 1st,2nd and 3rd year respectively
Amount = P (1+ (R1/100)) *(1+ (R2/100)) * (1+ (R3/100))

• Clocks Concepts:

When the minute hand passes over the 60 minutes spaces, the hour hand goes over the 5 minute spaces. That is in 60 minutes, the minute hand gains 55 minutes over the hour hand or 55/60 minute spaces in one minute.

In one minute, minute hand moves 6 degree. In one hour, hour hand moves 30 degree. Hence in one minute hour hand moves ½ degree.

In every hour the hands coincide once.

In every hour hands are twice at right angles (90 degree apart) and in these positions hands are 15 minute spaces apart.

In every hour the hands point in opposite directions (180 degree apart) once and in this position, the hands are 30 minute spaces apart.

The hands are in same straight line when they are coincident or opposite to each other. The hands coincide 11 times in every 12 hours. Between 11 and 1 o clock there is only one common position at 12 o clock. Hence hands coincide 22 times a day.

If both the hands start moving together from the same position, both the hands will coincide after 360*2/11 = 65 5/11 minutes.

The hands of a clock are at right angles twice in every hour, but in 12 hours they are at right angles 22 times since there are two common positions in every 12 hours.

Angle between minute hand and hour hand when time is x hours and y minutes is 30x – 11y/2 degree.

• Upstream and Downstream Concepts:

Downstream: In water the direction along the stream
Upstream: In water the direction against the stream

Let the speed of a boat in still water is x km/hr and the speed of the stream is y km/hr.
Downstream speed, d = (x+y) km/hr
Upstream speed, u = (x-y) km/hr
Speed of boat in still water, x = 1/2(d+u) km/hr
Speed of stream , y = 1/2(d-u) km/hr

• Time Speed & Distance Concepts:

If a body travels d1,d2..dn distances with speeds s1,s2..sn in time t1,t2…tn respectively, then the average speed of the body through the total distance is given by
Average speed = Total distance travelled / Total time taken
= (d1+d2…..dn)/(t1+t2+….tn)
= (d1+d2+…..dn)/(d1/s1+d2/s2+….dn/sn)

While travelling a certain distance d, if a man changes his speed in the ratio m:n, then the ratio of time taken becomes n:m.

If a certain distance (d) say from A to B is covered at ‘a’ km/hr and same distance is covered from B to A in ‘b’ km/hr then average speed during the whole journey is given by
Average speed = 2ab/(a+b) km/hr

If two persons A and B start at the same time in opposite directions from two points and arrive at the two points in ‘a’ and ‘b’ hours respectively after having met, then
A’s Speed/B’s speed = root(b)/root(a)

Time taken by a moving object ‘x’ meters long in passing a stationary object of negligible length from the time they meet is same as the time taken by moving object to cover x meters with its own speed.

Time taken by a moving object x meters long in passing a stationary object y meters long from the time they meet, is same as the time taken by moving object to cover x+y meters which its own speed.

If two objects of length ‘x’ and ‘y’ meters move in the same direction at ‘a’ and ‘b’ m/s then the time taken to cross each other from the time they meet = Sum of their length/Relative speed
= (x+y)/(a-b) if a>b -> else (x+y)/(b-a)

If two objects of length ‘x’ and ‘y’ meters move in the opposite direction at ‘a’ and ‘b’ m/s then the time taken to cross each other from the time they meet = Sum of their length/Relative speed = (x+y)/(a+b)

• Concept : If the ball rebounds to a/b th of the height then total distance covered = H * (a + b)/(a - b)

Q - After striking the floor a ball rebounds to 4/5th of the height from which it was fallen. What is the total distance that it travels before coming to rest if it is gently dropped of height 120m?

Distance = 120 * 9/1 = 1080 m

• 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.

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