Gyan Room : Arithmetic  Concepts & Shortcuts

Gyan Room  There's always room to learn more!
This thread is reserved for sharing concepts, short cuts and good questions from Arithmetic topic.Happy Learning, Stay MBAtious!

#PnL
If a number N is successively changed by x%, y%, c% ... then the final value = N (1 + x/100)(1 + y/100)(1 + z/100) ...
Successive Discount : If the first discount is x% and 2nd discount is y% then, Total discount = x + y  (xy/100) %
Example : What is the single discount rate equivalent to successive discounts of 20%, 25% and 10% ?
20 and 25, equivalent is 20 + 25  5 = 40
Now for 40 and 10, equivalent is 50  4 = 46%Similarly, if we make a successive increase of x% and y% then, total increase = x + y + (xy/100)
If we increase x% first and then decrease y%, then total change = x  y  (xy/100)
In the above formula, if we put y = x i.e, if we increase x% and then decrease x%, then we have a net decrease of x^2/100 %
If SP is same for two articles and one is sold at x% profit and other one is sold at x% loss, net is (x^2/100)% loss
Example: A bought 2 shares and sold them for 96 rupees each. If he had a profit of 20% on the sale of one of the shares but a loss of 20% on the sale of the other share, then on the sale of both shares A had
Ans  a loss of x^2/100 = 20^2/100 = 4%

#SICI
Rule of 72  An easy way to approximately calculate how long it will take to double an investment with a given rate of return.
Take your annual interest rate, and divide it by 72. Simple, right?
For example, if your interest rate is 12% then it would take 72/12 = 6 years to double your investment.
Rule of 115
To triple your investment, divide 115 by your annual interest rate.
So, if you're getting a return of 10% per year, it will take about 11.5 years for your investment to triple in size.

#Calendar
Zeller's Rule
f = k + [(13 * m1)/5] + D + [D/4] + [C/4]  2 * C.
Let's use January 29, 2064 as an example
k is the day of the month. For our date, k = 29.
m is the month number. Months have to be counted specially for Zeller's Rule: March is 1, April is 2, and so on to February, which is 12. (This makes the formula simpler, because on leap years February 29 is counted as the last day of the year.) Because of this rule, January and February are always counted as the 11th and 12th months of the previous year. In our example, m = 11.
D is the last two digits of the year. Because in our example we are using January (see previous bullet) D = 63 even though we are using a date from 2064.
C stands for century: it's the first two digits of the year. In our case, C = 20.Now let's substitute our example numbers into the formula.
f = k + [(13 * m1)/5] + D + [D/4] + [C/4]  2 * C
= 29 + [(13 * 111)/5] + 63 + [63/4] + [20/4]  2 * 20
= 29 + [28.4] + 63 + [15.75] + [5]  40
= 29 + 28 + 63 + 15 + 5  40
= 100.Once we have found f, we divide it by 7 and take the remainder. Note that if the result for f is negative, care must be taken in calculating the proper remainder. Suppose f = 17. When we divide by 7, we have to follow the same rules as for the greatest integer function; namely we find the greatest multiple of 7 less than 17, so the remainder will be positive (or zero). 21 is the greatest multiple of 7 less than 17, so the remainder is 4 since 21 + 4 = 17. Alternatively, we can say that 7 goes into 17 twice, making 14 and leaving a remainder of 3, then add 7 since the remainder is negative, so 3 + 7 is again a remainder of 4.
A remainder of 0 corresponds to Sunday, 1 means Monday, etc. For our example, 100 / 7 = 14, remainder 2, so January 29, 2064 will be a Tuesday.
source : mathforum.org

harris last edited by
NMIMS, Mumbai (Gold Medallist)  AsiaPacific Champion – CFA Global Research Challenge
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^2Sum 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)/3Sum 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.

harris last edited by zabeer
NMIMS, Mumbai (Gold Medallist)  AsiaPacific Champion – CFA Global Research Challenge
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.

harris last edited by harris
NMIMS, Mumbai (Gold Medallist)  AsiaPacific Champion – CFA Global Research Challenge
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 = 16If 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 fruitseller 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
 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.
 If Length is decreased by 25% then breadth should be increase by 20% to keep the area same.
 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 these10% 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 / KgIt 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/nm) 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))^NWhen interest is compounded half yearly,
Amount = P (1+(R/2)/100)^2NWhen interest is compounded quarterly,
Amount = P (1+(R/4)/100) ^4NWhen 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 streamLet 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 = (xy) km/hr
Speed of boat in still water, x = 1/2(d+u) km/hr
Speed of stream , y = 1/2(du) 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/hrIf 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)/(ab) if a>b > else (x+y)/(ba)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

harris last edited by harris
NMIMS, Mumbai (Gold Medallist)  AsiaPacific Champion – CFA Global Research Challenge
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 : 96Water : Milk = 1:6
So the mixture has 6 x 18 = 108 Litres of Milk.

harris last edited by harris
NMIMS, Mumbai (Gold Medallist)  AsiaPacific Champion – CFA Global Research Challenge
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:
 (Liquid A left after nth operation)/(Initial quantity of A in the vessel) = (ab)^n/(a)^n
 (Liquid A left after nth operation )/(Liquid B left after nth operation) = ((ab)^n/(a)^n )/(1 (ab)^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