Mathematical shortcuts

Mathematical Shortcuts
for
Aptitude
C.S.VEERARAGAVAN
Aptitude Trainer

Percentage
If the value of something is

Example – 1
• The Salary of a worker is first increased by
10% and thereafter decreased by 5%.
• What is the overall change in percent.
• overall change = + 10 - 5 + (+10)(-5)/100
• = + 4.5 % (increased)
• use (+) sign for increment and (-) sign for
decrement.

Example – 2
• A shopkeeper marks the price of goods 20%
more than the real price. He allowed a
discount of 10%.What profit or loss did he
get?
• Profit or loss the shopkeeper get:
• 20 - 10 + (+20)(-10)/100 = + 8% (profit)

Example 3
• The side of a square is increased by 30%. Find
the percentage increase in area.
• Increase in area = 30 + 30 + (30)(30)/100 =
69%.

Example 4
• If the radius of a circle is decreased by 20%.
What percent change in area?
• percent change in area:
• - 20 - 20 + (-20)(-20)/100 = -36% (decreased)

Example 5
• If the price of a shirt is increased by 15% than
decreased by 15%. What is percent change?
• percent change :
• + 15 - 15 + (+15)(-15)/100 = - 2.25 %
(decreased)

Example 6
• The length of a rectangle is increased by 40%
and breadth is decreased by 40%. Find change
in area.
• percent change in area:
• 40 - 40 + (-40)(+40)/100 = -16%(decreased)

Example 7
• if a number is increased by 20% and again
increased by 20%. By what percent should the
number increased.
• percent increased:
• 20 + 20 + (+20)(+20)/100 = 44%(increased)

S.I & C.I
• If the difference between Simple Interest and
Compound Interest on a certain sum of money
for 2 years at R% rate is given then
If the difference between simple interest and compound interest on a certain sum
of money at 10% per annum for 2 years is Rs 2 then find the sum.

S.I & C.I
• If the difference between Simple Interest and
Compound Interest on a certain sum of money
for 3 years at R% is given then
If the difference between simple interest and compound interest on a certain sum
of money at 10% per annum for 3 years is Rs 2 then find the sum.
Sol:

COMPOUND INTEREST
• If sum A becomes B in T1 years at compound
interest, then after T2 years
Rs 1000 becomes 1100 after 4 years at
certain compound interest rate. What will
be the sum after 8 years?
Sum:
Here A = 1000, B = 1100
T1 = 4, T2 = 8

Partnership – 1
If two partners are investing their money C1 and C2 for
equal period of time and their total profit is P then
their shares of profit are
If these partners are investing their money for
different period of time which is T1 and T2, then their
profits are

Partnership – 2
Jack and Jill start a business by investing $ 2,000 for 8 months and $
3,000 for 6 months respectively. If their total profit si $ 510 and then
what is profit of Jill?
Let’s Say C1 = 2000, T1 = 8
C2 = 3000, T2 = 6
P = 510

Partnership – 3
If n partners are investing their money C1, C2, …, Cn for
equal period of time and their total profit is P then their
shares of profit are
If these partners are investing their money for different
period of time which is T1, T2,… , Tn then their profits are

Example 1
Raju, Kamal and Vinod start a business by investing Rs 5,000 for 12 months,
Rs 8,000 for 9 months and Rs 10,000 for 6 months. If at the end of the year
their total profit is Rs 2000 then find the profit of each partner.
Let’s Say C1 = 5000, T1 = 12
C2 = 8000, T2 = 9
C3 = 10000, T3 = 6
P = 2000

Time & Distance 1
• If different distance is travelled in different
time then,
If a car travels 50 Km in 1 hour, another 40 Km in 2 hour and another 70 Km in
3 hour then what is average speed of car.
Total Distance Covered = 50 + 40 + 70 = 160 Km
Total Time Taken = 1 + 2 + 3 = 6 hours.

Time & Distance 2
• If equal distance is travelled at different speed.
• If equal distance is travelled at the speed of A and B
then,
A boy goes to his school which is 2 Km away in 10
minutes and returns in 20 mins then what is boy’s
average speed.
Let’s say A = 2/10 = 0.2 km/min
And B = 2/20 = 0.1 km/min
If the same distance is covered at two different speeds S1 and S2
and the time taken to cover the distance are T1 and T2, then the
distance is given by

Time & Distance 3
• If equal distance is travelled at the speed of A,
B and C then,
If a car divides its total journey in three equal parts and
travels those distances at speed of 60 kmph, 40 kmph and 80
kmph then what is car’s average speed?
Let’s say A = 60, B = 40 and C = 80, then
= 55.38

Decimal Equivalent of Fractions
• With a little practice, it's not hard to recall the decimal equivalents of
fractions up to
• 10/11!
• First, there are 3 you should know already:
• 1/2 = .5
• 1/3 = .333...
• 1/4 = .25
• Starting with the thirds, of which you already know one:
• 1/3 = .333...
• 2/3 = .666...
• You also know 2 of the 4ths, as well, so there's only one new one to learn:
• 1/4 = .25
• 2/4 = 1/2 = .5
• 3/4 = .75

• Fifths are very easy. Take the numerator (the number on top), double it, and stick a
• decimal in front of it.
• 1/5 = .2
• 2/5 = .4
• 3/5 = .6
• 4/5 = .8
• There are only two new decimal equivalents to learn with the 6ths:
• 1/6 = .1666...
• 2/6 = 1/3 = .333...
• 3/6 = 1/2 = .5
• 4/6 = 2/3 = .666...
• 5/6 = .8333...
• What about 7ths? We'll come back to them at the end. They're very unique.
• 8ths aren't that hard to learn, as they're just smaller steps than 4ths. If you have
trouble with any of the 8ths, find the nearest 4th, and add .125 if needed:

• 1/8 = .125
• 2/8 = 1/4 = .25
• 3/8 = .375
• 4/8 = 1/2 = .5
• 5/8 = .625
• 6/8 = 3/4 = .75
• 7/8 = .875
• 9ths are almost too easy:
• 1/9 = .111...
• 2/9 = .222...
• 3/9 = .333...
• 4/9 = .444...
• 5/9 = .555...
• 6/9 = .666...
• 7/9 = .777...
• 8/9 = .888...
• 10ths are very easy, as well. Just put a
decimal in front of the numerator:
• 1/10 = .1
• 2/10 = .2
• 3/10 = .3
• 4/10 = .4
• 5/10 = .5
• 6/10 = .6
• 7/10 = .7
• 8/10 = .8
• 9/10 = .9
• Remember how easy 9ths were? 11th are
easy in a similar way, assuming you know
your multiples of 9:
• 1/11 = .090909...
• 2/11 = .181818...
• 3/11 = .272727...
• 4/11 = .363636...
• 5/11 = .454545...
• 6/11 = .545454...
• 7/11 = .636363...
• 8/11 = .727272...
• 9/11 = .818181...
• 10/11 = .909090...
• As long as you can remember the pattern for
each fraction, it is quite simple to work out

• Oh, I almost forgot!
We haven't done
7ths yet, have we?
• One-seventh is an
interesting number:
• 1/7 =
.1428571428571428
57...
• For now, just think of
one-seventh as:
.142857
• See if you notice any
pattern in the 7ths:
• 1/7 = .142857...
• 2/7 = .285714...
• 3/7 = .428571...
• 4/7 = .571428...
• 5/7 = .714285...
• 6/7 = .857142...
• Notice that the 6
digits in the 7ths
ALWAYS stay in the
same order and the
starting digit is the
only thing that
changes!
• If you know your
multiples of 14 up to
6, it isn't difficult to
work out where to
begin the decimal
number. Look at this:
• For 1/7, think "1 *
14", giving us .14 as
the starting point.
• For 2/7, think "2 *
the starting point.
• For 3/7, think "3 *
the starting point.
• For 4/14, 5/14 and
6/14, you'll have to
adjust upward by 1:
• For 4/7, think "(4 *
14) + 1", giving us .57
as the starting point.
• For 5/7, think "(5 *
14) + 1", giving us .71
• For 6/7, think "(6 *
14) + 1", giving us .85
• Practice these, and
you'll have the
decimal equivalents
of everything from
1/2 to 10/11 at
• your finger tips!

Rule of Alligation
C D
D–M M–C
Mean
Price (m)

• 2.In what ratio must a grocer mix two varieties
of pulses costing `.15 and `.20 per kg
respectively so as to get a mixture worth
`.16.50 kg?
First type Mean Price Second type
15 16.50 20
3.50 1.50

• 4. A jar full of whisky contains 40% alcohol. A
part of this whisky is replaced by another
containing 19% alcohol and now the
percentage of alcohol was found to be 26%.
The quantity of whisky replaced is:
First Mean Price Second type
40% 26% 19%
7 14

0 1
Ratio of water and milk = 1:6.

0 1

• If n different vessels of equal size are filled
with the mixture of P and Q in the ratio p1 : q1,
p2 : q2, ……, pn : qn and content of all these
vessels are mixed in one large vessel, then

• Three equal buckets containing the mixture of milk and water are mixed
into a bigger bucket.
• If the proportion of milk and water in the glasses are 3:1, 2:3 and 4:2
then find the proportion of milk and water in the bigger bucket.
Sol:
• Let’s say P stands for milk and Q stands for water,
So, p1:q1 = 3:1
p2:q2=2:3
p3 : q3=4:2
So in bigger bucket,
Milk : Water = 109 : 71

• If n different vessels of sizes x1, x2, …, xn are
filled with the mixture of P and Q in the ratio
p1 : q1, p2 : q2, ……, pn : qn and content of all
these vessels are mixed in one large vessel,
then

• Three buckets of size 2 litre, 4 litre and 5 litre containing the
mixture of milk and water are mixed into a bigger bucket. If the
proportion of milk and water in the glasses are 3:1, 2:3 and 4:2
then find the proportion of milk and water in the bigger bucket.
Sol:
• Let’s say P stands for milk and Q stands for water,
So, p1:q1 = 3:1 , x1 = 2
p2:q2=2:3 , x2 = 4
p3 : q3=4:2 x3 = 5, so
So in bigger bucket,
Milk : Water = 193 : 137

• p gram of ingredient solution has a% ingredient in it.
• To increase the ingredient content to b% in the solution

• 125 litre of mixture of milk and water contains
25% of water. How much water must be added
to it to make water 30% in the new mixture?
Sol:
• Let’s say p = 125, b = 30, a = 25
So from the equation
Quantity of water need to be added = 8.92 litre.

Tuesday, October 08, 2013
C.S.VEERARAGAVAN, APTITUDE
TRAINER,veeraa1729@gmail.com
65
x ad bc
c a b

LCM – MODEL 1
• Any number which when divided by p,q or r leaving the
same remainder s in each case will be of the form
• k (LCM of p, q and r)+ s where k = 0,1,2…
• If we take k = 0, then we get the smallest such number.
• Example:
• The least number which when divided by 5,6,7 and 8 leaves
a remainder 3, but when divided by 9 leaves no remainder,
is:
• K(LCM of 5,6,7 and 8) + 3 = 840k + 3
• Least value of k for which 840k + 3 divisible by 9 is
• K=2.
• Required number is 1683.

LCM MODEL 2
• Any number which when divided by p,q or r leaving
respective remainders of s, t and u where (p–s) = (q – t)
= ( r – u) -= v (say) will be of the form
• K(LCM OF P, Q AND R) – V
• The smallest such number will be obtained by
substituting k = 1.
• Example:
• Find the smallest number which when divided by 4 and
7 gives remainders of 2 and 5 respectively.
• LCM OF 4 AND 7 IS 28.
• HENCE 28 – 2 = 26.

LCM MODEL 3
• Find the smallest number which when divided by
7 leaves a remainder of 6 and when divided by 11
leaves a remainder of 8.
• The required number will be 11k + 8
• When divided by 7 leaves a remainder 6.
• Subtracting 6 from 11k + 8 we have 11k + 2 which
should be multiple of 7.
• By trial, when k =3, we get 35.
• Hence Required number is when k = 3, 11k+8 =
41.

HCF MODEL 1
• The largest number with which the numbers p,q
or r are divided giving remainder of s, t and u
respectively will be the HCF of the three numbers
of the form (p – s), (q – t) and (r – u)
• Example
• Find the largest number with which when 906
and 650 are divided they leave remainders 3 and
5.
• The HCF of (906 – 3) and ( 650 – 5).
• HCF of 903 and 645 is 129.

HCF MODEL 2
• The largest number which when we divide by the
numbers p,q and r , the remainders are the same
is
• HCF of (p – r) and (q – r) where r is the smallest
among the three.
• Example
• Find the greatest number that will divide 43, 91
and 183 so as to leave the same remainder in
case.
• Required number = H.C.F of (91 – 43), and (183 –
43) = H.C.F of 48 and 140 is 4.

LAST DIGIT OF ANY POWER
• Last digit of 21 is 2

Digit
s
Powers
1 1 2 3 4 5 6 7 8 9
2 2 4 8 6 2 4 8 6 2
3 3 9 7 1 3 9 7 1 3
4 4 6 4 6 4 6 4 6 4
5 5 5 5 5 5 5 5 5 5
6 6 6 6 6 6 6 6 6 6
7 7 9 3 1 7 9 3 1 7
8 8 4 2 6 8 4 2 6 8
9 9 1 9 1 9 1 9 1 9
For every digit unit place digits of increasing powers repeat after 4th power.
This means unit place digit for power=5 is same as unit place digit for power=1 for
every number.
2) For digits 2, 4 & 8 any power will have either 2 or 4 or 6 or 8 at unit place.
3) For digits 3 & 7 any power will have either 1 or 3 or 7 or 9 at unit place.
4) For digit 9 any power will have either 1 or 9 at unit place.
5) And for digits 5 & 6 every power will have 5 & 6 at unit place respectively.

LARGEST POWER OF A NUMBER IN N!
• Find the largest power of
5 that can divide 216!
without leaving any
remainder. (or)
• Find the largest power of
5 contained in 216!
• Add all the quotients to
get 43 + 8 + 1 = 52.
• Therefore 552 is the
highest power of 5
contained in 216!
5 216  Number given
5 43  Quotient 1
5 8  Quotient 2
1  Quotient 3
Please note that this method is
applicable only when the number
whose largest power is to be found out
is a prime number.
If it is not a prime number, then split
the number as product of primes and
find the largest power of each factor.
Then the smallest amoung the largest
poser of these relative factors of the
given number will the largest power
required.

an – bn
• It is always divisible by a – b.
• When n is even it is also divisible by a + b.
• When n is odd it is not divisible by a + b.

an + bn
• It is never divisible by a – b.
• When n is odd it is also divisible by a + b.
• When n is even it is not divisible by a + b.

• There are three departments having students
64,58,24 .In an exam they have to be seated in
rooms such that each room has equal number
of students and each room has students of
one type only (No mixing of departments. Find
the minimum number rooms required ?
• The HCF is 2. Hence 32 + 29 + 12= 73.

Odd Days
• We are supposed to find the day of the week
on a given date.
• For this, we use the concept of 'odd days'.
• In a given period, the number of days more
than the complete weeks are called odd days.

Leap Year
• (i). Every year divisible by 4 is a leap year, if it is not a century.
• (ii). Every 4th century is a leap year and no other century is a leap
year.
• Note: A leap year has 366 days.
• Examples:
• Each of the years 1948, 2004, 1676 etc. is a leap year.
• Each of the years 400, 800, 1200, 1600, 2000 etc. is a leap year.
• None of the years 2001, 2002, 2003, 2005, 1800, 2100 is a leap
year.
• Ordinary Year:
• The year which is not a leap year is called an ordinary years. An
ordinary year has 365 days.

Counting of odd days
• 1 ordinary year = 365 days = (52 weeks + 1 day.)
• 1 ordinary year has 1 odd day.
• 1 leap year = 366 days = (52 weeks + 2 days)
• 1 leap year has 2 odd days.
• 100 years = 76 ordinary years + 24 leap years
• = (76 x 1 + 24 x 2) odd days
• = 124 odd days.
• = (17 weeks + days) 5 odd days.
• Number of odd days in 100 years = 5.
• Number of odd days in 200 years = (5 x 2) 3 odd days.
• Number of odd days in 300 years = (5 x 3) 1 odd day.
• Number of odd days in 400 years = (5 x 4 + 1) 0 odd day.
• Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc.
has 0 odd days.

Day of the Week Related to Odd Days
No. of
days:
0 1 2 3 4 5 6
Day: Sun. Mon. Tues. Wed. Thurs. Fri. Sat.

• If 6th March 2005 is Monday, what was the day of the
week on 6th March, 2004?
• The year 2004 is a leap year. So, it has 2 odd days.
• But Feb 2004 not included because we are calculating
from March 2004 to March 2005.
• So it has 1 odd day only.
• The day on 6th March, 2005 will be 1 day beyond the
day on 6th March, 2004.
• Given that, 6th March, 2005 is Monday, 6th March, 2004
is Sunday
• (1 day before to 6th March, 2005.)

• On what dates of April, 2001 did Wednesday fall?
• We shall find the day on 1st April,2001.
• 1st April 2001 = (2000 years + Period from 1.1.2001 to
1.4.2001)
• Odd days in 2000 years = 0
• Jan,Feb,Mar,Apr
• (31+28+31+1) = 91 days = o odd days.
• Total number of odd days = 0.
• On 1st April 2001 it was Sunday.
• In April 2001, Wednesday falls on 4th,11th,18th and 25th.

• The last day of a century cannot be?
• 100 years contain 5 odd days.
• Last day of 1st century is Friday.
• 200 years contain (5x2) 3 odd days.
• Last day of 2nd century is Wednesday.
• 300 years contain (5x3) 1 odd day.
• Last day of 3rd century is Monday.
• 400 years contain 0 odd day
• Last day of 4th century is Sunday.
• This cycle is repeated.
• Hence Last day of a century cannot be Tuesday or Thursday
or Saturday.

• On 8th Feb, 2005 it was Tuesday.
• What was the day of the week on 8th Feb
2004?
• The year 2004 is a leap year. It has 2 odd days.
• The day on 8th Feb 2004 is 2 days before the
day on 8th Feb.2005.
• Hence this day is Sunday.

JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC
0 3 3 6 1 4 6 2 5 0 3 5
Step1: Ask for the Date. Ex: 23rd June 1986
Step2: Number of the month on the list, June is 4.
Step3: Take the date of the month, that is 23
Step4: Take the last 2 digits of the year, that is 86.
Step5: Find out the number of leap years. Divide the last 2 digits of the year by 4, 86
divide by 4 is 21.
Step6: Now add all the 4 numbers: 4 + 23 + 86 + 21 = 134.
Step7: Divide 134 by 7 = 19 remainder 1.
The reminder tells you the day.
No. of
days:
0 1 2 3 4 5 6
Day: Sun. Mon. Tues. Wed. Thurs. Fri. Sat.
1600/2000 1700/2100 1800/2200 1900/2300
0 6 4 2

Boats – 1
• A man can row certain distance downstream
in t1 hours and returns the same distance
upstream in t2 hours. If the speed of stream is
y km/h, then the speed of man in still water is
given by

Boats – 2
• A man can row in still water at x km/h. In a
stream flowing at y km/h, if it takes him t
hours to row to a place and come back, then
the distance between two places is given by

Boats – 3
stream flowing at y km/h, if it takes t hours
more in upstream than to go downstream for
the same distance, then the distance is given
by

Boats – 4
stream flowing at y km/h, if he rows the same
distance up and down the stream, then his
average speed is given by

Pipes & Cistern
• Pipe and Cistern problems are similar to time and work
problems. A pipe is used to fill or empty the tank or cistern.
• Inlet Pipe: A pipe used to fill the tank or cistern is known as
Inlet Pipe.
• Outlet Pipe: A pipe used to empty the tank or cistern is
known as Outlet Pipe.
• Some Basic Formulas
• If an inlet pipe can fill the tank in x hours, then the part
filled in 1 hour = 1/x
• If an outlet pipe can empty the tank in y hours, then the
part of the tank emptied in 1 hour = 1/y
• If both inlet and outlet valves are kept open, then the net
part of the tank filled in 1 hour is

Pipes & Cistern – 1
• Two pipes can fill (or empty) a cistern in x and
y hours while working alone. If both pipes are
opened together, then the time taken to fill
(or empty) the cistern is given by

Pipes & Cisterns – 2
• Three pipes can fill (or empty) a cistern in x, y
and z hours while working alone. If all the
three pipes are opened together, the time
taken to fill (or empty) the cistern is given by

• If a pipe can fill a cistern in x hours and another
can fill the same cistern in y hours, but a third
one can empty the full tank in z hours, and all of
them are opened together, then

• A pipe can fill a cistern in x hours. Because of a
leak in the bottom, it is filled in y hours. If it is full,
the time taken by the leak to empty the cistern is

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