Mathematical shortcuts1

APTITUDE PROGRAMME
MAY 8TH
KONGUNADU COLLEGE
COMPUTER SCIENCE
C.S.VEERARAGAVAN
TRAINER
07-04-2014 1

Percentage
• increased by x % then
decreased by y % then
resulting effect in percent
• +x – y –
𝑥𝑦
100
• Increased by x% then again
increased by y% then
• +x+y+
𝑥𝑦
100
• Decreased by x% then
increased by y%
• –x+y –
𝑥𝑦
100
• Decreased by x% then again
decreased by y%
• – x – y +
𝑥𝑦
100
If the value of something is
If the price of a shirt is increased by 15%
than decreased by 15%. What is the
percent change?
+ 15 - 15 +
+15 −15
100
= - 2.25 %

Example – 2
• 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 – 3
• 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 4
• 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 5
• 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 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)

Percentage – 2
• If the price of the commodity
is decreased by r% then
increase in the consumption
so as not to decrease the
expenditure :
•
𝑟
100+𝑟
𝑋100
If the price of the sugar fall down by 10%. By how much percent
must the householder increase its consumption so as not to
decrease the expenditure.
increase in consumption :
=
10
100 − 10
%
=
100
9
= 11.11%

Example 2
• If the price of petrol is increased by 30
percent, by how much petrol a car owner
must reduce his consumption in order to
maintain the same budget.
• reduction in consuption :
•
30
130
% =
300
13
= 23.07%

Example 3
• If A's salary is 25% more than B then how
much percent the B's salary is less than that of
A.
• B's salary is less than that of A :
•
25
125
% =
100
5
= 20%

Percentage – 3
• x% of a quantity is taken by the first, y% of the
remaining is taken by the second and z% of the
remaining is taken and so on, Now if A is the amount
left then find the initial amount :
initial amount = A X
100𝑋100𝑋100
100−𝑥 100−𝑦 100−𝑧
… where
A is the left amount.
After reducing 10% from a certain sum and then 20% from the
remaining , there is Rs3600 left then find the original sum.
original sum =3600𝑋
100𝑋100
100−10 100−20
=3600 X
10000
90𝑋80
= Rs.5000.

Example 2
• In a library 20% of the books are in hindi , 50%
of the remaining are in English 30% of the
remaining are in french, the remaining 6300
books are in regional language, total no of
books would be?
• Total no of books =
6300X
100
80
𝑋
100
50
𝑋
100
70
=22500

Percentage – 4
• A candidate scoring x% in an examination fails by 'a'
marks, while another candidate who scores y%
marks gets 'b' marks more than then the minimum
required pass marks.
• Then the maximum marks for the examination are
• 100𝑋
𝑎+𝑏
𝑦−𝑥
A candidate scores 25% and fails by 30 marks, while another
candidate who scores 50% marks, gets 20 marks more than the
minimum required marks to pass the examination. Find the
maximum marks for the examination.
Maximum marks = 100X
30+20
50−25
=200.

PROFIT / LOSS
• If a man purchase 11 orange
for Rs 10 and sell them 10 for
Rs 11. How much profit or
loss did he made?
• Profit or loss made =
11𝑋11−10𝑋10
10𝑋10
X100 =21%
profit.
• If 'a' articles are bought for
Rs 'b' and sell them 'c' for Rs
'd'. Then profit or loss made
by the vendor:
•
𝑎𝑑−𝑏𝑐
𝑏𝑐
𝑥100
USUAL METHOD
C.P of 11 oranges Rs.10
C.P of 1 orange is Rs.
10
11
.
S.P of 10 oranges Rs.11
S.P of 1 orange is Rs.
11
10
Profit or loss made
=
𝑆.𝑃 −𝐶.𝑃
𝐶.𝑃
𝑋 100
=
11
10
−
10
11
10
11
𝑋 100
=
11𝑋11 −10𝑋10
110
𝑋
11
10
𝑋100
=
11𝑋11−10𝑋10
10𝑋10
𝑋100
07-04-2014 18

Example 2
• A boy buys orange 9 for Rs 16 and sell them
11 for Rs 20. Find profit or loss percent.
• profit or loss percent made :
9𝑋20−11𝑋16
11𝑋16
𝑥100 =
25
11
% 𝑝𝑟𝑜𝑓𝑖𝑡

Selling Price
• By selling a horse for Rs
570 a trader loses 5%. At
what price must he sell it
to gain 5%.
• Selling Price(SP)
• = given(Rs) X
100±𝑄
100±𝑔𝑖𝑣𝑒𝑛%
• Selling Price
• = 570 X
100+5
100−5
=Rs.630
USUAL METHOD
S.P = 570 and loss 5%
C.P =
100
100−𝑙𝑜𝑠𝑠
𝑥𝑆. 𝑃
=
100
95
𝑥570
Profit = 5%
So new S.P
=
100+𝑃𝑅𝑂𝐹𝐼𝑇
100
𝑋𝐶. 𝑃
=
105
100
𝑥
100
95
𝑋570=Rs.630
07-04-2014 20

Cost Price
• Mahesh sold a book at a
profit of 12%. Had he sold
it for Rs 18 more , 18%
would have been gained.
Find CP.
• Cost Price(CP)
• =
𝑀𝑜𝑟𝑒 𝐴𝑚𝑜𝑢𝑛𝑡
𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑃𝑟𝑜𝑓𝑖𝑡
X 100
• Cost Price
• =
18
6
𝑋100=Rs.300.
USUAL METHOD
Profit is 12%
Both S.P and C.P not given.
Let C.P be x.
Then S.P =
100+𝑝𝑟𝑜𝑓𝑖𝑡
100
𝑥𝐶. 𝑃 =
112𝑥
100
NEW S.P =
112𝑥
100
+ 18.
PROFIT %=
𝑁𝐸𝑊 𝑆.𝑃 −𝐶.𝑃
𝐶.𝑃
𝑋100 = 18
=
112𝑋
100
+18−𝑋
𝑋
𝑋100 = 18
=
112𝑋+1800−100𝑋
100𝑋
X100= 18
=12X + 1800 = 18X
SO X =
1800
18−12
=
1800
6
=300
07-04-2014 21

Example 2
• A man sold a horse at a loss of 7%. Had he be
able to sell it at a gain of 9%. It would have
fetch 64 more. Find Cost Price.
• Here difference in Percent is 9 - (-7) = 16
• Cost Price =
64
16
𝑋100=Rs.400

Profit & Loss
• There are two shopkeepers having
shops side by side.
• The first shopkeeper sells bicycles.
• He sells a bicycle worth $30 for $45.
• One day a customer comes and buys a
bicycle.
• He gives a $50 note to the shopkeeper.
• The shopkeeper doesn't have change so
he goes to the second shopkeeper, gets
the change for $50, and gives $5 and
the bicycle to the customer.
• The customer goes away.
• The next day the second shopkeeper
comes and tells the first shopkeeper
that the $50 note is counterfeit and
takes his $50 back.
• Now, how much does the first
shopkeeper lose?
From the second shopkeeper he took
$50 and gave back $50 so there was no
profit no loss. To the customer he gave
$5 + $30 bicycle. Therefore, his total
loss is $35

Simple Interest – Principal
• A sum was put at SI at a
certain value for 2 years,
had it been put at 3% higher
rate, it would have fetch 300
more , find the sum.
• Principal =
𝑀𝑜𝑟𝑒 𝑀𝑜𝑛𝑒𝑦 𝑋 100
𝑀𝑜𝑟𝑒 𝑟𝑎𝑡𝑒 𝑋 𝑇𝐼𝑀𝐸
• Principal =
300𝑋100
2 𝑋 3
• =Rs.5000.
USUAL METHOD
Let the Rate be R%
Let the Principal be P.
Then S.I =
𝑃𝑋2𝑋𝑅
100
GIVEN
𝑃𝑥2𝑥(𝑅 + 3)
100
−
𝑃𝑥2𝑥𝑅
100
= 300
2𝑃(𝑅 + 3 − 𝑅)
100
= 300
2𝑃𝑥3
100
= 300
𝑃 =
300𝑥100
2𝑥3
07-04-2014 27

Simple Interest – Time
• Time = given time X
𝑛2−1
𝑛1−1
• A sum of money double
itself in 4 years , in how
many years will it become
8 times of itself ?
• Time = 4 X
8−1
2−1
=28 years
Let the Principal be P
Let the Rate of Interest be R
S.I after 4 years =
4𝑃𝑅
100
=
𝑃𝑅
25
Amount after 4 years = 2P
P + S.I = 2P
That is S.I after 4 years = P
𝑃𝑅
25
= P  R = 25.
S.I after n years = 7P
25𝑃𝑁
100
= 7𝑃
𝑃𝑁
4
= 7𝑃
N = 28 YEARS.
07-04-2014 28

Simple interest
• A sum of money becomes n times in t years at simple interest,
then rate of interest will be
• R= 100∗
𝑛−1
𝑇
%
07-04-2014 29
At what rate percent of simple interest will a sum of money double
itself in 12 years?
a) 8
1
4
% b) 8
1
3
% c) 8
1
2
% d) 9
1
2
%
R = 100*
1
12
% = 8
1
3
%

• A sum of money becomes n times at R% per annum at
simple interest, then
• Time= 100
𝑛−1
𝑅
years
• A sum of money gets doubled at 12% per annum. Then
find the time?
• solution:
• time= 100(n-1)/R
• =100(2-1)/12
• =100/12
• = 25/3 years
07-04-2014 30

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?
Here A = 1000, B = 1100
T1 = 4, T2 = 8

How to divide the number when the
divisor ends in digit 9?
• 87÷129 =?
• The first step is
to round the denominator
to the next digit.
Long Division Method Shortcut Technique
07-04-2014 35

Divide 95÷159.
• Reduce 95÷159 as follows :
Now divide 9.5 by 16 like normal division,
instead of putting a zero in each step, you
replace it with the last quotient to continue
the division.
07-04-2014 36

Divide 436÷159.
Reduce 436÷159 as follows :
Now, there is a small difference compared to
the previous 2 examples.
In the previous 2 examples, the numerator was
less than the denominator.
However, in this case, we have the numerator
which is greater than the denominator.
In this division, we need to add the quotient
with the rest of the digit (6) and put this digit
next to the reminder.
07-04-2014 37

Divide 3462÷179.
Notice that numerator is greater
than the denominator.
There is a slight variation
compared to the previous example
Notice the carry-forward in step 2
of the division. So, therefore,
07-04-2014 38

Rule of Alligation
• If two ingredients are mixed, then
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑐ℎ𝑒𝑎𝑝𝑒𝑟
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑑𝑒𝑎𝑟𝑒𝑟
=
𝐶.𝑃 𝑜𝑓 𝑑𝑒𝑎𝑟𝑒𝑟 −𝑀𝑒𝑎𝑛 𝑃𝑟𝑖𝑐𝑒
𝑀𝑒𝑎𝑛 𝑝𝑟𝑖𝑐𝑒 −𝐶.𝑃.𝑜𝑓 𝑐ℎ𝑒𝑎𝑝𝑒𝑟
• C.P of a unit quantity of cheaper = (c)
• C.P of a unit quantity of dearer = (d)
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
3.50
1.50
=
35
15
=
7
3
MIXTURES

• 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
The ratio is 7:14 = 1:2.
Hence quantity replaced =
2
3
MIXTURES

• 5.In what ratio must water be mixed with milk to gain
16
2
3
% on selling the mixture at cost price?
• Let C.P. of 1 litre milk be Re. 1.
• S.P. of 1 litre of mixture = Re.1,
• C.P. of 1 litre of mixture =
100
100+
50
3
𝑋1 =
300
350
=
6
7
0 6
7
1
1
7
6
7
Ratio of water and milk = 1:6.

MIXTURES
• A milkman has 10 litres of pure milk. How
many litres of water have to be added to the
milk so that the milk man gets a profit of 150%
by selling at cost price?
•
150
100
=
3
2
.
•
1
1+
3
2
=
2
5
.
• The ratio of water and milk is 3:2.
• Hence 15 litres of water should be added.
0 2
5
1
3
5
2
5

MIXTURES
• 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

MIXTURES
• 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 2:1
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=2:1
So in bigger bucket,
Milk : Water = 109 : 71
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑀𝑖𝑙𝑘
𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑊𝑎𝑡𝑒𝑟
=
3
3 + 1
+
2
2 + 3
+
2
2 + 1
1
3 + 1
+
3
2 + 3
+
1
2 + 1
=
109
71

MIXTURES
• 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 2:1
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=2:1 x3 = 5, so
So in bigger bucket,
Milk : Water = 193 : 137
=
3𝑋2
3 + 1
+
2𝑋4
2 + 3
+
2𝑋5
2 + 1
1𝑋2
3 + 1
+
3𝑋4
2 + 3
+
1𝑋5
2 + 1
=
193
137

MIXTURES
• Suppose a container contains x of liquid from
which y units are taken out and replaced by
water. After n operations, the quantity of pure
liquid = 𝑥 1 −
𝑦
𝑥
𝑛
units.
• 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?
• Milk = 40 1 −
4
40
3
= 29.16 𝑙𝑖𝑡𝑟𝑒𝑠.

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

MIXTURES
• 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.

Mixtures
• In what ratio metal A at Rs.68 per kg be mixed
with another metal at Rs.96 per kg so that
cost of alloy (mixture) is Rs.78 per kg?
• 5:8 4:7 3:7 9:5
68 78 96
18 10
Ratio is 18:10 or 9:5

Divisors are different from 9’s but Close to TEN
• 108÷ 7
• Leave one digit on the right since
divisor has a single digit.
• Complement of 7 is 3.
• Put the dividend 1 below, multiply it
by 3 and place the product below 0.
• Add 0 and 3 and place it below.
• Multiply this by 3 and place the
product below 8 and
add to obtain 17.
• Now R = 17 > 7 (divisor). 17 contains
2 times 7. Place 2 below 3 and the
product below 17 (with – sign).
• Add the results to get
Q = 15 and R = 3
7 1 0 8
3 3 9
1 3 17
2 –14
1 5 3
07-04-2014 55

• 863 ÷7
• Put the complement of 7 below
7.
• Bring down 8 of the dividend
• multiply it by 3 to get 24
• Add 6 to 24 to get 30
• Put 0 on the Q–line and 3 on
the carry over line.
• Multiply 30 by 3 and write 90
on the RHS
• add it to 3 to get 93.
• As 93 = 7x13 +2 put 13 as
shown and the product 91 also.
• Q = 110 + 13 = 123
• R = 93 – 91 = 2
Divisors are different from 9’s but Close to TEN
7 8 6 3
3 24 90
8 0 93
3
11 0 93
+1 3 –91
12 3 2
07-04-2014 56

Divisors are different from 99 but Close to 100
87 2 59
13 26
2 85
259 ÷ 87 = ?
40854 ÷ 878 = ?
878 4 0 854
122 4 88
4 4 222
2
4
244
6 466
488
07-04-2014 57

Ratio / Proportion 2
• If A is x% of C and B is y% of C then A is
𝑥
𝑦
x100% of B
• Two numbers are respectively 20% and 25% of
a third number, what percentage is the first of
the second.
•
20
25
𝑥100=80%

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
Average Speed =
3𝑋60𝑋40𝑋80
60𝑋40 + 40𝑋80 +(80𝑋60)
=
576000
2400+3200+4800
=
576000
10400
= 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!

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.
645) 903(1
645
––––
258
258)645(2 129)258(2
516
–––––
129
258
–––––
0

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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