Mathematical shortcuts1

APTITUDE PROGRAMME MAY 8TH

KONGUNADU COLLEGE

COMPUTER SCIENCE

C.S.VEERARAGAVAN

TRAINER

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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+𝑟𝑋100

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 X100𝑋100𝑋100

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 =3600𝑋100𝑋100

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 =

6300X100

80𝑋

100

50𝑋

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

• 100𝑋𝑎+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

10𝑋10X100 =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

𝑏𝑐𝑥100

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.𝑃𝑋 100

= 11

10−

10

1110

11

𝑋 100

= 11𝑋11 −10𝑋10

110𝑋

11

10𝑋100

= 11𝑋11−10𝑋10

10𝑋10𝑋100

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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

11𝑋16𝑥100 =

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−𝑙𝑜𝑠𝑠𝑥𝑆. 𝑃

=100

95𝑥570

Profit = 5% So new S.P

=100+𝑃𝑅𝑂𝐹𝐼𝑇

100𝑋𝐶. 𝑃

= 105

100𝑥

100

95𝑋570=Rs.630

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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 𝑃𝑟𝑜𝑓𝑖𝑡X 100

• Cost Price

• = 18

6𝑋100=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

100𝑥𝐶. 𝑃 =

112𝑥

100

NEW S.P = 112𝑥

100+ 18.

PROFIT %= 𝑁𝐸𝑊 𝑆.𝑃 −𝐶.𝑃

𝐶.𝑃𝑋100 = 18

= 112𝑋

100+18−𝑋

𝑋𝑋100 = 18

=112𝑋+1800−100𝑋

100𝑋X100= 18

=12X + 1800 = 18X

SO X = 1800

18−12=

1800

6 =300

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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

16𝑋100=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

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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.

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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 %

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At what rate percent of simple interest will a sum of money double itself in 12 years?

a) 81

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

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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

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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.

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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.

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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,

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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

162

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=

33 + 1 +

22 + 3 +

22 + 1

13 + 1

+3

2 + 3+

12 + 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




3𝑋23 + 1 +

2𝑋42 + 3 +

2𝑋52 + 1

1𝑋23 + 1

+3𝑋4

2 + 3+

1𝑋52 + 1



193

137

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




33 + 1 +

22 + 3 +

22 + 1

13 + 1

+3

2 + 3+

12 + 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




3𝑋23 + 1 +

2𝑋42 + 3 +

2𝑋52 + 1

1𝑋23 + 1

+3𝑋4

2 + 3+

1𝑋52 + 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

𝑛units.

• 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

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• 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

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Ratio / Proportion 2

• If A is x% of C and B is y% of C then A is 𝑥

𝑦x100% of B

• Two numbers are respectively 20% and 25% of a third number, what percentage is the first of the second.

•20

25𝑥100=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 = .142857142857142857...

• 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 * 14", giving us .28 as the starting point.

• For 3/7, think "3 * 14", giving us .42 as

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.



• 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.

Calendar

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

• A man can row in still water at x km/h. In a 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

• A man can row in still water at x km/h. In a 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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