Apt maths shortcuts

METHODS
Eight Techniques
1
APPLICATIONS
In Commerce,
Geometry etc.,
2
LINKS
Profit & Loss &
other links
3
PERCENTAGE
SIX PHRASE – VEERARAGAVAN C.S
http://182.72.85.213:8080/AssessmentTool http://www.facebook.com/SixPhrase
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PRESENTER NAME SIX PHRASE – The Finishing School 1 I

X% of a Number
An x % increase
An x% decrease
40% OF 20 MEANS
40
100
𝑋20 = 8.
1) If 20% of a = b, then b% of 20 is the same as:
A.4% of a B. 5% of a C.20% of a D. None of these
20% of a = b 
20
100
𝑋𝑎 = 𝑏
b% of 20 =
𝑏
100
𝑋20
b% of 20 =
20
100
𝑋𝑎𝑋
1
100
𝑋20 =
4
100

X% of a Number
An x % increase
An x% decrease
Increase by 40% means 140% of that number.
III.The price of a product becomes `54 after it increases by 35%.
What was the original price of that product? 1)30 2)36 3)40 4)45
Increase by 35% means 135% of the price is 54.
135
100
𝑋 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑝𝑟𝑖𝑐𝑒 = 54 Hence the original price is 54 𝑋
100
135
= 40.

X% of a Number
An x % increase
An x% decrease
Decrease by 40% means (100–40)60% of that number.
V. The price of a product is reduced to Rs.72 after it is decreased by
10%. What is the original price of the product? 1)84 2)90 3)96 4)80
Decreased by 10% means 90% of that price.
Hence the original price is 72 𝑋
100
90
= 80.
90
100
𝑋𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑝𝑟𝑖𝑐𝑒 = 72

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
27-06-2014
VEERARAGAVAN C S, APTITUDE TRAINER,
veeraa1729@gmail.com
5
If the value of something is
I.If the price of a shirt is increased by 10%
than decreased by 10%. What is the
percent change?
+ 10 - 10 –
10 10
100
= - 1%
The best way to speed up calculations is to
eliminate the need of calculations.

Example – 2
• II.The price of a product is decreased by 10% and
then increased by 10%. What is the change in price?
• overall change = –10 + 10 –
10𝑋10
100
• = – 1 % (decreased)
• use (+) sign for increment and (-) sign for decrement.
27-06-2014
6
Most problems are actually meant to
test your ability to look at a complex
scenario and find simple solutions to
it.

Example – 3
• IV. The salaries of two persons are equal.
• If the salary of one person is increased by 10% and
that of the other is decreased by 10% then what is
the change in the total salary of the two persons?
• A)increases by 1% B)decreases by 1%
• C)No change in the total salary D)None of these.
27-06-2014
7
You are a human resource to the company.
Aren’t You? They invest in you & expect more
returns. Give them more & Take home more.

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%.
27-06-2014
8
The most
important step in
solving a problem
is to ‘study’ the
problem, NOT
READ IT.
Every second that
you spend
assimilating the
problem
statement will
save you minutes
when you start
solving it.

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)
27-06-2014
9
Do not carry all the
luggage on your head.
Storing every single
piece of information in
our heads is more
painful (and ridiculous)
than traveling on a
train with the luggage
on our heads.

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)
27-06-2014
10
There is a scope
for “speeding
up” at every
single step of the
solution.
Every second
counts.

Example 7
• if a number is increased
by 20% and again
increased by 20%. By
what percent is the
number increased.
• percent increased:
• 20 + 20 + (+20)(+20)/100
= 44%(increased)
27-06-2014
11
If you catch
yourself doing
several
unnecessary
computations
and
complicated
arguments,
rest assured –
the method is
ineffective.
There has to
be a better
method than
that.

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
27-06-2014
12
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
• VIII. The price of the
refrigerator is increased by
25%. By what percent
should the price of
refrigerator be decreased
to bring back to the original
price?
•
25
125
𝑋100% = 20%
27-06-2014
13
There is
sufficient
simplicity in
each problem,
waiting to
reveal itself.
Don’t insult the
beauty by
ignoring it and
rushing past
through it.

Example 3
• X. In an examination
Pradeep got 10% more
marks than Suresh. By
what percentage are the
marks of Suresh less
than that of Pradeep?
•
10
110
𝑋100% = 9.09%
27-06-2014
14
You need not be a
mathematical
genius but
somebody who
should be
proficient at
efficient utilization
of resources.

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.
27-06-2014
15
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
27-06-2014
16

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
27-06-2014
17
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
27-06-2014
18

27-06-2014
19

27-06-2014
20

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?
27-06-2014
21
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

27-06-2014
22

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
27-06-2014
23
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

Selling Price
• XIV. A shop keeper sells an item at Rs.36 and
incurs a loss of 10%. At what price should the
Shopkeeper sell it to gain 30%? 1)45.5 2) 52
3)58.5 4) 65.
• Selling Price = 36 X
130
90
=52
27-06-2014
24

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.
27-06-2014
25
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

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
27-06-2014
26

C.S.VEERARAGAVAN
SIXPHRASE
sixphrase.veeraa1729@gmail.com

A can complete a piece of work working alone in 10
days. B can complete the work working alone in 15
days. If they work on alternate days, the work would be
completed in the
1)Least number of days if A starts the work.
2)Least number of days if B starts the work.
3) In the same time irrespective of who starts the work.
4) None of these.
If A starts the work,
Then 2 days work =
1
10
+
1
15
=
3+2
30
=
5
30
=
1
6
If B starts the work,
Then 2 days work =
1
10
+
1
15
=
3+2
30
=
5
30
=
1
6
In both cases work will be completed in 12 days. So
choice 3.

P can complete a piece of work, working alone
in 3 days. Q can complete the work working
alone in 5 days. If they work on alternative
days, the work would be completed in the
1)Least number of days if P starts the work.
2)Least number of days if Q starts the work.
3)In the same time irrespective of who starts
the work.
4) None of these.

P can complete a piece of work, working alone
in 3 days. Q can complete the work working
alone in 5 days. If they work on alternative
days, the work would be completed in the
If P starts the work,
2 days work =
1
3
+
1
5
=
8
15
. Remaining work =
7
15
3rd day P will work
1
3
=
5
15
.Remaining work =
2
15
4th day Q can complete the work.

Suppose two persons work on alternate days to
complete a work.
If the work done by both together in 1 day is
1
𝑛
th of the total work and n is an integer, then, the
work will be completed in the same time
irrespective of who starts the work.
However if n is not integer, then the exact time
to complete the work depend upon who starts
the work.

A and B working together can complete a piece of
work in 10 days. B and C working together can
complete the work in 15 days. Who is the slowest of
the three worker?
1) A 2)B 3)C 4) Cannot be determined
Time taken when B work with A is less than when B
work with C. Hence A takes less time than C.
A can finish the work by working together earlier
than time taken by C working together.
Hence C is the slowest of the three workers.

Time &
Distance
C.S.VEERARAGAVAN
SIXPHRASE
The Finishing School

To convert from km/hr to m/sec Multiply by
𝟓
𝟏𝟖
To convert from m/sec to Km/hr Multiply by
𝟏𝟖
𝟓

Rule 2
» 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 =
𝑺 𝟏 𝑺 𝟐
𝑺 𝟏−𝑺 𝟐
𝑻 𝟏 − 𝑻 𝟐

Speed
» 7.Robert is travelling on his cycle and has calculated to reach point
A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon
if he travels at 15 kmph. At what speed must he travel to reach A at
1 P.M.?
» A. 8 kmph B. 11 kmph
» C. 12 kmphD. 14 kmph
» Let the distance travelled by x km.
» Then,
𝑥
10
−
𝑥
15
=2
» 3x – 2x = 60.
» X = 60.
» Time taken to travel 60 km at 10 km/hr =
60
10
=6hrs.
» So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
» Required speed =
60
5
=12 kmph.
Short cut
S1=15 and S2=10 and T2–T1=2.
Hence Distance =
2𝑋15𝑋10
5
= 60 km

AVERAGE SPEED
» If same distance travelled in two different speeds then the
average speed during the whole journey is given by
Average speed =
𝟐𝑿𝑷𝒓𝒐𝒅𝒖𝒄𝒕 𝒐𝒇 𝑺𝒑𝒆𝒆𝒅𝒔
𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝑺𝒑𝒆𝒆𝒅𝒔
» 4. A man on tour travels first 160 km at 64 km/hr and the
next 160 km at 80 km/hr. The average speed for the first 320
km of the tour is:
» A. 35.55 km/hr B. 36 km/hr
» C. 71.11 km/hr D. 71 km/hr
» Average speed =
2𝑋64𝑋80
64+80
=71.11 km/hr.
» If same distance travelled in three different speeds A,B & C,
then,
» Average Speed =
𝟑𝑨𝑩𝑪
𝑨𝑩+𝑩𝑪+𝑪𝑨

Average Speed
V. A person travelled from Hyderabad to
Mahabubnagar at an average speed of 30 kmph
and then came back to Hyderabad at an average
speed of 90 kmph along the same route. What is
the average speed for the entire journey?
1)75kmph 2)60kmph 3)45kmph 4)Cannot be
determined
Average speed =
2𝑋30𝑋90
30+90
= 45 kmph.

Average speed
Dheeraj travelled for 2 hours at an average speed
of 48 kmph and travelled for another 2 hours at
an average speed of 52 kmph. What is the average
speed?( in kmph) 1)49 2)50 3)51 4)52
As the duration of journey at different speeds is
constant, the average speed of the journey is
average of the speeds.
Average speed =
48+52
2
= 50 kmph.

Meeting Time
Vijay left P for Q at 10:00
am. At the same time Ajay
left Q for P. After their
meeting at a point on the
way, Vijay took 24 minutes
to reach Q and Ajay took
54 minutes to reach P. At
what time did they meet?
1) 10:37 am 2) 10:36 am
3) 10:39 am 4) 10:38 am.
Meeting time =
54 𝑋 24 = 36.
Hence they meet at 10:36
am.
P C Q
vijay ajay54 min 24 min
Let Vijay and Ajay take t minutes to meet at C.
Hence Vijay takes t minutes to cover PC.
But Ajay takes 54 minutes to cover PC.
As distance is same, speeds are in inverse ratio of
times.
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑣𝑖𝑗𝑎𝑦
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑗𝑎𝑦
=
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 𝑏𝑦 𝑎𝑗𝑎𝑦
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 𝑏𝑦 𝑣𝑖𝑗𝑎𝑦
=
54
𝑡
=
𝑡
24
Hence t2 = 54 x 24.
T = 36.

 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
27-06-2014 42
VEERARAGAVAN C S, APTITUDE
TRAINER, veeraa1729@gmail.com

 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.27-06-2014 43

 A sum of money becomes n times in t years at
simple interest, then rate of interest will be
 R= 100∗
𝑛−1
𝑇
%
27-06-2014 44
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
27-06-2014 45

• 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

• 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

• 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
First type Mean Price Second type
0 6
7
1
1
7
6
7
Ratio of water and milk = 1:6.

• 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.

• p gram of ingredient solution has a% ingredient in it.
• To increase the ingredient content to b% in the solution
• Quantity of ingredient need to be added =
𝑝 𝑏−𝑎
100−𝑏
• 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
125 30−25
100−30
=
125𝑋5
70
• =
125
14
=8.92 litres.

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