aptitude techniques are easy to follow. enjoy mathematics

1. APTITUDE AT EASE
C.S.VEERARAGAVAN
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2. PROFIT / LOSS
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3. Selling Price
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4. Cost Price
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5. 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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6. 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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7. 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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8. 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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9. Divisors are different from 9’s but Close to TEN
7 1 0 8
3 3 9
1 3 17
2 –14
1 5 3
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10. 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
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11. Divisors are different from 99 but Close to 100
87 2 59
13 26
2 85
878 4 0 854
122 4 88
4 4 222
2
4
244
6 466
488
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12. Simple Interest – Principal
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13. Simple Interest – Time
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14. Alligation / mixtures
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15. Alligation / Mixtures
• 15. Two equal glasses filled with mixture of milk and water
in the propositions of 2:1 and 1:1 respectively are emptied
into a third glass. What is the proposition of milk and water
in the third glass?
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16. Tip No.1
• The best way to speed up calculations is to eliminate
the need of calculations.
• Most problems are actually meant to test your ability
to look at a complex scenario and find simple solutions
to it.
• You are a human resource to the company. Aren’t You?
• You need not be a mathematical genius but somebody
who should be proficient at efficient utilization of
resources.
• In most problems, a bulk of calculations can be
avoided.

17. Tip 1 – Avoiding more calculations
• Consider the following problem.
• S is a 6–digit number that begins with 1. when
the digit “1” is moved from the left–most to
the right–most place, the resultant number is
three times the original number. The sum of
the digits of S is?
• (1)6 (2) 15 (3) 24 (4) 27

18. Tip 1 – contd(avoid calculations)
• What we have to find is the sum of the digits of S.
• Note that the digits of S remains the same, only
the order changes.
• Therefore, the sum of the digits of S and the
second number (3S) are the same.
• Since the second number is a multiple of 3, the
first number also should be multiple of 3.
• Hence both numbers should be multiple of 9.
• Can we find the answer now?

19. Tip No.2
• 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.

20. Tip 2 – Study. Don’t read
• Consider the following problem.
• Ram and Shyam start from a point A, move to
B which is 5 Kms away from A and then travels
back to A.
• Ram starts at 9 A.M traveling at 5 Km/hr while
Shyam starts at 9.45 A.M traveling at 10
Km/hr.
• when do they meet first?

21. Tip 2 –contd – no read, study.
• Let us explore what Ram does. He starts from the point A at 9 AM at 5
Km/Hr.
• Therefore, he reaches B at 10 AM and back to A at 11 AM.
• What about Mr Shyam ? He starts from the point A at 9:45 AM at 10
Km/Hr.
• Therefore, he reaches B at 10:15 AM and back to A at 10:45 AM.
• What happens at 10 AM ?
• At 10 AM, Ram has started moving from point B back to point A at 5
Km/Hr
• At 10 AM, Shyam is 2.5 Kms away from B and is moving towards it at 10
Km/Hr
• Now we have a new problem to solve :
• At 10 AM, A and B are 2.5 Kms apart, they are moving in the opposite
directions at 5 Km/Hr and 10 Km/Hr respectively.
• When do they meet ?

22. Tip 2 – studied?
• Did the original problem really have so much
information in it ? Yeeeeesssssss ---
• only if you go down deep enough.
• As an exercise, try finding the time when Ram
and Shyam meet for the second time.
• It is IMPOSSIBLE to get to solutions like these,
if you tend to rush through to the solution
using the most familiar ways known to you.

23. Tip No 3
• Do not carry all the luggage on your head.
• Imagine if you were traveling in a train with your suitcase
on your head, move to your seat and sit down comfortably.
• Now it feels so lazy to take off that damn suitcase off your
head. So, you leave it right there – anyway what harm is it
going to do ?
• Funny, isn’t it ? But that’s exactly what most of us do while
solving a problem !!!
• 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.

24. TIP 3 – Contd… (no luggage)
• Well, the solution is easy. Do not be lazy and do not be
presumptuous (O, I can remember all that !!!) .
• Write down all the bit of information in front of you with sufficient
clarify.
• Represent the problem well.
• Draw diagrams if necessary.
• Mark everything that you know on the diagram.
• If you make any inferences, mark that too.
• At any point, whatever you have known so far about the problem –
transfer that out of your head to a piece of paper in front on you.
• Keep your mind FREE – FREE,
• so that it can think, be creative and solve the problem, rather than
working hard at just keeping a tab on all the darn data.

25. Tip 3 eg:GEOMETRY
• Consider the following problem :
• An equilateral triangle ABE is drawn inside the
square ABCD. Find the angle CED.
So, we have this square ABCD and the
triangle AEB.
Try solving the problem with just this
diagram in front of you.
Try for a while before proceeding forward.
We have a square and an equilateral
triangle – so we have got lots of equal sides
( and equal angles) to manage.
Can you figure out which all sides are equal
and which all angles are equal ?
Not in your head –
Mark them ALL on the diagram.

26. Tip 3. – no luggage in your head.
• For most students, it is impossible NOT to
solve this problem beyond this point.
• You just can’t help seeing that BE = EC and
so angles BEC = BCE,.
• the angle EBC is 30 degrees and the rest of
the argument follows simply.
• Did that help ?
• For most students, just this single step –
“Transferring every bit of information
from your head to a piece of paper in
front of you” creates the difference
between solving the problem and not
reaching about it anywhere.
• Genius could sometimes be very simple –
isn’t it ?

27. Tip No. 4
• There is a scope for “speeding up” at every
single step of the solution.
• Every second counts.
• For most students, the toughest part of the
process of solving is the first step –
• getting the right “idea” about the problem.
• And once they have it, they rush through to the
end of the solution for their life, without waiting
for a breather.

28. Tip 4 – speed up
• Remember – every step of the solution is a sub-problem in
itself – and there is ample scope to speed up every single
step of the solution.
• You need a “habit” and a probing “eye” for areas to
simplify.
• Unless you have mastered this habit, you will be
performing a number of steps which are utmost wasteful.
• Consider multiplying 11 x 14 on step 2 and later dividing
that result by 14 to get 11 on step 5.
• How would you want to speed this up – by learning how
to multiply and divide faster, or
• by simply eliminating these two unnecessary steps ?

29. Tip 4 – speed, speed, speed…
• Find the remainder of 23015 with 17.
• The general solution of the problem would be to observe that
• 24 = 16 gives a remainder of -1 with 17.
• Hence, 28 would give a remainder of 1 with 17.
• Therefore, because of this cycle of 8, you would express 3015 = 8 x
376 + 7
• Hence 23015 = 2(8 x 376) * 27.
• Now, the first term would give a remainder 1 with 17, so we can
ignore it.
• 27 = 128 which gives a remainder 9 with 17.
• Pretty simple, isn’it ?
• Could we have done it better ?

30. Tip 4 – speed…
• Observe that all we need to worry about is the remainder
that 3015 leaves with 8.
• Can we do it faster than the way we did it before ?
• We know that 3000 is a multiple of 8, as any number that
ends with 3 zeroes is a multiple of 8.
• So, 3015 would give the same remainder with 8 as 15 ,
which is 7.
• Time taken for this process : 1 second ☺
• There are hundreds of ways you can speed up calculating
certain parts of a problem.
• Watch our for a possible speed-up at every single step and
you will save several minutes.

31. Tip 5
• Watch out for simplicity
• There is a simple test for whether a solution is
good enough – is it simple ? All elegant solutions
have an inherent simplicity and beauty to it.
• 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.
• This is a natural corollary of Tip No 2 – study the
problem carefully, watch out for simplifying
patterns.

32. Tip 5 – simplify
• Perform the following binary addition :
• 1 + 11 + 111 + 1111 + 11111 (all these numbers are binary).
• There are several “quick-fix” solutions to these :
• i) convert these numbers to decimal, add them and convert
the result back to binary
• ii) perform the brute-force binary addition.
• However, the solution to this could be much simpler.
• Consider – how would the problem look slightly differently
and be extremely simple to solve ?
• Would 1 + 10 + 100 + 1000 + 100000 be simple enough ?
What, then, makes the given problem complicated ?
• Can we find some simplifying patterns in it ?

33. Simplify
• What if we could represent the last three
numbers as
• 111 = (1000 – 1)
• 1111 = (10000 – 1)
• 11111 = (100000 – 1)
• Then the sum would be
• 1 + 11 + (1000 – 1) + (10000 – 1) + (100000 – 1)
• = 1 + 11 + (111000 – 11) (Note that 3 in decimal is
11 in binary)
• = 1 + 111000 = 111001

34. Simplify
• The sum of the ‘n’ terms of an arithmetic progression is
(an + bn2). Find the common difference between the terms.
• The most straightforward way to solve this problem is to
use the formula for the ‘n’ terms.
• Consider this :
• The sum of 1 term of the AP is the same as the first term of
the AP , which is ( a + b).
• The sum of 2 terms of the AP is the sum of the first and the
second terms and it equals (2a + 4b).
• So, what is the second term ? (2a + 4b) - (a + b) = a + 3b.
• So, we know the first term, we know the second term, can
we find the common difference between the terms ?

35. Enjoy Maths!
• There is sufficient simplicity in each problem,
waiting to reveal itself.
• Don’t insult the beauty by ignoring it and
rushing past through it.
• Stop.
• Relax.
• Enjoy the beauty
• and your journey will be much simpler.