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1. CALCULATING FASTER
2. VEDIC MATHS
VEDIC MATHS
Base Method
This is very suitable when numbers
are close to a base like 10, 100, 1000 or so on. Let's
take an example:
106 × 108
Here the base is 100 and the 'surplus' is 6 and 8 for
the two numbers. The answer will be found in two parts,
the right-hand should have only two digits (because
base is 100) and will be the product of the surpluses.
Thus, the right-hand part will be 6 × 8, i.e. 48. The
left-hand part will be one multiplicand plus the surplus
of the other multiplicand. The left part of the answer
in this case will be 106 + 8 or for that matter 108
+ 6 i.e. 114. The answer is 11448.
12 X 14.
10 would the most suitable base. In
the current example, the surplus numbers are +2 and
+4.
If 8x7 were to be performed and base
of 10 were chosen, then -2 and -3 would have been the
deficit numbers.
Try the following numbers
(a) 13 X 16 (b)
16 X 18 (c)
18 X 19 (d)
22 X 24
Once you get comfortable, do not use
any paper or pen.
27 X 28 322
#9; #9; 23
X 18 46
X 48 5255
582
53 X 57 622
382
42
X 46 #9; #9; 9698
92 X 93
99 X 99 #9; #9; 102 X 105
98 X 107 112X113
1082
123 X 127
USING OTHER BASES
In 46 X 48, the base chosen is 50 and multiplication
of 44 by 50 is better done like this: take the half
of 44 and put two zeros at the end, because 50 is same
as 100/2. Therefore, product will be 2200. It would
be lengthy to multiply 44 by 5 and put a zero at the
end. In general, whenever we want to multiply anything
by 5, simply halve it and put a zero.
Multiply 32 by 25. Most of the students
would take 30 as the base. The method is correct but
nonetheless lengthier. Better technique is to understand
that 25 is same as one-fourth. Therefore, one-fourth
of 32 is 8 and hence the answer is 800.
An application of Base Method to learn
multiplications of the type 3238, where unit's digit
summation is 10 and digits other than unit's digit are
same in both the numbers. In the above example, 2 +
8 = 10 and 3 in 32 is same as 3 in 38. Therefore method
can be applied. The method is simple to apply. The group
of digits other than unit's digit, in this case 3, is
multiplied by the number next to itself. Therefore,
3 is multiplied by 4 to obtain 12, which will form the
left part of the answer. The unit's digits are multiplied
to obtain 16 (in this case), which will form the right
part of the answer. Therefore, the answer is 1216.
Try these now
53 X 57 91
X 99 106
X 104 123
X 127
The rule for squares of numbers ending
with 5. e.g., 652. This is same as 65 X 65
and since this multiplication satisfies the criteria
that unit's digit summation is 10 and rest of the numbers
are same, we can apply the method. Therefore, the answer
is 42 / 25 = 4225.
Try these:
352 952
1252
2052
CUBING
Finding the cubes of numbers close
to the powers of 10. e.g., cubes of 998, 1004, 100012,
10007, 996, 9988, etc. Some of the numbers are in surplus
and others are in deficit. Explain the method as given
below.
Find (10004)3
Step
(I) : Base is 10000. Provide three spaces
in the answer.The base contains 4 zeros. Hence,
the second and third space must contain exactly 4 digits.
1
0 0 0 4 = —/ —/ —
Step
(II) : The surplus is (+4). If surplus
is written as 'a', perform the operation '3a' and
add to the base 10000 to get 10012. Put this in the
1st space.
1 0 0 0 4 = 1 0 0 1 2 /—/—
Step
(III) :
The new surplus is (+12). Multiply the new surplus by
the old surplus, i.e. (+4)(+12) = (+48). According to
the rule written in the step (I), 48 is written as 0048.
1 0 0 0 4 = 1 0 0 1 2 / 0 0 4 8 /—
Step
(IV) : The last space will be filled
by the cube of the old surplus (+4). Therefore, 43
= 64, which is written as 0064.
1 0 0 0 4 = 1 0 0 1 2 / 0 0 4 8 /
0 0 6 4
Therefore, the answer is 1001200480064.
Find (998)3
Step
(I) : Base
= 1000. Hence, exactly 3 digits must be there in the
2nd and 3rd space.The deficit = (+2)
9
9 8 = —/—/—
Step
(II) : Multiply the deficit by 3 and
subtract (because this is the case of deficit) from
the base.
9 9 8 = 9 9 4 /—/—
Step
(III) : (old deficit) x (new deficit)
= 2 x 6 = 12
9
9 8 = 9 9 4 / 0 1 2 /—
Step
(IV) : The cube of the old deficit =
8. Since it is the case of deficit, -8 should be written.
All that you need to do to write the negative number
in the third space is to find the complement of the
number, in this case 8. But since the third space must
have exactly 3 digits, the complement of 008 must be
calculated. The complement of 008 is 992. Don't forget
to reduce the last digit of the second space number
by 1
9
9 8 = 9 9 4 / 0 1 2 / 9 9 2
-
1
————————————
9
9 4 / 0 1 1 / 9 9 2
Therefore, the answer is 994011992
As an exercise, try the following
:
999943 = 9 9 9 8 2 / 0
0 1 0 8 / 0 0 2 1 6 = 99982/00107/99784
100053 = 1 0 0 1 5 / 0
0 7 5 / 0 1 2 5 = 10015/0075/0125
1000253 = 1 0 0 0 7 5 /
0 1 8 7 5 / 1 5 6 2 5 = 100075/01875/15625
99999883 = 9 9 9 9 9 6
4 / 0 0 0 0 4 3 2 / 0 0 0 1 7 2 8
=
9999964/0000431/9998272
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