APTITUDE PREPARATION: Numerical Ability


Introduction:

In a fraction, the number in the top is called a numerator and the number in the bottom is the Denominator
That is, Numerator/Denominator
The number to be divided is called dividend and the number by which it is divided is called a divisor
The number which tells how many times the divisor is contained in the dividend is called quotient

Some important results:

ax= x times a. Eg. a3 = a x a x a
√16 = ± 4
ax. ay = a(x+y)
ax/ ay = a(x-y)
(ax)y = a(xy)
a to the power of (xy) … find (xy) first and then evaluate a to the power of (xy)
If ax = ay then, x =y
If ax = bx then, a = b
ax = 1 then, x = 0 for all values of x (except zero)
1x = 1 for all values of x (except zero)
(1/ ax) = a-x
a+0 = a-0 = a and, (a x 1) = (a/1) = a
a x 0 = 0
(a/b)(-x/y) = (b/a)(x/y)
(a/b)/(c/d) = (a/b) x (d/c)

Division by zero is an operation for which you cannot find an answer, so it is disallowed.

Simplification:

For arithmetic simplification, BODMAS rule is used (from left to right):
B for Bracket
O for Of … (Eg. % of)
D for Division
M for Multiplication
A for Addition
S for Subtraction
In case of nested brackets, start simplifying from the inner bracket.

Divisibility Rules:

A number is divisible by:
(a). 2 if the last digit is even or zero
(b). 3 if the sum of all the digits is divisible by 3
(c). 4 if the number formed by last two digits is divisible by 4
(d). 5 if the last digit is 5 or zero
(e). 6 if the number is divisible by both 2 and 3
(f). 7: Remove the last digit, double it, subtract it from the truncated original number and continue doing this until only one digit remains. If this is 0 or 7, then the original number is divisible by 7.
Example: 1603 -> 160-2(3)=154 -> 15-2(4)=7, so 1603 is divisible by 7.
(g). 8 if the number formed by last three digits is divisible by 8
(h). 9 if the sum of all the digits is divisible by 9
(i) 10 if the last digit is zero
(j). 11 if the difference of the alternating sum of digits of the number is zero or a multiple of 11
Example: 2343 is divisible by 11 because (2+4) – (3+3) = 0
(k) 12 if the number is divisible by both 3 and 4
(l). 13: Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary.
Example: 50661–>5066+4=5070–>507+0=507–>50+28=78 and 78 is 6 x 13, so 50661 is divisible by 13.



Power of Numbers:

Any single digit number when raised to various powers exhibit a cycle pattern
They follow a specific pattern and repeat themselves after a fixed number of times.

For example,

to find the unit digit (last digit from left) of 326:
For 3, the cycle gets repeated after 4 powers (refer the table)
Divide 26 by 4, the reminder is 2
Hence, the last digit of the number will be the same as the last digit of 32
Second from left is 9.

Example 1:

Simplify: (27/216)(-4/3)

Solution:

=> (27/216)(-4/3)
=> (216/27)(4/3)
=> (6 x 6 x 6/3 x 3 x 3)(4/3)
=> (63/33)(4/3)
=> (6/3)3(4/3)
=> (2)(3×4/3)
=> 24
=> 16.

Example 2:

If 25(x+1) = (125/5x), find the value of x.

Solution:

=> 25(x+1) = (125/5x)
=> 52(x+1) = (53/5x)
=> 52(x+1) = 5(3-x)
=> 2(x+1) = (3-x)
=> 3x = 1
=> x = (1/3).

Example 3:

If 1245a42 is divisible by 11, find the value of a.

Solution:

=> (1+4+a+2) – (2+5+4) = 0 (or) 11 (As per Divisibility Rule of 11 sated above)
=> (a+7) – (11) = 0 … (must be zero and cannot be 11, as a ≤ 9)
=> a = 4.



Example 4:

Find the last digit of 126735

Solution:

Consider the last number 735
For 7, the cycle gets repeated after 4 powers (refer the table)
Divide 35 by 4, the remainder is 3
Hence, the last digit of the number will be the same as the last digit of 73 (refer the table)
Third from left is 3 which is the last digit.

Example 5:

Simplify: 25 – 48 ÷ 6 + 12 × 2

Solution:

=> 25 – 48 ÷ 6 + 12 × 2
=> 25 – 8 + 12 × 2, (‘division’ 48 ÷ 6 = 8)
=> 25 – 8 + 24, (‘multiplication’ 12 × 2 = 24)
=> 17 + 24, (‘subtraction’ 25 – 8 = 17)
=> 41, (‘addition’ 17 + 24 = 41)

Example 6:

Simplify: 78 – [5 + 3 of (25 – 2 × 10)]

Solution:

It is a problem of nested brackets.
=> 78 – [5 + 3 of (25 – 2 × 10)] => 78 – [5 + 3 of (25 – 20)], (‘multiplication’ 2 × 10 = 20)
=> 78 – [5 + 3 of 5], (‘subtraction’ 25 – 20 = 5)
=> 78 – [5 + 3 × 5], (‘of’)
=> 78 – [5 + 15], (‘multiplication’ 3 × 5 = 15)
=> 78 – 20, (‘addition’ 5 + 15 = 20)
=> 58, (‘subtraction’ 78 – 20 = 58)



Example 7:

Find the rightmost non-zero integer of the expression: 430342 + 470367.

Solution:

Consider 3342

For 3, the cycle gets repeated after 4 powers (refer the table)
Divide 342 by 4, the reminder is 2
Hence, the last digit of the number will be the same as the last digit of 32
Second from left is 9 which is the last non-zero digit.

Consider 7367

For 7, the cycle gets repeated after 4 powers (refer the table)
Divide 367 by 4, the reminder is 3
Hence, the last digit of the number will be the same as the last digit of 73
Third from left is 3 which is the last non-zero digit.
The last non-zero digit of the expression = (9+3) = 12 = 2.

Example 8:

If the unit digit of the product (49 x 46 x 28a x 44) is 0, the digit represented by a is:

Solution:

Multiplying the unit digits of all numbers = 9 x 6 x a x 4 = 216 x a
6 when multiplied by a yields 0 in its unit digit.
Hence, a should be 5.



Example 9:

If the largest 3 digit number is subtracted from the smallest 5 digit number, What is the reminder?

Solution:

Smallest 5 digit number = 10000
Largest 3 digit number = 999
Reminder = 10000 – 999 =9001

Example 10:

A two digit number is seven times the sum of its digits. If each digit is increased by 2, the resultant number is more than 6 times the sum of its digits by 4. What is the number?

Solution:

Let the number is: ab
10a + b = 7(a+b) … Number ab = 10a + b (applying the position values)
3a = 6b
a = 2b … (i)
Each digit is increased by 2:
10(a+2) + (b+2) = 6[(a+2)+(b+2)] + 4
10a + b + 22 = 6a + 6b + 28
4a = 5b + 6 … (ii)
Using equation (i) in (ii);
4(2b) = 5b + 6
b = 2 and a = 4.
The number is 42.

About Author

Dr. BASKAR .A,

a former Research Scientist Government of India, presently working as a Professor in the Dept of Mechanical Engineering, Authored books on Kinematics and Dynamics of Machinery, Also Published Numerous Research papers moreover a Geek in Mathematics

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