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’BODMAS’ Rule : This rule depicts the correct sequence in which the operations are to be executed,so as to find out the value of a given expression.

Here, ‘B’ stands for ’bracket’ ,’O’ for ‘of’ , ‘D’ for’ division’ and ‘M’ for ‘multiplication’, ‘A’ for ‘addition’ and ‘S’ for ‘subtraction’. Thus, in simplifying an expression, first of all the brackets must be removed, strictly in the order (), {} and [].

After removing the brackets, we must use the following operations strictly in the order: (1)division (2) multiplication (3)addition (4)subtraction.

Modulus of a real number : Modulus of a real number is defined as |5| = 5 and |-5| = -(-5) = 5. So modulus always gives positive value.

Virnaculum (or bar): When an expression contains Virnaculum, before applying the ‘BODMAS’ rule, we simplify the expression under the Virnaculum.

Q.Find the value of x in 3648.24 + 364.824 + x – 36.4824 = 3794.1696

1.12

2.2.2

3.0.2

4.2

Q.find value of 4−51+13+12+14

1.1/8

2.1/7

1/28

1/18

Q.

Q.If a/b=3/4 and 8a+5b=22,then find the value of a

1.1/2

2.3/2

3.5/2

4.7/2

Q.If 2x+3y+z=55,x-y=4 and y – x + z=12,then what are the values of x , y and z?

1.7,11,8

2.17,11,8

3.7,13,8

4.7,11,9

Q.A board 7ft. 9 inches long is divided into 3 equal parts . What is the length of each part?

1.2ft. 8 inches

2.2ft. 9 inches

3.2ft. 7 inches

4.1ft. 7 inches

*(a + b)2 = a2 + 2 ab + b2*>*(a – b)2 = a2 – 2ab + b2**(a + b)3 = a3 + 3a2b + 3ab2 + b3**(a – b)3 = a3 – 3a2b + 3 ab2 – b3**a2 – b2 = (a – b) (a + b)**a3 – b3 = (a – b)3 + 3 a b (a – b)*

The above formulae can be used for solving

Sometimes we have to calculate the powers of very large numbers like 13^35 which looks impossible but the answer is quite simple .. Recognize the cycle.

Suppose we need to find last digit of 2^ ^{2012}

2^1 = 2

2^2= 4

2^3 = 8

2^4 = 16 = 6

2^ 5 = 32 = 2

2^ 6 = 64 =4

Hence there is a cycle of size **4** like 2-4-8-6-2-4-8-6… so all powers of 2 in multiples of 4 like 4,8,12,16… have last digit 6 since 2012 also satisfies this 2^2012 has last digit 6.

So generalizing for k > 4 if 2^^{k} = 2^^{(4n+1)} then last digit is 2, if 2^^{(4n+2)} = 4, 2^^{(4n+3)} = 8 , 2^^{(4n)} = 6.

Similarly we can solve for all digits:

**For 3 cycle is:** 3-9-7-1-3-9-7-1… here too the cycle is 4 and So generalizing for k > 4 if 3^^{k} = 3^^{(4n+1)} then last digit is 2, if 3^^{(4n+2)} = 4, 3^^{(4n+3)} = 8 , 3^^{(4n)} = 6.

**For 4 cycle is: **4-6-4-6-4-6-… hence cycle is 2 So generalizing for k > 3 if 4^^{k} = 4^^{(2n+1)} then last digit is 6, if 4^^{(2n)} = 4.

Similarly we can obtain cycles for all digits till 9 any digit above this doesn’t matter as 13^n is same as 3^n and 17^^{n} is same as 7^^{n} etc.

Here ages of two people shall be given as a ratio. Then their ages before x years or after x years shall be given. This ratio would be equal to a value. The question would then be to calculate their present ages.

Methodology: Assume age of the lesser entity as ‘x’. The older entity become ‘kx’. Then their ages ‘b’ years ago would be ‘x-b’ and ‘kx-b’ respectively. This would be equal to some value say 10.

Obtaining the linear equation: (x-b) + (kx-b) = 10; When we solve this we get value of x.**Q1. **One year ago Jaya was four times as old as her daughter. Six years hence, Jaya’s age will exceed her daughter’s age by 9 years. The ratio of the present ages of Jaya and her daughter is :**Ans:** daughter’s age 1 yr ago= x , Jaya’s age 1 year ago = 4x

6 years hence: (4x+7) – (x+7) = 9 , we solve this to get x;**Q2. **Five years ago, the total of the ages of a father and his son was 40 years. The ratio of their present ages is 4 : 1. What is the present age of the father ?**Ans.**Present age: son = x , father = 4x;

Five years ago : sons age = x-5 , fathers age = 4x-5;

linear equation: (x-5)+(4x-5)=40;

(x+4) : (y+4) = 7 : 8 so 8x+32 = 7y+28 and we get 8x-7y=-4 solving both we get x = 24, y=28.

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Q1: What is 64^3000

1.4

2.6

3.0

4.1

Q2: Sum and number of even factors of 240:

1.720 , 16

2.744 , 20

3.736 , 8

4.700 , 10

Q3: sum and number of odd factors of 240

1.24,4

2.30,5

3.25,5

4.20,4

Q4: Finding the number of zeroes in a factorial of 127

1.30

2.31

3.32

4.33

Q5: One year ago Jaya was four times as old as her daughter. Six years hence, Jaya’s age will exceed her daughter’s age by 9 years. The ratio of the present ages of Jaya and her daughter is :

1.4,13

2.3,12

3.6,18

4.7,21