’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
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;
Q. One year ago, the ratio of Gaurav’s and Sachin’s age was 6: 7 respectively. Four years hence, this ratio would become 7: 8. How old is Sachin ?
A. let the age be ‘x’ and ‘y’ respectively. x : y = 6 : 7 so 7x – 6y = 0 and
(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.
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