Mathematics Model for Postal Assistant Exam

Mathematics

Some Important Maths Equations 

1. a(b+c) = ab + ac

2. (a+b)2 = a2 + 2ab + b2

3. (a-b)2 = a2-2ab + b2

4. (a+b)2 = (a-b)2 + 4ab

5. (a-b)2 = (a+b)2 - 4ab

6. (a-b) (a+b) = a2 - b2

7. (a+b)3 = a3 + 3ab(a+b) + b3

8. (a-b)3 = a3 - 3ab(a-b) - b3

9. a3 + b 3 = (a+b) (a2 - ab + b2)

10. a3 - b 3 = (a-b) (a2 + ab + b2)

11. (a3 - b3) / (a2 + ab + b2) = a-b

12. (a3 + b3) / (a2 - ab + b2) = a+b

How to use these equations in exams.

1. (4.75 x 4.75 x 4.75 + 1.25 x 1.25 x 1.25) / ( 4.75 x 4.75 + 1.25 x 1.25 - 4.75 x 1.25) = ?
If you saw a question like this, you can use these equations to find its answer. If try to solve these question without using equation you gonna take too much time , thats what the examiners want.

= (4.753 + 1.253) / (4.752 + 1.253 - 4.75 x 1.25)

now it is same like in the 12th equation (a3 + b3) / (a2 - ab + b2) = a+b

ie. = 4.75 + 1.25
 = 6

Mathematics question from Train

1. A 250 m long train is running at a speed of 55 Km/hr. It crossed a platform of length 520 m in ? (Its a Previously asked on SSC 2012 exam) (*without a train question it is impossible to conduct an exam, I think you got, what I mean)

a) 40.4 sec                                           b) 40.5 sec

b) 50.4 sec                                           c) 50.5 sec

Solution
Speed = 55 Km/hr
(to convert km/hr in to M/s)
= 55 x 5/18 M/s

Distance = 250 m + 520 m ( If questions is about train crossing a post you need to consider only the length of Train, )

= 770 m

Time = Distance / Speed

= 770 x 18 / (5 x 55)

= 50.4 sec

Answer: b) 50.4 sec

Mathematics question from Work and Time

1. A can do a piece of work in 10 days and B alone can do it in 15 days. How much time will both take to finish the work ?

Solution
This question can be solved by different methods. We need to conserve time in exams so solving this problem using equations is the good idea.

Time taken to finish the job = XY / (X + Y)

= 10 x 15 / (10 + 15)

= 150 / 25

= 6 days


2. A and B together can do a piece of work in 10 days. B alone can finish it in 20 days. In how many days can A alone finish the work ?

Solution
 Time taken by A to finish the work = XY/(Y-X)

= 10 x 20 / (20-10)

=  200/10

= 20 days

Mathematics question from Simple & Compound Interest

1. Find the difference between the Simple Interest and Compound Interest for an amount of 50,000 at 10% of interest rate for 2 years ?

Solution
If you try to solve this problem without using Equation, You gonna spend too much time for this question.

Difference between SI and CI for 2 years = P (R/100)2

                                                                = 50000 x 10/100 x 10/100

                                                                = 500


Equation to find Difference between SI and CI for 3 years = P (R/100)2 ( 3 + R/100 ) Equation to find Difference between SI and CI for 4 years = P (R/100)2 ( 4 + R/100 )
Equation to find Difference between SI and CI for 5 years = P (R/100)2 ( 5 + R/100 )

Mathematics question from LCM & HCF

1. The LCM and HCF of two numbers are 8 and 48 respectively. If one of them is 24, Find the other ?

Solution
HCF x LCM = Product of Numbers

8 x 48 = 24 x the other number

other number = (8 x 48) / 24

other number = 16

 Previously asked maths questions from numbers for competitive exams

1. Sum of two number = 16 and product = 48, find the difference of the two number ??

To solve this question, you can use the equation,
Solution
(X-Y)2  = (X+Y)2 - 4 XY

(X-Y)2 = (16)2 -  4 x 48

(X-Y)2 = 256 - 192

(X-Y)2 =  64

X-Y = 8