Mathematics/Numerical ability shortcuts for all competitive exams

Thanks to everyone for your support and encouragement

I would try to complete the portions within two weeks.

As Sbi PO exams are appearing, I would try to complete numerical ability as fast as possible.

Next lesson would be on Average, AGE, Time and Distance, Train problems.

You could score 1 mark from each section.

For all those who are appearing for competitive exams, please memorize 1 to 30 tables.

As this would be easy for you to answer, and you would not multiply or take time to think about it.

Learning 1 to 30 tables would help you in a lot ways. You would be able to answer quickly as well there is no need to spend time in calculating 23 * 25 on rough work.

 

Exercise 7: Average

Formulas:

1) Average = sum of quantities / No of quantities

Vice Versa [ 2) number of quantities = sum of quantities/ Average]

3) Sum of quantities = Average x no of quantity.

1) The average of 25 students in a class is 12 years. If the teacher's age is included the average increases by one. What is the age of teacher's?

so, here we apply average * no of quantity formulae.

25 * 12 = 300

In the given sum they have said teacher's age is included the average increases by one. So, we have to add 1 to teachers age and average

(25 + 1) * (12+1) = 26 * 13 = 338

Therefore, the age of teacher's = 338 - 300 = 38 years

2) The average salary of 15 persons is 5500. If the salary of one person is added, the average increases to 5700. What is the salary of this one person?

15 * 5500 = 8 2500
16 * 5700 = 91200 ( 15 + 1 = 16, because salary of one person is added)

91200-82500 = 8700

the solution for this sum is 8700.

3) The average of a man and his son is 33 years. The ratio of their ages is 8 : 3 respectively. What is the man's age?

2 * 33 = 66 ( a man and son, so 2 members)

8 + 3 = 11 ===> 66/11 = 6

So they have asked man's age , 8 * 6 = 48.

The solution is 48

Suppose if they have asked son's age

It would be 6 * 3 = 18.


Exercise for your practice:

1) The average age of a class of 20 students is 11 years. If the age of teacher is added the average age is increased by one year. what is the age of teacher?

2) The average monthly salary of 30 employees is RS 650. If the manager's salary is added, the average monthly salary increases by RS 50. The salary of the managers is ?

3)Find the average of following sets of scores.

178, 863, 441, 626, 205, 349, 462, 820

Always read the questions clearly twice. Because some of you tend to skip the additional details or tricky sentences asked in the exam.

 

Thanks for sharing your experience. These short tricks are not much into practice nowadays. This is ancient mathematics. Like these tricks, we have lost so many things without knowing how it was being done. All my credits goes to my 83 yr old grandfather. I loved his book assets more, but he refuses to share it with anyone.

 

Exercise 8 :- Problems on Ages

Question asked on this section would be one, and it would be asked mostly in bank exams, Intelligence officer (rarely, but in 2012 they have asked)

In bank clerical or PO, IBPS and other competition you could expect it.

* Whenever it is past tense = ago, then it would be subtraction

* In present tense = hence, will be, these words would be used so it would be addition

1) The ratio of the ages of Uma and Sita is 4 : 3. The sum of their ages is 28 years. The ratio of their ages after 8 years will be

Let the Uma and Sita age be 4x, 3x

STep 1) First we would find X value, because they have given sum of their ages

4x + 3x = 28 ===> 7x = 28 ===> x = 4

STep 2) Now substitue X value in 4x, 3x, so it would be 16, 12

STep 3) They have asked ratio after 8 years, so we add 8 to 16 and 12

24(16+ 8) : 20( 12+8)

The solution is 6:5

Eg:-2 The ratio of the ages of Manju and Sanju is 4 :5 . Twelve years hence, their ages will be in the ratio of 5 : 6. What will be Sanju's age after 6 years?

Explanation :

So for explanation I am elaborating briefly, in exam time, don't waste your time by putting all these steps , do it as soon as possible.

Let Manju's and Sanju;s present age be 4x and 5 x

4 x + 12 = 5
------------
5x+12 = 6


Now cross multiply, and find x value, X = 12

Therefore, they have asked Sanju's age after 6 years, so = 5x+6 = 66 years.

Suppose , if they have asked Manju's age after 6 years, then it would be

4 ( 12) + 6 = 54 years.

 

Exercise 8 :- Problems on Ages ( continuation)

3) The sum of their ages of Raju and his mother is 49 years. Also, 7 years ago, the mother's age was 4 times Raju's age. Find the present age of Aruna's mother

SO, Raju's age 7 years ago is X, mother's age 7 years ago = 4x

Sum of their ages = 49===? (x+7 ) +(4x+7) = 49 ===> x = 7

Therefore, the present Mother's age = ( 4x + 7) ==> 35 years


SUppose if they have asked Raju's present age, then it would be ( x + 7)==>14

2) A man is twice as old as his son. 20 years ago, the man was 12 times the age as his son. The present age of the man in years is

Son's age X, Man's age = 2X

12( x-20)= (2x-20)===> X= 22

Therefore man's present age = 2x = 44 yrs

Sum for practise


1) A father is four times as old as his son. Five years ago, the age of father was nine times as old as his son. What is the present age of father?

Solution of this sum : 32 yrs ( same example sum which I did in previous )

2) 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.

Same model which I did in 3rd sum

 

Exercise 9: Problems on Numbers:

Suppose subtracting or multiplying by 888888 or 999999 would be difficult for us.

In exam times, there is no need to spend time on multiplying , instead of that apply these formulas.

In this class, we would see various problems on Numbers.

Basic formulae:


1) (a+b)²=a²+b²+2ab

2)(a-b)²=a²+b²-2ab

3)(a+b)²-(a-b)²=4ab

4) (a+b)²+(a-b)²=2(a²+b²)

5) a²-b²=(a+b)(a-b)

6) (a-+b+c)²= a²+b²+c²+ 2(ab+bc+ca)

7) a³+b³=(a+b)(a²+b²-ab)

8) a³-b³=(a-b)(a²+b²+ab)

9) a³+b³+c³-3a b c=(a+b+c)(a²+b²+c²-ab-b c-ca)

Problem sums:

1) 8359233 * 9999

Solution: ==> 8359233 * ( 10,000 -1) [ we r writing 9999 as 10,000-1]


= 83592330000 - 8359233 = 83583970767

2) 321 * 99 = 300 * ( 100 - 1)

= 32100-321 = 31779

Practise with more combinations, I would teach you more problems on Numbers. Its a lengthy subject.

 

Exercise 9 : Problems on Numbers continuation

1) 657 * 657 * 657 + 434 * 434 * 434
-----------------------------------------
657 + 657 - 657 *434 + 434 * 434

Suppose, if they are giving problem like this, don't find it difficulty, Immediately think about the above given forumalas on Exercise 9

So , the formula is a^3 + B^3/a^2 -ab + B^2 = ( a+ B)

So the answer is ( 657 + 434 = 1091)


All you need to remember is forumales.

These basic formulas are not difficult for most of them, but for competitive exams these simple tricks helps in scoring 1 mark / for logic thinking.