Ekiras

Concepts Examples Properties of Averages Average of N things/quantities is equal to the sum of all the things/quantities divided by number of things/quantities. Average of N number of things/quantities always lies between the lowest and the highest quantities. If each quantity is increased by a certain value K, then the new average is increased by K. If each quantity is decreased by a certain value K, then the new average is decreased by K. If each quantity is multiplied by a certain value K, then the new average becomes K times the the original average. If each quantity is divided by a certain value K, then the new average becomes $\frac{1}{K}$ 1 K   times the the original average. Question 1 Average number of chocolates with 10 children is 5. If a person distributes all 20 chocolates he has to the children then what is the average number of chocolates with the children.

Concepts Examples There are three types of questions that are based on LCM with remainders. They are as follows. When the remainders are same for all the divisors In this case the required number will be the LCM × N + remainder, where N is any natural number N=1, will give smallest such number N=2, will give second smallest such number and so on When the remainders are different for different divisors but the respective difference between the divisors and the remainders remain constant. In this case the required number will be LCM × N +difference of any (divisor - remainder) refer to example 2. When neither the divisors are same nor the respective difference between the divisors and the remaniders remain constant. In this case solve the question by forming equations and solving them. Question 1 What is the least possible

Concepts Examples There can be direct questions on finding last two digits of any power of a natural number and this is very important for eliminating options in questions. Following are the four simple cases with which we will be able to solve almost every problem. Rules for Finding Last Two Digits of square of a number In order to find the last two digits of square of any two digit number number, we can write it as difference or sum from 50 or 100, whichever is closer. In general the last two digits of a number (50±k) 2 or (100-k) 2 will be determined by square of k . e.g 47 2 can be written as (50-3) 2 , Now, the last two digits will be determined by square of 3 that is 09 Rules for Finding Last Two Digits of any power of a number For a natural number N k , where N is a natural number ending with 0 and K is a natural number greater than 2, Last two digits will always be 00 . For

Concepts Examples To find the last digit i.e the unit digit you need to be aware of the cyclicity of the numbers and their powers 2 1 =2 3 1 =3 4 1 =4 2 2 =4 3 2 =9 4 2 =16 2 3 =8 3 3 =27 4 3 =64 2 4 =16 3 4 =81 4 4 =256 2 5 =32 3 5 =243 4 5 =1024 2 6 =64 3 6 =729 4 6 =4096 2 7 =128 3 7 =2187   2 8 =256 3 8 =6561   We can see that unit's digit of 2 1 ,2 5 ,2 9 is 2 and so on. Therefore, after every four powers of 2, the units digit of the number starts repeating.Thus we can say that cyclicity of unit's digit of higher powers of 2 is 4. Similarily unit digit of power of 4 starts reeating after 2, thus its cyclicity is 2. Unit digit follows a periodic pattern that is after a particular period it repeats in a cyclic form, this is called cyclicity. Unit digits of numbers ending with 0,1,5,6 is always the same irrespective of their powe

Concepts Examples We can use the following rules to find the remainder when a number is divided by another. Rules for Finding Remainders Rule 1 : x is divided by y When we have to divide a number by another number, then we can write x in terms of y, i.e. (y×a)+b. This means that the remainder is b. e.g. When 66 is divided by 10, we can write 66 as 10×6+6, thus the remainder is 6. Rule 2 : x ± y is divided by z When we have to divide the sum or difference of two number by another number, then we write both the numbers as the product of the number with which we have to divide. e.g. When we have to divide 45+23 by 8, then we write it like 8×5+5 + 8×2+7=8(5+2)+ 12 Since 8(5+2) is divisible by 8 thus we will divide 12 by 8, and we know the remainder will be 4. Similarily we can solve for 45-23 divided by 8. Rule 3 : x × y is divided by z When we have to divide the product of two numbers by anoth

Concepts Examples To find the number of zeros at the end of a factorial or a product, find the highest power of 2 and 5 in the factorial or the product, number of zeros will be decided by the limiting factor . ( one who's power is smaller will be the limiting factor) Limiting factor is the one who's power is less among 2 and 5 e.g. Finding number of Zeros in 150 150 = 2 × 3 × 5 × 5, ∴ Highest power of 2 = 1 ∴ Highest power of 5 = 2 ∴ Limiting factor = 1 ∴ Number of zeros in 150 = 1 e.g. Finding number of Zeros in 10! ∴ Highest power of 2 = 8 ∴ Highest power of 5 = 2 ∴ Limiting factor = 2 ∴ Number of zeros in 10! = 2 Question 1 Find the number of zeros at the end of 33 × 128 × 125 × 50 × 45