Divisibility Rules

A divisibility rule is a shorthand way of determining whether a given number is divisible by a fixed divisor without performing the division, usually by examining its digits. Although there are divisibility tests for numbers in any radix, and they are all different, this article presents rules and examples only for decimal numbers.

Divisibility rule for '2' : First, take any number (for this example it will be 376) and note the last digit in the number, discarding the other digits. Then take that digit (6) while ignoring the rest of the number and determine if it is divisible by 2. If it is divisible by 2, then the original number is divisible by 2.
Example:

1. 376 (The original number)
2. 37 6 (Take the last digit)
3. 6 ÷ 2 = 3 (Check to see if the last digit is divisible by 2)
4. 376 ÷ 2 = 188 (If the last digit is divisible by 2, then the whole number is divisible by 2)

( OR ) We can easily conclude a number divisible by '2' or not. By seeing the last digit of number if it  is even number (0,2,4,6,8). Then, the number is divisible by '2'.

Divisibility rule for '3' or '9 ': First, take any number (for this example it will be 492) and add together each digit in the number (4 + 9 + 2 = 15). Then take that sum (15) and determine if it is divisible by 3. The original number is divisible by 3 (or 9) if and only if the sum of its digits is divisible by 3
 (or 9). If a number is a multiplication of 3 consecutive numbers then that number is always divisible by 3. This is useful for when the number takes the form of (n × (n − 1) × (n + 1)).
Example:

1. 492 (The original number)
2. 4 + 9 + 2 = 15 (Add each individual digit together)
3. 15 is divisible by 3 at which point we can stop. Alternatively we can continue using the same method if the number is still too large:
4. 1 + 5 = 6 (Add each individual digit together)
5. 6 ÷ 3 = 2 (Check to see if the number received is divisible by 3)
6. 492 ÷ 3 = 164 (If the number obtained by using the rule is divisible by 3, then the whole number is divisible by 3)

Example:

1. 336 (The original number)
2. 6 × 7 × 8 = 336
3. 336 ÷ 3 = 112

Divisibility rule for '4' :The basic rule for divisibility by 4 is that if the number formed by the last two digits in a number is divisible by 4, the original number is divisible by 4^. this is because 100 is divisible by 4 and so adding hundreds, thousands, etc. is simply adding another number that is divisible by 4. If any number ends in a two digit number that you know is divisible by 4 (e.g. 24, 04, 08, etc.), then the whole number will be divisible by 4 regardless of what is before the last two digits.
Alternatively, one can simply divide the number by 2, and then check the result to find if it is divisible by 2. If it is, the original number is divisible by 4. In addition, the result of this test is the same as the original number divided by 4.
General rule and example

1. 2092 (The original number)
2. 20 92 (Take the last two digits of the number, discarding any other digits)
3. 92 ÷ 4 = 23 (Check to see if the number is divisible by 4)
4. 2092 ÷ 4 = 523 (If the number that is obtained is divisible by 4, then the original number is divisible by 4)

Example:

1. 1720 (The original number)
2. 1720 ÷ 2 = 860 (Divide the original number by 2)
3. 860 ÷ 2 = 430 (Check to see if the result is divisible by 2)
4. 1720 ÷ 4 = 430 (If the result is divisible by 2, then the original number is divisible by 4)

Divisibility rule for '5' : Divisibility by 5 is easily determined by checking the last digit in the number (475), and seeing if it is either 0 or 5. If the last number is either 0 or 5, the entire number is divisible by 5. If the last digit in the number is 0, then the result will be the remaining digits multiplied by 2. For example, the number 40 ends in a zero (0), so take the remaining digits (4) and multiply that by two (4 × 2 = 8). The result is the same as the result of 40 divided by 5(40/5 = 8).
If the last digit in the number is 5, then the result will be the remaining digits multiplied by two (2), plus one (1). For example, the number 125 ends in a 5, so take the remaining digits (12), multiply them by two (12 × 2 = 24), then add one (24 + 1 = 25). The result is the same as the result of 125 divided by 5 (125/5=25).
If the last digit is 0

1. 110 (The original number)
2. 11 0 (Take the last digit of the number, and check if it is 0 or 5)
3. 11 0 (If it is 0, take the remaining digits, discarding the last)
4. 11 × 2 = 22 (Multiply the result by 2)
5. 110 ÷ 5 = 22 (The result is the same as the original number divided by 5)

If the last digit is 5

1. 85 (The original number)
2. 8 5 (Take the last digit of the number, and check if it is 0 or 5)
3. 8 5 (If it is 5, take the remaining digits, discarding the last)
4. 8 × 2 = 16 (Multiply the result by 2)
5. 16 + 1 = 17 (Add 1 to the result)
6. 85 ÷ 5 = 17 (The result is the same as the original number divided by 5)

Divisibility rule for '6' : A number can be divisible by 6. If the number is divisible by '2' and '3'.
Example: 1,458: 1 + 4 + 5 + 8 = 18, so it is divisible by 3 and the last digit is even, hence the number is divisible by 6.

Divisibility rule for '7' : If you double the last digit and subtract it from the rest of the number and the answer is: 0, or divisible by 7.
Example:  i.672 (Double 2 is 4, 67-4=63, and 63÷7=9) Yes
                ii. 905 (Double 5 is 10, 90-10=80, and 80÷7=11 ) No
                                  OR
 Multiply each digit (from right to left) by the digit in the corresponding position in this pattern (from left to right): 1, 3, 2, -1, -3, -2 (repeating for digits beyond the hundred-thousands place). Then sum the results.
Example: 483595 - (4 × (-2)) + (8 × (-3)) + (3 × (-1)) + (5 × 2) + (9 × 3) + (5 × 1) = 7.
Divisibility rule for '8' : The last three digits are divisible by 8.
Example  i. 109816 (816÷8=102) Yes
               ii. 216302 (302÷8=37 ) No

Divisibility rule for '10' : The given number is divisible by '10' or not can easily predict by seeing last digit of the number. If it is '0' then, the number is divisible.

Divisibility rule for '11' : If you sum every second digit and then subtract all other digits and the answer is 0, or divisible by 11
Ex: i. 1364 ((3+4) -(1+6) = 0) Yes
     ii. 3729 ((7+9) -(3+2) = 11) Yes
     iii.25176 ((5+7) -(2+1+6) = 3) No

Divisibility rule for '13' : Multiply each digit (from right to left) by the digit in the corresponding position in this pattern (from left to right): -3, -4, -1, 3, 4, 1 (repeating for digits beyond the hundred-thousands place). Then sum the results.
Example : 30,747,912: (2 × (-3)) + (1 × (-4)) + (9 × (-1)) + (7 × 3) + (4 × 4) + (7 × 1) + (0 × (-3)) + (3 × (-4)) = 13.

Calendar

Today, I am going to explain about Zeller's Rule. Zeller’s Rule will help us find the day of the week for any date quickly.With this technique named after its founder Zeller, you can solve any ‘Dates and Calendars’ problems. Calculate the day of the week for any date. Zeller’s Rule can be used to find the day on any particular date in the calendar in the history. All you have to know is the formula given below and how to use it.
Zeller’s Rule Formula:
F = K + [(13xM - 1)/5] + D + [D/4] + [C/4] – 2C
where, K = Date, M = Month, C = The first two digits year and D = Last two digits of the year
* In Zellers rule, months start from March. March = 1, April = 2, May = 3 and so on… till Dec = 10,
Jan = 11 Feb. = 12
* Also remember that when you have to find day of the first or second month of any year, then Year=Given year-1 i.e., When you want to find Day of 15-2-1990., K=15, Month=12, D=Given Year-1=1990-1=1989=89
Ex: find the day of the 27-08-2014 ?
Sol: K=27,M=6,C=20 and D=14
Replacing the values in the formula, we get F = 27 + [{(13 x 6)- 1}/5] + 14 + 14/4 + 20/4 – (2 x 20)
Therefore, F = 27 + 77/5 + 14 + 14/4 + 20/4 –40
Which gives.. F =27 + 15.5 + 14 + 3.5 + 5 – 40
[ We have to Consider only the integral value and ignore the value after decimal. So, the equation changes a bit as shown below. We have just removed value after decimal ]
F = 27 + 15 + 14 + 3 + 5 – 40 Therefore, F = 3. Now that you have a numerical value for the day, divide the number by 7. We need the remainder only. For example, in this case, the remainder is 3.
Now, match the remainder with the chart below:
1 = Monday 2 = Tuesday 3 = Wednesday 4 = Thursday 5 = Friday 6 = Saturday 7 = Sunday
Here, 3 represents Wednesday .
So by Zeller’s rule, 27th of August, 2014 was on a Wednesday . So,Today is wednesday.
This formula will help you a lot in any Calendar question that you may encounter in Quant or DI. Remember that it is necessary to know the formula properly or else, even a little mistake can render the answer incorrect.
PS: It is natural for all of us to consider January as the first month of the year. However, we request you to please note that March should be treated as first month for using this formula.
With this, you are all set to rock any calendar question that comes in front of you.
The most important thing in this concept is to remember the formula as an incorrect interpretation can lead to a false answer.
Best of luck.

Above zeller rule is applicable to find exact day of given date. If you to find the day of the given date in a problem where a day is given on a specified date. then we have to follow the given below procedure:
MODULE OBJECTIVE:
The module calendar is used to find many problems related to odd days, leap year, and counting of odd days and many. The module clock is used to find many problems related to find angle between hour and minute hand of a clock, at what time the hands of clock will be together and many.
We are supposed to find the day of the week on a given date. Odd Days:For this, we use the concept of 'odd days'. In a given period, the number of days more than the complete weeks are called odd days.
Leap Year:
(i). Every year divisible by 4 is a leap year, if it is not a century.
(ii). Every 4th century is a leap year and no other century is a leap year.
Note: A leap year has 366 days.
Examples:
i. Each of the years 1948, 2004, 1676 etc. is a leap year.
ii. Each of the years 400, 800, 1200, 1600, 2000 etc. is a leap year.
iii. None of the years 2001, 2002, 2003, 2005, 1800, 2100 is a leap year.
Ordinary Year: The year which is not a leap year is called an ordinary years. An ordinary year has 365 days.

Counting of Odd Days:
1.1 ordinary year = 365 days = (52 weeks + 1 day.)1 ordinary year has 1 odd day.
2.1 leap year = 366 days = (52 weeks + 2 days)
    so,1 leap year has 2 odd days.
3.100 years = 76 ordinary years + 24 leap years= (76 x 1 + 24 x 2) odd days = 124 odd days.
= (17 weeks + days)5 odd days.
Number of odd days in 100 years = 5.
Number of odd days in 200 years = (5 x 2),3 odd days.
Number of odd days in 300 years = (5 x 3),1 odd day.
Number of odd days in 400 years = (5 x 4 + 1) 0 odd day.
Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc. has 0 odd days.

Some codes to remember the months and weeks of number of odd days :
a) Week
Day             -       Code
Sunday       –         1
Monday      –         2
Tuesday      –         3
Wednesday –         4
Thursday    –         5
Friday         –         6
Saturday     –         0

b) Month
jan     –1
july    –0
feb     –4
Aug   –3
Mar   –4
Sep    –6
Apr    –0
Oct     –1
May   –2
Nov    –4
june    –5
Dec    –6

Example: It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
A. Sunday                    B. Saturday                  C. Friday                    D. Wednesday
Answer: Option 'C'
Solution: On 31st December, 2005 it was Saturday. Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days. On 31st December 2009, it was Thursday.Thus, on 1st Jan, 2010 it is Friday.

Example: Today is Monday. After 61 days, it will be ?
A. Wednesday          B. Saturday               C. Tuesday               D. Thursday
Answer: Option 'B'
Solution: Each day of the week is repeated after 7 days. So, after 63 days, it will be Monday.After 61 days, it will be Saturday.

Example: On what dates of April, 2001 did Wednesday fall ?
A. 1st, 8th, 15th, 22nd, 29th
B. 2nd, 9th, 16th, 23rd, 30th
C. 3rd, 10th, 17th, 24th
D. 4th, 11th, 18th, 25th
Answer: Option 'D'
Solution: We shall find the day on 1st April, 2001.1st April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001). Odd days in 1600 years = 0. Odd days in 400 years = 0
Jan.+ Feb.+ March+ April = (31 + 28 + 31 + 1)= 91 days - 0 odd days. Total number of odd days = (0 + 0 + 0) = 0. On 1st April, 2001 it was Sunday. In April, 2001 Wednesday falls on 4th, 11th, 18th and 25th. 

Number Series

What is Number Series ?
Number series is a arrangement of numbers in a certain order,where some numbers are wrongly put into the series of numbers and some number is missing in that series,we need to observe and find the accurate number to the series of numbers.
Anything we learn in our school days was basics and that is well enough for passing our school exams. Now the time has come to learn for our competitive exams. For this we need our basics but also we have to learn something new. That’s where shortcut tricks are comes into action.
In competitive exams number series are given and where you need to find missing numbers. The number series are come in different types. At first you have to decided what type of series are given in papers then according with this you have to use shortcut tricks as fast as you can.
Different types of Number Series There are some format of series which are given in Exams.
Perfect Square Series:
This Types of Series are based on square of a number which is in same order and one square number is missing in that given series.
Ex: 441, 484, 529, 576, ?
Sol:Here, 21*21=441
                22*22=484....
So,missing number is 25*25=625
Perfect Cube Series:
This Types of Series are based on cube of a number which is in same order and one cube number is missing in that given series
Ex: 1331, 1728, 2197, ?
Sol:Here, the answer is 13^3=
Prime Series :
In which the terms are the prime numbers in Order
Ex : 2, 3, 5, 7, 11, 13, _ , 19
Sol: Here,missing term is 17.
Geometric Series:
This type of series are based on ascending or descending order of numbers and each successive number is obtain by multiplying or dividing the previous number with a fixed number.
Ex: 5, 45, 405, 3645, ?
Sol:Here,5*9=45,
             45*9=405,
           405*9=3645
         3645*9=32805
  missing term is 32805
Two stage Type Series:
A two stage Arithmetic series is one in which the differences of successive numbers themselves form an arithmetic series.
Ex: i. 3, 9, 18, 35, 58,——
     ii. 6, 9, 17, 23,——
Mixed Series:
This type of series are more then one different order are given in a series which arranged in alternatively in a single series or created according to any non-conventional rule.
Wrong Number Series:
Sometimes, a question is asked to find out the wrong number in a series. This is a bit difficult task. This is because from the point where you get a wrong number, all other succeeding numbers would also be look wrong. You can only solve such a question comfortably only if you have expertise in handling the questions on series very well.
Ex: Find the wrong number in the following series.
        10 21 43 85 175
Sol: Here, the logic involved is  (previous number*2) + 1 only.But the error lies where instead of
(43*2) + 1 = 87, 85 has been given. So 85 is the wrong number.
Twin Series:
This is a newly included type of series in which two series are given, and below that the starting point of another series is given. Now you are supposed to analyze the logic from the first series and apply the same logic to form the second series.
Ex: 5 8 12 17 23 30
       9 a  b   c   d  e
Which number should come in place of c?
Sol:In this question, it canbe analyzed that the difference between the numbers is 3, 4, 5, 6 and 7.
So youhave to apply the same logic and start the series with 9.
The first numberafter 9 would be
9 + 3 = 12 (this would replace letter a),
the second numberwould be
12 + 4 = 16 (this would replace letter b)
and the third number wouldbe 16 + 5 = 21 (this would replace the letter c).
Some other common types of number series involve the following types of logic
Type-1:
Alternate Primes :
Ex: 2, 5, 11, 17, 23, _, 41
Type-2:
Every Third number can be the sum of the preceding two numbers:
Ex : 3, 5, 8, 13, ?
Sol:Here, 3+5=8,
                5+8=13
The next term is 13+8=21
Type-3:
Every Third number can be the product of the preceeding two numbers
Ex : 1, 2, 2, 4, 8, 32. _
Sol:Here,1*2=2,
               2*2=4,
               2*4=8,
               4*8=32,
Next term is   32*8=256
Type-4:
The difference of any term from its succeding term is constant (either increasing series or decreasing series )
Ex : 4, 7, 10, 13, 16, 19, _, 25
Sol:Here, 7-4=3,
              10-7=3,
            13-10=3.... Similarly
Missing term is    19+3=22
Type-5:
The difference between two consecutive terms will be either increasing or decreasing by a constant number :
Ex : 2, 10, 26, 50, 82, _
Sol: Here, 10-2=8,   (8*1=8)
               26-10=16, (8*2=16)
               50-26=24, (8*3=24)
               82-50=32, (8*4=32)
Next number is 82+(8*5=40)=122
Type-6:
The difference between two numbers is power of  constant number  where power will increases continuously
Ex : 15, 16, 19, 28,
Sol:Here,16-15=1, (3^0)
               19-16=3, (3^1)
               28-19=9, (3^2)
Next term is 28+(3^3)=55
Type-7:
The difference can be multiplied by numbers which will be increasing by a constant number :
Ex : 2, 3, 5, 11, 35, _
Sol: The difference between two numbers are
3 - 2 = 1
5 - 3 = 2
11 - 5 = 6
35 -11 = 24
Here, the differences are multiplied by numbers which are in increasing order .
Differences are
1 x 2 = 2
2 x 3 = 6
6 x 4 = 2 4
So, the next difference wil be
24 x 5 = 120.
So , the answer is 35 + 120 = 155 .
Type-8:
Every succeeding term is got by multiplying the previous term by a constant number or numbers which follow a special pattern.
Ex : 5, 15, 45, 135, _
Sol: Here, 5 x 3 = 15
                15 x 3 = 45
                 45 x 3 = 135
So, the answer is 135 x 3 = 405 .
Type-9:
In certain series the terms are formed by various rule (miscellaneous rules). By keen observation you have to find out the rule and the appropriate answer.
Ex : 4, 11, 31, 90, _
Sol:Terms are,
    4 x 3 -1 = 1 1
   11 x 3 -2 = 3 1
   31 x 3 -3 = 9 0
So, the answer is (90 x 3) -4 = 2 6 6
Ex : 3, 7, 23, 95, _
Sol:Terms are,
        (3 x 2) + 1 = 7
        (7 x 3) + 2 = 23
      (23 x 4),+ 3 = 95
   So, the answer will be
       (95 x 5)+ 4 = 479

Aptitude Questions

Some Questions to practice :
1. Adam can do a job in 15 days, Eve can do the same job in 20 days. If they work together for 4 days on this job. What fraction of job is incomplete ?
a. 1/4.   b. 7/15    c.1/10     d. 8/15
2. 15 men can do a piece of work in 24 days.how many days will required to complete the work by 10 men ?
a. 35      b. 36       c. 37       d. 30
3. 25 men can do a piece of work 8 hours per day in 24 days.how many days will required to complete the work by 16 men, are working 6 hours per day ?
a. 55      b. 50       c. 75       d. 100
4. A can do a piece of work in 75 days & B can do a piece of work in 50 days. how many days will required to complete the work by A and B ?
a. 30      b. 25       c. 125       d. 100
5. A can do a piece of work in 30 days & A&B can do a piece of work in 50 days. how many days will required to complete the work by  B alone ?
a. 30      b. 25       c. 75       d. 150
6. A can do a piece of work in 18 days, B can do a piece of work in 15 days & C can do a piece of work in 12 days. how many days will required to complete the work by A, B and C  ?
a. 540/37            b. 1080/37   
c. 270/37            d.  135/37
7. A can do a piece of work in 30 days, B can do a piece of work in 20 days & A,B &C combinely can do a piece of work in 60 days. how many days will required to complete the work by C alone ?
8. A and B can do a piece of work in 33 days, B and C can do a piece of work in 22 days, C and A can do a piece of work in 11 days. how many days will required to complete the work by A, B and C  ?
a.11         b. 6          c.12           d.60
9. Certain men complete a work in 10 days.if there are 5 men are less,it could be finished in 5 days more.then,number of men originally worked ?
a.10         b.15        c.20     d. 25
10. Certain men complete a work in 10 days.if there are 5 men are more,it could be finished in 5 days less.then,number of men originally worked ?
11. A can complete a work in 20 days.but, A worked for 5 days. Remaining work completed by B in 5 days.so, Both A and B can complete's the work in ?
12. A can complete a work in 20 days, B can complete a work in 15 days. but, A worked for 5 days. Remaining work completed by B alone in how days ?
a. 45/4       b. 11      c. 44       d
13. A can complete a work in 20 days.but, A worked for 5 days. A and B can do  in 25 days.so, B can complete's the work in ?
a. 3       b. 5       c. 15        d. 8
14. A can do a piece of work in 20 days & B can do a piece of work in 15 days. After 13 days B joined with A.how many days will required to complete the work  ?
a. 20      b. 16       c. 15       d. 18
15. A can do a piece of work in 50 days & B can do a piece of work in 25 days. Both are worked 10 days, A left the work.how many days will required to complete the remaining work B alone ?
a. 20      b. 30       c. 10       d. 15
16.1 men or 2 women or 3 boys can do a piece of work in 10 days.then, 1 men+1 women+1 boy can do same piece of work in how many days ?
a. 30/11  b. 60/11  c. 120/11  d.60
17. 5 men or 10 women can do a piece of work in 18 days.6 men and 6 women can complete the work in how many days ?
a. 10     b. 20      c. 30     d. 40
18. The product of two numbers is 125 and hc.f is 5.what is the LCM of two numbers ?
19. If 4 digit smallest number divisible by 9,15,21. What is that number ?
20. If a 3 digit smaller number divisible by 4,12.and 18 and leaves remainder 3 in each case.what is that
number ?
21. If 5 digit biggest number divisible by 4,15 and 21.what is the number ?
22. Find the cars each originally had. A man bought a horse and a cart. If he sold the horse at 10 % loss and the cart at 20 % gain, he would not lose anything; but if he sold the horse at 5% loss and the cart at 5% gain, he would lose Rs. 10 in the bargain. The amount paid by him was Rs._______ for the horse and Rs.________ for the cart.
23. It was calculated that 75 men could complete a piece of work in 20 days. When work was scheduled to commence, it was found necessary to send 25 men to another project. How much longer will it take to complete the work?
24. A dishonest shopkeeper professes to sell pulses at the cost price, but he uses a false weight of 950gm. for a kg. His gain is …%.
25. A software engineer has the capability of thinking 100 lines of code in five minutes and can type 100 lines of code in 10 minutes. He takes a break for five minutes after every ten minutes. How many lines of codes will he complete typing after an hour?
26. A man was engaged on a job for 30 days on the condition that he would get a wage of Rs. 10 for the day he works, but he have to pay a fine of Rs. 2 for each day of his absence. If he gets Rs. 216 at the end, he was absent for work for ... days.
27. A contractor agreeing to finish a work in 150 days, employed 75 men each working 8 hours daily. After 90 days, only 2/7 of the work was completed. Increasing the number of men by ________ each working now for 10 hours daily, the work can be completed in time.
28. 40. The average speed of a train is 1 3/4 times the average speed of a car. The car covers a distance of 588 km in 6 hours. How much distance will the train cover in 13 hours? 

HCF and LCM

Here, some tricks to solve HCF and LCM of related problems....
Concept :
* Highest Common Factor (HCF)
If two or more number are broken into their prime factors (as explained in 2.3), then the product of the maximum common prime factors in the given numbers is the H.C.F. of the numbers. In other words, the HCF of two or more numbers is the greatest number (divisor) that divides all the given numbers exactly.
So, HCF is also called the Greatest Common Divisor (GCD).
Ex : Find the HCF of 72, 60, 96. Here, we first find the prime factors of each given number ?
Sol: Here 72 = 2 × 2 × 2 × 3 × 3
         60 = 2 × 2 × 3 × 5
         96 = 2 × 2 × 2 × 2 × 2 × 3
and so HCF = product of maximum common prime factors
                    = 2 × 2 × 3 = 12
Note : The common factors in the given numbers have been encircled.
* LCM (Lowest Common Multiple)
The LCM of two or more than two numbers is the product of the highest powers of all the prime factors that occur in these numbers.
Ex : Find the LCM of 36, 48, 64 and 72 ?
2| 36, 48, 64, 72
2| 18, 24, 32, 36
2| 9, 12, 16, 18
2| 9, 6, 8, 9
3| 9, 3, 4, 9
3| 3, 1, 4, 3
  | 1, 1, 4, 1
∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 4 = 576.
Some shortcuts :
Trick-1:
Product of Two Numbers
∴ HCF of numbers × LCM of numbers = Product of numbers
Ex: If the product of two numbers is 120 and LCM of two numbers is 40.find the HCF ?
Sol: 120 = 40 * HCF
        HCF = 3
Trick-2:
Find the GREATEST NUMBER that will exactly divide x, y and z. Required number = HCF of x, y & z
Ex: the greatest number that exactly divide 120,90 and 40 ?
Sol: = HCF of 120,90 and 40
       ∴ 10 is greatest number divides exactly 120,90 and 40
Trick-3:
Find the GREATEST NUMBER that will divide x, y and z leaving remainders a, b and c respectively. Required nimber  = HCF of (x - a), (y - b) and (z -c).
Ex: the greatest number that exactly divide 125,94 and 43 leaves remainders 5,4 and 3 respectively ?
Sol:
   = HCF of (125-5),(94-4) and (43-3)
       ∴ 10 is greatest number divides exactly 120,90 and 40
Trick-4:
Find the LEAST NUMBER which is exactly divisible by x, y and z. Required number = LCM of x, y and z (least dividend)
Ex: Least number which when divided by 35,45,55  ?
Sol: = LCM of ( 35,45,55)
       = 3465
Trick-5:
Find the LEAST NUMBER which when dividend by x, y and z leaves the remainders a, b and c respectively.
     Then, it is always observed that (x - a) = (y -b) = (z - c) = k (say)
∴ Required number = (LCM of x, y and z) - (K)
Ex: Least number which when divided by 35,45,55 and leaves remainder 18,28,38 ?
Sol: Here the difference between every divisor and remainder is same i.e. 17.
Therefore, required number
  =   LCM  of (35,45,55)-17
  =  (3465-17)= 3448.
Trick-6:
Find the LEAST NUMBER which when dividend by x, y and z leaves the same remainder 'r' each case. Required number = (LCM of x, y and z) + r.
Ex: Least number which when divided by 5,6,7,8 and leaves remainder 3 ?
Sol: = LCM ( 5,6,7,8 ) + 3
       = 843
Trick-7:
Find the GREATEST NUMBER that will divided x, y and z leaving the same remainder in each case. Required number = HCF of (x - y), (y -z) and (z -x)
Trick-7:
Find the GREATEST NUMBER that will divided x, y and z leaving the same remainder R in each case. Required number = HCF of (x - R), (y -R) and (z -R)
Trick-8:
Find the n-digit GREATEST NUMBER which when divided by x, y and z.
leaves no remainder(i.e exactly divisible)
Remainder 'R' =[ n-digit greatest number/ LCM of (x,y and z)
∴ Required number = (n-digit greatest number) - R
Ex: Greater number of 4 digits which is divisible by each one of 12,18,21 and 28 is?
Sol: LCM of 12,18,21,28 = 252 Therefore, required number must be divisible by 252.
Greatest four digit number = 9999 On dividing 9999 by 252 remainder = 171
Therefore, 9999-171 = 9828.
Trick-9:
Find the n-digit GREATEST NUMBER which when divided by x, y and z.
leaves remainder K  in each case
Remainder 'R' =[ n-digit greatest number/ LCM of (x,y and z)
∴ Required number = [ (n-digit greatest number) - R ] + K
Ex: Greater number of 4 digits which is divisible by each one of 12,18,21 and 28.leaves 3 as remainder in each case ?
Sol: LCM of 12,18,21,28 = 252 Therefore, required number must be divisible by 252.
Greatest four digit number = 9999 On dividing 9999 by 252, remainder = 171
Therefore, 9999-171 + 3= 9831.
Trick-10:
Find the n-digit SMALLEST NUMBER which when divided by x, y and z.leaves no remainder(i.e exactly divisible)
Remainder 'R' =[ n-digit smallest number/ LCM of (x,y and z)
∴ Required number = (n-digit smallest number) + [L - R ]
Ex: Smallest number of 4 digits which is divisible by each one of 12,18,21 and 28 is?
Sol: LCM of 12,18,21,28 = 252 Therefore, required number must be divisible by 252.
Smallest four digit number = 1000 On dividing 1000 by 252  remainder = 244
Therefore, 1000 + (252-244) = 1008.
Trick-11:
Find the n-digit SMALLEST NUMBER which when divided by x, y and z.
leaves remainder K  in each case
Remainder 'R' =[ n-digit greatest number/ LCM of (x,y and z)
∴ Required number = [ (n-digit greatest number) +(L - R) ] + K
Ex: Smallest number of 4 digits which is divisible by each one of 12,18,21 and 28.leaves 3 as remainder in each case ?
Sol: LCM of 12,18,21,28 = 252 Therefore, required number must be divisible by 252.
Smallest four digit number = 1000 On dividing 1000 by 252, remainder = 171
Therefore,1000+(252-244)+3= 1011.
Trick-12:
Find the HCF of fractions
= HCF of numerators / LCF of denominators
Trick-13:
Find the LCM of fractions
= LCM of numerators / HCM of denominators
Trick-14:
Find the HCF of decimal numbers ?
Step-1: Find the HCF of the given numbers without decimal.
Step-2: Put the decimal point (in the HCF of step 1) from right to left according to the MAXIMUM decimal places among the given numbers.
Trick-15:
Find the LCM of decimal numbers ?
Step-1: Find the LCM of the given numbers without decimal.
Step-2: Put the decimal point (in the LCM of step 1) from right to left at the place equal to the MINIMUM decimal places among the given numbers

Clock

Hey guys i am back with another interesting Post.This post is about clock problems which we normally asked in aptitude exams.Aptitude exams play a major role in any interviews.Without clearing Aptitude you cannot even into any interview session.
we need to get couple of basic facts clear:
* Speed of the hour hand = 0.5 degrees per minute (dpm) {The hour hand completes a full circle or 360 degrees in 12 hours or 720 minutes}
* Speed of the minute hand = 6 dpm {The minute hand completes a full circle in 60 minutes}
* At ‘n’ o’ clock, the angle of the hour hand from the vertical is 30n
Clock problems can be broadly classified in two categories:
a) Problems on angles
b) Problems on incorrect clocks
Problems on angles
Finding the angle between the hands category clock problems  major questions which is quite time taking and difficult to solve.
Here, Some easy tricks to solve
Trick-1:
The time is X h : Y min.so, the angle between hour and minute hand is given by
Angle between X and Y
=|(X*30)-((Y*11)/2)| or
  30 * [HRS - (MIN/5)] + (MIN/2)
Ex : Angle between hands at 5:30 ?
a. 15      b. 20       c. 25        d. 30
Sol : Angle between 5 and 30 =|(5*30)-((30*11)/2)|
=|150-165|
= 15 degrees
Thus, angle between hands at 5:30 is 15 degrees.
Trick-2:
Right angles at each other by hands of clock between one hour is given by
  = | (5x± 15) * (12/11) |
Ex : At what time do the hands of a clock between 7:00 and 8:00 form 90 degrees?
Sol: there is two possibles
Case-1:
    = (5*7 - 15) * (12/11)
    = 240/11
    = 21 minutes 9/11 of a minute
Case-2: 
    = (5*7 + 15) * (12/11)
    = 600/11
    = 54 minutes 6/11 of a minute
So, the hands of the clock are at 90 degrees at the following timings:
    7 : 21 : 9/11 and 7 : 54 : 6/11
Trick-3:
For coinciding the hands between one hour is
           =  (5x) * (12/11)
Ex: At what time do the hands of the clock meet between 7:00 and 8:00 ?
Sol: = (60*7)/11
       = 420/11
       = 38 minutes 2/11 minute
Hands of the clock meet at
     7 : 38 : 2/11
Opposite Direction , (5x - 30) * (12/11)
Trick-4:
Opposite to each other by hands of clock between one hour is given by
  = | (5x - 30) * (12/11) |
Ex : At what time do the hands of a clock between 7:00 and 8:00 form 180 degrees?
Sol:
    = | (5*7 - 30) * (12/11) |
    = 60/11
    = 5 minutes 5/11 of a minute
* Some other results which might be useful:
· Hands of a clock meet at a gap of 65 5/11 minutes.
      The meetings take place at 12:00:00, 1:05:5/11, 2:10:10/11 … and so on.
. A clock makes two right angles between 2 hours.The clock does not makes 48 right angles in 24 hours( 1 day)
· Hands of a clock meet 11 times in 12 hours and 22 times in a day.
· Hands of a clock are perfectly opposite to each other (i.e. 180 degrees) 11 times in 12 hours and 22 times a day. {Same as above}
· Any other angle is made 22 times in 12 hours and 44 times in a day
Number of right angles
Note:  Between 2 -4 and 8-10 there are 3 right angles and not 4.The first right angle between 3 - 4 (9-10) and second right angle bw:
2-3(8-9) are the same.
Problems on incorrect clocks
Such sort of problems arise when a clock runs faster or slower than expected pace. When solving these problems it is best to keep track of the correct clock.
Type-1: 
Ex: A watch gains 5 seconds in 3 minutes and was set right at 8 AM. What time will it show at 10 PM on the same day?
Ans: The watch gains 5 seconds in 3 minutes => 100 seconds in 1 hour.
From 8 AM to 10 PM on the same day, time passed is 14 hours.
In 14 hours, the watch would have gained 1400 seconds or 23 minutes 20 seconds.
So, when the correct time is 10 PM, the watch would show 10 : 23 : 20 PM
Type-2:
Ex: A watch gains 5 seconds in 3 minutes and was set right at 8 AM. If it shows 5:15 in the afternoon on the same day, what is the correct time?
Sol: The watch gains 5 seconds in 3 minutes => 1 minute in 36 minutes
From 8 AM to 5:15, the incorrect watch has moved 9 hours and 15 minutes = 555 minutes.
When the incorrect watch moves for 37 minutes, correct watch moves for 36 minutes.
=> When the incorrect watch moves for 1 minute, correct watch moves for 36/37 minutes
=> When the incorrect watch moves for 555 minutes, correct watch moves for (36/37)*555 = 36*15 minutes = 9 hours
=> 9 hours from 8 AM is 5 PM.
=> The correct time is 5 PM.
Note: I am sure you would have heard the proverb that even a broken clock is right twice a day. However, a clock which gains or loses a few minutes might not be right twice a day or even once a day. It would be right when it had gained / lost exactly 12 hours.
Type-3:
Ex: A watch loses 5 minutes every hour and was set right at 8 AM on a Monday. When will it show the correct time again?
Ans: For the watch to show the correct time again, it should lose 12 hours.
It loses 5 minutes in 1 hour
=> It loses 1 minute in 12 minutes
=> It will lose 12 hours (or 720 minutes) in 720*12 minutes = 144 hours = 6 days
=> It will show the correct time again at 8 AM on Sunday.

Compound Interest

In the first article on interests, we learned about Simple Interest. Well, simple interest is not the only interest mechanism used by lenders and borrowers. There is another method, one which actually runs majority of business operations and goes by the name of Compound Interest. The basic concept operating behind compound interest is very simple. Let us take a small example to understand its rudiments.
For example, Sham borrows a sum of Rs. 100 from Ghansham for a period of two years. The rate of interest for this loan is 10%. At the end of year one, the amount due is the principal and 10% interest on it, that is a total of Rs. 110. Now, effectively the principal value of the loan for the second year is no longer Rs. 100, it is in fact Rs. 110. That is what Ghansham would say and believe. According to him, for the 2^nd year, he has lent Rs. 110 as that was the amount he would have had if he taken back the money at the end of year 1. Now for the 2^nd year, the interest becomes Rs. 11 (10% of Rs. 110) and the total amount Ghansham would get would be Rs. 121.
If the same calculation was done using the logic of simple interest, you would see that the interest due for two years would be Rs. 20 (10% of Rs. 100 for two years). Thus, replace a S with a C and there is such a big difference in the calculations carried out.
Effectively, for compound interest, the 2^nd term of interest is actually the sum total of the principal and the interest for the first term.
Compound Interest Tool tip 1: The Definitions
Principal (P): The original sum of money loaned/deposited. Also known as capital.
Interest (I): The amount of money that you pay to borrow money or the amount of money that you earn on a deposit.
Time (T): The duration for which the money is borrowed. The duration does not necessarily have to be years. The duration can be semi-annual, quarterly or any which way deemed fit.
Rate of Interest (R):  The percent of interest that you pay for money borrowed, or earn for money deposited
Compound Interest Tool tip 2: The Basic Formula
Amount Due at the end of the time period, A =  P (1+r/100)^t
Where:
P: Principal (original amount)
R: Rate of Interest (in %)
T: Time period (yearly, half-yearly etc.)
Compound Interest (CI) =  A- P = P (1+r/100)^t -P
= P {(1+r/100)^t -1}
Compound Interest Tool tip 3: Basic Problems to explain the concept
Example 1: Maninder took a loan of Rs. 10000 from Prashant . If the rate of interest is 5% per annum compounded annually, find the amount received by Prashant by the end of three years
Solution:
The following is the data given:
Principal, P= 10000
Rate = 5%
Time =3 years
Using the formula for Compound Interest:
A = P(1+R/100)^t
So A= 10000(1+5/100)³
A = 10000(1+1/20)³
A = 10000 x 21/20 x 21/20 x21/20 =11576.25
So the total amount paid by Maninder at the end of third year is Rs.11576.25
Example 2: Richa gave Rs. 8100 to Bharat at a rate of 9% for 2 years compounded annually. Find the amount of money which she gained as a compound interest from Bharat at the end of second year.
Solution:
Principal value = 8100
Rate = 9%
Time = 2 years
So the total amount paid by Bharat
= 8100(1+9/100)²
=Rs.  9623.61
The question does not probe the amount, rather, it wants to know the CI paid, that the difference between the total amount and original principal.
The Compound Interest = 9623.61 – 8100 = 1523.61
Compound Interest Tooltip 4: Multiple Compounding in a year
Amount Due at the end of the time period

Where:
A = future value
P = principal amount (initial investment)
r = annual nominal interest rate
n = number of times the interest is compounded per year
t = number of years money borrowed
Amount for Half Yearly Compounding, A = P {1+(R/2)/ 100}^2T
(compound interest applied two times an year).
Like Half Yearly Compound Interest, we can calculate the amount for Quarterly Compounding:
A = P {1+(R/4)/ 100}^4T
Example 3: Sona deposited Rs. 4000 in a bank for 2 years at 5% rate. Find the amount received at the end of year by her from the bank when compounded half yearly.
Solution:
Principal value = Rs. 4000
Rate = 5%
Time = 2 years
Since the interest is compounded half yearly so 2 years = 4 times in two years
So we have = A = P {1+(R/2)/ 100}^2T
A= 4000{1+ (5/2)/100}⁴
A = 4000 x 41/40 x 41/40 x 41/40 x 41/40
A = Rs. 4415.2
So, Sona received Rs. 4415.2 from the bank after two years
Example 4: Manpreet lent Rs 5000 to Richa at 10% rate for 1 year. But she told her that she will take her money on compound interest. So find the amount of interest received by Manpreet when compounded quarterly?
Solution:
Principal value = Rs. 5000
Rate = 10%
Time = 1 year
Since the interest is compounded quarterly, that is 4 times in 1 year
Using the formula= A = P {1+(R/4)/ 100}^4T
A= 5000{1+ (10/4)/100}⁴
A = 5000 x 41/40 x 41/40 x 41/40 x 41/40
A = Rs. 5519.064
So, Manpreet received Rs. 4415.2 from bank after two years
And the total amount of interest received by her is 5519.064 – 5000 = Rs. 519.06
Compound Interest Tooltip 5: Difference between Simple Interest and Compound Interest
In case the same principle P is invested in two schemes, at the same rate of interest r and for the same time period t, then in that case:
Simple Interest = (P x R x T)/100
Compound Interest = P [(1+R/100)^T – 1]  So, the difference between them is
= PRT/100 – P[(1+R/100)^T -1] = P [(1+r/100)^T -1-RT/100]
Two shortcuts which we can use:
Difference between CI and SI when time given is 2 years = P(R/100)²
Difference between CI and SI when time given is 3 years = P[(R/100)³ + 3(R/100)²]
Example 5:
The difference between compound interest and simple interest is 2500 for two years at 2% rate, then find the original sum.
Solution:
Given Interest is = 2500
So, Simple Interest = (P X R X T)/100
Compound Interest = P [(1+r/100)^t – 1] So the difference between both of them is
= PRT/100 – P [(1+R/100)^T -1] = P [(1+r/100)^T -1-RT/100] So the sum is 2500 = P [{(1+2/100)²-1}-4/100] On simplification this equation the sum will be = Rs. 6250000
We can check it by our shortcut method
When time given is 2 years = P(R/100)²
Since we are given by the difference so
2500 = P (2/100)²
=> 2500 = P (1/50)²
=> 2500= P (1/2500)
=> 6250000=P
So the sum is Rs.6250000.
Similarly we can conclude the sum when time given is 3 years.

In the first article of our series, we covered the basics of Compound interest, including what does it stand-for. We gave a few tooltips and formulas that are important for the concept and help you grasp the topic better. In the second article of the series, we help you master the topic by providing a series of tooltips providing various formulae and tools that you can use to solve Compound Interest questions.
Compound Interest Tool tip 6: Formula for compound interest when compounding in a year but time is in fraction
The formula to calculate compound interest when the time given is in fractions is as follows:
A =P[(1+R/100)^{real part}{1+(Fraction part x R/100)}]
Where
A: Amount at the end of the time period
P: Principal amount
R: Rate of interest
Real Part and Fraction part: For example, the time given is two and half years, then real part would be two and half years.
Example 1:
Manpreet gave Rs. 1000 to Richa for 1 year 6 months. Then find the amount paid by Richa to Manpreet after this duration if rate of interest is 5% per annum compounded annually?
Solution :
We have a time of 1 years and 6 months = (1 year + 1/2 years)
So to find the amount we will use
A =P[(1+R/100)^{real part}{1+(Fraction part x R/100)}] A = 1000[(1+5/100)¹{1+{(6/12 x 5)/100] A= P[(21/20) x (41/40)] A = Rs. 1076.25
Compound Interest Tool tip 7: Concept of Equal Installments
How does the concept of equal installments work for Compound interests?
Well, in this case, the problem basically tells us that a certain sum of money is borrowed on compound interest for a certain period and it is returned with the help of equal installments. Lets us derive a formula for these installments.
Let us derive a formula where the amount is returned in two equal installments for a time period of two years.
Assume P to be the principal and r the rate of interest.
Step 1: P(1+r/100)=P1  (Amount for one year)
Step 2: New Principal
Now let X be the first installment. After giving the first installment, the principal value will change and the new principal will be = P1 – X
P2 = P1 –X (1+R/100)
Step 3: Amount and Interest for the second year
Now the interest charged will be charged on this amount.
Amount at the end of second year: [P(1+r/100)-X][1+r/100] Step 4: Since the installments are equal, this new amount has to equal X.
Hence,
[P(1+r/100)-X][1+r/100]=X
On solving, we have
P [(1+R/100)²-X (1+R/100)] = X
P [(1+R/100)²] = X+X (1+R/100)
Divide both sides by (1+r/100)²
So we left with
P= X/(1+r/100)² + X/(1+r/100)
Generalizing the formula for EQUAL INSTALLMENTS
P= X/(1+r/100)ⁿ + …………………….X/(1+r/100)
Where x is the installment and n is number of installment
Example 2:
Richa borrowed a sum of Rs. 4800 from Ankita as a loan . She promised Ankita that she will pay it back in two equal installments .If the rate of Interest be 5% per annum compounded annually, find the amount of each installment
Solution:
Given that principal value = 4800
Rate =5%
Two equal installments annually = 2 years
Appling the formula = X/(1+r/100)ⁿ…………………….X/(1+r/100)
So we have here two equal installments so
P= X/(1+r/100)² + X/(1+r/100)
4800=X/(1+5/100)² + X/(1+5/100)
On simplifying
We have x= Rs. 2581.46
So the amount of each installment is 2581.46
Compound Interest Tooltip 8: Application of Compound Interest for concepts of population growth.
Case 1: When population growing in a constant rate
If the rate growth of population increased with a constant, rate then the population after T years will be = P (1+R/100)^t
In fact, this is nothing else but an application of the fundamentals of compound interest.
It is actually similar to finding the compound amount after time T years
Net population after T years = = P (1+R/100)^t
Net population increase = P [(1+R/100)^t– 1]
Example 3:
The population of Chandigarh is increasing at a rate of 4% per annum. The population of Chandigarh is 450000, find the population in 3 years hence.
Solution:
P = 450000
Rate of increasing = 4%
Time =3 years
Therefore, the total population will be
=> T = P (1+R/100)³
=> T = 450000(1+4/100)³
=> T = 506188
Case 2: When the population growing with different rates and for different intervals of time
If the rate growth of population increased with different, rate and for different intervals of time then the population after T years will be =
P (1+R1/100)^t1 x  (1+R2/100)^t2……………………………..  (1+RN/100)^tn
Let us take an example for this concept.
Example 4:
The population of Chandigarh increased at a rate of 1% for first year, the rate for second year is 2%, and for third year, it is 3%. Then what will be the population after 3 years if present population of Chandigarh is 45000?
Solution:
Since the rate growth of population increased with different rates for the three difference years, the population after T years will be =
P (1+R1/100)^t1 x  (1+R2/100)^t x (1+R/100)^tn
45000(1+1/100)¹ x (1+2/100)¹ x (1+3/100)¹=47749.77
Case 3: When the population is decreasing with rate R
Population after a time period of T years=P (1-R/100)^t
Where T is the total population
R is rate at which the population is decreasing
Example 5:
The population of Chandigarh is increases at a rate of 1% for first year, it decreases at the rate of 2% for the second year and for third year it again increases at the rate of 3%. Then what will be the population after 3 years if present population of Chandigarh is 45000.
Solution:
Since the rate growth of population is increasing first and then decreasing for the second year and again it increases for third year, then the population after T years will be =
Present Population: P (1+1/100)^t1 x  (1-2/100)^t x (1+3/100)^t3
Present Population:  45000(1+1/100)¹ x (1-2/100)¹ x (1+3/100)¹=45877.23
Compound Interest Tooltip 8: Negative Compound Interest
As we can see from the last case above, it is not necessary that there is always an increase in any quantity or amount. There can also be a reduction in the amount of something. This reduction is actually called the rate of depreciation, especially in the financial world. In this case, we do nothing else but take the interest rate to be negative. The formula for this is as follows:
Let P be the original amount.
Let P1 be the new amount at the end of t years.
P1 = P (1-R/100)^t
Here R is the rate of interest (negative rate).
Always remember, the rate of depreciation is nothing else but negative rate of interest.
Example 6:
Manpreet bought a new car. The value of the car is Rs. 45000. If rate of depreciation is 10% per annum then what will be the value of the car after 2 years
Solution:
P = 45000
Rate of depreciation = 10%
T = 2 years
Therefore, the value will be after 2 years
= P (1 – R/100)^t
= 45000(1-10/100)²
= 36450

 Aplications
Case-i: City’s Population
i. Decline/Decrease
A city has 10,000 residents. Its population declines at the rate of 10% per annum, what’ll be its total population after 5 years?
Sol: After 5 years population
       = P*(1-R/100)^5
       = 10000*(9/10)^5
       =  5904
ii. Incline/Increases
A city has 10,000 residents. Its population increases at the rate of 10% per annum, what’ll be its total population after 5 years?
Sol: After 5 years population
       = P*(1+R/100)^5
       = 10000*(11/10)^5
       =  16105
Case-ii: Wine Replacement (Adulteration)
i.Remaining volume
A butler steals 10 ml of whiskey from 100 ml bottle and replaces it with water. He repeats this process 5 more times, how much % whisky is left in the bottle?
Sol: After 5 times remaining whisky
       = P*(1-R/100)^5
       = 100*(9/10)^5
       = 59.05 ml
ii. Adulteration→Finding Original volume
A tanker is full of milk, 25% of the liquid is stolen and replaced with water. If this process is repeated 4 times and ultimate mixture contains 810 litres of milk, what is the total capacity of this tanker?
Sol: Original volume
                   = P*(1-R/100)^5
            810 = O.V*(75/100)^4
                   = 2560 ml
Trick-1:
Pascal’s Triangle for computing Compound Interest (CI)
The traditional formula for computing CI is
CI = P*(1+R/100)^N –P
Using Pascal’s Triangle, Number of Years (N) -------------------
1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
..........................
Eg: P = 7000, R=5 %, and N=4 years. What is CI & Amount?
Sol: Time is 4 years and so we will focus on 4 th line of the Pascal triangle: i.e. 1 4 6 4 1
Amount = 7000 x [ ist pascal number on 4 th line i.e. "1"] + (7000 x 5% =350 x [ IInd number on 4 th line i.e. "4"] +
(350 x 5 %= 17.5 x [ III rd number on 4 th line i.e. "6"] +
(17.5 x 5% = 0.875 x
[ IV th number on 4 th line i.e. "4"] + (0.875 x 5% =0.044 x
[ V th number on 4 th line i.e. "1"] implying compounded amount is : (7000 x 1 + 350 x 4 + 17.5 x 6 + 0.875 x 4 + 0.044 x 1) = 8508.55 Using traditional formula we get (7000*(1+5%)^4) = 8508.55 Compound interest = 8508.55-7000 = 1508.55
* If compounded half yearly.
Then,make R half and T double.
* If compounded quarterly.
Then,make R one by fourth and T four times.