1.Sum of n natural numbers:
The sum of first n natural numbers = n (n+1)/2
The sum of squares of first n natural numbers is= n (n+1)(2n+1)/6
The sum of first n even numbers= n (n+1)
The sum of first n odd numbers= n^2
2.Finding Squares of numbers
To find the squares of numbers near numbers of which squares are known
To find 41^2 , Add 40+41 to 1600 =1681
To find 59^2 , Subtract 60^2-(60+59) =3481
3.The Divisibility Rules
| 2 | The last digit is even (0,2,4,6,8) | 128 is 129 is not |
| 3 | The sum of the digits is divisible by 3 |
381 (3+8+1=12, and 12÷3 = 4) Yes
217 (2+1+7=10, and 10÷3 = 3 1/3) No
|
| 4 | The last 2 digits are divisible by 4 |
1312 is (12÷4=3)
7019 is not |
| 5 | The last digit is 0 or 5 | 175 is 809 is not |
| 6 | The number is divisible by both 2 and 3 | 114 (it is even, and 1+1+4=6 and 6÷3 = 2)Yes 308 (it is even, but 3+0+8=11 and 11÷3 = 3 2/3) No |
| 7 | If you double the last digit and subtract it from the rest of the number and the answer is:
|
672 (Double 2 is 4, 67-4=63, and 63÷7=9)Yes
905 (Double 5 is 10, 90-10=80, and 80÷7=11 3/7) No
|
| 8 | The last three digits are divisible by 8 |
109816 (816÷8=102) Yes
216302 (302÷8=37 3/4) No
|
| 9 | The sum of the digits is divisible by 9 (Note: you can apply this rule to that answer again if you want) |
1629 (1+6+2+9=18, and again, 1+8=9) Yes
2013 (2+0+1+3=6) No
|
| 10 | The number ends in 0 |
220 is
221 is not |
| 11 |
If you sum every second digit and then subtract all other digits and the answer is:
|
1364 ((3+4) - (1+6) = 0) Yes
3729 ((7+9) - (3+2) = 11) Yes
25176 ((5+7) - (2+1+6) = 3) No
|
| 12 | The number is divisible by both 3 and 4 |
648
(By 3? 6+4+8=18 and 18÷3=6 Yes. By 4? 48÷4=12 Yes) Yes
524
(By 3? 5+2+4=11, 11÷3= 3 2/3 No. Don't need to check by 4.) No |
4.Simple Interest
S.I = PRT/100
Where
P = Amount
R= Rate of interest
T=Time(duration in months and years)
5.Compound Interest Shorcut
S.I = PRT/100
Where
P = Amount
R= Rate of interest
T=Time(duration in months and years)
5.Compound Interest Shorcut
A good shortcuts on Compound Interest.We know the traditional formula to compute interest...
CI = P*(1+R/100)^N - P
The calculation get very tedious when N>2 (more than 2 years). The method suggested below is elegant way to get CI/Amount after 'N' years. You need to recall the good ol' Pascal's Triange in following way:
Code:
Number of Years (N)
-------------------
1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
. 1 .... .... ... ... 1
Example: P = 1000, R=10 %, and N=3 years. What is CI & Amount?
Step 1: 10% of 1000 = 100, Again 10% of 100 = 10 and 10% of 10 = 1
We did this three times b'cos N=3.
Step 2:
Now Amount after 3 years = 1 * 1000 + 3 * 100 + 3 * 10 + 1 * 1 = Rs.1331/-
The coefficents - 1,3,3,1 are lifted from the pascal's triangle above.
Step 3:
CI after 3 years = 3*100 + 3*10 + 3*1 = Rs.331/- (leaving out first term in step
2)
If N =2, we would have had, Amt = 1 * 1000 + 2 * 100 + 1 * 10 = Rs. 1210/-
CI = 2 * 100 + 1* 10 = Rs. 210/-
This method is extendable for any 'N' and it avoids calculations involving higher
powers on 'N' altogether!
A variant to this short cut can be applied to find depreciating value of some
property. (Example, A property worth 100,000 depreciates by 10% every year, find
its value after 'N' years).
CI = P*(1+R/100)^N - P
The calculation get very tedious when N>2 (more than 2 years). The method suggested below is elegant way to get CI/Amount after 'N' years. You need to recall the good ol' Pascal's Triange in following way:
Code:
Number of Years (N)
-------------------
1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
. 1 .... .... ... ... 1
Example: P = 1000, R=10 %, and N=3 years. What is CI & Amount?
Step 1: 10% of 1000 = 100, Again 10% of 100 = 10 and 10% of 10 = 1
We did this three times b'cos N=3.
Step 2:
Now Amount after 3 years = 1 * 1000 + 3 * 100 + 3 * 10 + 1 * 1 = Rs.1331/-
The coefficents - 1,3,3,1 are lifted from the pascal's triangle above.
Step 3:
CI after 3 years = 3*100 + 3*10 + 3*1 = Rs.331/- (leaving out first term in step
2)
If N =2, we would have had, Amt = 1 * 1000 + 2 * 100 + 1 * 10 = Rs. 1210/-
CI = 2 * 100 + 1* 10 = Rs. 210/-
This method is extendable for any 'N' and it avoids calculations involving higher
powers on 'N' altogether!
A variant to this short cut can be applied to find depreciating value of some
property. (Example, A property worth 100,000 depreciates by 10% every year, find
its value after 'N' years).
http://www.scribd.com/doc/111364745/FCI-Assistant-Grade-III-Exam-Previous-Paper-Free-Download-Guide4BankExams
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