## How to Quickly find Rate of Simple Interest

• Simple interest is calculated only on the principal amount, or on that portion of the principal amount that remains.
• It excludes the effect of compounding.
• Simple interest can be applied over a time period other than a year, e.g. every month.

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Simple Interest FormulaA = P(1 + rt)

Where:

• A = Total Accrued Amount (principal + interest)

• P = Principal Amount

• I = Interest Amount

• r = Rate of Interest per year in decimal; r = R/100

• R = Rate of Interest per year as a percent; R = r * 100

• t = Time Period involved in months or years

From the base formula, A = P(1 + rt) derived from A = P + I and I = Prt so A = P + I = P + Prt = P(1 + rt)

#### Examples:

Suppose you give Rs100 to a bank which pays you 5% simple interest at the end of every year. After one year you will have Rs105, and after two years you will have Rs110. This means that you will not earn an interest on your interest. Your interest payments will be Rs5 per year no matter how many years the initial sum of money stays in a bank account.

#### Problems:

1.A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:

##### Explanation:

S.I. for 1 year = Rs. (854 – 815) = Rs. 39.

S.I. for 3 years = Rs.(39 x 3) = Rs. 117.

Principal = Rs. (815 – 117) = Rs. 698.

2.Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?

##### Explanation:

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 – x).

$then\;,\left(\frac{x\times14\times2}{100}\right)+\left(\frac{(13900-x)\times11\times2}{100}\right)=3508$

28x – 22x = 350800 – (13900 x 22)

6x = 45000

x = 7500.

So, sum invested in Scheme B = Rs. (13900 – 7500) = Rs. 6400.

3.  How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?

##### Explanation:

$Time=\left(\frac{100\times81}{450\times4.5}\right)years=4years$

Topics covered in this e-book are :

1. Simple Interest
2. Compound Interest

/quickest-way-solve-compound-interest-problems/

/solving-simple-interest-problems/

## Shortcut of Combination and Probability

Shortcut 99:    nCr=n!/[r!(n-r)!] Question: In how many ways 2 shirts and 3 pants can be selected from 5 shirts and 7 pants? Answer: Selecting 2 shirts out of 5 = 5C2 = 5!/[2!(5 – 2)!] = 10 Selecting 3 shirts out of 7 = 7C3 = 7!/[3!(7 – 3)!] = 35 Since the two events are dependent, Total ways of selecting = 10 x 35 = 350 Probability Shortcut 100:                   Probablity=Expected number of result / Total number of result Question: What is the probability of selecting 2 spades from a pack of 52 cards? Answer: Total number of ways of selecting 2 cards from 52 = 52C2 = 1275 Total ways of selecting 2 spades from 13 spades = 13C2 = 78 Probability = 78/1275]]>

## Shortcut of Permutation-2

Shortcut 95: Repeated elements      n!/(a!*b!) Question: In how many ways the letters of the word “ENVIRONMENT” can be arranged? Answer: n = 11 Let, a = 2 (E is repeated twice) Let, b = 3 (N is repeated thrice) Number of arrangements = 11!/(2! x 3!) Shortcut 96: Circular arrangement   (n-1)! Question: In how many ways 6 persons can be arranged in a circle? Answer: n = 6 Number of arrangements = (6 – 1)! = 5! = 120 Shortcut 97: Circular arrangement with elements occurring together   2!*(n-2)!        3!*(n-3)! Question: In how many ways 8 persons can be seated around a circular table with two persons always sitting together? Answer: 2! x (8 – 2)! = 2 x 720 = 1440 Note: If three persons are sitting together, then 3! x (8 – 3)! If four persons are sitting together, then 4! x (8 – 4)! Shortcut 98: Arranging a necklace with beads (n-1)!/2 Question: In how many ways a necklace with 8 beads can be arranged? Answer: n = 8 Number of arrangements = (8 – 1)!/2 = 2520]]>

## Shortcut of Permutation-1

Shortcut 91: Events   Dependent=multiplication   Independent =addition Question: There 3 busses from city A to city B and there are 5 busses from city B to city C. In how many ways a person can travel from city A to C through B? Answer: Choosing a bus from city B depends on choosing a bus from city A. Number of ways = 3 x 5 = 15 —————————————————————– Question: There 3 busses from city A to city B and there are 5 busses from city A to city C. In how many ways a person can travel from city A to C or B? Answer: Choosing a bus to city B or C are not dependent. Number of ways = 3 + 5 = 8 Shortcut 92: Arrangement with repetition $n^n;n^r$ Question: In how many ways the letters of the word “ORANGE” can be arranged with repetition? Answer: n = 6   (n = total number of elements) Since all the elements are taken, Number of arrangements = 66 —————————————————————- Question: In how many ways three letters from the word “ORANGE” can be arranged with repetition? Answer: n = 6; r = 3    (r = number of elements taken for arrangement) Number of arrangements = 63 = 216 Shortcut 93: Arrangement without repetition      $n!;nP_r=n!/(n-r)!$ Question: In how many ways the letters of the word “MANGO” can be arranged without repetition? Answer: n = 5 Since all the elements are taken for arrangement, Number of elements = n! = 5! = 120 Question: In how many ways any three letters of the word ‘MANGO” can be arranged without repetition? Answer: n= 5; r = 3 nPr = 5!/(5 – 3)! = 120/2 = 60 ways Shortcut 94: Elements occurring together    2!*(n-1)    3!*(n-2)  Question: In how many ways letters of the word “ORANGE” arranged so that the vowels will always occur together? Answer: n = 6 Three letters ‘O, A and E’ should occur together. Since three letters occur together 3! x (6 – 2)! = 6 x 24 = 144 Note: If there are 4 letters occurring together, then 4!(6 – 3)! If there are 5 letters occurring together, then 5!(6 – 4)!]]>

## Shortcut of Calendar and Clocks

Shortcut 87:
Using the odd days
Question:
Find the day of birth of MS Dhoni (7 July, 1981), if 31 Dec 1999 is Friday
Days after 7/7/1981 in 1981 = 177
Odd days by normal years from 1982 to 1999 = 14
Odd days by leap years = 4 x 2 = 8
Total odd days = 177 + 14 + 8 = 199
Remainder of 199/7 = 3
Since last day is Friday, MSD’s birth day is 3 days before Friday,
MSD’s day of birth = Tuesday.

Clocks
Shortcut 88:
Minute hand and Hour hand overlapping

$\frac{60h}{11}$

Question:
At what time between 3pm and 4pm, the hour hand and the minute hand will be overlapping?
h = starting time = 3
Time at which the hands will be overlapping,
= (60 x 3)/11
= 180/11
= 16.36 minutes after 3pm
= 03h 16m 21s

Shortcut 89:
Two hands at angleƟ
12/11(5h$\pm$0/6)
Question:
At what times between 1pm and 2pm the hour hand and minute hand will be 30 degrees apart?
h = 1
Ɵ = 30
Substitute the values in the above formula, we get
(12/11)[(5 x 1) ± (30/6)]
= (12/11)(5 + 5) , (12/11)(5 – 5)
= 10.9 minutes past 1, 0 minutes past 1
= [01h 10m 54s] and [01h 00m 00s]

Shortcut 90:
Two hands M minutes apart
$\frac{12}{11}(5h\pm M)$
Question:
At what times between 5pm and 6pm the hour hand and minute hand will be 10 minutes apart?
h = 5
M = 10
Substitute the values in the above equation, we get
(12/11)[(5 x 5) ± 10]
= (12/11)(25+10) and (12/11)(25 – 10)
= 38.18 minutes past 5 and 16.36 minutes past 5
= [05h 38m 10s] and [05h 16m 20s]

## Shortcut of Heights and Distances

Shortcut 85:
Using tanƟ
Ɵ                                tan

0                                    0

30                            $1/\sqrt3$

45                                  1

60                             $\sqrt3$

90                              $\infty$
Question:
A man on a boat saw a tower of height 100m at an angle of 60o. What is the distance between the man and the tower?

tan 60 = 100/x
√3 = 100/x
x= 100/√3

Shortcut 86:
Using Pythagoras theorem
$hyp^2=opp^2+adj^2$
Question:
A man looks at the top of a tower which is 400m height. The minimum distance between him and top of the tower is 500m. What is the distance between him and the base of the tower?

5002 = 4002 + x2
x2 = 90000
x = 300 m

## Shortcut of LCM and HCF

Shortcut 82:
Maximum number of items per group
Take HCF
Question:
There are 40, 45 and 50 balls in three different colors. They have to be arranged in boxes such that all the boxes will have equal number of balls and same colored balls. What is the maximum number of balls that can be arranged in each box?
Take HCF of 10,45, 50
HCF = 5
5 balls per box can be arranged in the above condition.

Shortcut 83:
Minimum number of groups
Total/HCF
Question:
There are 40, 45 and 50 balls in three different colors. They have to be arranged in boxes such that all the boxes will have equal number of balls and same colored balls. What is the minimum number of boxes required?
HCF = 5
Total number of balls = 40 + 45 + 50 = 135
Minimum number of boxes required = 135/5 = 27

Shortcut 84:
Ringing Bells
Take LCM
Question:
There are three different bells. The first bell rings every 4 hour, the second bell rings every 6 hours and the third bell rings every 15 hours. If all the three bells are rung at same time, how long will it take for them to ring together?
Take LCM of 4, 6, 15
LCM (4,6,15) = 60
All the three bells will ring after 60 hours

## Shortcut of Number system

Shortcut 77:
Divisibility test
Divisibleby                                    Rule

2                            Last digit of a number should be 0,2,4,6, or 8

3                            Sum of the digit should be divisible by 3

4                            Last two digits of a number should be divisible by 4

5                            Last digit of a number should be 0 or 5

6                             Number should be divisible by 2 and 3

8                             Last three digit of a number shoul be divisible by 8

9                             Sum of the digit should be divisible by 9

10                           Last three digit of a number shoul be 0

11                           Difference between sum of digits in even places shoud be 0 or 11
Question:
Which of the following numbers is divisible by both 5 and 9?
1115, 11115, 111115, 1111115
11115 – Its is divisible by 5 and 9

Shortcut 78:
Finding unit digit
Base Value                                            Power Value

Unit Place                                              Unit Place

0                                                               0

1                                                               1

5                                                               5

6                                                               6
Question:
Find the units place in the result of the expression:
240234 + 345344 x 1011009 – 36211
Unit’s place of 240234 = 0
Unit’s place of 345344 = 5
Unit’s place of 1011009 = 1
Unit’s place of 36211 = 6
Units place of the expression
= 0 + 5 x 1 – 6 = 9
(When subtracting 5 from 6 in unit’s place we will borrow 1 for 5, 15 -6 = 9)

Shortcut 79:
Finding unit digit
Odd           Even

4                       4                6

9                       9                 1
Question:
Find the unit’s digit of 3449 and 4934.
3449 = 4 power odd number – 4 in unit’s place
So, unit digit of 3449 = 4
4934 = 9 power even number – 1 in unit’s place
So, unit’s digit of  4934 = 1

Shortcut 80:
Finding unit digit

Question:
Find the unit digit of the expression
1211 + 1312 + 1718 + 1817
Unit’s place of 1211 = 8
Unit’s place of 1312 = 1
Unit’s place of 1718 = 9
Unit’s place of 1817 = 8
Sum of unit’s places = 8 + 1 + 9 + 8 = 26
Unit’s digit = 6

Shortcut 81:
Finding remainder
R[(a*b*c)/p]=R(a/p)*R(b/p)*R(c/p)
Question:
Find the remainder when 212 x 313 x 414 is divided by 5.
Using the above concept we can find the remainder for each term and then we can multiply.
R(212/5) = 2
R(313/5) = 3
R(414/5) = 4
2 x 3 x 4 = 24
The value we got is still greater than 5, so again divide it and find the remainder.
R(24/5) = 4
The remainder of the expression is = 4

## Shortcut of Simple and Compound Interest

Shortcut 75:
Simple interest
$SI=\frac{PNR}{100}$
P = Principle amount invested or borrowed
R = Rate of interest per term
N = Number of terms
Term = duration for which interest is calculated
Question:
A man invested Rs. 10000 at 4% per annum simple interest. What is the interest he will get at the end of 3 years?
P = 10000; N = 3; R = 4
Substitute the values in the above equation, we get
SI = 10000 x 3 x 4/100
= 1200

Shortcut 76:
Compound interest
$CI=\frac R{100}P+\frac R{100}[Interest till last term]$
Question:
What is the compound interest obtained for Rs. 16000 at 2% per annum for three years?
CI for three years = CI on first year + CI for second year + CI for third year
CI for first year = RP/100 = (3×16000)/100 = 480
CI for second year = (RP/100) + (3/100)480 = 494.4
CI for third year = (RP/100)+(3/100)(489.4+480) = 509.23
Total interest = 480 + 494.4 + 509.2
= 1483.6

## Shortcut of Pipes and Cisterns

Shortcut 73:
One pipe fills and another empties
$\frac{AB}{-A+B}$
Question:
Pipe A can fill a tank in 4 hours and pipe B can empty the tank in 5 hours. If both pipes are opened together, how long will they take to fill a complete tank?
A = 4
B = 5
Substitute the values in the above equation, we get
Time taken to fill the tank = (4 x 5)/(-4 + 5)
= 20 hours

Shortcut 74:”
Two pipes fill and one empties the tank
$\frac{ABC}{-AB+BC+AC}$
Question:
Pipes A and B can fill a tank in 10 hours and 12 hours respectively. Pipe C can empty a tank in 20 hours. If all the pipes are opened together, how long will they take to fill a complete tank?