Simple and Compound Interest

Important Formula: 

1. SI= P*R*T/100

2. Amount after 't' year compounded annually at rate 'r' pa.

                      A   = P[1+(R/100)]^t 

3.  When , Difference between CI and SI is given,such that CI-SI= D

for 2 year

P=  D [100/r]^2

for 3 year

P=  [ 100/ (300+r)]*D [100/r]^2 

4. A sum of money becomes x times in n1year  and y times in n2 year based on  Compound Interest : 

( x times)^ 1/n1   =  ( y times )^ 1/n2

5.A sum of money becomes x times in n1year, then time required to become  y times  based on  Simple  Interest 

TIME, n2  = [ y times - 1/ x times- 1]* n1

6. For Simple  Interest 
Rate*time = 100( n-1), here 'n' is how much times money has become.

Solved Example 

Q1. Rs. 800 amounts to Rs 920 in 3 year at certain interest rate of SI. If rate is increased by 3 % then find the amount after 3 year.

Solution:
When interest rate is increased by 3% then
extra earned to be earned=
3%  of 800 Rs for 3 year = (800*3/100)*3 = 72
Hence,  72 Rs extra interest will be earned. Hence amount will be 920+72 =992.

Q2. A certain sum of money amounts to Rs 1680 in 3 year and it becomes 1920 in 7 year at certain rate of Simple Interest.  Calculate Principal amount?
Solution:

Extra interest earned in (7-4=3 ) 3 year is  (1920-1680= 240) is 240 Rs.
hence , Interest earned in 1 year is  = 240/4 = 60
therefore ,Interest earned for first 3 year is  = 60*3 = 180 rs.
Now, use amount and interest for first 3 year only
Principal = Amount- Interest  = 1680-180 = 1500
Principal = 1500.

Q3. Under SI condition a sum of money becomes 4 times of itself in 30 year . find rate?
Solution :
100..............to ..............400  in 30 year
It means Interest earned in 30 year is = 300 Rs.
Thus, interest earned every year is 300/30= 10.
Now 10 Rs interest is earned on 100 Rs , hence rate % =( 10/100)*100 = 10%

Q4. A sum of money become 2 times in 5 year at a certain rate of SI. Find the time in which the same amount will be 8 times at same rate of SI.

Solution : 

Use the following equation:
Method 1
A sum of money becomes x times in n1year, then time required to become  y times  based on  Simple  Interest 

TIME, n2  = [ y times - 1/ x times- 1]* n1

  TIME, n2/5 = [ 8-1/ 2-1]
               n2  = 35

Method 2 :

equation 1
100................SI= 100 in 5 yr.....................200
equation 2
100.................SI= 700 in 'n2' yr ................800

Now, In order to earn 7 times interest [ SI is 100 and 700 respectively in two cases] time should be 7 times [ keeping rate constant]

time, n2 = 7* n1 = 7*5 = 35 year.

Q5. In a certain time  a sum of money becomes  3 times at 5 %. At what rate of SI , the same sum  will be 6 times in same duration.

equation 1
100................SI= 200 in 't' yr @ 5%.....................300
equation 2
100.................SI= 500 in 't' yr @ 'r%'................600

Now, In order to earn 2.5 times interest [ SI is 200 and 500 respectively in two cases] rate  should be 2.5 times [ keeping rate constant]

rate, r = 5%* 2.5 times  = 12.5%.

Please note, CI can be calculated easily without using any formula.
Just follow the concept,that CI will contain SI and Extra Interest (i.e.interest on interest )
Q6. P = 10000, rate= 10% per annum.  calculate CI for 2 year?
solution:

for yr 1 st :  Interest earned = 10 % of 10000=  1000
for yr 2nd : Interest earned = 1000  + interest on interest for yr 1.
                                       = 1000 + [ 10% of 1000]
                                        = 1000 +100

Total CI = (1000+1000+100) = 2100

Q6. P = 15000, rate= 10% per annum.  calculate CI for 3 year?

solution:

for yr 1st :  Interest earned = 10 % of 15000=  1500
for yr 2nd : Interest earned = 1500  + interest on interest for yr 1.
                                       = 1500 + [ 10% of 1500]
                                        = 1500 +150

for yr 3rd : Interest earned= 1500+[ 10% of 1500 for year 1]+[ 10% of 1500 for year 2]+ [ 10% of 150 for year 2]
for yr 3rd : Interest earned= 1500+150+150+15

Total CI = 1500+ (1500+150)+(1500+150+150+15)
              =4500+450+15 = 4965
Total CI = 4965.


Q7. P = 20000, rate= 20% per annum compounded half yearly .  calculate CI for 1.5 year?

solution:In this case, actually we have to calculate CI for 3 year at the rate 10 % per annum compounded annually.

for yr 1st :  Interest earned = 10 % of 20000=  2000
for yr 2nd : Interest earned = 2000  + interest on interest for yr 1.
                                       = 2000 + [ 10% of 2000]
                                        = 2000 +200

for yr 3rd : Interest earned= 2000+[ 10% of 2000 for year 1]+[ 10% of 2000 for year 2]+ [ 10% of 200 for year 2]
for yr 3rd : Interest earned= 2000+200+200+20

Total CI = 2000+ (2000+200)+(2000+200+200+20)
              =6000+600+20 = 6620
Total CI = 6620.


Q8. If diferrence between SI and CI for 3 year is Rs 31 at rate 10% per annum. Calculate Principal?
solution:
We know that, 

When , Difference between CI and SI is given, 

such that CI-SI= D

for 3 year

P=  [ 100/ (300+r)]*D *[100/r]^2 
p= [ 100/ (300+10)]*31* [100/10]^2
  = [100/310]*31*100   =  1000
p = 1000

Q9. At what rate % per annum, a sum of money becomes doubles itself in 12 years.

For Simple  Interest 

Rate*time = 100( n-1), here 'n' is how much times money has become.
rate *12 = 100(2-1)
rate = 100/12   = 8.33

Q9.In how much time , a sum of money becomes 17/5 times  itself   at the rate  of 12% per annum.

For Simple  Interest 

Rate*time = 100( n-1), here 'n' is how much times money has become.
12 * time = 100*[(17/5)-1]
12*time = 100*[12/5]   = 240
time = 240/12  = 20
time = 20 year.

Machine Input-output

Machine Input-output or simply input-output reasoning questions are one of the most famous variety and most frequently asked question in the competitive exam and specifically in banking exams.

Myth: Input-output reasoning questions are very difficult and time consuming.

Reality: These questions may appear complex but actually they are easy to solve. If you are able to crack the logical step for output quickly and you are only solving selective question (without wasting much time on difficult and time consuming) there is high chance that you can score at least 80% in input-output reasoning questions.

Tips: In order to solve these questions, please use the following golden rules:

1. Look at the last step (Output) without bothering about intermediate step.

2. See the input

3. Now, Identify the logic behind the rearrangement i.e. try to find in which pattern the elements of Input is rearranged to get the given output.

Model Example:

Input:   gas, nice, 52, 38, all 17, fruit again, 65.

Step1:  65, gas, nice, 52, 38, 17, fruit again.

Step2: 65, again, gas, nice, 52, 38, 17, fruit.

Step3: 65, again, 52, gas, nice, 38, 17, fruit.

Step4: 65, again, 52, fruit, gas, nice, 38, 17.

Step5: 65, again, 52, fruit, 38, gas, nice, 17.

Step6: 65, again, 52, fruit, 38, gas, 17, nice.

Step6 is the out-put for the given input.

Explanation:

By using the ‘golden rule’ discussed above, we can find that:

1. Output contains Numbers and words arranged in alternate order.

Numbers are arranged in decreasing order from left to right.

Words are arranged in ascending alphabetical order from left to right.

2. Output contains Numbers and words arranged in random order.

3. Only one element (either number or word) is moved in each step. The arrangement eventually leads to out as discussed in point1.

‘Frequently asked variety’ of question:

Variety 1: how many steps are required to complete the arrangement?

Using the above Model Example, find out how many steps are required to complete the arrangement for the following input.

Input:  hello, talk, 39,24,86,44, rice, oven.

     hello,  talk,    39,  24    ,86,      44,      rice,      oven.

      86   hello  44 oven   39  rice  24    talk

Explanation : here you can count that there are 6 movements ( as shown in different colour)   to get the desired output, we can say that total 6 steps are required.

Please note that, hello and talk are already arranged.

( If an element is already arranged, we simply move on next place without disturbing it.)

Variety 2:  What will be the definite input for the given step?

Example:

Step 3:  94 cat, 86, wild, sound, 52, 31, horse.

What will be the definite input for the given Step 3?

Solution: we cannot determine the definite previous step.

Step 3:  94 cat, 86, wild, sound, 52, 31, horse.

94 cat, 86,  wild,    sound,    52,     31,      horse.

94 cat, 86, wild, 86sound, 86 52, 86 31, 86horse. 86

Explanation: Here in step3, 86 has been brought from some other position [different possible position are shown as 86] , hence we are not sure from where 86 has been shifted to the current position as given in step 3. So, we cannot determine the definite previous step.

Remember: For any given step, definite previous step cannot be determined.

Variety3:   When some intermediary step are given:

Example:

Step2 : 76 apple, 12,32, ball, orange, joker, 42.

How many more steps are required to complete the arrangement as given in Model Example (refer first paragraph in this post)

Solution:

Already arranged in step2 :  76, apple

76   apple,12, 32,  ball, orange,joker, 42.

76, apple,42,ball, 32 , joker, 12, over.

Here 4 movements (as shown in different colour) are required, hence 4 more steps are required to complete the rearrangement.

Please, note that  “12 , over “   are already  auto arranged  as per the requirement, hence they are not disturbed.

Variety 4: when we are required to find out what exactly a particular step will look like for a given input.  

Example:

Input: Boy, wine, tiger, 52,38, 43, doll, 12 .  What will be step4 for this input as given in Model Example (refer first paragraph in this post).

Boy, wine, tiger, 52, 38, 43, doll, 12

 52 boy, 43, doll,    38  ,

1      1a    2      3       4    

[here 1, 1a, 2, 3, 4 indicates the step. Also, 1a signifies that the element ( boy) isalready  arranged ].

 4 movement (as shown in different colour)   = 4 step

Hence output will look like:

Step 4: 52 boy, 43, doll,    38  , wine , tiger 12

Please note that,” boy  wine , tiger 12” “    are already  auto arranged  as per the requirement, hence they are not disturbed.

Hope, this post is helpful for you all.
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Probablity

Basic Probability Questions:

Q1.You draw a card at random from a deck that contains  3 black cards and 7 red cards.Find probablity of getting a black card?

Solution:

Total outcome = 3 black +7 red = 10

favorable outcome = 3

probability = 3/10 = 0.3

Q2. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5? 

Solution:

S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/n(S) = 9/20.

Q3.What is the probability of getting a sum 9 from two throws of a dice?

Solution:   In 2 throws of dice, total outcome  S = 6 x 6  =36
Event of getting sum of 9( our favorable outcome)  is = (3,6),(6,3),(4,5),(5,4)
P(E) = n(E)/n(S) = 4/36 = 1/9

Q4. There are 5 marbles in a bag: 4 are blue, and 1 is red. What is the probability that a blue marble gets picked?Solution: 

P(E) = n(E)/n(S) = 4/5 = 0.8

Q5.A die is thrown once. What is the probability that the score is a factor of 6?
Solution:

The factors of six are 1, 2, 3 and 6, so the Number of ways it can happen = 4
There are six possible scores when a die is thrown, so the Total number of outcomes = 6
So the probability that the score is a factor of six = 4/6 = 2/3

Q6.A card is chosen at random from a deck of 52 playing cards. There are 4 Queens and 4 Kings in a deck of playing cards.What is the probability it is a Queen or a King?
Solution: 

There are 4 Queens and 4 Kings, so the Number of ways it can happen = 8
There are 52 cards altogether, so the Total number of outcomes = 52

Set theory: Application

Set theory: Application

Q1. Let A and B be two finite sets such that n(A) = 20, n(B) = 28 and n(A ∪ B) = 36, find n(A ∩ B).

Solution: 

 formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B).

 n(A ∩ B) = n(A) + n(B) - n(A ∪ B)

                   = 20 + 28 - 36 = 12

                 

Q2. If n(A - B) = 18, n(A ∪ B) = 70 and n(A ∩ B) = 25, then find n(B).

Solution: 

 formula n(A∪B) = n(A - B) + n(A ∩ B) + n(B - A)

                              70 = 18 + 25 + n(B - A)

                              70 = 43 + n(B - A)   = 27

                   
Now n(B) = n(A ∩ B) + n(B - A)

             = 25 + 27

             = 52

Q3. In a group of 60 people, 27 like cold drinks and 42 like hot drinks and each person likes at least one of the two drinks. How many like both coffee and tea?

Solution: 

Let A = Set of people who like cold drinks.

     B = Set of people who like hot drinks.


(A ∪ B) = 60            n(A) = 27       n(B) = 42 then;

Formula: n(A ∩ B) = n(A) + n(B) - n(A ∪ B)

            = 27 + 42 - 60 =   9.0

          Therefore, 9 people like both tea and coffee. 

Q4. There are 35 students in art class and 57 students in dance class. Find the number of students who are either in art class or in dance class,  When two classes meet at different hours and 12 students are enrolled in both activities.

Solution: 

Let A be the set of students in art class. 
B be the set of students in dance class.

n(A) = 35,       n(B) = 57,       n(A ∩ B) = 12


(i) When 2 classes meet at different hours n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

                                                                    = 35 + 57 - 12  =   80

                                                                                 


  Q5. In a group of 100 persons, 72 people can speak English and 43 can speak French. How many can speak English only? How many can speak French only and how many can speak both English and French?

Solution: 

Let A be the set of people who speak English.

B be the set of people who speak French.

A - B be the set of people who speak English and not French.

B - A be the set of people who speak French and not English.

A ∩ B be the set of people who speak both French and English.

Given,

n(A) = 72       n(B) = 43       n(A ∪ B) = 100

Now, n(A ∩ B) = n(A) + n(B) - n(A ∪ B)

                   = 72 + 43 - 100

                   = 115 - 100

                   = 15

Therefore, Number of persons who speak both French and English = 15

n(A) = n(A - B) + n(A ∩ B)

⇒ n(A - B) = n(A) - n(A ∩ B)

               = 72 - 15

               = 57

and n(B - A) = n(B) - n(A ∩ B)

                 = 43 - 15

                 = 28

Therefore, Number of people speaking English only = 57

Number of people speaking French only = 28

Q6.. Each student in a class of 40 plays at least one indoor game chess, carrom and scrabble. 18 play chess, 20 play scrabble and 27 play carrom. 7 play chess and scrabble, 12 play scrabble and carrom and 4 play chess, carrom and scrabble. Find the number of students who play (i) chess and carrom. (ii) chess, carrom but not scrabble.

Solution: 

Let A be the set of students who play chess

B be the set of students who play scrabble

C be the set of students who play carrom

Therefore, We are given n(A ∪ B ∪ C) = 40,

n(A) = 18,         n(B) = 20         n(C) = 27,

n(A ∩ B) = 7,     n(C ∩ B) = 12    n(A ∩ B ∩ C) = 4

We have

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

Therefore, 40 = 18 + 20 + 27 - 7 - 12 - n(C ∩ A) + 4

40 = 69 – 19 - n(C ∩ A)

40 = 50 - n(C ∩ A) n(C ∩ A) = 50 - 40

n(C ∩ A) = 10

Therefore, Number of students who play chess and carrom are 10.

Also, number of students who play chess, carrom and not scrabble.

= n(C ∩ A) - n(A ∩ B ∩ C)

= 10 – 4

= 6

Therefore, we learned how to solve different types of word problems on sets without using Venn diagram.

Profit and Loss

Profit and Loss

Important Formula:

1. Profit = SP-CP

2. Loss  = CP-SP

3. Profit% = (Profit/ CP)*100

4.Loss % = ( Loss/CP)*100

5. For Dishonest shopkeeper :

% profit or % loss = (Error *100) / True Value

6. SP= MARKED PRICE - DISCOUNT
[ Whenever discount is given , it is given on Marked Price, hence in such case Marked Price should be treated as 100%]

Important  Points : 

1.Always consider CP as 100% and then solve the question. Try to use all the variables in percentage form only.
2. CP should be the base value for calculating  Profit or Loss .

3. For   Successive Discount:

Net Discount =  a+ b -(ab/100)

4. When CP1 = CP2 and profit% 1st = Loss% on 2nd,
then : NO PROFIT & NO LOSS in overall transaction.

5. When SP1 = SP2 and profit% 1st = Loss% on 2nd,
then : LOSS in overall transaction.
Loss% = (rate*rate)/100

Solved Examples 

Q1. A man sold a book at 480 Rs and he gained 20%. Find CP ?

Solution: 

CP =  100% and SP= 120%

Here,

120% = 480

100% = x (i.e. CP)

Use cross multiplication and solve further,

x = 400

Q2.

If  Mark Price of a TV set  is 20% more than its CP. Shopkeeper  gives discount of 5%. Find profit% ?

Solution:

CP = 100%

MP= 120%

SP = 120% - ( 5% of 120%)  = 120% - 6%  = 114%

Profit% = 114% - 100% = 14%. 

Q3. 

If CP is 96 % of SP. Find Profit %?

Solution: 

Given that, CP= 96%  and SP = 100% 

profit = SP-CP = 100% - 96%  = 4% 

Profit% = (Profit/ CP)*100  = (4%/ 96%)*100   =  400/96 = 4.16 .

Q4. A Boy bought a bag with 15% discount on labelled price. He sold the bag for Rs 2880 with 20% profit on labelled price. At which price did he bought the bag?

Solution: 

CP of bag = 100% of labelled price - 15%  = 85% of   labelled price is CP

120 % of labelled Price  =  2880

85% of   labelled price   =  x ( i.e. CP)

Use cross multiplication and solve further,

x = (2880*85%)/ 120%  = Rs 2040.

Q5. A customer gets 2  discount of 40% and 20% . This discount is equivalent to how much  discount?

solution:

Net Discount =  a+ b -(ab/100)

Net discount = 40+20 - (40*20/100) = 60-8 = 52%.

alternate method:

100  - (40% of 100) - ( 20% of 40% of 100) 

100-40-8 = 52% 

Q6. A shopkeeper give 2 successive discount  of 20% and 15 % on purchase of an article. Ia a boy pays Rs 1156 for that article then find the Marked price ?

solution

Net Discount =  a+ b -(ab/100)

                      = 20+15- (20*15/ 100)  =  35-3 = 32%

It means the Boy has paid 68 %  [ as he get 32% discount]

now, 

68% = 1156

100%  = x ( i.e. marked price)

x = 1156*100 / 68 = Rs 1700.

Q7. A shopkeeper sell his good at CP., but he uses a weight of 960 gram instead of 1kg . Find his profit % ?

solution:

% profit  = (Error *100) / True Value 

               =(40 *100)/960 = 4000/960 = 4.16.

Q8. A man purchases some lemon at 6 lemon per 5 Rs. and sell all the lemon at 4 lemon per 3 Rs.

      Find Profit or loss % ?

solution:

              Quantity      Price

CP             6                5

SP              4                3

Please note , the TRICK is cross multiplication  we will use SP as  6*3 [ from left top to right bottom]   and CP  as 4*5  [ from left bottom  to right top ] 

SP = 18

CP =  20

loss = 2

loss % = 2*100/20   = 2% loss.

Q9. A person bought two articles for the same price and sold them at a profit of 10% on one and a loss of 10% on the other. What is the overall profit or loss % made by him?

here,
 CP1= CP2
 PROFIT% = LOSS % = 10%
When CP1 = CP2 and profit% on 1st = Loss% on 2nd,
then : NO PROFIT &NO LOSS in overall transaction.

Q10.If a person sold two articles at same price and realized 10% profit on one and 10% loss on the other article. What net profit/loss % does he make?
here,

 SP1= SP2
 PROFIT% = LOSS % = 10%
LOSS in overall transaction.
Loss% = (rate*rate)/100 = (10*10)/100 = 1%.

Q11.A shopkeeper marks his goods in such a way that after allowing a discount of 10%, he gains 17%. How much percent above C.P. is the marked price?

Solve it using ,
Net percentage change =  a+ b -(ab/100)

profit% = % change in MP (a)+ discount% (b)- ab/100
17  = a-10-(10a/100)
1700  = 100a-1000-10a
1700+1000 = 100a-10a
2700= 90a
a= 30
Marked price is 30% more than CP.

Simple and Compound Interest

Simple and Compound Interest

Important Formula: 

1. SI= P*R*T/100

2. Amount after 't' year compounded annually at rate 'r' pa.

                         = P[1+(R/100)]^t 

3.  When , Difference between CI and SI is given, 

such that CI-SI= D

for 2 year

P=  D [100/r]^2

for 3 year

P=  [ 100/ (300+r)]*D [100/r]^2 

4. A sum of money becomes x times in n1year  and y times in n2 year based on  Compound Interest : 

( x times)^ 1/n1   =  ( y times )^ 1/n2

5.A sum of money becomes x times in n1year, then time required to become  y times  based on  Simple  Interest 

TIME, n2  = [ y times - 1/ x times- 1]* n1

6. For Simple  Interest 
Rate*time = 100( n-1), here 'n' is how much times money has become.

Solved Example 

Q1. Rs. 800 amounts to Rs 920 in 3 year at certain interest rate of SI. If rate is increased by 3 % then find the amount after 3 year.

Solution:
When interest rate is increased by 3% then
extra earned to be earned=
3%  of 800 Rs for 3 year = (800*3/100)*3 = 72
Hence,  72 Rs extra interest will be earned. Hence amount will be 920+72 =992.

Q2. A certain sum of money amounts to Rs 1680 in 3 year and it becomes 1920 in 7 year at certain rate of Simple Interest.  Calculate Principal amount?
Solution:

Extra interest earned in (7-4=3 ) 3 year is  (1920-1680= 240) is 240 Rs.
hence , Interest earned in 1 year is  = 240/4 = 60
therefore ,Interest earned for first 3 year is  = 60*3 = 180 rs.
Now, use amount and interest for first 3 year only
Principal = Amount- Interest  = 1680-180 = 1500
Principal = 1500.

Q3. Under SI condition a sum of money becomes 4 times of itself in 30 year . find rate?
Solution :
100..............to ..............400  in 30 year
It means Interest earned in 30 year is = 300 Rs.
Thus, interest earned every year is 300/30= 10.
Now 10 Rs interest is earned on 100 Rs , hence rate % =( 10/100)*100 = 10%

Q4. A sum of money become 2 times in 5 year at a certain rate of SI. Find the time in which the same amount will be 8 times at same rate of SI.

Solution : 

Use the following equation:
Method 1
A sum of money becomes x times in n1year, then time required to become  y times  based on  Simple  Interest 

TIME, n2  = [ y times - 1/ x times- 1]* n1

  TIME, n2= [ 8-1/ 2-1]*5
                 = 35

Method 2 :

equation 1
100................SI= 100 in 5 yr.....................200
equation 2
100.................SI= 700 in 'n2' yr ................800

Now, In order to earn 7 times interest [ SI is 100 and 700 respectively in two cases] time should be 7 times [ keeping rate constant]

time, n2 = 7* n1 = 7*5 = 35 year.

Q5. In a certain time  a sum of money becomes  3 times at 5 %. At what rate of SI , the same sum  will be 6 times in same duration.


equation 1
100................SI= 200 in 't' yr @ 5%.....................300
equation 2
100.................SI= 500 in 't' yr @ 'r%'................600

Now, In order to earn 2.5 times interest [ SI is 200 and 500 respectively in two cases] rate  should be 2.5 times [ keeping rate constant]

rate, r = 5%* 2.5 times  = 12.5%.

Please note, CI can be calculated easily without using any formula.
Just follow the concept, that CI will contain SI and Extra Interest (i.e. interest on interest )

Q6. P = 10000, rate= 10% per annum.  calculate CI for 2 year?
solution:

for yr 1 st :  Interest earned = 10 % of 10000=  1000
for yr 2nd : Interest earned = 1000  + interest on interest for yr 1.
                                       = 1000 + [ 10% of 1000]
                                        = 1000 +100

Total CI = (1000+1000+100) = 2100

Q6. P = 15000, rate= 10% per annum.  calculate CI for 3 year?

solution:

for yr 1st :  Interest earned = 10 % of 15000=  1500
for yr 2nd : Interest earned = 1500  + interest on interest for yr 1.
                                       = 1500 + [ 10% of 1500]
                                        = 1500 +150

for yr 3rd : Interest earned= 1500+[ 10% of 1500 for year 1]+[ 10% of 1500 for year 2]+ [ 10% of 150 for year 2]
for yr 3rd : Interest earned= 1500+150+150+15

Total CI = 1500+ (1500+150)+(1500+150+150+15)
              =4500+450+15 = 4965
Total CI = 4965.


Q7. P = 20000, rate= 20% per annum compounded half yearly .  calculate CI for 1.5 year?
solution:
In this case, actually we have to calculate CI for 3 year at the rate 10 % per annum compounded annually.



for yr 1st :  Interest earned = 10 % of 20000=  2000
for yr 2nd : Interest earned = 2000  + interest on interest for yr 1.
                                       = 2000 + [ 10% of 2000]
                                        = 2000 +200

for yr 3rd : Interest earned= 2000+[ 10% of 2000 for year 1]+[ 10% of 2000 for year 2]+ [ 10% of 200 for year 2]
for yr 3rd : Interest earned= 2000+200+200+20

Total CI = 2000+ (2000+200)+(2000+200+200+20)
              =6000+600+20 = 6620
Total CI = 6620.


Q8. If diferrence between SI and CI for 3 year is Rs 31 at rate 10% per annum. Calculate Principal?
We know that, 

When , Difference between CI and SI is given, 

such that CI-SI= D

for 3 year

P=  [ 100/ (300+r)]*D *[100/r]^2 
p= [ 100/ (300+10)]*31* [100/10]^2 
  = [100/310]*31*100   =  1000
p = 1000

Q9. At what rate % per annum, a sum of money becomes doubles itself in 12 years.

For Simple  Interest 

Rate*time = 100( n-1), here 'n' is how much times money has become.
rate *12 = 100(2-1)
rate = 100/12   = 8.33

Q9.In how much time , a sum of money becomes 17/5 times  itself   at the rate  of 12% per annum.

For Simple  Interest 

Rate*time = 100( n-1), here 'n' is how much times money has become.
12 * time = 100*[(17/5)-1]
12*time = 100*[12/5]   = 240
time = 240/12  = 20
time = 20 year.