#dailyquizadda #Imptrick #SI&CI
After reading this, you will be able to solve all the questions that are asked by SSC from this topic.
(1) The basic concept of CI and SI
Let's say you have Rs. 30000 and you keep this money in three different banks for 2 years(Rs. 10000 each). The three banks have different policy :
a) Bank A keeps your money at simple interest and offers you 5% interest
b) Bank B keeps your money at compound interest and offers you 5% interest. The interest is compounded annually.
Bank C keeps your money at compound interest and offers you 5% interest. The interest is compounded half-yearly.
After 2 years, which bank will give you most interest?
Let us calculate
Case (A)
Simple interest is calculated simply as (P*R*T)/100
Here P= 10000, r = 5% and T = 2 years
T = 2 years. Let's divide this period in two equal intervals of 1 year each
Hence SI received for the period 0 to 1 = 10000*5*1/100 = Rs. 500
SI received for the period 1 to 2 = 10000*5*1/100 = Rs. 500
So after 2 years, you will get Rs. 10000 + 500 + 500 = Rs. 11000
Note : Simple Interest is proportional. The interest received is same each year. So in the above example where SI was Rs. 500 for 1 years, that will mean the SI for 3 years is Rs. 1500, the SI for 5 years is Rs. 2500 and so on.
Case (B)
Compounded annually means whatever interest you will earn on first year, that interest will be added to the principal to calculate the interest for 2nd year. Let us see how
We know the CI formula is, Amount = P(1 + r/100)^t (where Amount = P + CI)
CI received for the period 0 to 1 = Amount - Principal = 10000(1 + 5/100)^1 - 10000= Rs. 500
Now the amount received after 1 year will act as the Principal for calculating the Amount for next year
For calculating the amount for second year, you won't take P as 10000, but as Rs. 10500. So unlike SI where the interest was same each year, in CI the interest increases every year (because the principal increases every year)
CI received for the period 1 to 2 = Amount - Principal = 10500(1 + 5/100)^1 - 10500 = Rs. 525
Total interest received after two years = Rs. 500 + Rs. 525 = Rs. 1025
Total amount received after two years = Rs. 11025
Note : In Case (b), to calculate the amount received after 2 years, I had divided the calculation into 2 intervals. It was done just for the sake of explanation. You can calculate the amount received after 2 years directly by 10000(1 + 5/100)^2
Case (C)
Just like case (b), where Principal was getting updated every year, in case (c) we will update the Principal every 6 months (half-year)
Since I have given the explanation in case (b), so in this case I will directly apply the formula
Amount received after 2 years = 10000(1 + 2.5/100)^4 = Rs. 11038 approx.
So sum it up
Case A - amount received after two years= Rs. 11000
Case B - amount received after two years= Rs. 11025
Case C - amount received after two years= Rs. 11038
Case C is giving the maximum return and rightly so because in Case (C) principal is increasing every 6 months.
Important formulas for Compund Interest -
(2) A sum of money becomes x times in T years. In how many years will it become y times?
The approach to solve such questions is different for SI and CI
For SI : Formula = [(y - 1)/(x - 1)] * T
Q. 1) A sum of money becomes three times in 5 years. In how many years will the same sum become 6 times at the same rate of simple interest?
Solution : [(6 - 1)/(3 - 1)] * 5 = 5/2 * 5 = 12.5 years
Answer : 12.5 years
For CI : Formula = (logy/logx) * T
Now dont worry, I wont be asking you to study logarithms :)
But just remember one property of logs and that is enough to solve the questions
log(x y) = y.log(x)
Hence log(8) = log(23) = 3.log(2)
Q. 2) A sum of money kept at compound interest becomes three times in 3 years. In how many years will it be 9 times itself?
Solution : (log9/log3) * 3 ... (1)
log9 = log(32) = 2.log(3)
Put this value in (1)
= 2.log(3)/log(3) * 3
= 2 * 3 = 6 years
Answer : 6 years
(3) Interest for a number of days
Here P = 306.25
R = 15/4 %
T = Number of days/365
Number of days = Count the days from March 3rd to July 27th but omit the first day, i.e., 3rd March
= 28 days(March) + 30 days(April) + 31 days(May) + 30 days(June) + 27 days(July)
= 146 days
We know SI = (P * r * t)/100
Answer : Rs. 4.59
(4) Annual Instalments
This is the most dreaded topic of CI-SI. Before giving you the direct formula, I would like to tell you what actually is the concept of annual instalments(if you only want the formula and not the explanation, you can skip this part. But I would like you to read it)
Suppose you want to purchase an iPhone and its price is Rs. 100000 but you dont have Rs. 1 lakh as of now. What would you do? You have two options - either you can sell your kidney (which most the iphone buyers do :D), or you can go for instalments. But if you want to buy the iPhone through this instalment route, the seller will incur a loss. How? Had you paid Rs. 1 lakh in one go, the seller would have kept that money in his savings account and earned some interest on it. But you will pay this Rs. 1 lakh in instalments and that means the seller will get his Rs. 1 lakh after several years. So the seller is incurring a loss. The seller will compensate for this loss and will charge interest from you.
Now coming to the questions. There are two types of questions and they
are bit confusing. In one type, the Amount is given and in another type,
Principal is given
Type 1(Amount is given):
Q. 4) What annual installment will discharge a debt of Rs.6450 due in 4 years at 5% simple interest?
When the language the question is like "what annual payment will discharge a debt of ...", it means the Amount is given in the question.
In this question, the Amount(A) is given, i.e., Rs. 6450. So we can apply the formula directly
Here A = 6450, r = 5%, t = 4 years
Solution : 100*6450/[100*4 + 5*4*3/4]
Answer : 1500
Type 2 (Principal is given) :
Q. 5) A sum of Rs. 6450 is borrowed at 5% simple interest and is paid back in 4 equal annual installments. What is amount of each installment?
"Sum borrowed" means Principal.
This question is of Compound Interest and hence we cant apply the direct formula. We will solve this question with the help of the equation we derived earlier.
P(1+r/100)^2 = x(1+r/100) + x
Q. 7) What annual instalment will discharge a loan of Rs. 66000, due in 3 years at 10% Compound Interest?
Solution : Here again the question is of "Compound Interest" and hence we will solve it by equation :
Let each annual instalment be of Rs. x. Note that in this question, amount is given
Amount = x(1 + 10/100)^2 + x(1 + 10/100)^1 + x
66000 = x (1.21 + 1.1 + 1)
So x = Rs. 19939.58
Q. 8) What annual instalment will discharge a loan of Rs. 66000, due in 3 years at 10% Simple Interest?
I have just converted Q.7 into Simple Interest
Now we can either solve it by direct formula, or by equation
By Equation method :
66000 = (x + x*10*2/100) + (x + x*10*1/100) + x
66000 = x(3 + 0.2 + 0.1)
x = Rs. 20000
By Direct formula method :
A = 66000, t = 3, r = 10%
x = 100A/[100t + t(t-1)r/2]
x = 100*66000/[100*3 + 3*2*10/2]
x = 6600000/(300 + 30)
x = Rs. 20000
Q. 1) If a sum of money becomes 3 times itself in 20 years at simple interest. What is the rate of interest?
In such questions apply the direct formula-
Rate of interest = [100*(Multiple factor - 1)]/T
So R = 100*(3 - 1)/20
Answer : 10%
Note : With this formula you can find Rate if Time is given and Time if rate is given.
Q. 2)
In such questions, just write this line :
1st part : 2nd part : 3rd part = 1/(100+T1 * r) : 1/(100+T2 * r) : 1/(100+T3 * r)
= 1/(100+2 * 5) : 1/(100+3 * 5) : 1/(100+4*5)
= 1/110 : 1/115 : 1/120
= 23*24 : 22*24 : 23*22
Hence 1st part = (23*24)/ (23*24 + 22*24 + 23*22) * 2379
Answer : 828
Note : Surprisingly, such questions when asked mostly have this same data, i.e., R=5% and T1, T2, T3 = 2, 3, 4 years, respectively. Only the Principal is changed. So it would be wise if you can just mug this line :
1st part : 2nd part : 3rd part = 23*24 : 22*24 : 23*22
Based on the above line, you would be able to solve such questions in a jiffy.
But note that it will only work if the question is on Simple Interest. Like the below question appeared in SSC CGL Tier 2-
1st part : 2nd part : 3rd part = 23*24 : 22*24 : 23*22
A received = (23*24)/ (23*24 + 22*24 + 23*22) * 7930
Answer : Rs. 2760
Q. 5) If a certain sum of money P lent out for a certain time T amounts to P1 at R1% per annum and to P2 at R2% per annum, then
Apply the formula:
R = (650-600)*100/600*6 - 650*4
R = 5%
Alternative Method :
Difference between CI and SI
If t=3 years then difference between CI and SI can be given by two formulas-
P = Rs. 32000
Q.16) The difference between CI and SI on a certain sum at 10% per annum for 4 years is Rs
1282. Find the sum.
SI for 4 years @10% = 40P/100 = 0.4P
CI for 4 years @10% = P(1.1^4 - 1) = 0.4641P
0.4641P - 0.4P = 1282
P = Rs. 20000
Q.17) A sum of money lent at compound interest for 2 years at 20% per annum would fetch
Rs. 482 more of interest was payable half yearly than if it was payable annually. The sum is ?
When interest is compounded half-yearly
A = P(1 + 10/100)^4
When interest is compounded annually
A = P(1 + 20/100)^2
Given, P(1 + 10/100)^4 - P(1 + 20/100)^2 = 482
Solve for P
Answer : Rs. 20000
Note : In the question, they have given the difference between the interest but we have
taken the difference between the Amounts because the difference in amount is equal to the
difference in interest.
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After reading this, you will be able to solve all the questions that are asked by SSC from this topic.
(1) The basic concept of CI and SI
Let's say you have Rs. 30000 and you keep this money in three different banks for 2 years(Rs. 10000 each). The three banks have different policy :
a) Bank A keeps your money at simple interest and offers you 5% interest
b) Bank B keeps your money at compound interest and offers you 5% interest. The interest is compounded annually.
Bank C keeps your money at compound interest and offers you 5% interest. The interest is compounded half-yearly.
After 2 years, which bank will give you most interest?
Let us calculate
Case (A)
Simple interest is calculated simply as (P*R*T)/100
Here P= 10000, r = 5% and T = 2 years
T = 2 years. Let's divide this period in two equal intervals of 1 year each
Hence SI received for the period 0 to 1 = 10000*5*1/100 = Rs. 500
SI received for the period 1 to 2 = 10000*5*1/100 = Rs. 500
So after 2 years, you will get Rs. 10000 + 500 + 500 = Rs. 11000
Note : Simple Interest is proportional. The interest received is same each year. So in the above example where SI was Rs. 500 for 1 years, that will mean the SI for 3 years is Rs. 1500, the SI for 5 years is Rs. 2500 and so on.
Case (B)
Compounded annually means whatever interest you will earn on first year, that interest will be added to the principal to calculate the interest for 2nd year. Let us see how
We know the CI formula is, Amount = P(1 + r/100)^t (where Amount = P + CI)
CI received for the period 0 to 1 = Amount - Principal = 10000(1 + 5/100)^1 - 10000= Rs. 500
Now the amount received after 1 year will act as the Principal for calculating the Amount for next year
For calculating the amount for second year, you won't take P as 10000, but as Rs. 10500. So unlike SI where the interest was same each year, in CI the interest increases every year (because the principal increases every year)
CI received for the period 1 to 2 = Amount - Principal = 10500(1 + 5/100)^1 - 10500 = Rs. 525
Total interest received after two years = Rs. 500 + Rs. 525 = Rs. 1025
Total amount received after two years = Rs. 11025
Note : In Case (b), to calculate the amount received after 2 years, I had divided the calculation into 2 intervals. It was done just for the sake of explanation. You can calculate the amount received after 2 years directly by 10000(1 + 5/100)^2
Case (C)
Just like case (b), where Principal was getting updated every year, in case (c) we will update the Principal every 6 months (half-year)
Since I have given the explanation in case (b), so in this case I will directly apply the formula
Amount received after 2 years = 10000(1 + 2.5/100)^4 = Rs. 11038 approx.
So sum it up
Case A - amount received after two years= Rs. 11000
Case B - amount received after two years= Rs. 11025
Case C - amount received after two years= Rs. 11038
Case C is giving the maximum return and rightly so because in Case (C) principal is increasing every 6 months.
Important formulas for Compund Interest -
(2) A sum of money becomes x times in T years. In how many years will it become y times?
The approach to solve such questions is different for SI and CI
For SI : Formula = [(y - 1)/(x - 1)] * T
Q. 1) A sum of money becomes three times in 5 years. In how many years will the same sum become 6 times at the same rate of simple interest?
Solution : [(6 - 1)/(3 - 1)] * 5 = 5/2 * 5 = 12.5 years
Answer : 12.5 years
For CI : Formula = (logy/logx) * T
Now dont worry, I wont be asking you to study logarithms :)
But just remember one property of logs and that is enough to solve the questions
log(x y) = y.log(x)
Hence log(8) = log(23) = 3.log(2)
Q. 2) A sum of money kept at compound interest becomes three times in 3 years. In how many years will it be 9 times itself?
Solution : (log9/log3) * 3 ... (1)
log9 = log(32) = 2.log(3)
Put this value in (1)
= 2.log(3)/log(3) * 3
= 2 * 3 = 6 years
Answer : 6 years
(3) Interest for a number of days
Q. 3)
Here P = 306.25
R = 15/4 %
T = Number of days/365
Number of days = Count the days from March 3rd to July 27th but omit the first day, i.e., 3rd March
= 28 days(March) + 30 days(April) + 31 days(May) + 30 days(June) + 27 days(July)
= 146 days
We know SI = (P * r * t)/100
Answer : Rs. 4.59
(4) Annual Instalments
This is the most dreaded topic of CI-SI. Before giving you the direct formula, I would like to tell you what actually is the concept of annual instalments(if you only want the formula and not the explanation, you can skip this part. But I would like you to read it)
Suppose you want to purchase an iPhone and its price is Rs. 100000 but you dont have Rs. 1 lakh as of now. What would you do? You have two options - either you can sell your kidney (which most the iphone buyers do :D), or you can go for instalments. But if you want to buy the iPhone through this instalment route, the seller will incur a loss. How? Had you paid Rs. 1 lakh in one go, the seller would have kept that money in his savings account and earned some interest on it. But you will pay this Rs. 1 lakh in instalments and that means the seller will get his Rs. 1 lakh after several years. So the seller is incurring a loss. The seller will compensate for this loss and will charge interest from you.
Let the annual instalment be Rs. x. and you pay it for 4 years.
After 1 year you will pay Rs. x and the seller will immediately put this
money in his savings account (or somewhere else) to earn interest. He
will earn interest on this Rs. x at the rate of r% for 3 years (because
the total duration is 4 years and 1 year has already passed)
Hence the amount which the seller will get from this Rs. x instalment = x(1 + r/100)^3
After 2nd year, you will again pay Rs. x and the seller will earn interest on this Rs. x for 2 years.
The amount which the seller will get from this Rs. x instalment = x(1 + r/100)^2
After 3rd year, you will again pay Rs. x and the seller will earn interest on this Rs. x for 1 year.
The amount which the seller will get from this Rs. x instalment = x(1 + r/100)^1
After 4th year, you will pay Rs. x and your debt would be paid in full (no interest on this Rs. x)
The amount which the seller will get from this Rs. x instalment = x
Now let's add all the above four amounts to get the total amount the seller would get from all the instalments =
x(1 + r/100)^3 + x(1 + r/100)^2 + x(1 + r/100)^1 + x ... (1)
x(1 + r/100)^3 + x(1 + r/100)^2 + x(1 + r/100)^1 + x ... (1)
Now, had you paid Rs. 1 lakh in one go (without going for the instalment
route), then the amount received by the seller after 4 years would have
been = 100000(1+r/100)^4 ... (2)
Now (1) should be equal to (2) because only then the two routes
(instalment route and direct payment route) will give the same return
and seller would have no problem in giving you the iPhone in
instalments.
100000(1+r/100)^4 = x(1 + r/100)^3 + x(1 + r/100)^2 + x(1 + r/100)^1 + x
[Remember the above equation for solving questions of compound interest]
P + P*r*4/100 = (x + x*r*3/100) + (x + x*r*2/100) + (x + x*r*1/100) + x
[Remember the above equation for solving questions of simple interest]
Although for Simple Interest, we have a direct formula-
The annual instalment value is given by-
[Remember the above equation for solving questions of compound interest]
P + P*r*4/100 = (x + x*r*3/100) + (x + x*r*2/100) + (x + x*r*1/100) + x
[Remember the above equation for solving questions of simple interest]
Although for Simple Interest, we have a direct formula-
The annual instalment value is given by-
Type 1(Amount is given):
When the language the question is like "what annual payment will discharge a debt of ...", it means the Amount is given in the question.
In this question, the Amount(A) is given, i.e., Rs. 6450. So we can apply the formula directly
Here A = 6450, r = 5%, t = 4 years
Solution : 100*6450/[100*4 + 5*4*3/4]
Answer : 1500
Type 2 (Principal is given) :
Q. 5) A sum of Rs. 6450 is borrowed at 5% simple interest and is paid back in 4 equal annual installments. What is amount of each installment?
Here the sum is given. Sum means Principal.
But our formula requires Amount(A)
So we will calculate Amount from this Principal
A = P + SI = 6450 + 6450*5*4/100 = Rs. 7740
But our formula requires Amount(A)
So we will calculate Amount from this Principal
A = P + SI = 6450 + 6450*5*4/100 = Rs. 7740
Now put the values in the formula
A = 7740, r = 5%, t = 4
Annual instalment = 100*7740/(100*4 + 5*4*3/2)
A = 7740, r = 5%, t = 4
Annual instalment = 100*7740/(100*4 + 5*4*3/2)
Answer : Rs. 1800
Q. 6) 
"Sum borrowed" means Principal.
This question is of Compound Interest and hence we cant apply the direct formula. We will solve this question with the help of the equation we derived earlier.
P(1+r/100)^2 = x(1+r/100) + x
P(1 + 5/100)^2 = 17640(1 + 5/100) + 17640
Solve for P, you will get P = Rs.32800
Answer : (B)
Answer : (B)
Q. 7) What annual instalment will discharge a loan of Rs. 66000, due in 3 years at 10% Compound Interest?
Solution : Here again the question is of "Compound Interest" and hence we will solve it by equation :
Let each annual instalment be of Rs. x. Note that in this question, amount is given
Amount = x(1 + 10/100)^2 + x(1 + 10/100)^1 + x
66000 = x (1.21 + 1.1 + 1)
So x = Rs. 19939.58
Q. 8) What annual instalment will discharge a loan of Rs. 66000, due in 3 years at 10% Simple Interest?
I have just converted Q.7 into Simple Interest
Now we can either solve it by direct formula, or by equation
By Equation method :
66000 = (x + x*10*2/100) + (x + x*10*1/100) + x
66000 = x(3 + 0.2 + 0.1)
x = Rs. 20000
By Direct formula method :
A = 66000, t = 3, r = 10%
x = 100A/[100t + t(t-1)r/2]
x = 100*66000/[100*3 + 3*2*10/2]
x = 6600000/(300 + 30)
x = Rs. 20000
Q. 1) If a sum of money becomes 3 times itself in 20 years at simple interest. What is the rate of interest?
In such questions apply the direct formula-
Rate of interest = [100*(Multiple factor - 1)]/T
So R = 100*(3 - 1)/20
Answer : 10%
Note : With this formula you can find Rate if Time is given and Time if rate is given.
Q. 2)
In such questions, just write this line :
1st part : 2nd part : 3rd part = 1/(100+T1 * r) : 1/(100+T2 * r) : 1/(100+T3 * r)
= 1/(100+2 * 5) : 1/(100+3 * 5) : 1/(100+4*5)
= 1/110 : 1/115 : 1/120
= 23*24 : 22*24 : 23*22
Hence 1st part = (23*24)/ (23*24 + 22*24 + 23*22) * 2379
Answer : 828
Note : Surprisingly, such questions when asked mostly have this same data, i.e., R=5% and T1, T2, T3 = 2, 3, 4 years, respectively. Only the Principal is changed. So it would be wise if you can just mug this line :
1st part : 2nd part : 3rd part = 23*24 : 22*24 : 23*22
Based on the above line, you would be able to solve such questions in a jiffy.
But note that it will only work if the question is on Simple Interest. Like the below question appeared in SSC CGL Tier 2-
Q. 4
Here the data is same. i.e., R=5% and T1, T2, T3 = 2, 3, 4 years, respectively. So we will write directly -
1st part : 2nd part : 3rd part = 23*24 : 22*24 : 23*22
A received = (23*24)/ (23*24 + 22*24 + 23*22) * 7930
Answer : Rs. 2760
Q. 5) If a certain sum of money P lent out for a certain time T amounts to P1 at R1% per annum and to P2 at R2% per annum, then
The above formula is for calculating the Time, if the question asks
the rate, then just interchange the rate and time. Hence the formula
will become
R = (P1 - P2)*100/P2T1 - P1T2
R = (650-600)*100/600*6 - 650*4
R = 5%
Alternative Method :
You can solve such questions quickly without mugging the above formula. How?
The sum amounts to Rs. 600 in 4 years and Rs. 650 in 6 years. This means
the simple interest is Rs. 50 for 2 years (because the amount increased
from Rs. 600 to Rs. 650 in 2 years)
So the SI for 4 years is Rs. 100 (we have seen earlier than SI is
proportional. So if SI = 100 for 2 years, then SI = 150 for 3 years, SI =
250 for 5 years and so on)
Now SI = Rs. 100; P = 600-100 = Rs. 500; t = 4 years
R = 100*SI/(P*t) = 10000/2000
Answer: 5%
For CI, the formula is different
Difference between CI and SI
This topic is very important from examination point of view. Note the following things-
If t=1 year, then SI = CI
If t=2 years then difference between CI and SI can be given by two formulas-
If t=3 years then difference between CI and SI can be given by two formulas-
In all the above formulas we have assumed that the interest is compounded annually
Let us solve some CGL questions
Q. 6
A = P(1+r/100)^t
Given, A=1.44P t = 2 years
1.44P = P(1 + r/100)^2
r = 20%
Answer : (D)
Q. 7
Here the interest is compounded half yearly, so the formulas we mugged
earlier are of no use here. We will have to solve this question manually
SI = P*10*1.5/100 = 0.15P
CI = P(1 + 5/100)^3 - P = P(1.05^3 - 1)
Given CI - SI = 244
P(1.05^3 - 1) - 0.15P = 244P = Rs. 32000
Answer : (C)
Q. 8
Time = 2 years
Hence apply the formula: Difference(D) = R*SI/200
CI - SI = R*SI/200
CI - SI = (12.5/200)*SI
510 = 1.0625*SI [Since CI = Rs. 510]
SI = Rs. 480
Answer : (D)
Q. 9
CI for 1st year = 10% of 1800 = Rs. 180
CI for 2nd years = 180 + 10% of 180 = Rs. 198
Total = 180+198 = Rs. 378
Hence time = 2 years
Or you can apply the formula
A = P(1+r/100)^t
Answer : (B)
Q. 10
2.5 = P*R*2/100 - P*r*2/100
2.5 = 10R - 10r
R - r = 0.25
Answer : (D)
2.5 = 10R - 10r
R - r = 0.25
Answer : (D)
Q. 11
CI for 1st year = 5% of P = 0.05P
CI for 2nd year = 5% of P + 5% of (5% of P) = 0.05P + 0.0025P = 0.0525P
Total CI = 0.05P + 0.0525P = 0.1025P
Given, 0.1025P = 328
P = Rs. 3200
Answer : (C)
Note : You can solve this question by the formula A = P(1+r/100)^t as well
Q. 12
Note that in this question the CI for 2 years in not given, but the CI for the 2nd year is given.
CI for 2nd year = 10% of P + 10% of (10% of P) = 0.1P + 0.01P = 0.11P
Given, 0.11P = 132
P = Rs. 1200
Answer : (D)
Q. 13
Interest = Re. 1 per day = Rs. 365 for 1 year
SI = P*r*t/100
t=1, r=5%, SI = Rs. 365
So, P = 365*100/5 = Rs. 7300
Answer : (A)
Q. 14
We know
Difference = P(r/100)^2(r/100 + 3)
P = Rs. 10000, r = 5%, t = 3 years
Hence D = Rs. 76.25
Answer : (C)
Q. 15
We know, R = [(y/x)^(1/T2 - T1) - 1]*100
= [(1587/1200)^1/(3 - 1) - 1]*100
= [(1587/1200)^1/2 - 1]*100
= 3/20 * 100
= 15%
Answer : (B)
Q.16) The difference between CI and SI on a certain sum at 10% per annum for 4 years is Rs
1282. Find the sum.
SI for 4 years @10% = 40P/100 = 0.4P
CI for 4 years @10% = P(1.1^4 - 1) = 0.4641P
0.4641P - 0.4P = 1282
P = Rs. 20000
Q.17) A sum of money lent at compound interest for 2 years at 20% per annum would fetch
Rs. 482 more of interest was payable half yearly than if it was payable annually. The sum is ?
When interest is compounded half-yearly
A = P(1 + 10/100)^4
When interest is compounded annually
A = P(1 + 20/100)^2
Given, P(1 + 10/100)^4 - P(1 + 20/100)^2 = 482
Solve for P
Answer : Rs. 20000
Note : In the question, they have given the difference between the interest but we have
taken the difference between the Amounts because the difference in amount is equal to the
difference in interest.
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