## Chapter 13: SIMPLE INTEREST

### Introduction

Simple interest is calculated as
P * R * T / 100
Where P = Principal or the amount borrowed
R = Rate of interest
T = years
Note: The day on which money is deposited isn't counted but date on which money is withdrawn is counted.

• Relation Among Principal, Time, Rate Percent of Interest Per Annum and Total Interest :

• Amount = Principal + Total interest

• A = P + $$\frac{P * t * r}{100}$$

• Time = $$\frac{\text{Total interest}}{\text{Interest on the principal in one year}}$$. Thus, if we have the total interest as  300 and the interest per year is  50, then we can say that the number of years is 300/50 = 6 years.

COMPOUND INTEREST

• Let principal = P, time = n years and rate = r% per annum and let A be the total amount at the end of n years, then A = P[$$1 + \frac{r}{100}$$]n

• When compound interest is reckoned half-yearly. If the annual rate is r% per annum and is to be calculated for n years. Then in this case, rate = (r/2)% half-yearly and time = (2n) half-years then A = P[$$1 + \frac{r/2}{100}$$]2n

• When compound interest is reckoned quarterly. In this case, rate = (r/4)% quarterly and time = (4n) quarter years. \ As before, then A = P[$$1 + \frac{r/4}{100}$$]4n

• The difference between the compound interest and the simple interest over two years is given by $$\frac{P * r^2}{100^2}$$

Ans .

5

1. Explanation :

Let principal = x, time = t years
Then interest = x/4, rate = t%
Now, using the SI formula, we get
Interest = (Principal × Rate × Time)/100
so x/4 = (x × t × t)/100
so t2 = 25
and we get t = 5%

Ans .

912

1. Explanation :

Whenever it is not mentioned whether we have to assume SI or CI we should assume SI.
For any amount, interest for the 1st three years @ 6% SI will be equal to 6 × 3 = 18%
Again, interest for next 4 years will be equal to 7 × 4 = 28%.
And interest for next 4 years (till 11 years) –7.5 × 4 = 30%
So, total interest = 18 + 28 + 30 = 76%
So, total interest earned by him = 76% of the amount = (76*1200 / 100) = 912

Ans .

8.33%

1. Explanation :

Let principal = x, then interest = x, time = 12 years.
Using the formula, Rate = (Interest × 100)/Principal × Time
= (x × 100)/(x × 12) = 8.33%

Ans .

640

1. Explanation :

Let the principal be  x and rate = r%.
Then, difference in between the interest of 5 years and of 2 years equals to
 800 –  704 =  96
So, interest for 3 years =  96
Hence, interest/year =  96/3 =  32
So, interest for 2 years Æ 2 ×  32 =  64
So, the principal =  704 –  64 =  640

Ans .

4000

1. Explanation :

Let the rate be y% and principal be  x and the time be 3 years.
Then according to the question = (x(y + 4) × 3)/100 – (xy × 3)/100 = 480
so xy + 4x – xy = 160 × 100
so x = (160 × 100)/4 =  4000


Ans .

16

1. Explanation :

It trebles itself in 8 years, which makes interest equal to 200% of principal.
So, 200% is added in 8 years.
Hence, 400%, which makes the whole amount equal to five times of the principal, which will be added in
16 years.

Ans .

500

1. Explanation :

Using the formula, amount = Principal (1 + rate/100)time
605 = p(1 + 10/100)2 = p(11/10)2
p = 605(100/121) =  500

Ans .

5000 

1. Explanation :

Simple interest and compound interest for the first year on any amount is the same.
Difference in the second year’s interest is due to the fact that compound interest is calculated over the first
year’s interest also.
Hence, we can say that  72 = Interest on first year’s interest Æ 12% on first year’s interest =  72.
Hence, first year’s interest =  600 which should be 12% of the original capital. Hence, original capital =
5000 (this whole process can be done mentally).

Ans .

12,540 

1. Explanation :

Population at the end of 1 year will be Æ 10,000 + 10% of 10,000 = 11,000
At the end of second year it will be 11,000 + 20% of 11,000 = 13,200
At the end of third year it will be 13,200-5% of 13,200 = 12,540.

Ans .

6 

1. Explanation :

Let the rate of interest be = r%
Then, interest earned from  1200 at the end of year = (1200r)/100 =  12r
Again, interest earned from  1800 at the end of year = (1800/100) × (8/12) × 2r =  24r
So, total interest earned = 36r, which equals 216
fi r = 216/36 = 6%

Ans .

1380 

1. Explanation :

The annual interest would be  60. After 3 years the total value would be 1200 + 60 × 3 = 1380

Ans .

10

1. Explanation :

The interest earned per year would be 1500/3=500. This represents a 10% rate of interest

Ans .

2315.25

1. Explanation :

2100 + 5% of 2100 = 2100 + 105 = 2205 (after 1 year). Next year it would become:
2205 + 5% of 2205 = 2205 +110.25 = 2315.25

Ans .

1400, 10%

1. Explanation :

1400 increased by 10% gives 1540 increased by 10% gives  1694.


Ans .

5

1. Explanation :

Simple Interest for 2 years = 100 + 100 = 200.
Compound interest for 2 years: Year 1 = 5% of 2000 = 100.
Year 2: 5% of 2100 = 105 Æ Total compound interest =  205.
Difference between the Simple and Compound interest = 205 – 200 =  5

Ans .

10

1. Explanation :

Interest in 2 years =  240.
Interest per year =  120
Rate of interest = 10%

Ans .

4

1. Explanation :

12500 @ 10% simple interest would give an interest of  1250 per annum. For a total interest of 5000, it would take 4 years.

Ans .

5 

1. Explanation :

5% for 3 years (SI) = 15% of the amount; At the same time 4% SI for 4 years means 16% of the amount. The difference between the two is 1% of the amount. 1% of 500 =  5

Ans .

273

1. Explanation :

8% @ 700 =  56 per year for 3 years
7.5% @ 700 =  52.5 per year for 2 years
Total interest = 56 × 3 + 52.5 × 2 = 273

Ans .

cant say

1. Explanation :

8% of 800 for 4 years + 6% of 800 for 4 years = 64 × 4 + 48 × 4 = 256 + 192 = 448. However, we do not know the rate of interest applicable in the 5
th year and hence cannot determine the exact
simple interest for 9 years

Ans .

400

1. Explanation :

Simple interest @ 23% = 4600 × 2 = 9200
Compound interest @ 20%
20000 increase of 20% 24000 increase of 20% 28800
so 8800 compound interest.
Difference = 9200 – 8800 =  400

Ans .

331

1. Explanation :

1000 increase of 10% gives 1100 increase of 10% gives 1210 increase of 10% gives 1331.
Compound interest = 1331 – 1000 =  331

Ans .

2000 

1. Explanation :

Solve using options. Thinking about option (a):
2000 gives 2200 (after 1 year) gives 2420 (after 2 years) which gives us an interest of 420 as required
in the problem. Hence, this is the correct answer

Ans .

144

1. Explanation :

P × 7/6 × 7/6 = 196 gives P = (196 × 6 × 6)/7 × 7 = 144

Ans .

9.09

1. Explanation :

1331 × 1.090909 × 1.090909 × 1.090909 = 1331 × 12/11 × 12/11 × 12/11 = 1728. Hence, the rate of compound interest is 9.09%.

Ans .

6

1. Explanation :

Since compounding is half yearly, it is clear that the rate of interest charged for 6 months would be 3%. Thus we get 3300 gives 3% increase and we get 3399

Ans .

144000

1. Explanation :

The value of the van would be 196000 × 6/7 × 6/7 = 144000

Ans .

20

1. Explanation :

Solve through options:
10000 with 20% increase gives 12000 with 20% increase gives 14400 with 20% increase gives 17280.

Ans .

9274.2

1. Explanation :

12% per annum compounded quarterly means that the amount would grow by 3% every 3 months. Thus, 8000 gives 8000 + 3% of 8000 = 8240 after 3 months gives 8240 + 3% of 8240 = 8487.2 after 6
months and so on till five3 month time periods get over. It can be seen that the value would turn out to be 9274.2.


Ans .

5

1. Explanation :

For the last 5 years, the interest earned would be: 30% of 360 = 108. Thus, interest earned in the first 4 years would be  72 and   18 every year on an amount of  360- which means that the rate of
interest is 5%

Ans .

29100 

1. Explanation :

He will get 20000 + 45.5% of 20000 = 29100.
[Note: In this case we can take 13% simple interest compounded half yearly to mean 6.5% interest
getting added every 6 months. Thus, in 42 months it would amount to 6.5 × 7 = 45.5%

Ans .

66911.27 

1. Explanation :

50000 gives with a 6% interest per annum 53000 gives with a 6% interest per annum 56180 gives with a 6% interest per annum 59550.8 gives with a 6% interest per annum 63123.84 gives with a 6% interest per annum
66911.27

Ans .

126247.69 

1. Explanation :

100000 + 6% of 100000 (after the first 6 months) = 106000.
After 1 year: 106000 + 6% of 106000 = 112360
After 1 ½ years: 112360 + 6% of 112360 = 119101.6
After 2 years: 119101.6 + 6% of 119101.6 = 126247.69

Ans .

12.6 

1. Explanation :

(73/365) × 0.09 × 700 =  12.6.
(Since the time period is 73 days)

Ans .

1200

1. Explanation :

The average rate of interest he pays is 186 × 100/1500 = 12.4%.
The average rate of interest being 12.4%, it means that the ratio in which the two amounts would
be distributed would be 4:1 (using alligation). Thus, the borrowing at 12% would be  1200

Ans .

9 

1. Explanation :

If it doubles in 3 years, it would become 4 times in 6 year and 8 times in 9 years

Ans .

0.5 

1. Explanation :

The difference in Simple interest represents 1% of the amount invested. Since this difference has occurred in 2 years, annually the difference would be 0.5%.

Ans .

10 

1. Explanation :

It would take another 5 years to double again. Thus, a total of 10 years to become four fold

Ans .

30 

1. Explanation :

The sum becomes 4 times Æ the interest earned is 300% of the original amount. In 10 years the interest is 300% means that the yearly interest must be 30%

Ans .

400 

1. Explanation :

A × (1.02) + A = 808 × (1.02)
2 Æ A =  400

Ans .

100

1. Explanation :

The value would increase by 4% per year. To go to 5 times it’s original value, it would require an increment of 400%. At 4% SI it would take 100 years

Ans .

5

1. Explanation :

Total effective amount lent for 1 year
=  400 × 2 +  100 × 4 =  1200
Interest being  60, Rate of interest 5%

Ans .

375

1. Explanation :

Solve using options. If we try 500 (option b) for convenience, we can see that the difference between the two is  64 (as the SI would amount to 300 and CI would amount to 100 + 120 + 144
= 364). Since, we need a difference of only  48 we can realize that the value should be 3/4
th of 500. Hence, 375 is correct

Ans .

3725

1. Explanation :

The average Rate of interest is 6.8825%. The ratio of investments would be 1.1175: 1.8825 (@5% is to 8%). The required answer = 10000 × 1.1175/3 = 3725

Ans .

2

1. Explanation :

Interest per year =  25. Thus, an interest of  50 would be earned in 2 years.

Ans .

6 years and 8 months

1. Explanation :

42% on 2500 =  1050. The required answer would be: 1050/157.5 = 6 years and 8 months.

Ans .

700 

1. Explanation :

The difference would amount to 8% of the value borrowed. Thus 56 = 0.08 × sum borrowed in each case becomes Sum borrowed =  700.

Ans .

1640

1. Explanation :

882 × (1.05) + 882 = P × (1.05)2
Solve for P to get P = 1640

Ans .

data insufficient

1. Explanation :

The data is insufficient as we do not know the time period involved.


Ans .

1500

1. Explanation :

Based on the information we have, we can say that there would have been  30 extra interest per year. For 2% of the principal to be equal to  30, the principal amount should be  1500`

### Problems

Q. Find the simple interest on Rs. 3000 at rate of interest 6 1/4% for 73 days

A. 73 days = 73/365 = 1/5 yrs.

SI = (3000 * 25/4 * 1/5) / 100

Q. An amount loaned at interest rate 13.5% per annum becomes Rs. 2502.50 after 4 years. Find the sum.

A. P + SI = P + P * 13.5 * 4 / 100 (Adding P on both sides)

2502.5 = P ( 1 + 54/100)

2502.5 = P (1.54)

P = 2502.5 / 1.54 = 1625

Q. A borrowed money at interest rate 6% for first two years, 9% for next three years and 14% for period beyond 5 years. If he pays a total interest of Rs. 11400 after 9 yrs how much did he borrow?

A. 11400 = (P *6*2/100) + (P*9*3/100) + (P*14*4/100)

11400 = (12+27+56)P / 100

11400 * 100 / 95 = P

P = 12000

Q. A certain sum of money amounts to Rs 1008 in 2 years and Rs. 1164 in 3.5 yrs.Find sum and RI

A. SI for 1.5 yrs = 1164 - 1008 = 156

SI for 1 yrs = 156 * 2 / 3 = 104 and two years = 208.

Principal = 1008 - 208 = 800

Use simple interest formula to get RI.

Q. A sum of 1500 is lent in 2 parts where one is at 8% and second is at 6%. If the total annual income is Rs. 106, find money lent at each rate.

A. (x*1*8/100) + ((1500-x) * 1 * 6 / 100 ) = 106

8x/100 + (9000 - 6x)/100 = 106

8x +9000 - 6x = 10600

2x = 1600

x=800

### Compound Interest

When P = principal, n = years and R = rate of interest compounded annually

The Amount = P (1+R/100)^n

When P = principal, R = rate of interest compounded half yearly the
Amount = P ( 1 + (R/2) / 100)^2n

When P = principal, R = rate of interest compounded quarterly the
Amount = P ( 1 + (R/4) / 100)^4n

When P = principal, R = rate of interest compounded annually but the time is in fraction then like 3 2/5 yrs
Amount = P ( 1 + R / 100)^3 * (1 + (2R/5)/100)

When P = principal, R = rate of interest compounded annually but is different for first year R1, second year R2 and third year R3 then
Amount = P ( 1 + R1 / 100) * ( 1 + R2 / 100) * ( 1 + R3 / 100)

Present worth of Rs. x due n years hence is
Present worth = x / (1 + R/100)^n

### Solved Problems

Q. Find CI on Rs.7500, compounded annually at RI of 4% for 2 years.

A. Amount = 7500 * (1 + 4/100)^2

Then amount - 7500 gives CI.

Q. CI is compounded half yearly, principal = Rs. 10000 in rate 4% for 2 years.

A.  Amount = 10000( 1+ 2/100)^4

Amount = 10824.32

CI = Amount - principal = 10824.32 - 10000 = 824.32

Q. Difference between SI and CI accrued on an amount of Rs. 18000 in 2 years is Rs.405. What is the RI.

A. P( 1+ (R/100) )^n - P*R*T/100 = 405

{18000 ( 1 + (R/100))^2 - 18000}  - (18000 * R * 2 / 100) = 405

Solving this you can get R.

Q. Divide 1301 between A and B such that the amount of A after 7 years is equal to amount in B after 9 years. Interest is compounded at 4%.

A. Let the amount be 'x' and 1301 - x.

x(1+4/100)^7 = (1301-x)(1+4/100)^9

Solving this we can get 'x'.

Q. A sum of money amounts to Rs.6690 after 3 years and Rs. 10035 after 6 years on CI. Find the sum.

A. P(1+R/100)^3 = 6690 ; P(1+R/100)^6 = 10035

Dividing first eqn by second eqn we get (1+R/100)^3 = 10035/6690 = 3/2

Substituting this value in first equation we get P = 6690 * 2 / 3 = 4460.

Q. A sum doubles itself in 9 years how many will it take to become 8 times.

A. P(1+R/100)^9 = 2P

(1+R/100)^9 = 2

Now finding P(1+R/100)^n = 8P

we need to get (1 + R/100)^n = 8 but 8 = 2^3

(1+R/100)^n = (1 + R/100)^ 9 ^3

we know that A^b^c = A^b*c

so (1+R/100)^n = (1 + R/100)^ 9*3 = (1 + R/100)^27

n = 27

### CAT Problems

Q.A sum of money compounded annually becomes Rs.625 in two years and Rs.675 in three years. The rate of interest per annum is

1. 7
2. 8
3. 6
4. 5

Ans.b

### Quiz

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