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# Quantitative Aptitude > Simple and Compound Interest

Interest is earned when money is given to someone else to use, for example if it is deposited with a bank
or loaned to an individual. There are two common ways of calculating interest.

Simple Interest

Simple Interest is interest earned on the principal only. In other words, interest income will be distributed
to investors and no further interest will be earned from interest income. The equation for finding simple interest is:

A = P + (P x r x t)
where
A = Total amount due to the investor
P = Principal
r = Annual interest rate (decimal)
t = Number of years (time)

Bozo put As. 1,000 in the bank for 6 months at a 10% interest rate. Find how much money will he have in
his account at the end of six months.
A = P + (P x r x t)
A = 1,000 + (1,000 x 0.1 x 0.5)
A = 1,000 + (100 x 0.5)
A = 1 ,000 + 50
A = As. 1 ,050
The amount of money Bozo will have in his account at the end of six months is Rs. 1,050

Compound Interest

In simple interest, you earn interest payments from the principal only. In Compound Interest, you reinvest
the interest payments and earn interest on the interest amount as well.

Let us say you are investing Rs. 1,000 at 10% compound interest. After a specified period of time, called
compounding period (let us assume this is 1 year for this problem), you calculate the simple interest on the principal, i.e. 10% of Rs. 1000 = Rs. 100 and add it to the principal giving you Rs. 1,000 + Rs. 100 = Rs. 1,100. This new amount will be the principal for the next time period of1 year and so on. Thus for the second year, the interest will be 10% of Rs. 1100 = Rs. 110.

The period of compounding is important because as the period of compounding becomes lesser, the interest amount joining the capital becomes higher.

• If an amount is invested under compound interest for a time period that is less than the compounding
period, then it is equivalent to investing the money under simple interest at the same interest rate.

• An amount invested under compound interest will always earn more interest than the same amount
invested under simple interest at the same interest rate, provided the time it is invested for is more than the compounding period.

• Intuitively, we can see that a lower compounding period means a, higher totaI amount due to the investor since we are adding the interest to the principal more often.

The equation for finding compound interest is:

A = P (1 + r/k) kxn
Where
A = Amount due to the investor
P = Principal
r = Annual rate (in decimal or per unit)
k = number of times compounding per year
n = number or years
John deposited Rs. 1,000 at a 10% interest rate for 5 years. How much more money will John have in his
account from compound interest than from simple interest if it is compounded semi-annually?

Simple Interest:

A = P + (P x r x t)
A = 1,000 + (1,000 x 0.1 x 5)
A = 1,000 + (100 x 5)
A = 1,000 + 500
A = Rs. 1,500

Compound Interest:

A = ( 1 + r/k ) kxn

A = ( 1 + 0.1/2 ) 2*5

A = 1,000 (1 + 0.05) 10

A = 1,000 x 1.62889
A = Rs. 1,628.89
Rs. 1,628.89 (compound interest) – Rs.1,500 (simple interest) = Rs. 128,89. Thus, John will make Rs.
128.89 more from compound interest than from simple interest.

THE RULE OF 72

The rule of 72 is a quick way to show how long it will take to double your money under

Compound interest.
The equation for the rule of 72 is:
Number of years for money to double = 72 / Anual Interest rate
At 8% interest, it will take 72/8 = 9 years for your money to double.

Here are more examples:

At 6%, it will take 12 years ( 72/6 = 12 )

At 12%, it will take 6 years ( 72/12 = 6 )

The rule of 72 is a short cut to estimate the magic of compound interest that makes your money grow.

•  Remember that the rule of 72 is an approximation and its accuracy reduces as the interest rate becomes high.

Compound Interest Under Complex Conditions

For a principal amount P, we have:

Net Present Value

This is an important concept used in financial analysis. The basic principle underlying this concept is that
a rupee earned or paid today is not of the same value as a rupee earned or paid after a time period of say n years.

The money received or paid today can be invested by the receiver to earn interest.

Suppose Ram had to pay Re1 to Shyam after an year, he could also pay Re 0.91 to Shyam now if 10%
p.a. rate of interest is acceptable to both the parties.

If a company is expected to earn Rs A1 in year 1, Rs A2 in year 2, Rs A3 in year 3…. and Rs An in year
n, and we assume interest rate is r% p.a. the net present value of earnings of this company is:

If the earnings are same every year (A1 = A2 = …………..An = A)

This is used to calculate monthly instalments in instalment schemes offered by dealers.
If the value of goods (minus down payment if any) = V

Then V = (EMI) [(1+r) n-1] / [r] [1+r] n

So EMI = Vr(1+ r) n/ (1+ r) n-1

Interest Rate On Instalment
If R is the rate of interest % p.a., then R = (24Ix 100)/(N(F+L))
I = instalment charge
N = Number of instalments
F = Principal left during the first month = Cash price – Down payment if any
L = principal left during the last month = Cash rice = Down payment if any – (N – 1) x I

POPULATION FORMULA

If population is P and annual increase is r%, then population at the end of t years is P (1 + r/100) t
If population is P and annual decrease is r%, then population at the end of t years is P (1 - r/100) t

SOLVED EXAMPLES

QUESTION

Find the simple interest on Rs 400 for 5 years at 6 per cent

SOLUTION

SI = 400 x 5 x 6/100 = Rs 120

QUESTION

The simple interest on a sum of money is 1/9 of the principal, and the number of year is equal to the rate per cent per annum. Find the rate per cent.

SOLUTION

Let Principal = P, time = t years, rate = t
Then P t t/100 = P/9

So t = 100/9  So t = 10/3 = 3 1/3%

QUESTION

A sum of money doubles itself in 10 years at simple interest. What is the rate of interest?

SOLUTION

Let the sum be Rs 100.
After 10 years it becomes Rs 200.

Then rate = (100 x 100) / (100 x 10) = 10%

QUESTION

A sum of money doubles itself in 7 yrs. In how many years will it become fourfold?

SOLUTION

Rate = (100 * 100 ) / Pt = 100 / 7 So Time = 100 ( 4-1 ) / 100 / 7 = 21 years

QUESTION

A lent Rs 600 to B for 2 years, and Rs 150 to C for 4 years and received altogether from both Rs 90 as interest. Find the rate of interest, simple interest being calculated.

SOLUTION

Rs 600 for 2 years = Rs 1200 for one year.
And Rs 150 for 4 years = Rs 600 for one year.
Int. = Rs 90

So Rate = (90 * 100) / (1800 * 1) = 5%

QUESTION

The rate of interest for the first 2 yrs is 3% per annum, for the next 3 years is 8% per annum and for the period beyond 5 years 10% per annum. If a man gets Rs. 1520 as a simple interest for 6 years, how much money did he deposit?

SOLUTION

Let his deposit = Rs 100
Interest for first 2 yrs = Rs 6
Interest for next 3 yrs = Rs 24
Interest for the last year = Rs 10
Total interest = Rs 40
When interest is Rs 40, deposited amount is Rs 100
So 100/40 x 1520 = Rs 3800

QUESTION

A sum of money triples itself in 20 yrs at SI. Find the rate of interest.

SOLUTION

Rate = 100 (3 –1)/20 =10%

Note: A generalised form can be shown as:
If a sum of money becomes ‘x’ times in ‘t’ years at SI, the rate of interest is given by 100 (x –1) / t%

QUESTION

In what time does a sum of money become four times at the simple interest rate of 5% per annum?

SOLUTION

Using the formula,
Time = 100 (Multiple number of prinicipal –1) / Rate
= 100 (4 –1) / 5  = 60 yrs

QUESTION

A certain sum of money amounts to Rs 756 in 2 yrs and to Rs 873 in 3.5 yrs. Find the sum and the rateof interest.

SOLUTION

P + SI for 3.5 yrs = Rs 873
P + SI for 2 yrs = Rs 756
On subtracting, SI for 1.5 yrs = Rs 117
Therefore, SI for 2 yrs = Rs (117/1.5) x 2= Rs 156
So  P = 756 – 156 = Rs 600
and rate = (100 x 156) / ( 600 x 2 ) = 13% per annum.

QUESTION

A sum was put at SI at a certain rate for 2 yrs. Had it been put at 3% higher rate, it would have fetched Rs 300 more. Find the sum.

SOLUTION

Let the sum be Rs x and the original rate be y% per annum. Then, new rate = (y + 3)% per annum.
So  x (y +3) x (2/100) – x(y) x (2/100) = 300
xy + 3x - xy = 15,000 or, x = 5000
Thus, the sum = Rs 5000

QUESTION

At a certain rate of simple interest Rs 800 amounted to Rs 920 in 3 years. If the rate of interest be increased by 3%, what will be the amount after 3 years?

SOLUTION

First rate of interest = (120 x 100) / ( 800x3 ) = 5%
New rate = 5 + 3 = 8%
So  New interest = (800 x 3 x 8) / 100 = Rs 192
So  New amount = 800 + 192 = Rs 992.

QUESTION

Rs 4000 is divided into two parts such that if one part be invested at 3% and the other at 5%, the annual interest from both the investments is Rs 144. Find each part.

SOLUTION

Let the amount lent at 3% rate be Rs x, then
3% of x + 5% of (4000 – x) = 144
or, (3x + 5) x (4000 – 5x) = 14400
or, 2x = 5600 So x = 2800
Thus, the two amounts are Rs 2800 and Rs (4000 – 2800) or Rs 1200.

QUESTION

A sum of Rs 2600 is lent out in two parts in such a way that the interest on one part at 10% for 5 years is equal to that on another part at 9% for 6 years. Find the two sums.

SOLUTION

Each Interest = (1st Part) x (5) x (10/100) = (2nd Part) x (6) x (9/100)
or, 1st Part /2nd Part = (6) x (9/5) x (10) = 27/25 = 27 : 25
So  1st part = (2600) / (27 + 25) x 27 = Rs 1350
and 2nd part = 2600 – 1350 = Rs 1250

QUESTION

A certain sum of money amounts to Rs 2613 in 6 yrs at 5% per annum. In how many years will it amount to Rs 3015 at the same rate?

SOLUTION

Use the formula:
Principal = (100 x Amount) / (100 + rt) = (100 x 2613) / (100 + 30) = (100 x 2613) / 130 = Rs 2010

Again by using the same formula:

2010 = (100 x 3015) / (100 + 5t )
or, 100 + 5t = (100 x 3015) / 2010
So  t = 1/5 [{(100 x 3015) – (100 x 2010)} / 2010]
= 100 x (3015 – 2010) / (2010 x 5 ) = (100 x 1005) / (2010 x 5) = 10 years.

QUESTION

A person lent a certain sum of money at 4% simple interest; and in 8 years the interest amounted to Rs 340 less than the sum lent. Find the sum lent.

SOLUTION

Let the sum be Rs. x.
So  Interest = (x) x (8) x (4/100) = 32x / 100
(x) – (32x / 100) = 68x / 100
When interest is 68x / 100 less, the sum is Rs x.
So when interest is 340 less, the sum is (x / 68x)  x (100) x (340) = Rs 500

QUESTION

The difference between the interest received from two different banks on Rs 500 for 2 yrs is Rs 2.5. Find the difference between their rates.

SOLUTION

I 1 = (500) x (2) x (r 1/100) = 10 r 1
I 2 = (500) x (2) x (r 2/100) = 10 r 2
I 1 – I 2 = 10r 1 – 10r 2 = 2.5
or, r 1 – r 2 = 2.5/10 = 0.25%

QUESTION

The simple interest on a certain sum of money at 4% per annum for 4 yrs is Rs 80 more than the interest on the same sum for 3 yrs at 5% pr annum. Find the sum.

SOLUTION

Let the sum be Rs X , then at 4% rate for 4 yrs the simple interest  = (x) x (4) x (4) / 100

= Rs 4x / 25

At 5% rate for 3 yrs the simple interest =
(x) x (5) x (3) / 100 = Rs 3x / 20
Now, we have,
(4x / 25) – (3x / 20) = 80
or,
(16x - 15x) / 100 = 80
So  x = Rs 8000

QUESTION

Ramesh borrows Rs 7000 from a banks at SI. After 3 yrs he paid Rs 3000 to the bank and at the end of 5 yrs from the date of borrowing he paid Rs 5450 to the bank to settle the account. Find the rate of interest.

SOLUTION

Any sum that is paid back to the bank before the last instalment is deducted from the principal and not from the interest. Thus, Total interest = (Interest on Rs 7000 for 3 yrs) + Interest on (Rs 7000 – Rs 3000 =) Rs 4000 for 2 yrs.
Or, (5450 + 3000 – 7000) = {(7000) x (3) x (r/100)} + {(4000) x (2) x (r/100)}
or, 1450 = 210r + 80r  So  r = 1450/290 = 5%.

QUESTION

At what rate per cent compounded yearly will Rs 80,000 amount to Rs 88,200 in 2 yrs?

SOLUTION

We have 80,000 ( 1 + r/100 ) 2 = 88,200

Or, ( 1+ r/100 ) 2 = 88200 / 80000 = 441 / 400

= ( 21 / 20 ) 2

Or, (1 + r/100) = 21/20    So r = 5%

QUESTION

A sum of money placed at compound interest doubles itself in 4 yrs. In how many years will it amount to eight times itself?

SOLUTION

We have P (1 + r/100) 4 = 2P       So (1 + r/100) 4 = 2
Cubing both sides, we get
(1 + r/100) 12 = 2 3 = 8           or, P (1 + r/100) 12 = 8P
Hence the required time is 12 yrs.

QUESTION

A sum of money at compound interest amounts to thrice itself in three years. In how many years will it be 9 times itself?

SOLUTION

Detail Method:
Suppose the sum = Rs x
Then, we have
3x = x ( 1+ r/100 ) 3 or 3 = ( 1+ r/100 ) 3
Squaring both sides

(3) 2 = {( 1 + r/100 ) 3} 2

or 9 = ( 1 + r/100 ) 6

So the sum x will be 9 times in 6 years.

QUESTION

At what rate per cent compound interest does a sum of money become nine-fold in 2 years?

SOLUTION

Detail method: let the sum be Rs x and the rate of compound interest be r% per annum; then

QUESTION

In what time will Rs 390625 amount to Rs 456976 at 4% compound interest?

SOLUTION

QUESTION

Find the least number of complete years in which a sum of money at 20% CI will be more than doubled

SOLUTION

QUESTION

The CI on a certain sum is Rs 104 for 2 yrs and SI is Rs 100. What is the rate per cent ?

SOLUTION

Difference in CI and SI = 104 – 100 = Rs 4.
Therefore by using the formula
Rate  = (2) x (Diff) x (100) / SI = (2) x (4) x (100) / 100
= 8%

QUESTION

An amount of money grows upto Rs 4840 in 2 yrs and upto Rs 5324 in 3 yrs on compound interest. Find the rate per cent.

SOLUTION

We have,

P + CI of 3 yrs = Rs 5324———(1)
P + CI of 2 yrs = Rs 4840———(2)
Subtracting (2) from (1), we get
CI of 3rd year = 5324 – 4840 = Rs 484.
Thus, the CI calculated in the third year which is Rs 484 is basically the amount of interest on the amount
generated after 2 years which is Rs 4840.
So  r = 484 x 100/4840 x 1 = 10%

QUESTION

A certain amount of money at compound interest grows upto Rs 51168 in 15 yrs and upto Rs 51701 in 16 yrs. Find the rate per cent per annum.

SOLUTION

Rate =  (51701 - 51168) x 100 / 51168 = (533 x 100) / 51168

= 100 / 96 = 25 / 24 = 1 1/24 %

QUESTION

Find the compound interest on Rs 18,750 in 2 yrs, the rate of interest being 4% for the first year and 8% for the second year.

SOLUTION

After first year the amount = 18750 {(1 + 4) / 100} = 18750 (104/100)

After 2nd year the amount = 18750 (104/100) (108/100)
= 18750 (26/25) (27/25) = Rs 21060
So  CI = 21060 – 18, 750 = Rs 2310.

QUESTION

Rs 4800 becomes Rs 6000 in 4 years at a certain rate of compound interest. What will be the sum after 12 years?

SOLUTION

The above equation shows that Rs 4800 becomes Rs 9375 after 12 years.

QUESTION

A man borrows Rs 3000 at 10% compound rate of interest. At the end of each year he pays back Rs 1000. How much amount should he pay at the end of the third year to clear all his dues?

SOLUTION

The general formula for the above question may be written as: If a man borrows Rs P at r% compound interest and pays back Rs A at the end of ach year, then at the end of the nth year he should pay Rs

In the above case:

QUESTION

A sum of money is lent out at compound interest rate of 20% per annum for 2 years. It would fetch Rs 482 more if interest is compounded half-yearly. Find the sum.

SOLUTION

Suppose the sum is Rs P.

QUESTION

The simple interest on a sum at 4% per annum for 2 yrs is Rs 80. Find the compound interest on the same sum for the same period.

SOLUTION

Rate = (2 xDiff inCI& SI) / (SI) x 100
or, diff in CI and SI = Rate x SI
So  Difference in CI and SI = (4) x (80/2) x (100) = 1.6
So  CI = 80 + 1.6 = Rs 81.6

QUESTION

The compound interest on a certain sum of money for 2 years at 10% per annum is Rs 420. Find the simple interest at the same rate and for the same time.

SOLUTION

QUESTION

The compound interest on a certain sum for 2 yrs is Rs 40.80 and simple interest is Rs 40.00. Find the rate of interest per annum and the sum.

SOLUTION

A little reflection will show that the difference between the simple and compound interests for 2 yrs is the interest on the first year’s interest.
First year’s SI = Rs 40/2 = Rs 20
CI – SI = Rs 40.8 – Rs 40 = Re 0.80
Interest on Rs 20 for 1 year = Rs 0.80
So  Interest on Rs 100 for 1 yr = Rs 80 x {100 / (100x20)} = Rs 4
So  Rate = 4%
Now, principal P is given by
P = (100) x I / (tr) = (100) x {40 / (2 x 4} = Rs 500

QUESTION

If the CI on a certain sum for 2 yrs at 3% be Rs 101.50, what would be the SI?

SOLUTION

CI on 1 rupee = ( 1+ 3/100 ) 2 - 1 = (103/100) 2 – 1 = Re 609/10000

SI on 1 rupee = Rs (2) x (3 /100) = Re 6/100
So SI/CI = (6100) x (10000/609) = 200/203
So SI = 200/203 of CI = (200/203) x (101.5) = Rs 100

QUESTION

If the simple interest on a certain sum of money for 3 yrs at 5% is Rs 150, find the corresponding CI.

SOLUTION

Whenever the relationship between CI and SI is asked for 3 yrs of time, we us the formula:

QUESTION

The difference between the compound interest and the simple interest on a certain sum of money at 5% per annum for 2 years is Rs 1.50. Find the sum.

SOLUTION

Sum = 1.5 (100 / 5) 2 = 1.5 x 400 = Rs 600

QUESTION

Divide Rs 3903 between A and B, so that A’s share at the end of 7 yrs may equal B’s share at the end of 9 yrs, compound interest being at 4%.

SOLUTION

We have, (A’s share at present) (1 + 4 /100) 7
= (B’s share at present) (1 + 4 / 100) 9
So A’s share at present /B’s share at present = (1 + 4 / 100 ) 2 = (26/25) 2 = 676/625
Dividing Rs 3903 in the ratio 676 : 625
A’s present share = {(676) / (676 + 625)} x 3903 = Rs 2028
B’s present share = Rs 3903 – Rs 2028 = Rs 1875

QUESTION

The difference between the simple and the compound interests on a certain sum of money for 2yrs at 4% per annum is Re 1. Find the sum.

SOLUTION

Sum = Difference (100/r) 2 = 1(100/4) 2 = Rs 625

QUESTION

If the difference between CI and SI on a certain sum of money for 3yrs at 5% p.a. is Rs 122, find the sum

SOLUTION

Sum = [(122) x (100) x (100)x (100) / {5 2 (300 + 5)}] = Rs 16,000

QUESTION

Find the difference between CI and SI on Rs 8000 for 3 yrs at 2.5% p.a

SOLUTION

Difference = (Sum) x (r 2) (300 + r) / (100) 3

= 8000 x 2.5 x 2.5 (300 + 2.5) / 100 x 100 x 100
= (8 x 25 x 25 x 3025) / (100 x 100 x 100) = 121/8 = Rs 15 1/8

QUESTION

Find the compound interest on Rs 10000 for 3 years if the rate of interest is 4% for the first year, 5% for the second year and 6% for the third year.

SOLUTION

The Compound Interest on Rs x in ‘t’ years if the rate of interest is r1% for first year, r2% for the second year …. .and rt% for the tth year is given by