###
__Quantitative Aptitude Notes: Compound Interest__

__Quantitative Aptitude Notes: Compound Interest__

**Introduction:**

**Compound Interest:**The difference between the amount and the money borrowed is called the compound interest for given period of time.

**Important Formula:**

**Case 1:**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+ (r/100)]**

^{n};**CI = P*{[1+ (r/100)]**

^{n}-1};**Case 2:**When compound interest reckoned half yearly, then r% become r/2% and time n become 2n;

**A= P*[1+ (r/2*100)]**

^{2n};**Case 3:**for quarterly,

**A= P*[1+ (r/4*100)]**

^{4n};
The difference
between compound interest and simple interest over two years is given by

**Pr**

^{2}/100^{2}or P(r/100)^{2};
The
difference between compound interest and simple interest over three years is
given by

**P(r/100)**

^{2}*{(r/100)+3}
When Rates
are different for different years, say R1%, R2%, R3% for 1st, 2nd and 3rd year respectively,
Then total amount is given by

**P [(1 + R1)/100][(1 + R2)/100][(1 + R3)/100]**

Present
worth of Rs. x due n years hence is given by

**x/(1 + R/100)**

**Example:**If the difference between the simple interest and the compound interest on the same sum at 5% per annum for two years is Rs. 25, what is the sum?

**Solution:**

Given,
difference, d= Rs. 25; R = 5%; P =?

Difference=
P(r/100)2 => 25 =P (5 /100)2 => P = (25*100*100)/(5*5) => P = Rs.
10000.

**Depreciation of Value:**

The value
of machine or any other article subject to wear and tear decreases with the
time. This decrease is called its depreciation. Thus if V0 is the value at a
certain time and r% per annum is the rate of depreciation per year, then the value

**V1 at the end of t years V1 = V0 *[1-(r/100)]**

^{t}**Value of machine t years ago = V0 / [1-(r/100)]**

^{t}**Population:**

Concept of
Depreciation and Population is the just expansion of concept of compound
interest.

i) Let the
population of a town be P now and suppose it increases at the rate of R% per
annum, then:

**a) Population after n years = P*[1+ (r/100)]**

^{n};**b) Population n years ago = P/[1+ (r/100)]**

^{n};
ii) If the
present population P decreases at the rate of R% per annum, then:

**a) Population after n years = P*[1- (r/100)]**

^{n};**b) Population n years ago = P/[1- (r/100)]n;**

iii) If the
present population P increase at the rate of R1% for the first year and
decreases at the rate with rate of R2 for second year and again increases with
the rate of R3% for the third year, then:

**a) Population after 3 years = P[1+(R1/100)]*[1-(R2/100)]*[1+(R3/100)]**

**b) If the same is reversed from now, then: Population 3 years ago,= P/ {[1+ (R1/100)]*[1- (R2/100)]*[1+ (R3/100)]}.**

__MEMORY BASED SOLVED PROBLEMS BASED ON VARIOUS TYPES__**Type 1 – Difference between S.I and D.I**

**1. A person lent out a certain sum on simple interest and the same sum on compound interest at a certain rate of interest per annum. He noticed that the ratio between the difference of compound interest and simple interest of 3 years and that of 20 years is 25: 8. The rate of interest per annum is:**

**Solution:**

Let the
principal be Rs.P and rate of interest be R% per annum.

Difference
of C.I and S.I for 2 years = [P × (1 + R/100)

^{2}– P] – (PT2/100) = PR^{2}/10^{4}
Difference
of C.I and S.I for 2 years = [P × (1 + R/100)

^{3}– P] – (PT3/100) = PR^{2}/10^{4}[(300 + R)/100]
Therefore,
PR2/104 * ((300 + R) / 100) = 25/8 => (300 + R) / 100 = 25/8

=> R =
100/8 = 12 ½ %

**2. The difference between simple interest and compound interest on a sum for 2 years at 8%, when the interest is compounded annually Rs.16. If the interest was compounded half-yearly, the difference in two interests would be nearly:**

**Solution:**

For 1st year,
S.I = C.I

Thus, Rs.16
is the S.I on S.I for 1 year, which at 8% is thus Rs.200

i.e. S.I on
the principal for 1 year is Rs.200

Principal =
Rs. (100×2008×1)=Rs.2500

Amount for
2 years, compounded half-yearly = Rs. [2500×(1+4100)4]=Rs.2924.64

C.I =
Rs.424.64

Also, S.I =
Rs.(2500×8×2100)=Rs.400

Hence,
[(C.I)-(S.I)] = Rs. (424.64-400)=Rs.24.64

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

**Solution:**

Let the sum
be Rs . 100 then

S.I. =
Rs.(100 x 5 x 2/100)= Rs. 10

C.I = Rs.
[{100 x (1 + 5/100)2} - 100] = Rs. 41/4

∴ Difference
between C.I and S.I = Rs. (41/4 - 10) = Re. 0.25

⇒ 0.25: 150:
: 100: P

∴ P = (1.50
x 100) / 0.25 = Rs. 600

**4. The compound interest on a sum for 2 years is Rs. 832 and the simple interest on the same sum for the same period is Rs. 800. The difference between the compound and simple interest for 3 years will be ?**

**Solution:**

S.I. for
first year = Rs. 400

S.I. on Rs
.400 for 1 year =Rs. 32

∴ Rate =
(interest x 100) / (principle x time) = (100 x 32)/(400 x 1) = 8%

Hence, the
difference for 3rd year is S.I. on Rs. 832

= Rs.(832 x
8/100) = Rs. 66.56

∴ Total
difference = Rs. (32 + 66.56) = Rs. 98.56

**5. What is the difference between the compound interest and simple interest calculated on an amount of Rs. 16200 at the end of 3 yr at 25% pa? (Rounded off to two digits after decimal)**

**Solution:**

Let
required difference be Rs. D.

By formula,

D = P x
(R/100)2 x [(3 + R)/100]

= 16200 x
(25/100)2 x [(3 + 25)/100]

= 16200 x
625/10000 x 13/4

= 162 x 625
x 13/4 x 100 = Rs. 3290.63

**6. If the difference between the compound interest and simple interest on a sum of money at 5% pa for 2 yr is Rs. 16, then find the simple interest?**

**Solution:**

Given, CI -
SI = 16, R = 5%, n = 2 yr and SI = ?

According
to the formula,

CI - SI =
(SI x R)/200

⇒ 16 = (SI x
5)/200

⇒ 16 = SI/40

∴ SI = Rs.
640

**Type 2 – Finding P, R and T**

**1. A bank offers 5% compound interest calculated on half-yearly basis. A customer deposits Rs.1600 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by way of interest is**

**Solution:**

Amount =
Rs.[1600 × (1 + {5/2 × 100})

^{2}+ 1600 × (1 + {5/2 × 100})] = Rs.[1600 × 41/40 × 41/40 + 1600 × 41/40]
= Rs.[1600 ×
41/40 (41/40 + 1)] = Rs. [(1600 × 41 × 81)/(40 × 40)] = Rs.3321

Therefore
CI = 3321 – 3200 = Rs.121

**2. Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is:**

**Solution:**

C.I.=Rs.
4000(1+10/100)^2 – 40 = Rs. 840

Sum= Rs.
(420 × 100)/(3×8) = Rs. 1750

**3. The compound interest on a sum of Rs. 4000 becomes Rs. 630.50 in 9 months. Find the rate of interest, if interest is compounded quarterly.**

**Solution:**

Given, P =
Rs. 4000, n = 9 months = 3/4 yr and C.I = Rs. 630.50

Amount = P
+ CI = 4000 + 630.50 = Rs. 4630.50

According
to the formula,

⇒ Amount =
P[(1+R/(100 x 4)]

^{4n}
⇒ 4630.50 =
4000(1+R/400)

^{4 x 3/4}
⇒ 4630.50 =
4000[(400 + R)/400]

^{3}
⇒ 4630.50/4000
= [(400 + R)/400]

^{3}
⇒ 9261/8000
= [(400 + R)/400]

^{3}
⇒ (21/20)3 =
[(400 + R)/400]

^{3}
⇒ [(400 +
R)/400]/400 = 21/20

⇒ 400 + R =
21 x 20 = 420

∴ R = 420 -
400 = 20%

**4. An amount is invested in a bank at compound rate of interest. The total amount, including interest, after first and third year is Rs. 1200 and Rs. 1587, respectively. What is the rate of interest?**

**Solution:**

Let amount
be Rs. P and rate of interest be R % annually.

According
to the question, Amount after 1st yr = Rs. 1200

⇒ P * (1 +
R/100) = 1200 ... (i)

Amount
after 3rd yr = 1587

⇒ P * (1 +
R/100)

^{3}= 1587 ...(ii)
On dividing
Eq. (ii) from Eq. (i), we get

(1 + R/100)

^{2}= 1587/1200 = 529/400
⇒ 1 + R/100
= 23/20 ⇒ R/100 =
3/20 => ∴ R = 15 %

**5. If the compound interest on a certain sum for 2 years at 12.5% per annum is 170, the simple interest is?**

**Solution:**

Let the principle
is P

so compound
interest = P x ( 1 + (12.5/100))

^{2}- P = 170
⇒ P x
(112.5/100) x (112.5/100) - P = 170

⇒ P
[12656.25 - 10000] = 170 x 10000

∴ P = (170 x
10000) / 2656.25

Simple
interest SI = (P x T x R)/100 = [{(170 x 10000) / 2656.25} x 2 x 12.5] / 100 =
160

**6. A sum of money amounts to Rs. 10648 in 3 years and Rs. 9680 in 2 years. The rate of interest is?**

**Solution:**

Let P be
the principle and R% per annum be the rate. Then,

P(1 +
R/100)

^{3}= 10648 ....(i)
and P(1 +
R/100)

^{2}= 9680 .....(ii)
on dividing
(i) by (ii), we have

⇒ (1 +
R/100) = 10648/9680

⇒ R/100
=968/9680 = 1/10

∴ R =100/10
= 10%

**Type 3 – Money Doubling – up**

**1. The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is?**

**Solution:**

∵ P x (1 +
20/100)

^{t}> 2P
⇒ (6/5)

^{n}> 2
Now, (6/5)
x (6/5) x (6/5) x (6/5) > 2

⇒ 1296/615
> 2

∴ t = 4
years

**2. A sum of money placed at compound interest doubles itself in 5 years. It will amount to eight times itself in?**

**Solution:**

Let the
principal be P and rate be r%. Then

2P = P*(1 +
r/100)

^{5}⇒ (1 + r/100)^{5}= 2
Let it be 8
times in t years, then

∵ 8P = P(1 +
r/100)

^{t}⇒ (1 + r/100)^{t}= 8
⇒ (2)

^{3}⇒ (1 + r/100)^{15}
∴ t =15
year.

**3. If a sum of money placed at compound interest becomes 3 times of itself in 3 years. In how many years will it be 9 times at the same rate of interest?**

**Solution:**

Let the
principle be P then amount be 3P

∵ 3P =P(1 +
r/100)

^{3}
⇒ 3 = (1 +
r/100)

^{3}
On squaring
on both the sides,

⇒ (3)

^{2}= {(1 + r/100)^{3}}^{2}
∴ 9 = (1 +
r/100)

^{6}
Hence, it
will become 9 times in 6 years.

**4. A sum of Rs. 2400 deposited at CI, Double after 5 yr. After 20 yr it will become?**

**Solution:**

2P = P (1 +
R/100)

^{5}
⇒ 2

^{4}P = P * (1 + R/100)^{20}
⇒ The amount
becomes 2

^{4}times
= 16 times
and required amount = 2400 x 16 = Rs. 38400

**Type 4 – Different rates**

**1. Mr. Dua invested money in two schemes A and B offering compound interest @8 p.c.p.a and 9 p.c.p.a respectively. If the total amount of interest accrued through two schemes together in two years was Rs. 4818.30 and the total amount invested was Rs. 27,000, what was the amount invested in Scheme A?**

**Solution:**

Let the
investment of scheme A be Rs.x

Then,
investment in scheme B = Rs. (27000 – x)

Therefore
{x × [(1 + 8/100)

^{2}– 1] + (27000 – x)[(1 + 9/100)^{2}– 1]} = 4818.30
=> (x ×
104/625) + [(1881 (27000 – x))/10000] = 48183/10

=> 1664x
+ 1881 (27000 – x) = 48183000

=> 217x
= 2604000 => x = 12000

**2. Income of Shantanu was Rs. 4000. In the first 2 yr. his income decreased by 10% and 5% respectively but in the third year, the income increased by 15%. What was his income at the end of third year?**

**Solution:**

Given, P =
Rs. 4000

R1= 10%
(decreased), R2 = 5% (decreases) and R3 = 15% (growth)

∴ According
to the formula,

Income at
the end of third year = P*(1 - R1/100)(1 - R2/100)(1 + R3/100)

= 4000(1-
10/100) (1 - 5/100) (1 + 15/100)

= 4000 x
(9/10) x (19/20) x (23/20) = 9 x 19 x 23 = Rs. 3933

**Type 5 – Based on present worth of Rs. X due in T years**

**1. A sum of Rs. 550 was taken as a loan. This is to be repaid in two equal annual instalments. If the rate of interested be 20% compound annually then the value of each instalment is ?**

**Solution:**

Let the
value of each instalment be Rs. x

Then, {x/(1
+ 20/100) + x/(1 + 20/100)

^{2}} = 550
⇒ 5x / 6 +
25x / 36 =550

⇒ 55x / 36 =
550

∴ x = 360

**2. Divided Rs. 3903 between A and B, so that A's share at the end of 7 years may equal to B's share at the end of 9 years, compound interest being at 4 per cent?**

**Solution:**

We have
(A's present share) (1 + 4/100)

^{7}= (B's present share) (1 + 4/100)^{9}
∴ A's
present share/B's present share = (1 + 4/100)

^{2}= (26/25)^{2}= 676/625
Dividing
Rs. 3903 in the ratio of 676: 625

∴ A's
present share = 676/(676 + 625) of Rs .3903 = Rs. 2028

B's present
share = Rs. 3903 - Rs. 2028 = Rs. 1875

**3. The value of a machine depreciates every year at the rate of 10% on its value at the beginning of that years. If the present value of the machine is Rs. 729. Its worth 3 years ago was?**

**Solution:**

∵ P (1 -
10/100)

^{3}= 729
∴ P =
Rs.(729 x 10 x 10 x 10)/(9 x 9 x 9) = Rs. 1000

**4. A loan was repaid in two annual instalments of Rs. 121 each. if the rate of interest be 10% per annum. compounded annually, the sum borrowed was?**

**Solution:**

Principal =
(P.W. of Rs. 121 due 1 year later) + (P.W. of Rs. 121 due 2 years later)

= Rs. [121
/(1 + 10/100) + (121 / (1 + 10/100)

^{2}]
= Rs. 210

**5. A father divided his property between his two sons A and B. A invests the amount at compound interest of 8% per annum and B invests the amount at 10% per annum simple interest. At the end of 2 yr, the interest received by B is Rs. 1336 more than the interest received by A. Find the share of A in the father's property of Rs. 25000?**

**Solution:**

Suppose A
gets Rs. P and B gets Rs. (25000 - P).

Interest
received by A at the rate of 8% per annum C.I

= P * (1 +
8/100)

^{2}- P
= P *
(27/25)

^{2}- P
= 104P/625

Interest
received by B at the rate of 10% per annum SI

= [(25000 -
P) x 10 x 2]/100 = (25000 - P)/5

According
to the question,

(25000 -
P)/5 = 104P / (625 + 1336)

∴ P = 10000

**6. The cost price of an LCD TV set is Rs. 100000. If its price value depreciates at the rate of 10% pa. Then what will be the price at the end of 3 yr?**

**Solution:**

Given P =
Rs. 100000, R = 10%, n = 3 yr

According
to the formula,

Price of
LCD TV set after 3 yr. = P(1 - R/100)

^{n}
= 100000 (1
- 10/100)

^{3}
=
100000(90/100)

^{3}
= 100000 x
(9/10) x (9/10) x (9/10)

= Rs. 72900

**7. The population of a city increases at the rate of 5% pa. If the present population of the city is 185220, then what was its population 3 yr ago?**

**Solution:**

Given, P =
185220 R = 5% (increases) and n = 3 yr

According
to the formula,

Population
n yr ago = P/(1 + R/100)

^{n}
= 185220/(1
+ 5/100)

^{3}
=
185220/[(21/20) x (21/20) x (21/20)]

= [185220 x
20 x 20 x 20] / [21 x 21 x 21]

= 20 x 20 x
20 x 20 = 160000

**Type 6 – Miscellaneous**

**1. If Rs. 3000 amounts to Rs.4329 at compound interest in a certain time, then Rs. 3000 amounts to what inhalf of the time?**

**Solution:**

Let rate =
R% and time = n yr

Then, 4320
= 3000(1+R/100)

^{n}
⇒ (1+R/100)

^{n}= 4320/3000 = 1.44
∴ (1 +
R/100)

^{n/2}= √1.44 = 1.2
∴ Required
amount for n/2 yr

= 3000(1+
R/100)

^{n/2}
= 3000 x
1.2 = Rs. 3600

**2. If the simple interest on certain sum of money be Rs. 40 for 2 years and the compound interest on the same sum at the same rate and for the same time be Rs. 45. What is the principle?**

**Solution:**

S.I. for 2
year = Rs. 40

S.I. for 1
year = Rs. 20

Rate =
{(C.I. for 2 year - S.I. for 2 year) / (S.I. for 1 year)} x 100

= {(45 -
40) / 20} x 100 = 25%

∴ Principal
= (S.I. x 100)/(Rate x Time)

= (40 x
100)/(25 x 2) = Rs. 80

**3. Find the effective annual rate of 5 per cent per annum compound interest paid half yearly?**

**Solution:**

The amount
of Rs. 100 in one year at compound interest at 5% per annum payable half
yearly.

= Rs. 100(1
+ 5/2 /100)

^{2}
=Rs.
100(102.5/100)

^{2}
=Rs.
100(1.025)

^{2}
=Rs.
105.0625

Thus, the
nominal rate of 5% payable half yearly has the same effect as the rate of
5.0625 per cent would have, if payable yearly.

Hence
5.0625 per cent is called the effective annual rate 5% per annum payable half
yearly.