## Quantitative Aptitude Notes: Simple Interest

### Quantitative Aptitude Notes: Simple Interest

INTRODUCTION:
Amount (A): The sum of interest and principal is called Amount.
Principal (P): The sum borrowed is called the principal.
Interest (I): Interest is the money paid for the use of money borrowed.
Amount (A) = Principal (P) + Interest (I)
Simple Interest (S.I.): If the interest on a sum borrowed for certain period is reckoned uniformly, then it is called simple interest.

Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then
S.I = (P × T × R)/100
Example: A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?
Solution:
Principal = Rs. (100 × 4016.25)/9 × 5 = 401625/45 = Rs.8925
Note: If the rate of interest is specified in terms of 6- monthly rate, we take time in terms of 6 months. Also, the half-yearly rate of interest is half the annual rate. That is if the interest is 10% per annum is to be charged six-monthly, we have to add interest in every six month @ 5%.

### Memory Based Problems based of various types

Type 1 – Basic Problems
1. Arun took a loan of Rs. 1400 with simple interest for as many years as the rate of interest. If he paid Rs.686 as interest at the end of the loan period, what was the rate of interest?
Solution:
Let rate = R% Then, Time, T = R years; P = Rs.1400; SI = Rs.686
SI = PTR/100 => 686 = 1400 x R x R/100 => R2 = 49 => R = 7
Therefore, Rate of Interest was 7%
2. A certain sum of money amounts to Rs.1008 in 2 years and to Rs.1164 in 3.5 years. Find the sum and rate of interest.
Solution:
S.I. for 1 ½ years = Rs (1164 - 1008) = Rs 156.
S.I. for 2 years = Rs (156 x 2) / (1 ½) = Rs (156 x 2 x 2/3) = Rs 208.
Principal = Rs (1008 - 208) = Rs 800.
Now, P = 800, T= 2 and S.I. = 208.
Rate = (100 x S.I.) / (P x T) = [ (100 x 208)/(800 x 2)]% = 13%
3. In what time will the simple interest on Rs 400 at 10% per annum be the same as the simple interest on Rs 1000 for 4 year at 4 % per annum?
Solution:
Here , P= Rs 1000, T= 4 yrs, R= 4 %
where, P= Principal, T= Time, R= Rate
Since , Simple Interest on Rs 1000=(1000 × 4 × 4)/100 = Rs 160
now, simple interest=Rs 160
P = Rs 400
R = 10 %
then, T=(100 × SI)/P × R = (100 × 160)/(400 × 10) = 4 yr
4. The difference of 13% per annum and 12% of a sum in 1 year is Rs 110.Then the sum is?
Solution:
Let the sum be 'y'
Then, [(y x13 x 1)/100 ]- [( y x 12 x 1) / 100]=110
Since (y / 100) = Rs 110
Y = Rs 11000
5. Harsha makes a fixed deposit of Rs. 20000 in Bank of India for a period of 3 yr. If the rate of interest be 13% SI per annum charged half - yearly, what amount will he get after 42 months?
Solution:
Give, time = 42 months.
= (42/12) yr = 31/2 yr
= 7 half - yr,
rate = 13/2% half - yearly
S.I = (20000 x 13 x 7)/(100 x 2) = Rs. 9100
Amount (A) = 20000 + 9100 = Rs. 29100
6. A moneylender finds that due to a fall in the rate of interest from 13% to 121/2% his yearly income diminishes by Rs. 104. His capital is?
Solution:
Let capital = Rs. P
Then, S.I1 – S.I2 = 104.
(P x 13 x 1)/100 - (P x 25/2 x 1) /100 = 104
13P/100 - P/8 = 104
26P -25P = (104 x 200) P = 20800
Capital = Rs. 20800
7. The difference between the interest received from two different bank on Rs. 500 for 2 year is Rs. 2.50. The difference between their rates is?
Solution:
Let the rates be R1% and R2%.
Then, (500 x R1 x 2)/ 100 - (500 x R2 x 2)/ 100 = 2.5
10(R1 - R2) = 2.5
Req difference = R1 - R2 = 0.25%
Type 2 – Money Doubling – up or Tripling
1. At what rate percent per annum will a sum of money double in 8 yr?
Solution:
Let Sum = P, Then S.I=P
As Amount A = 2 × P
Rate R = (100 × SI)/(P × T) = (100 × P)/(P × 8) % = 12.5 %
Alternative Method:
RT = (n – 1) * 100; where n – no. of times i.e. doubling or tripling
Here the money is doubling. Therefore, n = 2;
R * 8 = (2-1 ) * 100 => 8R = 100 => R = 100/8 = 12.5 %
2. If a certain sum is doubled in 8 yr on simple interest, in how many years will it is four times?
Solution:
Let the sum be Rs 'y' , so amount = 2y
Simple interest =Rs y
Let R be the rate of interest,
R= (100 x S.I)/(P x T) = (100 x y) / (y x 8) = 12.5 %
now, the needed amount = Rs 4y
since SI = Rs (4x-x)= Rs 3y
since T= (100 x S.I)/(P x R)
= (100 x 3y)/(y x 125)= 24 yr
3. At what rate percent per annum simple interest, will a sum of money triple itself in 25 year?
Solution:
Let principal amount = P
As amount =3P, T=25 yr
S.I = 3P - P = 2P
Rate R = (100 x S.I) / (P x T) = (100 x 2P)/(P x 25)=8%
4. At the certain rate of simple interest, a certain sum doubles itself in 10 years. It will treble itself in?
Solution:
Let principal = P. Then, S. I = P. and Time = 10 years
Required time = [(n - 1) x t] / (m - 1)
= [(3 - 1) x 10] / (2 - 1) = 20 years
5. In how many years will a sum of money double itself at 12% per annum?
Solution:
Let principal = P.
Then, S.I = P,
Rate (R) = 12%
Time = (100 x SI) / (R x P) = (100 x P) / (P x 12) years
= 25/3 years
= 8 years 4 months
Type 3 – Installments and Investments at different rates
1. A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?
Solution:
Let the sum of Rs.725 is lent out at rate R% for 1 year
Then, at the end of 8 months, ad additional sum of 362.50 more is lent out at rate 2R% for remaining 4 months (1/3 year)
Total Simple Interest = 33.50
=> (725 × R × 1)/100 + (362.5 × 2R × 1/3)/100 = 33.50
=> 725R/100 + 725R/300 = 33.50 => 725R × 4/100 = 33.50 => R = 3350 × 4/725 = 3.46%
2. Mr. Mani invested an amount of Rs. 12000 at the simple interest rate of 10% per annum and another amount at the simple interest rate of 20% per annum. The total interest earned at the end of one year on the total amount invested became 14% per annum. Find the total amount invested.
Solution:
P1 = Rs. 12000, R1 = 10%; P2 = ?, R2 = 20%; R = 14%
14 = (1200 × 10 + P2 × 20)/(12000 + P2)
=>  12000 × 14 + 14P2 = 120000 + 20P2 => 6P2 = 14 × 12000 – 120000 = 48000 => P2 = 8000
3. The rate of interest for the first 2 years is 5% for the next 3 years is 8% and beyond this it is 10% per annum. If the simple interest for 8 years is Rs . 1280. What is the principal?
Solution:
r1 = 5%, r2 = 8%, r3 = 10%,
t1 = 2 years, t2 = 3 years, t3 = 8 - (2 + 3) = 3 years
Principal = (Interest x 100) / [(r1 x t1) + (r2 x t2) + (r3 x t3)]
= (1280 x 100) / (5 x 2 + 8 x 3 + 10 x 3)
= 128000/(10 + 24 + 30)
= 128000/64 =Rs. 2000
4. Pratap borrowed some money from Arun at simple interest. The rate of interest for the first 3 years was 12% for the next 5 years was 16% and beyond this it was 20%. If the simple interest for 11 years was more than the money borrowed by Rs. 6080. What was the money borrowed?
Solution:
Let the sum be P.
SI = SI for first 3 years + SI for next 5 years + SI for next 3 years
P + 6080 = (P x 12 x 3) / 100 + (P x 16 x 5) / 100 + (P x 20 x 3) / 100
P + 6080 = (36P + 80P + 60P) / 100
100 x (P + 6080) = 176P
P = 608000 / 76 = 8000
5. The annual payment of Rs. 160 in 5 yr at 5 % per annum simple interest will discharge a debt of?
Solution:
Given, annual payment = Rs. 160
R = 5%, T = 5 yr debt, P = ?
According to the formula.
Annual payment = 100*P / [100 x T + {RT (T - 1)/2}]
160 = 100P / [5 x 100 + {(5 x 4 x 5)/2}]
160 = 100P/550
P = (550 x 160) / 100
= 55 x 16 = Rs. 880
6. A sum was put at simple interest at a certain rate for 2 years . Had it been put at 3% higher rate, it would have fetched Rs 300 more. The sum is
Solution:
Let the sum be P.
And the original rate be y% per annum.
Then new rate=(y+3)% per annum
According to question, [(P × (y+3) × 2)/100]=[(P × y × 2)/100]=300
[(Py + 3P)/100]=[Py/100] = 150
Py+ 3P - Py=15000
3P=15000 => P= 5000
Thus, the sum is Rs 5000
Type 4 - Miscellaneous
1. Divide Rs. 2379 into 3 parts so that their amount after 2,3 and 4 years respectively may be equal, the rate of interest being 5% per annum at simple interest. The first part is
Solution:
Let the parts be x, y and z; R = 5%
x + interest on x for 2 years = y + interest on y for 3 years = z + interest on z for 4 years
2. A person closer his account in an investment scheme by withdrawing Rs. 10000. One year ago, he had withdrawn Rs. 6000. Two years ago, he had withdrawn Rs. 5000. Three years ago, he had not withdrawn any money. How much money had he deposited approximately at the time of opening the account 4 yr ago, if the annual simple interest is 10%?
Solution:
Suppose the person had deposited Rs. P at the time of opening the account.
P + (P x 10 x 1)/100 = Rs. 11P/10
11P/10 + (11P/10 x 10 x 1)/100 = Rs. 121P/100 ...(i)
After withdrawn Rs. 5000 from Rs. 121P/100, the balance = Rs. (121P - 500000)/100
(121P - 500000)/100 + [(121P - 500000)/100 x 10 x 1]/100 = 11(121P - 500000)/100 ... (ii)
After withdrawn Rs. 6000 from amount (ii) the balance = (1331P/1000 - 11500)
After 4 yr, he had Rs. (1331P - 5500000)/1000 + 10% of Rs. (1331P - 5500000)/1000 = Rs. (11/10) x (1331P/1000 - 11500) ... (iii)
After withdrawn Rs.10000 from amount (iii) the balance =0
11/10(1331P/1000 - 11500) - 10000 = 0
P = Rs.15470
3. In 4 yr, Rs. 6000 amounts to Rs. 8000. In what time at the same rate, will Rs. 525 amount to Rs. 700?
Solution:
Amount = Rs. 8000, Time (T) = 4 yr; Principal (P) = ₹ 6000
Simple interest (SI) = Amount - Principal = 8000 - 6000 = Rs. 2000
According to the formula. S.I = (P x R x T)/100
2000 = (6000 x R x 4)
R = (6000 x 100)/(6000 x 4) = (25/3) %
Now, again Amount (A) = Rs. 700
Principle (P) = Rs. 525, Rate (R) = 25/3%
Simple interest = A – P 700 - 525 = Rs. 175
Using formula, S.I = (P x R x T) / 100
175 = [525 x (25/3) x T] / 100
T = (175 x 100 x 3) / (525 x 25) = 4 yr
4. Reena had Rs. 10000 with her, out of this money she lent some money to Akshay for 2 yr at 15% simple interest. She lent remaining money to Brijesh for an equal number of years at the rate of 18%. After 2 yr Reena found that Akshay had given her 360 more as interest as compared to Brijesh. The amount of money which Reena had lent to Brijesh must be?
Solution:
Let the money lent to Akshay = Rs. P
Then, money lent to Brijesh = Rs. (10000 - P) [as total amount = Rs. 10000]
S.I for Akshay = (P x 15 x 2)/100 = 3P/10
S.I for Brijesh = {(10000 - P) x 18 x 2}/100 = 9/25 (10000 - P)
According to the given condition, (3P/10) - [(9/25) x (10000 - P ) = 360
[as S.I (Akshay) – S.I (Brijesh) = 360]
(3P/10) - 3600 + 9P/25 = 360
3P/10 + 9P/25 = 360 + 3600 = 3960
33P/50 = 3960 P = 3960 x 50/33 P = 6000
The amount of money lent to Brijesh
= 10000 - 6000 = Rs. 4000
5. Rajnish invested certain sum in three different schemes P, Q and R with the rates of interest 10% per annum, 12% per annum and 15% per annum, respectively. If the total interest accrued in 1 yr was ₹ 3200 and the amount invested in scheme R was 150% of the amount invested in scheme Q. what was the amount invested in scheme Q?
Solution:
Let a, b and c be the amount invested in schemes P, Q and R, respectively.
Then, according to the question,
[(a x 10 x 1)/100] + [(b x 12 x 1)/100] + [(c x 15 x 1)/100] = 3200
10a + 12b + 15x = 320000 ..... (i)
Now, c = 240% of b = 12b/5 .... (ii)
and c = 150% of a = 3a/2
a = 2c/3 = (2/3 x 12/5) b = 8b/5 .....(iii)
From Eqs. (i), (ii) and (iii), we get
16b + 12b + 36b = 320000
64b = 320000 => b = 5000
Sum invested in scheme Q = Rs. 5000