MATHEMATICS AND QUATITUATIVE APTITUDE – SIMPLE INTEREST
Introduction
Money is not free and it costs to borrow the money. Normally, the borrower has to pay an extra amount in addition to the amount he had borrowed. i.e, to repay the loan, the borrower has to pay the sum borrowed and the interest.
Lender and Borrower
The person giving the money is called the lender and the person taking the money is the borrower.
Principal (sum)
Principal (or the sum) is the money borrowed or lent out for a certain period. It is denoted by P.
Interest
Interest is the extra money paid by the borrower to the owner (lender) as a form of compensation for the use of the money borrowed.
Simple Interest (SI)
If the interest on a sum borrowed for certain period is calculated uniformly, it is called simple interest(SI).
Amount (A)
The total of the sum borrowed and the interest is called the amount and is denoted by A
 The statement “rate of interest 10% per annum” means that the interest for one year on a sum of Rs.100 is Rs.10. If not stated explicitly, rate of interest is assumed to be for one year.
 Let Principal = P, Rate = R% per annum and Time = T years. Then
Simple Interest, SI = PRT/100
 From the above formula , we can derive the followings
P=100×SI/RT
R=100×SI/PT
T=100×SI/PR
Some Formulae
 If a sum of money becomes n times in T years at simple interest, then the rate of interest per annum can be given be R = 100(n−1)/T %
 The annual instalment which will discharge a debt of D due in T years at R% simple interest per annum =100D/ (100T+RT(T1)/2)
 If an amount P_{1} is lent out at simple interest of R_{1}% per annum and another amount P_{2} at simple interest rate of R_{2}% per annum, then the rate of interest for the whole sum can be given by
R=(P1R1+P2R2)/ (P1+P2)  If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R_{1}, R_{2}, … , R_{n} respectively and time periods are T_{1}, T_{2}, … , T_{n} respectively, then the ratio in which the sum will be divided in n parts can be given by (1/R1T1):(1/R2T2):⋯(1/RnTn)
 If a certain sum of money P lent out for a certain time T amounts to P_{1} at R_{1}% per annum and to P_{2}at R_{2}% per annum, then P = (P2R1−P1R2)/ (R1−R2) and T = (P1−P2) ×100 years / (P2R1−P1R2)
SOLVED EXAMPLES
LEVEL 1


A. 8% 
B. 6% 
C. 4% 
D. 7% 
Ans. Let rate = R%
Then, Time, T = R years
P = Rs.1400
SI = Rs.686
SI= PRT/100 ⇒ 686 = 1400 × R × R/100 ⇒ 686=14 R x R ⇒ 49=R x R ⇒ R=7
i.e.,Rate of Interest was 7%. (D)


A. 2 years 
B. 3 years 
C. 1 year 
D. 4 years 
Ans. P = Rs.900
SI = Rs.81
T = ?
R = 4.5%
T= 100×SI/PR = 100×81/(900×4.5) = 2 years (A)


A. Rs. 700 
B. Rs. 690 
C. Rs. 650 
D. Rs. 698 
Ans. Simple Interest (SI) for 1 year = 854815 = 39
Simple Interest (SI) for 3 years = 39 × 3 = 117
Principal = 815 – 117 = Rs.698 (D)


A. Rs. 2323 
B. Rs. 1223 
C. Rs. 2563 
D. Rs. 2353 
Ans. SI = Rs.929.20
P = ?
T = 5 years
R = 8%
P = 100×SI/RT=100×929.20/(8×5) = Rs.2323 (A)


A. 3 : 2 
B. 1 : 3 
C. 2 : 3 
D. 3 : 1 
Solution 1
Let Principal = P
Rate of Interest = R%
Required Ratio = (PR×5/100)/ (PR×15/100) = 1:3 (B)
Solution 2
Simple Interest = PRT100
Here Principal(P) and Rate of Interest (R) are constants
Hence, Simple Interest ∝ T
Required Ratio = Simple Interest for 5 years Simple Interest for 15 years=T1T2=515=13=1:3 (B)


A. 15% 
B. 12% 
C. 8% 
D. 5% 
Ans. Simple Interest for 3 years = (Rs.12005 – Rs.9800) = Rs.2205
Simple Interest for 5 years = 22053×5=Rs.3675
Principal (P) = (Rs.9800 – Rs.3675) = Rs.6125
R = 100×SI/PT=100×3675/(6125×5) =12% (B)


A. 5% 
B. 10% 
C. 7% 
D. 8% 
Ans. Let the rate of interest per annum be R%
Simple Interest for Rs. 5000 for 2 years at rate R% per annum +Simple Interest for Rs. 3000 for 4 years at rate R% per annum = Rs.2200
⇒5000×R×2/100+3000×R×4/100=2200
⇒100R + 120R=2200⇒220R=2200⇒R=10
i.e, Rate = 10%. (B)


A. 4 years 
B. 6 years 
C. 8 years 
D. 9 years 
Ans. Let Simple Interest for Rs.150 at 6% for n years = Simple Interest for Rs.800 at 4½ % for 2 years
150×6×n/100 = 800×4.5×2/100
150×6×n = 800×4.5×2
LEVEL 2


A. Rs. 6400 
B. Rs. 7200 
C. Rs. 6500 
D. Rs. 7500 
Ans. Let the investment in scheme A be Rs.x
and the investment in scheme B be Rs. (13900 – x)
We know that SI = PRT/100
Simple Interest for Rs.x in 2 years at 14% p.a. = x×14×2100=28x100Simple Interest for Rs.(13900 – x) in 2 years at 11% p.a. = (13900−x)×11×2/100 =22(13900−x)/100
Total interest =Rs.3508
Thus, 28x/100+22(13900−x)/100 = 3508
28x+305800−22x=350800
6x = 45000
x=45000/6=7500
Investment in scheme B = 13900 – 7500 = Rs.6400 (A)


A. 45 years 
B. 60 years 
C. 40 years 
D. 50 Years 
Solution 1
Let the principal = Rs.x
and time = y years
Principal,x amounts to Rs.400 at 10% per annum in y years
Simple Interest = (400x)
Simple Interest = PRT/100
⇒ (400−x) = x×10×y/100
⇒ (400−x) = xy/10— (equation 1)
Principal,x amounts to Rs.200 at 4% per annum in y years
Simple Interest = (200x)
Simple Interest = PRT/100
⇒ (200−x) = x×4×y/100
⇒ (200−x) = xy/25— (equation 2)
(equation 1)/(equation2)
⇒ (400−x) / (200−x) = (xy/10)/(xy/25)
⇒ (400−x)/ (200−x) =25/10
⇒ (400−x)/ (200−x) =52
⇒ 800−2x = 1000−5x
⇒200=3x
⇒x =200/3 Substituting this value of x in Equation 1, we get,
(400−200/3) = (200y/3)/10
⇒ (400−200/3) = 20y/3
⇒ 1200−200=20y
⇒1000=20y
y=1000/20=50 years (D)
Solution 2
If a certain sum of money P lent out for a certain time T amounts to P_{1} at R_{1}% per annum and to P_{2} at R_{2}% per annum, then
P = (P2R1−P1R2)/ (R1−R2)
T = (P1−P2)x 100 years/(P2R1−P1R2)
R_{1} = 10%, R_{2} = 4%
P_{1} = 400, P_{2} = 200
T = (P1−P2)x 100 / (P2R1−P1R2) = (400−200)x 100 / (200×10−400×4)
=200 x 100/ (2000−1600) =200 ×100/400 = 12×100=50 years (D)


A. Rs. 25000 
B. Rs. 15000 
C. Rs. 10000 
D. Rs. 20000 
Ans. If an amount P_{1} is lent out at simple interest of R_{1}% per annum and another amount P_{2} at simple interest rate of R_{2}% per annum, then the rate of interest for the whole sum can be given by
R= (P1R1+P2R2)/(P1+P2)
P_{1} = Rs. 12000, R_{1} = 10%
P_{2} =? R_{2} = 20%
R = 14%
14 = (12000×10+P2×20)/ (12000+P2)
12000×14+14P2 =120000+20P2
6P2=14×12000−120000=48000
⇒P2=8000
Total amount invested = (P_{1} + P_{2}) = (12000 + 8000) = Rs. 20000 (D)


A. Rs. 200 
B. Rs. 600 
C. Rs. 400 
D. Rs. 500 
Ans. This means, simple interest at 4% for that principal is Rs.120
P=100×SI/ RT=100×120/ (4×6) =100×30/6 = 100×5 = 500 (D)


A. Rs. 27.30 
B. Rs. 22.50 
C. Rs. 28.80 
D. Rs. 29 
Ans. Time, T = (22 + 30 + 21) days = 73 days = 73/365 year=1/5 year
Rate, R = 7.5%=15/2%
SI = PRT/100 = 1820× (15/2) × (1/5)/100 = 1820 × (3/2)/100 = 910 × 3/100
= 2730/100 = 27.30 (A)


A. Rs.1200 
B. Rs.1400 
C. Rs.2200 
D. Rs.2800 
Ans. If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R_{1}, R_{2}, … , R_{n} respectively and time periods are T_{1}, T_{2}, … , T_{n} respectively, then the ratio in which the sum will be divided in n parts can be given by
1/R1T1:1/R2T2:⋯1/RnTn
T_{1} = 1 , T_{2} = 2, T_{3} = 3
R_{1} = 5 , R_{2} = 5, R_{3} = 5
Share of Vikas : Share of Vijay : Share of Viraj
= (1/5×1) : (1/5×2) : (1/5×3) = 1/1:1/2:1/3 = 6:3:2
Total amount is Rs. 7700
Share of Vikas = 7700×6/11=700×6 = 4200
Share of Viraj = 7700×2/11=700×2=1400
Share of Vikas is greater than Share of Viraj by (4200 – 1400) = Rs. 2800 (D)


A. Rs.5000 
B. Rs.2000 
C. Rs.6000 
D. Rs.3000 
Ans. Let x, y and x be his investments in A, B and C respectively. Then
Then, Interest on x at 10% for 1 year
+ Interest on y at 12% for 1 year
+ Interest on z at 15% for 1 year
= 3200
x×10×1/100+y×12×1/100+z×15×1/100=3200
⇒10x+12y+15z=320000−−−(1)
Amount invested in Scheme C was 240% of the amount invested in Scheme B
=>z=240y/100 = 60y/25=12y/5−−−(2)
Amount invested in Scheme C was 150% of the amount invested in Scheme A
=>z=150x/100=3x/2
=>x=2z/3=2/3×12y/5=8y/5−−−(3)
From(1),(2) and (3),
10x + 12y + 15z = 320000
10(8y/5)+12y+15(12y/5)=320000
16y+12y+36y=320000
64y=320000
y=320000/64=10000/2=5000
i.e.,Amount invested in Scheme B = Rs.5000 (A)