10. Simple Interest, Quantitative Aptitude, Civil Service Examination,RPSC | RAS RPSC Prelims Preparation - Notes, Study Material & Tests - RPSC RAS (Rajasthan) PDF Download

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

1. 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 %
2. The annual instalment which will discharge a debt of D due in T years at R% simple interest per annum =100D/ (100T+RT(T-1)/2)
3. If an amount P₁ is lent out at simple interest of R₁% per annum and another amount P₂ at simple interest rate of R₂% per annum, then the rate of interest for the whole sum can be given by 
   R=(P1R1+P2R2)/ (P1+P2)
4. 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 R₁, R₂, ... , R_n respectively and time periods are T₁, T₂, ... , 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)
5. If a certain sum of money P lent out for a certain time T amounts to P₁ at R₁% per annum and to P₂at R₂% per annum, then P = (P2R1−P1R2)/ (R1−R2) and T = (P1−P2) ×100 years / (P2R1−P1R2)

SOLVED EXAMPLES
 

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)
 

Ans. P = Rs.900

SI = Rs.81

T = ?

R = 4.5%

T= 100×SI/PR = 100×81/(900×4.5) = 2 years (A)
 

Ans. Simple Interest (SI) for 1 year = 854-815 = 39

Simple Interest (SI) for 3 years = 39 × 3 = 117

Principal = 815 - 117 = Rs.698 (D)
 

Ans. SI = Rs.929.20

P = ?

T = 5 years

R = 8%

P = 100×SI/RT=100×929.20/(8×5) = Rs.2323 (A)
 

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)
 

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)
 

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)
 

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

n = 8 years (C)
 

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)
 

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 = (400-x)

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 = (200-x)

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 P₁ at R₁% per annum and to P₂ at R₂% per annum, then

P = (P2R1−P1R2)/ (R1−R2)

T = (P1−P2)x 100 years/(P2R1−P1R2)

R₁ = 10%, R₂ = 4%

P₁ = 400, P₂ = 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)