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

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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(T-1)/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

- 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?
| |

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)

- How much time will it take for an amount of Rs. 900 to yield Rs. 81 as interest at 4.5% per annum of simple interest?
| |

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 sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is :
| |

A. Rs. 700 | B. Rs. 690 |

C. Rs. 650 | D. Rs. 698 |

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)

- A sum fetched a total simple interest of Rs. 929.20 at the rate of 8 p.a. in 5 years. What is the sum?
| |

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)

- What will be the ratio of simple interest earned by certain amount at the same rate of interest for 5 years and that for 15 years?
| |

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 sum of money amounts to Rs.9800 after 5 years and Rs.12005 after 8 years at the same rate of simple interest. The rate of interest per annum is
| |

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 lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is:
| |

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)

- In how many years, Rs. 150 will produce the same interest at 6% as Rs. 800 produce in 2 years at 4½% ?
| |

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

- Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
| |

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 certain sum in invested for T years. It amounts to Rs. 400 at 10% per annum. But when invested at 4% per annum, it amounts to Rs. 200. Find the time (T).
| |

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 = (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 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)

- 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.
| |

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 sum of money is lent at S.I. for 6 years. If the same amount is paid at 4% higher, Arun would have got Rs. 120 more. Find the principal
| |

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)

- The simple interest on Rs. 1820 from March 9, 2003 to May 21, 2003 at 7
^{1}⁄_{2}% rate is
| |

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 sum of Rs. 7700 is to be divided among three brothers Vikas, Vijay and Viraj in such a way that simple interest on each part at 5% per annum after 1, 2 and 3 years respectively remains equal. The Share of Vikas is more than that of Viraj by
| |

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)

- David invested certain amount in three different schemes A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in Scheme C was 150% of the amount invested in Scheme A and 240% of the amount invested in Scheme B, what was the amount invested in Scheme B?
| |

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)

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