# Solved Example

**Find S.I. on Rs. 68000 at 16 2/3% per annum for 9 months ?**

S.I = (P x R x T) /100 = (68000 x (50/3) x (3/4) x (1/100) ) =Rs. 8500

**Find S.I. on Rs. 3000 at 18% per annum for the period from 4th Feb to 18th April 2009 ?**

Time = (24 + 31 + 18) days = 73 days = 73/365 = 1/5 years

P = Rs. 3000

R = 18% p.a.

S.I. = (P Ã— R Ã— T) /100 = (3000 Ã— 18 Ã— 1/5 Ã— 1/100) = Rs. 108

**Note:** The day on which money is deposited is not counted while the day on which money is withdrawn is counted.

**In how many years will a sum of money becomes triple at 10% per annum ?**

S.I. = 2P

S.I. = (P Ã— T Ã— R)/100

2P = (P Ã— T Ã— 10)/100

T = 20 years

**Note:**

- Total amount = Principal + S.I.
- If sum of money becomes double means Total amount or Sum = Principal + S.I. = P + P = 2P

**A sum at Simple interest at 13 1/2% per annum amounts to Rs. 2502.50 after 4 years. Find the sum?**

Let Sum be S. then,

S.I. = (P Ã— T Ã— R)/100 = ((S Ã— 4 Ã— 27)/(100 Ã— 2)) = 54S/100

Amount = (S + (54S)/100) = 77S/50

77S/50 = 2502.50

S = (2502.50 Ã— 50)/77 S = 1625 Sum = Rs. 1625

**A sum of money becomes double of itself in 4 years, in 12 years it will become how many times at the same rate?**

12 years = ?

(12/4) Ã— P = 3P

Amount or Sum = P + 3P = 4 times

**A Sum was put at S.I. at a certain rate for 3 years Had it been put at 2% higher rate, it would have fetched Rs. 360 more. Find the Sum?**

Let Sum = P

Original rate = R

T = 3 years

If 2% is more than the original rate, it would have fetched 360 more i.e., R + 2

(P Ã— (R + 2) Ã— 3/100) - (P Ã— R Ã— 3)/100 = 360 = >3PR + 6P - 3PR = 36000 = >6P = 36000 = >P = 6000

**Rs. 800 amounts to Rs. 920 in 3 years at S.I. If the interest rate is increased by 3%, it would amount to how much?**

Rate = (100 Ã— 120)/(800 Ã— 3) = 5%

New Rate = 5 + 3 = 8%

Principal = 800

Time = 3 yrs

S.I. = (800 Ã— 8 Ã— 3)/100 = 192

New Amount = 800 + 192 = Rs. 992

**Prabhat took a certain amount as a loan from bank at the rate of 8% p.a. S.I. and gave the same amount to Ashish as a loan at the rate of 12% p.a.. If at the end of 12 yrs, he made a profit of Rs. 320 in the deal, What was the original amount?**

Let the original amount be Rs. S.

T = 12

R1 = 8%

R2 = 12%

Profit = 320

P = S

(P Ã— T Ã— R2)/100 - (P Ã— T Ã— R1)/100 = 320

(S Ã— 12 Ã— 12)/100 - (S Ã— 8 Ã— 12)/100 = 320

S = 2000/3

S = Rs. 666.67

**Simple Interest on a certain sum at a certain rate is 9/16 of the sum. If the number representing rate percent and time in years be equal, then the rate is.**

Then,

S.I. = 9S/16

Let time = n years & rate = n%

n = 100 Ã— 9S/16 Ã— 1/S Ã— 1/n = >n Ã— n = 900/16 = > n = 30/4 = 7 1/2%

**A certain sum of money amounts to 1680 in 3 years & it becomes 1920 in 7 years. What is the sum?**

3 years - > 1680

7 years - > 1920

then, 4 years - > 240

1 yr - > ?

(1/4) Ã— 240 = 60

S.I. in 3 years = 3 Ã— 60 = 180

Sum = Amount - S.I = 1680 â€“ 180 = 1500 we get the same amount if we take S.I. in 7 yrs

i.e., 7 Ã— 60 = 420, Sum = Amount - S.I. = 1920 â€“ 420 = 1500

**A Person takes a loan of Rs. 200 at 5% simple Interest. He returns Rs. 100 at the end of 1 yr. In order to clear his dues at the end of 2 yrs, he would pay:**

**A Man borrowed Rs. 24000 from two money lenders. For one loan, he paid 15% per annum and for other 18% per annum. At the end of one year, he paid Rs. 4050. How much did he borrow at each rate?**

Let the Sum at 15% be Rs. A

& then at 18% be Rs. (24000 - A)

P1 = A, R1 = 15

P2 = (24000 - A), R2 = 18

At the end of ine year T = 1

(P1 Ã— T Ã— R1)/100 + (P2 Ã— T Ã— R2)/100 = 4050

(A Ã— 1 Ã— 15)/100 + ((24000 - A) Ã— 1 Ã— 18)/100 = 4050

15A + 432000 â€“ 18A = 405000 = > A = 9000

Money borrowed at 15% = 9000

Money borrowed at 18% = (24000 - 9000) = 15000

**What annual instalment will discharge a debt of Rs. 1092 due in 3 years at 12% Simple Interest?**

(A + (A Ã— 12 Ã— 1)/100) + (A + (A Ã— 12 Ã— 2)/100) + A = 1092

= > 28A/25 + 31A/25 + A = 1092 => (28A + 31A + 25A) = (1092 Ã— 25) = > 84A = 1092 25

= > A = (1092 25) /84 = 325 Each instalment = Rs. 325

**A Sum of Rs. 1550 was lent partly at 5% and partly at at 8% p.a. Simple interest. The total interest received after 3 years was Rs. 300. The ratio of the money lent at 5% to that lent at 8% is:**

Let the Sum at 5% be Rs. A

at 8% be Rs. (1550 - A)

(A Ã— 5 Ã— 3)/100 + ((1500 - A) Ã— 8 Ã— 3)/100 = 300 = > 15A + 1500 Ã— 24 â€“ 24A = 30000 = > A = 800

Money at 5%/ Money at 8% = 800/(1550 - 800) = 800/750 = 16/15

**A Man invests a certain sum of money at 6% p.a. Simple interest and another sum at 7% p.a. Simple interest. His income from interest after 2 years was Rs. 354. One fourth of the first sum is equal to one fifth of the second sum. The total sum invested was?**

R1 = 6, R2 = 7, T = 2

(P1 Ã— R1 Ã— T)/100 + (P2 Ã— R2 Ã— T )/100 = 354 = > (A Ã— 6 Ã— 2)/100 + (B Ã— 7 Ã— 2)/100 = 354 = > 6A + 7B = 17700

also one fourth of the first sum is equal to one fifth of the second sum

A/4 = B/5 = > 5A â€“ 4B = 0

By solving the equations we get,

A = 1200, B = 1500

Total sum = Rs. (1200 + 1500) = Rs. 2700

**Rs. 2189 are divided into three parts such that their amounts after 1, 2 & 3 years respectively may be equal, the rate of S.I. being 4% p.a. in all cases. The Smallest part is?**

Let these parts be A, B and [2189 - (A + B)] then, (A Ã— 1 Ã— 4)/100 = (B Ã— 2 Ã— 4)/100 = (2189 - (A + B)) Ã— 3 Ã— 4/100 = > 4A/100 = 8B/100 = > A = 2B

By substituting values

(2B Ã— 1 Ã— 4)/100 = (2189 - 3B) Ã— 3 Ã— 4/100 = > 44B = 2189 Ã— 12 = > B = 597

Smallest Part = Rs. 597

**A man invested 1/3 of his capital at 7%, 1/4 at 8% and the remainder at 10%.If his annual income is Rs. 561. The capital is:**

7A/30 0 + A/50 + A/24 = 561

51A = 561 Ã— 600 thus, A = Rs. 6600

**Find CI on Rs. 6250 at 16% per annum for 2 years, compounded annually?**

Rate R = 16, n = 2, Principal = Rs. 6250

Amount = P[1 + (R/100)]^{n}

= 6250[1 + (16/100)]^{2} = Rs. 8410

C.I. = Amount - P

= 8410 - 6250

= Rs. 2160

**Find C. I on Rs. 16000 at 20% per annum for 9 months compounded quarterly?**

Rate for 3 months = 20/4 = 5%

For 9 months, there are '3' 3months

Amount = P[1 + (R/100)]

^{n }= 160 00[1 + (5/100)

^{3}= Rs. 18522

C.I. = 18522 - 16000 = Rs. 2522

**The difference between C.I. and S.I. on a certain sum at 10% per annum for 2 years is Rs. 631. Find the sum?**

Let the sum be Rs. P, Then

C.I. = P[1 + (10/100)]^{2 }â€“ P = 21P/100

S.I. = (P Ã— 10 Ã— 2) /100 = P/5

C.I. - S.I. = 21P/100 - P/5

= P/100

Given CI - SI = Rs. 631

= > P/100 = 631 = > P = Rs. 63100

**If C. I on a certain sum for 2 years at 12% per annum is Rs. 1590. What would be S.I?**

^{2}â€“ P = > 784P/625 - P = 1590 = > P = Rs. 6250

S.I. = (6250 Ã— 12 Ã— 2)/100 = Rs. 1500