**Important Facts and Formulae: **

**Principal or Sum**:- The money borrowed or lent out for a certain period is called Principal or the Sum.

**Interest**:- Extra money paid for using others money is called Interest.

**Simple Interest**:- If the interest on a sum borrowed for a certain period is reckoned uniformly,then it is called Simple Interest.

**Formulae:**

Principal = P

Rate = R% per annum

Time = T years. Then,

(i)Simple Interest(S.I)= (P*T*R)/100

(ii) Principal(P) = (100*S.I)/(R*T)

Rate(R) = (100*S.I)/(P*T)

Time(T) = (100*S.I)/(P*R)

**Simple Problems**

**1.Find S.I on Rs68000 at 16 2/3% per annum for 9months. **

Sol:- P=68000

R=50/3% p.a

T=9/12 years=4/3 years

S.I=(P*R*T)/100

=(68000*(50/3)*(3/4)*(1/100))

=Rs 8500

Note:If months are given we have to converted into

years by dividing 12 ie., no.of months/12=years

**2.Find S.I on Rs3000 at 18% per annum for the period from 4th Feb to 18th April 1995 **

Sol:- Time=(24+31+18)days

=73 days

=73/365=1/5 years

P= Rs 3000

R= 18% p.a

S.I = (P*R*T)/100

=(3000*18*1/5*1/100)

=Rs 108

Remark:- The day on which money is deposited is not

counted while the day on which money is withdrawn is

counted.

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

Sol:- Let principal =P

S.I = 2P

S.I = (P*T*R)/100

2P = (P*T*10)/100

T = 20 years

Note:

(1) Total amount = Principal + S.I

(2) If sum of money becomes double means Total amount

or Sum

= Principal + S.I

= P + P = 2P

**Medium Problems**

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

Sol:- Let Sum be x. then,

S.I = (P*T*R)/100

= ((x*4*27)/(100*2))

= 27x/100

Amount = (x+(27x)/100)

= 77x/50

77x/50 = 2502.50

x = (2502.50*50)/77

= 1625

Sum = 1625

**2. A some of money becomes double of itself in 4 years in 12 years it will become how many times at the same rate. **

Sol:- 4 yrs – – – – – – – – – P

12 yrs – – – – – – – – – ?

(12/4)* P =3P

Amount or Sum = P+3P = 4 times

**3. 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. **

Sol:- Let Sum =P

original rate = R

T = 3 years

If 2% is more than the original rate ,it would have

fetched 360 more ie., R+2

(P*(R+2)*3/100) – (P*R*3)/100 = 360

3PR+ 6P-3PR = 36000

6P = 36000

P = 6000

Sum = 6000.

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

Sol:- S.I = 920 – 800 = 120

Rate = (100*120)/(800*3) = 5%

New Rate = 5 + 3 = 8%

Principal = 800

Time = 3 yrs

S.I = (800*8*3)/100 = 192

New Amount = 800 + 192

= 992

**5. 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? **

Sol:- Let the original amount be Rs x.

T = 12

R1 = 8%

R2 = 12%

Profit = 320

P = x

(P*T*R2)/100 – (P*T*R1)/100 =320

(x*12*12)/100 – (x*8*12)/100 = 320

x = 2000/3

x = Rs.666.67

**6. Simple Interest on a certail 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.**

Sol:- Let Sum = x .Then,

S.I = 9x/16

Let time = n years & rate = n%

n = 100 * 9x/16 * 1/x * 1/n

n * n = 900/16

n = 30/4 = 7 1/2%

**Complex Problems**

**1. A certain sum of money amounts t 1680 in 3yrs & it becomes 1920 in 7 yrs .What is the sum. **

Sol:- 3 yrs – – – – – – – – – – – – – 1680

7 yrs – – – – – – – – – – – – – 1920

then, 4 yrs – – – – – – – – – – – – – 240

1 yr – – – – – – – – – – – – – ?

(1/4) * 240 = 60

S.I in 3 yrs = 3*60 = 18012

Sum = Amount – S.I

= 1680 – 180

= 1500

we get the same amount if we take S.I in 7 yrs

I.e., 7*60 =420

Sum = Amount – S.I

= 1920 – 420

= 1500

**2. 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 2yrs ,he would pay:**

Sol:- Amount to be paid

= Rs(100 + (200*5*1)/100 + (100*5*1)/100)

= Rs 115

**3. 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
borrowed at each rate?**

Sol:- Let the Sum at 15% be Rs.x

& then at 18% be Rs (24000-x)

P1 = x R1 = 15

P2 = (24000-x) R2 = 18

At the end of ine year T = 1

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

(x*1*15)/100 + ((24000-x)*1*18)/100 = 4050

15x + 432000 – 18x = 405000

x = 9000

Money borrowed at 15% = 9000

Money borrowed at 18% = (24000 – 9000)

= 15000

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

Sol:- Let each instalment be Rs x

(x + (x * 12 * 1)/100) + (x + (x * 12 * 2)/100) + x = 1092

28x/25 + 31x/25 + x =1092

(28x +31x + 25x) = (1092 * 25)

84x = 1092 * 25

x = (1092*25)/84 = 325

Each instalement = 325

**5.If x,y,z are three sums of money such that y is the simple interest on x,z is the simple interest on y for the same time and at the same rate of interest ,then we have:**

Sol:- y is simple interest on x, means

y = (x*R*T)/100

RT = 100y/x

z is simple interest on y,

z = (y*R*T)/100

RT = 100z/y

100y/x = 100z/y

y * y = xz

**6.A Sum of Rs.1550 was lent partly at 5% and partly at 5% and partly 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: **

Sol:- Let the Sum at 5% be Rs x

at 8% be Rs(1550-x)

(x*5*3)/100 + ((1500-x)*8*3)/100 = 300

15x + 1500 * 24 – 24x = 30000

x = 800

Money at 5%/ Money at 8% = 800/(1550 – 800)

= 800/750 = 16/15

**7. 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:**

Sol:- Let the sums be x & y

R1 = 6 R2 = 7

T = 2

(P1*R1*T)/100 + (P2*R2*T)/100 = 354

(x * 6 * 2)/100 + (y * 7 * 2)/100 = 354

6x + 7y = 17700 â€”â€”â€”(1)

also one fourth of the first sum is equal to one

fifth of the second sum

x/4 = y/5 => 5x – 4y = 0 â€”â€” (2)

By solving 1 & 2 we get,

x = 1200 y = 1500

Total sum = 1200 +1500

= 2700

**8. 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:**

Sol:- Let these parts be x,y and[2189-(x+y)] then,

(x*1*4)/100 = (y*2*4)/100 = (2189-(x+y))*3*4/100

4x/100 = 8y/100

x = 2y

By substituting values

(2y*1*4)/100 = (2189-3y)*3*4/100

44y = 2189 *12

y = 597

Smallest Part = 597

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

Sol:- Let the capital be Rs.x

Then, (x/3 * 7/100 * 1) + ( x/4 * 8/100 * 1)

+ (5x/12 * 10/100 * 1) = 561

7x/300 + x/50 + x/24 = 561

51x = 561 * 600

x = 6600

Rs.100000 at the rate of Rs.1500 per annum.Find how many intrest we get in 15 days?

61.5/-

divide rs 2500 into two parts such that the s.i on one at 4% for 5 years is double at 5% for 3 years.

A company accepts deposits and pays 15% interest per year , if the deposit is made for a minimum of 5 years.

A borrowed Rs 400 from B @ 6.25% pa on 1st February 2006. As soon as the interest became Rs 5, A repaid with interest. Find the date on which Ajay had repaid the money?

rs1500 is invested at a rate of 10 simple interest and interest is added to the principal after every 5 years.in how many days will it amount to rs2500?