*Some may find it difficult to solve the problems based on Simple Interest and Compound Interest. This is due to lack of interest. Anyway, three or four questions are frequently asked from this area. So it is a must to study and practice this portion. *Interest is basically of fee charged for borrowing the money, ie it is the extra money a borrower has to pay in addition the sum borrowed or loan taken. Two types of Interest are Simple Interest (SI) and Compound Interest (CI). Definitions of certain terms used here are as follows.
1.PRINCIPAL SUM (P)
It is the money borrowed for a certain period.
2.AMOUNT (A)
It is the sum of Principal and Interest. A=PI
3.TIME (T)
It is the period for which the money is borrowed.
4.RATE OF INTEREST(R)
It is the money paid per Rs. 100 per year.
5.SIMPLE INTEREST
It is the interest calculated on the original principal at any rate of interest for any period of time. It will be the same every year for a fixed amount of Principal ie, the interest amount per year will not increase over years. SIMPLE INTEREST, SI =PRT/100 SI=ART/100RT P=100A/100RT
EXAMPLES
1.
Find the simple Interest on Rs. 5380 at 10% per annum for 2 years. Also find the amount. Principal (P)= Rs. 5380 Rate percent (R) = 10% Time period (T) = 2 years Simple Interest (SI) =PRT/100 =5380*2*10/100 = Rs. 1076 Amount = Rs. 5380 Rs. 1076 = Rs.6456
2.
A sum becomes two times in 6 years at a certain rate of interest. Find the time in which the same amount will be 4 times at the same rate of interest. If Sum = p, then for 6 years, SI = 2P-P = P SI = P=PxRx6/100= 3PR/50 => R = 50/3% Now, For another time (T), SI = 4P - P = 3P 3P= P*50*T/3*100 => T =18 .'.Time = 18 years
3.
What will be the simple interest for 1 year and 6 months on a sum of Rs. 50,000 at the rate of 12% per annum. Here, P = Rs. 50,000 R = 12% T = 1 yr 6 months =[16/12] years =[11/2]=3/2 yrs SI=PRT/100 =50,000*12*3 /100x2=9000 .'.Simple Interest = Rs. 9000
4.
A sum was put at simple interest at a certain rate for 2 years. Had it been put at 5% higher rate, it would have fetched Rs.92 more. Find the sum. Let the sum be Rs. x and original rate be R% Then, x(R5)*2/100- x*R*2/100= 92 2Rx 10x - 2Rx = 9200 10x = 9200 x = 920 .'.Sum is Rs. 920
5.
Simple Interest on a certain sum is 9/25 of the sum. Find the rate percent if the rate percent and time are equal. SI=PRT/100 9/25P=P*R*T/100(R=T) =>R2=900/25 R=/900/25=30/5=6% .'.Rate percent is 6%
COMPOUND INTEREST
It is the method in which the amount of interest is added to the Principal each year and the further interest is calculated on the new principal amount for the next year, ie the amount after first unit of time becomes the Principal for the second unit of time and so on. It is calculated at fixed interval of time, ie at the end of a year, half yearly, quarterly, etc. The Principal as well as the interest is increasing every year in componud interest
COMPOUND INTEREST, Cl =P [1R/100]^T -P
EXAMPLES
1.
Find the Compound Interest on Rs.8000 at 2% per annum for 2 years CI=P[1R/100]^T -P = 8000[1 2/100]^2 -8000 8000x102/100x102/100-8000 = 8323.2-8000 = Rs. 323.2 .'.Compound Interest = Rs.323.2
2.
In what time will the compound interest of Rs. 4000 at 10% per annum be 840 Let T be the time period Compound Interest, P [1R/100]^T -P=840 4000[110/100]^T-4000=840 4000[110/100]^T=4840 [110/100]^T=[121/100] ie,[11/10]^T =[11/10]^2 .'.Time period is 2 years
TYPE-I
*When the interest is compounded annually, Amount=P[1R/100]^T *When the interest is compounded half yearly, Amount=P[1R/200]^2T *When the interest is compounded quarterly Amount -P[1R/400]4T
EXAMPLES
1.
Find compouned interest on Rs. 3000 at 10% per annum for l 1/2 years compounded half yearly. Principal = Rs. 3000 R = 10% per annum = 5% half-yearly Time period = 1 1/2 years = 3 years half yearly Amount=P[1R/100]^T =3000[15/100]^3 = 3000x21/20*21/20*21/20 = 3472.88 Compound Interest = Rs. 3472.88-Rs. 3000 = Rs. 472.88
2.
Find the compound interest on Rs. 20,000 at 8% per annum for 2 years and 3 months - Here, P = Rs. 20,000, R = 8%,T= 2 1/4 yrs 1 CI=P[1R/100]^T {14/4*R/100]-P =20000[18/100]^2 [11/4*8/100]-20000 = 20,000[108/100*108/100][12/100]-20 000 = 23794.56 - 20,000 = 3794.56 .'.Compound Interest = Rs. 3794.56
TYPE-II
The difference between compound interest and simple interest for an amount 'P' deposited for 2 years at an interest rate of R% is:
CI-SI =P(R/100)^2
EXAMPLES
1.
If the difference between Simple Interest and Compound Interest for 3 years on a sum of money lent at 4% is Rs. 8, them the sum is __ Difference=P[R/100]^2=8 P=8*100*100/4*4=5000 .'.Difference = Rs. 5000
2.
If the difference between Simple Interest and Compound Interest for 2 yrs on a sum of money lent at 6% is Rs. 12, the sum is__ Difference=P(R/100)2=12 P=12x100x100 /6x6 =833.33 .'. Difference = Rs.833.33
TYPE III
Whe the rate interest is R1%,R2% and R3% for 1st year,2nd year and third year respectively,the amount= P(1R1/100)*(1R2/100)*(1R3/100)
EXAMPLES
1.
The rate of interest for the first 2 years is 2%per annum,for the next 3 years is 4% per annum and for the period beyond 5 years is 10% per annum.If a man gets 1500 as simples interest for 6 years,how much money did he deposit Let the deposit be Rs.X *Interest for 2 years =x*2*2/100 =Rs.4x/100 *Interest for 3 years =x*4*4/100 Rs.16x/100 *Interest for 1 yr=x*1*10/100 =Rs.10x/100 *Total interest =Rs.1500 4x/10016x/10010x/100=1500 30x/100=1500
EXAMPLES
1.
If the area of a triangle is 39cm2 and the side is 13cm, find its height A = 1/2bh 39 = 1/2 x 13 x h .'.h=39x2/13=6 .'.Height of the triangle = 6cm
2.
If one side of an equilateral triangle is 6cm, find its height and area Height =/3 /2a =/3/2*6 = 3 /3 cm Area =/3/4 a2 =/3/4*6*6 = 9 3 cm2
3.
If the sides of a triangle are 13cm, 14 cm and 15cm, find its area. S=abc/2 = 13 14 15/2 = 42/2=21 A=/s(s-a)(s-b)(s-c) = /21x8x7x6 =/7x3x2x2x2x7x3x2 =7 x 3 x 2 x 2 = 84 cm2
4.
The perimeter of an equilateral triangle is 150cm. Find its area. Perimeter = 150 cm 3a = 150 a = 50 cm .'.Area =3/4 a2 =3/4*50*50 =625 /3 cm2
5.
The area of a right-angled triangle is 40cm2. If its perpendicular is equal to 10cm, find its base. Area = 40cm2 40 = 1/2 * b * 10 .'.base = 40*2/10 = 8 cm
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