Q. Find the simple interest on Rs. 3000 at rate of interest 6 1/4% for 73 days
A. 73 days = 73/365 = 1/5 yrs.
SI = (3000 * 25/4 * 1/5) / 100
Q. An amount loaned at interest rate 13.5% per annum becomes Rs. 2502.50 after 4 years. Find the sum.
A. P + SI = P + P * 13.5 * 4 / 100 (Adding P on both sides)
2502.5 = P ( 1 + 54/100)
2502.5 = P (1.54)
P = 2502.5 / 1.54 = 1625
Q. A borrowed money at interest rate 6% for first two years, 9% for next three years and 14% for period beyond 5 years. If he pays a total interest of Rs. 11400 after 9 yrs how much did he borrow?
A. 11400 = (P *6*2/100) + (P*9*3/100) + (P*14*4/100)
11400 = (12+27+56)P / 100
11400 * 100 / 95 = P
P = 12000
Q. A certain sum of money amounts to Rs 1008 in 2 years and Rs. 1164 in 3.5 yrs.Find sum and RI
A. SI for 1.5 yrs = 1164 - 1008 = 156
SI for 1 yrs = 156 * 2 / 3 = 104 and two years = 208.
Principal = 1008 - 208 = 800
Use simple interest formula to get RI.
Q. A sum of 1500 is lent in 2 parts where one is at 8% and second is at 6%. If the total annual income is Rs. 106, find money lent at each rate.
A. (x*1*8/100) + ((1500-x) * 1 * 6 / 100 ) = 106
8x/100 + (9000 - 6x)/100 = 106
8x +9000 - 6x = 10600
2x = 1600
x=800
When P = principal, n = years and R = rate of interest compounded annually
The Amount = P (1+R/100)^^{n}
^{ }
When P = principal, R = rate of interest compounded half yearly the
Q. Find CI on Rs.7500, compounded annually at RI of 4% for 2 years.
A. Amount = 7500 * (1 + 4/100)^^{2}
Then amount - 7500 gives CI.^{ }
Q. CI is compounded half yearly, principal = Rs. 10000 in rate 4% for 2 years.
A.^{ }Amount = 10000( 1+ 2/100)^4
Amount = 10824.32
CI = Amount - principal = 10824.32 - 10000 = 824.32
Q. Difference between SI and CI accrued
on an amount of Rs. 18000 in 2 years is Rs.405. What is the
RI.
A. P( 1+ (R/100) )^^{n} - P*R*T/100 = 405
{18000 ( 1 + (R/100))^^{2 }- 18000}^{ }- (18000 * R * 2 / 100) = 405
Solving this you can get R.
Q. Divide 1301 between A and B such that the amount of A after 7 years is equal to amount in B after 9 years. Interest is compounded at 4%.
A. Let the amount be 'x' and 1301 - x.
x(1+4/100)^^{7} = (1301-x)(1+4/100)^^{9}
Solving this we can get 'x'.^{ }
^{ }
Q. A sum of money amounts to Rs.6690 after 3 years and Rs. 10035 after 6 years on CI. Find the sum.
A. P(1+R/100)^3 = 6690 ; P(1+R/100)^6 = 10035
Dividing first eqn by second eqn we get (1+R/100)^3 = 10035/6690 = 3/2^{ }
Substituting this value in first equation we get P = 6690 * 2 / 3 = 4460.^{ }
^{ }
Q. A sum doubles itself in 9 years how many will it take to become 8 times.
A. P(1+R/100)^9 = 2P
(1+R/100)^9 = 2
Now finding P(1+R/100)^^{n} = 8P
we need to get (1 + R/100)^^{n }= 8 but 8 = 2^^{3}
(1+R/100)^n = (1 + R/100)^ ^{9 ^3}
we know that A^^{b^c }= A^^{b*c}
so (1+R/100)^^{n} = (1 + R/100)^ ^{9*3} = (1 + R/100)^^{27}
n = 27
Q.A sum of money compounded annually becomes Rs.625 in two years and Rs.675 in three years. The rate of interest per annum is
Ans.b
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