S.I & C.I Quiz 1

Solution:
SI = \frac{8000\times 4 \times 2}{100} = 640[/latex<br /> CI = 8000(1 +\frac{1}{100})^{8} - 8000= 662.85

Difference = 662.85 - 640 = Rs. 22.85

Solution:
Case 1
P = 10000
R = 10%
N = 3
CI = P(1 +\frac{R}{100})^{N} - P
= 10000(1 +\frac{10}{100})^{3} - 10000
=3310
Case 2
P = 10000
R = 15%
N = 2
CI = P(1 +\frac{R}{100})^{N} - P
= 10000(1 +\frac{15}{100})^{2} - 10000
=3225

Difference = 3310 - 3225 = Rs. 85

Solution:
P = 225
R = 10 %
n = 2 yr 6 month = 2\tfrac{6}{12}
Amount = 225(1 + \frac{10}{100})^{2}(1 + \frac{\frac{6}{12}\times10}{100}) = 285.86
CI = A - P = 285.86 - 225 = Rs. 60.86

Solution:
Case 1
P = 5000
R = 10%
T = 3
SI = \frac{PRT}{100} = \frac{5000 \times 10 \times 3}{100} = 1500
Case 2
P = 5000
R = 15%
T = 2
SI = \frac{PRT}{100} = \frac{5000 \times 15 \times 2}{100} = 1500

So difference = 1500 - 1500 = 0

Solution:
P = 100
R = 8%
T = 3

CI = P(1 +\frac{R}{100})^{T} - P
= 100(1 +\frac{8}{100})^{3} - 100
= Rs 25.97
SI = \frac{PRT}{100} = \frac{100 \times 8 \times 3}{100} = Rs.24

Difference = 25.97 - 24 = Rs.1.97

Solution:
P=15000,
r = 24%
As compounded quarterly R= 6%
T = 9 = 3 quarters
CI = P(1 +\frac{R}{100})^{T} - P
= 15000(1 +\frac{6}{100})^{3} - 15000
= Rs. 2865.24

Solution:
An amount get doubled at compound interest after \frac{72}{R} years
Where R = Compound interest rate.
Here R = 9.5%
So T = \frac{72}{9.5} = 7.57 years
So minimum complete years required = 8 years

Solution:
Let x be Messi’s share. Then Ronaldo gets 1850 - x
Interest rate R = 5%
CI = P(1 +\frac{R}{100})^{T} - P
We have,
( x(1 +\frac{5}{100})^{4} - x) = ((1850-x)(1 +\frac{5}{100})^{6} -(1850-x))
Solving,
x = Rs. 1132.3

Solution:
The salary grew from 3600 to 4800 in 3 years.
So,
3600(1 +\frac{R}{100})^{3} = 4800
Solving, R = 10.06%
After 3 years,
4800(1 +\frac{10.06}{100})^{3} = 6399.26

Hence, the salary three years from now will be = Rs. 6400

Solution:
P = 15000
R = 20 %
N = 3

Amount = P(1 +\frac{R}{100})^{T}
= 15000(1 +\frac{20}{100})^{3}
=Rs. 25920