2586.The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is:
3
4
5
6
Explanation:
| P$ \left(1 +\dfrac{20}{100} \right)^ n $> 2P $\Rightarrow$ $ \left(\dfrac{6}{5} \right)^n $> 2. |
| Now,$ \left(\dfrac{6}{5} \times\dfrac{6}{5} \times\dfrac{6}{5} \times\dfrac{6}{5} \right) $> 2. |
So, $ n $ = 4 years.
2588.Albert invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at compound interest rate 5 p.c.p.a. How much amount will Albert get on maturity of the fixed deposit?
Rs. 8600
Rs. 8620
Rs. 8820
None of these
Explanation:
| Amount | = Rs.$ \left[8000 \times\left(1 +\dfrac{5}{100} \right)^2\right] $ |
| = Rs.$ \left(8000 \times\dfrac{21}{20} \times\dfrac{21}{20} \right) $ | |
| = Rs. 8820. |
2589.What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12 p.c.p.a.?
Rs. 9000.30
Rs. 9720
Rs. 10123.20
Rs. 10483.20
Explanation:
| Amount | = Rs.$ \left[25000 \times\left(1 +\dfrac{12}{100} \right)^3\right] $ |
| = Rs.$ \left(25000 \times\dfrac{28}{25} \times\dfrac{28}{25} \times\dfrac{28}{25} \right) $ | |
| = Rs. 35123.20 |
$\therefore$ C.I. = Rs. (35123.20 - 25000) = Rs. 10123.20
2592.The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period in years is:
2
2$ \dfrac{1}{2} $
3
4
Explanation:
Amount = Rs. (30000 + 4347) = Rs. 34347.
Let the time be $ n $ years.
| Then, 30000$ \left(1 +\dfrac{7}{100} \right)^n $= 34347 |
| $\Rightarrow \left(\dfrac{107}{100} \right)^ n$ =$ \dfrac{34347}{30000} $=$ \dfrac{11449}{10000} $=$ \left(\dfrac{107}{100} \right)^2$ |
$\therefore n $ = 2 years.
44150.Find the compound amount and compound interest on the principal Rs.20,000 borrowed at 6% compounded annually for 3 years.
3820.32
3080.23
1500.00
3100.32
Explanation:
A=P(1+r)n
=20000(1+.06)3=23820.32
The compound interest
=23820.32−20000=3820.32
Let P = 20000, r = 6%, n = 3
using formulaA=P(1+r)n
=20000(1+.06)3=23820.32
The compound interest
=23820.32−20000=3820.32
44152.Find the compound amount which would be obtained from the interest of Rs.2000 at 6% compounded quarterly for 5 years.
3256.97
2567.90
1693.17
2693.71
Explanation:
=2000(1+.015)20
=2693.71
Let principal = 2000, r=6%=64×100=.015, n=5×4=20quarters
∴A=P(1+r)n=2000(1+.015)20
=2693.71
44154.If the present value of my investment is $1,000 and the rate of interest is 6% compounded annually, what will the value be after 10 years?
$1,600
$1,771.56
$1,790.85
$1,898.30
Explanation:
Use the formula:
FV = PV × (1 + r)n
Substitute PV = 1,000, r = 6% = 0.06, and n = 10
FV = 1,000 × (1 + 0.06)10 = 1,000 × (1.06)10 = 1,000 × 1.790847...
= 1,790.847
... The value after 10 years = 1,790.85
FV = PV × (1 + r)n
Substitute PV = 1,000, r = 6% = 0.06, and n = 10
FV = 1,000 × (1 + 0.06)10 = 1,000 × (1.06)10 = 1,000 × 1.790847...
= 1,790.847
... The value after 10 years = 1,790.85
44156.If the present value of my investment is $9,000 and the rate of interest is 3½% compounded annually, what will the value be after 4 years?
$10,176.87
$10,260
$10,324.27
$10,327.71
Explanation:
Use the formula: FV = PV × (1 + r)n
Substitute PV = 9,000, r = 3½% = 0.035 and n = 4
FV = 9,000 × (1 + 0.035)4 = 9,000 × (1.035)4
= 9,000 × 1.14752...
= 10,327.707...
So the value after 6 years = 10,327.71
Substitute PV = 9,000, r = 3½% = 0.035 and n = 4
FV = 9,000 × (1 + 0.035)4 = 9,000 × (1.035)4
= 9,000 × 1.14752...
= 10,327.707...
So the value after 6 years = 10,327.71
44158.Your goal is to have $2,000 in 6 years. The rate of interest is 10% compounded annually, so how much should you start with?
$800
$1,026.32
$1,116.79
$1,128.95
Explanation:
Use the formula:PV=$\dfrac{FV}{(1+r)^n}$
substitute FV=2,000 ,r=10% =0.10 and n=6
PV=$\dfrac{2,000}{(1+0.10)^6}$
=$\dfrac{2000}{(1.10)^6}$
=$\dfrac{2000}{1.771561}$
=$1,128.947
so you should start with 1,128.95
substitute FV=2,000 ,r=10% =0.10 and n=6
PV=$\dfrac{2,000}{(1+0.10)^6}$
=$\dfrac{2000}{(1.10)^6}$
=$\dfrac{2000}{1.771561}$
=$1,128.947
so you should start with 1,128.95
44160.Your goal is to have $3,500 in 10 years. The rate of interest is 3% compounded annually, so how much should you start with?
$2,126.16
$2,450
$2,604.33
$2,629.60
Explanation:
Use the formula:PV=$\dfrac{FV}{(1+r)^n}$
substitute FV=3,500 ,r=3% =0.03 and n=10
PV=$\dfrac{3500}{(1+0.03)^10}$
=$\dfrac{3500}{(1.03)^10}$
=$\dfrac{3500}{1.343916}$
=2,604.328
so you should start with 2,604.33
substitute FV=3,500 ,r=3% =0.03 and n=10
PV=$\dfrac{3500}{(1+0.03)^10}$
=$\dfrac{3500}{(1.03)^10}$
=$\dfrac{3500}{1.343916}$
=2,604.328
so you should start with 2,604.33