Question

Recall that in the problem involving compound interest, the balance A for P dollars invested at rate r for t years compounded n times per year can be obtained by A = P 1 + r n nt Consider the following situations:________.
(a) P = $2, 500, r = 5%, t = 20 years, n = 4. Find A.
(b) P = $1, 000, r = 8%, t = 5 years, n = 2. Find A.
(c) A = $10, 000, r = 6%, t = 5 years, n = 4. Find P.
(d) A = $50, 000, r = 7%, t = 10 years, n = 12. Find P.
(e) A = $100, 000, r = 10%, t = 30 years, compounded monthly. Find P.
(f) A = $100, 000, r = 7%, t = 20 years, compounded quarterly. Find P.

Answer 1

Answer:

(a)∴A=$6753.71.

(b)∴A=$1480.24

(c) ∴P=$7424.70

(d)∴P=$49759.62

(e)∴P=$5040.99

(f) ∴P=$2496.11

Step-by-step explanation:

We use the following formula

$A=P(1+\frac rn)^{nt}$

A=amount in dollar

P=principal

r=rate of interest

(a)

P=$2,500, r=5%=0.05,t =20 years , n= 4

$A=\ 2500(1+\frac{0.05}{4})^{(20\times 4)$

   =$6753.71

∴A=$6753.71.

(b)

P=$1,000, r=8% =0.08,t =5 years , n= 2

$A=\ 1000(1+\frac{0.08}{2})^{(5\times 2)$

   =$1480.24

∴A=$1480.24

(c)

A=$10,000, r=6% =0.06,t =5 years , n= 4

$10000=P(1+\frac{0.06}{4})^{(5\times 4)}$

$\Rightarrow P=\frac{10000}{(1+0.015)^{20}}$

$\Rightarrow P=7424.70$

∴P=$7424.70

(d)

A=$100,000, r=6% =0.06,t =10 years , n= 12

$100000=P(1+\frac{0.07}{12})^{(10\times 12)}$

$\Rightarrow P=\frac{100000}{(1+\frac{0.07}{12})^{120}}$

$\Rightarrow P=49759.62$

∴P=$49759.62

(e)

A=$100,000, r=10% =0.10,t =30 years , n= 12

$100000=P(1+\frac{0.10}{12})^{(30\times 12)}$

$\Rightarrow P=\frac{100000}{(1+\frac{0.10}{12})^{360}}$

$\Rightarrow P=5040.99$

∴P=$5040.99

(f)

A=$100,000, r=7% =0.07,t =20 years , n= 4

$100000=P(1+\frac{0.07}{4})^{(20\times 4)}$

$\Rightarrow P=\frac{100000}{(1+\frac{0.07}{4})^{80}}$

$\Rightarrow P=2496.11$

∴P=$2496.11