Question

A man wishes to invest $35000. He can buy savings bonds which pays simple interest at the rate of 12% per annum or he can start a savings account which pays compound interest at the same rate. Calculate the difference in the amounts of the two investments at the end of 6 years

Answer 1

Answer:

$8,750.

Step-by-step explanation:

After reading question carefully, we came to know that, we've to calculate simple and compound interest and we've to find the difference of amounts of both interests.

Calculating Amount of Simple Interest

We can calculate amount of simple interest by two methods.

Principal = P = $35,000.

Rate = R = 12% p.a.

Time = T = 6years.

Method 1: By amount formula,

A = P{1+ (R×T)/(100)}

⇒ A = 35000{1+ (12×6)/(100)}

⇒ A = 35000{1+ (72)/(100)}

⇒ A = 35000{1 + 0.72}

⇒ A = 35000 × 1.72

⇒ A = 60200.

∴ Amount = $60,200.

Method 2: By formulas

S.I. = (P×R×T)/(100)

A = P + S.I.

By simple interest formula,

S.I. = (P×R×T)/(100)

⇒ S.I. = (35000×12×6)/(100)

⇒ S.I. = 350 × 12 × 6

⇒ S.I. = $25,200.

A = P + S.I.

⇒A = 35000 + 25000

⇒ A = 60200

∴ Amount = $60,000.

Calculating Amount of Compound Interest

Principal = P = $35,000.

Rate = R = 12% p.a.

Time = n = 6years.

By amount formula,

A = P{1+ (R/100)}^n

⇒ A = 35000{1+ (12/100)}⁶

⇒ A = 35000{1 + (3/25)}⁶

⇒ A = 35000{28/25}⁶

⇒ A = 35000{1.12}⁶

⇒ A =35000{1.12×1.12×1.12×1.12×1.12×1.12}

⇒ A = 35000 × 1.97(aprrox.)

⇒ A = 68950

∴ Amount = $68,950.

Now,

Difference of amounts = 68950 - 60,200

= 8750.

∴ Difference in the amounts of the two investments will be $8,750.