Question

Use the following compound interest formula to complete the problem. a = p (1 startfraction r over n endfraction) superscript n superscript t sandra has two credit cards, p and q. card p has a balance of $726.19 and an interest rate of 10.19%, compounded semiannually. card q has a balance of $855.20 and an interest rate of 8.63%, compounded monthly. assuming that sandra makes no purchases and no payments with either card, after four years, which card’s balance will have increased by more, and how much greater will that increase be? a. card q’s balance increased by $7.22 more than card p’s balance. b. card q’s balance increased by $6.69 more than card p’s balance. c. card p’s balance increased by $3.43 more than card q’s balance. d. card p’s balance increased by $0.80 more than card q’s balance. please select the best answer from the choices provided a b c d

Answer 1

The difference between the amount in two cards of Sandra after 4 years will be $125.58 approx.

How to find the compound interest?

If n is the number of times the interested is compounded each year, and 'r' is the rate of compound interest annually, then the final amount after 't' years would be:

$a = p(1 + \dfrac{r}{n})^{nt}$

For this case, Sandra has got 2 credit cards p and q. Calculating the final amount both card will have after 4 years:

• Case 1: For card p:

Initial amount = p = $726.19

The rate of interest is r = 10.19% = 10.19/100 = 0.1019 (converted percent to decimal)

Years for which amount is compounded = 4 years, Thus, t = 4

The interest is compounding semiannually, that means twice per year, thus, n = 2

Thus, the final amount in card p after 4 years would be:

$a = p(1 + \dfrac{r}{n})^{nt}\\\\a = 726.19(1+0.1019/2)^{2 \times 4} \\\\a= 726.19 \times (1.05095)^8 \approx 1080.70\: \rm (in \: dollars)$

• Case 2: For card q:

Initial amount = p = $855.20

The rate of interest is r = 8.63% = 8.63/100 = 0.0863 (converted percent to decimal)

Years for which amount is compounded = 4 years, Thus, t = 4

The interest is compounding monthly, that means twelve times per year, thus, n = 12

Thus, the final amount in card q after 4 years would be:

$a = p(1 + \dfrac{r}{n})^{nt}\\\\a = 855.20(1+0.0863/12)^{12 \times 4} \\\\a= 855.20\times (1.0071916)^{48} \approx 1206.28\: \rm (in \: dollars)$

Clearly card q has more amount than card p at the end of 4 years.

The difference is: 1206.28 - 1080.70 = 125.58 (in dollars)

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