1 A 910 r 16 t 13 weeks P 2 A 14560 P 13000 t 4 mont

1. A = $910; r = 16%; t = 13 weeks; P = ?

2. A = $14,560; P = $13,000; t = 4 months; r = ?

3.A =$736; P = $640; r = 15%; t = ?

4. I = Prt; for r

5. I = Prt; for P

6. A = P + Prt; for P

7. A = P + Prt; for r

8. A = P (1+ rt); for t

9. I = Prt; for t

Solution

There is no mention of compounding. Therefore, we will presume that simople interest is being charged.

1. A = P( 1 + rt) here, A = 910, r = 0.16/52 and t = 13 . Therefore, 910 = P( 1 + 0.16/52*13) = P ( ! + 0.04) = P (1.04) so that P = 910/1.04 = $ 875.

2. Here, 14560 = 13000[ 1 + ( r/1200)*4] = (13000 /300) ( 300 + r) = (130/3) (300 +r) or, 300 +r = 14560*3/130 336. Therefore, r = 6 % per annum

3.Here, 736 = 640( 1 + 0.15t) Therefore, 1 + 1.15t = 736/640 = 1.15 so that 0.15t = 1.15- 1 = 0.15. Then t = 10

4. Here, I = Prt, therefore, r = I/ Pt

5. Here, I = Prt, therefore P = I/rt

6. A = P+ Prt = P(1 + rt), therefore, P = 1/(1 +rt)

7. A = P + Prt, therefore Prt = A - P so that r = (A - P)/ Pt

8. A = P (1 + rt) = P + Prt or, Prt = A - P, so that t = (A - P)/ Pr

9. I = Prt, so that t = !/Pr