MIGUEL LEAVES SHREVEPORT LA AND TRAVEL NORTH AT 85 KILOMETER

MIGUEL LEAVES SHREVEPORT. LA, AND TRAVEL NORTH AT 85 KILOMETERS PER HOUR. TRAVIS LEAVES AT THE SAME TIME AND TRAVEL SOUTH AT 95 KILOMETERS PER HOUR. HOW LONG WILL IT TAKE BEFORE THEY ARE 315 KILOMETERS APART?

SAM HAS $ 40,000 TO INVEST. HE WILL PUT PART OF THE MONEY IN AN ACCOUNT PAYING 4% SIMPLE INTEREST, AND THE REMAINDER INTO STOVKS PAYING 6% SIMPLE INTEREST. HIS ACCOUNTANT TELLS HIM THE TOTAL ANNUAL INCOME FROM THESE INVESTMENTS SHOULD BE $ 2040. HOW MUCH SHOULD HE INVEST AT EACH RATE?

Solution

1. Let Miguel and Travis be 315 km apart x hours after leaving Shreveport.In x hours. Miguel travels 85x km ( @ 85 kph). Also in x hours, Travis travels 95x km ( @ 95kph). Since Miguel and Travis are travelling in opposite directions, we have 85x + 95x = 315 or, 180x = 315 or, x = 315/180 = 1.75 hours or, 1 hour, 45 minutes.

2. Let the amount invested by SAM in the fixed deposit paying 4 % simple interest be x. Then, the amount invested in the stocks yielding 6 % simple interest is 40000 - x. The annual interests on these two investments are 0.04x and 0.06( 40000 - x) respectively. Since the total annual interest is $ 2040, we have, 0.04x + 0.06( 40000 - x) = 2040. On multiplying both the sides by 100, we have 4x + 6(40000 - x) = 204000 or, 4x + 240000 - 6x = 20400 -2x = 204000 - 240000 or, -2x = - 36000 or, 2x = 36000 , so that x = 18000. Then 40000 - x = 40000 - 18000 = 22000. Thus SAM invests $ 18000 in the fixed deposit paying 4 % simple interest and $ 22000 in the stocks yielding 6 % simple interest.