Question

A boat traveled upstream 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average speed of (v+3) miles per hour. If the trip upstream took a half hour longer than the trip downstream, then how many hours did it take the boat to travel downstream?
A. 2.5.

B. 2.4.

C. 2.3.

D. 2.2.

E. 2.1.

Answer 1

Answer:

A) 2.5

Explanation:

Extracting vital information from the question;

speed upstream = (v-3)mile/hr, distance traveled = 90 mile, speed downstream = (v+3), time for downstream in hours = t while time for upstream = t + 0.5 hr since the upstream journey is half hour longer.

speed = distance / time = 90 / (t + 0.5)

( v - 3) = 90 / ( t + 0.5)

cross multiply

(v-3) (t + 0.5) = 90 equation (1) for upstream motion

( v+3) = 90 / t

cross multiply

t(v+3) = 90 equation (2) for downstream motion

make v subject of the formula in equation 2

vt + 3t = 90

vt = 90 - 3t

divide both side by t

vt/t = (90 - 3t) / t

v = (90 - 3t) / t

 substitute for t in equation 1

(( 90 - 3t) / t) - 3) (t + 0.5) = 90

solve through finding l.c.m ( lowest common multiple)

(90 - 3t - 3t)/ t (t + 0.5) = 90

(( 90 - 6t ) / t )(t + 0.5) = 90

open the brackets and cross multiply

90t + 45 - 6t² - 3t = 90 t

rearrange  and collect the like terms

- 6t² - 3t + 45 = 90t - 90t

- 6t² - 3t + 45 = 0 multiply both side by -1

6t² + 3t - 45 = 0

divide both side by 3

2t² + t - 15 =  0

factorize the expression by multiplying - 15 by 2t² = - 30t²

find factors of - 30t² that adds to t  = 6t × (-5t)

replace t with (+6t - 5t) in the equation

2t²+ 6t - 5t - 15 = 0

factorize

2t ( t + 3) - 5 ( t + 3) =0

(2t -5)(t + 3) = 0  

2t - 5  = 0 or t + 3 = 0

2t = 5 or t = -3

divide through 2

2t / 2 = 5/ 2 = 2.5 or t = -3 since time cannot be negative

them t = 2..5 seconds