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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.

User Sergiomse
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1 Answer

6 votes

Answer:

A) 2.5

Step-by-step 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

User John Breakwell
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