Question

A consultant traveled 255 miles to attend a meeting, traveling 45 mph hours for the first part of the trip, then increasing to a speed of 60 mph for the second part. If the entire trip took 5 hours, how far did the consultant travel at the faster speed?

Answer 1

The consultant traveled 120 miles at the faster speed.

Step-by-step explanation:

Given,

Total distance = 255 miles

Speed at first part = 45 mph

Speed at second part = 60 mph

Time taken = 5 hours

Let,

x be the time for the distance of first part.

y be the time for the distance of second part.

According to given statement;

x+y=5    Eqn 1

45x+60y=255    Eqn 2

Multiplying Eqn 1 by 45

$45(x+y=5)\\45x+45y=225\ \ \ Eqn\ 3$

Subtracting Eqn 3 from Eqn 2

$(45x+60y)-(45x+45y)=255-225\\45x+60y-45x-45y=30\\15y=30$

Dividing both sides by 15

$\frac{15y}{15}=\frac{30}{15}\\y=2$

Distance covered at faster speed = Speed*Time = 60*2 = 120 miles

The consultant traveled 120 miles at the faster speed.

Keywords: linear equation, elimination method

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