Question

Two cars leave towns 1000 kilometers apart at the same time and travel toward each other. One car's rate is 16 kilometers per hour less than the other's. If they meet in 5 hours, what is the rate of the slower car? Do not do any rounding.

Answer 1

Answer:

The slower car travels at 90km/hr.

Step-by-step explanation:

D = total distance both cars cover

t = time passed from starting point to when the cars meet

Va=the speed of the slower car

Vb=the speed of the faster car

DA=the distance travelled by the slower car.

DB=the distance travelled by the faster car.

 

The sum of the distances travelled by each car equals the total distance between the two starting points.

D = DA + DB

 

The distance travelled for each car is the speed multiplied by the time elapsed.

 

DA = VAt

DB = VBt

 

From the problem we know, VB = VA + 20

 

Putting all of this together,

 

D = VAt + (VA + 20)t,

 

D/t = 2VA + 20

 

VA = D/2t - 10

 

plugging in all the values: VA = 90km/hr.

 

the faster automobile travels at VB = 110km/hr

Answer 2

Answer: the rate of the slower car is 92 kilometers/ hour

Step-by-step explanation:

Initially, the two cars were 1000 kilometres apart before they started travelling towards each other.

Let the speed of the faster car be y km/ h

Let the speed of the slower car be x km/h

One car's rate is 16 kilometers per hour less than the other's. This means that

y = x + 16

x = y - 16

If they meet in 5 hours, it means that both cars have covered a total distance 1000 miles.

Distance = speed × time

Distance travelled by the faster car in 5 hours is

y × 5 = 5y kilometers

Distance travelled by the slower car in 5 hours is

x × 5 = 5x kilometers

Therefore,

5x + 5y = 1000

x + y = 200 - - - - - - - - -1

Substituting y = x + 16 into equation, it becomes

x + x + 16 = 200

2x = 184

x = 184/2 = 92 kilometers/ hour.

y = + 16 + x

y = 16 + 92. = 208 kilometers