Question

If Mr. Cruz will increase his usual rate by 10 kilometers per hour, it will take him one hour shorter to cover a distance of 200 kilometers. What is his usual speed in driving?

Answer 1

Let V be his usual rate, and T the usual time.

V*T = 200 km

We also know that if he speeds up by 10 km/h, the time goes down by an hour for the same distance.

(V+10 km/h)(T-1 hr) = 200 km

You should be able to solve that by substitution.

Answer 2

You'll get a quadratic equation with two roots, but only one will make sense here.

Answer 3

What's your answer?

Answer 4

I still don't get it.

Answer 5

okay i just saw this question so what dont you get?

Answer 6

what equation I can make out of the problem.

Answer 7

So the questions asks for an equation?

Answer 8

no but i would need it to get the answer. I don't know how to get the answer.

Answer 9

well just one question what grade of math is this?

Answer 10

what do you mean?

Answer 11

ike is it 6th grade algebra 7th 8th etc

Answer 12

9th grade

Answer 13

basically just take this (V+10 km/h)(T-1 hr) = 200 km and substitute the variables. Or is that what you can't do?

Answer 14

yes, it is.

Answer 15

well this is a bit out of my hands Im still in 8th but try bumping your question someone might see it.

Answer 16

okay.. Thanks!

Answer 17

\[(V + 10)(T-1) = 200\]
\[VT - V + 10T -10 = 200\]
But we know that VT = 200 — that's the usual time and speed.
\[200 - V + 10T -10 = 200\]Subtract 200 from both sides
\[10T - V -10 =0\]To solve this, we use VT=200 and solve it for V
[\V = 200/T\]
\[10T - (200/T) - 10 = 0\]Multiply both sides by T to eliminate fraction
\[10T^2 - 200 - 10T = 0\]
Now use the quadratic formula with a = 10, b = -10, c = -200
\[T = \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-(-10)\pm\sqrt{(-10)^2-4(10)(-200)}}{20}\]
\[T=\frac{10\pm\sqrt{100-(-8000)}}{20} = \frac{10\pm\sqrt{8100}}{20} = \frac{10\pm 90}{20} = 5, -40\]
The only solution that makes sense here is T = 5, because he can't very well be taking -40 hours to drive somewhere.

Now you know T, find his original velocity from V*T = 200. Then find the new velocity and time from (V + 10) and (T - 1) and make sure the solution works!