# solving equations in English

• Dec 22nd 2006, 08:36 PM
shenton
solving equations in English
Systems of equations can be easily solved through the matrix. But when English is used, the difficulty of the problem is compounded. I am stumped by this question, I cannot figure out how to solve this question. Please help.

Jack drives from her country home to the city, a distance of 380 km. He drives part of the way on back roads, at an average speed of 50 km/h, and part on highways, at an average speed of 100 km/h. The trip takes 5 hours. How far does he drive on each type of road?

Thanks.
• Dec 22nd 2006, 10:01 PM
CaptainBlack
Quote:

Originally Posted by shenton
Systems of equations can be easily solved through the matrix. But when English is used, the difficulty of the problem is compounded. I am stumped by this question, I cannot figure out how to solve this question. Please help.

Jack drives from her country home to the city, a distance of 380 km. He drives part of the way on back roads, at an average speed of 50 km/h, and part on highways, at an average speed of 100 km/h. The trip takes 5 hours. How far does he drive on each type of road?

Thanks.

Translate this into an equation. Let $x \mbox{ km}$ be the distance travelled on back roads. Then $380-x\mbox{ km}$ is the distance travelled on highways. Then the time taken is:

$t=x/50+(380-x)/100=5 \mbox{ hours}$

Now solve this equation for $x$.

RonL
• Dec 27th 2006, 07:05 AM
Soroban
Hello, shenton!

I use baby-talk with such word problems . . . it works for me.

Quote:

Jack drives from her country home to the city, a distance of 380 km.
He drives part of the way on back roads, at an average speed of 50 km/h,
and part on highways, at an average speed of 100 km/h.
The trip takes 5 hours.
How far does he drive on each type of road?

There are a number of approaches to this problem.
If you prefer a system of equations, we can do it like this . . .

Let $x$ = number of km on back roads.
Let $y$ = number of km on highways.
. . We know that: . $x + y \:=\:380$ (1)

He drove $x$ km at 50 km/hr.
. . This took him $\frac{x}{50}$ hours.
He drove $y$ km at 100 km/hr.
. . This took him $\frac{y}{100}$ hours.
Since his total time was 5 hours, we have: . $\frac{x}{50} + \frac{y}{100} \:=\:5$
. . Multiply by 100: . $2x + y \:=\:500$ (2)

Solve the system of equations: . $\begin{array}{cc}(1) \\ (2) \end{array}\begin{array}{cc} x + y & =\:380 \\ 2x + y & =\:500\end{array}$

Subtract (1) from (2): . $\boxed{x \,= \,120}$

Substitute into (1): . $120 + y\:=\:380\quad\Rightarrow\quad\boxed{y \,=\,260}$

Therefore, Jack drove: . $\begin{array}{cc} 120\text{ km} & \text{on back roads} \\ 260\text{ km} & \text{on highways}\end{array}$

• Dec 27th 2006, 12:56 PM
shenton
Thanks, Soroban for showing the 2 equations and the detailed workings.

This question and the quarters and dimes questions are one of the most difficult word problems.