In driving from town A to town D, you pass through town B and then through town C. It is twice as far as from A to b as from B to C, and 10 miles more from C to D than from B to C. If it is 58 miles from A to D, how far is it from town A to C?
If BC is x, then
2x + x + x+10 = 58
x = 12
AC = AB+BC = 2x+x = 36
To solve this problem, we need to break it down and use equations to represent the given information. Let's assign variables to the distances between the towns:
Let x represent the distance from A to B.
Let y represent the distance from B to C.
Let z represent the distance from C to D.
According to the given information, we have the following relationships:
1. It is twice as far from A to B as from B to C: x = 2y
2. It is 10 miles more from C to D than from B to C: z = y + 10
3. The total distance from A to D is 58 miles: x + y + z = 58
Now, let's solve these equations to find the values of x, y, and z.
From the first equation (x = 2y), we can substitute 2y for x in the third equation:
2y + y + z = 58
Simplifying this equation, we get:
3y + z = 58
Next, we substitute (y + 10) for z in the above equation, using the second equation:
3y + y + 10 = 58
Combining like terms:
4y + 10 = 58
Subtracting 10 from both sides:
4y = 48
Dividing both sides by 4:
y = 12
Using this value of y, we can find x by substituting it in the first equation:
x = 2y
x = 2 * 12
x = 24
Now that we have the values of x and y, we can find z using the second equation:
z = y + 10
z = 12 + 10
z = 22
Therefore, the distance from town A to C is x + y:
Distance from A to C = x + y
= 24 + 12
= 36 miles
So, it is 36 miles from town A to C.