Let's Get Some Rest

Let $D$ be the downhill distance, $P$ be the plane surface distance and $U$ be the uphill distance, all of them in kilometers.

We can extract from the problem that

$\frac{D}{100} + \frac{P}{80} + \frac{U}{60} =2$ $\frac{D}{60} + \frac{P}{80} + \frac{U}{100} =3$

because $\frac{\text{Distance}}{\text{Speed}}= \text{Time}$ .

Plugging in $P = 8$ and adding together the two equations, we get that $\frac{D}{100} + \frac{D}{60} + \frac{16}{80} + \frac{U}{60}+ \frac{U}{100} = 5$ $\Rightarrow \frac{16D}{600} + \frac{16U}{600} = \frac{2880}{600}$ which leads us to $D + U = 180$ . Adding $P$ , we find out that the total distance $D+P+U$ is $\boxed{188.}$

While going towards beach

distance travelled downhill = x

distance travelled uphill = y

so time : (8/80) + (x/100) + (y/60) = 2

first eqn : 3x + 5y = 570

while travelling back

distance travelled downhill = y

distance travelled uphill = x

so time : (8/80) + (x/60) + (y/100) = 3

second eqn : 5x + 3y = 870

so solving 2 eqns

x=165

y=15

so total distance = 165 + 15 + 8

=188

Let the distance of downhill, plane surfaces, and uphill be $x$ , $y$ , and $z$ respectively. Now we have the system of equations

$\begin{cases} \frac{x}{100}+\frac{y}{80}+\frac{z}{60}=2\cdots(1)\\ \frac{x}{60}+\frac{y}{80}+\frac{z}{100}=3\cdots(2)\\y=8\cdots(3)\\ \end{cases}$

and we want to find the value of $x+y+z$ . We add $(1)$ and $(2)$ to get

$\frac{(60+100)(x+z)}{600\cdot10}+\frac{y}{40}=5 \Rightarrow \frac{160(x+z)}{6000}=\frac{24}{5} \Rightarrow x+z=180$

Hence $x+y+z=180+8=\boxed{188}$ .

First, I determined that the plane travel of 8km would be traveled in 1/10 hr.

Then,I looked at the outbound trip and decided that if x=uphill travel in km (outbound only) it would be traveled in A hours, and y=downhill travel in km(outbound only) it would be traveled in B hours, I could formulate the following equations:

1/10+A+B=2hr x=60A y=100B 8+x+y=J km J being the solution to the problem-designating the total km

Next, I looked at inbound travel (from the beach back to the city), x and y would remain the same but the speed would be inverted and I represented the time asthe distance y being traveled in C hr. and distance x being traveled in D hr. as such:

1/10+D+C=3 hr x=100B y=60C and still J would equal 8+x+y

The distances represented by x and y will be the same no matter which direction of travel, therefore the x values and the y values can be set equal to each other:

60A=100D 100B=60C

From these equations I determined that outbound travel could be represented by:

8+60A+100B=J

And inbound travel:

8+100D+60C=J

Since the distance is the same these equations can be set equal to each other:

8+60A+100B=8+100D+60C

Using all these equations I decided to start by putting everything I could in terms of A.

Therefore, 60A=100D becomes: D=3/5 A

and

1/10+A+B=2 becomes: B=19/10 - A

I plugged both of these into my equation which set the distances equal to each other:

8+60A+100B=8+100D+60C 8+60A+100(19/10 - A)=8+100(3/5 A)+60C

Now I can solve the aforementioned equation for C:

C=19/6 - 5/3 A

which means I can use the equation for inbound travel (1/10 + D +C=3) and substitute both D and C for solutions in terms of A:

1/10 + 3/5 A + (19/6 - 5/3 A)=3

This gives the a value for A of 1/4

Using this value I solved for B using the following equation:

1/10 +A +B=2 1/10+ (1/4) + B=2

This gives me a value of B of 33/20

I continued this process of plugging in values to determine C=55/20 and D=3/20

Since I had values for A, B, C, and D I checked my values with both equations (inbound and outbound) values for x and y : x=15 y=165

Then I went back to my equation for distance J:

8+x+y=J 8+(15)+(165)=J

For a solution of 188km

The key to this question is the fact that the distance uphill when travelling from the city to the beach is equal to the distance downhill when travelling from the beach to the city. (obviously, if you go uphill from A to B, you gotta go downhill from B to A). Similarly, the distance downhill when travelling from the city to the beach is equal to the distance uphill when travelling from the beach to the city.

Let x be the distance uphill when travelling from the city to the beach and, thus, the distance downhill when travelling from beach to city.

Let y be the distance downhill when travelling from the beach to the city and, thus, the distance uphill when travelling from the beach to the city.

Case 1: Travelling from city to the beach

We need to travel x km uphill at 60 km/hr. Therefore, the time taken to do this is: (Acc. to distance = speed x time taken)

$t_1 = \frac {x}{60}$

We need to travel y km downhill at 100 km/hr. Therefore, the time taken to do this is:

$t_2 = \frac {y}{100}$

We need to travel 8 km along a plane surface at 80 km/hr. Therefore, the time taken to do this is:

$t_3 =\frac {8}{80}$

The total time to travel from the city to the beach will be $t_1 + t_2 + t_3$ . This total time is given as 2 hours. Therefore:

$t_1 + t_2 + t_3 = 2$

$\frac {x}{60} + \frac {y}{100} + \frac {8}{80} = 2$

Simplifying the above, we get:

$5x + 3y = 570$ ------------ ( Equation 1)

Case 2: Travelling from beach to city

We need to travel y km uphill at 60 km/hr. Therefore, the time taken to do this is:

$t_4 = \frac {y}{60}$

We need to travel y km downhill at 100 km/hr. Therefore, the time taken to do this is:

$t_5 = \frac {x}{100}$

We need to travel 8 km along a plane surface at 80 km/hr. Therefore, the time taken to do this is:

$t_6 =\frac {8}{80}$

The total time to travel from the city to the beach will be $t_4 + t_5 + t_6$ . This total time is given as 3 hours. Therefore:

$t_4 + t_5 + t_6 = 3$

$\frac {y}{60} + \frac {x}{100} + \frac {8}{80} = 3$

Simplifying the above, we get:

$3x + 5y = 870$ ------------ ( Equation 2)

Solving Equations 1 and 2, we get x=15 and y=165. x + y = 180. 180 does not include distance along plane surface which is 8km.

Therefore, total distance required = x + y + 8 = 188

Simple two equations X/100 + Y/60 = 2-8/80 and X/60 +Y/100 = 3-8/80 solving these two equations results in 165 km downhill, 15 km uphill and 8 km plane while driving from city to beach

There are a lot of ways to answer this question.nut i prefer to use this way.

we note that there are three part of places which are down hill uphill and plane

so we minus both duration with 0.1 which i get from plane road

so we got a+b=the distant,1.9=a/s1+b/s2,2.9=a/s2+b/s1

we plus the second and third equation and we got 28800=160(a+b)

finally we got a+b=180 but the answer still not perfect before you add 8

the finnal answer is 188

X/100+8/80+y/60=2:x/60+8/80+y/100=3:solve this two system of equations .we'll get answer

2 equations two unknowns

Concept-the distances travelled from city to beach and beach of city are sum...but time taken are different....when one is going from city to beach ,the uphill and downhill are downhill and uphill respectively while coming from beach to city.... Thus L/100+X/60+8/80=>2 where X,L are distances travelled on uphill and downhill while going from city to beach.....now while coming from beach to city X becomes distance travelled on downhill and L on uphill.....therefore total time= 3=L/60+X/100+8/80...solving this 2 equation we get value of X and L....thus adding all we get total distance.....here 8 km is the distance travelled on plane surface....

Let x is the length of the road downhill, y is the length of the road uphill. So, x/100 + y/60 + 8/80 = 2 and x/60 + y/100 + 8/80 = 3 <=> x =165, y =15 <=> The distance in kilometers between the city and the beach is 165 + 15 + 8 = 188