Far Away From School

Let time be 'x'.. First case: Eq: speed=dist/time..... 5=d/x-10 Second case: 4=d/x+5 Domt forget to.convert minutes to hours..... Find d.

We count the time in minute and the distance is the same, so..

$s=s$

$5 \times (t-10) = 4 \times (t+5)$

$5t-50 = 4t+20$

$t=70 minutes$

Then the distance..

$s=v \times t$

$s=5 \times (70-10)$

$s=5 \times 60$

Change the time into hour..

$s=5 \times 1$

$s= \boxed {5 miles}$

he is saying that the difference in time will be 15 min or 1/4 hr so let the distance be n miles now, n/4 - n/5 = 1/4 or n=5 miles

X is the time remaining until the start of class, where you will not be late or early arriving at this time. This is measured in hours. The minutes early or late must be converted.

10 minutes are 0,16666... hours 5 minutes are 0,0833... hours

We have:

$5$ is to $x-0.1666...$

$4$ is to $x+0.08333...$

But the time and the speed are inversely proportional.

Now we have:

$5$ is to $x+0.08333...$ $4$ is to $x-0.1666...$

Making a cross multiplication

$4x+0,333...=5x-0,8333...$

Solving:

$4x-5x=-0,8333...-0,333...$

$-x=-1,1666....$

$x=1,1666....$

This value is in hours. In minutes is 70.

The distance formula is $V=\frac { S }{ T }$

$5=\frac { S }{ 70-10 }$

$5=\frac { S }{ 60 }$

60 minutes is 1 hour. The velocity is in miles per hour, the time and the distance must be measured in hours and miles respectively.

$5=\frac { S }{ 1}$

$5\times 1=S$

$S=5$

The distance from Paul's house to school is 5 miles.

let time be 'x', as we know D=VxT then for first case(V=5,T=X) D=5X.... for second case (V=4, T= X+1/4 hours) D=4(X+1/4)..... solving simultaneously X=1 then Distance(D)=5miles

Let the distance be d.

When he's on time, he walks with speed x, hence t = d/x.

When he's early, the time t1 is d/5.

When he's late, the time t2 is d/4.

d/x - d/5 = 1/6 ( 10 min difference )

d/4 - d/x = 1/12 ( 5 min difference )

The distance follows. :)

d/5 = t-1/6 (equation 1) note: 1/6 is 10 minutes converted to hrs. d/4 = t + 1/12 (equation 2) note 1/12 is 5 minutes converted to hrs., then solve for d. so, d=5.

Let the distance be $x$

Time (1)= $\dfrac{x}{5}$

Time (2)= $\dfrac{x}{4}$

The difference between their time is $10+ 5= 15$ minutes $= 1/4$ hour.

Clearly,

$\dfrac{x}{5} + \dfrac{1}{4} = \dfrac{x}{4}$

$\dfrac{x}{20} = \dfrac{1}{4}$

$\large\boxed{x=5}$

Hence, the distance is 5 miles.

d=5(t-10)/60=4(t+5)/60

5t-50=4t+20

t=70

d=5(t-10)/60

d=5(70-10)60

d=5

If Paul walks at 5 miles per hour, he reaches 10 minutes early, but if he walks at 4 miles per hour, he reaches 5 minutes late. Therefore, the time difference to cover the same distance is 15 minutes. Now,

$t_{1} = \frac {d}{s_{1}}$ ,

where $t_{1}$ is the time taken when walking at $s_{1} = {5} {miles} {per} {hour}$ .

Similarly,

$t_{2} = \frac {d}{s_{2}}$ ,

where $t_{2}$ is the time taken when walking at $s_{2} = 4 miles per hour$ .

We know that,

$t_{2} - t_{1} = 15$ .

Substituting the values and solving for d, we get

$\boxed {d = 5 miles}$ .

Hey yo,

For 5mph, 10 minutes earlier, by applying v = s/t ,

5 = s / (t - 10/60)

5 = s / ( t - 1/6)

5( 6t - 1) = 6s

30t - 5 = 6s

30t - 5 = 2(3s) (1st),

For 4mph, 5 minutes late,

4 = s / ( t + 5/60 )

4 = s / ( t + 1/12 )

4(12t + 1) = 12s

12t + 1 = 3s (2nd),

By substituting (2nd) into (1st),

30t - 5 = 2( 12t + 1)

6t = 7

t = 7/6 hours,

therefore,

s = 12(7/6) + 1 / 3 = 15/3 = 5 miles...

thanks...

distance = 5/60 (t - 10) = 4/60 (t+5) -> t = 70 mins -> distance = 5 miles.

Separate everything into rate time and distance

t= on time

5 (t-1/6) = 4 (t+1/12)

t= 7/6

d= 5

Time for 1st: Distance/Speed = x/5,

Time for 2nd : x/4,

Slower Time is x/4 and faster is x/5,

Equation: slower time - faster time = difference of time,

x/4 - x/5 = 15/60

Take LCM

5x- 4x = 20*3/12

x = 5 miles (distance travelled)

Let time taken usually be T.

Distance D =5(T-10/60) = 4(T+5/60)

Solving T = 7/6.

D= 5(7/6-10/60) = 5