Alicia's Commute

very easy assume that the correct time reqd is t and distance is d so, acc to given cond time diff in case of 45 miles per hour=5min=5/60 hours 40(t+1/20)=d; 45(t-1/12)=d; solve for d; voila ans is 48

$d=v \times t$ $$ $d$ is distance, $v$ is speed and $t$ is time. $$ ) On Tuesday, speed is $40$ miles per hour and the time is the sum of time of work (t) and $\frac{3}{60}$ (3 minutes in unity of hour. So: $$ $d=40 \times (t+0,05)=40t+2$ $$ On Wednesday, speed is $45$ miles per hour and the time is the sum of time of work(t) and $\frac{10}{60}$ and $\frac{-15}{60}$ . Because she left early and arrived early. So: $$ $d=45 \times (t-\frac{5}{60})=45t-3.75$ $$ $d=d$ $$ $40t+2=45t-3.75$ $$ $t=1,15$ $$ $d=48$ $$

first, we call + time to go to work is t(min) + the distance is s(miles) so, on Tuesday, we have : s = 40(t+3) (1) (because she late 3 min) On wed,she left early 10min and go to work early 15min => She only early (15 - 10) = 5min than usual. => s=45(t-5) (2) From (1) and (2), we have 40(t+3) = 45(t-5) => 40t + 120 = 45t - 225 => 5t = 345 => t= 69(min) As we know, s = 40(t+3), but speed is miles per hour, our time is min, so we need to divide 60 => s= [40(t+3)]/60 = 48 miles!

First, solve the equation 40 (t+3)=45 (t-5). You'll find t=69 minutes. You can calculate the distance either with 40 (t+3) or with 45 (t-5). If you substitute t=69 into 40 (t+3), you'll find it as 40 [(69+3)/60]=48 miles.

Let the distance be d When she travels at 40mph she reaches at 8:03 Then she leaves 10min earlier and reaches at 7:45 Had she left at her regular time she would have reached 5 mins earlier at 7:55 Difference between 8:03 & 7:55 is 8 minutes Equation can be d/40 - d/45 = 8/60 Solve for d which comes to 48 miles

Let the distance be x. x/40 - x/45=8/60 x=48

3 min late Tuesday and 5 min early on Wednesday= 8 min x/40-x/45=8/60 45x-40x=40 45 8/60 5x=240 x=48 distance of route=48 miles

We first take a variable t for time, a variable d for distance, and a variable s for speed. The second sentence gives us the values s =40 mph, t =T+3 where T is the original time Alicia takes to get to work, when she reaches on time. We now use these values in the equation $speed= \frac{distance}{time}$ to get $\frac{40 mph}{60 min} = \frac{*d* miles}{(T+3) min}$ . Note that we divide the speed by 60 to balance out the units, as the speed is in miles per hour and the time is in minutes, and 1 hour = 60 minutes.This is our first equation. Now, the third sentence gives us the values s =45 mph, t =T+10 as she leaves 10 minutes early. Now, the next sentence tells us that we have to correct the value of t to t =T+10-15=T-5 as she reached 15 minutes earlier than expected. We now get our second equation as $\frac{45 mph}{60 min} = \frac{*d* miles}{(T-5) min}$ . We now modify both equations to get the value of d in either equation, and then equate these two values. This gives us $\frac{40 mph}{(T+3) min} = \frac{45 mph}{(T-5) min}$ . Solving this equation, we get T=69. We substitute this value in the first equation to get d = 48 miles.

Let alicia take $t$ hours to go to work normally. The first time, she takes $t + \dfrac{3}{60}$ hours going at a speed of $40$ miles per hour.

Therefore distance travelled by her is $40 \times \left( t + \dfrac{3}{60} \right)$

The second time, she leaves $10$ minutes early and reaches $15$ minutes early. If she had left had the normal time, she would have reached $5$ minutes early.

Therefore, she takes $t - \dfrac{5}{60}$ hours going at a speed of $45$ miles per hour.

Therefore distance travelled by her is $45 \times \left( t - \dfrac{5}{60} \right)$

Equating those, as distance travelled is equal,

$40 \times \left( t + \dfrac{3}{60} \right) = 45 \times \left( t - \dfrac{5}{60} \right)$

We get $t = \dfrac{23}{20}$

Substituting the value of $t$ in $40 \times \left( t + \dfrac{3}{60} \right)$ , we get distance equal to $\boxed{48}$

Let the distance of the route to work be d and the time required to get to work at 8 am be t .

Note that all times given in minutes are converted into hours and since she let 10 minutes earlier on Wednesday and arrived 15 minutes early, this is the same as saying she arrived 5 minutes early at work .

Thus on Tuesday, ${d}=40(t+\frac{1}{20}).....(1)$

and on Wednesday ${d}=(t-\frac{1}{12}).....(2)$

solving, $40(t+\frac{1}{20})=45(t-\frac{1}{12})$

i.e. ${40t}+{2}={45t}-\frac{45}{12}$

From which ${t}=\frac{23}{20}hours$

Substituting for t into.....(1) we get ${d}=({\frac{23}{20}+\frac{1}{20}})$

and ${d}={40}\times\frac{6}{5}$

Hence ${d}=\boxed{48 miles}$