M28-24

Question

Typically, Holly leaves home for school at 9:00. However, today she left home 20 minutes later than usual. To make it to school on time, she increased her regular speed by 20% but still arrived at school 15 minutes later than usual. What is her usual travel time from home to school?

A. \(15\) minutes
B. \(20\) minutes
C. \(25\) minutes
D. \(30\) minutes
E. \(210\) minutes

Show Answer

Official Answer

D

A.  15  minutes
B.  20  minutes
C.  25  minutes
D.  30  minutes
E.  210  minutes

Bunuel · Accepted Answer

Official Solution:

Typically, Holly leaves home for school at 9:00. However, today she left home 20 minutes later than usual. To make it to school on time, she increased her regular speed by 20% but still arrived at school 15 minutes later than usual. What is her usual travel time from home to school?

A. \(15\) minutes
B. \(20\) minutes
C. \(25\) minutes
D. \(30\) minutes
E. \(210\) minutes

Show Answer

Official Answer

D

A.  15  minutes
B.  20  minutes
C.  25  minutes
D.  30  minutes
E.  210  minutes

Let Holly's usual speed be represented by  s  and her usual travel time by  t  minutes. Since the distance covered remains the same, we have the following equation:  st = 1.2s(t-20+15) . Solving this equation gives us  t = 30  minutes as her usual travel time from home to school.
 
To provide more context, Holly usually leaves home for school at 9:00 and arrives at school by 9:30. Today, she left home 20 minutes later than usual, at 9:20, and arrived at school 15 minutes later than usual, at 9:45. Therefore, her travel time today was  t - 20 + 15 = 30 - 20 + 15 = 25  minutes.

Answer: D

psavalia · Answer

Bunuel,

Can you please explain why we subtract 20 and add 15 in  st=1.2s*(t-20+15)  ? The stem says she left 20 minutes late and arrived 15 minutes later than usual.. so shouldn't we add 15 and 20 to the usual time of t? Or am I missing something here?

Bunuel · Answer

Let me ask you what time she spent today to reach school? It's t-5 minutes, no? Which is t - 20 + 15.

LostinNY · Answer

How does the math work out that you get t= 30? I understand how the formula is set up though.

Bunuel · Answer

st=1.2s∗(t−20+15)

st=1.2s∗(t−5)

t=1.2∗(t−5)  (reduce by s)

t=1.2t−6

0.2t=6

t=30

Hope it's clear.

Prices · Answer

The way I'm calculating it, I'm getting answer  C .

Normal time to school: 25 min
20% of 25 = 5
Time to school at increased speed = 20 min

Time to school today = 20 minute delay + 20 min at increased speed = 40 min

40 (Time Today) - 25 (Normal Time) = 15

Am I thinking of the speed vs time increase incorrectly?

chetan2u · Answer

Hi,
If you are picking values from the choices and take 25, the method is wrong thereafter..
  It is the speed that is 20% more and you cannot CONVERT it into TIME the way you have..

If your choice is 25.. 
speed = 1:1.2
so TIME= 1.2:1..
therefore 1.2x= 25 and x= 25/1.2 and thereafter you can find the answer, which will not be correct..

Lets try with 30.. 
 speed = 1:1.2
so TIME= 1.2:1..
therefore 1.2x= 30 and x= 30/1.2=25
so if Holly starts at 9:20, she will reach at 9:20 + 25 minutes= 9:45..
NORMAL time= 9:00+30 minutes= 9:30 ..
difference= 9:45-9:30=15 minutes CORRECT

oanhnguyen · Answer

Hi, 
Can you please explain why the new time is (t-20+15)? I don't get it

Diya52 · Answer

If you leave 20 mins later than usual then travelling at your usual speed you are supposed to reach 20 mins later but since the speed got increased you are reaching 15 mins later which means you saved 5 mins and hence travelled for 5 mins less which is (t-5)

ShivamoggaGaganhs · Answer

This questions is excellent one, kudos...

RenanBragion · Answer

I think this is a  high-quality  question and I agree with explanation.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

rajtare · Answer

Actually there is another logical way to look at the solution,

since she started 20 minute late, if she would have traveled at her usual speed she would have been late by 20 minutes, but she is actually late by 5 minutes, so this five minutes difference occur because of the increase in speed by 6/5 of usual speed.

since speed is increase by 6/5 , time will decrease by 5/6 of usual time

Let usual time be t, then we have
t-(5/6)t= 5
or t=30

phmahi1997 · Answer

I like the solution - it’s helpful.

phmahi1997 · Answer

I like the solution - it’s helpful.

Mantrix · Answer

Please find attached image.

hope that may help for the more clear solution and understanding the (t+15-20) term

siddhantvarma · Answer

Great question. Got it wrong on the test simply because I didn't understand the question correctly.

Break down each sentence one by one:

Typically, Holly leaves home for school at 9:00. However, today, she left home 20 minutes later than usual => This means she left home at 9:20

To make it to school on time, she increased her regular speed by 20% but still arrived at school 15 minutes later than usual. 
If her original speed is s, her new speed is 1.2s.

What is her usual travel time from home to school?

Let's say it takes Holly time t to reach school when travelling at a speed s covering the distance d. This is her usual travel time from home to school

We can setup the equation as:
s = d/t

But now, she is 20 minutes late, so to reach on time, she needs to cover the same distance in t-20 minutes.

But we're told despite a speed 1.2s (20% greater than regular speed), she arrives 15 minutes later than usual. So, instead of reaching in t-20, she reaches in t-20+15 => t-5 minutes.

We can now setup another equation as:
1.2s = d/t-5
Substitute d=s*t from earlier equation;

1.2s = s*t/t-5 => 1.2*(t-5) = t => 1.2t - 6 = t => 0.2t = 6 => t = 30

Her usual travel time from home to school is 30 minutes. (D)

vishwaraje · Answer

I like the solution - it’s helpful.

arjunsathish · Answer

We can use the formula (Speed OLD)/(Speed NEW) = TIME(NEW)/TIME(OLD) . Say she takes T minutes normally in her normal speed , If she went in her normal speed she would have reached 20 minutes late because she left 20 minutes late , but she reached 15 minutes late. SO we can understand that SHE SAVED 5 MINUTES BY GOING 20 PERCENT FASTER , if old speed is 5x new speed would be 6x , SO we get 5x/6x=T-5/5. which equals 6T-30 = 5T giving T=30 .
Using this formula you can easily solve similar questions in under a minute.

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