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Mr. Alex usually starts at 9:00am and reaches his office jus

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Mr. Alex usually starts at 9:00am and reaches his office jus  [#permalink]

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New post Updated on: 02 Aug 2013, 13:11
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Mr. Alex usually starts at 9:00am and reaches his office just in time, driving at his regular speed. Last Wednesday, he started at 9:30am and drove 25% faster than his usual speed. Did he reach the office in time?

(1) Last Monday, he started to his office 20 minutes early, drove 20% slower than his regular speed, and reached his office just in time.
(2) Last Tuesday, he started to his office 10 minutes early, and reached the office 10 minutes early driving at his regular speed.

Spoiler: :: "user's solution"
Solution:
Statement (2) is nothing but an ordinary saying. At same speed if you start 10 minutes early you will reach 10 minutes early and it’s not a surprise. So it’s nothing but to waste time. So not sufficient.

Before discussing over statement (1), one thing we have to realize clearly.

If we start late and then still want to reach in time then we have to move faster and save time in respect to the regular duration.

Formula, S = Vt , where S=distance , V= velocity and t = time.

For regular speed, journey starts from just 9am. We have the original equation,
S = Vt
Or, t = S/V (Regular time)

Last Wednesday (starting at 9:30am), drove 25% faster over the same distance,
so now the time, T= S/1.25V (here , 25% faster velocity = 1.25V)

After making 30 minutes late Mr. Alex still wants to reach in time, so he has to move faster and save at least 30 minutes in respect of the regular time (because he is 30 minutes late already, he has to cover this time) .

Now he can save time = S/V - S/1.25V = S/5V .
This time value S/5V has to be more than 30 minutes. Because this is his savings and he is already 30 minutes late. If S/5V > 30 minutes then only it will be possible for him to reach in time.

Now S/5V > 30
Or, S/V > 150 ………………………………..(eq1)

From statement (1) ,we know he started 20 minutes early {REMIND IT: HE STARTED 20MINUTES EARLY, SO HE DROVE 20 MINUTES LONGER HERE IN THIS CASE}, drove 20% slower than his regular speed and reached in time.



So , time = S/0.8V (more time than the regular) (20% slower speed = 80% of V = 0.8V)

Now the time difference between the regular and the above (S/0.8V) time
is = S/0.8V – S/V = S/4V = 20 minutes.
Or, S/V = 80 minutes.

From eq1, we know in order to reach in time S/V has to be greater than 150 ,
which is S/V > 150 .
But from statement(1) we found that S/V=80 .
So Mr. Alex didn’t reach office in time that day.

Statement(1) is sufficient alone. Answer is (A) .

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Originally posted by Asifpirlo on 02 Aug 2013, 12:53.
Last edited by Zarrolou on 02 Aug 2013, 13:11, edited 1 time in total.
Edited the question.
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Re: Mr. Alex usually starts at 9:00am and reaches his office jus  [#permalink]

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New post 03 Aug 2013, 11:08
2
from 1:

let x be the speed and t be the time.

as distance traveled is equal
x*t=.8x(t+1/3)
on solving gives
t=4/3 or 80 minutes

Now for the question
as again distance traveled is equal
1.25x*t1=4/3*x

t1=64 minutes. Therefore if he started at 9:30 he did not reach on time. As he had to reach by 10:20

2) does not give us an equation
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Re: Mr. Alex usually starts at 9:00am and reaches his office jus  [#permalink]

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New post 03 Aug 2013, 21:43
1
Hi,
let, d = distance between his office and home
s = regular speed
t = regular time taken to reach office.
\(d = s * t\)
Now, it is given that on one day, Alex starts at 9.30 (30 mins late) for office with 25% higher speed.
\(d' = 1.25s * t'\)

What we are really asked is, does Alex travel the same distance on Wednesday, as he does usually:
i.e. is \(s * t = 1.25s * t'\) ?

Statement 1: Last Monday, he started to his office 20 minutes early, drove 20% slower than his regular speed, and reached his office just in time.
this statement simply means the following:
\(s * t = 0.8s * (t +20)\)
Upon solving for t, we get,
t = 80 mins, usual time he takes to reach office.

Using this,
\(d = s * 80\) (time he reaches office usually is 10.20)....(i)
Since on Wednesday, he leaves office at 9.30, to be able to reach his office in time, time taken by him should be 80-30 = 50 mins,
so \(d' = 1.25s * 50 = 62.5s\)....(ii)
From (i) and (ii) we can conclude that Alex does not reach office on time. So Statement 1 is sufficient.

Statement 2: Last Tuesday, he started to his office 10 minutes early, and reached the office 10 minutes early driving at his regular speed.
This statement just states gives no additional information than already given. So Statement 2 is insufficient.

Correct Ans: A :)
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Re: Mr. Alex usually starts at 9:00am and reaches his office jus  [#permalink]

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New post 04 Aug 2013, 07:51
ramannanda9 wrote:
from 1:

let x be the speed and t be the time.

as distance traveled is equal
x*t=.8x(t+1/3)
on solving gives
t=4/3 or 80 minutes

Now for the question
as again distance traveled is equal
1.25x*t1=4/3*x

t1=64 minutes. Therefore if he started at 9:30 he did not reach on time. As he had to reach by 10:20

2) does not give us an equation


nice work boy......
but t1 = 56.25 here..... how did you get 64min ?
he was already 30 min late and now he is taking 56.25 min more.
so the total time he spares = 30 + 56.25 = 86.25min.
so from 9am he requires 86.25 min to reach, but as usual from 9am he requires 80min.
finally he is 6.25min late.... 1 is sufficient alone, 2 is nothing here. so A
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Re: Mr. Alex usually starts at 9:00am and reaches his office jus  [#permalink]

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New post 04 Aug 2013, 08:11
Asifpirlo wrote:
ramannanda9 wrote:
from 1:

let x be the speed and t be the time.

as distance traveled is equal
x*t=.8x(t+1/3)
on solving gives
t=4/3 or 80 minutes

Now for the question
as again distance traveled is equal
1.25x*t1=4/3*x

t1=64 minutes. Therefore if he started at 9:30 he did not reach on time. As he had to reach by 10:20

2) does not give us an equation


nice work boy......
but t1 = 56.25 here..... how did you get 64min ?
he was already 30 min late and now he is taking 56.25 min more.
so the total time he spares = 30 + 56.25 = 86.25min.
so from 9am he requires 86.25 min to reach, but as usual from 9am he requires 80min.
finally he is 6.25min late.... 1 is sufficient alone, 2 is nothing here. so A


As per the equation 80/1.25=64 mins
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Re: Mr. Alex usually starts at 9:00am and reaches his office jus  [#permalink]

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New post 04 Aug 2013, 09:15
ramannanda9 wrote:
Asifpirlo wrote:
ramannanda9 wrote:
from 1:

let x be the speed and t be the time.

as distance traveled is equal
x*t=.8x(t+1/3)
on solving gives
t=4/3 or 80 minutes

Now for the question
as again distance traveled is equal
1.25x*t1=4/3*x

t1=64 minutes. Therefore if he started at 9:30 he did not reach on time. As he had to reach by 10:20

2) does not give us an equation


nice work boy......
t1 = 64 here....
he was already 30 min late and now he is taking 64 min more.
so the total time he spares = 30 + 64 = 94min.
so from 9am he requires 94 min to reach, but as usual from 9am he requires 80min.
finally he is 14min late.... 1 is sufficient alone, 2 is nothing here. so A


As per the equation 80/1.25=64 mins

yes u r right...... :( :(
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Re: Mr. Alex usually starts at 9:00am and reaches his office jus  [#permalink]

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New post 18 Jan 2014, 16:13
Asifpirlo wrote:
Mr. Alex usually starts at 9:00am and reaches his office just in time, driving at his regular speed. Last Wednesday, he started at 9:30am and drove 25% faster than his usual speed. Did he reach the office in time?

(1) Last Monday, he started to his office 20 minutes early, drove 20% slower than his regular speed, and reached his office just in time.
(2) Last Tuesday, he started to his office 10 minutes early, and reached the office 10 minutes early driving at his regular speed.

Spoiler: :: "user's solution"
Solution:
Statement (2) is nothing but an ordinary saying. At same speed if you start 10 minutes early you will reach 10 minutes early and it’s not a surprise. So it’s nothing but to waste time. So not sufficient.

Before discussing over statement (1), one thing we have to realize clearly.

If we start late and then still want to reach in time then we have to move faster and save time in respect to the regular duration.

Formula, S = Vt , where S=distance , V= velocity and t = time.

For regular speed, journey starts from just 9am. We have the original equation,
S = Vt
Or, t = S/V (Regular time)

Last Wednesday (starting at 9:30am), drove 25% faster over the same distance,
so now the time, T= S/1.25V (here , 25% faster velocity = 1.25V)

After making 30 minutes late Mr. Alex still wants to reach in time, so he has to move faster and save at least 30 minutes in respect of the regular time (because he is 30 minutes late already, he has to cover this time) .

Now he can save time = S/V - S/1.25V = S/5V .
This time value S/5V has to be more than 30 minutes. Because this is his savings and he is already 30 minutes late. If S/5V > 30 minutes then only it will be possible for him to reach in time.

Now S/5V > 30
Or, S/V > 150 ………………………………..(eq1)

From statement (1) ,we know he started 20 minutes early {REMIND IT: HE STARTED 20MINUTES EARLY, SO HE DROVE 20 MINUTES LONGER HERE IN THIS CASE}, drove 20% slower than his regular speed and reached in time.



So , time = S/0.8V (more time than the regular) (20% slower speed = 80% of V = 0.8V)

Now the time difference between the regular and the above (S/0.8V) time
is = S/0.8V – S/V = S/4V = 20 minutes.
Or, S/V = 80 minutes.

From eq1, we know in order to reach in time S/V has to be greater than 150 ,
which is S/V > 150 .
But from statement(1) we found that S/V=80 .
So Mr. Alex didn’t reach office in time that day.

Statement(1) is sufficient alone. Answer is (A) .


Is (t-1/2)(5/4r)>=rt?

Is t>=2.5?

Statement 1

(t+1/3)(4/5r)>=rt

t<=4/3

Sufficient

Statement 2

(t+1/6)(r) >= (t+1/6)(r)

Doesn't tell us anything

Insufficient

Hence answer is A

Is this clear?

Cheers!
J :)
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Re: Mr. Alex usually starts at 9:00am and reaches his office jus  [#permalink]

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New post 29 Sep 2014, 02:28
can anybody plz explain the second option..
isn`t it similar to the already stated line in the question..

also,
isn`t there any problem in the question since without using any of the statements we can check for the answer.

usual time=t1
usual speed=s1

last wednesday...

speed=1.25s1
time=t2(let)

we need to calculate..t1=t2?

since distance is equal...

t1*s1=1.25s1*t2

which implies..t1=1.25t2

and hence ..no..and hence,sufficient.

kindly correct me..
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Re: Mr. Alex usually starts at 9:00am and reaches his office jus  [#permalink]

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New post 30 Sep 2014, 02:19
2
1
shreygupta3192 wrote:
can anybody plz explain the second option..
isn`t it similar to the already stated line in the question..

also,
isn`t there any problem in the question since without using any of the statements we can check for the answer.

usual time=t1
usual speed=s1

last wednesday...

speed=1.25s1
time=t2(let)

we need to calculate..t1=t2?

since distance is equal...

t1*s1=1.25s1*t2

which implies..t1=1.25t2

and hence ..no..and hence,sufficient.

kindly correct me..


The question stem alone is not sufficient to answer the question.

Mr. Alex usually starts at 9:00am and reaches his office just in time, driving at his regular speed. Last Wednesday, he started at 9:30am and drove 25% faster than his usual speed. Did he reach the office in time?

Whether he reached on time or not depends on how much time he usually takes to reach office.

Say, when he starts at 9, he usually reaches office at 9:35 am. Now, if he starts at 9:30, increasing his speed by 25% will not be enough to reach office on time. Hence he will not reach office on time.

On the other hand, if he usually reaches at 11:30 when he starts at 9:00, increasing his speed by 25% will be enough to reach office on time when he starts at 9:30. In this case, he will reach office on time.
Usual speed:Increased speed = 4:5
Time taken usually:Time taken now = 5:4 (since the same distance has to be covered)
The difference of 1 in time is .5 hrs. So usual time taken must be .5*5 = 2.5 hrs
So when he starts at 9:00, he reaches at 11:30.
Last Wednesday, time taken = 0.5*4 = 2 hrs. Since he started at 9:30, he must have reached at 11:30.

So whether he reached office on time or not depends on the time he usually reaches office.So we need the statements:

(1) Last Monday, he started to his office 20 minutes early, drove 20% slower than his regular speed, and reached his office just in time.

Usual Speed:Last Monday's Speed = 5:4
Usual Time taken: Time taken last Monday = 4:5
The difference of 1 accounts for 20 mins i.e. (1/3)hr. so usual time taken = 4*(1/3) = 80 mins
So when he starts at 9, he reaches at 10:20. Hence, on Wednesday, he did not reach on time. Sufficient alone.

(2) Last Tuesday, he started to his office 10 minutes early, and reached the office 10 minutes early driving at his regular speed.
This adds no new information. If he goes at his usual speed, he will take the same time as always. So if he starts 10 min early, he will reach 10 mins early... Not sufficient alone.

Answer (A)
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Mr. Alex usually starts at 9:00am and reaches his office jus  [#permalink]

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New post 01 Oct 2014, 02:55
VeritasPrepKarishma wrote:

The question stem alone is not sufficient to answer the question.

Mr. Alex usually starts at 9:00am and reaches his office just in time, driving at his regular speed. Last Wednesday, he started at 9:30am and drove 25% faster than his usual speed. Did he reach the office in time?

Whether he reached on time or not depends on how much time he usually takes to reach office.

Say, when he starts at 9, he usually reaches office at 9:35 am. Now, if he starts at 9:30, increasing his speed by 25% will not be enough to reach office on time. Hence he will not reach office on time.

On the other hand, if he usually reaches at 11:30 when he starts at 9:00, increasing his speed by 25% will be enough to reach office on time when he starts at 9:30. In this case, he will reach office on time.
Usual speed:Increased speed = 4:5
Time taken usually:Time taken now = 5:4 (since the same distance has to be covered)
The difference of 1 in time is .5 hrs. So usual time taken must be .5*5 = 2.5 hrs
So when he starts at 9:00, he reaches at 11:30.
Last Wednesday, time taken = 0.5*4 = 2 hrs. Since he started at 9:30, he must have reached at 11:30.

So whether he reached office on time or not depends on the time he usually reaches office.So we need the statements:

(1) Last Monday, he started to his office 20 minutes early, drove 20% slower than his regular speed, and reached his office just in time.

Usual Speed:Last Monday's Speed = 5:4
Usual Time taken: Time taken last Monday = 4:5
The difference of 1 accounts for 20 mins i.e. (1/3)hr. so usual time taken = 4*(1/3) = 80 mins
So when he starts at 9, he reaches at 10:20. Hence, on Wednesday, he did not reach on time. Sufficient alone.

(2) Last Tuesday, he started to his office 10 minutes early, and reached the office 10 minutes early driving at his regular speed.
This adds no new information. If he goes at his usual speed, he will take the same time as always. So if he starts 10 min early, he will reach 10 mins early... Not sufficient alone.

Answer (A)


Hi Karishma, I just want to make sure I am thinking about this correctly

On Monday he left 20 minutes early and travelled 20% slower
\(Sm:S=\frac{4}{5}\) therefore \(tm:t=\frac{5}{4}\). This means that 20% slower speed leads to 25% more time.
Now, since he left 20 mins early, the 25% more time = 20 mins. This means normal time = 80 mins
On Monday he left at 840 am and arrived at 1020am. Normally he leaves at 9am and arrives at 1020am.

Back to the stem: On Wednesday, he left at 930am and drove 25% faster
\(Sw:S=\frac{5}{4}\) therefore \(tw:t=\frac{4}{5}\). This means that 25% faster speed leads to 20% less time.
80% of 80mins is 64mins
On Wednesday he arrived 14mins late

Question: Can we always make the assumption that if speed increases by 25%, then time taken will reduces by 20% and vice-versa, provided distance remains the same ? I have never really thought about it like that.
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New post 05 Oct 2014, 16:37
Anybody have any insights into the question I asked?
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Re: Mr. Alex usually starts at 9:00am and reaches his office jus  [#permalink]

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New post 06 Oct 2014, 01:53
1
gooner wrote:
VeritasPrepKarishma wrote:

The question stem alone is not sufficient to answer the question.

Mr. Alex usually starts at 9:00am and reaches his office just in time, driving at his regular speed. Last Wednesday, he started at 9:30am and drove 25% faster than his usual speed. Did he reach the office in time?

Whether he reached on time or not depends on how much time he usually takes to reach office.

Say, when he starts at 9, he usually reaches office at 9:35 am. Now, if he starts at 9:30, increasing his speed by 25% will not be enough to reach office on time. Hence he will not reach office on time.

On the other hand, if he usually reaches at 11:30 when he starts at 9:00, increasing his speed by 25% will be enough to reach office on time when he starts at 9:30. In this case, he will reach office on time.
Usual speed:Increased speed = 4:5
Time taken usually:Time taken now = 5:4 (since the same distance has to be covered)
The difference of 1 in time is .5 hrs. So usual time taken must be .5*5 = 2.5 hrs
So when he starts at 9:00, he reaches at 11:30.
Last Wednesday, time taken = 0.5*4 = 2 hrs. Since he started at 9:30, he must have reached at 11:30.

So whether he reached office on time or not depends on the time he usually reaches office.So we need the statements:

(1) Last Monday, he started to his office 20 minutes early, drove 20% slower than his regular speed, and reached his office just in time.

Usual Speed:Last Monday's Speed = 5:4
Usual Time taken: Time taken last Monday = 4:5
The difference of 1 accounts for 20 mins i.e. (1/3)hr. so usual time taken = 4*(1/3) = 80 mins
So when he starts at 9, he reaches at 10:20. Hence, on Wednesday, he did not reach on time. Sufficient alone.

(2) Last Tuesday, he started to his office 10 minutes early, and reached the office 10 minutes early driving at his regular speed.
This adds no new information. If he goes at his usual speed, he will take the same time as always. So if he starts 10 min early, he will reach 10 mins early... Not sufficient alone.

Answer (A)


Hi Karishma, I just want to make sure I am thinking about this correctly

On Monday he left 20 minutes early and travelled 20% slower
\(Sm:S=\frac{4}{5}\) therefore \(tm:t=\frac{5}{4}\). This means that 20% slower speed leads to 25% more time.
Now, since he left 20 mins early, the 25% more time = 20 mins. This means normal time = 80 mins
On Monday he left at 840 am and arrived at 1020am. Normally he leaves at 9am and arrives at 1020am.

Back to the stem: On Wednesday, he left at 930am and drove 25% faster
\(Sw:S=\frac{5}{4}\) therefore \(tw:t=\frac{4}{5}\). This means that 25% faster speed leads to 20% less time.
80% of 80mins is 64mins
On Wednesday he arrived 14mins late

Question: Can we always make the assumption that if speed increases by 25%, then time taken will reduces by 20% and vice-versa, provided distance remains the same ? I have never really thought about it like that.


Yes, this is correct.
And, it is not an assumption. If distance stays the same, 25% increase in speed, reduces the time by 20%. It's only logical. Here is a post that discusses this relation in detail:
http://www.veritasprep.com/blog/2011/03 ... os-in-tsd/
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Re: Mr. Alex usually starts at 9:00am and reaches his office jus  [#permalink]

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New post 06 Oct 2014, 20:29
VeritasPrepKarishma wrote:
gooner wrote:
VeritasPrepKarishma wrote:

The question stem alone is not sufficient to answer the question.

Mr. Alex usually starts at 9:00am and reaches his office just in time, driving at his regular speed. Last Wednesday, he started at 9:30am and drove 25% faster than his usual speed. Did he reach the office in time?

Whether he reached on time or not depends on how much time he usually takes to reach office.

Say, when he starts at 9, he usually reaches office at 9:35 am. Now, if he starts at 9:30, increasing his speed by 25% will not be enough to reach office on time. Hence he will not reach office on time.

On the other hand, if he usually reaches at 11:30 when he starts at 9:00, increasing his speed by 25% will be enough to reach office on time when he starts at 9:30. In this case, he will reach office on time.
Usual speed:Increased speed = 4:5
Time taken usually:Time taken now = 5:4 (since the same distance has to be covered)
The difference of 1 in time is .5 hrs. So usual time taken must be .5*5 = 2.5 hrs
So when he starts at 9:00, he reaches at 11:30.
Last Wednesday, time taken = 0.5*4 = 2 hrs. Since he started at 9:30, he must have reached at 11:30.

So whether he reached office on time or not depends on the time he usually reaches office.So we need the statements:

(1) Last Monday, he started to his office 20 minutes early, drove 20% slower than his regular speed, and reached his office just in time.

Usual Speed:Last Monday's Speed = 5:4
Usual Time taken: Time taken last Monday = 4:5
The difference of 1 accounts for 20 mins i.e. (1/3)hr. so usual time taken = 4*(1/3) = 80 mins
So when he starts at 9, he reaches at 10:20. Hence, on Wednesday, he did not reach on time. Sufficient alone.

(2) Last Tuesday, he started to his office 10 minutes early, and reached the office 10 minutes early driving at his regular speed.
This adds no new information. If he goes at his usual speed, he will take the same time as always. So if he starts 10 min early, he will reach 10 mins early... Not sufficient alone.

Answer (A)


Hi Karishma, I just want to make sure I am thinking about this correctly

On Monday he left 20 minutes early and travelled 20% slower
\(Sm:S=\frac{4}{5}\) therefore \(tm:t=\frac{5}{4}\). This means that 20% slower speed leads to 25% more time.
Now, since he left 20 mins early, the 25% more time = 20 mins. This means normal time = 80 mins
On Monday he left at 840 am and arrived at 1020am. Normally he leaves at 9am and arrives at 1020am.

Back to the stem: On Wednesday, he left at 930am and drove 25% faster
\(Sw:S=\frac{5}{4}\) therefore \(tw:t=\frac{4}{5}\). This means that 25% faster speed leads to 20% less time.
80% of 80mins is 64mins
On Wednesday he arrived 14mins late

Question: Can we always make the assumption that if speed increases by 25%, then time taken will reduces by 20% and vice-versa, provided distance remains the same ? I have never really thought about it like that.


Yes, this is correct.
And, it is not an assumption. If distance stays the same, 25% increase in speed, reduces the time by 20%. It's only logical. Here is a post that discusses this relation in detail:
http://www.veritasprep.com/blog/2011/03 ... os-in-tsd/


Fantastic! Thanks and Kudos
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Mr. Alex usually starts at 9:00am and reaches his office jus  [#permalink]

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New post 31 Jul 2018, 09:36
Asifpirlo wrote:
Mr. Alex usually starts at 9:00am and reaches his office just in time, driving at his regular speed. Last Wednesday, he started at 9:30am and drove 25% faster than his usual speed. Did he reach the office in time?

(1) Last Monday, he started to his office 20 minutes early, drove 20% slower than his regular speed, and reached his office just in time.
(2) Last Tuesday, he started to his office 10 minutes early, and reached the office 10 minutes early driving at his regular speed.


Regular day:
Let r = the regular rate and t = the regular time, implying that the distance to the office = rt.

Last Wednesday:
Since Mr. Alex drives 25% faster -- 5/4 of his regular speed -- the rate = (5/4)r.
Since Mr. Alex leaves 30 minutes late -- reducing the time by 30 minutes -- the time = t-30.
Thus, the distance traveled between 9:30am and the desired arrival time = (5/4)(r)(t-30).
For Mr. Alex to arrive at the office in time, the distance traveled by the desired arrival time must be equal to the distance to the office:
(5/4)(r)(t-30) = rt
(5/4)(t-30) = t
5(t-30) = 4t
5t - 150 = 4t
t = 150.
Question stem, rephrased:
Does t = 150?

Statement 1:
Since Mr. Alex drives 20% slower -- 4/5 of his regular speed -- the rate = (4/5)r.
Since Mr. Alex leaves 20 minutes early -- increasing the time by 30 minutes -- the time = t+20.
Thus, the distance traveled between 9:30am and the desired arrival time = (4/5)(r)(t+20).
Since Mr. Alex arrives at the office in time, the distance traveled by the desired arrival time is equal to the distance to the office:
(4/5)(r)(t+20) = rt
(4/5)(t+20) = t
4(t+20) = 5t
4t + 80 = 5t
80 = t.
Thus, the answer to the rephrased question stem is NO.
SUFFICIENT.

Statement 2:
Since he started 10 minutes early and reached his office 10 minutes early, Mr. Alex traveled at his usual speed last Tuesday.
No new information about last Wednesday.
INSUFFICIENT.


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Mr. Alex usually starts at 9:00am and reaches his office jus &nbs [#permalink] 31 Jul 2018, 09:36
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Mr. Alex usually starts at 9:00am and reaches his office jus

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