A car travels from town A to town B at constant speed. If it increases speed by 20%, it will arrive 1 hour ahead of schedule.If it increased by 25% after travelling the first 120 km at usual speed,it will arrive 36 minutes ahead.Find the distance from A to B.

a.205
b.210
c.230
d.220
e.250

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Given:
1. A car travels from town A to town B at constant speed.
2. If it increases speed by 20%, it will arrive 1 hour ahead of schedule.
2. If it increased by 25% after travelling the first 120 km at usual speed,it will arrive 36 minutes ahead.

Asked: Find the distance from A to B.

Let the distance from A to B be x km and normal speed of the car be v kmh

x/1.2v = x/v - 1 (1)
x/v = 6 => 1/v = x/6
x = 6v (1a)
\(\frac{x-120}{1.25v} = \frac{x-120}{v} - \frac{36}{60}\) (2)
(x-120)/v = 36*5/60 = 3
x -120 = 3v = x/2
x = 240

Distance from A to B = 240 km

None of the choices match the correct answer.
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Thank You very much. I really appreciate

When distance travelled is the same, ratio of speeds is inverse of ratio of time taken.

Case 1: When car increases its speed by 20% over the entire distance.
Ratio of speeds = 5:6 (20% increase in speed), ratio of time taken = 6:5
The actual difference in time taken is 1 hr so in usual case, the car takes 6 hrs but with increased speed, it will take 5 hrs.

Case 2: When car increases its speed by 25% over second part of the distance (ignore the first 120 kms for the time being)
Over the distance that the car increases its speed by 25%, ratio of speeds = 4:5 (increase of 25%)
Then ratio of time taken = 5:4 (a difference of 1 on ratio scale)
The actual difference in time taken is 36 mins = 36/60 hrs = 3/5 hrs. So normally, the car takes 5*(3/5) hrs = 3 hrs over this distance.

Now note that for the entire distance, the car usually takes 6 hrs. For the second part distance, the car usually takes 3 hrs. So this second part distance must be exactly 1/2 of total distance.
Hence total distance must be 2*120 km = 240 km
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Hi VeritasKarishma
When I have a PS problem such as \(3^{-2x} + 3^{-x}\) - 12 = 0 how should I approach this problem?

Second, related to the above, I have realized that I am oftentimes confused by \(2^{2+x}\) = \(2^2*2^x\) and \(2^{2x}\) = (\(2^{x}\))*(\(2^{x}\)). Usually, what I do is I make up numbers and variables to see how it works. However, this takes considerable time and is not exactly the best technique under pressure. Thus, I am wondering if you can recommend some drill or daily exercise that would help me to solidify these differences.

Thank you very much in advance. I really appreciate your help.

Put 3^{-x} = a and then take it from there.

a^2 + a - 12 = 0
(a + 4)(a - 3) = 0
a = -4 or 3

3^{-x} cannot be negative because the base 3 is positive. So no matter what the exponent, it will not make a negative number.

3^{-x} = 3 = 3^1
-x = 1
x = -1

Also check the following posts on our blog:
https://www.veritasprep.com/blog/2011/0 ... eparation/
https://www.veritasprep.com/blog/2011/0 ... ration-ii/
https://www.veritasprep.com/blog/2011/0 ... s-applied/
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Practice questions on exponents (we have some drills in our skill builder if you have access to Veritas curriculum). Also, you can search the forum by selecting exponents tag and practice those questions. With a bit of practice, you will not get confused. Checking with numbers is a huge waste of time.
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VeritasKarishma thank you very much for both the detailed explanation and the tips for exercises. I really appreciate it. Thank you!

A bus stop is served by express buses, which run every hour at ten minutes past the hour, and local buses, which run every quarter of an hour, starting at five past the hour. What is the probability that a passenger arriving at the stop at random will have to wait more than six minutes for a bus?

A) 1/2
B) 31/60
C) 29/60
D) 1/15
E) 11/20

A bus stop is served by express buses, which run every hour at ten minutes past the hour, and local buses, which run every quarter of an hour, starting at five past the hour. What is the probability that a passenger arriving at the stop at random will have to wait more than six minutes for a bus?

A) 1/2
B) 31/60
C) 29/60
D) 1/15
E) 11/20
My confusion is that
Say , If he arrives at 6:14 , then he will have to wait for 6 minutes as the next bus is at 6:20.
Why is then 6:14 being included ? We need those time-instants for which wait-period is more than 6 minutes....right ?

Please help.

Express: 6:10, 7:10, 8:10, 9:10 ...
Local: 6:05, 6:20, 6:35, 6:50, 7:05, 7:20, 7:35, 7:50 ...

If passenger arrives in these time slots between 6:00 to 7:00, he will need to wait more than 6 mins.
6:10 to 6:14 (arrives just a microsecond before the clock turns 6:14 so the minute of 6:13 to 6:14 is also added)
6:20 to 6:29
6:35 to 6:44
6:50 to 6:59

This adds up to 31 mins out of the total 60 min in an hr so the probability = 31/60.
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Hi VeritasKarishma,
That is what my question is.
Why are you considering the time-instant 6:14?
From 6:14 to 6:20 ,it is 6 minutes..right ? And at 6:20 , local bus is available. I am okay with 6:13 as from 6:13 to 6:20 it's more than 6 minutes (7 minutes actually).
Can you please explain a bit ? I apologise because of my inability to get your explanation.

Posted from my mobile device

Read again the highlighted part above.
6:10 to 6:14 gives us 4 mins - from exact 6:10 to a microsecond before 6:11 (imagine the face of a clock with the second hand. The second hand starts from 12 at 6:10 and completes a full circle and just before it touches 12 again, a minute is over)
from 6:11 to just before 6:12
from 6:12 to just before 6:13
from 6:13 to just before 6:14

This adds up to 4 mins. The moment of 6:14 is not included. 6:14 is a moment, not a minute.

I have explained this on the question link too on which you tagged me. Check that out too.
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Read again the highlighted part above.
6:10 to 6:14 gives us 4 mins - from exact 6:10 to a microsecond before 6:11 (imagine the face of a clock with the second hand. The second hand starts from 12 at 6:10 and completes a full circle and just before it touches 12 again, a minute is over)
from 6:11 to just before 6:12
from 6:12 to just before 6:13
from 6:13 to just before 6:14

This adds up to 4 mins. The moment of 6:14 is not included. 6:14 is a moment, not a minute.

I have explained this on the question link too on which you tagged me. Check that out too.[/quote]

VeritasKarishma maa'm, Why do you say "........from exact 6:10 ..." ?
Why are you considering the "6:10 " moment....? There is an express bus at 6:10 ...right ?
I apologize if I am over-analyzing.
VeritasKarishma

It's the moment right after 6:10. Say a nano second after 6:10. It is infinitesimally small.
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can u help me with this doubt:
https://gmatclub.com/forum/if-x-and-y-a ... l#p2377000

Done here: https://gmatclub.com/forum/if-x-and-y-a ... l#p2378680
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hi ,plz clear this doubt,

https://gmatclub.com/forum/how-many-peo ... l#p2379587

thanks a lot

Hi Karishma,

Please can you help solve this question.
Let S be the set of all positive integers having at most 4 digits and such that each of the digits is 0 or 1. What is the greatest prime factor of the sum of all the numbers in S ?

A. 11
B. 19
C. 37
D. 59
E. 101

PS41661.01