Buses leave town B at 3 pm and every 10 hours after that.

This was a tough one!
Took me 4 mins still not sure if the approach is correct

Here's my method
Let Bus B make x trips and Bus C make y trips b4 they start at the same time.
The time when they will meet is
Remainder(15+10x)/24 .... B
Remainder(16+15y)/24 .... C

These two must be equal
i.e.
Remainder(15+10x)/24=Remainder(16+15y)/24

Hence I assume we should have integer values of x,y such that
15+10x=16+15y or 10x=15y+1
no integral (x,y) combo exist for this equation

Hence I guess Ans is E

Anyone has a better approach to this problem?

i think the giveaway is the 3pm vs 4pm start .... if they left at the same time they would eventually meet up (10hrs vs 15hrs common divisor or somesuch) but the offset means they will never meet up anytime soon, if at all.

tough one for me and E.
Take a & b as the numbers of buses which leave town B & town C after the first ones. a & b must be positive integers.
We got: 3 + 10a = 4 + 15b
---> 10a = 1 + 15b
We see that: (1 + 15 x an positive integer) will never evenly divide to 10
SO E is the ans

Hey, I don't understand this explanation:

Are you saying that the times will end in 5s? I mean, yes, the number of hours elapsed will always end in a 5 or 0, but that doesn't say much about the time, other than demonstrating that the first bus must leave, on a 24-hour clock, at times of 3, 13, 23, 9, 19, 5, 15, 1, 11, 21... and that the second bus must leave at times of 4, 19, 10, 1, 16, 7, 22..

Yes, there is a pattern that is created, but in my opinion, this is not trivial and does not follow easily from the 'number of hours elapsed ending in 5 or 0'.

Any clarification would be appreciated. As of now, I still don't know how to solve this question in a proper way.

I think this one should be E.
It took me 4.44mins to manually calculate the whole thing. It turns out they never meet on the same time!
This one was a tough one...

Buses B 3,13,23,33 etc....(pattern ending in 3 always)
Buses C 4,19,34,49,54...(pattern ends only in 4 and 9).

Thus E is the correct answer

10 and 15 are both multiple of 5 . The minimum difference between any multiple of 10 and 15 is always 0 and the next difference is 5. For example , 10 and 15 , 40 and 45.
The offset of their starting time is 1 hour. We can never have account for this 1 hour difference since the difference that we can accommodate is 0 or 5.
Had the departure of Bus C be (3pm + multiple of 5) then there was a possibility of buses leaving at the same time

Just for interest. To calculate how many days it takes for the buses to leave at their original departures of 3 and 4 pm:

Bus A: Leaves at 15 and every 10 hours afterwards. After x repetitions of 10 and y days the bus leaves at 15 again and the following formula would apply:

(10 x + 15)/24 = y + 15/24
10 × = 24 y
10 * (uncommon factors of 24 with 10) = 24*(uncommon factors of 10 with 24)
× = 12
y = 5

After 120 hours/5 days and 15/24ths of a day the bus will leave at 15 again.

Bus B: Leaves at 16 and every 15 hours afterwards. After x repetitions of 15 and y days the bus leaves at 16 again and the following formula would apply:

(15 x + 16)/24 = y + 16/24
15 × = 24 y
15 * (uncommon factors of 24 with 15) = 24*(uncommon factors of 15 with 24)
× = 8
y = 5

After 120 hours/5 days the bus will leave at 16 again.

Another straightforward approach would be to just list out the time that has elapsed from the start for each bus that departs. If ever the time elapsed for two buses is the same, then the buses have departed at the same time.

To make things simpler, we can say that the first Bus B leaves at time t=0. Then the times of all Bus B departures will be 0, 10, 20, 30, etc., (always a multiple of 10). Note we don't need to deal with a 24-hour clock since all we are interested in is how many hours have passed since the beginning for each departure.

The first departure time for Bus C will be one hour after the first departure time for Bus B, so at 1 hour. All subsequent departure times for Bus C will add 15 hours onto the previous departure time: 1, 16, 31, 46, 61, etc.

We can see that the times for Bus C will never be a multiple of 10, so the buses will never depart at the same time.

Answer E.

if we look at the question then it is just asking for a number which is of the form: 10x+3 = 15y+4
or x = (5y+1)/10
no integer values exist for x. Hence, E.

I think the simplest approach would be to check if any term of the two series below have a common term.

A(for buses leaving at 3pm and every 10hrs after that) =3,13,23,33,43....
B(for buses leaving at 4pm and every 15 hrs after that)=4,19,34,49,64...

Thus,since there cannot be any common term in both the series,hence the buses will never leave at the same time from the two cities.

Mmmm this question is much easier than the discussions

B: If we put 3 PM at 0 hours,

All buses leave at intervals of 10n

C: Buses start at 4 PM : i.e; 00 hrs + 1 hr

So at C : 1 + 15t

From equation

10n = 1 + 15t

or

10n - 15t = 1

For any multiples of 10 and 15, the result will be a multiple of 5. Never 1.

E

Let us have time relative to 12 pm,
so starting time of buses at Town B = \(3 + 10x\) (where x is integer)
starting time of buses at Town C = \(4 + 15y\) (where y is integer)

when they start at same time => \(3 + 10x = 4 + 15y\)
=> \(x = (15y + 1)/10\)
=> from above, to make x integer , 15y + 1 has to be multiple of 10
=> 15y must have unit digits 9, but only unit digits of 15 is 0, 5
=> so not possible to have x as integer => buses can never start at same time

Hi All,

This question is based primarily on pattern-matching, and here's how the pattern-matching approach can save you a lot of effort:

Buses leave B at 3pm and every 10 hours thereafter….
Buses leave C at 4pm and every 15 hours thereafter…

3pm is the 15th hour of the day. 4pm is the 16th hour of the day. We're looking to see if the start times EVER occur at the same time.

Town B busses: 15, 25, 35, 45, 55, etc.

Notice how EVERY starting hour "ends in a 5"

Town C busses: 16, 31, 46, 61, 76, 91, etc.

Notice how EVERY starting hour EITHER "ends in a 6 or a 1"

This means that the two busses will NEVER leave at the same time.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

very well explained.. thanks .. can you please share trick to solve similar kind of problems with another example.

very well explained.. can you please share similR KIND OF PROBLEMS for further understanding??

I used the Logic behind Remainder Concepts and Divisibility to get to the Answer


Ignoring the Times for now, we can Assume that Time just keeps having Hours added on to it (in a world where 254 hours is a Time)

how many hours will it take before the buses depart at the Same Time?

Bus A: 1st Leaves at 3 -----> every 10 hours thereafter

Bus B: 1st Leaves at 4 -----> every 15 hours thereafter


thus, we are looking for a Number of Hours, H, where:

10a + 3 = H = 15b + 4

*where a and b = No. of 10 hour and 15 hour Intervals, respectively*

10a + 3 = 15b + 4

10a = 15b +1

a = (15b + 1) / 10 = Integer No. of Intervals


Using Number Properties to Analyze the NUM and DEN:

we need a (+)Pos. Multiple of 15 + 1 that will end in a Units Digit of 0 -----> such that the NUM will be Divisible by 10


However, every Multiple of 15 will end in a Units Digit of 5 or 0

(...5) + 1 = Units Digit of 6 ------> NOT Divisible by 10

(...0) + 1 = Units Digit of 1 -----> NOT Divisible by 10


Therefore, there will never be a specific Number of Hours that Pass in which the 2 Buses will leave at the Same Time.

-E-