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A car is travelling at a uniform speed. The driver sees a milestone showing a 2-digit number.
After travelling for an hour the driver sees another milestone with the same digits in reverse order.
After another hour the driver sees another milestone containing the same 2 digits.
What is the average speed of the driver?

can anyone help me solve this one .i cant fiugure it out as yet

45 mph:
16 miles + 45 miles = 61 miles
61 miles + 45 miles = 106 miles

Eero R?nik wrote: 45 mph:
16 miles + 45 miles = 61 miles
61 miles + 45 miles = 106 miles

Beaten to it.
You're lucky I had a phonecall ::)

can you please give me a step by step explanation. it woul really help.

Ok, let's get all mathematical about this...
In what follows A, B and C are natural numbers, wherefore: 0 < A,B,C < 9 (in other words, they're digits).

So you've got three milestones:
M1-M2-M3

It is given that:
M1 = 10*A + B and
M2 = 10*B + A and (switch digits)
|M1M2| = |M2M3| (both distances take an hour, speed is uniform)

M1 < M2
=> A can only be 1,2,3,4 or 5
=> B can only be 5,6,7,8 or 9

|M1M2| = (10*B + A) - (10*A + B) = 9*B - 9*A (*)

It's pretty easy to see that 100 < M3 < 200

We've got four possibilities for M3:
M3 = 100 + 10*C + B and A = 1 and C = 0,1,...,8 or 9
OR
M3 = 100 + 10*B + C and A = 1 and C = 0,1,...,8 or 9
OR
M3 = 100 + 10*A + B
OR
M3 = 100 + 10*B + A
(four possibilities, not six, because B can't be 1)

Let's try what the first possibility gives us:
|M1M2| = |M2M3|
=> 9*B - 9 = (100 + 10*C + B) - (10*B + 1) ((*) and A = 1)
=> 9*B - 9 = 99 + 10*C - 9*B
=> 18*B - 10*C = 108

=> The last digit of 18*B = The last digit of 108 = 8
Combine that with: B can only be 5,6,7,8 or 9
=> B = 6

We were lucky on our first try!
If you want to, you can try solving it for the other 3 possibilities for M3, and you'll see that -given the restrictions for A, B and C- you won't be able to find a solution.

I hope that helped. Any questions left?

M1 < M2
=> A can only be 1,2,3,4 or 5
=> B can only be 5,6,7,8 or 9

how come?

What about a==7 and b==8 and such?

I got it,
His car is actually broken down and is thusly travewing at 0 mph.

...His car broke down at either the 11, 22, 33, 44, 55, 66, 77, 88, or 99 mile marker. After traveling for one hour at 0 mph he looks up and sees the same two numbers reversed, after another hour of "traveling" he looks up and sees a sign containing the same two digits.

Another quick one

1. A man rides on a bike and covers 1/3rd of a distance. Then his bike gets punctured and walks the remaining distance. Walking time is twice that of time took when he rides on a bike.How much faster he is when rides on a bike.

isn't the answer to this 0 ?
i mean the walking and riding speeds are equal aren't they?

isn't the answer to this 0 ?
i mean the walking and riding speeds are equal aren't they?

Well it's 1 not 0.
it looks indeed as they are equal

S- distance, v- velovity, t-time
S=vt
1/3s= V1 * t
2/3s= V2 * 2t

Adek336 wrote:

isn't the answer to this 0 ?
i mean the walking and riding speeds are equal aren't they?

Well it's 1 not 0.

The difference is 0 exactly, the speeds are any number larger than 0.

There is a 5digit no. 3 pairs of sum is eleven each.
Last digit is 3 times the first one.
3 rd digit is 3 less than the second.
4 th digit is 4 more than the second one.

Is the answer to this 69252 or 25296 ?

Last digit is 3 times the first one.

so if logically taken...2 x 3 = 6

so it should be 69252

but I have no idea if I'm right or wrong thought

25296...
Last digit is 3 times the first one, so it's obviously bigger.

hehe, so much of my logical thinking... ::)

I just knew I was wrong, because I'm never right in maths or anything like that.

but as always...
tatatatataat

Eero saves the day!!

Eero R?nik wrote: 25296...
Last digit is 3 times the first one, so it's obviously bigger.

isn't the units digit the first one?