Problem Solving

1) If a 3-digit integer is selected at random from the integers 100 through 199, inclusive, what is the probability that the first digit and the last digit of the integer are each equal to one more than the middle digit?

and

2) . A certain ball was dropped from a window 8 meters above a sidewalk. On each bounce it rose straight up exactly one-half the distance of the previous fall. After the third bounce the ball was caught when it reached a height of exactly 1 meter above the sidewalk. How many meters did the ball travel in all?

i am a little confused with the first question. On the second i thikn i am right, but the answers says it should be 21m. what do you guys think?

My answer to the first one is a "guess" I suppose. Based on the question, there is only one 3-digit integer between 100-199 where the first and last digits are each one greater than the middle digit, and that is 101 (I think). Since you only have one such occurance out of the 100 possible 3-digit integers, the probability is 1/100?

For the second one, I will try to explain it:
On the first drop, the ball travels 8m down.
It then bounces up 4m before traveling 4m back down.
It again bounces up 2m before traveling down 2m.
On the third bounce, it travels up 1m.
So, 8 + 4 + 4 + 2 + 2 + 1 = 21m.

I hope this helps.

Number of digits where first and last digits are one or more than the middle digit are
101, 102, 103,104,105,106,107,108 and 109
Hence the probability is 9/100

Second problem answer I agree with gklick

based on the answer given, gklick you are right. the probability is 1/100.
what i dont understand is : when they ask the first and last digit is one greater than the middle, are they talking about the sum of the first and last digit or just being one greater than the middle digit.

the second answer is right. thanks . very useful stuff.

1) If a 3-digit integer is selected at random from the integers 100 through 199, inclusive, what is the probability that the first digit and the last digit of the integer are each equal to one more than the middle digit?

Ans: The question says to find the number in which "the first digit and the last digit of the integer are each equal to one more than the middle digit"
and not that number in which "the first digit and the last digit of the integer are each equal to one more OR MORE than the middle digit"
So, the only number to satisfy this condition would be "101" and since there are 100 number to choose from, the probability will be 1/100.


2) . A certain ball was dropped from a window 8 meters above a sidewalk. On each bounce it rose straight up exactly one-half the distance of the previous fall. After the third bounce the ball was caught when it reached a height of exactly 1 meter above the sidewalk. How many meters did the ball travel in all?

Ans: This is a common physics phenomenon (if you guys remember co-efficient of restitution).
In such questions if you observe a G.P is formed along with distance of first fall and can be used to determine the total distance that the ball shall travel if let free to fall.
i.e. S = A + 2 ( a (r^(n-1)) / (1-r) ) where,
S = Total distance travelled,
A = Distance covered in first fall
a = distance covered in first jump after first fall
r = the ratio by which the bounce happens (in this case r = 1/2)
and n = the number of bounces.

In this case, we have to find the distance travelled by the ball till third bounce. So, it becomes
8 + 2 (4 + 2 ) + 1 (Since the ball is caught after third bounce, return distance is not counted)
= 21

wow! the coefficient of restitution! with no real math training, i've never heard of it. i drew a silly little picture to figure it out.

thanks for posting these problems and answers. they are helping me get a handle on different approaches to one problem.

Rishabh wrote: 2) . A certain ball was dropped from a window 8 meters above a sidewalk. On each bounce it rose straight up exactly one-half the distance of the previous fall. After the third bounce the ball was caught when it reached a height of exactly 1 meter above the sidewalk. How many meters did the ball travel in all?

Ans: This is a common physics phenomenon (if you guys remember co-efficient of restitution).
In such questions if you observe a G.P is formed along with distance of first fall and can be used to determine the total distance that the ball shall travel if let free to fall.
i.e. S = A + 2 ( a (r^(n-1)) / (1-r) ) where,
S = Total distance travelled,
A = Distance covered in first fall
a = distance covered in first jump after first fall
r = the ratio by which the bounce happens (in this case r = 1/2)
and n = the number of bounces.

In this case, we have to find the distance travelled by the ball till third bounce. So, it becomes
8 + 2 (4 + 2 ) + 1 (Since the ball is caught after third bounce, return distance is not counted)
= 21

Frankly speaking, my approach of the GMAT is that it's certainly not worth trying to remember this formula. If you already know it from your background, fine. But otherwise, the 2-min quota per question will give you more than enough time to do the calculations by hand. There's no point in trying to know each and every formula you might potentially use in a problem; the GMAT is about solving rapidly problems, whatever the method you use.

Thoughts ?

Rishabh wrote: 2) . A certain ball was dropped from a window 8 meters above a sidewalk. On each bounce it rose straight up exactly one-half the distance of the previous fall. After the third bounce the ball was caught when it reached a height of exactly 1 meter above the sidewalk. How many meters did the ball travel in all?

Ans: This is a common physics phenomenon (if you guys remember co-efficient of restitution).
In such questions if you observe a G.P is formed along with distance of first fall and can be used to determine the total distance that the ball shall travel if let free to fall.
i.e. S = A + 2 ( a (r^(n-1)) / (1-r) ) where,
S = Total distance travelled,
A = Distance covered in first fall
a = distance covered in first jump after first fall
r = the ratio by which the bounce happens (in this case r = 1/2)
and n = the number of bounces.

In this case, we have to find the distance travelled by the ball till third bounce. So, it becomes
8 + 2 (4 + 2 ) + 1 (Since the ball is caught after third bounce, return distance is not counted)
= 21

Frankly speaking, my approach of the GMAT is that it's certainly not worth trying to remember this formula. If you already know it from your background, fine. But otherwise, the 2-min quota per question will give you more than enough time to do the calculations by hand. There's no point in trying to know each and every formula you might potentially use in a problem; the GMAT is about solving rapidly problems, whatever the method you use.

Thoughts ?

i agree piren.

If a 3-digit integer is selected at random from the integers 100 through 199, inclusive, what is the probability that the first digit and the last digit of the integer are each equal to one more than the middle digit?

let abc be the required 3 digit number.
condition: a=c=b+1
thus once we select b, a and c are automatically selected.
also b cannot be equal to 9.
If b=9,a=c=10 which is not possible
Thus b can range from 0-8
Numbers satisfying the condition=9
Total=999-100+1 = 900
probability = 9/900 = [spoiler]1/100[/spoiler]

A certain ball was dropped from a window 8 meters above a sidewalk. On each bounce it rose straight up exactly one-half the distance of the previous fall. After the third bounce the ball was caught when it reached a height of exactly 1 meter above the sidewalk. How many meters did the ball travel in all?

Before first bounce, distance traveled = 8 meters
distance traveled between first and second bounce = 8 meters (4 meters rise and 4 meters fall)
distance traveled between second and third bounce = 4 meters (2 meters rise and 2 meters fall)
distance traveled after third bounce = 1 meter (after that ball was caught)
total = 8+8+4+1 = 21 meters

My 2 cents:
Instead of visualizing this kind of ball dropping sequence, which is very time consuming - as already noted - i did as follows:

piren wrote: Frankly speaking, my approach of the GMAT is that it's certainly not worth trying to remember this formula. If you already know it from your background, fine. But otherwise, the 2-min quota per question will give you more than enough time to do the calculations by hand. There's no point in trying to know each and every formula you might potentially use in a problem; the GMAT is about solving rapidly problems, whatever the method you use.

Thoughts ?

Definitely true. You rarely want to use something complex when something simple will do: physics to cope with arithmetic is a bit much!