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# TOUGH & TRICKY SET Of PROBLEMS

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Manager
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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13 Oct 2009, 12:13
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Bunuel wrote:
SOLUTION:
5. RACE:
A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20

Let x be the speed of B.
Write the equation:

(440-48)/x (time of B for first heat) - 6 (seconds, time B lost to A first heat) = TIME OF A (in both heats A runs with constant rate, so the time for first and second heats are the same)=(440-144)/x (time of B for second heat) + 2 (seconds, time B won to A second heat)

(440-48)/x-6=(440-144)/x+2
x=12

Equation is formed with 440 whereas the question talks about 480m race. Also the equation doesnt give x=12. if I subtitute for x in equation I get 392/6 = 296/14 which is not correct
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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16 Oct 2009, 20:03
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This is just great Bunuel - keep em coming .. Kudos
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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18 Oct 2009, 11:33
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thanks for the post . and 2 kudous to you for posting such gud questions
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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26 Oct 2009, 20:17
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9. PROBABILITY OF INTEGER BEING DIVISIBLE BY 8:
If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?

A. 25%
B 50%
C 62.5%
D. 72.5%
E. 75%

Ans C 62.5%

Given, n is an integer from 1 to 96.
Case 1:
If n is even then n*(n+1)*(n+2) is divisible by 8.
reason: n is even then n is divisible by 2 and n+2 is divisible by 4. Hence the product should be divisible by 8.
ex: 2*3*4.

1 to 96 there are 48 even integers. Hence 48 possibilities divisible by 8.

If n is odd, then n*(n+1)*(n+2) is divisible by 8 when the even integer n+1 is divisible by 8.
There are 12 possible cases here. Write now multiple of 8s you will get till 96, 12 possibilities.

Add 48+12 and divide by total 96 = .625. Hence, 62.5% is correct answer.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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02 Apr 2010, 02:36
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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17 Apr 2010, 11:50
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Bunuel wrote:
SOLUTION:
3. LEAP YEAR:
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5

Probability of a randomly selected person NOT to be born in a leap year=3/4
Among 2 people, probability that none of them was born in a leap = 3/4*3/4=9/16. The probability at least one born in leap = 1- 9/16=7/16<1/2
So, we are looking for such n (# of people), when 1-(3/4)^n>1/2
n=3 --> 1-27/64=37/64>1/2

Thus min 3 people are needed.

Well, I have a different approach, if youhave time crunch and u want to try GUESS. As the question asked -
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year

So, we need an odd number of people to select from so that the prob > 50%. This leds to choice A, C and E. 1 can't be the number. So, if we take 3 (minimum is needed) then we will have 2 people
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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12 May 2010, 03:47
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Bunuel wrote:
7. THE DISTANCE BETWEEN THE CIRCLE AND THE LINE:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0

This problem can be solved much quicker using the right triangles.
Note that line y = 3/4*x - 3, x-axis and y-axis form a 3:4:5 type triangle.
now, there are 2 quick ways to solve it from here:
1) using similar triangles - perpendicular to line y = 3/4*x - 3 creates two triangles similar to the original one , hence the length of the perpendicular can be calculated
or 2) using formula for the area of the triangle (a=1/2hieght x base) - we know area is 6 (4x3/2=6) and we know base is 5, so plug this into the formula and we get 12/5 or 2,4 for the height, which is also the length of the perpendicular we were looking for.
Hope this helps
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Re: TOUGH & TRICKY SET Of PROBLMS Question 9 [#permalink]

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07 Dec 2010, 03:32
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Expert's post
mmcooley33 wrote:
9. PROBABILITY OF INTEGER BEING DIVISIBLE BY 8:
If n is an integer from 1 to 96 (inclusive), what is the probability for n*(n+1)*(n+2) being divisible by 8?

A. 25%
B 50%
C 62.5%
D. 72.5%
E. 75%

N=n*(n+1)*(n+2)

N is divisible by 8 in two cases:
When n is even:
No of even numbers (between 1 and 96)=48
AND
When n+1 is divisible by 8. -->n=8p-1 --> 8p-1<=96 --> p=12.3 --> 12 such nembers

Total=48+12=60

Probability=60/96=0.62

so if I were to write out n(n+1)(n+2) = n^3 + 3n^2 + 2n meaning that every even number will be divisible by 8 because every even number will have at least three 2's as factors, also taking care of the n+2 because it would be even as well, then add the 12 numbers divisible by 12 when you add 1. Is the way I am thinking correct?

Expanding is not a good idea. Below is a solution for this problem:

If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

$$n(n+1)(n+2)$$ is divisible by 8 in two cases:

1. When $$n$$ is even:
$$n=2k$$ --> $$n(n+1)(n+2)=2k(2k+1)(2k+2)=4k(2k+1)(k+1)$$ --> either $$k$$ or $$k+1$$ is even so 8 is a multiple of $$n(n+1)(n+2)$$.

# of even numbers between 1 and 96, inclusive is $$\frac{96-2}{2}+1=48$$ (check this: totally-basic-94862.html?hilit=last%20first%20range%20multiple)

AND

2. When $$n+1$$ is divisible by 8. --> $$n+1=8p$$ ($$p\geq{1}$$), $$n=8p-1$$ --> $$8p-1\leq{96}$$ --> $$p\leq{12.1}$$ --> 12 such numbers.

Also note that these two sets have no overlaps, as when $$n$$ and $$n+2$$ are even then $$n+1$$ is odd and when $$n+1$$ is divisible by 8 (so even) then $$n$$ and $$n+2$$ are odd.

Total=48+12=60

Probability: $$\frac{60}{96}=\frac{5}{8}=62.5%$$

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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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21 Feb 2011, 09:35
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Expert's post

it is time of B - time of A = 6

changed the order ..

thnx

Check some other rate problems here: search.php?search_id=tag&tag_id=64

Hope it helps.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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03 Apr 2012, 10:32
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Expert's post
theamwan wrote:
Bunuel wrote:

That's not correct out of 4 years 1 is leap, so the probability of a randomly selected person NOT to be born in a leap year is 3/4.

But the question doesn't say that it's a set of 4 consecutive years. What if the years chosen are 2011, 2010, 2009, 2007?
IMO, a person's birth year to fall on a leap year should be a binary value - Y or N, hence a 50% probability.

Again that's not correct. Let me ask you a question: what is the probability that a person in born on Monday? Is it 1/2? No, it's 1/7.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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30 Jun 2012, 04:17
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Expert's post
dianamao wrote:
Bunuel wrote:
SOLUTION:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

This is with replacement case (and was solved incorrectly by some of you):

$$P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}$$

We are multiplying by 2 as there are two possible wining scenarios RW and WR.

The question says that red and white balls are selected in two Successive draws. Doesn't this imply that white is selected AFTER red? Thus no need for x2?

Thanks,
Diana

No, in that case we would be asked "what is the the probability of the first ball being red and the second ball being white?"
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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21 Dec 2012, 04:40
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Expert's post
Archit143 wrote:
I think i made mistake since 6 can be also a probable number but is not a multiple of 4..............Bunuel can u pls help me solve this one using multiple principle

If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 25%
B. 50%
C. 62.5%
D. 72.5%
E. 75%

$$n(n + 1)(n + 2)$$ is divisible by 8 in two cases:

A. $$n=even$$, in this case $$n+2=even$$ too and as $$n$$ and $$n+2$$ are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8;
B. $$n+1$$ is itself divisible by 8;

(Notice that these two sets have no overlaps, as when $$n$$ and $$n+2$$ are even then $$n+1$$ is odd and when $$n+1$$ is divisible by 8 (so even) then $$n$$ and $$n+2$$ are odd.)

Now, in EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8=0.625.

Similar question: divisible-by-12-probability-121561.html

Hope it helps.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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13 Oct 2009, 13:00
Bunuel wrote:
asterixmatrix wrote:
Bunuel wrote:
SOLUTION:
5. RACE:
A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20

Let x be the speed of B.
Write the equation:

(440-48)/x (time of B for first heat) - 6 (seconds, time B lost to A first heat) = TIME OF A (in both heats A runs with constant rate, so the time for first and second heats are the same)=(440-144)/x (time of B for second heat) + 2 (seconds, time B won to A second heat)

(440-48)/x-6=(440-144)/x+2
x=12

Equation is formed with 440 whereas the question talks about 480m race. Also the equation doesnt give x=12. if I subtitute for x in equation I get 392/6 = 296/14 which is not correct

First of all thanks for pointing out the typo. Edited the post above. Second it's funny but the equation with typo also gives the correct answer x=12:

392/x-6=296/x+2 --> 96/x=8 x=12
432/x-6=336/x+2 --> 96/x=8 x=12

since we subtract from a common value thats y we get the ans as 12 even with incorrect value

Asterix
Maths was better with 1 and 0s
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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16 Oct 2009, 09:17
Thanks ..
Keep them coming .. i enjoyed a lot solving them .
GMAT quant is getting tougher ... there questions will surely train us . thx ..
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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19 Oct 2009, 04:42
jax91 wrote:
1.) (E) 159

4.) AB + CD = AAA

adding 2 2 digit numbers is giving a 3 digit number. So hunderds digit of the 3 digit number has to be 1

so it becomes 1B + CD = 111

B cannot be 1, CD cannot be 99 (they are distinct)
so CD can take any value from 98 to 92.

So (E) Cannot be determined

That not true... the question is on C and not CD --> answer is 9.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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26 Oct 2009, 20:05
Awesome questions and explanations. I could solve 80%, dist from circle to line was the best and I had no clue...
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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31 Oct 2009, 06:43
Bunuel wrote:
SOLUTION:
1. THE SUM OF EVEN INTEGERS:
The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?

(A) 79
(B) 80
(C) 81
(D) 157
(E) 159

The solution given in the file was 79, which is not correct. Also the problem was solved with AP formula thus was long and used the theory rarely tested in GMAT. Here is my solution and my notes about AP. Some may be useful:

The number of terms in this set would be: n=(k-1)/2 (as k is odd)
Last term: k-1
Average would be first term+last term/2=(2+k-1)/2=(k+1)/2
Also average: sum/number of terms=79*80/((k-1)/2)=158*80/(k-1)
(k+1)/2=158*80/(k-1) --> (k-1)(k+1)=158*160 --> k=159

How is n = (k-1)/2??

If the last term is K, and the 1st term is 1, the no. of terms should be k-1+1
Hence, here the no. of terms should be =>(k-1+1)/2 => k/2 since there should be half as many even numbers...
Am I missing something?
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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18 Dec 2009, 11:53
A very good (ofcourse tough and tricky) set of problems. But when i was prcticing the tests from book that time i felt they are quite easier. Are these the actual type of questions that come in GMAT?
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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22 Dec 2009, 21:28
9. PROBABILITY OF INTEGER BEING DIVISIBLE BY 8:
If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?

A. 25%
B 50%
C 62.5%
D. 72.5%
E. 75%

Ans C 62.5%
--------------------------------------------------------------
I solve it in an opposite side of view.
How about n is odd and n,n+2 are both odds, which means only if n+1 is indivisible by 8 can satisfy the product of n*(n+1)*(n+2) is indivisible.
Let's take a look at n+1 which is even.
There are 48 even integers from 1 to 96. 96/2 = 48, and among these 48 even integers, 96/8=12 of them are divisible by 8. So 48-12=36 even integers are indivisible. 1 - 36/96 = 0.625
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]

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23 Dec 2009, 15:35
Vyacheslav wrote:
7. THE DISTANCE BETWEEN THE CIRCLE AND THE LINE:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0

I find useful to roughly draw such questions. You can instantly recognize that line y=3/4*x-3 and axises form a right triangle with sides 3 and 4, on Y-axis and X-axis respectively. The hypotenuse lies on the line y=3/4*x-3 and equals 5 (Pythagorean Triple). The closest distance from any point to line is perpendicular from that point to line. Perpendicular to the hypotenuse (from origin in this case) will always divide the triangle into two triangles with the same properties as the original triangle. So, lets perpendicular will $$x$$, then

$$\frac{x}{3}=\frac{4}{5}$$ and $$x=3*\frac{4}{5}=\frac{12}{5}=2,4$$

To find distance from point on circle to line we should substract lenth of radius from lenth of perpendicular: 2,4-1=1,4

Unfortunately, I cannot attach illustration. But if you draw this problem (even roughly), you will find it really easy to solve.

Excellent! +1.

For those who don't know the distance formula (which is in fact very rarely tested) this is the easiest and most elegant solution.
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Re: TOUGH & TRICKY SET Of PROBLMS   [#permalink] 23 Dec 2009, 15:35

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