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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
07 Mar 2010, 06:18

Gents,

An easier way to solve this prob is thru basic geometry if anyone interested. An illustration will explain better but if you draw a (perpendicular) line between the center of the circle to the line y = 3/4*x - 3, you get a 3,4,5 right triangle where the longest edge is 4cm long. 4/5*3 minus r of the circle (1) is equal to 1.4, the least possible distance.

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
02 Apr 2010, 01:36

1

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Many thanks for your post. _________________

But there’s something in me that just keeps going on. I think it has something to do with tomorrow, that there is always one, and that everything can change when it comes. http://aimingformba.blogspot.com

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
17 Apr 2010, 10:50

1

This post received KUDOS

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.

Answer: C.

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 _________________

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
12 May 2010, 02:47

1

This post received KUDOS

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

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
23 Sep 2010, 13:03

Bunuel wrote:

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

So, the answere is A.

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.

Bunuel, Could you please provide the formula that is rarely tested? I want to be prepared in case I face such a question and the trick with triangles does not work due to some mistery

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
23 Sep 2010, 13:14

Expert's post

Financier wrote:

Bunuel wrote:

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

So, the answere is A.

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.

Bunuel, Could you please provide the formula that is rarely tested? I want to be prepared in case I face such a question and the trick with triangles does not work due to some mistery

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
02 Oct 2010, 00:09

Bunuel wrote:

SOLUTION OF 8-10

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

Answer: C

Thanks /bunuel Your posts really help strengthen ones fundas!

I have a query regarding the above question. This is the way I approached it and I got a different answer

Probability that a given no. is divisible by 8(in 1 to 8) is 1/8 (because only 8 is divisible by 8 in values ranging from 1 to 8) So, for (n)*(n+1)*(n+2) (For 1 to 8) For n=6,7,8, the above sequence will be divisible by 8, therefore probability that given sequence to be divisible by 8 for first 8 numbers is 3/8 Now this will hold good for first 16 numbers also it will be 6/16 similarly for first 96 numbers it should be 36/96 which is 3/8 ...37.5% That is my answer 3/8. or 37.5%.....But it jus t doesnt seem to be working at all any help on this?

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
02 Oct 2010, 00:17

gowrits wrote:

Bunuel wrote:

SOLUTION OF 8-10

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

Answer: C

Thanks /bunuel Your posts really help strengthen ones fundas!

I have a query regarding the above question. This is the way I approached it and I got a different answer

Probability that a given no. is divisible by 8(in 1 to 8) is 1/8 (because only 8 is divisible by 8 in values ranging from 1 to 8) So, for (n)*(n+1)*(n+2) (For 1 to 8) For n=6,7,8, the above sequence will be divisible by 8, therefore probability that given sequence to be divisible by 8 for first 8 numbers is 3/8 Now this will hold good for first 16 numbers also it will be 6/16 similarly for first 96 numbers it should be 36/96 which is 3/8 ...37.5% That is my answer 3/8. or 37.5%.....But it jus t doesnt seem to be working at all any help on this?

Okay, got my mistake. I am missing out on cases like 4*5*6 which is divisible by 8 indirectly. This one truly was tricky! Thanks

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
13 Oct 2010, 08:13

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

my doubt is total even numbers is 48 + 12 , but the 12 numbers divisible by 8 already included in 48 . so why de we need to add 12

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
13 Oct 2010, 08:20

2

This post received KUDOS

Expert's post

anilnandyala 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

my doubt is total even numbers is 48 + 12 , but the 12 numbers divisible by 8 already included in 48 . so why de we need to add 12

thanks in advance

See the bold part in the end of the solution:

\(n(n+1)(n+2)\) 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 \(n(n+1)(n+2)\).

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.

It would be easily founded if we could us calculators but we dont. This means that k should be a number ending at 9 or 1, so we eliminate answer D- 157 So we have three possibilities. 79, 81, 159 Also we could see without calculating that k2>79.79 as well as k2>81 so the right answer is E. It's not the most rational way, but it was the first one that came to my mind.

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
27 Oct 2010, 22:39

was just wondering if these questions can be combined together into one document... or maybe in an online test format. Is it possible ? _________________

Regards, If you like my post, consider giving me akudo. THANKS!

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
03 Nov 2010, 21:06

Bunuel wrote:

SOLUTION: 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 is tough: First note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).

Now we can do this by finding the equation of a line perpendicular to given line \(y=\frac{3}{4}*x-3\) (we know it should cross origin and cross given line, so we can write the formula of it), then find the croos point of these lines and then the distance between the origin and this point. But it's very lengthy way.

There is another, shorter one. Though I've never seen any GMAT question requiring the formula used in it.

We know the formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\): \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\) BUT there is a formula to calculate the distance between the point (in our case origin) and the line:

DISTANCE BETWEEN THE LINE AND POINT: Line: \(ay+bx+c=0\), point \((x_1,y_1)\)

\(d=\frac{|ay_1+bx_1+c|}{\sqrt{a^2+b^2}}\)

DISTANCE BETWEEN THE LINE AND ORIGIN: As origin is \((0,0)\) -->

\(d=\frac{|c|}{\sqrt{a^2+b^2}}\)

So in our case it would be: \(d=\frac{|3|}{\sqrt{1^2+(\frac{3}{4})^2}}=\frac{12}{5}=2.4\)

So the shortest distance would be: \(2.4-1(radius)=1.4\)

Answer: A.

OR ANOTHER APPROACH:

In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that: \((x-a)^2+(y-b)^2=r^2\)

If the circle is centered at the origin (0, 0), then the equation simplifies to: \(x^2+y^2=r^2\)

So, the circle represented by the equation \(x^2+y^2 = 1\) is centered at the origin and has the radius of \(r=\sqrt{1}=1\).

Then note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).

So we should find the length of perpendicular, or the height of the right triangle formed by the X and Y axis and the line \(y = \frac{{3}}{{4}}x-3\).

The legs would be the value of x for y=0 (x intercept) --> y=0, x=4 --> \(leg_1=4\). and the value of y for x=0 (y intercept) --> x=0, y=-3 --> \(leg_2=3\).

So we have the right triangle with legs 4 and 3 and hypotenuse 5. What is the height of this triangle (perpendicular from right angle to the hypotenuse)? As perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle: \(\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}\) --> \(\frac{height}{3}=\frac{4}{5}\) --> \(height=2.4\).

\(Distance=height-radius=2.4-1=1.4\)

Answer: A.

You can check the link of Coordinate Geometry below for more.

co-ordinates of line are (1,0) and (0,-3/4). This line will be a secant to the circle, the least distance then should le less than 1. Please correct me if I m wrong somewhere.

gmatclubot

Re: TOUGH & TRICKY SET Of PROBLMS
[#permalink]
03 Nov 2010, 21:06

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