Last visit was: 19 Nov 2025, 16:17 It is currently 19 Nov 2025, 16:17
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2   3   4   
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
TOUGH & TRICKY SET Of PROBLEMS 09 Jan 2010, 05:49
Kudos
Add Kudos
Bookmarks
Bookmark this Post
angel2009
Bunuel
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*3/9*2/9=4/27

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

Answer: D.

I've some problem to understand this ; as the two events are mutually exclusive , independent.
So why not the answer is 3/9*2/9 = 2/27
Show more

Winning scenario consists of TWO cases RW and WR. Probability of each case is: 3/9*2/9, so 3/9*2/9+3/9*2/9=2*3/9*2/9.

Only 3/9*2/9 would be correct answer if we were asked to determine the probability of FIRST ball being red and SECOND white, OR FIRST white and SECOND red. These are the cases in which order of drawing matters.
User avatar
GMATMadeeasy
User avatar
Joined: 25 Dec 2009
Last visit: 17 Nov 2011
Posts: 65
Own Kudos:
1,904
Given Kudos: 3
Posts: 65
Kudos: 1,904
TOUGH & TRICKY SET Of PROBLEMS 19 Jan 2010, 06:43
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
SOLUTION:
2. THE PRICE OF BUSHEL:
The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of 5x cents per day while the price of wheat is decreasing at a constant rate of 2^1/2*x-x cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat?

(A) $4.50
(B) $5.10
(C) $5.30
(D) $5.50
(E) $5.60

Note that we are not asked in how many days prices will cost the same.
Let y # of days when these two bushels will have the same price.
First 2^1/2*x-x=0.41x
3.2+5xy=5.8-0.41xy solving for xy=0.48
The cost of a bushel of corn=3.2+5*0.48=5.6

Answer: E.
Show more

How do you make this Bunuel , I dont get it :(

Quote:
First 2^1/2*x-x=0.41x
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
TOUGH & TRICKY SET Of PROBLEMS 19 Jan 2010, 07:19
Kudos
Add Kudos
Bookmarks
Bookmark this Post
GMATMadeeasy
Bunuel
SOLUTION:
2. THE PRICE OF BUSHEL:
The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of 5x cents per day while the price of wheat is decreasing at a constant rate of 2^1/2*x-x cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat?

(A) $4.50
(B) $5.10
(C) $5.30
(D) $5.50
(E) $5.60

Note that we are not asked in how many days prices will cost the same.
Let y # of days when these two bushels will have the same price.
First 2^1/2*x-x=0.41x
3.2+5xy=5.8-0.41xy solving for xy=0.48
The cost of a bushel of corn=3.2+5*0.48=5.6

Answer: E.

How do you make this Bunuel , I dont get it :(

Quote:
First 2^1/2*x-x=0.41x
Show more

Yes it's kind of confusing, partly because I didn't use the formula formatting. Edited the original solution. So here it is:


Note that we are not asked in how many days prices will cost the same.

Let \(y\) be the # of days when these two bushels will have the same price.

First let's simplify the formula given for the rate of decrease of the price of wheat: \(\sqrt{2}*x-x=1.41x-x=0.41x\), this means that the price of wheat decreases by \(0.41x\) cents per day, in \(y\) days it'll decrease by \(0.41xy\) cents;

As price of corn increases \(5x\) cents per day, in \(y\) days it'll will increase by \(5xy\) cents;

Set the equation: \(3.2+5xy=5.8-0.41xy\), solve for \(xy\) --> \(xy=0.48\);

The cost of a bushel of corn in \(y\) days (the # of days when these two bushels will have the same price) will be \(3.2+5xy=3.2+5*0.48=5.6\).

Answer: E.

Hope it's clear. Tell me if it needs more clarification.
User avatar
GMATMadeeasy
User avatar
Joined: 25 Dec 2009
Last visit: 17 Nov 2011
Posts: 65
Own Kudos:
1,904
Given Kudos: 3
Posts: 65
Kudos: 1,904
TOUGH & TRICKY SET Of PROBLEMS Updated on: 19 Jan 2010, 08:44
Originally posted by GMATMadeeasy on 19 Jan 2010, 07:43.
Last edited by GMATMadeeasy on 19 Jan 2010, 08:44, edited 1 time in total.
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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):

Quote:
P=2*3/9*2/9=4/27

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

Answer: D.
Show more

The probability of drawing a red and a white ball has to be( 2/9)*(3/9) and then the solution follows. How come it is (2*3/9) / (2/9), i dont get it again .

Will continue again with rest of the questions.

P.S. I have never liked questions who have just been tough but no brainerbut the one in the list here are some of the best questions to train on many tricks and fundamentals I believe. More I see the stuff on quant , more I admire the work of the people engaged with the site. Bravo.. and cheers from cold Paris but grâce à GMAT for keeping me hot :)
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
TOUGH & TRICKY SET Of PROBLEMS 19 Jan 2010, 11:52
Kudos
Add Kudos
Bookmarks
Bookmark this Post
GMATMadeeasy
Bunuel
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):

Quote:
P=2*3/9*2/9=4/27

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

Answer: D.

The probability of drawing a red and a white ball has to be( 2/9)*(3/9) and then the solution follows. How come it is (2*3/9) / (2/9), i dont get it again .

Will continue again with rest of the questions.

P.S. I have never liked questions who have just been tough but no brainerbut the one in the list here are some of the best questions to train on many tricks and fundamentals I believe. More I see the stuff on quant , more I admire the work of the people engaged with the site. Bravo.. and cheers from cold Paris but grâce à GMAT for keeping me hot :)
Show more

We want out of two balls one to be red (R) and another white (W).

There are two ways this could happen: we can draw FIRST ball red and SECOND white OR FIRST ball white and SECOND red.

Probability of each case is: \(\frac{3}{9}*\frac{2}{9}\), so \(P=\frac{3}{9}*\frac{2}{9}+\frac{2}{9}*\frac{3}{9}=2*\frac{3}{9}*\frac{2}{9}\).

One more thing worth of mentioning: \(\frac{3}{9}*\frac{2}{9}\) would be correct answer if we were asked to determine the probability of FIRST ball being red and SECOND white, OR FIRST white and SECOND red. These are the cases in which order of drawing matters.

Hope it's clear.
User avatar
GMATMadeeasy
User avatar
Joined: 25 Dec 2009
Last visit: 17 Nov 2011
Posts: 65
Own Kudos:
1,904
Given Kudos: 3
Posts: 65
Kudos: 1,904
TOUGH & TRICKY SET Of PROBLEMS 19 Jan 2010, 13:08
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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):

Quote:
P=2*3/9*2/9=4/27
Show more

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

Answer: D.[/quote]

The probability of drawing a red and a white ball has to be( 2/9)*(3/9) and then the solution follows. How come it is (2*3/9) / (2/9), i dont get it again .

Will continue again with rest of the questions.

P.S. I have never liked questions who have just been tough but no brainerbut the one in the list here are some of the best questions to train on many tricks and fundamentals I believe. More I see the stuff on quant , more I admire the work of the people engaged with the site. Bravo.. and cheers from cold Paris but grâce à GMAT for keeping me hot :)[/quote]

We want out of two balls one to be red (R) and another white (W).

There are two ways this could happen: we can draw FIRST ball red and SECOND white OR FIRST ball white and SECOND red.

Probability of each case is: \(\frac{3}{9}*\frac{2}{9}\), so \(P=\frac{3}{9}*\frac{2}{9}+\frac{2}{9}*\frac{3}{9}=2*\frac{3}{9}*\frac{2}{9}\).

One more thing worth of mentioning: \(\frac{3}{9}*\frac{2}{9}\) would be correct answer if we were asked to determine the probability of FIRST ball being red and SECOND white, OR FIRST white and SECOND red. These are the cases in which order of drawing matters.

Hope it's clear
Perfect. That is how I solved it.

Next attack : Inequality : Will be going through your links to flex "inequality muscles" . Can't thank enough for the efforts you guys have put.
avatar
gaddam506
avatar
Joined: 31 Jan 2010
Last visit: 01 Jan 2011
Posts: 2
Own Kudos:
28
 [2]
Given Kudos: 1
Posts: 2
Kudos: 28
 [2]
TOUGH & TRICKY SET Of PROBLEMS 04 Mar 2010, 11:06
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
@ Bunuel

Q7 isn't that tough. we need a better approach.
Here is what I did.
Draw a circle with radius 1 and center as origin. Now draw the line y=(3/4)x-3. This has y intercept -3 and x-intercept =4. Now draw a line(blue line) from (0,0) perpendicular to the line. What we need is the length of red line. See the attached Diagram
Attachment:
q7.JPG
q7.JPG [ 12.01 KiB | Viewed 26735 times ]
Area of the right triangle with 3 as base is same as that with 5 as base
(1/2)*3*4 = (1/2)*5*(blue line) ==> blue line = 2.4 units
Hence red line = 2.4 - 1(radius of circle)
= 1.4

Hope this helps.
User avatar
ykaiim
User avatar
Joined: 25 Aug 2007
Last visit: 21 Aug 2012
Posts: 519
Own Kudos:
5,901
 [1]
Given Kudos: 40
WE 1: 3.5 yrs IT
WE 2: 2.5 yrs Retail chain
Posts: 519
Kudos: 5,901
 [1]
TOUGH & TRICKY SET Of PROBLEMS 17 Apr 2010, 11:50
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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.
Show more

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 :wink:
avatar
gunnersrule
avatar
Joined: 11 May 2010
Last visit: 05 May 2011
Posts: 1
Own Kudos:
1
 [1]
Posts: 1
Kudos: 1
 [1]
TOUGH & TRICKY SET Of PROBLEMS 12 May 2010, 03:47
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel

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
Show more


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
User avatar
anilnandyala
User avatar
Joined: 07 Feb 2010
Last visit: 19 Jun 2012
Posts: 101
Own Kudos:
4,711
Given Kudos: 101
Posts: 101
Kudos: 4,711
TOUGH & TRICKY SET Of PROBLEMS 13 Oct 2010, 09:13
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
 [2]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
 [2]
TOUGH & TRICKY SET Of PROBLEMS 13 Oct 2010, 09:20
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
anilnandyala
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
Show more

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

# 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: \(\frac{60}{96}=\frac{5}{8}\)

Hope it's clear.
avatar
sap
avatar
Joined: 22 Sep 2010
Last visit: 26 Nov 2012
Posts: 19
Own Kudos:
8
Given Kudos: 5
Posts: 19
Kudos: 8
TOUGH & TRICKY SET Of PROBLEMS 03 Nov 2010, 22:06
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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.
Show 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.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
TOUGH & TRICKY SET Of PROBLEMS 03 Nov 2010, 22:24
Kudos
Add Kudos
Bookmarks
Bookmark this Post
sap
Bunuel
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.
Show more

I think you mean not the coordinates but X and Y intercepts of the line. If the X and Y intercepts of the line \(y=\frac{3}{4}*x-3\) indeed were (1,0) and (0,-3/4), you would be right: the line would intersect the circle and the least possible distance between a point on the circle and a point on the line would be zero.

But the the X and Y intercepts of the line \(y=\frac{3}{4}*x-3\) are (4, 0) and (0, -3) respectively.

Hope it's clear.
avatar
sap
avatar
Joined: 22 Sep 2010
Last visit: 26 Nov 2012
Posts: 19
Own Kudos:
8
Given Kudos: 5
Posts: 19
Kudos: 8
TOUGH & TRICKY SET Of PROBLEMS 03 Nov 2010, 22:36
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I beg to ask again ( not too good at maths :-( ) the equation of the line is x/1 - y/(-3/4) = 1. What is the difference between co-ordinates and intercepts ? I know I m missing something but not able to identify what :-) . I m still in a position that it is a secant to the circle. Can you please explain a little more, if possible with another example ?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
TOUGH & TRICKY SET Of PROBLEMS 03 Nov 2010, 22:48
Kudos
Add Kudos
Bookmarks
Bookmark this Post
sap
I beg to ask again ( not too good at maths :-( ) the equation of the line is x/1 - y/(-3/4) = 1. What is the difference between co-ordinates and intercepts ? I know I m missing something but not able to identify what :-) . I m still in a position that it is a secant to the circle. Can you please explain a little more, if possible with another example ?
Show more

There is no such thing as "co-ordinate of a line".

As for X and Y intercepts: X-intercept is the point of intersection of a curve (line) and X-axis, so the point (x, 0), similarly Y-intercept is the point of intersection of a curve (line) and Y-axis, so the point (0, y).

So, X and Y intercepts of the line \(y=\frac{3}{4}*x-3\) are (4, 0) and (0, -3) respectively and the line does not intersect the circle \(x^2+y^2=1\).

For more on this issues please check Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.html

Hope it helps.
avatar
sixsigma1978
avatar
Joined: 01 Dec 2010
Last visit: 03 Jan 2011
Posts: 4
Own Kudos:
8
Posts: 4
Kudos: 8
TOUGH & TRICKY SET Of PROBLEMS 02 Dec 2010, 11:00
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
SOLUTION:
2. THE PRICE OF BUSHEL:
The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of \(5x\) cents per day while the price of wheat is decreasing at a constant rate of \(\sqrt{2}*x-x\) cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat?

(A) $4.50
(B) $5.10
(C) $5.30
(D) $5.50
(E) $5.60

Note that we are not asked in how many days prices will cost the same.

Let \(y\) be the # of days when these two bushels will have the same price.

First let's simplify the formula given for the rate of decrease of the price of wheat: \(\sqrt{2}*x-x=1.41x-x=0.41x\), this means that the price of wheat decreases by \(0.41x\) cents per day, in \(y\) days it'll decrease by \(0.41xy\) cents;

As price of corn increases \(5x\) cents per day, in \(y\) days it'll will increase by \(5xy\) cents;

Set the equation: \(320+5xy=580-0.41xy\), solve for \(xy\) --> \(xy=48\);

The cost of a bushel of corn in \(y\) days (the # of days when these two bushels will have the same price) will be \(320+5xy=320+5*48=560\) or $5.6.

Answer: E.
Show more

In the equation above- can someone explain why cents is not being converted to $?
The two sides are being added without /100
so shouldn't the equations read :

3.2 + 0.05xy --- LHS
and 5.8 - 0.0041xy -- RHS?
Apples to apples?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
TOUGH & TRICKY SET Of PROBLEMS 02 Dec 2010, 11:23
Kudos
Add Kudos
Bookmarks
Bookmark this Post
sixsigma1978
Bunuel
SOLUTION:
2. THE PRICE OF BUSHEL:
The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of \(5x\) cents per day while the price of wheat is decreasing at a constant rate of \(\sqrt{2}*x-x\) cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat?

(A) $4.50
(B) $5.10
(C) $5.30
(D) $5.50
(E) $5.60

Note that we are not asked in how many days prices will cost the same.

Let \(y\) be the # of days when these two bushels will have the same price.

First let's simplify the formula given for the rate of decrease of the price of wheat: \(\sqrt{2}*x-x=1.41x-x=0.41x\), this means that the price of wheat decreases by \(0.41x\) cents per day, in \(y\) days it'll decrease by \(0.41xy\) cents;

As price of corn increases \(5x\) cents per day, in \(y\) days it'll will increase by \(5xy\) cents;

Set the equation: \(320+5xy=580-0.41xy\), solve for \(xy\) --> \(xy=48\);

The cost of a bushel of corn in \(y\) days (the # of days when these two bushels will have the same price) will be \(320+5xy=320+5*48=560\) or $5.6.

Answer: E.

In the equation above- can someone explain why cents is not being converted to $?
The two sides are being added without /100
so shouldn't the equations read :

3.2 + 0.05xy --- LHS
and 5.8 - 0.0041xy -- RHS?
Apples to apples?
Show more

Even though it didn't affect the final answer I've still changed it to: \(320+5xy=580-0.41xy\), so now everything is in cents.
User avatar
mmcooley33
User avatar
Joined: 31 Oct 2010
Last visit: 27 Feb 2011
Posts: 24
Own Kudos:
351
Given Kudos: 25
Posts: 24
Kudos: 351
TOUGH & TRICKY SET Of PROBLEMS 07 Dec 2010, 03:17
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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

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?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
 [1]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
 [1]
TOUGH & TRICKY SET Of PROBLEMS 07 Dec 2010, 03:32
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mmcooley33
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

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?
Show more

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: \(\frac{60}{96}=\frac{5}{8}=62.5%\)

Answer: D.
User avatar
yogesh1984
User avatar
Joined: 12 Dec 2010
Last visit: 13 Nov 2021
Posts: 176
Own Kudos:
113
Given Kudos: 23
Concentration: Strategy, General Management
Schools: HBS '15 (D) Booth '15 (D) Insead '13 (D) ISB '14 (D) Stanford '15 (S) Stern '15 (S)
GMAT 1: 680 Q49 V34
GMAT 2: 730 Q49 V41
GPA: 4
WE:Consulting (Other)
Products:
My Reviews
Schools: HBS '15 (D) Booth '15 (D) Insead '13 (D) ISB '14 (D) Stanford '15 (S) Stern '15 (S)
GMAT 2: 730 Q49 V41
Posts: 176
Kudos: 113
TOUGH & TRICKY SET Of PROBLEMS 15 Jan 2011, 11:51
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
4. ADDITION PROBLEM:
AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1

(B) 3

(C) 7

(D) 9

(E) Cannot be determined

AB and CD are two digit integers, their sum can give us only one three digit integer of a kind of AAA it's 111.
So, A=1. 1B+CD=111
C can not be less than 9, because no to digit integer with first digit 1 (mean that it's<20) can be added to two digit integer less than 90 to have the sum 111 (if CD<90 meaning C<9 CD+1B<111).
C=9

Answer: D.
Show more
Bunuel I have doubt here-- consider following AB as 25 CD 86 which sums to 111 and ||ly
if we consider AB as 15 & CD as 96 then also we end up with 111 so I am doubtfull with
the answer provided above....(please do correct if Am missing something :| )
   1   2   3   4   
Moderators:
Bunuel
Math Expert
105390 posts
Krunaal
Tuck School Moderator
805 posts