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Re: TOUGH & TRICKY SET Of PROBLEMS
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09 Jan 2010, 05:49
angel2009 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*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 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.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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19 Jan 2010, 06:43
Bunuel wrote: 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*xx 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*xx=0.41x 3.2+5xy=5.80.41xy 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*xx=0.41x



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Re: TOUGH & TRICKY SET Of PROBLEMS
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19 Jan 2010, 07:19
GMATMadeeasy wrote: Bunuel wrote: 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*xx 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*xx=0.41x 3.2+5xy=5.80.41xy 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*xx=0.41x 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: \(\sqrt{2}*xx=1.41xx=0.41x\), 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: \(3.2+5xy=5.80.41xy\), 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.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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Updated on: 19 Jan 2010, 08:44
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): 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
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.



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Re: TOUGH & TRICKY SET Of PROBLEMS
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19 Jan 2010, 11:52
GMATMadeeasy 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): 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 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.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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19 Jan 2010, 13:08
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.[/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.



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Re: TOUGH & TRICKY SET Of PROBLEMS
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04 Mar 2010, 11:06
@ 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)x3. This has y intercept 3 and xintercept =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 [ 12.01 KiB  Viewed 23325 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.



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Re: TOUGH & TRICKY SET Of PROBLEMS
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17 Apr 2010, 11:50
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 > 127/64=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
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Re: TOUGH & TRICKY SET Of PROBLEMS
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12 May 2010, 03:47
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, xaxis and yaxis 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 PROBLEMS
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13 Oct 2010, 09:13
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=8p1 > 8p1<=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



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Re: TOUGH & TRICKY SET Of PROBLEMS
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13 Oct 2010, 09:20
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=8p1 > 8p1<=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 \(n(n+1)(n+2)\). # of even numbers between 1 and 96, inclusive is \(\frac{962}{2}+1=48\) (check this: totallybasic94862.html?hilit=last%20first%20range%20multiple) AND 2. When \(n+1\) is divisible by 8. > \(n+1=8p\) (\(p\geq{1}\)), \(n=8p1\) > \(8p1\leq{96}\) > \(p\leq{12.1}\) > 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.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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03 Nov 2010, 22:06
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 \(y=\frac{3}{4}*x3\) (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 \((x_2,y_2)\): \(d=\sqrt{(x_1x_2)^2+(y_1y_2)^2}\) 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: \(2.41(radius)=1.4\)
Answer: A.
OR ANOTHER APPROACH:
In an xy Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that: \((xa)^2+(yb)^2=r^2\)
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 \(y = \frac{{3}}{{4}}x3\).
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: \(\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}\) > \(\frac{height}{3}=\frac{4}{5}\) > \(height=2.4\).
\(Distance=heightradius=2.41=1.4\)
Answer: A.
You can check the link of Coordinate Geometry below for more. coordinates 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.



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Re: TOUGH & TRICKY SET Of PROBLEMS
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03 Nov 2010, 22:24
sap wrote: 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 \(y=\frac{3}{4}*x3\) (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 \((x_2,y_2)\): \(d=\sqrt{(x_1x_2)^2+(y_1y_2)^2}\) 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: \(2.41(radius)=1.4\)
Answer: A.
OR ANOTHER APPROACH:
In an xy Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that: \((xa)^2+(yb)^2=r^2\)
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 \(y = \frac{{3}}{{4}}x3\).
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: \(\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}\) > \(\frac{height}{3}=\frac{4}{5}\) > \(height=2.4\).
\(Distance=heightradius=2.41=1.4\)
Answer: A.
You can check the link of Coordinate Geometry below for more. coordinates 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. I think you mean not the coordinates but X and Y intercepts of the line. If the X and Y intercepts of the line \(y=\frac{3}{4}*x3\) 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 \(y=\frac{3}{4}*x3\) are (4, 0) and (0, 3) respectively. Hope it's clear.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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03 Nov 2010, 22:36
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 coordinates 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 ?



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Re: TOUGH & TRICKY SET Of PROBLEMS
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03 Nov 2010, 22:48
sap wrote: 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 coordinates 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 ? There is no such thing as "coordinate of a line". As for X and Y intercepts: Xintercept is the point of intersection of a curve (line) and Xaxis, so the point (x, 0), similarly Yintercept is the point of intersection of a curve (line) and Yaxis, so the point (0, y). So, X and Y intercepts of the line \(y=\frac{3}{4}*x3\) 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: mathcoordinategeometry87652.htmlHope it helps.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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02 Dec 2010, 11:00
Bunuel wrote: 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 \(\sqrt{2}*xx\) 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: \(\sqrt{2}*xx=1.41xx=0.41x\), 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: \(320+5xy=5800.41xy\), 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?



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Re: TOUGH & TRICKY SET Of PROBLEMS
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02 Dec 2010, 11:23
sixsigma1978 wrote: Bunuel wrote: 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 \(\sqrt{2}*xx\) 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: \(\sqrt{2}*xx=1.41xx=0.41x\), 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: \(320+5xy=5800.41xy\), 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? Even though it didn't affect the final answer I've still changed it to: \(320+5xy=5800.41xy\), so now everything is in cents.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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07 Dec 2010, 03:17
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=8p1 > 8p1<=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?



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Re: TOUGH & TRICKY SET Of PROBLEMS
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07 Dec 2010, 03:32
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=8p1 > 8p1<=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? 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 \(\frac{962}{2}+1=48\) (check this: totallybasic94862.html?hilit=last%20first%20range%20multiple) AND 2. When \(n+1\) is divisible by 8. > \(n+1=8p\) (\(p\geq{1}\)), \(n=8p1\) > \(8p1\leq{96}\) > \(p\leq{12.1}\) > 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.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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15 Jan 2011, 11:51
Bunuel wrote: 4. ADDITION PROBLEM: AB + CD = AAA, where AB and CD are twodigit 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. 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 )
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Re: TOUGH & TRICKY SET Of PROBLEMS
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