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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
30 Dec 2009, 03:25

there is an errata on solution of 4. the answer requested is the value of C not CD.

Thus the answer is 9, option D

Megan

Bunuel wrote:

As asked I'm combining all the problems from Tough & tricky problems in single thread. Here are the first ten questions. Next set and solutions to these ten will follow in couple of hours.

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

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

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

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

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

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

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

8. THE AVERAGE TEMPERATURE: The average of temperatures at noontime from Monday to Friday is 50; the lowest one is 45, what is the possible maximum range of the temperatures?

A. 20 B. 25 C. 40 D. 45 E. 75

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%

10. SUM OF INTEGERS: If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is:

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
09 Jan 2010, 04:10

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

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
09 Jan 2010, 04:49

Expert's post

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. _________________

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
19 Jan 2010, 05:43

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*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

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
19 Jan 2010, 06:19

Expert's post

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*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

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. _________________

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
19 Jan 2010, 06:43

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

Last edited by GMATMadeeasy on 19 Jan 2010, 07:44, edited 1 time in total.

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
19 Jan 2010, 07:42

Bunuel wrote:

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*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

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.

My problem was in understanding how to get .41x . It is crystal clear now. Thank you so much.

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
19 Jan 2010, 10:52

Expert's post

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.

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
19 Jan 2010, 12:08

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.

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]
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

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