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# TOUGH & TRICKY SET Of PROBLEMS

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TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

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Updated on: 12 Oct 2009, 13:22
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/the-sum-of-t ... 68732.html

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 OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/the-price-of ... 70101.html 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 OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/how-many-ran ... 89297.html 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 OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/ab-cd-aaa-wh ... 36903.html 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 OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/a-and-b-ran- ... 06921.html 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 OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/a-bag-contai ... -8350.html 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 OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/what-is-the- ... 85184.html 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 OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/the-average- ... 16784.html 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% OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-an-intege ... 26654.html 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: A. A+1 inquiry B. A+5 C A+25 D 2A E. 5A OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-the-sum-o ... 02768.html THE OA WITH SOLUTIONS WILL BE PROVIDED. _________________ Originally posted by Bunuel on 12 Oct 2009, 08:22. Last edited by Bunuel on 12 Oct 2009, 13:22, edited 1 time in total. ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 64322 Re: TOUGH & TRICKY SET Of PROBLEMS [#permalink] ### Show Tags 12 Oct 2009, 13:06 33 28 SOLUTION: 1. THE SUM OF EVEN INTEGERS: The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=? (A) 79 (B) 80 (C) 81 (D) 157 (E) 159 The solution given in the file was 79, which is not correct. Also the problem was solved with AP formula thus was long and used the theory rarely tested in GMAT. Here is my solution and my notes about AP. Some may be useful: The number of terms in this set would be: n=(k-1)/2 (as k is odd) Last term: k-1 Average would be first term+last term/2=(2+k-1)/2=(k+1)/2 Also average: sum/number of terms=79*80/((k-1)/2)=158*80/(k-1) (k+1)/2=158*80/(k-1) --> (k-1)(k+1)=158*160 --> k=159 Answer E. MY NOTES ABOUT AP: ARITHMETIC PROGRESSION Sequence a1, a2,…an, so that a(n)=a(n-1)+d (constant) nth term an = a1 + d ( n – 1 ) Sn=n*(a1+an)/2 or Sn=n*(2a1+d(n-1))/2 Special cases: I. 1+2+…+n=n(1+n)/2 (Sum of n first integers) II. 1+3+5+… (n times)=n^2 (Sum of n first odd numbers). nth term=2n-1 III. 2+4+6+… (n times)=n(n+1) (Sum of n first even numbers) nth term=2n SOLUTION WITH THE AP FORMULA: Sequence of even numbers First term a=2, common difference d=2 since even number Sum to first n numbers of AP: Sn=n*(a1+an)/2=n(2*2+2(n-1))/2=n(n+1)=79*80 n=79 (odd) Number of terms n=(k-1)/2=79 k=159 OR Sum of n even numbers n(n+1)=79*80 n=79 k=2n+1=159 _________________ ##### Most Helpful Community Reply Manager Joined: 01 Jan 2009 Posts: 66 Location: India Schools: LBS Re: TOUGH & TRICKY SET Of PROBLEMS [#permalink] ### Show Tags 12 Oct 2009, 09:15 10 1 1.) (E) 159 sum of numbers till n = n(n+1)/2 , sum of even numbers will be almost half of this i.e. n(n+1)/4 . so if S = 79.80, then n= 159. 2.) (E) 5.60 3.) (C) 3 , with 2 we will have exactly 50%, so for more than 50% should be 3. 4.) AB + CD = AAA adding 2 2 digit numbers is giving a 3 digit number. So hunderds digit of the 3 digit number has to be 1 so it becomes 1B + CD = 111 B cannot be 1, CD cannot be 99 (they are distinct) so CD can take any value from 98 to 92. So (E) Cannot be determined 5.) (A) 12 ( and A's comes to 18) 6.) (A) 2/27 7.) (A) 1.4 eqn of circle = x^2 + y^2 = 1, center = (0,0) radius = 1 min dist of line from circle = dist of line from the center - radius Make the distance of the line from the circle to be 0 and we see that it becomes a tangent to the circle. Now we know if we draw a line from the center to the point where the tangent touches the circle the line and the tangent are perpendicular to each other. So we need to find the equation of this line first. We can take the line back where it was now Since the lines are perpendicular m1 x m2 = -1 m of line = 3/4 so slope of the new line = -4/3 Since the line passes through the origin (center of circle) its eqn => y=-4/3x now we need to get the point of intersection of our two lines, which comes out to be (36/25,-48/25) now get the distance of this point from the origin and subtract the radius from it. Comes to 1.4 (may have made calculation errors ) So A. 8.) (B) 25 (values will be like 45,45,45,45,70) 9.) (A) 25% , to be divisible by 8 the number needs to have 3 2s in it. Only a multiple of 4 can provide that. Number of numbers divisible by 4 = 96/4 = 24. So P(8) = 24/96 = 25% . 10.) A + 25 , each of the new 5 consecutive numbers are going to be 5 more than the prev 5 consecutive numbers. great work Bunuel ##### General Discussion Math Expert Joined: 02 Sep 2009 Posts: 64322 Re: TOUGH & TRICKY SET Of PROBLEMS [#permalink] ### Show Tags 12 Oct 2009, 11:41 5 2 jax91 wrote: great work Bunuel Thanks! You did a great job too, though 2 answers are not correct. You can try to find them or wait for the solutions. _________________ Manager Joined: 01 Jan 2009 Posts: 66 Location: India Schools: LBS Re: TOUGH & TRICKY SET Of PROBLEMS [#permalink] ### Show Tags 12 Oct 2009, 13:17 1 Bunuel wrote: jax91 wrote: great work Bunuel Thanks! You did a great job too, though 2 answers are not correct. You can try to find them or wait for the solutions. got my mistakes, i always end up doing such clumsy mistakes. 4.) (D) 9 . I solved for CD 9.) Just simply forgot to take into consideration n+1 and n+2 ! So from 1-96, we get all even numbers = 48 + all numbers 1 less than a multiple of 8= 12 so 60/96 (C) 62.5% hope these are right! Math Expert Joined: 02 Sep 2009 Posts: 64322 Re: TOUGH & TRICKY SET Of PROBLEMS [#permalink] ### Show Tags 12 Oct 2009, 13:22 15 18 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.

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

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12 Oct 2009, 13:33
26
25
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.

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

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12 Oct 2009, 13:43
14
18
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

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

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Updated on: 13 Oct 2009, 11:49
20
21
SOLUTION:
5. RACE:
A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20

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

(480-48)/x (time of B for first heat) - 6 (seconds, time B lost to A first heat) = TIME OF A (in both heats A runs with constant rate, so the time for first and second heats are the same)=(480-144)/x (time of B for second heat) + 2 (seconds, time B won to A second heat)

(480-48)/x-6=(480-144)/x+2
x=12

_________________

Originally posted by Bunuel on 12 Oct 2009, 13:57.
Last edited by Bunuel on 13 Oct 2009, 11:49, edited 1 time in total.
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Posts: 64322
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

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12 Oct 2009, 14:21
11
6
SOLUTION:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

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

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

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

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

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Updated on: 13 Oct 2009, 10:01
11
16
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: $$2.4-1(radius)=1.4$$

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: $$\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}$$ --> $$\frac{height}{3}=\frac{4}{5}$$ --> $$height=2.4$$.

$$Distance=height-radius=2.4-1=1.4$$

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

Originally posted by Bunuel on 12 Oct 2009, 15:01.
Last edited by Bunuel on 13 Oct 2009, 10:01, edited 1 time in total.
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Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

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12 Oct 2009, 15:21
10
11
SOLUTION OF 8-10
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

Average=50, Sum of temperatures=50*5=250
As the min temperature is 45, max would be 250-4*45=70 --> The range=70(max)-45(min)=25

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

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:

A. A+1 inquiry
B. A+5
C A+25
D 2A
E. 5A

Sum=A, next 5 consecutive will gain additional 5*5=25, so sum of the next five consecutive integers in terms of A is: A+25

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

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13 Oct 2009, 11:13
1
Bunuel wrote:
SOLUTION:
5. RACE:
A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20

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

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

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

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

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13 Oct 2009, 11:53
2
asterixmatrix wrote:
Bunuel wrote:
SOLUTION:
5. RACE:
A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20

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

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

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

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

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

392/x-6=296/x+2 --> 96/x=8 x=12
432/x-6=336/x+2 --> 96/x=8 x=12
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Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

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26 Oct 2009, 19:17
1
9. PROBABILITY OF INTEGER BEING DIVISIBLE BY 8:
If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?

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

Ans C 62.5%

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

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

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

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

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31 Oct 2009, 05:43
Bunuel wrote:
SOLUTION:
1. THE SUM OF EVEN INTEGERS:
The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?

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

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

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

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

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

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31 Oct 2009, 10:45
3
vkb16 wrote:
Bunuel wrote:
SOLUTION:
1. THE SUM OF EVEN INTEGERS:
The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?

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

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

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

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

If the last term is K, and the 1st term is 1, the no. of terms should be k-1+1
Hence, here the no. of terms should be =>(k-1+1)/2 => k/2 since there should be half as many even numbers...
Am I missing something?

The problem is that the formula you proposed wont't give you the integer value as k is an odd number. Consider the list from 1 to 9, clearly there are 4 even numbers in the list(2,4,6,8):
if we use your formula k/2=9/2=4.5 --> incorrect;
if we use the formula proposed in solution (k-1)/2=(9-1)/2=4 --> correct.

Hope it's clear.
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Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

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23 Dec 2009, 14:04
3
1
7. THE DISTANCE BETWEEN THE CIRCLE AND THE LINE:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

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

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

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

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

Unfortunately, I cannot attach illustration. But if you draw this problem (even roughly), you will find it really easy to solve.
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Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

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23 Dec 2009, 14:35
Vyacheslav wrote:
7. THE DISTANCE BETWEEN THE CIRCLE AND THE LINE:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

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

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

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

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

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

Excellent! +1.

For those who don't know the distance formula (which is in fact very rarely tested) this is the easiest and most elegant solution.
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Posts: 160
Re: TOUGH & TRICKY SET Of PROBLEMS  [#permalink]

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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.

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 PROBLEMS   [#permalink] 09 Jan 2010, 04:10

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