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Re: TOUGH & TRICKY SET Of PROBLEMS [#permalink]
25 Aug 2013, 05:49

Expert's post

Jazmet wrote:

Hi Brunel,

Thank you for the questions and the explanations as well.

I have a question. I was only able to solve 3 out of 10. (Q8/9/10) - I found the other ones really hard. It has demotivated me. I am aiming for 49-50 in Maths. Please suggest.

The set is called TOUGH & TRICKY, so don't be demotivated if you solved only 3 out of 10. Practice will improve your results. _________________

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
31 Aug 2013, 23:12

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

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

Answer: D.

Hi Bunuel,

I always have problem in understand the probablity problems. If this is asked without replacement then how we can solve this problem:????

Can we solve the original question with nCr method???

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
02 Sep 2013, 01:11

Expert's post

rrsnathan 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*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}

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

Answer: D.

Hi Bunuel,

I always have problem in understand the probablity problems. If this is asked without replacement then how we can solve this problem:????

Can we solve the original question with nCr method???

Please clear my doubts.

Thanks in advance, Rrsnathan.

If we had without replacement case, then the answer would be 2*3/9*2/8. _________________

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
17 Oct 2013, 05:03

Bunuel wrote:

SOLUTION: 3. LEAP YEAR: How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?

A. 1 B. 2 C. 3 D. 4 E. 5

Probability of a randomly selected person NOT to be born in a leap year=3/4 Among 2 people, probability that none of them was born in a leap = 3/4*3/4=9/16. The probability at least one born in leap = 1- 9/16=7/16<1/2 So, we are looking for such n (# of people), when 1-(3/4)^n>1/2 n=3 --> 1-27/64=37/64>1/2

Thus min 3 people are needed.

Answer: C.

Hi Bunuel,

If I shall encounter this question or one like it on the GMAT, will I be given the information as to what the ratio of regular year to leap year is? If not, is there a list of little things like this that are important to know?

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
17 Oct 2013, 07:08

Expert's post

ronr34 wrote:

Bunuel wrote:

SOLUTION: 3. LEAP YEAR: How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?

A. 1 B. 2 C. 3 D. 4 E. 5

Probability of a randomly selected person NOT to be born in a leap year=3/4 Among 2 people, probability that none of them was born in a leap = 3/4*3/4=9/16. The probability at least one born in leap = 1- 9/16=7/16<1/2 So, we are looking for such n (# of people), when 1-(3/4)^n>1/2 n=3 --> 1-27/64=37/64>1/2

Thus min 3 people are needed.

Answer: C.

Hi Bunuel,

If I shall encounter this question or one like it on the GMAT, will I be given the information as to what the ratio of regular year to leap year is? If not, is there a list of little things like this that are important to know?

GMAC says that we must solve the questions using your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise). Guess that the meaning of a leap year is considered as everyday facts, though this question is still not perfect so I wouldn't worry about it at all. _________________

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
16 Dec 2013, 03:02

Dear Brunel ,

Please explain special cases from AP

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

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

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

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
16 Dec 2013, 03:08

Expert's post

archit wrote:

Dear Brunel ,

Please explain special cases from AP

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

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

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

Sum of first n positive integers: 1+2+...+n=\frac{1+n}{2}*n. For example, the sum of first 5 positive integers is 1+2+3+4+5=\frac{1+5}{2}*5=15

Sum of n first odd numbers: a_1+a_2+...+a_n=1+3+...+a_n=n^2, where a_n is the last, n_{th} term and given by: a_n=2n-1. Given n=5 first odd integers, then their sum equals to 1+3+5+7+9=5^2=25.

Sum of n first positive even numbers: a_1+a_2+...+a_n=2+4+...+a_n=n(n+1), where a_n is the last, n_{th} term and given by: a_n=2n. Given n=4 first positive even integers, then their sum equals to 2+4+6+8=4(4+1)=20.

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
24 Jan 2014, 14:52

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

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

HI Bunel, Thank you for the explanataion. N(n+1) makes it very easy. I was actually taking n =k which is not the right answer. Could you please explain Why is k 2n+1?

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
24 Jan 2014, 16:07

Bunuel wrote:

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

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

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

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

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

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

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

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

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

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

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

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

Answer: A.

OR ANOTHER APPROACH:

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

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

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

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

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

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

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

Distance=height-radius=2.4-1=1.4

Answer: A.

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

Thank you very much. I just want to know why did we subtract the radius from the distance to find the minimum distance

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
24 Jan 2014, 16:35

Bunuel wrote:

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

Answer: B.

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

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

Answer: C.

Hi Bunel, For the 10th q should we able to derive it also by using numbers.for some reasons if you try to figure it out with numbers it doesnt come to A+25

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
26 Jan 2014, 08:36

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}*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.

I am having some trouble understanding the second problem, THE PRICE OF BUSHEL. Since the text speaks about prices increasing at a constant rate (but a rate expressed in cents ??), the equation I set up is: 3.20*(1+0.05x)^t=5.80*(1-(\sqrt{2}x-x))^t, where t is the time it takes for a bushel of corn to cost the same as a peck of wheat. I do not understand how you set up your equation 320+5xy=580-0.41xy. I would set up an equation similar to yours if the increase (or decrease) were constant, not the rate.

gmatclubot

Re: TOUGH & TRICKY SET Of PROBLMS
[#permalink]
26 Jan 2014, 08:36