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gmatbusters' challenge questions
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07 Jun 2018, 20:17
There is a box containing Balls of Black and white colors. Each ball is numbered distinctively from 1 to 101. If three balls are chosen at random, what is the probability that the numbers on the balls drawn are in Arithmetic progression? A) \(\frac{50}{3333}\) B) \(\frac{51}{3333}\) C) \(\frac{52}{3332}\) D)\(\frac{53}{3232}\) E) \(\frac{54}{3355}\)
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07 Jun 2018, 20:53
Solution Given:• In a box, balls are there of color blue and black • Each ball is numbered distinctively from 1 to 101 • 3 balls are chosen at random To find:• The probability that the numbers on the chosen balls are in arithmetic progression Approach and Working: As there are total 101 balls, • Number of total ways of choosing 3 balls from 101 balls = \(^{101}C_3 = 33 * 101 * 50\) Now there can be multiple cases of arithmetic progression, with different common difference • Common difference = 1
o Progressions can be {1, 2, 3}, {2, 3, 4}, {3, 4, 5}, … , {99, 100, 101} o Total cases = 99 • Common difference = 2
o Progressions can be {1, 3, 5}, {2, 4, 6}, {3, 5, 7}, … , {97, 99, 101} o Total cases = 97 • Common difference = 3
o Progressions can be {1, 4, 7}, {2, 5, 8}, {3, 6, 9}, … , {95, 98, 101} o Total cases = 95 We can see number of total cases getting decreased by 2 every time Therefore, in the last scenario, • Common difference = 50
o Progression can be {1, 51, 101} o Total case = 1 Hence, total number of progressions possible = \(99 + 97 + 95 + … + 1 = 50 * 50\) Therefore, the probability of getting an arithmetic progression = \(\frac{50 * 50}{33 * 101 * 50} = \frac{50}{33 * 101} = \frac{50}{3333}\) Hence, the correct answer is option A. Answer: A
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Re: gmatbusters' challenge questions
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09 Jun 2018, 03:57
gmatbusters wrote: There is a box containing Balls of Black and white colors. Each ball is numbered distinctively from 1 to 101. If three balls are chosen at random, what is the probability that the numbers on the balls drawn are in Arithmetic progression? A) \(\frac{50}{3333}\) B) \(\frac{51}{3333}\) C) \(\frac{52}{3332}\) D)\(\frac{53}{3232}\) E) \(\frac{54}{3355}\) (PS: Solution will be posted in 12 hours) Given that the balls are distinctly numbered as \(1, 2, 3,.......101\) We have to select 3 balls at random. Total # of ways to select 3 balls out of 101 balls = \(101C3\) = \((101*100*99)/(1*2*3) = 101*50*33\) ways Now the # ways to select 3 balls, that form an arithmetic progression, can be done considering the ways in which the common difference "d" between the numbers can be obtained, till we run out of numbers to choose. For \(d = 1\), the numbers can be \((1,2,3) or (2,3,4) or (3,4,5)........or (99,100,101)\), so we get \(99\) such sequences. For \(d = 2\), the numbers can be \((1,3,5) or (2,4,6) or (3,5,7)........or (97,99,101),\) so we get \(97\) such sequences For \(d = 3\), the numbers can be \((1,4,7) or (2,5,8) or (3,6,9)........or (95,98,101)\), so we get \(95\) such sequences Similarly we can do do these for \(d = 4,5,6.....50\) & we will get number of sequences as \(93,91,89.....1\) So Total Sequences which are in arithmetic progression are = \(99+97+95+......1 = 50 * 50\) Hence the required probability = \((50 * 50) / (101*50*33) = 50/3333\) Answer A. Thanks, GyM
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Re: gmatbusters' challenge questions
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09 Jun 2018, 04:19
Given that the balls are distinctly numbered as \(1, 2, 3,.......101\) We have to select 3 balls at random. Total # of ways to select 3 balls out of 101 balls = \(101C3\) = \((101*100*99)/(1*2*3) = 101*50*33\) ways Let the three numbers be a, b, c. For a, b, c to be in AP a+c = 2b = EVEN hence either both a & c = even or both a & c = odd for each pair of a & c , b is unique. So we need to select only pair of a and c. Now 1,2,3...101 has 50 even and 51 odd numbers. So selection of 2 even numbers out of 50 even numbers = \(50C2\) selection of 2 odd numbers out of 51 odd numbers =\(51C2\) So, number of way of selecting a & c = \(50C2 + 51C2\) = 2500. this is the favorable number of ways. Probability = 2500/(101*50*33) = \(\frac{50}{3333}\) (A)
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Re: gmatbusters' challenge questions
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09 Jun 2018, 23:20
gmatbusters wrote: Given that the balls are distinctly numbered as \(1, 2, 3,.......101\)
We have to select 3 balls at random.
Total # of ways to select 3 balls out of 101 balls = \(101C3\) = \((101*100*99)/(1*2*3) = 101*50*33\) ways
Let the three numbers be a, b, c.
For a, b, c to be in AP
a+c = 2b = EVEN
hence either both a & c = even or both a & c = odd
for each pair of a & c , b is unique. So we need to select only pair of a and c.
Now 1,2,3...101 has 50 even and 51 odd numbers.
So selection of 2 even numbers out of 50 even numbers = \(50C2\) selection of 2 odd numbers out of 50 even numbers =\(51C2\)
So, number of way of selecting a & c = \(50C2 + 51C2\) = 2500. this is the favorable number of ways.
Probability = 2500/(101*50*33) = \(\frac{55}{3333}\) (A) gmatbusters correction in highlighted text. 51 Odd numbers Answer is \(50/3333\) Thanks, GyM
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Re: gmatbusters' challenge questions
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09 Jun 2018, 23:24
Thanks for pointing out the typo. I have removed the typo now. GyMrAT wrote: gmatbusters wrote: Given that the balls are distinctly numbered as \(1, 2, 3,.......101\)
We have to select 3 balls at random.
Total # of ways to select 3 balls out of 101 balls = \(101C3\) = \((101*100*99)/(1*2*3) = 101*50*33\) ways
Let the three numbers be a, b, c.
For a, b, c to be in AP
a+c = 2b = EVEN
hence either both a & c = even or both a & c = odd
for each pair of a & c , b is unique. So we need to select only pair of a and c.
Now 1,2,3...101 has 50 even and 51 odd numbers.
So selection of 2 even numbers out of 50 even numbers = \(50C2\) selection of 2 odd numbers out of 50 even numbers =\(51C2\)
So, number of way of selecting a & c = \(50C2 + 51C2\) = 2500. this is the favorable number of ways.
Probability = 2500/(101*50*33) = \(\frac{55}{3333}\) (A) gmatbusters correction in highlighted text. 51 Odd numbers Answer is \(50/3333\) Thanks, GyM
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Re: gmatbusters' challenge questions
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Re: gmatbusters' challenge questions
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10 Jun 2018, 10:09
Bunuel please help to validate my approach .. totat of 12 letters in the word CONSTITUTION of which we have single letters {c,s,u} and repeated letters {o2,n2,t3,i2} Position of C and N fixed ..left with 10 letters Say , set A {n1,s1,u1} and set B {o2,t3,i2} ...please note both Sets have 3 different letters We have to find how many different words can be formed such that word starts with C and ends with N C _ _ N ......need to select two letter from the sets and arrange .. selection can be of following types.. 1)Two letters from set A. ===3 ways 2)one letter from each set ..i.e one letter from set A and one from set B . =3 ways * 3 ways 3)Two letter from set B ,but letters are different = 3 ways 4)Two same letters from set B = 3 ways ... we can select any 2 of o , t & i Let us now find the number of different words which can be formed from above selection in each case . 1)Two letters from set A. ===3 ways ; 3*2!=6 2)one letter from each set ..i.e one letter from set A and one from set B . =3 ways * 3 ways ; 9*2!=18 3)Two letter from set B ,but letters are different = 3 ways ; 3*2! 4)Two same letters from set B = 3 ways ... we can select any 2 of o , t & i ;3 Answer should be =6+18+6+3=33 "B"



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Re: gmatbusters' challenge questions
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10 Jun 2018, 10:52
gmatbusters wrote: How many different 4letter words can be formed using the letters of the word CONSTITUTION such it starts with C and ends with N? A) 30 B) 33 C) 40 D) 42 E) 24 The word "CONSTITUTION" has count of each letter as follows C = 1 O = 2 N = 2 S = 1 T = 3 I = 2 U = 1 Now 4 letter words are to be formed, from these letters & each word has to be distinct. Also each letter has to start with C & end in N. So we have format which is C _ _ N, hence we are left with all the above letters, except C & one N. lets consider each of the remaining letters (O,N,S,T, I & U) in the second blank & calculate the possible arrangements for the third blank. Case 1  C O _ N The blank can be filled with 5 letters (N, S, T, I & U). The letters in 2nd blank & 3rd blank can be arranged among themselves in 2 ways. Hence total arrangements for Case 1 = 5 * 2 = 10 words Case 2  C N _ N The blank can be filled with 4 letters (S, T, I & U), since arrangements with O are already considered in case 1, we neglect those. The letters in 2nd blank & 3rd blank can be arranged among themselves in 2 ways. Hence total arrangements for Case 2 = 4 * 2 = 8 words Case 3  C S _ N The blank can be filled with 3 letters (T, I & U), since arrangements for O & N are already considered in Cases 1 & 2, we neglect those. The letters in 2nd blank & 3rd blank can be arranged among themselves in 2 ways. Hence total arrangements for Case 3 = 3 * 2 = 6 words Case 4  C T _ N The blank can be filled with 2 letters ( I & U), since arrangements for O, N & S are already considered in Cases 1,2 & 3, we neglect those. The letters in 2nd blank & 3rd blank can be arranged among themselves in 2 ways. Hence total arrangements for Case 4 = 2 * 2 = 4 words Case 5  C I _ N The blank can be filled with 1 letter (U), since arrangements for O, N, S & T are already considered in Cases 1,2,3 & 4, we neglect those. The letters in 2nd blank & 3rd blank can be arranged among themselves in 2 ways. Hence total arrangements for Case 5 = 1 * 2 = 2 words Now Case 6  When two same letters are selected C O O N C T T N C I I N Hence 3 arrangements Hence total # of different words formed = 10 + 8 + 6 + 4 + 2 + 3 =33 Answer B. I am sure gmatbusters will come up with an even simpler solution than this. Thanks, GyM
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Re: gmatbusters' challenge questions
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10 Jun 2018, 11:37
gmatbusters wrote: How many different 4letter words can be formed using the letters of the word CONSTITUTION such it starts with C and ends with N? A) 30 B) 33 C) 40 D) 42 E) 24 Well just realized an alternate simpler solution. The word "CONSTITUTION" has count of each letter as follows C = 1 O = 2 N = 2 S = 1 T = 3 I = 2 U = 1 Now 4 letter words are to be formed, from these letters & each word has to be distinct. Also each letter has to start with C & end in N. So we have format which is C _ _ N, where blanks can be filled with distinct letters or similar letters out of the remaining letters, to form distinct words. Now Total arrangements with selection of 2 distinct letters from 6 letters(O, N, S, T, I & U) = \(6C2 * 2\) = 30 words Total arrangements with selection 2 same letters (O, T & I) (COON, CTTN, CIIN) = 3 Hence the total # of different words with C _ _ N = 30 + 3 = 33 words Answer B. Thanks, GyM
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Re: gmatbusters' challenge questions
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10 Jun 2018, 17:04
gmatbusters wrote: How many different 4letter words can be formed using the letters of the word CONSTITUTION such it starts with C and ends with N? A) 30 B) 33 C) 40 D) 42 E) 24 In the questions involving rearrangement of letters, always first note the number of distinct letters. Here CONSTITUTION = C = 1 O = 2 N = 2 S = 1 T = 3 I = 2 U = 1 Now as C and N are fixed, we need to fill only the middle 2 gaps. Case1The 2 gaps can be filled by different letter.i.e 6C2*2!=30 Case 2The gaps can be filled by 2 identical letters. since we have at least two of the same letter word of 3 distinct letters(i.e 3Ts,2Os,2Is) Thus we 3C1*2!/2!=3 Total cases =30+3=33 (B)
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Re: gmatbusters' challenge questions
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Updated on: 11 Jun 2018, 03:55
With the provided information ,I guess for ABC to be equilateral triangle ... X coordinate of the third vertex must be (8(2))/2 ;5 unit distance from other X coordinates of the other two vertices i.e from 2 and 5 ...X for third vertex must be 3 Also , i can infer all sides need to be equal in length . So x&y coordinates of the third unknown vertex have to be even numbers .
Now let us see one by one what statements tell
1) X=4 ..No ..hence sufficient 2)X and y are coprime ..we have inferred X and Y are even numbers hence answer is no ...hence sufficient ..
Answer should be D ..l
Originally posted by sumitkrocks on 11 Jun 2018, 03:30.
Last edited by sumitkrocks on 11 Jun 2018, 03:55, edited 1 time in total.



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Re: gmatbusters' challenge questions
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11 Jun 2018, 03:48
gmatbusters wrote: Is the triangle ABC Equilateral? Vertices of the triangle ABC are A(2,2), B(8,2)& C(x,y).
1) x = 4.
2) x and y are coprime. Given Vertices A(2,2), B(8,2) & C(x,y) Points A & B lie on line y = 2, with distance between them as 10 units The midpoint of line AB is 5 units, hence it lies on line x = 3 For the triangle ABC to be an equilateral Triangle, the x coordinate of point C should be x=3 Now if we drop an angle bisector from point C, if triangle ABC is equilateral, then it will pass through M at 90 degree angle The triangle CMA will be a 306090 triangle, hence length of altitude, CM = \(5*\sqrt{3}\) Hence, y coordinate of point C has to be \(2+5\sqrt{3}\) or \(5\sqrt{3}2\) Lets look at the statements now. Statement 1 x = 4, Hence ABC is not an equilateral triangle. Statement 1 alone is Sufficient. Statement 2 x & y are coprimes Hence ABC is not an equilateral triangle. Statement 2 alone is Sufficient. Answer D. Thanks, GyM
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11 Jun 2018, 04:02
Hi GyMrAT , If my understanding of your explanation is correct ..Y coordinates of C should be 2Sq.rt(5) or 2+Sq.rt(5) Please help if there is any flaw in my understanding . Thanks



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Re: gmatbusters' challenge questions
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11 Jun 2018, 04:19
A (2,2) B  8,2 .. now for triangle to be equilateral the X coordinate of third vertex should be 3 {8+(2)/2} as in equilateral triangle a line drawn from every vertex cuts the opposide side in half.. Since the distance between the two sides is 10units thus when x=3 only then perpendicular drawn from C cuts AB in half... So in statement 1.. x=4 thus no it cant be equilateral triangle thus we can eliminate B,C,E in statement 2 it is given that x and y are coprimes.. by putting x=3 and y = various possible values in distance formula we find that none give us d=10 and at y=11 the value crosses 10 thus according to statement 2 its not equilateral triangle So the answer is D.. Sent from my iPhone using GMAT Club Forum mobile app



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Re: gmatbusters' challenge questions
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11 Jun 2018, 08:04
sumitkrocks wrote: Hi GyMrAT , If my understanding of your explanation is correct ..Y coordinates of C should be 2Sq.rt(5) or 2+Sq.rt(5) Please help if there is any flaw in my understanding . Thanks The distance between C & the midpoint of Line AB is \(5\sqrt{3}\). You can imagine the point C on both sides of X axis. Now if point C is above X axis, to find the Y coordinate, we have to add 2 units to \(5\sqrt{3}\) since the line AB is 2 units above X axis. A & B have y coordinate as 2. If Point C is below X axis, to find the Y coordinate, we have to subtract 2 units from \(5\sqrt{3}\), for the same above reason. I hope its clear. Thanks, GyM
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11 Jun 2018, 08:33
Explanation:St1: x=4 As we know for any isosceles triangle, the third vertex C would lie on the perpendicular bisector of AB. Hence x = (2+8)/2 =3. So it is sufficient to so that it is not an equilateral triangle. St2: x and y are coprime. this means x and y are integers. Now we know that " A triangle having all integral coordinates (even rational vertices) can never be an equilateral triangle"So it is sufficient to so that it is not an equilateral triangle. hence Answer is D(PS: proof: A triangle having integral coordinates can never be an equilateral triangle : Using contradiction. At first take that we have a triangle with integral coordinates then find its area by the determinant method ,where you would get an rational solution. Then find the area by the formula of area of equilateral triangle(i.e.,area=(√3÷4)×a²). Here you would get an irrational solution. TNow, you get two different areas for the same triangle.proved by contradiction.) this is only for those who are interested in proof. gmatbusters wrote: Is the triangle ABC Equilateral? Vertices of the triangle ABC are A(2,2), B(8,2)& C(x,y).
1) x = 4.
2) x and y are coprime.
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Re: gmatbusters' challenge questions
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23 Jun 2018, 01:52
Six blocks shown below are to be rearranged so that the letters are in alphabetical order, reading from left to right. What is the minimum number of blocks that must be moved to arrive at the desired arrangement? Attachment:
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A) 2 B) 3 C) 4 D) 5 E) 6
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23 Jun 2018, 02:45
In a certain orchestra, each musician plays only one instrument and the ratio of musicians play either the violin or the viola to musicians play neither instrument is 5 to 9. If 7 members of the orchestra play the violin and four times as many play the violin, how many play neither? (A) 14 (B) 28 (C) 35 (D) 63 (E) 72
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