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12 Easy Pieces (or not?) [#permalink]
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21 Jan 2012, 06:10
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After posting some 700+ questions, I've decided to post the problems which are not that hard. Though each question below has a trap or trick so be careful when solving. I'll post OA's with detailed solutions after some discussion. Good luck.1. There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four socks are picked at random what is the probability of getting two socks of the same color?A. 1/5 B. 2/5 C. 3/4 D. 4/5 E. 1 Solution: 12easypiecesornot126366.html#p10339192. If x is an integer and 9<x^2<99, then what is the value of maximum possible value of x minus minimum possible value of x?A. 5 B. 6 C. 7 D. 18 E. 20 Solution: 12easypiecesornot126366.html#p10339213. Fanny and Alexander are 360 miles apart and are traveling in a straight line toward each other at a constant rate of 25 mph and 65 mph respectively, how far apart will they be exactly 1.5 hours before they meet?A. 25 miles B. 65 miles C. 70 miles D. 90 miles E. 135 miles Solution: 12easypiecesornot126366.html#p10339244. If 3<x<5 and 7<y<9, which of the following represent the range of all possible values of yx?A. 4<yx<4 B. 2<yx<4 C. 12<yx<4 D. 12<yx<12 E. 4<yx<12 Solution: 12easypiecesornot126366.html#p1033925 5. The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true?I. c>a+b II. c^2>a^2+b^2 III. c/a/b=10/6/2 A. I only B. II only C. III only D. I and III only E. II and III only Solution: 12easypiecesornot126366.html#p10339306. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colors. How many different arrangements are possible?A. 30 B. 60 C. 120 D. 240 E. 480 Solution: 12easypiecesornot126366.html#p1033932 7. After 2/9 of the numbers in a data set A were observed, it turned out that 3/4 of those numbers were nonnegative. What fraction of the remaining numbers in set A must be negative so that the total ratio of negative numbers to nonnegative numbers be 2 to 1?A. 11/14 B. 13/18 C. 4/7 D. 3/7 E. 3/14 Solution: 12easypiecesornot126366.html#p10339338. There are 15 black chips and 5 white chips in a jar. What is the least number of chips we should pick to guarantee that we have 2 chips of the same color?A. 3 B. 5 C. 6 D. 16 E. 19 Solution: 12easypiecesornot126366.html#p10339359. Julie is putting M marbles in a row in a repeating pattern: blue, white, red, green, black, yellow, pink. If the row begins with blue marble and ends with red marble, then which of the following could be the value of M?A. 22 B. 30 C. 38 D. 46 E. 54 Solution: 12easypiecesornot126366.html#p103393610. If \(n\) is an integer and \(\frac{1}{10^{n+1}}<0.00737<\frac{1}{10^n}\), then what is the value of n?A. 1 B. 2 C. 3 D. 4 E. 5 Solution: 12easypiecesornot126366.html#p103393811. The numbers {1, 3, 6, 7, 7, 7} are used to form three 2digit numbers. If the sum of these three numbers is a prime number p, what is the largest possible value of p?A. 97 B. 151 C. 209 D. 211 E. 219 Solution: 12easypiecesornot12636620.html#p103393912. If \({\frac{1}{3}}\leq{x}\leq{\frac{1}{5}}\) and \({\frac{1}{2}}\leq{y}\leq{\frac{1}{4}}\), what is the least value of \(x^2*y\) possible?A. 1/100 B. 1/50 C. 1/36 D. 1/18 E. 1/6 Solution: 12easypiecesornot12636620.html#p1033949Please read the whole thread before posting a question. Chances are that your doubt has been already addressed.
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20 Oct 2012, 02:11
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Muki wrote: There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four socks are picked at random what is the probability of getting two socks of the same color? A. 1/5 B. 2/5 C. 3/4 D. 4/5 E. 1
This is from Bunuel's "12 Easy Pieces (or not?)" collection. I understand and agree with the simple explanation provided by Bunuel. However, I am wondering what should be the correct algebraic approach to find this probability using Combinations.
Please provide explanations, as the answer has already been provided by Bunuel. Here is one: There are \(3^4=81\) possibilities to choose the colors of the 4 socks. Here order counts  3 choices for the first sock, 3 for the second,...We have all the time 3 color choices, because we have at least 4 socks of the same color for each color. Let's count the number of possibilities to choose at least twice the same color: 2 socks of the same color and 1 of each of the other two colors  \(3\cdot\frac{4!}{2!}=36\)  3 choices for the color with 2 socks, then 4! permutations of the 4 socks, divide by 2! because 2 socks are of the same color 2 colors, 2 socks of each color  \(\frac{3\cdot{2}}{2}\cdot{\frac{4!}{2!2!}}=18\)  choose 2 colors out of 3, then 4! permutations ... divide... 3 socks of the same color, 1 of a different color  \(3\cdot{2}\cdot{\frac{4!}{3!}}=24\)  3 choices for the first color (with 3 socks), 2 choices for the other sock, 4! ... divide ... finally, all 4 socks of the same color  3 possibilities \(\frac{36+18+24+3}{81}=\frac{81}{81}=1\) Is this worth doing? If you think of the complementary event  no two socks of the same color: 1st sock  3 color choices 2nd sock  2 color choices 3rd sock  1 color choice 4th sock  0 choice, we don't have a fourth color So, number of choices for 4 socks of different colors is 0. Doesn't matter how many for the total number of possible choices, 0 divided by anything not zero is still 0!!! So the requested probability is 1  0 = 1.
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29 Apr 2013, 11:18
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Q4. x > 3 and x < 5 y > 7 and y < 9 yx < 9  (3) = 12 yx > 75 = 12
hence 12 < yx < 12 Ans D



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26 Nov 2013, 08:11



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12 Easy Pieces (or not?) [#permalink]
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09 Dec 2015, 22:39
Peltina wrote: chetan2u wrote: Hi,
in such questions, we look at the worst scenario.. here it would be getting different coloured socks , every time we pick socks..
but there are only three different coloured socks, so as a worst case , the first three picked up are different colours.. the fourth which you pick up has to be one of these three colours, thus ensuring that we have two socks of same colour.. hence prob is 1, that is it is sure to have two socks of atleast one colour.. 5 pairs of white, 3 pairs of black and 2 pairs of grey socks Thank you! I was looking more for the formula that we should use to get to the answer P=1 in case we fail to notice that the answer is obvious and needs no calculation. Hi, there cant be direct formulas for most of the questions but they are just a step to correct answer.. so if you want the answer the formula way, then it would be something like this... there are three different coloured socks.. let us first find the scenario where we do not get two of one kind... 1)the first can be picked up, say white, prob=5/10.. 2)the second can be picked up, say black, prob=3/9.. 3)the third can be picked up, the remaining grey, prob=2/8.. 4)the fourth picking up is 0 as no other colour is left prob=0/7.. now these four can be arranged in 4!/2! ways.. prob of two colours of one kind = 1 5/10 *3/9 * 2/8* 0/7 * 4!/2!=10=1
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22 Jan 2012, 11:17
Ans 1E 2A 3E 4D 5B 6C 7A 8A 9C 10D 11D 12B



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22 Jan 2012, 12:10



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22 Jan 2012, 15:37



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23 Jan 2012, 12:07
2d 3e 4d 5b 6d 7d 8c 9c 10b 11d 12d
Still working on the 1st one...plz check



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25 Jan 2012, 22:00
Bunuel wrote: 5. The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true? I. c>a+b II. c^2>a^2+b^2 III. c/a/b=10/6/2
A. I only B. II only C. III only D. I and III only E. II and III only
According to the relationship of the sides of a triangle: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Thus I and III can never be true: one side (c) can not be larger than the sum of the other two sides (a and b). Note that III is basically the same as I: if c=10, a=6 and b=2 then c>a+b, which can never be true. Thus even not considering the angles, we can say that only answer choice C (III only) is left.
Answer: C.
Now, if interested why III is true: as the angles in a triangle are x, 3x, and 5x degrees then x+3x+5x=180 > x=20, 3x=60, and 5x=100. Next, if angle opposite c were 90 degrees, then according to Pythagoras theorem c^2=a^+b^2, but since the angel opposite c is more than 90 degrees (100) then c is larger, hence c^2>a^+b^2. Bunnel: If I am correct you meant to say that II is the correct choice????



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26 Jan 2012, 02:50
subhajeet wrote: Bunuel wrote: 5. The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true? I. c>a+b II. c^2>a^2+b^2 III. c/a/b=10/6/2
A. I only B. II only C. III only D. I and III only E. II and III only
According to the relationship of the sides of a triangle: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Thus I and III can never be true: one side (c) can not be larger than the sum of the other two sides (a and b). Note that III is basically the same as I: if c=10, a=6 and b=2 then c>a+b, which can never be true. Thus even not considering the angles, we can say that only answer choice B (II only) is left.
Answer: B.
Now, if interested why II is true: as the angles in a triangle are x, 3x, and 5x degrees then x+3x+5x=180 > x=20, 3x=60, and 5x=100. Next, if angle opposite c were 90 degrees, then according to Pythagoras theorem c^2=a^+b^2, but since the angel opposite c is more than 90 degrees (100) then c is larger, hence c^2>a^+b^2. Bunnel: If I am correct you meant to say that II is the correct choice???? Yes, correct choice is c^2>a^2+b^2, as explained, so B.
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26 Jan 2012, 06:09
Got 3 answers wrong Bunnel thanks for the questions



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26 Jan 2012, 23:47
Got only 8 correct out of 12 questions. The most shameful part is that all of my incorrect responses are related to easier problems than those problems that are given correct responses by me. Sometimes, I just dive into doing maths rather than using the wits. Learning from this series of questions: GIVE YOURSELF A FEW SECONDS OF REFLECTION TIME BEFORE YOU START SOLVING A QUANT PROBLEM. MAYBE, THE SOLUTION DOESN'T EVEN REQUIRE DOING MATH (e.g. Question#1)! Thank you, Bunuel, as always, for this series of questions.



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06 Apr 2012, 07:51
Bunuel wrote: 6. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colors. How many different arrangements are possible? A. 30 B. 60 C. 120 D. 240 E. 480 .... as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.
Answer: B. I am little confused ...after this bold part.. since we have 5 slots available to be filled by 5 marbles and we can pick any marbles given our setting of marbles (without disturbing any conditions in the stem ?). now # arrangements in either case is going to be 5! (=120) so total # arrangements = 2*120= 240... what am I missing here
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06 Apr 2012, 09:09
Bunuel wrote: yogesh1984 wrote: Bunuel wrote: 6. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colors. How many different arrangements are possible? A. 30 B. 60 C. 120 D. 240 E. 480 .... as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.
Answer: B. I am little confused ...after this bold part.. since we have 5 slots available to be filled by 5 marbles and we can pick any marbles given our setting of marbles (without disturbing any conditions in the stem ?). now # arrangements in either case is going to be 5! (=120) so total # arrangements = 2*120= 240... what am I missing here THEORY. Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is: \(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\). Hope it's clear. How can I do this feels like slapping myself ! anyway seems time to turn over to the basics. Thanks for the explanation mate
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16 May 2012, 00:02
Bunuel a slightly different query, based on the difficulty level how much time would you give to solve these questions. Got Q no. 2 & 6 wrong!



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16 May 2012, 00:24
Bunuel wrote: 11. The numbers {1, 3, 6, 7, 7, 7} are used to form three 2digit numbers. If the sum of these three numbers is a prime number p, what is the largest possible value of p? A. 97 B. 151 C. 209 D. 211 E. 219
What is the largest possible sum of these three numbers that we can form? Maximize the first digit: 76+73+71=220=even, so not a prime. Let's try next largest sum, switch digits in 76 and we'll get: 67+73+71=211. Question is it a prime number? If you notice 210=2*3*5*7=the product of the first four primes. So, 210+1=211 must be a prime. For example: 2+1=3=prime, 2*3+1=7=prime, 2*3*5+1=31=prime.
Answer: D. Bunuel could you elaborate on the observation you presented. Is it that product of consecutive primes +1 is prime or is it something else?



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17 May 2012, 02:22
good work pal keep them comming....



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17 May 2012, 02:23
i wonder if any one got all 12 right....




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