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I'm posting the next set of medium/hard PS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000 Solution: baker-s-dozen-128782-20.html#p10575022. If y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}, then y is NOT divisible by which of the following?A. 6^4 B. 62^2 C. 65^2 D. 15^4 E. 52^4 Solution: baker-s-dozen-128782-20.html#p10575033. For the past k days the average (arithmetic mean) cupcakes per day that Liv baked was 55. Today Bibi joined and together with Liv they baked 100 cupcakes, which raises the average to 60 cupcakes per day. What is the value of k?A. 6 B. 8 C. 9 D. 10 E. 12 Solution: baker-s-dozen-128782-20.html#p10575044. What is the smallest positive integer k such that 126*\sqrt{k} is the square of a positive integer?A. 14 B. 36 C. 144 D. 196 E. 441 Solution: baker-s-dozen-128782-20.html#p10575055. There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble to remain in the jar?A. 460 B. 490 C. 493 D. 455 E. 445 Solution: baker-s-dozen-128782-20.html#p10575076. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool?A. \frac{yz}{x+y+z}B. \frac{yz}{yz+xz-xy}C. \frac{yz}{yz+xz+xy} D. \frac{xyz}{yz+xz-xy}E. \frac{yz+xz-xy}{yz}Solution: baker-s-dozen-128782-20.html#p10575087. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive?A. $60,000 B. $90,000 C. $100,000 D. $120,000 E. $180,000 Solution: baker-s-dozen-128782-20.html#p10575098. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A?A. -165 B. -175 C. -195 D. -205 E. -215 Solution: baker-s-dozen-128782-20.html#p10575129. If x and y are negative numbers, what is the value of \frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}?A. 1+y B. 1-y C. -1-y D. y-1 E. x-y Solution: baker-s-dozen-128782-20.html#p105751410. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x-2y?A. 7 B. 11 C. 13 D. 17 E. 19 Solution: baker-s-dozen-128782-20.html#p105751511. In an infinite sequence 1, 3, 9, 27, ... each term after the first is three times the previous term. What is the difference between the sum of 13th and 15th terms and the sum of 12th and 14th terms of the sequence?A. 10*3^11 B. 20*3^11 C. 10*3^12 D. 40*3^11 E. 20*3^12 Solution: baker-s-dozen-128782-40.html#p105751712. x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z?A. 12 B. 20 C. 24 D. 29 E. 33 Solution: baker-s-dozen-128782-40.html#p105751913. If x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}, what is the product of the tens and the units digits of \frac{x}{(8!)^3}-39?A. 0 B. 6 C. 7 D. 12 E. 14 Solution: baker-s-dozen-128782-40.html#p1057520
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[quote2] 9. If x and y are negative numbers, what is the value of \frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}?A. 1+y B. 1-y C. -1-y D. y-1 E. x-y Correction/suggestion/advice/confirmation --- needed Answer to 9th question If x and y are negative numbers, what is the value of \frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}?[/b] Sqrtx^2/x-sqrt(-y*|y|) = |x|/x-|y|= -1-y
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-y = +y and |y| no matter what, comes out as positive hence Sqrt (-y*|y| ) comes out as possitive y and sqrtx^2 =|x|= positive x , of course from your tone i get that i'm going wrong some where but
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Quote: 8. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A?
A. -165
B. -175
C. -195
D. -205
E. -215 x ;x+2;x+2*2;x+2*3....x+2*6 since the set has consecutive odd integers, then mean =median mean of 5 largest integers=-185/5= -37 median of 5 largest integers is the 3d of the largest or the 5th of the whole set A= x+2*4 x+8=-37 ;x=-45 median of the first 5 small integers =-45+2*2=-41 the sum of the first small odd integers=number of integers*mean (or median)=5*median=5*(-41)=-205 answ is
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Quote: 13. If x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}, what is the product of the tens and the units digits of \frac{x}{(8!)^3}-39?
A. 0
B. 6
C. 7
D. 12
E. 14
x=((8!)^6*((8!)^4-1))/((8!)^3*((8!)^2-1))=(8!)^3*((8!)^2+1)
((8!)^3*((8!)^2+1) /(8!)^3)) -39=(8!)^2+1)-39
note, that 8! has 5 and 2. so 8! ends with 0. then (8!)^2 will end by two 0s.
***00-38=62
6*2=12
answ is
4. What is the smallest positive integer k such that 126*\sqrt{k} is the square of a positive integer?
A. 14
B. 36
C. 144
D. 196
E. 441
126*\sqrt{k}=3*7*3*2*sqrt{k}=3^2*7*2
so we need one more 7*2 (14)
sqrootk=14
k=196
answ is
what an irony. all the 3 questions are heh
9. If x and y are negative numbers, what is the value of \frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}?
A. 1+y
B. 1-y
C. -1-y
D. y-1
E. x-y
-x/x-|y|=-1-(-y)=y-1 since y<0 ; x<0
is the answ
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Last edited by LalaB on 10 Mar 2012, 09:08, edited 1 time in total.
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12. x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z?A. 12 B. 20 C. 24 D. 29 E. 33 x/y the remainder is 3 . then the smallest x is 3 y/z the remainder is 8. then the smallest y is 8 and z is 9 3+8+9=20 answ is B
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10. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x-2y?A. 7 B. 11 C. 13 D. 17 E. 19 -9<x<9 -5<y<5 -9<x<9 -10<2y<10 x-2y<9-(-10) x-2y<19 the answer is 17. D again  the mystical letter for today heh 11. In an infinite sequence 1, 3, 9, 27, ... each term after the first is three times the previous term. What is the difference between the sum of 13th and 15th terms and the sum of 12th and 14th terms of the sequence?A. 10*3^11 B. 20*3^11 C. 10*3^12 D. 40*3^11 E. 20*3^12 1, 3, 9, 27 ok please pay attention to the fact that the term of sequence has 3 one less than the number of the term,i.e. the 3d term is 9, but the number has two 3s (one less than the term). the 4th term is 3*3*3 ,which has three 3s. ok, then we have the following- the q. asks us to find out - (13th term+15th term) -(12th term +14th term) (3^12+3^14)-(3^11-3^13)= 3^11(3-1)+3^13(3-1)= 2*3^11(1+3^2)= 20*3^11 the answer is B p.s. also note, that answer choices B and D are leaders of the day heh
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I am not so certain whether I am right but I would go with D in #9.
IF x<0 and y<0 then sqrt(x^2)/x=|x|/x, then -x/x=-1; because there is a minus in front of x, than -(-x) should be positive.
sqrt two negative y gives square root of y^2, than because y<1, y is negative as well. -y.
From the expression we've got -1+y, thus D should be a correct answer.
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LalaB wrote: Bunuel, could u please tell us which question is still unanswered or answered wrong by all of us? I want to think on such questions more, before u post solutions.
and btw, do u read our solutions or just only answers? I wonder whether my way of thinking was ok. in some cases I tried to use another method not to repeat others. I just wonder whether it worked, or it was just coincidence I do read all the solutions and award +1 Kudos if there is at least one correct explanation in the post. As for your other question: I don't see a solution for #7, so you can try this one.
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Bunuel wrote:
As for your other question: I don't see a solution for #7, so you can try this one.
ok my answer to the q7 is D , because D is the leader of the day hehe  kidding the answer is really D ,because- 2/3(L+A+W) =F 3/7(F+W+A)=L 4/11(F+W+L)=A total=360 (I deleted the 0s just to make a number easy) W=? 2/3(L+A+W) =F means F=2/3(360-F) =>(2/3)*360=5/3F=> F=144 3/7(F+W+A)=L means L=3/7(360-L) => L=108 4/11(F+W+L)=A means A=4/11(360-A)=>A=96 W=360- (L+A+F)=360-144-108-96=12 or W=120 000 (return the 0s which were deleted before)
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utkarshlavania wrote: [quote2]9. If x and y are negative numbers, what is the value of \frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}?
A. 1+y
B. 1-y
C. -1-y
D. y-1
E. x-y
Correction/suggestion/advice/confirmation --- needed
Answer to 9th question If x and y are negative numbers, what is the value of \frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}?[/b]
Sqrtx^2/x-sqrt(-y*|y|) = |x|/x-|y|= -1-y Responding to a pm: First of all, I would suggest you to put x = -2 and y = -3 and see what you get.
\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}
\frac{\sqrt{(-2)^2}}{-2}-\sqrt{-(-3)*|-3|}
\frac{2}{-2}-\sqrt{9} (because square root will always be positive)
-1 - 3 = -1 + y (since y = -3)
Coming to your solution:
I see you have everything correct till the last step: |x|/x-|y|
|x|/x will be -1 which is fine.
What about |y|?
Do you remember how we define mods?
We say:
|x| = x if x >= 0
|x| = -x if x < 0
Why? If x >= 0 e.g. say x = 5,
|x| = x = 5
Instead, if x < 0 e.g. say x = -6,
is |x| = x? No! mods are never negative. So |x| = -x = -(-6) = 6 (Since x itself is negative, mod of x is negative of x which becomes positive.)
Therefore, if we know that y is negative, what is the value of |y|?
|y| = -y
Therefore, |x|/x-|y| = -1 -(-y) = -1 + y _________________
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SOLUTIONS:1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000 Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit. # of passwords with three digit 6 is 9*9*C^3_5=810: each out of two other digits (not 6) has 9 choices, thus we have 9*9 and C^3_5 is ways to choose which 3 digits will be 6's out of 5 digits we have. P=\frac{favorable}{total}=\frac{810}{10^5}Answer: B.
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5. There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble to remain in the jar?A. 460 B. 490 C. 493 D. 455 E. 445 Total ways to select 8 marbles out of 7+5=12 is C^8_{12}; Ways to select 8 marbles so that zero red marbles is left in the jar is C^7_7*C^1_5; Ways to select 8 marbles so that zero blue marbles is left in the jar is C^5_5*C^3_7; Hence ways to select 8 marbles so that at least one red marble and at least one blue marble to remain the jar is C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455. Answer: D.
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Bunuel wrote: 13. If x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}, what is the product of the tens and the units digits of \frac{x}{(8!)^3}-39?
A. 0
B. 6
C. 7
D. 12
E. 14
Apply a^2-b^2=(a-b)(a+b): x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}=\frac{((8!)^{5}-(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}-(8!)^3}=(8!)^{5}+(8!)^3.
Next, \frac{x}{(8!)^3}-39=\frac{(8!)^{5}+(8!)^3}{(8!)^3}=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}-39=(8!)^2+1-39=(8!)^2-38.
Now, since 8! has 2 and 5 as its multiples, then it will have 0 as the units digit, so (8!)^2 will have two zeros in the end, which means that (8!)^2-38 will have 00-38=62 as the last digits: 6*2=12.
Answer: D. Although there is hardly anything left after your's explanations but still I'll post my approach, which might help few people out here (esp. on the # trailing 0's as you people have kept in the Math book)-- first I simplified the whole x value as it looks so clumsy/heavy: given x = (8!)^6 [(8!)^4 - 1]/ (8!)^3[(8!)^2 - 1]
= (8!)^3[(8!)^2 + 1][(8!)^2 - 1]/[(8!)^2 - 1]
so, (x/ (8!)^3) - 39 = [(8!)^2 + 1] -39
= (8!)^2 -38Now as per the Q we need to know unit & tenth digit of the above number- as we know factorials of numbers > 4 have trailing zero's so here we can determine how many trailing 0's in 8!. Since 8 < 5^2 so that means we have only 1 trailing 0 in here (alternately you can have manual calc in this case as well because it's small number). So (8!)^2 has 2 trailing 0's. Now it does not matter what hundredth position is becasue we need to subtract only 38- this would result in 6 in tenth pos and 2 in unit pos. so product is = 12... P.S.- just for refrence # trailing 0's in factorial N is determined by- N/5^1+N/5^2+....+N/5^k (where 5^k < N)for each division we take integer part (quotient) and then add them up. this should give # trailing 0's
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Bunuel wrote: 5. There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble to remain in the jar?
A. 460
B. 490
C. 493
D. 455
E. 445
Total ways to select 8 marbles out of 7+5=12 is C^8_{12};
Ways to select 8 marbles so that zero red marbles is left in the jar is C^7_7*C^1_5;
Ways to select 8 marbles so that zero blue marbles is left in the jar is C^5_5*C^3_7;
Hence ways to select 8 marbles so that at least one red marble and at least one blue marble to remain the jar is C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455.
Answer: D. I tried feagure out this way first but thought to calculate little differently- Since there are a toatl of 12 Marbels and after picking the required # marbels we are left with 4 marbels in the Jar- so our task is determine out of these 4 in how many ways we can pick R, B marbels so that there exists at least 1 marbel of each type- (R-1,B-3), (R-2,B-2),(R-3,B-1) So the required # ways = 7C1*5C3 + 7C2*5C2 + 7C3*5C1 = (7 * 10) + (21 * 10)+ (35 * 5) = 70 + 210 + 175 = 455 (D)
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Bunuel wrote: 6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the fraction of the pool which pump A filled?
A. \frac{yz}{x+y+z}
B. \frac{yz}{yz+xz-xy}
C. \frac{yz}{yz+xz+xy}
D. \frac{xyz}{yz+xz-xy}
E. \frac{yz+xz-xy}{yz}
With pumps A and B both running and the drain unstopped the pool will be filled in a rate \frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{yz+xz-zy}{xyz} pool/hour. So, the pool will be filled in \frac{xyz}{yz+xz-xy} hours (time is reciprocal of rate).
In \frac{xyz}{yz+xz-xy} hours pump A will do \frac{1}{x}*\frac{xyz}{yz+xz-xy}=\frac{yz}{yz+xz-xy} part of the job.
Answer: B. hi , i feel the ans to this Q is wrong. as we have to tell what fraction of the pool was filled up by pump A. it will not depend on the drain C as the filling was done only by pump A and B. ans shud be y/(x+y). the given ans would be correct if we were to find "the amount of water pumped by A in that duration"
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Bunuel wrote: 2. If y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}, then y is NOT divisible by which of the following?
A. 6^4
B. 62^2
C. 65^2
D. 15^4
E. 52^4
y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=3^4*(3^3-1)^2*5^8*(5^3-1)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2.
Now, if you analyze each option you'll see that only 52^4=2^4*13^4 is not a factor of y, since the power of 13 in it is higher than the power of 13 in y.
Answer: E. Pls correct the Typo 2^4 to 2^8
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mofasser08 wrote: Can any one please explain: why it is (9*9*1*1*1* 5C3) in Question no. 1 (password problem)? Why not : 9*9*5P3 ?? That's a tough 700+ problem. We are interested 666XX numbers. Now, two this can be arranged in 10 ways; 666XX; 66X6X; 6X66X; X666X; 66XX6; 6X6X6; X66X6; 6XX66; XX666; X6X66. So, basically with C^3_5=10 we are choosing which 3 places out of 5 will be occupied by X's. Or we could do the other way around: C^2_510 choosing which 2 places out of 5 will be occupied by 6's. Next, each X can take any value from 0 to 9, inclusive but 6, so total of 9 values. Notice that the first digit can be zero here since the 5-digit password unlike 5-digit number can start with 0. Total 9*9*C^3_5=810. Hope it's clear.
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alphabeta1234 wrote: Bunuel wrote: SOLUTIONS:
1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?
A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000
Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.
# of passwords with three digit 6 is 9*9*C^3_5=810: each out of two other digits (not 6) has 9 choices, thus we have 9*9 and C^3_5 is ways to choose which 3 digits will be 6's out of 5 digits we have.
P=\frac{favorable}{total}=\frac{810}{10^5}
Answer: B. Hey Bunuel, Sorry to trouble you. If you could, I would appreciate if you could clarify more about 5C3. So this is how I approached the problem: We have three - 6's. With two slots that can be filled by 9 numbers. 666 9 9 . Now the question is how many ways can we rearrange these codes of three 6's and 2 non 6's? Well, have 5 elements 3 identical (because we have 3 identical 6's and two non-6's) Hence we have 5!/(3!2!) ways of rearranging the numbers. Is this why we have 5C3 in the problem? But the two non-6's can be identical or different. For example 66611 66622 66633 66644 66655 ... 66699... etc all satisfy the condition of only three 6's. The number of ways to rearrange the letters {aaabb} is 5!/(3!2!). Moreover, we can also have the two non-6's be different, 66612 66654 66624, etc . the number of ways to rearrange the letters {aaabc} is 5!/(3!). What do we do in this situation? Do we look at this problem as (#6)(#6)(#6)(not #6)(not #6) therefore 5!/(3!2!). Three repeating elements of #6 and two repeating elements of not #6. OR Do we look at the problem as (#6)(#6)(#6)(x)(y), where x and y two numbers that are not 6, that may or may not be equal, therefore 5!/3! (3! for the 3 repeating 6's but we cannot say anything about whether x and y are equal to each other) Bunuel, as always, thank you so much!! And again, sorry to trouble you. In this post there are all cases listed: baker-s-dozen-128782-60.html#p1079496Please tell me what part of this post needs farther clarification.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
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