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Re: New DS set!!!
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09 Mar 2014, 07:26
damamikus wrote: Bunuel wrote: 4. Two machines, A and B, each working at a constant rate, can complete a certain task working together in 6 days. In how many days, working alone, can machine A complete the task?
Given that 1/A+1/B=1/6, where A is the time needed for machine A to complete the task working alone and B is the time needed for machine B to complete the task working alone.
(1) The average time A and B can complete the task working alone is 12.5 days. This statement implies that A+B=2*12.5=25. Now, since we don't know which machine works faster then even if we substitute B with 25A (1/A + 1/(25A) = 1/6) we must get two different answers for A and B: A<B and A>B. Not sufficient.
(2) It would take machine A 5 more days to complete the task alone than it would take for machine B to complete the task. A=B+5, so we have that 1/A+1/(A5)=1/6. From this we can find that A=2 (not a valid solution since in this case B will be negative) or A=15. Sufficient.
Answer: B. Hi Bunuel! Could you please tell me where I made a mistake in my solution to statement 1? statement 1: (1) A+B=25 > from the question stem we have that (1/A)+(1/B)=1/6 > [(A+B)/AB]=1/6 > [25/AB]=1/6 > 150=AB > B=150/A (2) insert 2 into 1: A+[150/A]=25 > A²25A 150=0 > (A30)(A+5)=0 > A=30 or (5), both answers obviously can't be right, since A+B=25 and no value can be negative here. But where exactly was my approach wrong? Thanks a lot for your help! Max From \(A+\frac{150}{A}=25\) > \(A^225A+150=0\) (not A^2  25A  150 = 0) > \((A10)(A15)=0\) > \(A=10\) and \(B=15\) OR \(A=15\) and \(B=10\). Hope it's clear.
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04 Sep 2014, 23:45
Bunuel wrote: 8. Two marbles are drawn from a jar with 10 marbles. If all marbles are either red of blue, is the probability that both marbles selected will be red greater than 3/5?
The question: is P(R and R)=R/10*(R1)/9>3/5? Is R(R1)>54? Is R>7? (By number plugging) So, the question asks whether the number of red marbles is more than 7 (8, 9, or 10).
(1) The probability that both marbles selected will be blue is less than 1/10. This implies that B/10*(B1)/9<1/10. So, we have that B(B1)<9, thus B<4, so the number of red marbles in the jar is 7, 8, 9, or 10. Not sufficient.
(2) At least 60% of the marbles in the jar are red. This implies that the number of red marbles is 6 or more. Not sufficient.
(1)+(2) From above we have that R>6. Not sufficient.
Answer: E. Hi, Could you please help me on this.. how R(R1)>54 turns R>7?



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05 Sep 2014, 04:56
luckyme17187 wrote: Bunuel wrote: 8. Two marbles are drawn from a jar with 10 marbles. If all marbles are either red of blue, is the probability that both marbles selected will be red greater than 3/5?
The question: is P(R and R)=R/10*(R1)/9>3/5? Is R(R1)>54? Is R>7? (By number plugging) So, the question asks whether the number of red marbles is more than 7 (8, 9, or 10).
(1) The probability that both marbles selected will be blue is less than 1/10. This implies that B/10*(B1)/9<1/10. So, we have that B(B1)<9, thus B<4, so the number of red marbles in the jar is 7, 8, 9, or 10. Not sufficient.
(2) At least 60% of the marbles in the jar are red. This implies that the number of red marbles is 6 or more. Not sufficient.
(1)+(2) From above we have that R>6. Not sufficient.
Answer: E. Hi, Could you please help me on this.. how R(R1)>54 turns R>7? Bu number plugging. R is an integer, if R = 7, then R(R  1) = 42 < 54 but if R = 8, then R(R  1) = 56 > 54, thus R must be greater than 7.
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Re: New DS set!!!
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14 Dec 2014, 07:04
doe007 wrote: 3. The length of the median BD in triangle ABC is 12 centimeters, what is the length of side AC? (1) ABC is an isosceles triangle (2) AC^2 = AB^2 + BC^2
1) There is no information on angles and which two sides are equal. Not sufficient.
2) It says that angle at B is right angle and AC is hypotenuse. If any rightangled triangle is inscribed in a circle, the hypotenuse of the triangle must be diameter of circle and the median extending to the hypotenuse of the triangle must be radius of the circle >Median is equal to half of the hypotenuse > AC = 24. Sufficient.
Answer is B. In a circle???? where is that in the question???



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15 Dec 2014, 08:02
Arnav180 wrote: doe007 wrote: 3. The length of the median BD in triangle ABC is 12 centimeters, what is the length of side AC? (1) ABC is an isosceles triangle (2) AC^2 = AB^2 + BC^2
1) There is no information on angles and which two sides are equal. Not sufficient.
2) It says that angle at B is right angle and AC is hypotenuse. If any rightangled triangle is inscribed in a circle, the hypotenuse of the triangle must be diameter of circle and the median extending to the hypotenuse of the triangle must be radius of the circle >Median is equal to half of the hypotenuse > AC = 24. Sufficient.
Answer is B. In a circle???? where is that in the question??? It says: IF ANY rightangled triangle is inscribed in a circle, the hypotenuse of the triangle must be diameter of circle. So, this is just a property for any right triangle inscribed in a circle.
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Re: New DS set!!!
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04 Mar 2017, 10:20
Bunuel wrote: 5. Set A={32x, 3x, 3, 3+x, 3+2x}, where x is an integer. Is the standard deviation of set A more than the standard deviation of set B={32x, 3x, 3, 3+x, 3+2x, y}
(1) The standard deviation of set A is positive. We know that the standard deviation of any set is more than or equal to zero. The standard deviation of a set is zero only when the set consists of identical elements. So, this statement implies that set A does NOT consists of identical elements or that x does not equal to zero. Still this statement is not sufficient to answer the question.
(2) y=3. The mean of set A is 3. Now, if \(x\neq{0}\) for example if x=1, then the standard deviation of B would be smaller that the standard deviation A, since the elements of B would be less widespread than the element of set A. But if x=0, then A={3, 3, 3, 3, 3} and B={3, 3, 3, 3, 3, 3}, so both will have the standard deviation of zero. Bot sufficient.
(1)+(2) Since from (1) \(x\neq{0}\), then adding a new element which equals to the mean will shrink the standard deviation, thus SD(A)>SD(B). Sufficient.
Answer: C. I have a doubt that in set A if i substitute x as 1 and y=3 i get (1,2,3,4,5) and for set B it is (1,2,3,3,4,5) now which will have a greater standard deviation or will it be same? Like the elements in B are more and the deviation in both sets from mean is same so B should have bigger SD but If use the logic adding nos same as mean decreases SD then Set A will have bigger SD



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05 Mar 2017, 03:23
jamescath wrote: Bunuel wrote: 5. Set A={32x, 3x, 3, 3+x, 3+2x}, where x is an integer. Is the standard deviation of set A more than the standard deviation of set B={32x, 3x, 3, 3+x, 3+2x, y}
(1) The standard deviation of set A is positive. We know that the standard deviation of any set is more than or equal to zero. The standard deviation of a set is zero only when the set consists of identical elements. So, this statement implies that set A does NOT consists of identical elements or that x does not equal to zero. Still this statement is not sufficient to answer the question.
(2) y=3. The mean of set A is 3. Now, if \(x\neq{0}\) for example if x=1, then the standard deviation of B would be smaller that the standard deviation A, since the elements of B would be less widespread than the element of set A. But if x=0, then A={3, 3, 3, 3, 3} and B={3, 3, 3, 3, 3, 3}, so both will have the standard deviation of zero. Bot sufficient.
(1)+(2) Since from (1) \(x\neq{0}\), then adding a new element which equals to the mean will shrink the standard deviation, thus SD(A)>SD(B). Sufficient.
Answer: C. I have a doubt that in set A if i substitute x as 1 and y=3 i get (1,2,3,4,5) and for set B it is (1,2,3,3,4,5) now which will have a greater standard deviation or will it be same? Like the elements in B are more and the deviation in both sets from mean is same so B should have bigger SD but If use the logic adding nos same as mean decreases SD then Set A will have bigger SD The standard deviation of A should be greater than that of B. That's because we add to A an element equal to the mean of A, thus making B less widespread than A. The SD of A = \(\sqrt{2} \approx 1.41\) and the SD of \(B = \sqrt{\frac{5}{3}} \approx 1.29\). Please check the links below to brush up fundamentals on SD and statistics: Hope it helps.
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Re: New DS set!!!
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23 Apr 2017, 02:54
Bunuel wrote: 9. Is x^2>2x?
Is x(x2)>0? > is x<0 or x>2. Basically if x is not 0, 1, or 2 we have an YES answer to the question.
(1) x is a prime number. If x=2 then the answer is NO but if x is some other prime, then the answer is YES. Not sufficient.
(2) x^2 is a multiple of 9. If x=0 then the answer is NO but if x=3, then the answer is YES. Not sufficient.
(1)+(2) Since from (1) x is a prime and from (2) x^2 is a multiple of 9, then x can only be 3. Therefore the answer to the question is YES. Sufficient.
Answer: C. Problem with statement 2. How can 0 be a multiple of 9?



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23 Apr 2017, 03:22
hannahkagalwala wrote: Bunuel wrote: 9. Is x^2>2x?
Is x(x2)>0? > is x<0 or x>2. Basically if x is not 0, 1, or 2 we have an YES answer to the question.
(1) x is a prime number. If x=2 then the answer is NO but if x is some other prime, then the answer is YES. Not sufficient.
(2) x^2 is a multiple of 9. If x=0 then the answer is NO but if x=3, then the answer is YES. Not sufficient.
(1)+(2) Since from (1) x is a prime and from (2) x^2 is a multiple of 9, then x can only be 3. Therefore the answer to the question is YES. Sufficient.
Answer: C. Problem with statement 2. How can 0 be a multiple of 9? 0 is not a divisor of any integer, but a multiple of every integer. So, 0 is divisible by 9: 0/9 = 0 = integer. ZERO:1. 0 is an integer. 2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even. 3. 0 is neither positive nor negative integer (the only one of this kind). 4. 0 is divisible by EVERY integer except 0 itself.Check more here: http://gmatclub.com/forum/numberproper ... 74996.html
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Re: New DS set!!!
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23 Apr 2017, 03:33
hannahkagalwala wrote: Bunuel wrote: 9. Is x^2>2x?
Is x(x2)>0? > is x<0 or x>2. Basically if x is not 0, 1, or 2 we have an YES answer to the question.
(1) x is a prime number. If x=2 then the answer is NO but if x is some other prime, then the answer is YES. Not sufficient.
(2) x^2 is a multiple of 9. If x=0 then the answer is NO but if x=3, then the answer is YES. Not sufficient.
(1)+(2) Since from (1) x is a prime and from (2) x^2 is a multiple of 9, then x can only be 3. Therefore the answer to the question is YES. Sufficient.
Answer: C. Problem with statement 2. How can 0 be a multiple of 9? Hi hannahkagalwala, 0 is a multiple of everything. Go thorough the following page: Multiple_(mathematics)Please read the section 4.1(Properties of Integers) from OG throughly. Hope this helps.



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30 Apr 2017, 23:24
hannahkagalwala wrote: 9. Is x^2>2x?Is x(x2)>0? > is x<0or x>2. Basically if x is not 0, 1, or 2 we have an YES answer to the question. (1) x is a prime number. If x=2 then the answer is NO but if x is some other prime, then the answer is YES. Not sufficient. (2) x^2 is a multiple of 9. If x=0 then the answer is NO but if x=3, then the answer is YES. Not sufficient. (1)+(2) Since from (1) x is a prime and from (2) x^2 is a multiple of 9, then x can only be 3. Therefore the answer to the question is YES. Sufficient. Answer: C. Problem with statement 2. How can 0 be a multiple of 9? 0 is not a divisor of any integer, but a multiple of every integer. So, 0 is divisible by 9: 0/9 = 0 = integer. ZERO:1. 0 is an integer. 2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even. 3. 0 is neither positive nor negative integer (the only one of this kind). 4. 0 is divisible by EVERY integer except 0 itself.Check more here: http://gmatclub.com/forum/numberproper ... 74996.htmlThank you Bunuel for explaining that to me. Coming back to this question, I can't figure out why x<0??? Hi hannahkagalwala, To solve any inequality, first find the critical point. Replace inequality by equality sign and solve it. \(x(x2) = 0 \Rightarrow x = 0 \textrm{ or } x = 2\) Please,refer the attached diagram. If you take any value less than 0, say x = 1 , x(x2) = (1)(1  2) = 3 (>0) . Similarly take any value greater than 2 , say x = 3 => x(x2) = 3(3  2) = 3 (>0) . This implies that x(x2) > 0 is possible only when x < 0 or x> 2. Any value between 0 and 2 will not satisfy the condition. For x = 1 , x(x2) = 1(1  2) = 1. Please go through the inequality theory. Refer following links inequalitiesmadeeasyinequalitiestrickHope this helps.
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New DS set!!!
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28 Mar 2018, 20:37
Bunuel wrote: 3. The length of the median BD in triangle ABC is 12 centimeters, what is the length of side AC?
(1) ABC is an isosceles triangle. Clearly insufficient.
(2) AC^2 = AB^2 + BC^2. This statement implies that ABC is a right triangle and AC is its hypotenuse. Important property: median from right angle is half of the hypotenuse, hence BD=12=AC/2, from which we have that AC=24. Sufficient.
Answer: B. Median of a right angled triangle is half of the hypotenuse only if it is an isosceles triangle. Should this not be option C??



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28 Mar 2018, 20:56
praveenbaranwal wrote: Bunuel wrote: 3. The length of the median BD in triangle ABC is 12 centimeters, what is the length of side AC?
(1) ABC is an isosceles triangle. Clearly insufficient.
(2) AC^2 = AB^2 + BC^2. This statement implies that ABC is a right triangle and AC is its hypotenuse. Important property: median from right angle is half of the hypotenuse, hence BD=12=AC/2, from which we have that AC=24. Sufficient.
Answer: B. Median of a right angled triangle is half of the hypotenuse only if it is an isosceles triangle. Should this not be option C?? The median on the hypotenuse of a right triangle ALWAYS equals onehalf the hypotenuse. To prove you can consider right triangle inscribed in a circle, where the median on the hypotenuse is radius and hypotenuse is a diameter, so the median = onehalf the hypotenuse OR you can consider any rectangle, where diagonals cut each other in half: because half of the diagonal is basically the median to the another diagonal.
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11 Jul 2018, 10:06
Bunuel wrote: praveenbaranwal wrote: Bunuel wrote: 3. The length of the median BD in triangle ABC is 12 centimeters, what is the length of side AC?
(1) ABC is an isosceles triangle. Clearly insufficient.
(2) AC^2 = AB^2 + BC^2. This statement implies that ABC is a right triangle and AC is its hypotenuse. Important property: median from right angle is half of the hypotenuse, hence BD=12=AC/2, from which we have that AC=24. Sufficient.
Answer: B. Median of a right angled triangle is half of the hypotenuse only if it is an isosceles triangle. Should this not be option C?? The median on the hypotenuse of a right triangle ALWAYS equals onehalf the hypotenuse. To prove you can consider right triangle inscribed in a circle, where the median on the hypotenuse is radius and hypotenuse is a diameter, so the median = onehalf the hypotenuse OR you can consider any rectangle, where diagonals cut each other in half: because half of the diagonal is basically the median to the another diagonal. Edit: Sorry silly question, got it.



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27 Aug 2018, 22:00
Hi Bunuel,
Sorry to bring back this question :
2. If x and y are both positive integers and x>y, what the remainder when x is divided by y?
if x and y are positive integers, there exist unique integers q and r, called the quotient and remainder, respectively, such that y=divisor∗quotient+remainder=xq+ry=divisor∗quotient+remainder=xq+r and 0≤r<x0≤r<x.
(1) y is a twodigit prime number. Clearly insufficient since we know nothinf about x.
(2) x=qy+9, for some positive integer q. It's tempting to say that this statement is sufficient and r=9r=9, since given equation is very similar to y=divisor∗quotient+remainder=xq+ry=divisor∗quotient+remainder=xq+r . But we don't know whether y>9y>9: remainder must be less than divisor.
For example: If x=10x=10 and y=1y=1 then 10=1∗1+910=1∗1+9, then the remainder upon division 10 by 1 is zero. If x=11x=11 and y=2y=2 then 11=1∗2+911=1∗2+9, then the remainder upon division 11 by 2 is one. Not sufficient.
(1)+(2) From (2) we have that x=qy+9x=qy+9 and from (1) that y is more than 9 (since it's a twodigit number), so we have direct formula of remainder, as given above. Sufficient.
Answer: C.
I have a doubt in the above part of the sentence which is in bold i.e.
if x and y are positive integers, there exist unique integers q and r, called the quotient and remainder, respectively, such that y=divisor∗quotient+remainder=xq+ry=divisor∗quotient+remainder=xq+r and 0≤r<x0≤r<x. Please correct me if I am wrong if x/y then x= y*q+r, were q = quotient, r=remainder, y = divisor, x = dividend
the rest of the part of the solution I understood, but the above sentence confused me . Can you please tell me what I am missing here?



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