January 17, 2019 January 17, 2019 08:00 AM PST 09:00 AM PST Learn the winning strategy for a high GRE score — what do people who reach a high score do differently? We're going to share insights, tips and strategies from data we've collected from over 50,000 students who used examPAL. January 19, 2019 January 19, 2019 07:00 AM PST 09:00 AM PST Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 52138

Re: New DS set!!!
[#permalink]
Show Tags
09 Mar 2014, 06:26
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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. ,11 Mixed Questions, 12 Fresh Meat 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., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 07 Apr 2014
Posts: 111

Re: New DS set!!!
[#permalink]
Show Tags
04 Sep 2014, 22:45
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?



Math Expert
Joined: 02 Sep 2009
Posts: 52138

Re: New DS set!!!
[#permalink]
Show Tags
05 Sep 2014, 03:56
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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. ,11 Mixed Questions, 12 Fresh Meat 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., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 05 Oct 2013
Posts: 37
Location: United States
Concentration: General Management, Strategy
WE: Design (Transportation)

Re: New DS set!!!
[#permalink]
Show Tags
14 Dec 2014, 06:04
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???



Math Expert
Joined: 02 Sep 2009
Posts: 52138

Re: New DS set!!!
[#permalink]
Show Tags
15 Dec 2014, 07:02
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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. ,11 Mixed Questions, 12 Fresh Meat 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., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 05 Sep 2016
Posts: 13

Re: New DS set!!!
[#permalink]
Show Tags
04 Mar 2017, 09:20
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



Math Expert
Joined: 02 Sep 2009
Posts: 52138

Re: New DS set!!!
[#permalink]
Show Tags
05 Mar 2017, 02:23
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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. ,11 Mixed Questions, 12 Fresh Meat 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., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 29 Nov 2016
Posts: 9

Re: New DS set!!!
[#permalink]
Show Tags
23 Apr 2017, 01:54
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?



Math Expert
Joined: 02 Sep 2009
Posts: 52138

Re: New DS set!!!
[#permalink]
Show Tags
23 Apr 2017, 02:22
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
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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. ,11 Mixed Questions, 12 Fresh Meat 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., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 17 May 2015
Posts: 249

Re: New DS set!!!
[#permalink]
Show Tags
23 Apr 2017, 02:33
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.



Manager
Joined: 17 May 2015
Posts: 249

Re: New DS set!!!
[#permalink]
Show Tags
30 Apr 2017, 22:24
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.
Attachments
Inequality.jpeg [ 3.11 KiB  Viewed 3357 times ]



Intern
Joined: 17 Feb 2018
Posts: 10
Location: New Zealand
Concentration: Strategy, Technology
WE: Consulting (Computer Software)

New DS set!!!
[#permalink]
Show Tags
28 Mar 2018, 19:37
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??



Math Expert
Joined: 02 Sep 2009
Posts: 52138

Re: New DS set!!!
[#permalink]
Show Tags
28 Mar 2018, 19:56
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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. ,11 Mixed Questions, 12 Fresh Meat 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., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 21 May 2018
Posts: 11

New DS set!!!
[#permalink]
Show Tags
11 Jul 2018, 09:06
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.



Intern
Joined: 14 Aug 2017
Posts: 14

New DS set!!!
[#permalink]
Show Tags
27 Aug 2018, 21:00
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?




New DS set!!! &nbs
[#permalink]
27 Aug 2018, 21:00



Go to page
Previous
1 2 3 4 5 6
[ 115 posts ]



