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The next set of medium/hard DS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers.
1. What is the product of three consecutive integers?
(1) At least one of the integers is positive (2) The sum of the integers is less than 6
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?
(1) The average time A and B can complete the task working alone is 12.5 days. (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
5. Set A={3-2x, 3-x, 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={3-2x, 3-x, 3, 3+x, 3+2x, y}
(1) The standard deviation of set A is positive (2) y=3
6. The ratio of the number of employees of three companies X, Y and Z is 3:4:8, respectively. Is the average age of all employees in these companies less than 40 years?
(1) The total age of all the employees in these companies is 600 (2) The average age of employees in X, Y, and Z, is 40, 20, and 50, respectively.
7. Was the average (arithmetic mean) temperature in city A in March less than the average (arithmetic mean) temperature in city B in March?
(1) The median temperature in City A in March was less than the median temperature in city B (2) The ratio of the average temperatures in A and B in March was 3 to 4, respectively
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?
(1) The probability that both marbles selected will be blue is less than 1/10 (2) At least 60% of the marbles in the jar are red
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*(R-1)/9>3/5? Is R(R-1)>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*(B-1)/9<1/10. So, we have that B(B-1)<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.
5. Set A={3-2x, 3-x, 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={3-2x, 3-x, 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.
7. Was the average (arithmetic mean) temperature in city A in March less than the average (arithmetic mean) temperature in city B in March?
(1) The median temperature in City A in March was less than the median temperature in city B. Clearly insufficient.
(2) The ratio of the average temperatures in A and B in March was 3 to 4, respectively. Temperatures can be negative, thus this statement is also not sufficient. Consider T(A)=3 and T(B)=4 AND T(A)=-3 and T(B)=-4.
(1)+(2) We have no additional useful info. Not sufficient.
1. (1) Insufficient. It could be {0,1,2} and the product is 0, or it could be {1,2,3} and their product is 6. (2) Insufficient. It could be {0,1,2} and the product is 0, or it could be {-3,-2,-1} and the product is -6
(1)+(2) Sufficient. Suppose the least number is n, so the second is (n+1) and the third is (n+2). Their sum must be less than 6: n+(n+1)+(n+2)<6 or 3n+3<6 or n<1. By first statement at least one of the integers must be positive, it means that the largest is positive: n+2>0 or n>-2. So n=-1 or n=0. Therefore, there are two possible sets {-1, 0, 1} or {0, 1, 2}. The product anyway is 0.
The correct answer is C.
2. (1) Insufficient. For x=11 and y=10 the remainder when x is divided by y is 1, for x=12 and y=2 the remainder when x is divied by y is 2. (2) Insufficient. The main point here is the remainder must be less than the divisor. We don't know is y>9 or y<=9. For y=1 the remainder when x is divided by y is 0, for y=2, x=2q+9 the remainder is 1 when x is divided by 2.
(1)+(2) Sufficient. If y is a two-digit number the when qy is divided by y there is remainder 0, and when 9 is divided by y the remainder is 9, since 9<y. So, x when divided by y gives the remainder 9.
The correct answer C _________________
I'm happy, if I make math for you slightly clearer And yes, I like kudos:)
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 \(y =divisor*quotient+remainder= xq + r\) and \(0\leq{r}<x\).
(1) y is a two-digit 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 \(r=9\), since given equation is very similar to \(y =divisor*quotient+remainder= xq + r\) . But we don't know whether \(y>9\): remainder must be less than divisor.
For example: If \(x=10\) and \(y=1\) then \(10=1*1+9\), then the remainder upon division 10 by 1 is zero. If \(x=11\) and \(y=2\) then \(11=1*2+9\), then the remainder upon division 11 by 2 is one. Not sufficient.
(1)+(2) From (2) we have that \(x=qy+9\) and from (1) that y is more than 9 (since it's a two-digit number), so we have direct formula of remainder, as given above. Sufficient.
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 25-A (1/A + 1/(25-A) = 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/(A-5)=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.
9. If x is an integer, is x^2>2x? \(x^2-2x>0\) \(x(x-2)>0\) The question asks is x in one of those intervals? \(x<0 , x>2\)
(1) x is a prime number. Sufficient All prime numbers are greater than 2 so \(x\geq{2}\) so we are in the right interval. What a stupid mistake!Not suff, x can be 2 (2) x^2 is a multiple of 9. Not sufficient. If x=0, x^2 =0 is a multiple of 9 but 0 isn't in the intervals, if x=3 x^2=9 is a multiple of nine and 3 is in our intervals.
C _________________
It is beyond a doubt that all our knowledge that begins with experience.
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 right-angled 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.
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? (1) The average time A and B can complete the task working alone is 12.5 days. (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
Assuming A and B takes x and y days respectively to finish the work alone. In 6 days, work done by A = 6/x, work done by B = 6/y, total work done by A and B together = 6/x + 6/y 6/x + 6/y = 1 --> 6x + 6y = xy …. (eq1)
1) (x+y)/2 = 12.5 --> x+y = 25 Substituting this in eq1, 6*25 = x(25-x) --> x^2 -25x + 150 = 0 --> (x-10)(x-15) = 0 --> x = 10 or 15 Not sufficient.
2) y = x-5 Substituting this in eq1, 6x + 6x – 30 = x(x-5) --> x^2 -17x + 30 = 0 --> (x-2)(x-15) = 0 --> x = 2 or 15 If x = 2, y become -3 which is impossible. --> x = 15. Sufficient.
1. What is the product of three consecutive integers?
(1) At least one of the integers is positive.
We can have three cases: (i) All three integers are positive. In this case the product will obviously be positive. (ii) Two of the integers are positive: {0, 1, 2}. In this case the product will be zero. (iii) Only one of the integers is positive: {-1, 0, 1}. In this case the product will be zero.
Not sufficient.
(2) The sum of the integers is less than 6. Clearly insufficient, consider {-1, 0, 1} and {-3, -2, -1}.
(1)+(2) The second statement implies that we cannot have case (i) from (1), since the least sum of three positive consecutive integers is 1+2+3=6. Thus we have either case (ii) or case (iii). Therefore the product of the integers is zero. Sufficient.
(1) No number in set A is less than the average (arithmetic mean) of set A.
Since no number is less than the average, then no number is more than the average, which implies that the list contains identical elements: A={x, x, x, ...}. From this it follows that (the average)=(the median). But we don't know the value of x, thus this statement is NOT sufficient.
(2) The average (arithmetic mean) of set A is equal to the range of set A.
Not sufficient: if A={0, 0, 0, 0}, then (the median)=0, but if A={1, 2, 2, 3}, then (the median)=2.
(1)+(2) From (1) we have that the list contains identical elements. The range of all such sets is 0. Therefore, from (2) we have that (the average)=(the range)=0 and since from (1) we also know that (the average)=(the median), then (the median)=0. Sufficient.
1. What is the product of three consecutive integers?
(1) At least one of the integers is positive Clearly Not sufficient. {0, 1, 2} product=0; {1, 2, 3} product = 6 (2) The sum of the integers is less than 6 Clearly Not sufficient. {0, 1, 2} product=0; {-1, -2, -3} product = 6
(1)+(2) From 2 we know that \(a+(a+1)+(a+2)<6\) so \(a<1\) The three integers are \(a,a+1,a+2\), and at least one is positive (knowing that a is an integer less than 1) : 1)a>0 this cannot be: there is no integer between 0 and 1; 2)a+1>0 a>-1 with a<1 there is one solution a=0 and the product is 0 {0,1,2};3) a+2>0 a>-2 here we have 2 solutions a=-1 and the numbers are {-1,0,1}=>prod=0 or a=0 and the product is 0. IMO C _________________
It is beyond a doubt that all our knowledge that begins with experience.
2. If x and y are both positive integers and x>y, what the remainder when x is divided by y?
(1) y is a two-digit prime number Not sufficient. ie 22/11 R=0, 23/11 R=1 (2) x=qy+9, for some positive integer q Not sufficient. ie x= 29 = 20+9 = 4*5+9 or 5*4+9 I don't know if 4 or 5 is y and this changes the R.
(1)+(2) If y is a two-digit prime number x=qy+9 for any q says that the R of x/y is 9. x/y=q+9/y IMO C _________________
It is beyond a doubt that all our knowledge that begins with experience.
6. The ratio of the number of employees of three companies X, Y and Z is 3:4:8, respectively. Is the average age of all employees in these companies less than 40 years?
(1) The total age of all the employees in these companies is 600 Not sufficient. ie totEmployees= 15 (3+4+8) => average age = 600/15=40 or totEmployees = 30 (2*(3+4+8))=> average age = 600/30=20. In the first case the answer is NO, in the second is YES (2) The average age of employees in X, Y, and Z, is 40, 20, and 50, respectively. \(\frac{(40*3+20*4+50*8)}{15}=\frac{120+80+400}{15}=\frac{600}{15}=40\) If you weight the average age with the ratios you find out that the average age is 40, which is NOT less than 40. Sufficient IMO B _________________
It is beyond a doubt that all our knowledge that begins with experience.
The next set of medium/hard DS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers.
1. What is the product of three consecutive integers?
(1) At least one of the integers is positive -Can not determine the values (2) The sum of the integers is less than 6- Can not determine the values
(1)+(2)-> at least one positive and <6 so can't be 1,2,3 -> so 0 must be part of the sequence .
IMO: C
Bunuel wrote:
2. If x and y are both positive integers and x>y, what the remainder when x is divided by y?
(1) y is a two-digit prime number - Can not determine the values (2) x=qy+9, for some positive integer q - Can not determine the values 1+2=>y>9 , as y is two digit ...so remainder 9
IMO: C
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 No idea about which sides are same (2) AC^2 = AB^2 + BC^2 D is mid point of AC , SO AD= DC=BD = 12 => AC
IMO:B
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?
(1) The average time A and B can complete the task working alone is 12.5 days.say A can complete in x , then b in = 12.5*2 - x hence \(\frac{1}{x}+\frac{1}{25-x}=\frac{1}{6}=> x= 15 or 10\)
(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 \(\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}=> x= 10\) So A 15 days [/color]
IMO: B
Bunuel wrote:
5. Set A={3-2x, 3-x, 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={3-2x, 3-x, 3, 3+x, 3+2x, y}
(1) The standard deviation of set A is positiveNot sufficient (2) y=3So we added 3 to the existing series , where 3 was mean=> SD will decrease
IMO: B
Bunuel wrote:
6. The ratio of the number of employees of three companies X, Y and Z is 3:4:8, respectively. Is the average age of all employees in these companies less than 40 years?
(1) The total age of all the employees in these companies is 600Can 't do anything (2) The average age of employees in X, Y, and Z, is 40, 20, and 50, respectively.can determine \(40*3x + 20*4x + 50 *8x = total age , now dividing by 15x we can get avg age\)
IMO: B
Bunuel wrote:
7. Was the average (arithmetic mean) temperature in city A in March less than the average (arithmetic mean) temperature in city B in March?
(1) The median temperature in City A in March was less than the median temperature in city BNo idea about avg (2) The ratio of the average temperatures in A and B in March was 3 to 4, respectivelyit means Avg(A) < Avg(B)
IMO: B
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?
(1) The probability that both marbles selected will be blue is less than 1/10 say there are b blue and 10-b red=> \frac{(bC2)}{(10C2) < 0.1=> b(b-1) < 9 Not sufficient} (2) At least 60% of the marbles in the jar are red So Red GE 6 and Blue LE 4 => for 6 red p(selecting 2 red out of 6) = 0.33 < \frac{3}{5} but if all are red P(red) = 1 > \frac{3}{5} so insufficient
Now adding 1+2 => blue can be 4 ,3,2,1,0 both blue should be less than 0.1 => so b can't be 4 it can only 3,2 So red balls can be 7 or 8-> no definite ans
IMO: E
Bunuel wrote:
9. If x is an integer, is x^2>2x?
(1) x is a prime number.not sufficient can be 2 or 3.. (2) x^2 is a multiple of 9. =>X is a multiple of 3 when we do squaring we multiply with a multiple of 3 and whem we do 2X we multiply with 2 so x^2>2x is always true except x= 0
adding 1 +2 => x can't be zero IMO: C
Bunuel wrote:
10. What is the value of the media of set A?
(1) No number in set A is less than the average (arithmetic mean) of set A. so all the numbers must be same =>median is any number of the set, sufficient to determine (2) The average (arithmetic mean) of set A is equal to the range of set A.not sufficient
IMO: A
Please let me know how many i got correct
Last edited by focus2k13 on 11 Apr 2013, 19:00, edited 1 time in total.
5. Set A={3-2x, 3-x, 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={3-2x, 3-x, 3, 3+x, 3+2x, y}
(1) The standard deviation of set A is positive Not sufficient.ie with x=1 A={1,2, 3, 4, 5} B={1, 2, 3, 4, 5, y} All depends on y. (2) y=3 Not sufficient.ie if x=0 A={3,3,3,3,3} B={3,3,3,3,3,3} STD of B is = STD of A if x=1 A={1,2, 3, 4, 5} B={1, 2, 3, 4, 5, 3} STD of B is < STD of A
(1)+(2) From 1 we know that \(x\neq{0}\) and from 2 that y=3. Sufficient. ie:x=1 : A={1,2, 3, 4, 5} B={1, 2, 3, 4, 5, 3} STD of B is < STD of A x=1000 : A={-1997,-997, 3, 1003, 2003} B={-1997,-997, 3, 1003, 2003, 3} STD of B is < STD of A A and B share 4 elements in common that are different from 3, but because B has one more 3 than X it will have a STD lesser than A _________________
It is beyond a doubt that all our knowledge that begins with experience.
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