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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

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 (2) x=qy+9, for some positive integer q

Sol: From St1, we get that x/y will be of the form

x/y= x/11, x/13, x/17, x/29, x/37...... Consider x = 12, Y =11, remainder 1 Consider x= 13, Y =11, remainder 2 So we have 2 ans and hence St 1 not sufficent alone (A & D ruled out)

St 2 x=qy+9, for some positive integer q

x/y= q+ 9/y ------> remainder will depend on value of y and hence st 2 not sufficient alone (Option B ruled out)

Combining both statement we get that remainder will always be 9 since Y is a 2 digit prime no

Hence answer should be C
_________________

“Many of life's failures are people who did not realize how close they were to success when they gave up.”

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

“Renew, release, let go. Yesterday’s gone. There’s nothing you can do to bring it back. You can’t “should’ve” done something. You can only DO something. Renew yourself. Release that attachment. Today is a new day!”

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

Sol:

We know SD >/ 0 and hence Set A is Positive does not tell us about y in statement B

So St 1 is alone not sufficent

St 2: y =3

Set A : SD \sqrt{10X^2/5}

Set B : SD is \sqrt{10x^2/6}

Since x is an integer we have SD of A as x \sqrt{2} and SD of B as x\sqrt{5/3}

Clearly SD of A is greater than that of B and hence ans should be B
_________________

“Many of life's failures are people who did not realize how close they were to success when they gave up.”

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

“Renew, release, let go. Yesterday’s gone. There’s nothing you can do to bring it back. You can’t “should’ve” done something. You can only DO something. Renew yourself. Release that attachment. Today is a new day!”

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.

Sol: From St 1 we have

600 = Average age of all employees (A)* No. of Employees(n) So Q asks is A<40

Clearly 1 alone is not sufficient

St 2 : Let the employees in the company be in the ratio 3c :4c: 8c where c is a positive integer

Therefore we have (3c*40+4c*20+ 5c* 50 )/15c = A If c =1 we have (120+80+250)/15 ----> (450/15) = 30 < 40 If c =2 we have ( 3*2*40+ 4*2*20+ 5*2*50/15*2), A = 900/30 < 40 If c= 3, we have (360+240+750)/45 ----> 1350/45 < 40 If c= 4, we have ( 480+320+1000)/60 -----> 1800/60 < 40

Therefore ans should be B
_________________

“Many of life's failures are people who did not realize how close they were to success when they gave up.”

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

“Renew, release, let go. Yesterday’s gone. There’s nothing you can do to bring it back. You can’t “should’ve” done something. You can only DO something. Renew yourself. Release that attachment. Today is a new day!”

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

Sol:

We know SD >/ 0 and hence Set A is Positive does not tell us about y in statement B

So St 1 is alone not sufficent

St 2: y =3

Set A : SD \sqrt{10X^2/5}

Set B : SD is \sqrt{10x^2/6}

Since x is an integer we have SD of A as x \sqrt{2} and SD of B as x\sqrt{5/3}

Clearly SD of A is greater than that of B and hence ans should be B

Notice that x can be 0 for (2), so this statement is NOT sufficient.

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.

Hi Bunnel Could you elaborate how this is true - median from right angle is half of the hypotenuse?? Thanks

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.

Hi Bunnel Could you elaborate how this is true - median from right angle is half of the hypotenuse?? Thanks

Sure.

Imagine a right triangle inscribed in a circle. We know that if a right triangle is inscribed in a circle, then its hypotenuse must be the diameter of the circle, hence half of the hypotenuse is radius. The line segment from the third vertex to the center is on the one hand radius of the circle=half of the hypotenuse and on the other hand as it's connecting the vertex with the midpoint of the hypotenuse it's median too.

Hi Bunuel.. How to be sure that A={0,0,0,0} and A={1,2,2,3} are the only sets possible from Statement2. Is there any quick method to find this. infact i was unable to find A={1,2,2,3}.

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.

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.

Hi Bunuel.. How to be sure that A={0,0,0,0} and A={1,2,2,3} are the only sets possible from Statement2. Is there any quick method to find this. infact i was unable to find A={1,2,2,3}.

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.

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.

Answer: C.

A={0, 0, 0, 0} and A={1, 2, 2, 3} are NOT the only sets possible. For example A={0, 0, 0} and A={1, 2, 3}. You can find these sets by trial and error.
_________________

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

Ans B ,

Given : Ta + Tb = 6 ( time for working together is 6 days) Please explain if wrong

Explanation as

1 As avg value given for individual working rate, (we want specific value for Ra) , as Ra can take multiple values in this case, stmt not sufficient.

2. Ta= Tb +5 & Ta + Tb =6 given in main stmt

therefore Ans B

Please correct if wrong
_________________

Correct me If I'm wrong !! looking for valuable inputs

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.

if ABC is isosceles triangle then all sides are equal. so AC=24. why not?
_________________

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Helpful Geometry formula sheet:best-geometry-93676.html I hope these will help to understand the basic concepts & strategies. Please Click ON KUDOS Button.

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.

if ABC is isosceles triangle then all sides are equal. so AC=24. why not?

(1) says that ABC is an isosceles triangle, not equilateral. Also, if ABC were equilateral AC would be \frac{24}{\sqrt{3}} not 24.

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.

Can anyone please explain how 1/A + 1/B = 1/6
_________________

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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.

Can anyone please explain how 1/A + 1/B = 1/6

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?

A is the time needed for machine A to complete the task working alone, thus the rate of A is 1/A job/day. B is the time needed for machine B to complete the task working alone, thus the rate of A is 1/B job/day.

Their combined rate is 1/A+1/B, which given to be equal to 1/6.

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.

Answer: C.

Hello Bunuel, the questions says x is divided by y.... So x will be dividend and y will be divisor.... the very first representation between y and x will confuse a better understanding of the latter explanation. Why don't you switch x and y in the first formula too Thank you.

(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.

Answer: C..

Bununel, what if Set A only contains one factor "1"?
_________________

(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.

Answer: C..

Bununel, what if Set A only contains one factor "1"?

The range of one element set is 0. If set A={1}, then it's range (0) does not equal to its mean (1). Thus this example contradicts the second statement and therefore is not valid.

(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.

Answer: C..

Bununel, what if Set A only contains one factor "1"?

The range of one element set is 0. If set A={1}, then it's range (0) does not equal to its mean (1). Thus this example contradicts the second statement and therefore is not valid.

Hope it's clear.

OMG, why can I forget this? thank you
_________________

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.

Can anyone please explain how 1/A + 1/B = 1/6

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?

A is the time needed for machine A to complete the task working alone, thus the rate of A is 1/A job/day. B is the time needed for machine B to complete the task working alone, thus the rate of A is 1/B job/day.

Their combined rate is 1/A+1/B, which given to be equal to 1/6.

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?

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.

Can anyone please explain how 1/A + 1/B = 1/6

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?

A is the time needed for machine A to complete the task working alone, thus the rate of A is 1/A job/day. B is the time needed for machine B to complete the task working alone, thus the rate of A is 1/B job/day.

Their combined rate is 1/A+1/B, which given to be equal to 1/6.

in question..A+b=6.. but in statement 1..its a+b=25? are these both not contradict with eachother?

Seems like m missing something :/

What are A and B in your first equation? What are A and B in your second equation?

If you don't pay attention what a variable represents you get each and every question wrong.

From the stem: 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.