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# New DS set!!!

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10 Apr 2013, 08:10
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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

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211902

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

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211903

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

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211904

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

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211906

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

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211907

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.

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211908

7. Was the average (arithmetic mean) temperature in degrees Celsius in city A in March less than the average (arithmetic mean) temperature in degrees Celsius in city B in March?

(1) The median temperature in degrees Celsius in City A in March was less than the median temperature in degrees Celsius in city B.

(2) The ratio of the average temperatures in degrees Celsius in A and B in March was 3 to 4, respectively.

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211909

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

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211910

9. If x is an integer, is x^2>2x?

(1) x is a prime number.
(2) x^2 is a multiple of 9.

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211911

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.
(2) The average (arithmetic mean) of set A is equal to the range of set A.

Solution: http://gmatclub.com/forum/new-ds-set-15 ... l#p1211912

Kudos points for each correct solution!!!
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14 Apr 2013, 09:02
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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?

Given that the ratio of the number of employees is 3x:4x:8x, for some positive multiple x.

The questions asks whether (average age)=(total age)/(# of employees)<40, or whether (total age)/(3x+4x+8x)<40, which is the same as: is (total age)<600x?

(1) The total age of all the employees in these companies is 600. The question becomes: is 600<600x? Or is 1<x. We don't know that: f x=1, then the answer is NO but if x>1, then the answer is YES. Not sufficient.

(2) The average age employees in X, Y, and Z, is 40, 20, and 50, respectively. (total age)=40*3x+20*4x+50*8x=600x, so the answer to the question is NO. Sufficient.

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14 Apr 2013, 09:03
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Was the average (arithmetic mean) temperature in degrees Celsius in city A in March less than the average (arithmetic mean) temperature in degrees Celsius in city B in March?

(1) The median temperature in degrees Celsius in City A in March was less than the median temperature in degrees Celsius in city B. Clearly insufficient.

(2) The ratio of the average temperatures in degrees Celsius 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.

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14 Apr 2013, 09:03
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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.

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14 Apr 2013, 09:04
9. Is x^2>2x?

Is x(x-2)>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.

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14 Apr 2013, 09:04
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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.

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14 Apr 2013, 23:14
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?

Given that the ratio of the number of employees is 3x:4x:8x, for some positive multiple x.

The questions asks whether (average age)=(total age)/(# of employees)<40, or whether (total age)/(3x+4x+8x)<40, which is the same as: is (total age)<600x?

(1) The total age of all the employees in these companies is 600. The question becomes: is 600<600x? Or is 1<x. We don't know that: f x=1, then the answer is NO but if x>1, then the answer is YES. Not sufficient.

(2) The average age employees in X, Y, and Z, is 40, 20, and 50, respectively. (total age)=40*3x+20*4x+50*8x=600x, so the answer to the question is NO. Sufficient.

------------------------------------

Hi Bunuel,

Doubt in this Q. Pls help & suggest where am I going Wrong in this Question ......

Q .. 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

Now, it gives the ratio of number of employees i.e.; 3:4:8 & then we can take this as real number of employees like 3 employees in Company X, 4 in Y & 8 in company Z ...therefore, we have 3+4+8=15 , 15 employees in total....& average age of employees in company X must be something, lets say its x for Company X, so the total age for company X will be 3x, similarly, for Company Y the average age is y therefore, total age for company Y will be 4y & for Company Z, the total age will be 8z. & the average of the total ages will be 3x+4y+8z/15 ...... & as the question asks, Is the average age of all employees in these companies less than 40 years ?? i.e.; 3x+4y+8z/15 <40 or 3x+4y+8z<600...... so we have to find out if 3x+4y+8z<600 ..... rephrasing complete.

now statement (1) The total age of all the employees in these companies is 600.
that means 3x+4y+8z=600 but we have to check if 3x+4y+8z<600. so it is clearly NO.

statement (2) The average age of employees in X, Y, and Z, is 40, 20, and 50, respectively
therefore, total age = 3*40+4*20+8*50 = 600. as we have to check if 3x+4y+8z<600. so it is also a clear NO. & hence , answer is D.

Now please let me know where am I going wrong. Your help will be appreciated. Thanks !!!!
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15 Apr 2013, 00:21
manishuol wrote:
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?

Given that the ratio of the number of employees is 3x:4x:8x, for some positive multiple x.

The questions asks whether (average age)=(total age)/(# of employees)<40, or whether (total age)/(3x+4x+8x)<40, which is the same as: is (total age)<600x?

(1) The total age of all the employees in these companies is 600. The question becomes: is 600<600x? Or is 1<x. We don't know that: f x=1, then the answer is NO but if x>1, then the answer is YES. Not sufficient.

(2) The average age employees in X, Y, and Z, is 40, 20, and 50, respectively. (total age)=40*3x+20*4x+50*8x=600x, so the answer to the question is NO. Sufficient.

------------------------------------

Hi Bunuel,

Doubt in this Q. Pls help & suggest where am I going Wrong in this Question ......

Q .. 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

Now, it gives the ratio of number of employees i.e.; 3:4:8 & then we can take this as real number of employees like 3 employees in Company X, 4 in Y & 8 in company Z ...therefore, we have 3+4+8=15 , 15 employees in total....& average age of employees in company X must be something, lets say its x for Company X, so the total age for company X will be 3x, similarly, for Company Y the average age is y therefore, total age for company Y will be 4y & for Company Z, the total age will be 8z. & the average of the total ages will be 3x+4y+8z/15 ...... & as the question asks, Is the average age of all employees in these companies less than 40 years ?? i.e.; 3x+4y+8z/15 <40 or 3x+4y+8z<600...... so we have to find out if 3x+4y+8z<600 ..... rephrasing complete.

now statement (1) The total age of all the employees in these companies is 600.
that means 3x+4y+8z=600 but we have to check if 3x+4y+8z<600. so it is clearly NO.

statement (2) The average age of employees in X, Y, and Z, is 40, 20, and 50, respectively
therefore, total age = 3*40+4*20+8*50 = 600. as we have to check if 3x+4y+8z<600. so it is also a clear NO. & hence , answer is D.

Now please let me know where am I going wrong. Your help will be appreciated. Thanks !!!!

Given that the ratio of the number of employees of three companies X, Y and Z is 3:4:8, respectively, so the number of employees could be:
3, 4, 8;
3*2=6, 4*2=8, 8*2=16;
3*3=9, 4*3=12, 8*3=24;
3*4=12, 4*4=16, 8*4=32;
...

Notice that the multiple is the same in each case. Thus the ratio of the number of employees is 3x:4x:8x, for some positive multiple x, not 3x:4y:8z.

Hope it's clear.
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15 Apr 2013, 10:21
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.

Hi Bunnel
Could you elaborate how this is true - median from right angle is half of the hypotenuse??
Thanks
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16 Apr 2013, 03:33
2
Dipankar6435 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.

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.

Hope it's clear.
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18 Apr 2013, 10:02
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.

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19 Apr 2013, 03:59
buffaloboy wrote:
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.

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.
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26 Nov 2013, 02:36
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.

if ABC is isosceles triangle then all sides are equal. so AC=24. why not?
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26 Nov 2013, 02:46
monirjewel 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.

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.

Hope it's clear.
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23 Dec 2013, 22:42
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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24 Dec 2013, 02:27
PUNEETSCHDV 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.

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.

All DS work/rate problems to practice: search.php?search_id=tag&tag_id=46
All PS work/rate problems to practice: search.php?search_id=tag&tag_id=66

Hope this helps.
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07 Jan 2014, 06:47
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.

Bununel, what if Set A only contains one factor "1"?
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07 Jan 2014, 06:52
1
envoy2210 wrote:
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.

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.
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08 Mar 2014, 07:50
Bunuel wrote:
PUNEETSCHDV 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.

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.

All DS work/rate problems to practice: search.php?search_id=tag&tag_id=46
All PS work/rate problems to practice: search.php?search_id=tag&tag_id=66

Hope this helps.

Bunuel..

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 :/
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08 Mar 2014, 07:57
sanjoo wrote:
Bunuel wrote:
PUNEETSCHDV 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.

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.

All DS work/rate problems to practice: search.php?search_id=tag&tag_id=46
All PS work/rate problems to practice: search.php?search_id=tag&tag_id=66

Hope this helps.

Bunuel..

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.

From (2): A+B=25.

Complete solution: new-ds-set-150653-60.html#p1211906

Hope it helps.
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09 Mar 2014, 06:38
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 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.

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 --> A²-25A-150=0 --> (A-30)(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
Re: New DS set!!!   [#permalink] 09 Mar 2014, 06:38

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