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# New DS set!!!

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10 Apr 2013, 07:10
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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: new-ds-set-150653-60.html#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: new-ds-set-150653-60.html#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: new-ds-set-150653-60.html#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: new-ds-set-150653-60.html#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: new-ds-set-150653-60.html#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: new-ds-set-150653-80.html#p1211908

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

Solution: new-ds-set-150653-80.html#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: new-ds-set-150653-80.html#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: new-ds-set-150653-80.html#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: new-ds-set-150653-80.html#p1211912

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15 Apr 2013, 01:08
Bunuel wrote:
manishuol wrote:
okay ..sir .. as you said in the first line of ur solution to this Q ...

Given that the ratio of the number of employees is 3x:4x:8x, for some positive multiple x.,, I think you have taken x as an integer so that it is flexible to say tat this ratio can be 3:4:8 or 3*2 : 4*2 : 8*2 or 6:8:16 or 9:12:24... I would like to know why can't we change this ratio of number employees to the ratio of weights of employees of the three companies...Can we change the Ratio of number of Employees to the Ratio of weights of Employees of three companies .... Pls Advise , Thanks in Advance !!

What is the ratio of weights?

Anyway, we need to find 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?

Does this makes sense?

--------------------------------------------------
I think there is some communication gap b/w us but still sir will try to make u understand what am I trying to say otherwise I will concentrate on ur solution & will try to understand....

First Ratio of # of Employees given as :::::: 3:4:8 ... so for me the # of employees in the three companies can be 3,4,8 or 6,8,16 or 9,12,24 any three numbers that satisfy the ratio 3:4:8. okay.

Second, in company X there must some average weight of employees, right ?? lets say that average weight is a1 okay.
so the total weight of employees in company X will be 3*a1 , if we take that there 3 employees in company X. or if u want you can take this as 6 but for that u must satisfy the ratio of # of employees accordingly okay.

now similarly, in company Y there must be some average weight of employees, right ?? lets say that average weight is a2 okay.
& therefore the total weight of employees in company Y will be 4*a2.

now similarly, in company Z there must some average weight of employees, right ?? lets say that average weight is a3 okay.
& therefore the total weight of employees in company Z will be 8*a3.

Now the total weight of employees in three companies will be 3*a1 + 4*a2 + 8*a3......... okay.
& if we are taking number of employees as 6,8,16, this will become 3*2*a1 + 4*2*a2 + 8*2*a3.... & in both the cases the # of employees will change but not the equation as if we take in first case

total weight of employees in three companies is 3*a1 + 4*a2 + 8*a3 so in this case the average weight will be 3*a1 + 4*a2 + 8*a3/3+4+8 ... or 3a1+4a2+8a3/15

& as the Q asks (average age)=(total age)/(# of employees)<40, or whether 3a1+4a2+8a3/15<40, or 3a1+4a2+8a3<15*40 or 3a1+4a2+8a3<600.

similarly for second scenario, if we take #of employees as 6,8,16 maintaining the same ratio of employees as 3:4:8.
in this case the total weight will be 3*2*a1 + 4*2*a2 + 8*2*a3 or 6a1+8a2+16a3 & the average will be 6a1+8a2+16a3/6+8+16
or 6a1+8a2+16a3/30

again as the Q asks is (average age)=(total age)/(# of employees)<40, or whether 6a1+8a2+16a3/30,
or 6a1+8a2+16a3<30*40 or 6a1+8a2+16a3<1200 or 2 (3a1+4a2+8a3) < 2*600 or
3a1+4a2+8a3<600 ..

but both the statements are disallowing it. so that's how I have this solution. if you think this can't be the way, so I'm really sorry to waste your time. As you are the Master of GMAT Math & you can't be wrong, I will review my logic again & concentrate on your solution to understand it better. Thanks for your precious time . Thanks !!!!
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15 Apr 2013, 02:55
manishuol wrote:
Bunuel wrote:
manishuol wrote:
okay ..sir .. as you said in the first line of ur solution to this Q ...

Given that the ratio of the number of employees is 3x:4x:8x, for some positive multiple x.,, I think you have taken x as an integer so that it is flexible to say tat this ratio can be 3:4:8 or 3*2 : 4*2 : 8*2 or 6:8:16 or 9:12:24... I would like to know why can't we change this ratio of number employees to the ratio of weights of employees of the three companies...Can we change the Ratio of number of Employees to the Ratio of weights of Employees of three companies .... Pls Advise , Thanks in Advance !!

What is the ratio of weights?

Anyway, we need to find 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?

Does this makes sense?

--------------------------------------------------
I think there is some communication gap b/w us but still sir will try to make u understand what am I trying to say otherwise I will concentrate on ur solution & will try to understand....

Second, in company X there must some average weight of employees, right ?? lets say that average weight is a1 okay.

Yes, your solution is not correct.

Also, we have the number of employees and age. What is weight???
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15 Apr 2013, 03:06
Sorry, I have made a mistake to write weight instead of age.
Anyways. Thanks for clearing my Doubts. Now, I realize, I was wrong & now will concentrate on ur solution....... I appreciate your help. Thanks a Lot Big Brother !!!!! Cheers .....

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15 Apr 2013, 05:20
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

Sol:Consider St 1, we get atleast one of the integer is positive ie. may be more than 1 integer are also postive.

Looking at possible option based on statement we have

Case 1: --+
Possible options : -2,-1 and 0 ( But Zero is neither positive nor negative so we can rule out this option)
Case 2: -++
Possible options: -1,0,1 (again 0 is neither postive nor negative and hence cannot be the option)
Case 3 :+++
1,2,3 or 3,4,5 ----Product can be anything and hence not sufficient

St 2: The sum of the integers is less than 6
Possible options

Case 1:-1,0,1
Case 2: 0,1,2
Case 3: -3,-2,-1

Again more than 1 answer is possible

Combining we get Sum< 6 and Atleast one integer is positve and hence Case 3 from St1 can be removed

But Case 3 of statement 2 is still valid and hence we can get 2 answers.

Ans should be E
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15 Apr 2013, 05:24
mridulparashar1 wrote:
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

Sol:Consider St 1, we get atleast one of the integer is positive ie. may be more than 1 integer are also postive.

Looking at possible option based on statement we have

Case 1: --+
Possible options : -2,-1 and 0 ( But Zero is neither positive nor negative so we can rule out this option)
Case 2: -++
Possible options: -1,0,1 (again 0 is neither postive nor negative and hence cannot be the option)
Case 3 :+++
1,2,3 or 3,4,5 ----Product can be anything and hence not sufficient

St 2: The sum of the integers is less than 6
Possible options

Case 1:-1,0,1
Case 2: 0,1,2
Case 3: -3,-2,-1

Again more than 1 answer is possible

Combining we get Sum< 6 and Atleast one integer is positve and hence Case 3 from St1 can be removed

But Case 3 of statement 2 is still valid and hence we can get 2 answers.

Ans should be E

There is no positive integer in case 3 (red).

Check correct solution here: new-ds-set-150653-60.html#p1211902

Hope it helps.
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15 Apr 2013, 05:32
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

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15 Apr 2013, 05:49
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
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15 Apr 2013, 05:59
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)

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
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15 Apr 2013, 09:21
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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18 Apr 2013, 09:02
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, 02:59
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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31 May 2013, 08:24
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?

(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)

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

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08 Jul 2013, 23:34
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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26 Nov 2013, 01:36
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, 01:46
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, 21:42
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, 01:27
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.

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26 Dec 2013, 14:14
Bunuel wrote:
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.

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.
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07 Jan 2014, 05:47
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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Kudos [?]: 1 [0], given: 28

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07 Jan 2014, 06:04
Bunuel wrote:
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.

OMG, why can I forget this? thank you
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Re: New DS set!!!   [#permalink] 07 Jan 2014, 06:04

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