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Math Expert V
Joined: 02 Sep 2009
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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 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

From $$A+\frac{150}{A}=25$$ --> $$A^2-25A+150=0$$ (not A^2 - 25A - 150 = 0) --> $$(A-10)(A-15)=0$$ --> $$A=10$$ and $$B=15$$ OR $$A=15$$ and $$B=10$$.

Hope it's clear.
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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*(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.

Hi,

how R(R-1)>54 turns R>7?
Math Expert V
Joined: 02 Sep 2009
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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*(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.

Hi,

how R(R-1)>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.
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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 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.

In a circle???? where is that in the question???
Math Expert V
Joined: 02 Sep 2009
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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 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.

In a circle???? where is that in the question???

It says: IF ANY right-angled 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.
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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 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.

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 V
Joined: 02 Sep 2009
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jamescath wrote:
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 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.

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.
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Bunuel wrote:
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.

Problem with statement 2. How can 0 be a multiple of 9?
Math Expert V
Joined: 02 Sep 2009
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hannahkagalwala wrote:
Bunuel wrote:
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.

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/number-proper ... 74996.html
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hannahkagalwala wrote:
Bunuel wrote:
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.

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)

Hope this helps.
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hannahkagalwala wrote:
9. Is x^2>2x?

Is x(x-2)>0? --> isx<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.

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/number-proper ... 74996.html

Thank 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. $$x(x-2) = 0 \Rightarrow x = 0 \textrm{ or } x = 2$$

Please,refer the attached diagram. If you take any value less than 0, say x = -1 , x(x-2) = (-1)(-1 - 2) = 3 (>0) .

Similarly take any value greater than 2 , say x = 3 => x(x-2) = 3(3 - 2) = 3 (>0) .

This implies that x(x-2) > 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(x-2) = 1(1 - 2) = -1.

inequalities-trick

Hope this helps.
Attachments Inequality.jpeg [ 3.11 KiB | Viewed 5788 times ]

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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.

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 V
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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.

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 one-half 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 = one-half 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.
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Bunuel - I am a little confused with your solution -

8. Two marbles are drawn from a jar with 10 marbles. If all marbles are either red or 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.
3/10 * 2/9 = 0.066
4/10 * 3/9 = 0.133 so clearly there can be only 3 Blue or less.
So as long as we prove that there are 7 Red, shouldn't it be sufficient? 7/10 is greater than 3/5
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TarPhi wrote:
Bunuel - I am a little confused with your solution -

8. Two marbles are drawn from a jar with 10 marbles. If all marbles are either red or 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.
3/10 * 2/9 = 0.066
4/10 * 3/9 = 0.133 so clearly there can be only 3 Blue or less.
So as long as we prove that there are 7 Red, shouldn't it be sufficient? 7/10 is greater than 3/5

If R = 7, then P(R and R) = R/10*(R-1)/9 = 7/10*6/9 = 7/15 < 3/5.
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Bunuel wrote:
SOLUTION:

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.

What is the problem with 0,1,2? isn't it also an answer and thus even both together can't answer the question
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Bunuel wrote:
SOLUTION:

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.

What is the problem with 0,1,2? isn't it also an answer and thus even both together can't answer the question

{0, 1, 2} is the only set possible --> the product is 0.
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Bunuel wrote:
Bunuel wrote:
SOLUTION:

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.

What is the problem with 0,1,2? isn't it also an answer and thus even both together can't answer the question

{0, 1, 2} is the only set possible --> the product is 0.

OHHHHHHH

Product.... is multiplying.

Thank god this word didn't come out when I had my GMAT
I always consider it as Adding instead of Multiplying. My Q49 would have been heavly damaged with this mistake.

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Bunuel

I got it now! Thank you  Re: New DS set!!!   [#permalink] 19 Feb 2020, 04:12

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