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Math Expert V
Joined: 02 Sep 2009
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Re: New DS set!!!  [#permalink]

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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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Re: New DS set!!!  [#permalink]

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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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Re: New DS set!!!  [#permalink]

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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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Re: New DS set!!!  [#permalink]

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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
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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
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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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Re: New DS set!!!  [#permalink]

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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
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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)

Please read the section 4.1(Properties of Integers) from OG throughly.

Hope this helps.
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Re: New DS set!!!  [#permalink]

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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.

Please go through the inequality theory. Refer following links

inequalities-trick

Hope this helps.
Attachments Inequality.jpeg [ 3.11 KiB | Viewed 5031 times ]

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New DS set!!!  [#permalink]

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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??
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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 wrote:
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.

Edit: Sorry silly question, got it.
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Hi Bunuel,

Sorry to bring back this question :-

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+ry=divisor∗quotient+remainder=xq+r and 0≤r<x0≤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=9r=9, since given equation is very similar to y=divisor∗quotient+remainder=xq+ry=divisor∗quotient+remainder=xq+r . But we don't know whether y>9y>9: remainder must be less than divisor.

For example:
If x=10x=10 and y=1y=1 then 10=1∗1+910=1∗1+9, then the remainder upon division 10 by 1 is zero.
If x=11x=11 and y=2y=2 then 11=1∗2+911=1∗2+9, then the remainder upon division 11 by 2 is one.
Not sufficient.

(1)+(2) From (2) we have that x=qy+9x=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.

I have a doubt in the above part of the sentence which is in bold i.e.

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+ry=divisor∗quotient+remainder=xq+r and 0≤r<x0≤r<x.
Please correct me if I am wrong if
x/y then x= y*q+r, were q = quotient, r=remainder, y = divisor, x = dividend

the rest of the part of the solution I understood, but the above sentence confused me .
Can you please tell me what I am missing here?
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