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Re: If one of the angles of a parallelogram is 120 degrees, [#permalink]

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22 Aug 2012, 13:40

Its simple actually. If one angle is 120, the adjacent angle has to be 60 (parallelogram rule).

Draw a parallelogram and draw diagonals. Apply 30-60-90 angle rules and you will get the answer. Catch is, you are asked ratio of the diagonal lengths, not the actual lengths.

Re: A train passes a man standing on a platform in 9 sec. and pa [#permalink]

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22 Aug 2012, 20:32

1

This post received KUDOS

What is the approximate area of a circle that has an equilateral triangle with an area of 4√3 inscribed in it?

Image attached O is the center of the circle and triangle ABC is equilateral. AD is the angle bisector and is perpendicular bisector of the other side...

So, AO:OD = 2:1 (Radius = 2x) In Triangle ADB, applying Pythagorean theorem we have a^2 = (3x)^2 + (a/2)^2

=> x = a/(2*(sqrt 3))

given that area of triangle = 4 sqrt3 = (sqrt 3)*a^2/4

so, a = 4

so, x = a/(2*(sqrt 3)) = 2/(sqrt 3) radius = 2x = 4/ (sqrt3)

Re: Two men plan to run in opposite directions around a park [#permalink]

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23 Aug 2012, 19:45

NYC5648 wrote:

Two men plan to run in opposite directions around a park 1 km long. The first man can run 15 meters in a minute and the second can run 25 meters per minute. How much further will the second runner have run at the point they cross paths?

Official answer is: 250m

Many thanks!!!

The distance covered by the first man is S and the one run by the second man is 1000-S (meters). We need to find out X=1000-2S. Their time is equal hence s/15 = (1000-s)/25 5s=3000-3s s=375 1000-2S=1000-750=250

The question is not correctly worded. See the attached drawing. \(x\) and \(y\) can be any positive numbers. If one of the angles is 120, the acute angle then is 60. We have 30-60-90 right triangles.

The longer diagonal squared is \((2x+y)^2+3x^2\) and the shorter one squared is \(3x^2+y^2.\) Unless you know the ratio \(x:y,\) you cannot find the ratio of the diagonals. In case \(x = y\), the requested ratio is \(\sqrt{\frac{4}{12}}=\sqrt{\frac{1}{3}}=1:\sqrt{3},\) which seems to be the given answer.

What is the source of this question? BTW, a parallelogram doesn't have diagonals outside itself...

Attachments

Parallelogram60-120.jpg [ 14.5 KiB | Viewed 1618 times ]

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PhD in Applied Mathematics Love GMAT Quant questions and running.

Working alone, Andy can do a job in 20 days and Beth can do [#permalink]

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25 Aug 2012, 11:37

Working alone, Andy can do a job in 20 days and Beth can do the same job in 30 days. When the two work together Andy has to leave 5 days before the job is finished. If Beth finishes the rest of the job herself, how long does the entire job take?

Mike and Tina can complete a job in 5 days working together. [#permalink]

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25 Aug 2012, 12:00

Mike and Tina can complete a job in 5 days working together. If Mike worked twice as efficiently as he did and Tina worked one third as efficiently as she did, the job would have been completed in only three days. How long does Tina need to do the job alone?

Re: Working alone, Andy can do a job in 20 days and Beth can do [#permalink]

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25 Aug 2012, 12:07

NYC5648 wrote:

Working alone, Andy can do a job in 20 days and Beth can do the same job in 30 days. When the two work together Andy has to leave 5 days before the job is finished. If Beth finishes the rest of the job herself, how long does the entire job take?

Can anybody help me out! I must do something wrong. Here is how I did it:

Andy: 1/20 Beth: 1/30

1/20x + 1/30x = 1

X= 1/12 (Rate of Andy and Beth)

5 days before the job is finished means that Andy and Beth worked 7 days together which still leaves 5/12 of the job to do.

Beth is working alone now which means:

1/30x = 5/12

Result: Additional 12,5

Unfortunately the wrong answer. Could somebody please help me?

Thanks!!!!

Your mistake is the assumption that the job took 12 days to finish. If Andy left 5 days before the job was finished and Beth had to work alone until the end, it means that the project took more than 12 days.

If you denote by T the time you are looking for, the following equation can be written: \(\frac{T-5}{12}+\frac{5}{30}=1,\) as \(T-5\) days they worked together at a constant common rate of 1/12, then Beth worked 5 more days to finish the job alone.

Solving, you get \(T=15.\)
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PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Mike and Tina can complete a job in 5 days working together. [#permalink]

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25 Aug 2012, 12:31

1

This post received KUDOS

NYC5648 wrote:

Mike and Tina can complete a job in 5 days working together. If Mike worked twice as efficiently as he did and Tina worked one third as efficiently as she did, the job would have been completed in only three days. How long does Tina need to do the job alone?

4 men and 3 women can do a job in 8 days. If 6 men and 8 w [#permalink]

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25 Aug 2012, 14:09

4 men and 3 women can do a job in 8 days. If 6 men and 8 women work on the same job, they will complete the job in half the time. How long will it take if one man and one woman work on the job?

Re: 4 men and 3 women can do a job in 8 days. If 6 men and 8 w [#permalink]

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25 Aug 2012, 23:08

NYC5648 wrote:

4 men and 3 women can do a job in 8 days. If 6 men and 8 women work on the same job, they will complete the job in half the time. How long will it take if one man and one woman work on the job?

The question is not correctly worded. See the attached drawing. \(x\) and \(y\) can be any positive numbers. If one of the angles is 120, the acute angle then is 60. We have 30-60-90 right triangles.

The longer diagonal squared is \((2x+y)^2+3x^2\) and the shorter one squared is \(3x^2+y^2.\) Unless you know the ratio \(x:y,\) you cannot find the ratio of the diagonals. In case \(x = y\), the requested ratio is \(\sqrt{\frac{4}{12}}=\sqrt{\frac{1}{3}}=1:\sqrt{3},\) which seems to be the given answer.

What is the source of this question? BTW, a parallelogram doesn't have diagonals outside itself...

The question is not correctly worded. See the attached drawing. \(x\) and \(y\) can be any positive numbers. If one of the angles is 120, the acute angle then is 60. We have 30-60-90 right triangles.

The longer diagonal squared is \((2x+y)^2+3x^2\) and the shorter one squared is \(3x^2+y^2.\) Unless you know the ratio \(x:y,\) you cannot find the ratio of the diagonals. In case \(x = y\), the requested ratio is \(\sqrt{\frac{4}{12}}=\sqrt{\frac{1}{3}}=1:\sqrt{3},\) which seems to be the given answer.

What is the source of this question? BTW, a parallelogram doesn't have diagonals outside itself...

Refer to the drawing attached to my previous post.

The longer diagonal is the one slanted to the right having one leg \(x+y+x\) (last \(x\) is dashed), other leg is the height of the parallelogram and is \(x\sqrt{3}\). Use Pythagoras. The other diagonal is slanted to the left, one leg, vertical, again the height of the parallelogram and the other leg is the side of the parallelogram \(y.\) Use again Pythagoras.
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PhD in Applied Mathematics Love GMAT Quant questions and running.

Oh, I checked the answer and I think this is what the formula should be. But the question must be what does \(a_{99}\) and \(a_1\) equal?

The given formula can be written as \(a_{n+1}=a_{n}-1\). It means that we have an arithmetic progression with a constant difference between consecutive terms of \(-1\). Or forget about the mathematical jargon, we have a sequence of evenly spaced numbers, the difference between each term in the sequence and the previous one being \(-1\). Then \(a_2=a_1-1, \, a_3=a_2-1=a_1-2\,\) ...\(a_{99}=a_1-98.\) The sum of any number of consecutive terms in such a sequence equals \(the \, average \, of \, the \, terms * the \, number \, of \, terms \, in \, the \,sum.\) And the average of the terms always equals the average of the first and the last term. Therefore, the sum being 99, we have \(\frac{a_1+a_{99}}{2}*99=\frac{a_1+a_1-98}{2}*99=99\) . Solving we get \(a_1=50\) so \(a_{99}=a_1-98=50-98=-48.\)
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PhD in Applied Mathematics Love GMAT Quant questions and running.

Can anyone please explain a step-by-step solution. I guess I am missing a concept here.

Many thanks!!!

use the equilateral triangle area formula to get the length of the sides sqrt(3)/4 * a^2 = 4sqrt(3) we find that a=4 Now the next step I drew a line bisecting the triangle to form a right triangle and solve for the hypotenuse which is the diameter. Using the 30-60-90 triangle numbers i get 8/sqrt(3) as the diameter and then solve for the area of the circle which comes to approximately 15.9 or round up to 16.

Apply Pythagoras starting with the triangle with sides 1 and 1. The hypotenuse is \(\sqrt{2}.\) In the next right triangle, the hypotenuse is \(\sqrt{3}\) ... the longest hypotenuse is \(\sqrt{12}=\sqrt{2^2\cdot3}=2\sqrt{3}.\)
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gmatclubot

Re: In the figure to the right what is the length of the longest
[#permalink]
28 Aug 2012, 13:42

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