GMAT prep question

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sarahw_gmat: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Fri Mar 12, 2010 1:39 pm; Thanked: 1 times

by sarahw_gmat » Thu Feb 03, 2011 8:31 pm

Could someone please help with below questions? Thanks.

Q. if ab <> 0 and points ( -a,b) and ( -b,a) are in the same quadrant of the xy-plane, is point ( -x,y) in this same quadrant?
1. xy > 0
2. ax > 0

and : C

Q. the lifetimes of all the batteries produced by a certain company in a year have a distrubityion that is symmetric about the
mean m. If the distribution has a standard deviation of d, what percent of the distribution is greater than m+d?

1. 68 percent of the distribution lies in the interval from m-d to m+d, inclusive.
2. 16 percent of the distribution is less than m-d.

Ans : D

Q. if x and y are positive, which of the following must be greater than 1/root( x+ y ) ?

1. root( x+y)/2x
2. ( root(x) + root(y) )/ x+y
3. (root(x) - root(y))/ x+y

Ans : II only

Q. If w+x < 0, is w-y > 0 ?

1. x+y < 0
2. y < x < w

Ans : B

Q. If x and y are integers and x>0, is y>0 ?
1. 7x-27 > 0
2. -y < x

And. E

Q In the figure shown, point O is the center of the semicircle and points B,C, and D lie on the semicircle. If the length of line segment AB is equal to the length of line segment OC, what is the degree measure of angle BAO?

1. The degree measure of angle COD is 60.
2. The degree measure of angle BCO is 40.

Ans : d( each sufficient )

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Source: Beat The GMAT — Data Sufficiency | Thu Feb 03, 2011 8:31 pm

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Thu Feb 03, 2011 10:41 pm

If ab â‰ 0 and points (-a, b) and (-b, a) are in the same quadrant of the xy-plane, is point (-x, y) in this same quadrant?
1. xy > 0
2. ax > 0

Refer to this post : https://www.beatthegmat.com/quadrant-pro ... tml#310294

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by Anurag@Gurome » Thu Feb 03, 2011 10:57 pm

sarahw_gmat wrote:The lifetimes of all the batteries produced by a certain company in a year have a distribution that is symmetric about the mean m. If the distribution has a standard deviation of d, what percent of the distribution is greater than m+d?

1. 68 percent of the distribution lies in the interval from m-d to m+d, inclusive.
2. 16 percent of the distribution is less than m-d.

As the "distribution is symmetric about mean", the number of data points above the mean is same as that below the mean. Also they are same at the same distance on the both side of the mean. Therefore the curve of the distribution will look something like this :

Hence,

1. [% distribution < (m - d)] = [% distribution > (m + d)]
2. [% distribution between (m - d) and m] = [% distribution between m and (m + d)]

Statement 1: 68% in the interval (m - d) to (m + d), inclusive.
Therefore, 32% lies outside this interval.
Hence, [% distribution < (m - d)] + [% distribution > (m + d)] = 32
=> 2*[% distribution > (m + d)] = 32
=> [% distribution > (m + d)] = 16

Sufficient

Statement 2: [% distribution < (m - d)] = 16
Hence, [% distribution > (m + d)] = 16

Sufficient

The correct answer is D.

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by Anurag@Gurome » Thu Feb 03, 2011 11:06 pm

sarahw_gmat wrote:If x and y are positive, which of the following must be greater than 1/âˆš(x + y) ?

1. âˆš(x + y)/2x
2. (âˆšx + âˆšy)/(x + y)
3. (âˆšx - âˆšy)/(x + y)

Let's see what happens when each of the given expressions is greater than 1/âˆš(x + y), i.e. let's find the conditions for which they will be great than 1/âˆš(x + y).

I. âˆš(x + y)/2x

.... âˆš(x + y)/2x > 1/âˆš(x + y)
=> (x + y) > 2x ...................... Cross-multiplication
=> y > x

y > x is a particular situation. Thus (I) is not always greater than 1/âˆš(x + y).

II. (âˆšx + âˆšy)/(x + y)

.... (âˆšx + âˆšy)/(x + y) > 1/âˆš(x + y)
=> (âˆšx + âˆšy) > (x + y)/âˆš(x + y)
=> (âˆšx + âˆšy) > âˆš(x + y)
=> (âˆšx + âˆšy)Â² > (x + y) ....................... Squaring both sides
=> (x + 2âˆš(xy) + y) > (x + y)
=> 2âˆš(xy) > 0
=> âˆš(xy) > 0

âˆš(xy) > 0 is always true as x and y are not zero. Thus (II) is always greater than 1/âˆš(x + y).

III. (âˆšx - âˆšy)/(x + y)

.... (âˆšx - âˆšy)/(x + y) > 1/âˆš(x + y)
=> (âˆšx - âˆšy) > (x + y)âˆš(x + y)
=> (âˆšx - âˆšy) > âˆš(x + y)
=> (âˆšx - âˆšy)Â² > (x + y) ....................... Squaring both sides
=> (x - 2âˆš(xy) + y) > (x + y)
=> -2âˆš(xy) > 0
=> âˆš(xy) < 0

But âˆš(xy) can't be less than zero as âˆš(xy) > 0 as x and y are not zero. Thus (III) is never greater than 1/âˆš(x + y).

Thus only (II) is a must be greater case.

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by Anurag@Gurome » Thu Feb 03, 2011 11:12 pm

sarahw_gmat wrote:If (w + x) < 0, is (w - y) > 0 ?

1. (x + y) < 0
2. y < x < w

Statement 1: (x + y) is a negative quantity.
Hence, -(x + y) is positive.
Now, (w - y) = (w + x) + [-(x + y)] = (Some negative) + (Some positive) = Can be negative or Zero or positive

Not sufficient

Statement 2: y is less than w.
Hence, (w - y) is greater than zero.

Sufficient

The correct answer is B.

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Thu Feb 03, 2011 11:18 pm

sarahw_gmat wrote:If x and y are integers and x > 0, is y > 0?
1. (7x - 27) > 0
2. -y < x

Statement 1: x > (27/7)
But no information about y.

Not sufficient

Statement 2: -y < x
Take x = 5

For y = 2 : -y < x
For y = -2 : -y < x

Not sufficient.

1 & 2 Together: Take the same examples as of statement 2.

Not sufficient

The correct answer is E.

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by Anurag@Gurome » Fri Feb 04, 2011 12:01 am

sarahw_gmat wrote:In the figure shown, point O is the center of the semicircle and points B,C, and D lie on the semicircle. If the length of line segment AB is equal to the length of line segment OC, what is the degree measure of angle BAO?

1. The degree measure of angle COD is 60.
2. The degree measure of angle BCO is 40.

Refer to the figure below,

OC is the radius of the circle.
Hence, AB = OC implies, AB = OC = OD = OB

Hence, triangle ABO is isosceles with AB = OB.
Hence, angle BAO = angle BOA = x (say)
Hence, angle ABO = (180 - 2x)

Now on straight line AC, angle ABO = (180 - 2x)
Hence, angle CBO = 180 - (180 - 2x) = 2x

Again triangle CBO is isosceles with OB = OC
Hence, angle BCO = CBO = 2x
Hence, angle BOC = (180 - 4x)

Now on straight line AD, (angle AOB + angle BOC + angle COD)= 180
Hence, (x + (180 - 4x) + angle COD) = 180
=> angle COD = 3x

Statement 1: angle COD = 3x = 60
Hence, angle BAO = x = 20

Sufficient

Statement 2: angle BCO = 2x = 40
Hence, angle BAO = x = 20

Sufficient

The correct answer is D.

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sarahw_gmat: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Fri Mar 12, 2010 1:39 pm; Thanked: 1 times

by sarahw_gmat » Mon Feb 07, 2011 3:47 pm

Thanks Anurag for all your time and awesome explanations.

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vaflaly: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Thu Oct 21, 2010 8:40 am

by vaflaly » Tue Feb 08, 2011 2:10 pm

Anurag@Gurome wrote:
sarahw_gmat wrote:In the figure shown, point O is the center of the semicircle and points B,C, and D lie on the semicircle. If the length of line segment AB is equal to the length of line segment OC, what is the degree measure of angle BAO?

1. The degree measure of angle COD is 60.
2. The degree measure of angle BCO is 40.
Refer to the figure below,

OC is the radius of the circle.
Hence, AB = OC implies, AB = OC = OD = OB

Hence, triangle ABO is isosceles with AB = OB.
Hence, angle BAO = angle BOA = x (say)
Hence, angle ABO = (180 - 2x)

Now on straight line AC, angle ABO = (180 - 2x)
Hence, angle CBO = 180 - (180 - 2x) = 2x

Again triangle CBO is isosceles with OB = OC
Hence, angle BCO = CBO = 2x
Hence, angle BOC = (180 - 4x)

Now on straight line AD, (angle AOB + angle BOC + angle COD)= 180
Hence, (x + (180 - 4x) + angle COD) = 180
=> angle COD = 3x

Statement 1: angle COD = 3x = 60
Hence, angle BAO = x = 20

Sufficient

Statement 2: angle BCO = 2x = 40
Hence, angle BAO = x = 20

Sufficient

The correct answer is D.

@Anurag@Gurome

Is it possible to use the formula:

BAO = 1/2 (COD-BOA)

also BAO=BOA=x

1----> COD=60 and
x=1/2 (60-x)
x=20

2-----> BCO=40 => BOA+COD=80 => COD=80-x
x=1/2(80-x-x)

x=20

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