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5 GmatPrep questions [#permalink]
13 Aug 2009, 04:45
1
This post was BOOKMARKED
00:00
A
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D
E
Difficulty:
(N/A)
Question Stats:
100% (01:42) correct
0% (00:00) wrong based on 3 sessions
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!
Re: 5 GmatPrep questions [#permalink]
13 Aug 2009, 05:10
Try uploading them as jpegs instead of docs.....that way members will be able to view the question in their explorer windows itself.... _________________
Re: 5 GmatPrep questions [#permalink]
13 Aug 2009, 05:41
Ibodullo wrote:
Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!
Q 1 .See the attached figure.
Let the sides of the triangle be a, b and c Clearly using simple trigo calculations we can see that a=b= 5 and c= 5\sqrt{2} Now lets draw perpendicular on C with the height h. This perpendicular will bisect the side C, So let X= 1/2 x C = 1/2 x 5\sqrt{2}
Re: 5 GmatPrep questions [#permalink]
13 Aug 2009, 05:47
Ibodullo wrote:
Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!
Q 2 See the attached figure.
Distance traveled = D = 2 pi R = 2pi = 2 x 3.14
where r = radius of the lake = 1 mile
Distance = speed x time Speed = 3 miles per hour time = Distance / speed =(2 x 3.14) / 3
Re: 5 GmatPrep questions [#permalink]
13 Aug 2009, 05:49
Ibodullo wrote:
Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!
Re: 5 GmatPrep questions [#permalink]
13 Aug 2009, 05:54
SOL for Q10)
Method 1: There is a coordinate geometry formula that uses matrices to give you the area of a triangle whose coordinates are known. The final formula is: If the coordinates of a triangle are A(x1,y1), B(x2,y2) & C(x3,y3), then the area of the triangle is given by, 1/2* [(x1-x2)(y2-y3) - (y1-y2)(x2-x3)]
Applying we get, A(PQR) = 1/2* [(4-0)(3-4) - (0-3)(0-7)] = 1/2 [(-4) - (21)] = 1/2 [-25] .......... ignore the '-' sign = 12.5 ANS: A
Note: This method is also useful to decide whether the given 3 points form a triangle or they are collinear. If the area turns out to be zero, then the points are collinear.
Method 2: Drop perpendiculars from point R to the X and Y axes at points S(7,0) and T(0,4) respectively. Thus OSRT is a rectangle.
Now A(PQR) = A(OSRT) - A(OQP) - A(QRT) - A(PRS) where all the three triangles are right triangles. A(PQR) = 7*4 - (3*4/2) - (7*1/2) - (3*4/2) = 12.5 ANS: A
Method 3: Find out the lengths of the three sides using distance formula and then use the formula that calculates the area of a triangle from its sides. _________________
Re: 5 GmatPrep questions [#permalink]
13 Aug 2009, 06:02
Ibodullo wrote:
Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!
The way u can represent the scores = 48, 48 , 70, 70, 70 , 80 , 80, 80, 80, 84 , 84 , 84, 84, 84, 84 , 84, 96, 96, 96,96
So clearly the Median = 84
See the figure for this question .
I don't know how to work on the 5th question may be some body can shed some light on this one.
Re: 5 GmatPrep questions [#permalink]
13 Aug 2009, 06:04
SOL for Q24
Its given that the student worked for 20 days and earned: 48$ - on 2 days 70$ - on 3 days 80$ - on 4 days 84$ - on 7 days 96$ - on 4 days
For Median of even no of amounts, we need to arrange the amounts in ascending order and take the avg of the amounts placed in 10th and 11th positions => (84 + 84)/2 = 84
Re: 5 GmatPrep questions [#permalink]
13 Aug 2009, 06:05
samrus98 wrote:
SOL for Q10)
Method 1: There is a coordinate geometry formula that uses matrices to give you the area of a triangle whose coordinates are known. The final formula is: If the coordinates of a triangle are A(x1,y1), B(x2,y2) & C(x3,y3), then the area of the triangle is given by, 1/2* [(x1-x2)(y2-y3) - (y1-y2)(x2-x3)]
Applying we get, A(PQR) = 1/2* [(4-0)(3-4) - (0-3)(0-7)] = 1/2 [(-4) - (21)] = 1/2 [-25] .......... ignore the '-' sign = 12.5 ANS: A
Note: This method is also useful to decide whether the given 3 points form a triangle or they are collinear. If the area turns out to be zero, then the points are collinear.
Method 2: Drop perpendiculars from point R to the X and Y axes at points S(7,0) and T(0,4) respectively. Thus OSRT is a rectangle.
Now A(PQR) = A(OSRT) - A(OQP) - A(QRT) - A(PRS) where all the three triangles are right triangles. A(PQR) = 7*4 - (3*4/2) - (7*1/2) - (3*4/2) = 12.5 ANS: A
Method 3: Find out the lengths of the three sides using distance formula and then use the formula that calculates the area of a triangle from its sides.
Hey Sameer,
I was actually wondering if there is a faster way of solving this one and u just came up with that ..
Re: 5 GmatPrep questions [#permalink]
13 Aug 2009, 06:13
SOL for Q32)
Mean = 8.1 SD = 0.3
Now we need to calculate the acceptable range for the observations measuring 1.5 SDs on either side of the mean => 8.1 + (0.3)*1.5 => 8.1 + 0.45 => 7.65 to 8.55
Except for 7.51 all the entries are within 1.5 SDs of the mean.
Re: 5 GmatPrep questions [#permalink]
27 Aug 2009, 15:27
nitishmahajan wrote:
Ibodullo wrote:
Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!