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GMAT Qs

by msd_2008 » Mon Jun 30, 2008 3:31 am
Hello,

Please help me with the following questions (with step by step explanation)

Q1. Right triangle PQR is to be constructed in a xy-plane so that the right angle is at P and PR is parallel to X axis. The x and y coordinates of P, Q, R are to be integers that satisfy the inequalities -4<=x<=5 and 6<=y<=16. How many different triangles with these properties could be constructed?

a. 110 b. 1,100 c. 9,900 d. 10,000 e. 12,100

Q2. John and Mary were each paid x dollars in advance to do a certain job together. John worked on the job for 10 hrs and Mary worked for 2 hours less than him. If Mary gave John y dollars of her payment so that they receive the same hourly wage, what was the dollar amount, in terms of y, that John was paid in advance?

a. 4y b.5y c.6y d.8y e.9y

Q3. Seed mixture X is 40% ryegrass and 60% bluegrass; seed mixture Y is 25% ryegrass and 75% fescue. If a mixture of X and Y contains 30% ryegrass, what percent of the weight of the mixture is X?

a. 10% b.33 1/3% c.40% d.50% e.66 2/3%

Q4. A certain junior class has 1000students and a senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior, 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?

a. 3/40,000 b.1/3,600 c.9/2,000 d.1/60 e.1/15

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by AleksandrM » Mon Jun 30, 2008 9:34 am
Q2. John and Mary were each paid x dollars in advance to do a certain job together. John worked on the job for 10 hrs and Mary worked for 2 hours less than him. If Mary gave John y dollars of her payment so that they receive the same hourly wage, what was the dollar amount, in terms of y, that John was paid in advance?

a. 4y b.5y c.6y d.8y e.9y

x + y/10 = x - y/8

8x + 8y = 10x - 10y

18y = 2x

x = 9y
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by AleksandrM » Mon Jun 30, 2008 9:40 am
Not sure about this one, but

Q3. Seed mixture X is 40% ryegrass and 60% bluegrass; seed mixture Y is 25% ryegrass and 75% fescue. If a mixture of X and Y contains 30% ryegrass, what percent of the weight of the mixture is X?

a. 10% b.33 1/3% c.40% d.50% e.66 2/3%

.4x + .25y = .3(x + y)

x = 5/10 or .50, so 50%
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by AleksandrM » Mon Jun 30, 2008 9:46 am
Not too sure, but here goes:

Q4. A certain junior class has 1000students and a senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior, 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?

a. 3/40,000 b.1/3,600 c.9/2,000 d.1/60 e.1/15

Since there are 60 pairs, that means there are 120 students so:

60/1000 x 60/800 = 6/100 x 6/80 = 36/8000 = 9/2000.
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by szapiszapo » Tue Jul 01, 2008 6:34 am
I believe the answer to Q1 is c).

First idea is that you have 4 different representations of a right triangle with the right angle being at P. Indeed, PR is parallel to X axis but either xP <xR> xR. Similarly for the Y axis, xP <xQ> xQ.


For a given triangle representation (hereunder defined by xP<XR and xP<xQ), you have the following:

Given that -4<=x<=5 , the PR length can have 9 different values, from a length of 1 to a length of 9.
For PR=1, that straight line can be positioned in 9 different places (i.e. xP=-4 and xR=-3, or xP=-3 and xR=-2, ...).
For PR=2, the straight line can be positioned in 8 different places, and so on.
Therefore, you have a total number of positions of 9+8+7+6+5+4+3+2+1 = 45

Given that 6<=y<=16, the PQ length can take 10 different values, from a length of 1 to a length of 10.
Reasoning in the same manner than above, you have a total number of positions of 10+9+8+7+6+5+4+3+2+1 = 55

Combining the two figures (each for one axis) gives you a total number of possible triangles of 45 * 55 = 2,475 triangles, under a single representation.



Multiplying this value by 4 gives you 2,475 x 4 = 9,900, answer c).


Hope I made myself understood...
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by beeparoo » Wed Jul 02, 2008 4:30 pm
msd_2008 wrote:Q3. Seed mixture X is 40% ryegrass and 60% bluegrass; seed mixture Y is 25% ryegrass and 75% fescue. If a mixture of X and Y contains 30% ryegrass, what percent of the weight of the mixture is X?

a. 10% b.33 1/3% c.40% d.50% e.66 2/3%
AleksandrM wrote:Not sure about this one, but

.4x + .25y = .3(x + y)

x = 5/10 or .50, so 50%
You forgot the last step, which is to determine the percentage (by weight) of x in the total mixture. Incidentally, you also expressed x incorrectly. Let's rephrase:

x = (5/10)*y or .50y

Thus, x/(x+y) = x/(x + 2x) = 1/3 or 33 1/3%
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Thanks for the solution but I still have a question....

by msd_2008 » Wed Jul 02, 2008 10:55 pm
szapiszapo wrote:I believe the answer to Q1 is c).

First idea is that you have 4 different representations of a right triangle with the right angle being at P. Indeed, PR is parallel to X axis but either xP <xR> xR. Similarly for the Y axis, xP <xQ> xQ.


For a given triangle representation (hereunder defined by xP<XR and xP<xQ), you have the following:

Given that -4<=x<=5 , the PR length can have 9 different values, from a length of 1 to a length of 9.
For PR=1, that straight line can be positioned in 9 different places (i.e. xP=-4 and xR=-3, or xP=-3 and xR=-2, ...).
For PR=2, the straight line can be positioned in 8 different places, and so on.
Therefore, you have a total number of positions of 9+8+7+6+5+4+3+2+1 = 45

Given that 6<=y<=16, the PQ length can take 10 different values, from a length of 1 to a length of 10.
Reasoning in the same manner than above, you have a total number of positions of 10+9+8+7+6+5+4+3+2+1 = 55

Combining the two figures (each for one axis) gives you a total number of possible triangles of 45 * 55 = 2,475 triangles, under a single representation.



Multiplying this value by 4 gives you 2,475 x 4 = 9,900, answer c).


Hope I made myself understood...
I havent understood the concept of 4 representations i.e what do you mean by 'Indeed, PR is parallel to X axis but either xP <xR> xR. Similarly for the Y axis, xP <xQ> xQ'.
Can you please explain in more simpler terms? I really wish to understand this solution correctly.....

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MSD
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Are you sure about number 2?

by evansbd » Thu Jul 03, 2008 5:23 am
:wink:

I've followed the CRACKING THE GMAT guide to questions like this by plugging in an original number then solving the answer choices for the solution given by your particular number.

For example.

John and Mary were each paid X dollars in advance.

Let x = 80$

John worked 10 hrs. Mary worked 2 hrs less (8 hrs)

John 80$ for 10hrs ---> 8$/hr

Mary 80$ for 8hrs ---> 10$/hr

Mary gives John Y dollars of her payment so they recieve the same hourly wage.

Mary ---> lower her wage to 8$/hr ---> 8$*8hrs ---> 64 dollars.

She would need to give John 80 - 64 = 16 dollars.

Find x (80) in terms of y (16). 80 = 16*5 . So the answer is 5y.
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Re: Thanks for the solution but I still have a question....

by szapiszapo » Thu Jul 03, 2008 5:47 am
Re. question 1

well there has been a mistake when submitting my post, I should have disabled HTML, BBCOde and Smilies, it then works better. Anyway
I meant:


xP being the position of point P on the X axis;
xR being the position of point R on the X axis;
yP being the position of point P on the Y axis;
yR being the position of point R on the Y axis;

PR is parallel to the X axis and xP can be < xR or > xR as we have no information of the position of the P and R points along the X axis. Visually, the right angle can right-sided or left-sided.

Similarly, PQ is parallel to the Y axis and yP can be <yQ or > yQ. Visually, the right angle can be up-sided or down-sided.

therefore we have 4 different possible representations of that right triangle PQR.
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by AleksandrM » Thu Jul 03, 2008 8:31 am
80 + 16/10 does not equal 80 - 16/8

96/10 = 9.6

64/8 = 8
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Re: GMAT Qs

by manju_ej » Fri Jul 04, 2008 3:03 am
Q4. A certain junior class has 1000students and a senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior, 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?

a. 3/40,000 b.1/3,600 c.9/2,000 d.1/60 e.1/15

There are 60 sibling pairs, that is , 60 juniors + 60 seniors .
So the probability of selecting one student from 1000 students in the junior class, which satisfies the condition that he has to be one in the sibling pair is
60 juniors from 1000 which is equal to 60/1000
So the probability of selecting the matching sibling from the senior class is 1 in 800 that is 1/800.

Therefor, the probability that the 2 students selected will be a sibling pair
60/1000*1/800=3/40000

So the answer is A
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Question 2....

by evansbd » Mon Jul 07, 2008 6:24 am
Can someone confirm or disprove that the answer to problem 2 is B?
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Confirm

by evansbd » Tue Jul 08, 2008 8:54 am
just looking for confirmation...
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Re: Confirm

by msd_2008 » Tue Jul 08, 2008 10:58 pm
evansbd wrote:just looking for confirmation...
The correct answer to Q.2 is (e) which is 9y

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Re: Confirm

by msd_2008 » Tue Jul 08, 2008 10:58 pm
evansbd wrote:just looking for confirmation...
The correct answer to Q.2 is (e) which is 9y

Regards
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