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Can somebody please help with solutions to these questions

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Re: Can somebody please help with solutions to these questio

by jayhawk2001 » Sun May 20, 2007 11:34 am
Ramesh2007 wrote:Please somebody help me with the solution to these questions.
Can you please post the questions individually in separate threads.

The discussion will benefit a few folks on this forum.
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Here are my questions

by Ramesh2007 » Sun May 20, 2007 12:25 pm
Thanks Jayhawk. I have them individually here.
I will post the rest of the questions in another thread
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by jayhawk2001 » Sun May 20, 2007 9:19 pm
1. Ans = 133

Compute num students 'x' as

5/x = 3/80
x = 133.33

Now, the ratio of TA to students should be > 3/80. So, in the fraction
5/133.333, either 5 should increase or 133.333 should decrease.

133 hence gives you the absolute max

------------------------------------------------------------------------------------
2. Ans = 10

tn = tn-1 - 3
tn = tn-2 - 2*3
...
tn = t1 - (n-1)*3

-4 = 23 - (n-1)*3
n-1 = 9
n= 10
------------------------------------------------------------------------------------

3. E

Use POE to solve this.

xy + z = xy + xz

Now, if z = 0, we have xy = xy which means x and y can be of any
value. Eliminate A and C

If x = 1, we have y+z = y+z. Again, y and z can be any value
to satisfy above equation. So, eliminate B and D

That leaves us with E.
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by Cybermusings » Sun May 20, 2007 11:31 pm
xy + z = xy + xz
If z = 0 then xz = z = 0 and LHS = RHS = xy
If x = 1 then z = xz and LHS = RHS = xy + z
Hence Choice E
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by Cybermusings » Sun May 20, 2007 11:42 pm
It's important to remember in Q3 that the ratio shoulbe be greater than 3:80; If the ratio asked was equal to 3:80 then for every teaching assistant there would be 80/3 = 26.67 students; Hence if there were 5 assistants there would be 80/3 * 5 or 133.333 students for every teaching assiatant. Since we require a ratio greater than 3:80 there should be lesser students per assistant (that's the only way we can increase the ratio); hence answer would be 133
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by Cybermusings » Sun May 20, 2007 11:48 pm
Perimeter = 16 + 16 * sqr rt. 2
For an isoceles right angle triangle the sides are in the ratio 1:1:sqr. rt 2
So the perimeter would be x + x + (x*sq rt. 2)
2x + x*sqr rt. 2 = 16 + 16sqr rt. 2
Taking x*sqr rt. 2 common
x*sqr rt. 2 (sqr rt. 2 + 1) = 16 (1 + sqr rt.2)
x*sqr rt. 2 = 16
Hence hypotenuese = 16 (since hypotenuese = x* sqr rt. 2)
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by Cybermusings » Mon May 21, 2007 1:01 am
Statement I : we can make out from this statement that none of the white balls had even numbers on them or P(x and y) = 0. However, this Statement fails to give us the total number of white balls...hence insufficient
Statement II : Say probability of ball being white = P(x); Probability that ball is even = P(y); P(x) - P(y) = .2
We need to Find Probability of P(x) U P(y)
P(x) U P(y) = P(x) + P(y) - P(x and y); Insufficient

Statement I and II: Even after combining the 2 we only know P(x and y); hence E
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by Cybermusings » Mon May 21, 2007 1:10 am
Statement I : From this we have two pairs of co-ordinates for line l; hence we can find the slops of line l
y = mx+c ; m = y2-y1 / x2-x1 = 2/1 = 2
So the slope of the other line should be -1/2 for product of the 2 slopes to be 0; hence insufficient (we don't know the slopes of line k)
Statement II: Co-ordinates of line k are (4,0) and (0,2). Hence slope = 2-0/0-4 = 2/-4 = -1/2; Hence insufficient (since we don't know the slope of line l)
Statement I and II : After combining the 2 we can deduce that the product of the slopes of the two lines is -1

Hence C
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by Cybermusings » Mon May 21, 2007 1:14 am
arithmetic mean = (6*2 + 7*4 + 8*7 + 9*9 + 10*3) / 25 = (12+28+56+81+30)/25 = 8.28
Hence 8.3 should be the answer
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by Cybermusings » Mon May 21, 2007 1:15 am
Different combinations = 4C2 * 3C2 = 6*3 = 18
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by Cybermusings » Mon May 21, 2007 1:18 am
Let length of shorter piece be x and that of the longer piece be y

Statement I : y = x + 20 ; Insufficient since we have two unknowns and 1 equation

Statement II : x = y/3 ; Insufficient since we have 2 unknowns and 1 equation

Statment I & II : We can solve the 2 equations simultaneously or even use substitution; Hence C should be the answer
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