SOLUTION OF 8-10
N is divisible by 8 in two cases:
When n is even:
No of even numbers (between 1 and 96)=48
AND
When n+1 is divisible by 8. -->n=8p-1 --> 8p-1<=96 --> p=12.3 --> 12 such nembers

Total=48+12=60

Probability=60/96=0.62

Answer: C

Hi Bunuel

I guess the question should be re-worded to restrict the number of draws to 2.

Winning scenario consists of TWO cases RW and WR. Probability of each case is: 3/9*2/9, so 3/9*2/9+3/9*2/9=2*3/9*2/9.

Only 3/9*2/9 would be correct answer if we were asked to determine the probability of FIRST ball being red and SECOND white, OR FIRST white and SECOND red. These are the cases in which order of drawing matters.

SOLUTION:
7. THE DISTANCE BETWEEN THE CIRCLE AND THE LINE:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0

This is tough:
First note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).

Now we can do this by finding the equation of a line perpendicular to given line y=\frac{3}{4}*x-3 (we know it should cross origin and cross given line, so we can write the formula of it), then find the croos point of these lines and then the distance between the origin and this point. But it's very lengthy way.

There is another, shorter one. Though I've never seen any GMAT question requiring the formula used in it.

We know the formula to calculate the distance between two points (x_1,y_1) and (x_2,y_2): d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} BUT there is a formula to calculate the distance between the point (in our case origin) and the line:

DISTANCE BETWEEN THE LINE AND POINT:
Line: ay+bx+c=0, point (x_1,y_1)

d=\frac{|ay_1+bx_1+c|}{\sqrt{a^2+b^2}}

DISTANCE BETWEEN THE LINE AND ORIGIN:
As origin is (0,0) -->

d=\frac{|c|}{\sqrt{a^2+b^2}}

So in our case it would be: d=\frac{|3|}{\sqrt{1^2+(\frac{3}{4})^2}}=\frac{12}{5}=2.4

So the shortest distance would be: 2.4-1(radius)=1.4

Answer: A.

OR ANOTHER APPROACH:

In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
(x-a)^2+(y-b)^2=r^2

If the circle is centered at the origin (0, 0), then the equation simplifies to:
x^2+y^2=r^2

So, the circle represented by the equation x^2+y^2 = 1 is centered at the origin and has the radius of r=\sqrt{1}=1.

Then note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).

So we should find the length of perpendicular, or the height of the right triangle formed by the X and Y axis and the line y = \frac{{3}}{{4}}x-3.

The legs would be the value of x for y=0 (x intercept) --> y=0, x=4 --> leg_1=4.
and the value of y for x=0 (y intercept) --> x=0, y=-3 --> leg_2=3.

So we have the right triangle with legs 4 and 3 and hypotenuse 5. What is the height of this triangle (perpendicular from right angle to the hypotenuse)? As perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle: \frac{height}{leg_1}=\frac{leg_2}{hypotenuse} --> \frac{height}{3}=\frac{4}{5} --> height=2.4.

Distance=height-radius=2.4-1=1.4

Answer: A.

You can check the link of Coordinate Geometry below for more.

SOLUTION:
7. THE DISTANCE BETWEEN THE CIRCLE AND THE LINE:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0

This is tough:
First note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).

Now we can do this by finding the equation of a line perpendicular to given line y=\frac{3}{4}*x-3 (we know it should cross origin and cross given line, so we can write the formula of it), then find the croos point of these lines and then the distance between the origin and this point. But it's very lengthy way.

There is another, shorter one. Though I've never seen any GMAT question requiring the formula used in it.

We know the formula to calculate the distance between two points (x_1,y_1) and (x_2,y_2): d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} BUT there is a formula to calculate the distance between the point (in our case origin) and the line:

DISTANCE BETWEEN THE LINE AND POINT:
Line: ay+bx+c=0, point (x_1,y_1)

d=\frac{|ay_1+bx_1+c|}{\sqrt{a^2+b^2}}

DISTANCE BETWEEN THE LINE AND ORIGIN:
As origin is (0,0) -->

d=\frac{|c|}{\sqrt{a^2+b^2}}

So in our case it would be: d=\frac{|3|}{\sqrt{1^2+(\frac{3}{4})^2}}=\frac{12}{5}=2.4

So the shortest distance would be: 2.4-1(radius)=1.4

Answer: A.

OR ANOTHER APPROACH:

In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
(x-a)^2+(y-b)^2=r^2

If the circle is centered at the origin (0, 0), then the equation simplifies to:
x^2+y^2=r^2

So, the circle represented by the equation x^2+y^2 = 1 is centered at the origin and has the radius of r=\sqrt{1}=1.

Then note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).

So we should find the length of perpendicular, or the height of the right triangle formed by the X and Y axis and the line y = \frac{{3}}{{4}}x-3.

The legs would be the value of x for y=0 (x intercept) --> y=0, x=4 --> leg_1=4.
and the value of y for x=0 (y intercept) --> x=0, y=-3 --> leg_2=3.

So we have the right triangle with legs 4 and 3 and hypotenuse 5. What is the height of this triangle (perpendicular from right angle to the hypotenuse)? As perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle: \frac{height}{leg_1}=\frac{leg_2}{hypotenuse} --> \frac{height}{3}=\frac{4}{5} --> height=2.4.

Distance=height-radius=2.4-1=1.4

Answer: A.

You can check the link of Coordinate Geometry below for more.

bro bunuel,
do u think we would need to knw these formulas and such a question can ever appear on gmat?
you mentioned that u have never seen any gmat question that requires us to know formula.

Hello,

Can you help me understand "Probability of a randomly selected person NOT to be born in a leap year=3/4"?

-Thanks,