math problem

[steven_ghoos]

Can anyone explain the solution to below math problem?

Stocks are designated by a 4-letter or 5-letter code using the 26 letters of the alphabet.

What are the maximum number of codes?

[Anurag@Gurome]

Stocks are designated by a 4-letter or 5-letter code using the 26 letters of the alphabet.

What are the maximum number of codes?

Possible number of four letter codes = 26^4
Possible number of five letter codes = 26^5

Total number of possible codes = (26^4 + 26^5) = 26^4(1 + 26) = 27*(26^4)

[steven_ghoos]

Thanks!

Could you also help me with the following questions:

Coordinate plane with circle which' centre, 0, is also the origin of the plane. There is a sector in the circle, P0Q, with central angle P0Q of 90°. P lies in quadrant II and is equal to (- V3, 1). Q lies in quadrant I and is equal to (s,t).

What is value of s? (ps: I used V to denote the square root)

Another one:

Square ABCD is inscribed in a circle, which is inscribed in another (bigger) square.

What is the ratio of the area of the larger square to the area of the smaller square?

Last one - data sufficiency:

Triangle ABC is inscribed in a circle.

Is the area of triangle ABC greater than 40 percent of the area of the circle?

a) Triangle ABC is an equilateral triangle
b) The perimeter of triangle ABC is 9

[Anurag@Gurome]

Coordinate plane with circle which' centre, 0, is also the origin of the plane. There is a sector in the circle, P0Q, with central angle P0Q of 90°. P lies in quadrant II and is equal to (- V3, 1). Q lies in quadrant I and is equal to (s,t).

What is value of s?

Refer to this post: https://www.beatthegmat.com/gmat-prep-qu ... tml#326094

[Anurag@Gurome]

Square ABCD is inscribed in a circle, which is inscribed in another (bigger) square.

What is the ratio of the area of the larger square to the area of the smaller square?

Refer to the image above.

In the small right-angle triangle with r as hypotenuse,

• r² = (a² + a²) = 2a²

Length of each side of the smaller square = 2a
Area of the smaller square = (2a)² = 4a²

Length of each side of the larger square = 2r
Area of the larger square = (2r)² = 4r²

Required ratio = (4r²/4a²) = (r²/a²) = (2a²/a²) = 2

[Anurag@Gurome]

Triangle ABC is inscribed in a circle. Is the area of triangle ABC greater than 40 percent of the area of the circle?
a) Triangle ABC is an equilateral triangle
b) The perimeter of triangle ABC is 9

Statement 1: Triangle ABC is an equilateral triangle
Refer to this image:


Clearly there is a relation between the sides of the triangle (a) and radius of the circle (r). Therefore there is also a relation between the area of the triangle and are of the circle. Thus we can easily determine whether the area of triangle is greater than 40 percent of the area of the circle or not.

Sufficient.

Statement 2: The perimeter of triangle ABC is 9
From this information we can only determine the radius of the circle.

Not sufficient.

The correct answer is A.

[steven_ghoos]

Thank you again! It's very helpfull to me.

For the coordinate plane question, I hope they will not ask many like this. I also still have a small question (in orange):

Point P and Q lies on the same circle with center at (0, 0).
Thus, (s² + t²) = (-√3)² + 1² = 3 + 1 = 4
Is this always true for points on a circle with center (0,0)? X1² + y1² = x2² + y2²

Again line segments OP and OQ are perpendicular.
Thus (slope of OP)*(slope of OQ) = -1

Slope of OP = 1/(-√3) = -(1/√3)
=> Slope of OQ = (t - 0)/(s - 0) = t/s = (-1)/(-1/√3) = √3
=> t = √3s

Thus, (s² + (√3s)²) = 4
=> (s² + 3s²) = 4
=> s² = 1
=> s = ±1

As point Q lies in the first quadrant s = 1.

[anshumishra]

Thank you again! It's very helpfull to me.

For the coordinate plane question, I hope they will not ask many like this. I also still have a small question (in orange):

Point P and Q lies on the same circle with center at (0, 0).
Thus, (s² + t²) = (-√3)² + 1² = 3 + 1 = 4
Is this always true for points on a circle with center (0,0)? X1² + y1² = x2² + y2²

Again line segments OP and OQ are perpendicular.
Thus (slope of OP)*(slope of OQ) = -1

Slope of OP = 1/(-√3) = -(1/√3)
=> Slope of OQ = (t - 0)/(s - 0) = t/s = (-1)/(-1/√3) = √3
=> t = √3s

Thus, (s² + (√3s)²) = 4
=> (s² + 3s²) = 4
=> s² = 1
=> s = ±1

As point Q lies in the first quadrant s = 1.

Answer to the question in orange :

Since those two points are on the circumference of the circle, the line joining the center of the circle with the points (x1,y1) and (x2,y2), is the radius. Hence they are same.

If instead of center being at (0,0), it is at some arbitrary point (a,b), then you'll have :

(x1-a)^2+(y1-b)^2 = (x2-a)^2+(y2-b)^2