Algebra probability and Geometry

[Tagne]

1. The area of triangle "a" is twice the area of triangle "b". let s be one side of b and S be the corresponding side in triangle "a"; in terms of s, S =? The angles of the triangles are the same x,y &z.
a. (Sqrt2/2)s
b. (Sqrt3/2)s
c. (Sqrt2)s
d. (SQRT3)s
e. 2s.

2. If x and y are positive, which of the following must be greater than 1/SQRT(x + y)
a. SQRT(x + y)/2x
b. (SQRT(X) + SQRT(Y))/(x + y)
c. SQRT(x) - SQRT(y)/ (X + Y)

3. Joshua and Jose work at an auto repair center with four others. For survey on health care insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and jose would be chosen?
a. 1/15
b. 1/12
c. 1/9
d. 1/6
e. 1/3

[Anurag@Gurome]

1. The area of triangle "a" is twice the area of triangle "b". let s be one side of b and S be the corresponding side in triangle "a"; in terms of s, S =? The angles of the triangles are the same x,y &z.
a. (Sqrt2/2)s
b. (Sqrt3/2)s
c. (Sqrt2)s
d. (SQRT3)s
e. 2s

For two similar triangles, the ratio of their areas is equal to the ratio of the square of the lengths of their corresponding sides.

Hence, (S/s)² = Area of the triangle a/Area of the triangle b = 2/1 = 2

=> S = (√2)s

The correct answer is C.

[Anurag@Gurome]

3. Joshua and Jose work at an auto repair center with four others. For survey on health care insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and jose would be chosen?
a. 1/15
b. 1/12
c. 1/9
d. 1/6
e. 1/3

Number of ways to choose 2 worker out of 6 = 6C2 = 15
Out of these 15 combinations, only one combination is our required combination, i.e Joshua and Jose.

Hence, required probability = 1/15

The correct answer is A.

[yellowho]

Number 2:

What is greater 3/4 or 5/7? 3*7 =? 5*4? 21>20 so 3/4 is greater.

I. sqrt(x+y)/2x =? 1/sqrt(x+y) => cross multiply => x+y =? 2x => y=x Not sufficient
II. (sqrt(x)+sqrt(y)) =? X+Y/sqrt(x+Y) => sqrt(x)+sqrt(y)= sqrt(x+y) => Squaring both sides
x+y+2sqrt(xy)=x+y; since x,y are positives the left side is greater hence the sufficient

III. sqrt(x)-sqrt(y)=x+y/sqrt(x+y)=> sqrt(x)-sqrt(y)= sqrt(x+y)=> squaring both sides
x-y-2sqrt(xy)=x+y; since x,y positive right since is larger. (ur subtract things to x on the left side and adding things to x on the right side). Sufficient.

II, III = ok.




[quote="Tagne"]1. The area of triangle "a" is twice the area of triangle "b". let s be one side of b and S be the corresponding side in triangle "a"; in terms of s, S =? The angles of the triangles are the same x,y &z.
a. (Sqrt2/2)s
b. (Sqrt3/2)s
c. (Sqrt2)s
d. (SQRT3)s
e. 2s.

2. If x and y are positive, which of the following must be greater than 1/SQRT(x + y)
a. SQRT(x + y)/2x
b. (SQRT(X) + SQRT(Y))/(x + y)
c. SQRT(x) - SQRT(y)/ (X + Y)

3. Joshua and Jose work at an auto repair center with four others. For survey on health care insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and jose would be chosen?
a. 1/15
b. 1/12
c. 1/9
d. 1/6
e. 1/3[/quote]