Math Team Questions

8/3/2019 Math Team Questions

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Math Team Questions and Solutions

1. Solve for xx

2- 3|x - 2| - 4x = - 6

A) 1,2,3,0

B) 2,-1,-2,1

C) 4,2,1,0

D) 3,0,4,1E) None of the Above

Solution:

A. x2 - 3|x - 2| - 4x = - 6 : given

B. Let Y = x - 2 which gives x = Y + 2

C. (Y + 2)2 - 3|Y| - 4(Y + 2) = - 6 : substitute in above equation

D. Y2 - 3|Y| + 2 = 0

E. Y2 = |Y|2 : note

F. |Y|2 - 3|Y| + 2 = 0 : rewrite equation as

G. (|Y| - 2)(|Y| - 1) = 0

H. |Y| = 2 , |Y| = 1 : solve for |Y|I. Y = 2, -2 , 1 , -1 : solve for Y

J. x = 4 , 0 , 3 , 1 : solve for x using x = Y + 2.

2. Find all points of intersections of the 2 circles defined by the equations(x - 2)

2+ (y - 2)

2= 4

(x - 1)2

+ (y - 1)2

= 4

A) (2.82,0.177),( 0.177,2.82)B) (3.14,3),(3,3.14)C) (1.46,0.189),(0.189,1.46)

D) (2.76,0.201),(0.201,2.76)

E) None of the Above

Solution:

A. x2 - 4x + 2 + y2 - 4y + 2 = 4 : expand equation of first circle

B. x2 - 2x + 1 + y2 - 2y + 1 = 4 : expand equation of second circle

C. -2x - 2y - 6 = 0 : subtract the left and right terms of the above equations

D. y = 3 - x : solve the above for y.

E. 2x2 - 6x + 1 = 0 : substitute y by 3 - x in the first equation, expand and group like terms.

F. (3/2 + sqrt(7)/2 , 3/2 - sqrt(7)/2) , (3/2 - sqrt(7)/2 , 3/2 + sqrt(7)/2) : solve the above for x

and use y = 3 - x to find y.

3. If 200 is added to a positive integer I, the result is a square number. If 276 is added toto the same integer I, another square number is obtained. Find I.

A) 123

B) 124


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C) 125

D) 126


Solution:

A. x2 - 4x + 2 + y2 - 4y + 2 = 4 : expand equation of first circle

B. x2 - 2x + 1 + y2 - 2y + 1 = 4 : expand equation of second circle

C. -2x - 2y - 6 = 0 : subtract the left and right terms of the above equations

D. y = 3 - x : solve the above for y.

E. 2x2 - 6x + 1 = 0 : substitute y by 3 - x in the first equation, expand and group like terms.

F. (3/2 + sqrt(7)/2 , 3/2 - sqrt(7)/2) , (3/2 - sqrt(7)/2 , 3/2 + sqrt(7)/2) : solve the above for x

and use y = 3 - x to find y.

4. Two boats on opposite banks of a river start moving towards each other. They first pass each other

1400 meters from one bank. They each continue to the opposite bank, immediately turn around and

start back to the other bank. When they pass each other a second time, they are 600 meters from the

other bank. We assume that each boat travels at a constant speed all along the journey. Find the

width of the river?

A) 3000 meters

B) 4600meters

C) 3600 metersD) 2600 meters


Solution:

A. S1*t1 = 1400 : S1 speed of boat 1, t1 : time to do 1400 meters(boat 1)

B. 1400 + S2*t1 = X : S2 speed of boat 2

C. S1*t2 = X + 600 : t2 time to do X + 600 (boat 2)


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D. S2*t2 = 2X - 600

E. S1 = 1400/t1

F. S2 = (X-1400)/t1

G. T = t2/t1 : definition

H. substitute S1, S2 and t2/t1 using the above expressions in equations 3 and 4 to obtain

I. 1400*T = X + 600

J. X*T - 1400*T = 2X - 600 : 2 equations 2 unknowns

K. Eliminate T and solve for X to obtain X = 3600 meters.

5. The right triangle ABC shown below is inscribed inside a parabola. Point B is also the maximum

point of the parabola (vertex) and point C is the x intercept of the parabola. If the equation of the

parabola is given by y = -x2

+ 4x + C, find C so that the area of the triangle ABC is equal to 32

square units.

A) 10

B) 15

C) 12D) 13


Solution:A. h = -b / 2a = 2 : x coordinate of the vertex of the parabola

B. k = -(2)2 + 4(2) + C = 4 + C : y coordinate of vertex

C. x = (2 + sqrt(4 + C)) , x = (2 - sqrt(4 + C)) : the two x intercepts of the parabola.

D. length of BA = k = 4 + C

E. length of AC = 2 + sqrt(4 + C) - 2 = sqrt(4 + C)

F. area = (1/2)BA * AC = (1/2) (4 + C) * sqrt(4 + C)

G. (1/2) (4 + C) * sqrt(4 + C) = 32 : area is equal to 32

H. C = 12 : solve above for C.

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Math Team Questions