Maths Problems with Solutions
Grade 11 maths problems with answers and solutions are presented.
Problems

An airplane flies against the wind from A to B in 8 hours. The same airplane returns from B to A, in the same direction as the wind, in 7 hours. Find the ratio of the speed of the airplane (in still air) to the speed of the wind.

Find the area between two concentric circles defined by
x^{2} + y^{2} 2x + 4y + 1 = 0
x^{2} + y^{2} 2x + 4y  11 = 0

Find all values of parameter m (a real number) so that the equation 2x^{2}  m x + m = 0 has no real solutions.

The sum an integer N and its reciprocal is equal to 78/15. What is the value of N?

m and n are integers so that 4^{m} / 125 = 5^{n} / 64. Find values for m and n.

Simplify: 3^{n + 4001} + 3^{n + 4001} + 3^{n + 4001}

P is a polynomial such that P(x^{2} + 1) =  2 x^{4} + 5 x^{2} + 6. Find P( x^{2} + 3)
 For what values of r would the line x + y = r be tangent to the circle x^{2} + y^{2} = 4?
Solutions to the Above Problems
Let x = speed of airplane in still air, y = speed of wind and D the distance between A and B. Find the ratio x / y
Against the wind: D = 8(x  y), with the wind: D = 7(x + y)
8x  8y = 7x + 7y, hence x / y = 15
Rewrite equations of circles in standard form. Hence equation x^{2} + y^{2} 2x + 4y + 1 = 0 may be written as
(x  1)^{2} + (y + 2) ^{2} = 4 = 2^{2}
and equation x^{2} + y^{2} 2x + 4y  11 = 0 as
(x  1)^{2} + (y + 2) ^{2} = 16 = 4^{2}
Knowing the radii, the area of the ring is π (4)^{2}  π (2)^{2} = 12 π 
The given equation is a quadratic equation and has no solutions if it discriminant D is less than zero.
D = (m)^{2}  4(2)(m) = m^{2}  8 m
We nos solve the inequality m^{2}  8 m < 0
The solution set of the above inequality is: (0 , 8)
Any value of m in the interval (0 , 8) makes the discriminant D negative and therefore the equation has no real solutions.
Write equation in N as follows
N + 1/N = 78/15
Multiply all terms by N, obtain a quadratic equation and solve to obtain N = 5.
4^{m} / 125 = 5^{n} / 64
Cross multiply: 64 4^{m} = 125 5^{n}
Note that 64 = 4^{3} and 125 = 5^{3}
The above equation may be written as: 4^{m + 3} = 5^{n + 3}
The only values of the exponents that make the two exponential expressions equal are: m + 3 = 0 and n + 3 = 0, which gives m =  3 and n =  3. 
3^{n + 4001} + 3^{n + 4001} + 3^{n + 4001} = 3(3^{n + 4001}) = 3^{n + 4002} 
P(x^{2} + 1) =  2 x^{4} + 5 x^{2} + 6
Let t = x^{2} + 1 which also gives x^{2} = t  1
Substitute x^{2} by t  1 in P to obtain: P(t) =  2 (t  1)^{2} + 5 (t  1) + 6 = 2 t ^{2} + 9t  1
Now let t =  x^{2} + 3 and substitute in P(t) above to obtain
P( x^{2} + 3) = 2 ( x^{2} + 3) ^{2} + 9 ( x^{2} + 3)  1 = 2 x ^{4} + 3 x ^{2} + 8 
Solve x + y = r for y: y = r  x
Substitute in the equation of the circle:
x^{2} + (r  x)^{2} = 4
Expand: 2 x^{2} 2 r x + r ^{2}  4 = 0
If we solve the above quadratic equation (in x) we will obtain the x coordinates of the points of intersection of the line and the circle. The 2 points of intersection "become one" and therefore the line and the circle become tangent if the discriminant D of the quadratic equation is zero. Hence
D = (2r)^{2}  4(2)(r^{2}  4) = 4(8  r^{2}) = 0
Solve for r to obtain: r = 2 √2 and r =  2√2
More References and links
High School Maths (Grades 10, 11 and 12)  Free Questions and Problems With AnswersMiddle School Maths (Grades 6, 7, 8, 9)  Free Questions and Problems With Answers
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