Problem 1:

Simplify:       

Solution:

We can rewrite the expression as follows,

By cross multiplication,

Since the denominator is an irrational number, we rationalize it by multiplying

the expression by 1 to preserve its identity.

Problem 2:

Evaluate:    

Solution:

By long method, we can convert negative exponents to fractions and simplify

it using fractions.

Recall that  

We do the same in the problem,

Simplify numerator and denominator as follows,

Problem 3:

Simplify:  

Solution:

We can just do it by using fractions.

Extracting the perfect squares,

Problem 4:

Simplify:     

Solution:

rewrite the expression as follows,

Rationalizing the denominator,

x-y cancels out leaving,

Problem 5:

If a, b, and c are real numbers such that               and       Find the

value of       .

Solution:

Given the ratio of b and a and b and c, we find ways that b must be of the

same value in both proportions. By multiplying a number that is equal to 1,

such condition can be attained.

  

  

  

For the other ratio:

Since we already have a uniform values for a, b, and c. we can say that

a:b:c=2:10:5

Solving for the value of    .

Problem:

If     and   , what is x?

Solution:

We solve for y for the second equation,

Now, we substitute this y to the first equation,

we equate this to 0,

Observe that this is a quadratic equation in x/9, thus by factoring we have

Solving for x’s:

Solving for another x:

Since x>0 is the given restriction in the problem, the only solution is x=9.

Problem 7:

If  , what is f(x-3)?

Solution:

We can rewrite f(x) in factor form as follows,

To find f(x-3), we let x=x-3,

Problem 8:

If   and  , for all real number x. For what real number

does 

Solution:

Solving for g(-a) and f(-a),

For f(-a),

Going back to the desired equation,

Solving for a:

Solving for another value of a:

Therefore, the answers are 0 and -1.

Problem 9:

Solve for x:   

Solution:

First step is to square both sides of equation to remove the outermost radical.

Squaring both sides again,

Problem 10:

Solve for x:  

Solution:

By cross multiplying the left side of equation we have,

let A=2x, B=x+2,

Simplifying,

By factoring,

Plugging in the real values of A and B we have,

Solving for x:

Problem 11:

If  , solve for x in 

Solution:

To solve equations, we always eliminate fractions. By multiplying the whole

equation by   we can remove all fractions.

Now, since the right side of equation is radical, we raise both sides by 2,

By factoring we have,

These factors will immediately tell us that the values of x are 1 and 9/4. By

quick check however, 9/4 fails to satisfy the original equation. Thus, the only

solution is 1.

Problem 12:

Evaluate  

Solution:

This is a nested radical   and I already have a tutorial on how to solve this

problem. If you’re used to see problem like this, you might be able to solve it

in less than 3 seconds. For real.

To solve this though, we let x be equal to 

Square both sides,

Now, recall that 

The square root of a positive number will always be positive. Thus, the answer

is 2. -1 is an extraneous root.

Problem 13:

Find the two consecutive integers whose product is 506.

Solution:

Let x be the smaller integer. Since we are looking for the two CONSECUTIVE

integers, the other number must be x+1.

It is stated that their product is 506, thus

Therefore, the numbers that we are looking for are are 22 and 23.

Problem 14:

If   and if  , what is the larger root of  

Solution:

Let x and y be the roots of equation.

By Vieta’s formula, we have

 *

 **

From the relationship of the coefficients, we solve for c in terms of a and b.

***

We substitute *** to c of *,

 ****

From ** we solve for b/a,

We substitute this to ****

By factoring,

Since the roots are equal, there is no greater root. The answer is -1.

Problem 15:

Solve for x in 

Solution:

We also provided the easiest way to solve this quadratic inequality   in

this site. But let us demonstrate the answer to this problem.

In this format, the answer is in the form of a<x<b, where b>a and a and b are

the roots of inequality.

The roots are supposed to be 3/2 and -2. Thus, the solution set is -2<x<3/2

or (-2,3/2)

Problem 16:

Solve for real numbers satisfying the inequality  

Solution:

By rearranging,

Solving for x:

Obviously, -2 is an extraneous root since we have square root of x in the left

side of equation and the square root of any number will always be greater

than 0. Thus,

Check this tutorial on how to deal with quadratic inequality flawlessly.

Problem 17:

Find the minimum value of   

Solution:

Minimum value of quadratic equation can be found using the following

formula,

where a,b, and c are the coefficients of quadratic equation.

By substitution,

Problem 18:

Find the smallest value of  

Solution:

By AM-GM Inequality theorem we have,

Applying the theorem,

Thus, the minimum value of x+5/x is 

Problem 19:

Solve for b in the equation:  

Solution:

By comparison, we can say that b=a+1. But a=3(constant part), hence

Problem 20:

Write the quadratic equation with integer coefficients whose roots are the

reciprocals of the roots of 

Solution:

This problem actually can be rigidly solved using Vieta’s formula but there is

a hippest way   to solve problem like this where we can solve it in just a

matter of three to 4 lines long.

Let y be the roots of the desired equation. Since the given equation has  a

root of x, we can establish the following relation based on the condition

given(reciprocal of the roots).

Now, we substitute this to the given equation,

Simplifying to have an integer coefficients we have,

Dropping y back to x,

Problem 21:

Compute the sum of all the roots of (x-2)(x+1)+(x-1)(x+4)=0

Solution:

To find the sum of the roots, we expand the equation,

Now, the sum of the roots   of this quadratic equation is just the negative

ratio of b and a. Thus, the sum is -1.

Problem 22:

If r and s are the roots of  , evaluate 

Solution:

The sum of the roots of quadratic equation is again the negative ratio of b and

a. Thus,

This means that  .

Actually, this problem doesn’t really make sense at all. I was expecting a twist

somewhere. They might mistyped this   from this  . The latter is

more challenging and adds flavor to the problem. Anyways, let’s proceed.

Problem 23:

For what value(s) of m are the roots of   equal?

Solution:

For a quadratic equation to have an equal roots, the discriminant must be

equal to 0. Thus,

Problem 24:

It is known that y varies as the square of x and y=8 when x=1. What is y when

x=8?

Solution:

Let k be the constant of proportionality. Since y varies as the square of x we

have,

Solving for k when  y=8 and x=1,

Solving for y when x=8,

Problem 25:

Suppose that x and y are inversely proportional and are positive quantities. By

what percent does y decrease if x is increased by 25%?

Solution:

We let k be the constant of proportionality and we establish the equation.

IF x in increased by 25%, the value of x will become 1.25x. Since the constant

of proportionality is constant. We equate the ks of two situations.

From 1, the value of y decreases to 0.8. Thus, it decreased by 20%.

Problem 26:

If 4 men can paint a house in 5 days, in how many days can 10 men paint the

same house?

Solution:

This problem is an application of variation. As the number of workers

increases, the number of days to complete the job decreases. Thus, they are

inversely proportional.

Let m be the number of people, d be the number of days, and k be the

proportionality constant.

From this, we can say that

Or we can rearrange this as follows,

In variation, the proportionality constant is always constant. Thus,

Thus, answer is 2 days.

Problem 27:

If y is proportional to the cube of y and x is proportional to the fourth power of

z, then y is proportional to which power of z?

Solution:

This problem is vague. y is proportional to cube of y doesn’t make sense. We

will verify this problem first before writing the solution.

Problem 28:

Running at uniform speed in a race, Alice can beat Ben by 20 m, Ben can

beat Carlo by 10 m and Alice can beat Carlo by 28 m. How long is the race?

Solution:

Let d be the distance of the race,

Let r1 be the rate of Alice

Let r2 be the rate of ben

Let r3 be the rate of Carlo

Formulating equation:

For Alice:

d=r1(t)

For Ben:

d-20=r2(t)

Since alice finished 20 m ahead of Ben and the time is equal. Equating time

we have,

   #

For “Ben can beat Carlo by 10 m”. For Ben’s rate equation.

d=r2t

For Carlo’s rate equation,

d-10=r3t

Equating the time,

We can rewrite this as follows,

   ##

For “Alice can beat Carlo by 28 m”

Alice’s rate equation:

d=r1t

For Carlo’s equation:

d-28=r3t

equating time,

    ###

Now, we observe these 3 equations. We equate the first equation and last

equation since the ratio of the distance and the rate of Alice is just the same.

We can rewrite this equation as follows,

But we have the ratio of Carlo’s rate to Bob’s rate in ##,

thus,

Voila! we have an equation in terms of distance of the race. By cross

multiplication,

Therefore, the distance of the race is 100 m.

Problem 29:

Find the measure of the vertex angle of an isosceles triangle whose base

angles measure 65◦.

Solution:

An isosceles triangle has two base angles and one vertex angle.

Let x be the measure base angle.

x+65+65=180

x+130=180

x=180-130

x=50

Answer is 50 degrees.

Problem 30:

Find x if the angles of a quadrilateral measure x ◦, (2x+10)◦, (3x+20)◦ and

(4x−30)◦.

Solution:

Since the sum of interior angles of quadrilateral is 360 degrees we have,

x+2x+10+3x+20+4x-30=360

10x=360

x=360/10

x=36

Problem 31:

An equilateral triangle and a square have the same perimeter. What is the

ratio of the length of a side of the triangle to the length of a side of the square?

Solution:

The perimeter of equilateral triangle can be found using the formula,

where t is the length of the side

The perimeter of square can be found using the formula,

where s is the length of the side

Since the perimeter is equal, we equate their perimeters,

Since we are asked to find the ratio of the side of triangle to the length of the

side of the square we find t/s.

Thus, the ratio is 4:3

Problem 32:

John cuts an equilateral triangular paper whose sides measure 2 cm into

pieces. He then rearranges the pieces to form a square without overlapping.

How long is the side of the square formed?

After cutting, the area of the figure is still the same. To find the area of an

equilateral triangle, we use the following formula,

Where a is the length of the side of an equilateral triangle.

Finding the area we have,

Now, since the area of this triangle is equal to the area of the square, we can

use the formula of the area of the square to find the length of the side.

Now, we equate the area of the triangle,

Since we are solving for s, we take the square root of both sides,

Or,

Problem 23:

The sides of a triangle are of lengths 5, 12 and 13 cm. What is the length of its

shortest altitude?

Solution:

Draw the triangle and label like shown below.

Since the triangle formed is a right triangle, 12 and 5 are also altitudes. If we

draw another altitude to the hypotenuse and label it x. Both 5 and 12 will

become the hypotenuse of the two new triangles formed. Thus, x is the

shortest altitude.

Now, take note that 

By similar triangles we can solve for the value of x.

Problem 34:

Each side of triangle ABC measures 8 cm. If D is the foot of the altitude drawn

from A to the side BC and E is the midpoint of AD, how long is segment BE?

Solution:

Draw the figure as shown below,

Since this is an equilateral triangle, triangle ADB is a 36-60-90 triangle. Using

the property of this triangle, the side of AD is square root of 3 times the length

of BD. Thus,

Since E is the midpoint of AD, 

Now, triangle BDE also forms a right triangle and we can solve for BE using

Pythagorean theorem.

Problem 35:

A point E is chosen inside a square of side 8 cm such that it is equidistant

from two adjacent vertices of the square and the side opposite these vertices.

Find the common distance.

Solution:

Draw the figure like shown below,

The only way that a point will be equidistant from two adjacent vertices, it

should lie in the symmetrical axis of the square. Label the figure accordingly. If

your figure is not correct, your solution will not be correct as well.

To find the value of x, we use the right triangle formed in the upper left side of

upper right side. Either of the two.

Using Pythagorean theorem we have,

We would like to thank MMC 2013 finalist Daniel James Molina for helping us

out solving this problem.

Problem 36:

A point E is chosen inside rectangle ABCD and its distances from A, B and C

are 2 cm, 3 cm and 4 cm. How far is E from D?

Solution:

If you draw the figure, it will give you thrill how to solve it because the solution

to this problem has never been teach for a regular grade 9 students unless

you studied the first tutorial we posted in this website or you love reading

olympiad math problems.

This problem can be solved using British Flag Theorem, better check that

tutorial first before you go through. This theorem states that if a point is

selected inside the rectangle, the sum of the square of the distances from a

point from opposite vertices are equal.

Thus,

By substitution,

Problem 37:

Find the area of a rectangle with diagonal 10 cm that is twice as long as it is

wide.

Solution:

The diagonal, length, and the width of the rectangle will form a right triangle.

Let x be the width and the length will be 2x as given.

Using Pythagorean theorem,

Thus, the length is 4√5

Finding the area:

Problem 38:

Triangle ABC has a right angle at C. If sin A=1/3,what is cos A?

Solution:

Draw and label the figure as shown below. Label your angle, label the

opposite and the hypotenuse since sine is the ratio of opposite side and

hypotenuse.

We are asked for the value of cosA which is the ratio of adjacent side to

opposite side. Since the adjacent side is unknown, we look for it using

Pythagorean theorem.

Solving for cosA:

Where a is the adjacent side to the angle A and H is the hypotenuse.

Problem 39:

A side of a triangle measures 3 cm. A line segment is drawn parallel to this

side, forming a trapezoid whose area is 2/3 of the area of the triangle. How

long is the line segment?

Solution:

Draw and label the triangle like shown below.

Since the segment is parallel to the side with length 3, two similar triangles will

be formed.

We let A be the area of the bigger triangle. Given that the area of the

trapezoid is 2/3 of A, the area of smaller traingle is 1-2A/3=A/3.

Using the side-area relationship of two similar polygons, we have

Where s1 and s2 are the length of the sides of bigger and smaller triangle

respectively. A1 and A2 are the corresponding areas.

Problem 40:

A cone has volume 64 m^3 vertex equal to 1/4. If the cone is cut parallel to the

base at a distance from the of the height of the cone, what is the volume of

the resulting cone?

Solution:

The figure would look like this.

The volume of the larger cone can be expressed as,

The volume of smaller cone can be expressed as,

Since the height is 1/4 of the bigger cone, the radius will follow this proportion

since the cones formed here are congruent.(Parallel cut)

Dividing the volumes we have,