FUN

ADVANCED PRACTICE PROBLEMS 

The following problems are more challenging than the typical BCA admission test problems. They are derived from some of my favorites on the BCA 8th grade math exams and from other sources. Enjoy!

VIDEO OF 3 QUESTION IQ TEST & INSIDE VIEW OF BCA LABS?

INSIGHT PUZZLE & STEVEN JOBS?




1. Calculate the value of the following infinite geometric sum:
\[\frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + .....\]
A. 3    B. 1/3    C. 2/3    D. 1/2    E. 1
 Correct Answer:  D. 1/2   Explanation: This problem is of the following form where x is a fraction less than 1 and we are trying to solve for N. \[\begin{array}{l}N = x + {x^2} + {x^3} + {x^4} + ...\\N = x\left( {1 + x + {x^2} + {x^3}...} \right)\\N = x\left( {1 + N} \right)\\N = x + xN\\N - xN = x\\N\left( {1 - x} \right) = x\\N = \frac{x}{{\left( {1 - x} \right)}}\end{array}\] In this case x=1/3 so N=(1/3)/(1-1/3)=(1/3)/(2/3)=(1/3) x (3/2)= 1/2.
2. Calculate the value of the following:
\[\sqrt {25 + 10\sqrt 6 }  + \sqrt {25 - 10\sqrt 6 } \]
\[\begin{array}{l}{\rm{A. }}2\sqrt 5 \\{\rm{B.}}\sqrt {55} \\{\rm{C. }}2\sqrt {15} \\{\rm{D. }}50\\{\rm{E. }}60\end{array}\]  Correct Answer:  C. 2xSQRT15   Explanation: \[\begin{array}{l}{\rm{Square the expression to get rid of the roots}}\\{\left( {\sqrt {25 + 10\sqrt 6 }  + \sqrt {25 - 10\sqrt 6 } } \right)^2} = \\{\rm{Note it is of the form (x + y}}{{\rm{)}}^2} = {x^2} + 2xy + {y^2}\\25 + 10\sqrt 6  + 2\left( {\left( {\sqrt {25 + 10\sqrt 6 } } \right)\left( {\sqrt {25 - 10\sqrt 6 } } \right)} \right) + 25 - 10\sqrt 6  = \\50 + 2\left( {\left( {\sqrt {25 + 10\sqrt 6 } } \right)\left( {\sqrt {25 - 10\sqrt 6 } } \right)} \right) = \\50 + 2\left( {\sqrt {\left( {25 + 10\sqrt 6 } \right)\left( {25 - 10\sqrt 6 } \right)} } \right) = \\50 + 2\left( {\sqrt {625 + 10\sqrt 6  - 10\sqrt 6  + 100 \times 6} } \right) = \\50 + 2\left( {\sqrt {625 - 600} } \right) = \\50 + 2\left( {\sqrt {25} } \right) = 50 + 2\left( 5 \right) = 60\\{\rm{Reversing the original square...}}\\\sqrt {60}  = \sqrt {4 \times 15}  = 2\sqrt {15} \end{array}\] 3. What is the maximum number of intersections between 10 lines and 2 circles? 
A. 100    B. 87    C. 1024    D. 900    E. 67
 Correct Answer:  B. 87   Explanation: This problem is an extended case of a problem on Exam #4. There are 3 types of intersections to count in this problem. Lines intersecting lines. Lines intersecting circles. Circles intersecting circles. From the previous problem....
Lines to Lines: Two lines intersect at most 1 time.  2: 1 If there are 3 lines then the 1st line can intersect the other two lines 2 times. The two remaining lines can intersect 1 time like the previous case. 3: 2+1 If there are 4 lines then the 1st line can intersect the others 3 times. The remaining 3 reverts to our previous case and has 2 intersections. The final two lines can intersect 1 time. 4: 3+2+1 and so on... 5: 4+3+2+1 .. ... 10: 9+8+7+6+5+4+3+2+1= (9 x 10) / 2= 45. {Note the sum of integers 1 to n= n(n+1)/2}
This approach suggests another way to look at the problem. We can count the possible intersections by choosing all combinations of 2 lines that each can make 1 intersection from a group of 10 lines. Order doesn't matter in an intersection. There is no concept of line B first and line A second. Only 1 intersection results. The combinations formula is as follows and the answer is the same.  \[{\rm{C}}\left( {\begin{array}{*{20}{c}}{10}\\2\end{array}} \right) = \frac{{n!}}{{r!\left( {n - r} \right)!}} = \frac{{10!}}{{2!\left( {10 - 8} \right)!}} = \frac{{10 \times 9}}{2} = 45\] The extended considerations in this problem begin with.... Lines to Circles: A single line will intersect one  circle at most 2 times. In the case of 10 lines and 2 circles it is simply 10 lines x 2 circles x 2 intersections/circle=40 Circles to Circles: One circle intersects another at most 2 times. There are only 2 circles in this problem so that creates only 2 intersections. Note that this is a simplified case and it should really be thought of as 2C2=2!/(0!2!)=1 arrangements of circles multiplied by the 2 intersections per circle. If there were 5 circles, for example, the circle to circle intersections would be 5C2 arrangements of two intersecting circles x 2 intersections maximum per two circles. 5C2=5!/(3!2!)=(5 x 4)/(2 x 1)=20/2=10 arrangements x 2 intersections maximum per two circles= 20 maximum circle to circle intersections in the case of 5 circles.  The combined intersections of all 3 types in this problem total 45 + 40 + 2=87 maximum intersections between 10 lines and 2 circles. This is a problem from the 2012 BCA Math Competition and a different answer is shown on-line, but I have confirmed this is correct. 

4. Dave makes a regular tetrahedron out of clay with side length square root of 2. He then turns the tetrahedron into a cube without removing or adding clay. What is the side length of the cube?

 Correct Answer:  D.    Explanation: The volume of a tetrahedron is 1/3(AREAabc)H where AREAabc equals the area of the triangle base and H equals the height shown below in red. If no material is lost then this volume equates to the volume of the resulting cube. First, we must calculate the tetrahedron volume.

Tetrahedron Volume: The first step is to calculate the base triangle ABC area. You may note this is a problem from the BCA Math Competition and the solution provided on-line presumes the reader knows the area equals s squared times the square root of 3 divided by 4. This was not obvious to me so I start with first principles. Area=1/2bh for a triangle. We know the base is b=s. We need to find h as shown in the figure on the right to calculate an area. This isn't difficult as h makes a right angle bisecting AC so Pythagoras' theorem with sides h, s/2, and hypotenuse s leads us to... \[\begin{array}{l}{h^2} + {\left( {\frac{s}{2}} \right)^2} = {s^2}\\{h^2} = {s^2} - \frac{{{s^2}}}{4} = \frac{{3s}}{4}\\h = \frac{{\sqrt 3 s}}{2}\\ARE{A_{ABC}} = \frac{1}{2}s\left( {\frac{{\sqrt 3 s}}{2}} \right) = \frac{{{s^2}\sqrt 3 }}{4}\end{array}\] which is of course the same as provided in the BCA solution, but hopefully this makes it a little more clear. The second step is to calculate the height of the tetrahedron, H, shown in red. Again we will make use of the fact that segment DO of length H is perpendicular to the base and segment AO forming another right triangle with hypotenuse AD. The BCA solution assumes the reader is aware that the length of AO is s divided by square root 3. Whomever crafted this problem clearly knows more than me because that fact does not seem obvious, but it is nice to know we can again derive the result on our own as follows.  \[\begin{array}{l}AO + RO = AR\\AR = h = \frac{{\sqrt 3 s}}{2}\end{array}\] If we can figure out the length r shown in red of segment RO and deduct that value from length h we know for AR then we will know the length of AO. We can derive the length r from the in-radius of the circle contained within ABC that is tangent to the midpoint of each line of ABC. In our case that is point R and the center is point O. From Heron's Formula, we know  AREAabc= r(semi-perimeter of ABC) The semi-perimeter of ABC with sides s is half the perimeter=(1/2)3s=3s/2 \[r = \frac{{ARE{A_{ABC}}}}{{{S_P}}} = \frac{{\left( {\frac{{{s^2}\sqrt 3 }}{4}} \right)}}{{\left( {\frac{{3s}}{2}} \right)}} = \frac{{s\sqrt 3 }}{6}\]Now we can solve for AO.  \[\begin{array}{l}AO = h - r = \frac{{s\sqrt 3 }}{2} - \frac{{s\sqrt 3 }}{6} = \frac{{3s\sqrt 3 }}{6} - \frac{{s\sqrt 3 }}{6} = \frac{{s\sqrt 3 }}{3}\\\end{array}\] ..and using Pythagoras' Theorem with this calculated length AO and hypotenuse, s, we can figure out the tetrahedron height, H which is the key to finding the tetrahedron volume. \[\begin{array}{l}{H^2} + A{O^2} = A{D^2}\\{H^2} + {\left( {\frac{{\sqrt 3 s}}{3}} \right)^2} = {s^2}\\{H^2} + \frac{1}{3}{s^2} = {s^2}\\{H^2} = \frac{2}{3}{s^2}\\H = \frac{{\sqrt 2 s}}{{\sqrt 3 }}\end{array}\] Finally we can calculate the volume of the tetrahedron and use that volume to solve for the cube side length. \[\begin{array}{l}V = \frac{1}{3}\left( {ARE{A_{ABC}}} \right)H\\V = \frac{1}{3}\left( {\frac{{{s^2}\sqrt 3 }}{4}} \right)\left( {\frac{{s\sqrt 2 }}{{\sqrt 3 }}} \right) = \frac{{{s^3}\sqrt 2 }}{{12}} = \frac{{{{\left( {\sqrt 2 } \right)}^3}\sqrt 2 }}{{12}} = \frac{4}{{12}} = \frac{1}{3}\\\\\end{array}\] The side length, X, of the cube is... \[\begin{array}{l}{V_{CUBE}} = {V_{TETRAHEDRON}}\\{X^3} = \frac{1}{3}\\X = \frac{1}{{\sqrt[3]{3}}}\end{array}\]