Bosco

In order not to have two yellows next to each other,  

the three yellows must occupy squares 1, 3, and 5. 

That leaves squares 2 and 4 to fill, and two colors  

to do it with. There are only four ways. 

Blue Blue     Red Red     Blue Red     Red Blue  

So, the answer is four ways.  

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There are three real numbers x that are not in the domain of f(x).     

What is the sum of those three numbers?     

x = 0   results in a zero in the denominator  3 / x.    

x = –3   results in a zero in the denominator  2 / (1 + 3 / x).   

x = –1   results in a zero in the denominator  1 + 2 / (1+ 3 / x)    

The sum of the three numbers is  – 4.    

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With each stroke, a pump removes 2/5 of the air in a container.    

After 3 strokes, what is the fraction of the original amount of air in the container that is left.    

Call "x" the original amount of air in the container.   

After the 1st stroke, (2/5) of x has been removed ... 3/5 of x is left.    

After the 2nd stroke, (2/5) • (3/5) of x has been removed    

                                 (2/5) • (3/5) = 6/25 removed .... 19/25 of x is left.    

After the 3rd stroke, (2/5) • (19/25) of x has been removed    

                                 (2/5) • (19/25) = 38/125 removed ... 87/125 of x is left.    

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A positive integer is called fine if it has exactly 2 positive divisors. Which numbers are fine?    

I'll take a stab at it.  I'll say the numbers that are the product of two prime numbers.    

For the purpose of my submission, I exclude counting 1 and the number itself as divisors.   

For example, 17 does not count as fine because 1 and 17 itself are disqualified as divisors.   

However, 34 would count as fine because its divisors, i.e., 2 and 17, both are prime numbers.    

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In a rhombus, all the side lengths are $12,$ and one of the angles is equal to $90^\circ.$ Find the area of the rhombus.   

In a rhombus, if one of the angles is 90^o then every angle is 90^o ... this makes the rhombus a square.   

The width is 12, the height is 12, and the Area = length x width ...   A  =  12 x 12  =  144 square units     

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The quadratic equation $x^2-mx+24 = 10$ has roots $x_1$ and $x_2$. If $x_1$ and $x_2$ are integers, how many different values of $m$ are possible?   

                                                 x^2-mx+24 = 10     

Combine terms and put in    

formatting as ax² + bx + c           x² – mx + 14 = 0      

The roots will be integers   

that multiply to 14                  The roots can be    +1 & +14    

                                                                            –1 & –14    

                                                                            +2 & +7    

                                                                     or    –2 & –7    

So, m has four possible values, namely +15, –15, +9, or –9    

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Let a1, a2, a3, ..., a10 be an arithmetic series. If a1 + a3 = -2 and a2 + a4 + a6 = 1, then find a1.   

call "d" the addition between successive terms

starting with a₁   then  a₂  =  a₁ + d    

                                    a₃  =  a₂ + d  =  a₁ + d + d    

                                    a₄  =  a₃ + d  =  a₁ + d + d + d    

                                    a₅  =  a₄ + d  =  a₁ + d + d + d + d    

                                    a₆  =  a₅ + d  =  a₁ + d + d + d + d + d    

given       a₁ + a₃  =  – 2    

                a₁ + a₁ + 2d  =  – 2    

                2a₁ + 2d  =  – 2    

                     a₁ + d  =  – 1                                   (eq 1)    

given        a₂ + a₄ + a₆  =  1    

                (a₁ + d) + (a₁ + 3d) + (a₁ + 5d)  =  1    

                3a₁ + 9d  =  1                                       (eq 2)    

multiply (eq 1) by – 9         – 9a₁ – 9d  =  9    

add (eq 2)                             3a₁ + 9d  =  1    

                                           – 6a₁          =  10    

                                                a₁          =  – 10 / 6    

                                                        a₁  =  – 5 / 3

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Find the ordered triple (p,q,r) that satisfies the following system:    

p - 2q = 3    

q - 2r = -2 + q    

p + r = 9 + p    

Look at the system    

q – 2r  =  –2 + q     ==>>     –2r = –2     ==>>    r = 1    

p + r  =  9 + p         ==>>                                   r = 9    

r can't be both 1 and 9, therefore the system is not valid and there is no solution.    

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At a cafeteria, Mary orders two pieces of toast and a bagel, which comes out to $3.15. Gary orders a bagel and a muffin, which comes out to $3.50. Larry orders a piece of toast, two bagels, and three muffins, which comes out to $8.15. How many cents does one bagel cost?    

                                        T + B = 315 ¢    ==>>    T = 315 – B    

                                        B + M = 350 ¢    ==>>    M = 350 – B    

                                         T + 2B + 3M = 815 ¢   

                                         (315 – B) + 2B + [3 • (350 – B)]  =  815    

                                         315 – B + 2B + 1050 – 3B  =  815    

                                          – 2B + 1365  =  815    

                                          – 2B  =  – 550    

                                               B  =  275 ¢   

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