Bosco

Seven years ago, Grogg's dad was 9  times as old as Grogg. Four years ago, Grogg's dad was 5 times as old as Grogg. How old is Grogg's dad currently?    

(D – 7)  =  9 • (G – 7)    

 D  =  9G – 63 + 7  =  9G – 56    

(D – 4)  =  5 • (G – 4)    

 D  =  5G – 20 + 4  =  5G – 16    

9G – 56  =  5G – 16    

4G  =  40                                 ==>>    Grogg's current age is 10    

D  =  9G – 56  =  (9)(10) – 56    

D  =  90 – 56                            ==>>    Dad's current age is 34    

check answer   

(34 – 7)  =?  9 • (10 – 7)    ==>    27 = 27  True    

(34 – 4)  =?  5 • (10 – 4)    ==>    30 = 30  True    

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In a row of five squares, each square is to be colored either red, yellow, or blue, so that no two consecutive squares have the same color. How many ways are there to color the five squares, if there must be at least three yellow squares?    

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.    

_.    

Find the largest integer k such that the equation 5x^2 - kx + 8 - 2x^2 + 25 =0 has no real solutions    

                                                              5x² – kx + 8 – 2x² + 25 = 0    

Combine like terms and arrange in    

standard notation ax² + bx + c = 0         3x² + (–k) x + 33 = 0    

The discriminant of a quadratic equation written in standard form is b² – 4ac.  

For the equation to have imaginary solutions, its discriminant has to be negative.   

so [ (–k)² – (4)(3)(33) ] has to be negative ... that is, k² has to be less than 396    

The square root of 396 is ≈ 19.9   

The largest integer less than 19.9 is 19, so k = 19 is the largest integer that will produce an imaginary root.    

_.    

You have a total supply of $1000$ pieces of candy, and an empty vat.  You also have a machine that can add exactly $5$ pieces of candy per scoop to the vat, and another machine that can remove exactly $3$ pieces of candy with a different scoop from the vat.  When these two machines are done, there is only one piece of candy left in the vat.  What is the smallest possible number of times the the first machine added candy to the vat?    

Let the 1^st machine cycle 2 times          2 x 5  =  10  pieces added       

Let the 2^nd machine cycle 3 times           3 x 3  =   9  pieces removed    

                                                                               1  piece remaining     

_.    

Real numbers a and b satisfy    
a + ab = 250    
a - ab = -240    
Enter all possible values of a, separated by commas.    

a + ab  =   250    

a – ab  = –240    

Just add the two equations     a + ab  =   250    

                                               a – ab  = –240    

                                             2a          =     10    

                                                        a  =  5    

_.    

Fill in the blanks to make a quadratic whose roots are $1$ and $-1.$    
x^2 + ___ x + ___    

x = + 1     ==>>   therefore the factor is (x – 1)    

x = – 1     ==>>   therefore the factor is (x + 1)    

(x + 1)(x – 1)  =  x² – 1    

so, fill in the blanks with 0 and – 1 respectively, thus:     x² +  0x  + (– 1)     

_.    

Fill in the blanks to make a quadratic whose roots are $1$ and $-1.$   
x^2 + ___ x + ___    

x = + 1     ==>>   therefore the factor is (x – 1)    

x = – 1     ==>>   therefore the factor is (x + 1)    

(x + 1)(x – 1)  =  x² – 1    

so, fill in the blanks with 0 and – 1 respectively, thus:     x² +  0x  + (– 1)     

_.    

What is the coefficient of x in (x^3 + x^2 + x + 1)(x^4 + 3x^3 + 4x^2 + 8x + 7)?     

Every term in one is to be multiplied by every term in the other.  

Consider only the pairs of terms that result in x with a coefficient.     

                       (x³ + x² + x + 1) • (x⁴ + 3x³ + 4x² + 8x + 7)    ==>>  7x    

                       (x³ + x² + x + 1) • (x⁴ + 3x³ + 4x² + 8x + 7)    ==>>  8x    

                                                                                           total   15x    

                       The coefficient of x is 15.   

_.    

A terminal zero is a $0$ that appears at the end of a number. For example, the number $3,800$ has two terminal zeros.    
How many terminal zeroes does $40 \cdot 6 \cdot 75 \cdot 12$ have?    

40 • 6 • 75 • 12  =  216,000    

There are three terminal zeros, as we have determined empirically.    

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