Bosco

Find the coefficient of y^4 in the expansion of (2y-5+3y^2)^6.   

That's way too much multiplication to try to do manually.  

Fortunately, a useful tool has been provided for us, at   

https://www.symbolab.com/solver/expand-calculator

Per the "expand" tool on that page, the coefficient of y⁴ is 375.   

But don't take my word for it, go there and check it for yourself.  :-)  

Is it considered cheating, to use a tool to avoid the drudgery of endless     

multiplications?  I don't consider it cheating to use any tool that has been  

made available.  The abacus, the slide rule, the calculator, the computer.      

_.

That's right, because I clumsily misread the problem. 

It would have been nice if you'd said how it was wrong. 

I should have said x is to 18 as 2 is to 12    

(should have included the blue ones in the known ratio)   

                                                         x          2     

                                                       –––  =  –––     

                                                        18        12               

cross multiply                                  12x  =  36              

                                                           x  =  3     

_.

Warehouse A has 5 times as many employees as Warehouse B, and Warehouse C has 3 times as many employees as Warehouse B. Warehouse A has how many times as many employees as Warehouse C?   

                                              A  =  5B      (eq 1)   

                                              C  =  3B      (eq 2)   

Multiply (eq 1) by 3               3A  =  15B    

Multiply (eq 2) by 5               5C  =  15B    

Therefore                             3A  =  5C   

                                               A  =  (5 / 3) C   

_.

Goldbach's conjecture states that every even number greater than two can be expressed as the sum of two prime numbers. For example, $2022 = 191 + 1831.$ How many ordered pairs of prime numbers have a sum of $100 \,?$   

You could list all the prime numbers up to 50. 

No need to go past 50.   I counted 6 of them.  

  3,  11,  17,  29,  41,  47  

97,  89,  83,  71,  59,  53

Can't count 2 because 98 isn't prime.  

ditto   5 ... 95  

ditto   7 ... 93

ditto 13 ... 87  

ditto 19 ... 81  

ditto 23 ... 77  

ditto 31 ... 69  

ditto 37 ... 63

ditto 43 ... 57  

edited to add ~  I'm not certain what an "ordered" pair is. 

                          So if the sequence matters, then it's 12.  

_.

Our badminton team has finished 75% of its season. So far, we have won 45% of the games we played. What percent of the remainder of our games must we win in order to finish the season with the same number of wins as losses?    

You've played 75% of the season.  So, you've played   

75% of the total number of games in the whole season.   

You have won 45% of 75% of the total number of games    

in the season.  That means you've won 33.75% of the total    

number of games in the season.  

To achieve that 50% winning season, you will need to win   

another 16.25% of the total number of games in the season.  

There is 25% of the season left, meaning there are 25%

of the total number of games in the season left to play. 

You will have to win 16.25% of the total number of games

in the season, but there's only a fourth of a season left to 

do it in, so you must win 4 times 16.25% = 65% to achieve   

that 50% winning season.   

You must win 65% of the remaining games in the season.    

45% of 75% of the games already played      =  33.75%  of all the games  

65% of 25% of the games remaining to play  =  16.25%  of all the games  

                                                                          ––––––   

                                                             total       50.00%  of all the games  

_.

>>> we have lost 55 games. To finish the season with the same number of wins as losses, we must win 55 games <<<

If you've lost 55 games and won 55 games, that comes to a total of 110 games. 

But, you based your calculations on the premise that the season is 100 games.   

_.

Our badminton team has finished 75% of its season.  So far, we have won 10% of the games we played.  What percent of the remainder of our games must we win in order to finish the season with the same number of wins as losses?  

You've played 75% of the season.  So, you've played   

75% of the total number of games in the whole season.   

You have won 10% of 75% of the total number of games    

in the season.  That means you've won 7.5% of the total    

number of games in the season.  

To achieve that 50% winning season, you will need to win   

another 42.5% of the total number of games in the season.  

However, since there's only 25% of the season left, then   

there are only 25% of the total number of games in the  

season left to play.  You don't have enough games left.    

It's impossible to achieve a 50% winning season.    

_.

Very good.  I'll just add that 0.285 minute is 17 seconds.   

If you're going to call it 2:54:30 you might as well call it 2:54:17       

and be precise, since this is just a math problem and theoretical.   

_.

jeeze, fella, (29+29+42)/2​ does NOT = 40  it's 50 

didnt you think something was wrong when you did 40-42  

In Linguistics 101, the ratio of the number of juniors to the number of seniors is 3:2.  When six more juniors join the class, and one senior drops the class, the ratio of the number of juniors to the number of seniors becomes 2:3.  How many students are in the class after these changes?   

I can tell you immediately that increasing the number     

of juniors will not invert the ratio.  But I'll go through    

the motions for you anyway.    

                                                       J          3   

                                                      ––   =   ––  

                                                       S          2     

                                                   J + 6          2   

                                                   ––––   =   –––   

                                                   S – 1          3        

Cross multiply both                            2J  =  3S            (eq 1)  

                                                  3J + 18  =  2S – 2      (eq 2)      

Get either J or S in terms of the   

other.  It doesn't matter which,   

but J would be simpler.                       J  =  3S/2      

Substitute this in eq 2                (3) • (3S/2) + 18  =  2S – 2    

                                                        9S/2 + 18  =  2S – 2     

Multiply both sides by 2                      9S + 36  =  4S – 4    

Subtract 4S from both sides               5S + 36  =  –4   

Subtract 36 from both sides                    5S  =  –40     

You cannot have a negative number of seniors    

so something is wrong somewhere.  The error     

is in the initial proposition.  This problem has ...... no solution.    

_.