+0  
 
+1
938
7
avatar+884 

1. Paula can paint a room in six hours. Paula starts painting the room at
9:00 AM. At 11:00 AM, Carlos accompanies Paula, and both of them,
working together, finish painting the room at 1:30 PM. Assuming both
painters always paint at a constant rate, how many hours can Carlos
paint the same room working by himself?
 

 

2. An alphagram is a string of letters, such as ABC or AABEGLR,
whose letters are in alphabetical order. Find the number of three-letter
alphagrams.

 

4.  The numbers 1, 2, 3, 4, . . . are written starting with 1, spiraling outward
in a counterclockwise direction, as shown below. What is the
sum of the four numbers directly adjacent to the number “2018” in
this spiral?
.................. 26
13 12 11 10 25
14 3 2 9 24
15 4 1 8 23
16 5 6 7 22
17 18 19 20 21

 

 

5. A frog starts at “0” on a number line. Each second, it flips a fair coin,
and moves forward 1 unit if the coin shows heads, and forward 3
units if the coin shows tails. What is the probability that the frog will
eventually land on “6?” Express your answer as a common fraction.
 

 Feb 19, 2018
 #1
avatar
+1

1)

6 - 2 = 4 hours left for Paula to finish by herself

1/4 + 1/C =2/5, where C = Carlos' time

C =20/3 = 6 2/3 - Carlos' time in painting 2/3 of the room!!!

So, to paint the whole room by himself, it would take him: 6 2/3 x 3/2 = 10 hours !!!!.

EDITED!!. I simply reversed the time they finished the room together!!. Oh, my!!

 Feb 19, 2018
edited by Guest  Feb 19, 2018
edited by Guest  Feb 19, 2018
edited by Guest  Feb 19, 2018
 #2
avatar+128407 
+2

1. Paula can paint a room in six hours. Paula starts painting the room at
9:00 AM. At 11:00 AM, Carlos accompanies Paula, and both of them,
working together, finish painting the room at 1:30 PM. Assuming both
painters always paint at a constant rate, how many hours can Carlos
paint the same room working by himself?

 

Paula paints 1/6 of the room every hour

By the time Carlos joins in, Paula has painted   2 (1/6) =  1/3 of the room

And after Carlos joins....she paints an additional 2.5 hrs

So...by  1:30 PM....she has painted  1/3  + (2.5)(1/6)  =  1/3 + (5/2)(1/6)  = 

1/3 + 5/12  =  4/12 + 5/12  = 9/12  =  3/4  of the room

 

So.....Carlos  must paint  the other  1/4 of the room in  2.5 hours

 

So...the time it would take him to paint the room by himself must be 4 times this  =

 

10 hrs

 

 

cool cool cool

 Feb 19, 2018
 #3
avatar+884 
+1

Thanks, CPhill! Amazing! Can you answer the others, too?

 Feb 19, 2018
 #4
avatar+128407 
0

Sorry, ant....I would just be guessing on the rest and that usually leads to wrong answers....maybe some other people know the solutions to the rest of these   !!!!

 

 

 

cool cool cool

 Feb 19, 2018
 #5
avatar+26367 
+1

4.  The numbers 1, 2, 3, 4, . . . are written starting with 1, spiraling outward
in a counterclockwise direction, as shown below.

What is the sum of the four numbers directly adjacent to the number “2018” in this spiral?


.................. 26
13 12 11 10 25
14 3 2 9 24
15 4 1 8 23
16 5 6 7 22
17 18 19 20 21

 

 

\(\begin{array}{cccc} \text{Start at $(0, 0) = 1$ }\\\\ && \boxed{2019\\\tiny{(22,16)}} \\ && \uparrow \\ & \boxed{1843\\\tiny{(21,15)}} \leftarrow & \boxed{2018\\\tiny{(22,15)}} & \rightarrow \boxed{2201\\\tiny{(23,15)}} \\ & & \downarrow \\ & & \boxed{2017\\\tiny{(22,14)}} \\ \end{array} \)

 

\(1843+2017+2019+2201 = 8080 \)

 

laugh

 Feb 19, 2018
 #6
avatar+128407 
0

heureka....can you explain how this was derived  ??

 

 

cool cool cool

CPhill  Feb 19, 2018
 #7
avatar+26367 
+2

Computing...

 

1.

\(\text{Numbers}:\\ \begin{array}{rrrrrrr} 31 & 30 & 29 & 28 & 27 & 26 & \\ 32 & 13 & 12 & 11 & 10 & 25 & \\ \ldots & 14 & 3 & 2 & 9 & 24 & \\ \ldots & 15 & 4 & 1 & 8 & 23 & \\ \ldots & 16 & 5 & 6 & 7 & 22 & \\ \ldots & 17 & 18 & 19 & 20 & 21 & \\ \end{array} \)

 

2.

\(\text{Coordinates of the lattice points}:\\ \begin{array}{rrrrrrr} \text{Start at $(0, 0) = 1$ }\\\\ (-3,3) & (-2,3) & (-1,3) & (0,3) & (1,3) & (2,3) & \\ (-3,2) & (-2,2) & (-1,2) & (0,2) & (1,2) & (2,2) & \\ \ldots & (-2,1) & (-1,1) & (0,1) & (1,1) & (2,1) & \\ \ldots & (-2,0) & (-1,0) & (0,0) & (1,0) & (2,0) & \\ \ldots & (-2,-1) & (-1,-1) & (0,-1) & (1,-1) & (2,-1) & \\ \ldots & (-2,-2) & (-1,-2) & (0,-2) & (1,-2) & (2,-2) & \\ \end{array}\)

 

3.

\(\text{Algorithem to get the lattice point coordinates:} \\ \begin{array}{|r|r|} \hline \text{lattice point number} & \text{coordinate }(x,y) \\ \hline 1.& (0,0) \\ 2.& (0,1) \\ 3.& (-1,1) \\ 4.& (-1,0) \\ 5.& (-1,-1) \\ 6.& (0,-1) \\ 7.& (1,-1) \\ 8.& (1,0) \\ 9.& (1,1) \\ 10.& (1,2) \\ 11.& (0,2) \\ 12.& (-1,2) \\ 13.& (-2,2) \\ 14.& (-2,1) \\ 15.& (-2,0) \\ 16.& (-2,-1) \\ 17.& (-2,-2) \\ 18.& (-1,-2) \\ 19.& (0,-2) \\ 20.& (1,-2) \\ 21.& (2,-2) \\ 22.& (2,-1) \\ 23.& (2,0) \\ 24.& (2,1) \\ 25.& (2,2) \\ 26.& (2,3) \\ 27.& (1,3) \\ 28.& (0,3) \\ 29.& (-1,3) \\ 30.& (-2,3) \\ 31.& (-3,3) \\ 32.& (-3,2) \\ \ldots \\ 1843.& (21,15) \\ \ldots \\ 2017.& (22,14) \\ \ldots \\ 2018.& (22,15) \\ \ldots \\ 2019.& (22,16) \\ \ldots \\ 2201.& (23,15) \\ \ldots \\ \hline \end{array}\)

 

4.

short algorithm to get (x,y) the index in a grid-net of the spiral in c++:

    int n_max = 2818;
    int y_coordinate = 0;
    int x_coordinate = 0;
    for(int i = 0; i < n_max; ++i) {
        // output Number i+1
        // output x_coordinate
        // output y_coordinate
        if(abs(y_coordinate) <= abs(x_coordinate) && (y_coordinate != x_coordinate || y_coordinate >= 0))
            y_coordinate += ((x_coordinate >= 0) ? 1 : -1);
        else
            x_coordinate += ((y_coordinate >= 0) ? -1 : 1);
    }

 

Find the lattice point coordinate of 2018.

We find (22,15).

 

The four numbers directly adjacent to the number “2018” in
this spiral are (21,15), (22,14), (22,16), (23,15).

The Numbers are 1843, 2017, 2019, 2201.

 

laugh

heureka  Feb 20, 2018

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