Help me with these problems, please

Question

1. a @ b=c means a= (3+3c)(8-2b). If 36 @ 3= c, find the value of c in simplest form.

 

2. Fifty-four (54) marbles were divided among 3 boys in the ratio 1: 2/3: 1/3. The boy receiving the most marbles received ____marbles.

 

3. If m is 500% of n, then n divided by ____will yield m.

 

4. Two number cubes each have 6 faces and each is labeled 1, 2, 4, 6, 7, 8. If the two number cubes are rolled and the numbers showing on the upper faces are added, what is the probability of rolling a sum of 10?

 

5. Let N₆ mean 1+2+3+4+5+6. Let O₆ mean 1+3+5+7+9+11=36. Note O_N =N² ;

meaning O₂ = 2² ; O₃=3² - - -; O₆= 6² - - - . Let E₆ mean 2+4+6+8+10+12= 42.

Furthermore, N₁₂= O₆+E₆. If N₄₀₁= 80, 601, find the value of E₂₀₀.

Answer 1

1. a @ b=c means a= (3+3c)(8-2b). If 36 @ 3= c, find the value of c in simplest form.

a @ b = c ==> a = (3+3c)(8-2b)
36 @ 3= c ==> 36 = (3+3c)(8-2*3) and solve for c.

2. Fifty-four (54) marbles were divided among 3 boys in the ratio 1: 2/3: 1/3. The boy receiving the most marbles received ____marbles.

x(1: 2/3: 1/3) = x : 2/3 x : 1/3 x

x + 2/3 x + 1/3 x = 54 and solve for x.

3. If m is 500% of n, then n divided by ____will yield m.

(m)(is)(500%)(of)(n)
m = (500/100)*n where * means multiply
m = 5n = n*(5/1) = n/?

4. Two number cubes each have 6 faces and each is labeled 1, 2, 4, 6, 7, 8. If the two number cubes are rolled and the numbers showing on the upper faces are added, what is the probability of rolling a sum of 10?

10 = 2 + 8 or 8 + 4
10 = 4 + 6 or 6 + 4

probability = 4/?

5. Let N6 mean 1+2+3+4+5+6.
Let O6 mean 1+3+5+7+9+11=36. ON = N^2: O2=2^2, O3=3^2, …, O6=6^2, ….
Let E6 mean 2+4+6+8+10+12= 42.
Furthermore, N12= O6+E6. If N401=80601, find the value of E200.

N400+401 = N401 = 80601; and
N400=O200+E200=200^2 + E200

Take it from there, Abaline.

Answer 2

1: 36=(3+3c)(8-2*3)

    c=5

 

2:  54/(1+2/3+1/3)=27

 

3:  m=5n   then n divided by 1/5.

 

4:  result should be 2/15.  Because there are only two possible groups to make 10, 2+8, 4+6. And you can get        15 summation group.

 

5: N401=N201+E200, so E200=80601-201*201=40200

Answer 3

1) a = ( 3 +3c) ( 8 -2b)

     36 = ( 3 +3c ) ( 8 - 2c) 

 

      36 = 24 + 24c - 6c -6 c^2

     

        12 = 18c - 6 c^2

 

         c^2 - 3c + 2 = 0

 

         ( c -2 ) ( C-1 ) =0

 

        c = 2  c = 1

 

2.

 

 

 

 

54/ 3  = 18

 

 18 *2 = 36    18*1 =18

 

    36 / 18 = (54 * 2/3 ) / ( 54 * 1/ 3) = (2/3)/ ( 1/ 3) .

 

   Algebraically:

 

      X / ( 54 - X ) = ( 2/3 ) / ( 1/ 3) = 2/1

 

     X = 108 - 2X

 

      3x = 108 

 

        X = 36       54- X = 54 -36 = 18

 

     3. 

         500% n = m

 

          5n = m

 

           n/ (1/5) = m

4.

 sum  10={ ( 6, 4) ,  ( 2, 8)}

 

       Probability of  { 6.4) U ( 2, 8) } = ( 1/6 * 1/6 ) + ( 1/6 * 1/6) = 1/18

5.

 

E ₂₀₀ = O_{100 +} E_{100 = 100}^{^2} + E _{100 =} 

 

    E ₁₀₀₌ 2 + 4 + 6 + .....+ 200 = 10100

 

 E_{200 = 10100 + 10000= 20100}