Very Hard Math Problem?

Question

* first person to give a step by step proccesds and the correct answer gets my vote for best answer

Some marbles in a bag are red and the rest are blue. If one red marble is removed, then one-seventh of the remianing marbles are red. If two blue marbles are removed instead of one red, then one fifth of the remaining marbles are red. How many marbles were in the bag originaly.

Answer 1

(Note the earlier version of the first answer was wrong!)

I hope you already finished your home work!

R is the number of red marbles, B is the number of blue marbles. We are looking for how many marbles were in the bag originally, or R+B.

R-1 / ((R-1)+B) = 1/7 (after removing 1 red, then 1/7th are red)

Solve for R:
R - 1 = ((R-1) + B) / 7
7R - 7 = R-1+B
6R - 6 = B

Now if removed 2 blue marbles we would have:

R /(R+B-2) = 1/5

Solve again for B
R = (R+B-2)/5
5R = R+B-2
B = 4R + 2

Now replace B from the first solution:
6R - 6 = 4R + 2
R2 = 8
R = 4

Using either solution for B:
B = 4*4 + 2 = 6*4 - 6 = 18

R+B = 22

22 marbles, 4 red and 18 blue

Checking the answer:
3/21 = 1/7 (after removing 1 red there are 21 marbles left)
4/20 = 1/5 (after removing 2 blue, 1/5 are red)

Answer 2

Let R be the initial number of red marbles, and let B be the initial number of blue marbles.

"If one red marble is removed, then one-seventh of the remaining marbles are red." This means that the number of remaining red marbles (R - 1) is equal to one seventh the total number of remaining marbles (R + B - 1).

R - 1 = (1/7)(R + B - 1).

"If two blue marbles are removed instead of one red, then one fifth of the remaining marbles are red." This means that the number of remaining red marbles (R) is equal to one fifth the total number of marbles remaining (R + B - 2).

R = (1/5)(R + B - 2).

Solve the system

R - 1 = (1/7)(R + B - 1)
R = (1/5)(R + B - 2).

Multiply the first equation by 7 and the second by 5:

7R - 7 = R + B - 1
5R = R + B - 2

Bring all the variables to the left side, and all the constants to the right side:

6R - B = 6
4R - B = -2

Subtract the second equation from the first:

2R = 8

Thus R = 4.

Since 4R - B = 2 and R = 4, then 4(4) - B = -2, so 16 - B = -2, so B = 18.

Initially, there were 18 blue marbles and 4 red marbles. (22 total)

Answer 3

Initially there are R reds and B blues.

Remove 1 red. (R - 1) red are left, 7 times this is the remaining number (R + B -1).

So 7 (R - 1) = R + B -1. ........(1)

Remove 2 blues, R red are still left, 5 times this is the remainiing number (R + B - 2).

So 5 R = R + B - 2. ........(2)

Subtract eqn (2) from (1): 2 R - 7 = 1, so 2 R = 8, i.e, R = 4.

Then from (2), B = 4 R + 2 = 18.

Check: this solution also satisfies eqn (1), so all conditions are satisfied.

There were originally 22 marbles in the bag, 4 reds and 18 blues.

Live long and prosper.

Answer 4

x = red marbles
y = blue marbles 10 m
(x-1)/(x-1+y) = 1/7 , 7x-7 = y + x - 1 ==> y = 6x -6
x/(x+y-2) = 1/5, 5x = x+y - 2 , 4x + 2 = y

substitute y into the second equation:

4x + 2 = 6x - 6 ,

2x = 8
x = 4

so y = 4x + 2 = 4(4) + 2 = 18

answer: 4 red and 18 blue marbles

Answer 5

Let there be R red marble
Let there be B blue marbles
(R-1)/R+B-1 = 1/7 or 7R-7 = R+B-1 or 6R = B+6
R/R+B-2 = 1/5 or 5R = R+B-2 or 4R = B-2

or 2R = 8
R= 4 and B = 18

Answer 6

Number of marbles is n with r ed and b lue ones.

Your two statements say:

(r-1)/(n-1) = 1/7

and

r/(n-2) = 1/5

You have two equations and two unknowns so eliminate r:

From the second r = (n-2)/5 so combined with the first:

(((n-2)/5)-1)/(n-1) = 1/7
7((n-2)/5)-1) = n - 1
7((n-2)-5) = 5(n - 1)
7n - 49 = 5n - 5
2n = 44
n = 22

Answer 7

either skip it or go to http://www.webmath.com/ hope it helped

Answer 8

I dont know.........Im sorry.