1. After 2/3 of the marbles are removed from a jar, 5 more than 1/6 of the marbles remain.
2. In a jar, there are 4 more than twice as many blue marbles as red marbles.
thats all, thanks to all who answered!!!
2007-07-23 12:45:31 · 8 answers · asked by Ann 3 in Science & Mathematics ➔ Mathematics
Hi,
1. After 2/3 of the marbles are removed from a jar, 5 more than 1/6 of the marbles remain.
x - 2/3x = 1/6x + 5 where x = original number of marbles
2. In a jar, there are 4 more than twice as many blue marbles as red marbles.
4 + 2B = R
I hope those help!! :-)
2007-07-23 12:52:15 · answer #1 · answered by Pi R Squared 7 · 0⤊ 0⤋
Use logic!
1. if a jar has X marbles & 2/3 are removed, what is left?
Ans 1/3X or X/3.
X/6 is one sixth of the marbles so:
X/3=X/6 + 5 need to add 5 to one sixth to make them equal.
Now solve to get: 2X=X+30 Multiply both sides by 6
X=30 Marbles.
2. Red = R Blue=B
R= 4 +2B
since there are 4 more red than twice the blue then add 4 to the blue to make them equal.
2007-07-23 19:50:21 · answer #2 · answered by 037 G 6 · 0⤊ 0⤋
1) Let j = number of marbles in jar (initially)
j - (2/3)j = 5 + (1/6)j
Read: start with number of marbles, subtract 2/3 of them. This equals 5 more than 1/6 of the marbles.
Solve:
Multiply by 6 (the LCM of the fractions' denominators):
6j - 4j = 30 + j
2j = 30 + j
j = 30
2) not solvable; you're missing some information
2007-07-23 19:54:05 · answer #3 · answered by Tony The Dad 3 · 0⤊ 0⤋
1. Let x = number of marbles
x - (2/3)x = (1/6)x + 5
2. Let x = number of red marbles
2x + 4 = number of blue marbles
2007-07-23 19:50:12 · answer #4 · answered by kousuke51 2 · 0⤊ 1⤋
1. 1/3x = 1/6 + 5
2. x = blue
red = 4 +2x
2007-07-23 19:50:42 · answer #5 · answered by tequilujan 2 · 0⤊ 0⤋
x=y-2/3
x=1/6(y)+5
2007-07-23 19:58:59 · answer #6 · answered by Anonymous · 0⤊ 0⤋
1. M -2M/3 = M/6 + 5.
2. B = 2R + 4.
2007-07-23 19:49:59 · answer #7 · answered by Anonymous · 0⤊ 0⤋
I will have to think on this one
2007-07-23 19:49:41 · answer #8 · answered by smittybo20 6 · 0⤊ 0⤋