Beaker A and Beaker B contains 320ml of water. 3 metal rods and 4 beads were put in to Beaker A and the volume is 433cm cube. 4 metal rods and 3 beads were put in to Beaker B and the volume is 438cm cube. What is the volume of each metal rod and bead?
2006-09-04 17:49:30 · 6 answers · asked by megal 1 in Science & Mathematics ➔ Mathematics
Oh i think i got it !
Rod = 19cm cube
Bead = 14cm cube
am i right ?
2006-09-04 17:55:18 · update #1
let r be the volume of rod and b be the volume of bead
3r + 4d = 113
4r + 3d = 118
7r + 7d = 231
r+d = 33
r-d = 5
r = 5+d
5+d+d = 33
2d = 28
d = 14
r = 5 +14 = 19
2006-09-04 18:00:40 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Let the volumes of a rod and a bead be r and b cm3 respectively.
3r + 4b = 433 - 320 = 113
4r + 3b = 438 - 320 = 118
Eqn1*4: 12r + 16b = 452
Eqn2*3: 12r + 9b = 354
Eqn3 - Eqn4: 7b = 98
Therefore, b=14
Substitute into Eqn1: 3r + 4*14 = 113
3r = 57
r=19
2006-09-04 17:55:46 · answer #2 · answered by klwh_88 2 · 0⤊ 0⤋
Let us use r = volume of rod and b = volume of bead, using the given, you can establish two equations:
(1) 320+3r+4b = 433
(2) 320+4r+3b = 438
You can establish an equation for r using the first equation, substitute this in the second equation and you will get the value of b which is 14, substitute this in your equation for r and you will get r=19. Thats it...volume of one rod = 19ml and volume of one bead = 14ml
2006-09-04 18:08:39 · answer #3 · answered by ken_masters 1 · 0⤊ 0⤋
Consider,
x=weight of a metal rod
y=weight of a beads. and 1ml=1 cu cm.
3 metal rods and 4 beads were put in to Beaker A and the volume is 433cm cube.
3x+4y=(433-320)cm cu. =113 .............................(1)
4 metal rods and 3 beads were put in to Beaker B and the volume is 433cm cube
4x+3y=438-320=118 cm cu. .............................(2)
From (1) and (2) we have
x-y=5
x=y+5
put this value in equation (1)
3(y+5)+4y=113cm cu.
3y+15+4y=113
7y=113-15=98
y=98/7
y=14 ml = .014 kg weight of a bead.
now we have ,
x=5+y=5+14=19 ml =.019kg weight of a metal rod.
2006-09-04 20:45:25 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Beaker A increased by 113 ml and B increased 118 ml. (As I recall, ml=cubic centimeter.) Let volume of rod = r and volume of bead = b. So 113 = 3r + 4b and 118 = 4r + 3b. You take it from there.
2006-09-04 17:59:51 · answer #5 · answered by banjuja58 4 · 0⤊ 0⤋
its good that u already got it!
2006-09-04 18:37:45 · answer #6 · answered by darLa 2 · 0⤊ 0⤋