A health-food shop sells three grades of fiber products, grade A for $1.80 a pound, grade B for $1.30 a pound, and grade C for 65 cents a pound. The shopkeeper uses all three grades of fiber products to make 21 pounds of a mixture that will sell for $1 a pound. He uses twice as many pounds of grade C fiber as grade B fiber. How many pounds of each grade of fiber does he use?
2006-12-14 05:13:44 · 12 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
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BTW it is 3, 6, 12.
2006-12-14 08:38:26 · answer #1 · answered by Anonymous · 0⤊ 0⤋
A = 3 pound
B = 6 pound
C = 12 pound
2006-12-14 05:25:27 · answer #2 · answered by AndyPandy 3 · 0⤊ 0⤋
I think we need to make two or more equations to solve the problem. its those equations using x and y thing.
One equation will be about all the pounds of A ,B,C added up together equals to 21.
---> let the pounds used for grade A be x
let the pounds used for grade B be y
let the pounds used for grade C be 2y
thus ,
x + y +2y = 21
-> x + 3y = 21 ..... equation 1
second equation would be of the total prices of all grades used divided by 21 pounds equals to one pound each.
---> 1.8x + 1.3y + 1.3y
______________ = 1
21
1.8x + 2.6y = 21 ....... ( 2nd equation)
thus, we have
x+3y = 21 .....(1)
1.8x + 2.6 y = 21 .....(2)
---> x+3y = 1.8x + 2.6y
3y - 2.6y = 1.8x - x
0.4y = 0.8x
y = 0.8x / 0.4
y = 2x ....... (3)
We substitute y with equation three in equation 1
---> x+3y = 21 .......(1) becomes
x + 3 (2x) = 21
x + 6x = 21
7x = 21
x = 3
so we can find the value of Y
--- > y = 2x
y = 2 . 3
= 6
THEREFORE,
pounds used for each grade =
A (x pound ) = 3 pounds
B (y pound) = 6 pounds
C (2y pound) = 12 pounds
is that correct?
hit me back, im curious. never practiced my maths for so long. thankyou.
2006-12-14 05:51:10 · answer #3 · answered by Lovedrug 2 · 0⤊ 0⤋
A 1.8 / lb
B 1.3 / lb
C 0.65/lb
To avoid using fraction lets use the given basis of 21 lbs
A+B+C = 21 lbs
A(1.8) + B(1.3) + C (0.65) = 21 (1)
B = C(2)
Eliminate B from the first two equations using the third equation.
Solve either of the equations for either A or B, substitute it into the other equation and solve for the last varible. You have 3 unknowns and three equations.
Its not as hard as you think.
2006-12-14 05:26:03 · answer #4 · answered by Roadkill 6 · 0⤊ 0⤋
A + B + C = 21 pounds
$1.8A + $1.3B + $.65C = $1(21)
2C=B
Substitute 2C for B into the first equation.
A + 2C + C = 21 so A = 21 - 3C
Substitute A = 21 - 3C and B = 2C into the second equation.
1.8(21-3C) + 1.3(2C) + .65C = 1(21)
so,
C = 12, then
B = 6, substitute B and C back into the original equation to get
A = 3
2006-12-14 05:29:03 · answer #5 · answered by slider 2 · 0⤊ 0⤋
12 pound of C at .65 is 7.80
6 pound of B at 1.30 is 7.80
3 pound of A at 1.80 is 5.40
21 pound = 21 dollar = 1 dollar per pound
2006-12-14 05:20:06 · answer #6 · answered by smitty4626 3 · 0⤊ 0⤋
enable d = the type of dogs enable b = the type of boys we may be able to write equations. First, write one for this sentence: there have been 40 heads in all. each canines and each boy has actually a million head, precise? d + b = 40 Write one for this one: there have been a finished of one hundred ft. each canines has 4 ft, so as it truly is 4d and each boy has 2 ft, so as it truly is 2b. 4d + 2b = one hundred d + b = 40 4d + 2b = one hundred -2(d + b = 40) 4d + 2b = one hundred -2d - 2b = -80 -------------------- 2d = 20 d = 10 b + d = 40 b + 10 = 40 b = 30 there have been 10 dogs and 30 boys. examine it. each canines has 4 ft, so as it truly is 40 ft. and each boy has 2 ft, so as it truly is 60 ft. 60 + 40 = one hundred, examine! and each canines has a million head, so as it truly is 10 heads. and each boy has a million head, so as it truly is 30 heads. 30 + 10 = 40, examine!
2016-10-18 07:11:50 · answer #7 · answered by Anonymous · 0⤊ 0⤋
If you meant weight of grade B is equal to twice of weight of grade C, then I couldn't find a positive answer.
However, if you meant weight of grade C is equal to twice of weight of grade B, then the answer I got is
A = 3 Ib and B = 6 Ib and C = 12Ib
2006-12-14 05:33:57 · answer #8 · answered by beastbelle 2 · 0⤊ 0⤋
No I'm not. This translates to:
A+B+C = 21
1.80A + 1.30B + 0.65C = 21
C=2B.
Solve for A, B, C.
2006-12-14 05:18:34 · answer #9 · answered by Anonymous · 1⤊ 0⤋
1.30 + (.65*2) = 2.60
(2.60 + 1.8X)/ (3+x) = 1
Solve for X
2006-12-14 05:26:56 · answer #10 · answered by Steve C 3 · 0⤊ 0⤋