Question 1103145: Hi
an iron bar is cut into 3 equal parts. One half of A , three eighths of B and one third of C had a mass of 58 kg. What was the original mass of the iron bar.
thanks
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Cut the bar into three parts: A, B, C
They are equal parts (as given in the instructions), so A = B = C. So we can refer to B as A. The same applies for C as well.
Half of A, plus three-eights of B, plus one-third of C gives us
(1/2)*A + (3/8)*B + (1/3)*C
We can replace B with A. We can replace C with A
(1/2)*A + (3/8)*A + (1/3)*A
Then combine like terms by adding fractions
Don't forget to get the LCD first before adding the fractions (the denominators must be the same)
(1/2)*A + (3/8)*A + (1/3)*A
(1/2+3/8+1/3)*A
(12/24+9/24+8/24)*A
((12+9+8)/24)*A
(29/24)*A
The expression (29/24)*A represents the mass of 58 kg, which is the other piece of given information. So we can set it equal to 58 and solve for A
(29/24)*A = 58
(24/29)*(29/24)*A = (24/29)*58 Multiply both sides by the reciprocal of 29/24
A = (24/29)*(58/1)
A = (24*58)/(29*1)
A = (24*29*2)/(29*1)
A = (24*2)/(1)
A = 48
Mass A is 48 kg. So is mass B and C.
Therefore, the total mass is A+B+C = 48+48+48 = 144 kilograms
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