Two Alloys of Iron have different percentages of Iron in them. The first of them weights 6 kg and the second one weights 12 kg. One piece of equal weight was cut off from both Alloys. The peice from the first was mixed with the leftover of the second Alloy and the peice from the second Alloy with the leftover part of the first alloy. As a result the percentage of iron became same in both the Alloys. What was the weight of each piece of cut off?
Answers
Answered by
10
Step-by-step explanation:
Given Two Alloys of Iron have different percentages of Iron in them. The first of them weights 6 kg and the second one weights 12 kg. One piece of equal weight was cut off from both Alloys. The piece from the first was mixed with the leftover of the second Alloy and the piece from the second Alloy with the leftover part of the first alloy. As a result the percentage of iron became same in both the Alloys. What was the weight of each piece of cut off?
- Let x and y be the percentage of aluminium in the 6 kg and 12 kg piece.
- So taking away m kg away from both we get
- So we have xm and ym amount of aluminium in both pieces.
- So from the remaining 6 – m and 12 – m pieces the amount of aluminium will be
- (6 – m) x + my / 6 = (12 – m) y + mx / 12
- Therefore we get
- (6 – m) x + my = (12 – m) y + mx / 2
- 2(6x – mx + my) = (12y – my + mx)
- 12x – 2mx + 2my = 12y – my + mx
- 12x – 2mx + 2my – 12y + my – mx
- 12 (x – y) – 3mx + 3my
- 12 (x – y) = 3mx – 3my
- 12(x – y) = 3m (x – y)
- So 3m = 12
- Or m = 4
- So the weight of each piece of cut off is 4 kg
Reference link will be
https://brainly.in/question/16067814
Answered by
0
Answer:
4 kg
Step-by-step explanation:
let K be the piece of alloy from both the same alloy
so. 6-k/6 = l/12
6-k = k/2
2(6-2k) = k
12-2k = k
12 = 3k
k=4kg
Similar questions