X and Y are two alloys of copper and aluminium prepared by mixing the metals in the ratio 6 : 5 and 5 : 9, respectively. If equal
quantities of the alloys are melted to form a third alloy Z, what ratio of copper and aluminium will be in Z?
This questions was previously asked in
SSC MTS 7th July 2022 Shift-2
154 : 163
Incorrect Answer165 : 139
Incorrect Answer139 : 154
Incorrect AnswerExplanation:
Let’s break down the problem:
- Alloy X has Cu : Al = 6 : 5. In 11 parts, 6 are Cu, 5 are Al.
- Alloy Y has Cu : Al = 5 : 9. In 14 parts, 5 are Cu, 9 are Al.
- Equal quantities are mixed. So, assume 11g of X and 14g of Y (the LCM of 11 and 14 is 154).
Copper:
- In 11g of X: 6g Cu,
- In 14g of Y: 5g Cu,
Total Cu = 6 + 5 = 11g
Aluminium:
- In 11g of X: 5g Al,
- In 14g of Y: 9g Al,
Total Al = 5 + 9 = 14g
However, we should scale both to 154g for equal proportions:
- 154g X: Cu = (6/11)*154 = 84, Al = (5/11)*154 = 70
- 154g Y: Cu = (5/14)*154 = 55, Al = (9/14)*154 = 99
- Total Cu = 84 + 55 = 139
- Total Al = 70 + 99 = 169
- Ratio = 139 : 169
So, the correct answer is:
Option 4, 139 : 169
![]()
By: santosh