Multiple choice

In a steel factory, an engineer prepared two alloys L and M of iron and carbon by mixing the metals in the ratios 8 : 3 and 8 : 13, respectively. If equal quantities of the alloys are melted to form a third alloy O, then what is the ratio of iron and carbon in alloy O?

  1. 128 : 103

  2. 51 : 43

  3. 49 : 13

  4. 19 : 21

  5. 15 : 28
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A Correct answer
Explanation

Alloy L has iron and carbon in ratio 8:3 (total 11 parts), so iron is 8/11 and carbon is 3/11. Alloy M has ratio 8:13 (total 21 parts), so iron is 8/21 and carbon is 13/21. Mixing equal quantities means taking the average of the fractions: Iron = (8/11 + 8/21)/2 = (168+88)/(231*2) = 256/462 = 128/231. Carbon = (3/11 + 13/21)/2 = (63+143)/(231*2) = 206/462 = 103/231. The ratio is 128:103.

AI explanation

To compare the alloys, the total parts for L and M are 11 and 21 respectively, so use a common multiple of 231 for equal quantities. Alloy L has 168 parts iron and 63 parts carbon, while alloy M has 88 parts iron and 143 parts carbon. Adding these together gives 256 parts iron and 206 parts carbon, which simplifies by dividing by 2 to a final ratio of 128 to 103.

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