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Some byes are hi(I) + All hi are long(A) ⇒ Some byes are long(I) + All long are far(A) ⇒ Some byes are far(I) ⇒ Conversion ⇒ Some far are byes(I). Hence conclusion I follows. Some hellos are byes(I) + Some byes are hi(I) ⇒ No conclusion. Hence conclusion II does not follow. Some byes are hi(I) + All hi are long(A) ⇒ Some byes are long(I) ⇒ Conversion ⇒ Some long are byes. (I). Hence conclusion III follows. All hi are long(A) + All long are far(A) ⇒ All hi are far(A) ⇒ Conversion ⇒ Some far are hi(I). Hence conclusion IV follows. ALTERNATE SOLUTION:
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