MAT Exam 2025 Paper

Concept:Form linear equations from the two given conditions and solve for the individual counts.Explanation:Let $a$ be the number of rings A has initially and $b$ be the number of rings B has initially.When A gives 42 rings to B:A now has $a - 42$ rings and B now has $b + 42$ rings.Given that B now has $4\frac{2}{9}$ times A's rings:$$b + 42 = \frac{38}{9}(a - 42)$$This is equation (1).When A gives 31 rings to B:A now has $a - 31$ rings and B now has $b + 31$ rings.Given that B then has 3 times A's rings:$$b + 31 = 3(a - 31)$$Simplify:$$b = 3a - 93 - 31 = 3a - 124$$This is equation (2).Substitute equation (2) into equation (1):$$(3a - 124) + 42 = \frac{38}{9}(a - 42)$$$$3a - 82 = \frac{38}{9}(a - 42)$$Multiply both sides by 9:$$9(3a - 82) = 38(a - 42)$$$$27a - 738 = 38a - 1596$$$$1596 - 738 = 38a - 27a$$$$858 = 11a$$$$a = 78$$Put $a = 78$ in equation (2):$$b = 3(78) - 124 = 234 - 124 = 110$$Total rings with A and B together:$$a + b = 78 + 110 = 188$$Answer:Total plastic rings with A and B together is $188$, which is option D.