SBI PO Prelims 8 Mar 2025 Shift 3 Paper

Concept:Form equations using speed-distance-time relations and relative speed concept, then solve a quadratic equation.Explanation:Let the speed of car A be $a$ km/h.Then speed of car B is $a + y$ km/h.First condition: $120$ km distance, time taken by A is 1 hour more than time taken by B.Equation: $\frac{120}{a} = \frac{120}{a + y} + 1$ ...(1)Second condition: Car B increases speed by 10 km/h, so new speed of B = $a + y + 10$ km/h.They are 160 km apart and moving in opposite directions. They meet after 2 hours.Relative speed = sum of speeds = $a + (a + y + 10) = 2a + y + 10$ km/h.Distance = relative speed × time: $160 = (2a + y + 10) \times 2$Thus $2a + y + 10 = 80$ → $2a + y = 70$ → $y = 70 - 2a$.Substitute $y$ into (1):$\frac{120}{a} = \frac{120}{(a + 70 - 2a)} + 1 = \frac{120}{70 - a} + 1$Multiply both sides by $a(70 - a)$:$120(70 - a) = 120a + a(70 - a)$$8400 - 120a = 120a + 70a - a^2$$8400 - 120a = 190a - a^2$$a^2 - 310a + 8400 = 0$Factorize: $(a - 30)(a - 280) = 0$Thus $a = 30$ or $a = 280$.Given both speeds less than 100 km/h, so $a = 30$ km/h.Answer:30 kmph (Option C)