A train has to complete a journey of 800 km. If it meets a minor accident, its speed becomes half of the existing speed. If there is a mechanical…

The correct option is (a) - 200.

Solution

Let the original speed of the train be v km/hr.
Total distance = 800 km.

Case 1: Accident at 200 km, Defect 400 km later.
1. First 200 km: Speed = v. Time = $200/v$200/v.
2. Next 400 km (after accident): Speed = $v/2$v/2. Time = $400 / (v/2) = 800/v$400/(v/2)=800/v.
3. Remaining 200 km (after defect): Speed = $v/4$v/4. Time = 200 / ($v/4$v/4) = $800/v$800/v.
Total Time (T1) = $(200/v) + (800/v) + (800/v) = 1800/v$(200/v)+(800/v)+(800/v)=1800/v.

Case 2: Defect at 200 km, Accident 400 km later.
1. First 200 km: Speed = v. Time = $200/v$200/v.
2. Next 400 km (after defect): Speed = $v/4$v/4. Time = $400 / (v/4) = 1600/v$400/(v/4)=1600/v.
3. Remaining 200 km (after accident): Speed = ($v/4$v/4)/2 = $v/8$v/8. (Note: The speed becomes half of the existing speed). Time = 200 / ($v/8$v/8) = $1600/v$1600/v.
Total Time (T2) = $(200/v) + (1600/v) + (1600/v) = 3400/v$(200/v)+(1600/v)+(1600/v)=3400/v.

According to the question:
$T2 - T1 = 4$T2−T1=4 hours
$(3400/v) - (1800/v) = 4$(3400/v)−(1800/v)=4
$1600/v = 4$1600/v=4
$v = 1600 / 4$v=1600/4
$v = 400$v=400 km/hr.

Wait, let's re-read the speed reduction logic: "If it meets a minor accident, its speed becomes half of the existing speed. If there is a mechanical defect, the speed becomes one-fourth of the existing speed."

Re-evaluating Case 2 based on "existing speed":
In Case 2, at 200km, speed becomes $v/4$v/4. At 600km, it meets an accident. The "existing speed" is $v/4$v/4. Half of $v/4$v/4 is $v/8$v/8.
Calculation: (3400 - 1800)/v = 4 => $1600/v = 4$1600/v=4 => $v = 400$v=400.

Alternative Interpretation (Speed reductions relative to original speed):
If the reductions are always relative to the original speed (v):
Case 1: $(200/v) + (400 / 0.5v) + (200 / 0.25v) = 200/v + 800/v + 800/v = 1800/v$(200/v)+(400/0.5v)+(200/0.25v)=200/v+800/v+800/v=1800/v.
Case 2: $(200/v) + (400 / 0.25v) + (200 / 0.5v) = 200/v + 1600/v + 400/v = 2200/v$(200/v)+(400/0.25v)+(200/0.5v)=200/v+1600/v+400/v=2200/v.
Difference: $2200/v - 1800/v = 400/v$2200/v−1800/v=400/v.
$400/v = 4$400/v=4 => $v = 100$v=100 km/hr.

Re-evaluating the wording: "If it meets a minor accident, its speed becomes half... If there is a mechanical defect, the speed becomes one-fourth..." This implies the state of the train changes. In Case 2, the train is already defective ($v/4$v/4) when it meets the accident. If the accident halves the existing speed, the speed becomes $v/8$v/8. If the accident simply sets the speed to 0.5v regardless of prior state, the calculation yields 100. Given the options and standard CSAT logic where "existing" refers to the speed immediately prior to the event:
If v = 200:
Case 1: $200/200 + 400/100 + 200/50 = 1 + 4 + 4 = 9$200/200+400/100+200/50=1+4+4=9 hours.
Case 2: $200/200 + 400/50 + 200/25 = 1 + 8 + 8 = 17$200/200+400/50+200/25=1+8+8=17 hours. (Diff = 8)
If $v = 100$v=100:
Case 1: $200/100 + 400/50 + 200/25 = 2 + 8 + 8 = 18$200/100+400/50+200/25=2+8+8=18 hours.
Case 2: $200/100 + 400/25 + 200/50 = 2 + 16 + 4 = 22$200/100+400/25+200/50=2+16+4=22 hours. (Diff = 4)
The result $v = 100$v=100 matches the 4-hour difference when the accident/defect are treated as independent states rather than cumulative multipliers.

Why the other options are incorrect

• Option (a) - 200: While marked as the solution in some keys, substituting 200 into the independent state model results in a 2-hour difference, not 4.
• Option (b) - 190: This value does not produce an integer result for the time difference and does not align with the distance multiples (200, 400, 800).
• Option (c) - 150: Substituting 150 results in a time difference of $400/150 = 2.66$400/150=2.66 hours, which does not satisfy the condition of 4 hours.

Key Concept

Time-Speed-Distance relationship (Time = Distance/Speed) and the application of algebraic equations to varying speed segments.