CAT 2025

Slot 1

QA

Question & Solution

1. Concept Used

• Topic: Arithmetic Progression (AP) applied to Time-Speed-Distance
• Formula: $$ \text{Distance} = \text{Speed} \times \text{Time} \quad \text{and} \quad S_n = \frac{n}{2}(2a + (n-1)d) $$

2. Calculation

Step 1: Find the time for each part.

The total travel time is 3 hours = 180 minutes. The time for each of the four parts forms an AP. Since the first part takes 30 minutes, let the common difference of the time AP be $$d_t$$ minutes. The four times are: $$30, \quad (30 + d_t), \quad (30 + 2d_t), \quad (30 + 3d_t)$$

Their sum equals 180 minutes: $$30 + (30 + d_t) + (30 + 2d_t) + (30 + 3d_t) = 180$$ $$120 + 6d_t = 180$$ $$6d_t = 60 \implies d_t = 10 \text{ minutes}$$

So the four times are: 30, 40, 50, 60 minutes.

Step 2: Find the speed for each part.

The total distance is 224 km = 224,000 metres. The first part speed is 960 m/min. Let the common difference of the speed AP be $$x$$ m/min. The four speeds are: $$960, \quad (960 + x), \quad (960 + 2x), \quad (960 + 3x) \text{ m/min}$$

Total distance = sum of (speed × time) for each part: $$960 \times 30 + (960 + x) \times 40 + (960 + 2x) \times 50 + (960 + 3x) \times 60 = 224000$$ $$960(30 + 40 + 50 + 60) + 40x + 100x + 180x = 224000$$ $$960 \times 180 + 320x = 224000$$ $$172800 + 320x = 224000$$ $$320x = 51200 \implies x = 160 \text{ m/min}$$

Step 3: Find the distance in the fourth part.

Speed in the fourth part: $$960 + 3 \times 160 = 960 + 480 = 1440 \text{ m/min}$$

Time for the fourth part = 60 minutes.

Distance in the fourth part: $$1440 \times 60 = 86400 \text{ metres}$$

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3. Solution

Answer = Option D ✅

The distance Shruti travels in the fourth part is 86,400 metres.