A person was working to transfer bricks from a place X to a place Y, where the distance between X and Y is 200 meters. At a time the person was carrying 15 bricks. Starting from the place X, after travelling 5.5 km while working, how many bricks were shifted to the place Y by the person ?
210
The person shuttles between X and Y: each load of 15 bricks is delivered after walking 200 m from X to Y, and the walk back to X adds another 200 m. The question is how many deliveries fall within 5.5 km of total walking.
Step 1 — Convert the distance: \(5.5 \text{ km} = 5500 \text{ m}\).
Step 2 — One full round trip X to Y to X is \(200 + 200 = 400 \text{ m}\) and delivers one load.
Step 3 — Number of complete round trips: \(\left\lfloor \frac{5500}{400} \right\rfloor = 13\), covering \(13 \times 400 = 5200 \text{ m}\) and delivering 13 loads.
Step 4 — Remaining distance: \(5500 - 5200 = 300 \text{ m}\). The first 200 m of this takes the person from X to Y with a 14th load; the last 100 m is part of the return walk with no bricks.
Step 5 — Bricks shifted: \(14 \times 15 = 210\).
The figure 195 equals \(13 \times 15\), counting only the 13 complete round trips and missing the 14th delivery made within the final 300 m. The figures 190 and 215 are not multiples of 15, so they cannot be the result of whole 15-brick loads.
210 option is correct.
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