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ATQ, As per the problem, The distance Person 'A' covers in the first 36 minutes is (36/60) × 120 = 72 km In the subsequent 36 minutes, the distance covered is = (36/60) × (120 – 5) = 69 km In the following 36 minutes, the speed reduces by another 5 km/hr, resulting in a distance of= (36/60) × (115 – 5) = 66 km/hr The distance travelled in the intervals of 36 minutes each forms an A.P. = (72, 69, 66…… n)km Where, a = 72, d = (69 – 72) = -3 Therefore, (n/2){2a + (n – 1)d} = 492 Or, (n/2){2 × 72 + (n – 1)(-3)} = 492 Or, 144n – 3n2 + 3n = 984 Or, n2 – 49n + 328= 0 Or, n2 – 8n – 41n + 328 = 0 Or, n(n – 8) – 41(n – 8) = 0 Or, n = 41, 8 However, n = 41 would result in a total time of (41 × 36 = 1476) minutes, which would exceed the distance of 492 km. Thus, the suitable value is (n=8),leading to a total time of 36 × 8 = 288 minutes
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