CBSE Class 10 Answered
We are given that the distance = 1500 km
Let the time taken by the aeroplane at its usual speed be T hours and its usual speed be S kmph
Case 1(usual speed):
T = D/S = 1500/S ....(1)
Case 2(speed was increased):
Speed = (S+100) kmph
As the aeroplane left half an hour late and still reached the destination in time, time taken = (T - 1/2) hours
Therefore we have:
T - 1/2 = D/(S+100) = 1500/(S+100)
From (1):
1500/S - 1/2 = 1500/(S+100)
⇒1500/S - 1500/(S+100) = 1/2
⇒1/S - 1/(S+100) = 1/3000
⇒(S+100-S)3000 = S(S+100)
⇒300000 = S^2 + 100S
⇒S^2 + 100S - 300000 = 0
⇒S^2+600S-500S-300000 = 0
⇒S(S+600)-500(S-600) = 0
⇒(S-500)(S+600) = 0
Therefore, either S = 500 or S = -600. Since speed of the aeroplane cannot be negative, we reject S = -600. Therefore, the usual speed of the aeroplane is 500 kmph.
2.
We know that the diagonal of a parallelogram divides it into 2 triangles of equal areas.
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