In racing over a given distance D at uniform speed,A can beat B by 30 meters,B can beat C by 20 meters and A can beat C by 48 meters.Find D in meters.
Try my new set Algebra.
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Sir, Can you tell me why this is wrong? This is my approach.
If the Race is 100 meters, Then A:B=100:70
B:C= 100:80
Therefore, A:B:C= 1000:700:560.
So, When the race is 100 meters long, A beats C by 44 meters. So, By unitary Method, For A to Beat C by 48 Meters, The Race has to be
4 4 1 0 0 ∗ 4 8 = 109.0909... Meters.
@Rajen Kapur Sir. Kindly help. Thanks! :)
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@Sravanth Chebrolu , @Nihar Mahajan Can you Explain what is Wrong with my solution?
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Let me try once more. See that 30 m. and 20 m. are with respect to a certain length of race (which works out to 300 in this case) and not for any assumed length as per your choice. If you assume that distance is 100 then also know that 30 m will become 10 m.
Obviously your assumption that distance is 100m is misplaced. It is like this, if I give someone 40 Rupees, 360 Rupees remain with me. So what was the original amount with me. Assume that I had 100Rs. After giving 40 Rupees 60 Rupees remain. Now if in the question 360 Rupees remain then I must have started with 600 Rupees. Richer man.
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Thanks Sir. But would it have worked if I took 1000 m instead of 100?
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@Mehul Arora – It is algebra, it works for x when x = 300. Only one answer.
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@Rajen Kapur – But what about the Unitary method thing?
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Assuming that B and C have speeds B and C, respectively. Time taken to cover a certain distance is = s p e e d d i s t a n c e . Equating the race time(s): B D − 3 0 = C D − 4 8 B D = C D − 2 0 Dividing out, D(D - 48) = (D - 30)(D - 20) or 2D = 600 or D = 300.