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mike runs four times as fast as jun. if mike gives jun a head start of 15 secs, how many secs will mike catch up with jun?
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when mike starts jun will be 15*x far away then : // x=speed of jun 4xt=15x+t*x // t=time in secs 3xt=15x t=5 then mike will catch up with jun after he runs for 5 seconds
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