Question

At t minutes past 2 pm the time needed by the minutes hand and a clock to show 3pm was found to be 3 min less than t^2/4 minutes. Find t.

Answer 1

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Given time needed by the minutes hand show (t2/4) – 3
Hence the quadratic equation becomes, (t2/4) – 3 = 60 – t
(t2 – 12)/4 = 60 – t
(t2 – 12) = 240 – 4 t
t2 – 12 – 240 + 4 t = 0
t2 + 4t – 252 = 0
t2 + 18t – 14t – 252 = 0
t(t + 18) – 14(t + 18) = 0
(t + 18)(t – 14) = 0
(t + 18) = 0 or (t – 14) = 0
∴ t = - 18 or t = 14
Since t cannot be negative, t = 14 

Answer 2

Hope it helps...!!!!

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Since minute hand is t min more than 2 PM minutes hand has completed t min after 2 PM.

i.e. to show 3 PM it will cover 60-t min more.

Therefore our equation becomes;

60-t=t^2/4-3

t^2/4+t-63=0

t^2+4t-252=0

t^2+18t-14t-252=0

(t+18)(t-14)=0

Therefore t=14 or t=-18

Since t can not be negative t = 14 min.

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