What number should be subtracted from both 31 and 40 so that their third proportional is 5 more than 43?
Solution
Correct Answer: Option C
Let the number subtracted be T.
⇒ Third proportional of (31-T) and (40-T) is 5 more than 43.
⇒ (31 – T)/(40 – T) = (40 – T)/48
⇒ 1488 – 48T = 1600 + T2 – 80T
⇒ T2 – 32T + 112 = 0
⇒ (T – 28)(T – 4) = 0
∴ T can be either 28 or 4, but among the options only option 4 is given. So, 4 must be subtracted from both 31 and 40 so that their third proportional is 5 more than 43.
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