Question

(18)
(
5 m ઊંચાઇએ રહેલા પાણીના નળ માંથી નિયત સમયે પાણીનાં બંદો
પડે છે.જયારે નળમાંથી પાણીનું ત્રીજું બંદ નીકળે છે, તે ક્ષણે પહેલું બુંદ
જમીનને સ્પર્શે છે, તો તે ટાણે પાણીનું બીજું બુંદ જમીનથી કેટલી
ઊંચાઇ એ હશે ? (1995)
(a) 3.75 777 (b) 4.0 int (c) 1.25 on (d) 2.5
(23) |ર
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Answer 1

Answer:

ANSWER EXPLANATION: There are two ways to solve this question. The faster way is to multiply each side of the given equation by ax−2 (so you can get rid of the fraction). When you multiply each side by ax−2, you should have:

24x2+25x−47=(−8x−3)(ax−2)−53

You should then multiply (−8x−3) and (ax−2) using FOIL.

24x2+25x−47=−8ax2−3ax+16x+6−53

Then, reduce on the right side of the equation

24x2+25x−47=−8ax2−3ax+16x−47

Since the coefficients of the x2-term have to be equal on both sides of the equation, −8a=24, or a=−3.

The other option which is longer and more tedious is to attempt to plug in all of the answer choices for a and see which answer choice makes both sides of the equation equal. Again, this is the longer option, and