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Question: A farmer planted louts in the village pond. One fine day, to the happiness of the villagers, one lot...

A farmer planted louts in the village pond. One fine day, to the happiness of the villagers, one lotus flower has bloomed.
The next day the two flowers were seen. The third day four flowers were seen. Everyday they saw double the number of flowers then the previous day. On the 30th day the villagers found that half of the pond was filled with lotus flowers. How many days more will it take to fill the pond with lotus flowers? In how many days the pond was quarter full with lotus flowers?

Explanation

Solution

Hint – This type of question comes under Geometric progression. Geometric progression is a sequence of numbers where each term is multiplied by the same factor to obtain the next number in sequence. In this question we used a simple method, we equate the day and numbers of flowers with other days and numbers of flowers and find that day.

Complete step-by-step answer:
In this question, it is given that;
The flowers double the previous day i.e. 1,2,4,8...
Let the total no of flowers in the pond be and total days are d.
So we can say that
At the dth{d^{th}} day no of flowers in the pond is a.
At the (d1)th{\left( {d - 1} \right)^{th}} day no of flowers in the pond isa2\dfrac{a}{2}.
At the (d2)th{\left( {d - 2} \right)^{th}} day no of flowers in the pond is a4\dfrac{a}{4} .
And so on...
From question,
\because On the 30th day half of the pond filled with lotus flowers.
That means at the 30th day no of flowers =a2 = \dfrac{a}{2}
So, (d-1) =30
d =30+1
d =31
Thus, the pond is fully fielded with lotus flowers at 31th{31^{th}} day.
And also quarter full i.e. a4\dfrac{a}{4} is at (d-2) =31-2=29
So, the pond is quarter fielded with lotus flowers at29th{29^{th}} day.

Note – To find the no of flowers on 31th{31^{th}} day, we use the formula sum of n term of GP issn=a(rn1)r1{s_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}} , where; a= first term of GP, r= common ratio between two successive terms, and sn{s_n} is sum of n term of GP. Put value and get the answers31=1(2311)311{s_{31}} = \dfrac{{1\left( {{2^{31}} - 1} \right)}}{{31 - 1}}. And also we find no flowers on any day. In GP the ratio of any two adjacent numbers is the same.