Question

there is a frog in the well which is 24m deep. the frog wants to jump out of the well. every time jumps 2/3 m and fall back by 1/4 m. in how many jumps will he come out of tha well?

Answer 1

$<b><i>Hey mate$

✨ Here's ur answer

Given, depth of the well = 24m

Upward distance covered by frog in one jump =
$\frac{2}{3}$
m

Downward distance moved when he fall back in one jump =

$\frac{1}{4}$
m

Therefore, total distance covered by frog

$= \frac{2}{3} m - \frac{1}{4} m \\ = \frac{8m - 3m}{12} \\ = \frac{5}{12} m$
of height required one jump of frog.

Therefore, 24m of height required $= \frac{ \frac{1}{5} }{12} \times 24 \\ \frac{12}{5} \times 24 \\ \frac{288}{5} = 57.6$

Here the jump comes in decimals which is not possible.

So in56th jump distance covered by frog
$\frac{5}{12} \times 56 = \frac{280}{12} m$

Now in 57th jump he moves,
$\frac{2}{3 } m$
Then falls back by
$\frac{1}{4} m$

Therefore height in 57th jump=

$\frac{280}{12} + \frac{2}{3} = \frac{288}{12} = 24m$

$<u>Hence, in 57th jump, the frog will move out of the well.</u>$

$\huge \red {Thanks!!!}$

Answer 2

Given data:

Depth of well = 24 m

In 1 jump, he goes up = 2/3 m

and falls back = 1/4 m
___________________

Net upward movement of frog in 1 jump:

$\frac{2}{3} - \frac{1}{4} = \frac{2 \times 4 - 1 \times 3}{12} = \frac{8 - 5}{12} = \frac{5}{12} \ m$
__________________

$\textbf{The last jump:}$

Consider the last jump. When it jumps for the last time, it will reach the top and stay there. It will not fall back by 1/4 meter. So in the last jump, he will only go up by 2/3 meter.

So remaining height to ascend is:

$24 - \frac{2}{3} = \frac{24 \times 3 - 2}{3} = \frac{72 - 2}{3} = \frac{70}{3} \: m$
__________________
$\textbf{remaining jumps:}$

the frog ascends 5/12 m in 1 jump.

To ascend remaining 70/3 meter, number of jumps will be:

$\frac{70}{3} \div \frac{5}{12} \\ \\ = \frac{70}{3} \times \frac{12}{5} \\ \\ = 14 \times 4 \\ \\ = 56 \: jumps$
__________________

$\textbf{Total jumps = 56 + 1 = } \boxed{ \textbf{57 \: jumps}}$