Question

A boy finds a magical pool. Every time he jumps in the pool, the width shrinks by1/3and length shrinks by 1/2. He jumps in the pool a total of 3 times. The final area of the pool is 4 square meters. The pool was originally 9m wide. What was the original length of the pool?

Answer 1

Given:

Original width of the pool = 9 m

Every time the boy jumps in the pool, the width shrinks by 1/3 and length shrinks by 1/2

Total no. of times he jumps into the pool = 3

Final area of the pool = 4 m²

To find:

The original length of the pool

Solution:

Formula to be used:

Area of the rectangle = length × breadth

Let "l" meters be the original length of the pool.

It is given that the width = 9 m of the pool shrinks by 1/3rd every time and he jumps for a total of 3 times. So, we will work out the final width of the pool

9 - (9/3) = 9 - 3 = 6

6 - (6/3) = 6 - 2 = 4

4 - (4/3) =8/3 m

∴ The final width of the pool after shrinking for 3 times becomes = $\frac{8}{3} \:m$

Also, the length = "l" m of the pool shrinks by 1/2 every time and he jumps for a total of 3 times. So, we will work out the final length of the pool

$l - \frac{l}{2} = \frac{l}{2} \\\\\frac{l}{2} - [\frac{l}{2}\times\frac{1}{2} ] = \frac{l}{2} - \frac{l}{4} = \frac{l}{4} \\\\\\\frac{l}{4} - [\frac{l}{4}\times\frac{1}{2} ] = \frac{l}{4} - \frac{l}{8} = \frac{l}{8}$

∴ The final length of the pool after shrinking for 3 times becomes = $\frac{l}{8} \:m$

Now, using the formula of the rectangle, we get

Final area of the pool = [Final length] × [Final width]

⇒ $4$ = [ $\frac{l}{8}$ ] × [ $\frac{8}{3}$ ]

⇒ $4$ = $\frac{l}{3}$

⇒ $l = 4\times3$

⇒ $l = \bold{12\:m}$

Thus, the original length of the pool was 12 m.

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Answer 2

Answer:

The original length of the pool was 12 m.

Step-by-step explanation: