Question

4/7 part of pole is underground.when 1/4 of this part is pulled out,5/2 of the pole is still under ground.what is the length of the pole​

Answer 1

Given :

Part of the pole underground

$= \frac{4}{7}$

The part pulled out from

$\frac{1}{4}$

$= \frac{5}{2}$

are still underground.

To Find :

What is the length of the pole ?

Solution :

Let us assume length of pole is L.Length of the pole in Mud is = L x 4/7 = 4L/7

Length of the Pole which is pulled out = L x 1/3 = L/3Length of

the pole is still in mud = Length of the pole in Mud - Length of the Pole which is pulled outLength of the pole is still in mud = 4L/7 - L/3

According to question,

Length of the pole is still in mud = 2504L/7 - L/3 = 250 (4L x 3 - L x 7)/7 x 3 = 250 12L - 7L = 250 x 215L

= 250 x 21L = 250 x 21/5L = 50 x 21

_______________

Answer 2

Answer :-

$\rm{Total \; length \; of \; the \; pole\; is =7.77cm}$

Explanation :-

Given :

• 4/7 part of the pole is in ground,when 1/4 part of the pole was pulled 5/2 of the pole still left.

To Find :

• Total length of the pole

Solution :

Let us assume length of pole be “x”

Length of the pole in Mud => 4x/7

Length of the Pole which is pulled out => 1x/4

Length of the pole is still in mud = Length of the pole in Mud - Length of the Pole which is pulled out

= 4x/7 - x/4

Length of the pole is still in mud = 5/2

According to question,

$\rm{\dfrac{4x}{7} - \dfrac{x}{4} = \dfrac{5}{2}}$

$\rm{ \dfrac{(4x \times 4 - x \times 7)}{28} = \dfrac{5}{2}}$

$\rm{\dfrac{(16x - 7x)}{28} = \dfrac{5}{2}}$

$\rm{\dfrac{9x}{28} = \dfrac{5}{2}}$

$\rm{9x\times 2 =5 \times 28}$

$\rm{18x =140}$

$\rm{x =\dfrac{140}{18}}$

$\rm{x =7.77}$

Total pole length is 7.77