Question

A certain height of pole is fixed in a lake whose one third is in mud, one fourth of the remaining in
sand and half of the remaining
in water. If rest 6 m is seen above water, find total length of the pole.​

Answer 1

Answer:

ans-24

Step-by-step explanation:

let height of the pole is x .

let height of the pole is x .so in mud x/3

let height of the pole is x .so in mud x/3remaining=2x/3

let height of the pole is x .so in mud x/3remaining=2x/3in sand=1/4×2x/3

let height of the pole is x .so in mud x/3remaining=2x/3in sand=1/4×2x/3 =x/6

let height of the pole is x .so in mud x/3remaining=2x/3in sand=1/4×2x/3 =x/6total=x/3+x/6

let height of the pole is x .so in mud x/3remaining=2x/3in sand=1/4×2x/3 =x/6total=x/3+x/6 =2x+x/6

let height of the pole is x .so in mud x/3remaining=2x/3in sand=1/4×2x/3 =x/6total=x/3+x/6 =2x+x/6 =3x/6

let height of the pole is x .so in mud x/3remaining=2x/3in sand=1/4×2x/3 =x/6total=x/3+x/6 =2x+x/6 =3x/6 =x/2

let height of the pole is x .so in mud x/3remaining=2x/3in sand=1/4×2x/3 =x/6total=x/3+x/6 =2x+x/6 =3x/6 =x/2remaining=x-x/2

let height of the pole is x .so in mud x/3remaining=2x/3in sand=1/4×2x/3 =x/6total=x/3+x/6 =2x+x/6 =3x/6 =x/2remaining=x-x/2 =x/2

let height of the pole is x .so in mud x/3remaining=2x/3in sand=1/4×2x/3 =x/6total=x/3+x/6 =2x+x/6 =3x/6 =x/2remaining=x-x/2 =x/2in water=1/2×x/2

let height of the pole is x .so in mud x/3remaining=2x/3in sand=1/4×2x/3 =x/6total=x/3+x/6 =2x+x/6 =3x/6 =x/2remaining=x-x/2 =x/2in water=1/2×x/2 =x/4

let height of the pole is x .so in mud x/3remaining=2x/3in sand=1/4×2x/3 =x/6total=x/3+x/6 =2x+x/6 =3x/6 =x/2remaining=x-x/2 =x/2in water=1/2×x/2 =x/4remaining=x-(x/2+x/4)

let height of the pole is x .so in mud x/3remaining=2x/3in sand=1/4×2x/3 =x/6total=x/3+x/6 =2x+x/6 =3x/6 =x/2remaining=x-x/2 =x/2in water=1/2×x/2 =x/4remaining=x-(x/2+x/4) =x-3x/4

let height of the pole is x .so in mud x/3remaining=2x/3in sand=1/4×2x/3 =x/6total=x/3+x/6 =2x+x/6 =3x/6 =x/2remaining=x-x/2 =x/2in water=1/2×x/2 =x/4remaining=x-(x/2+x/4) =x-3x/4 =x/4

let height of the pole is x .so in mud x/3remaining=2x/3in sand=1/4×2x/3 =x/6total=x/3+x/6 =2x+x/6 =3x/6 =x/2remaining=x-x/2 =x/2in water=1/2×x/2 =x/4remaining=x-(x/2+x/4) =x-3x/4 =x/4a/c to question

let height of the pole is x .so in mud x/3remaining=2x/3in sand=1/4×2x/3 =x/6total=x/3+x/6 =2x+x/6 =3x/6 =x/2remaining=x-x/2 =x/2in water=1/2×x/2 =x/4remaining=x-(x/2+x/4) =x-3x/4 =x/4a/c to question=>x/4=6

let height of the pole is x .so in mud x/3remaining=2x/3in sand=1/4×2x/3 =x/6total=x/3+x/6 =2x+x/6 =3x/6 =x/2remaining=x-x/2 =x/2in water=1/2×x/2 =x/4remaining=x-(x/2+x/4) =x-3x/4 =x/4a/c to question=>x/4=6=>x=24

Answer 2

Given : A certain height of pole is fixed in a lake whose one third is in mud, one fourth of the remaining in sand and half of the remaining

in water.   rest 6 m is seen above water,

To find : total length of the pole.​

Solution:

Let say pole height  =  12P  m

one third is in mud = (1/3)12P = 4P m

Remaining = 12P - 4P = 8P  m

one fourth of the remaining  in sand   =  (1/4) 8P  = 2P  m

Remaining = 8P - 2P   = 6P   m

half of the remaining in water  = (1/2)6P  = 3P  m

Remaining = 6P - 3P  = 3P  m

rest 6 m is seen above water

=> 3P  = 6

=> P = 2

pole height  =  12P  m  = 12 x 2  = 24 m

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