Question

one fourth of a herd of camels was in the forest. twice the square root of the herd had gone to mountains and the remaining 15camels were seen on the bank of river. find the total no of camels​

Answer 1

Let's assume that the total no. of camels is "x".

Then,

1/4 of x was in the forest.

2 (√x) went to the mountain.

No. of camels remaining = 15

15 + x/4 + 2√x = x

$\frac{60 + x + 8 \sqrt{x} }{4}$          = x

60 + x + 8√x = 4x

60 = 4x - x - 8√x

60 = 3x - 8√x

3x - 8√x - 60 = 0

Psst:- I tried my best in finding the answer, I'm not really sure of it.

Psst (2) :- Refer to @sathvikachowdary's answer :)

Answer 2

$\huge {\underline {\underline {\mathtt {\pink {Answer:-}}}}}$

Let the total number of camels in the herd be 'x'

${\green {\boxed {\mathtt {Given:-}}}}$

one fourth of herd of camels was in forest

=> no.of camels in forest = $\frac {x}{4}$

twice the square root of the herd had gone to mountains

=>no.of camels in mountains = $2\sqrt {x}$

no of camels on bank of river = 15

${\purple {\boxed {\mathtt {To \:find:-}}}}$

Total number of camels which is x

${\red{\boxed {\mathtt {Solution:-}}}}$

$\frac {x}{4}+2\sqrt {x}+15=x$

$2\sqrt{x}+15-\frac {3x}{4}=0$

$\frac {8\sqrt {x}+60-3x}{4}=0$

$8\sqrt {x}+60-3x=0$

$assume\: that\: x = y^2$

then

$y+60-3y^2=0$

$3y^2-8y-60=0$

$3y^2-18y+10y-60=0$

$3y(y-6)+10(y-6)$

$(3y+10)(y-6)=0$

According to zero product rule

3y+10=0 or y-6=0

=>y= -10/3 or 6

If y value is taken as -10/3

$then x = y ^2 => x=\frac {100}{9}$

If y value is taken as 6

$then x = 6^2=36$

the number of camels cannot be in fraction

so the answer is 36

${\blue {\boxed {\mathtt {number \: of \: camels \: =36}}}}$