Explanation:
100 persons can finish a work in=7 days
? persons can finish work in=35 days

$= \dfrac{100\times7}{35}$

$=20persons$

Explanation:
1 min =60 sec
1 hour=$60 \times 60=3600seconds$
therefore,Percentage of hour in 36 seconds= $\dfrac{36}{3600}\times100$=1%

Explanation:

$\dfrac{x^{m^{2}}\times x^{mn}}{x^{nm}\times x^{n^{2}}}\times \dfrac{x^{n^{2}\times x^{ln}}}{x^{ln}\times x^{l^{2}}}\times \dfrac{x^{l^{2}}\times x^{lm} }{x^{lm}\times x^{m^{2}} }=1$

Explanation:

$Given question becomes,\dfrac{1}{5}\times2\div\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}=?$

$=[\dfrac{1}{5}\times2\times\dfrac{5}{1}]+[\dfrac{1}{5}-\dfrac{1}{10}]$

$=2+[\frac{2-1}{10}]$

$=2+\frac{1}{10}$

$=2+0.1$

$=2.1$

Explanation:
By applying the formula,$(a^{3}-b^{3})=(a-b)(a^{2}+ab+b^{2})$

$\dfrac{(x-1)(x^{2}+x+1)}{x+3}\div \dfrac{x^{2}+x+1}{3(x+3)}$

For multiply the fraction will get reciprocal

$\dfrac{(x-1)(x^{2}+x+1)}{x+3}\times \dfrac{3(x+3)}{x^{2}+x+1}$

=3(x-1)

=3x-3

Explanation:

$Geometric progression series$

$r=\dfrac{x^{\dfrac{3}{2}}}{x}=x^{\dfrac{1}{2}}$

$t_{n}=a\times r^{n-1}=l$

$x \times [x^{\dfrac{1}{2}}]^{85-1}=x\times x^{\dfrac{1}{2}\times84} $

$=x\times x^{42}$

$=x^{43}$

Explanation:

$l^{2}=r^{2}+h^{2}$

$h^{2}=l^{2}-r^{2}$

$h=\sqrt{l^{2}-r^{2}}$

$V=\dfrac{1}{3}\pi r^{2}\sqrt{l^{2}-r^{2}}$

Explanation:

Arithmetic Progresson series

$t_{n}=a+(n-1)d=l$

a=100, n=20, d=-5

$ t_{n}=(100)\times(20)\times(-5)$

=100-100

=0

Explanation:
$\frac{2x^4-162}{\left(x^2+9\right)\left(2x-6\right)}$
$=2x^4-162:\quad 2\left(x^4-81\right)$
$=\mathrm{Factor}\:2x-6:\quad 2\left(x-3\right)$
$=\frac{2\left(x^4-81\right)}{2\left(x-3\right)\left(x^2+9\right)}$
$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1$
$\frac{x^4-81}{\left(x^2+9\right)\left(x-3\right)}$
$\:x^4-81:\quad \left(x+3\right)\left(x-3\right)\left(x^2+9\right)$
$x^4-81$
$x^4-9^2$
$\mathrm{Apply\:difference\:of\:squares\:rule:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$
$x^4-9^2=\left(x^2+9\right)\left(x^2-9\right)$
$=\left(x^2+9\right)\left(x^2-9\right)$
$\mathrm{Factor}\:x^2-9:\quad \left(x+3\right)\left(x-3\right)$
$=\left(x+3\right)\left(x-3\right)\left(x^2+9\right)$
$=\frac{\left(x+3\right)\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)\left(x^2+9\right)}$
$\mathrm{Cancel\:}\frac{\left(x+3\right)\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)\left(x^2+9\right)}:\quad x+3$
$\frac{\left(x+3\right)\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)\left(x^2+9\right)}$
$\mathrm{Cancel\:the\:common\:factor:}\:x-3$
$\frac{\left(x+3\right)\left(x^2+9\right)}{x^2+9}$
$\mathrm{Cancel\:the\:common\:factor:}\:x^2+9$
$=x+3$

Explanation:

LCM*HCF=Product of two numbers

$Z \times HCF=X \times Y$

$HCF=\dfrac{X \times Y}{2}$