Maths

Let x be the required percent.

Using the successive percentage change rule, we have

150 + x + $\frac{150x}{100}$ $\frac{}{}$ = 0

$\frac{}{}$ $\frac{250x}{100}$  = – 150

x = – 60%

so 60%

Required time t =

$\frac{1760}{2\pi r}=\frac{1760}{\left[2*\left(\frac{22}{7}\right)*\frac{3.5}{2}\right]}=160min.$

  $f\left(x\right)=[\begin{array}{l}-\frac{4}{3}{x}^3+2x^2+3x,x>0\\ 3x{e}^x,x\le 0\end{array}$

  $f\text{'}\left(x\right)>0forx<0e^x\left(x+1\right)>0\Rightarrow X>-1\Rightarrow x\in \left(-1,0\right)$            

for  x > 0  $-4x^2+4x+3>0$   

$\Rightarrow x\in \left(0,\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$  

Again    at  x = 0  f'(0) = 3 > 0

$\therefore f\left(x\right)isincreasingin\left(-1,\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$            

CP of 378 = SP of 525

$\begin{array}{l}\frac{378}{525}=\frac{SP}{CP}\\ \frac{SP}{CP}=0.72\end{array}$

So, 28% loss

$cose{c}^{2}xdy+2dx=\left(1+ycos2x\right)cose{c}^{2}xdx.$

$\Rightarrow \frac{dy}{dx}+2sin2x=1+ycos2x.$

$I.F.=e^{-{\displaystyle \int cos2xdx}}=e^{-\frac{sin2x}{2}}$

$\therefore Solutiony{e}^{-\frac{sin2x}{2}}={\displaystyle \int e^{-\frac{sin2x}{2}}.cos2xdx.Put\frac{sin2x}{2}=t\Rightarrow cos2xdx=dt}$

$y\left(0\right)=-1+e^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\Rightarrow {\left(y\left(0\right)+1\right)}^2={\left(e^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^2=e^{-1}$

$\frac{{(27+5)}^{332}}{9}=\frac{{(5)}^{332}}{9}=\frac{5^2*{(125)}^{110}}{9}=25*{(-1)}^{110}=7$

Let the speed of B = X kmph.

We have X * t = 240, (X – 15) * t = 180

or X = 60, X – 15 = 45

L : 2x + y = k.

$y=-2x+k.istangent\frac{x^2}{3}-\frac{y^2}{3}=1$                

$\Rightarrow k^2=3{\left(-2\right)}^2-3=9\Rightarrow k=3ask>0$              

y = -2x + 3 is also tangent to y² = $4\left(\frac{\alpha }{4}\right)x$  

$\Rightarrow 3=\frac{\alpha /4}{-2}$ -> a = -24

The work done by 12 boys in 20 days = $\frac{4}{7}$ $\frac{}{}$

So, remaining work need to be done in 6 days = $\frac{3}{7}$

$\frac{M_1*D_1}{W_1}=\frac{M_2*D_2}{W_2}$ $\frac{}{}$

$\frac{12*20}{\frac{4}{7}}=\frac{M_2*6}{\frac{3}{7}}$

We get, M2 = 30. Hence, 18 extra boys are required.