Maths

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.

Option (A)

              Option (B)

              Option (C)

              Option (D)

LCM of 70, 84, 280 is 840.

Let total work = 840 units.

Per hour work of P = $\frac{840}{70}=12\text{?}units/hour$ $$

Per hour work of Q = $\frac{840}{84}=10\text{?}units/hour$ $$

Per hour work of R = $\frac{840}{280}=3\text{?}units/hour$ $$

Number of hours they require together =

$\frac{}{}$ $\frac{840}{12+10+3}=\frac{840}{25}$  = 33 hours and 36 minutes

${\left[e^x\right]}^2+\left[e^x\right]-2=0$              

Let $\left[e^x\right]=t$   

->t² + t – 2 = 0 Þ (t + 2) (t – 1) = 0

t = -2 $\left[e^{-x}\right]$ -> = -2 not possible

$\therefore t=\left[e^x\right]=1\Rightarrow 1\le e^x<2$

$\Rightarrow 0\le x<ln2$