Maths

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

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

$\Rightarrow 0\le x<ln2$              

Five years ago the ratio of their ages was 5:3 with the total age being 64 years. Their individual ages will be 40 and 24 years.

Today their ages will be 45 years and 29 years and after 9 years their ages will be 54 and 38 years.

Required ratio will be 27 : 19.

${\displaystyle \underset{}{\overset{}{\int }}\frac{si{n}^{2}xdx}{e^{\left\{\frac{x}{\pi }\right\}}}=l}$

$f\left(x\right)=\frac{si{n}^{2}x}{e^{\left\{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$\pi $}\right.\right\}}}$ is periodic with period =

$l=100{\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{si{n}^{2}x}{e^{x/\pi }}dx.}$

$=50{\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{1-cos2x}{e^{x/\pi }}}dx=50{\displaystyle \underset{0}{\overset{\pi }{\int }}{e}^{-x/\pi }dx-50}{\displaystyle \underset{0}{\overset{\pi }{\int }}{e}^{-x/\pi }}cos2xdx$

$=50\pi \left(1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$e$}\right.\right)\left[1-\frac{1}{1+4{\pi }^2}\right]=\frac{50\left(1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$e$}\right.\right)4{\pi }^3}{1+4{\pi }^2}$

$\therefore \alpha =200\left(1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$e$}\right.\right)$

612 = 212 * 312. For perfect cubes, power must be multiple of 3.

Therefore, power of 2 equal 20, 23, 26, 29, 212 and power of 3 equals 30, 33, 36, 39, 312 So, number of divisors equal 5* 5 = 25.