Quantitative Aptitude Prep Tips for MBA

As per the problem, we have Length of the first train = 120 m

Length of the second train = 180 m

As per the problem,

$\frac{(120+180)}{(8+s)}=5$

8 + s = 60

s = 52 m/sec

6X + 5Y = 218 . (1)

5X - 3Y = 24 . (2)

Solve to get X = 18, Y = 22.

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 =

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

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

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

Let the speed of B = X kmph.

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

or X = 60, X – 15 = 45

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.