CAT 2020 Slot 3 Question Paper

Anil and Sunil will meet at a first point after LCM ( $$\frac{3}{15},\frac{3}{10}$$) = 3/5 hr

In the mean time, distance travelled by ravi = 8 * 3/5 = 4.8 km

Given, BC = DE = 4

CD = BE = 5

In triangle ADE, EAD=45^{0}$$

$$\tan\ 45\ =\ \frac{DE}{AE}$$ => AE = 4

Area of trapezium = Area of rectangle BCDE + Area of triangle AED

= 20 + 8 = 28

The required figure is a trapezium with vertices A(0,0), B(2,0), C(2,4) and D(0,6)

AB = 2 BC = 4 and AD = 6

Area of trapezium = $$\frac{1}{2}\left(sum\ of\ the\ opposite\ sides\right)\cdot height$$ = $$\frac{1}{2}\left(4+6\right)\cdot2\ =\ 10$$

The given function is equivalent to f(x) = $$a^x$$

Given, f(5) = 4

=> $$a^5=4=>\ a=4^{\frac{1}{5}}$$

=> f(x) = $$4^{\frac{x}{5}}$$

f(10) - f(-10) = 16 - 1/16 = 15.9375

$$\frac{\left(2\cdot4\cdot8\cdot16\right)}{\left(\log_22^2\right)^2\cdot\left(\log_{2^2}2^3\right)^3\cdot\left(\log_{2^3}2^4\right)^4}\cdot$$

= $$\frac{2^{10}}{4\cdot\left(\frac{3}{2}\right)^3\cdot\left(\frac{4}{3}\right)^4}=24$$

Equation of circle $$x^2+y^2+2gx+2fy+c=0$$

It passes through (0,0), (4,0) and (3,9). Substitute each point in the above equation:

=> On substituting the value (0,0) in the above equation, we obtain: $$c=0$$

=> On substituting the value (4,0) in the above equation, we obtain:  $$16+0+8g+0 = 0$$ ; $$g=-2$$

=> On substituting the value (3,9) in the above equation, we obtain: $$9+81-12+18f = 0$$ ; $$f= -13/3$$

Radius of the circle r = $$\sqrt{\ g^2+f^2-c}$$ => $$r^2=\frac{205}{9}$$

Therefore, Area = $$\pi\ r^2=\frac{205\pi\ }{9}$$