Hi Rupanshi, it has now been uploaded.

Thank you :)

its 55

56?

Hi Rupanshi,
Let's say breadth is b and length is l. Consider angle made by diagonals AC and BD with base as AB (=breadth (b)) is theta. so bcos($$\theta$$) = 16 and 20cos($$\theta$$) = b now solve both equation find the values of sin($$\theta$$) and cos($$\theta$$) then find l and b. Answer will be 160. Hope it helps. If you need further explanation please do let us know :)

Thanks Vikas Bansal. It seems i was committing a silly mistake. And yes the ans is 160 :)

Let Vishal's time be considered on the X-axis and Anirudh's time be considered on the Y-axis. The total time of consideration is a square of side 60.

If Vishal arrives at 0 minutes, Anirudh can arrive at any time before 15 min. If Vishal arrives at 1 min, Anirudh can arrive at any time before 16 min. If Vishal arrives at 15 min, then Anirudh can arrive at any time between 0 and 30 minutes. Similarly, if Vishal arrives at 16 min, Anirudh can arrive at any time between 1 and 31 min. If we draw the area of intersection carefully, we see that the required area is a hexagon with two adjacent sides of lengths 15 on X and Y axes respectively; 2 adjacent sides of 15 on the other two sides of the square and two sides of $$45\sqrt2$$ parallel to each other.

The required probability is the area of this hexagon divided by the area of the square = 1575/3600 = 63/144 = 7/16

1-9/16=7/16

I ma not sure .. pls have a look at it .. they reached 15 min early means car was driven for 7.5 min less den what it is driven daily. It shows lady travelled for 52.5 min and distacnce covered by her in 52.5 min = distance by car in 7.5 min so ratio of speed is 1:7 hence speed of car is 21kmph. and lady traveled for 52.5 km.

Question implies that Mrs Jha got picked up 7.5 mint. Early than any other day, means if on any other day she use to get car at 5 o'clock, on that she got the car at 4:52.5 and as she started walking earlier by one hour There fore she walked for 52.5 mint.

@srikanth thanks for correcting brother...bt shudnt it be 12c2(2^6-2)/12^6

(12c2*2^6)/12^6??

Differentiate both equations and equate them to zero.
Curve one: 2x-6=0 i.e. x=6
Curve two: -2x+b=0. For x=6 obtained from above equation, we get b=6

Hi Sujoy, as the coefficient of $$x^2$$ in both the equations is positive, both equations will have only minima. Hence for no value of b, the maxima of one curve can be equal to the minima of the other curve.
Please go through our CC50 concept on Maxima and Minima of Quadratic equations.
https://cracku.in/cat/cc50/maxima-minima-quad-equations

Given that $$x^2 - 12x + k = 0$$ has real roots. Hence $$ 144 - 4k \geq 0$$ => $$k \leq 36 $$ . Also |k| < 15. Hence the integral values for k are from -14 to 14 ie 29 in number

132 ?

@vikas bansal...approach pls?

Hi Priyanka,

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