PGDBA 2018

For Men's 1500 m, fastest time recorded= 206 seconds.

So, average speed = $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 1500}{206}=7.28\ metres\ per\ \sec.$$

For Women's 1500 m, fastest time recorded= 230.1

So, average speed= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 1500}{230.1}=6.52\ metres\ per\ \sec.$$

Therefore, Option A.

Using the given options, we create the following table:

We can observe that only in case of 100 meters is Women's record time is less than 110% of the men's record time. 

Hence, Option D.

Using the table as shown below:

We can infer that the ratio W/M is greatest for 2000 metres.

Using the formula Speed= Distance/ Time we can formulate the following table:

From the above table, we an infer that the highest speed is that in case of men 100 and 200 metres race. 

The second highest speed is in case of women 100 metre race.

Let us evaluate each of the options one by one.

In the first option, we need to check the speeds for 100 m and 200 m.

For 100 m, Average speed in women's category= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 100}{10.5}=9.52\ metres\ per\ \sec.$$

For 200 m, Average speed in women's category= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 200}{486.1}=9.39\ metres\ per\ \sec.$$

$$\alpha\ =1-\ \frac{\ 9.39}{9.52}=1-0.9863=0.0137$$

In the second option, we need to check the speeds for 1500 m and 3000 m.

For 1500 m, Average speed in women's category= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 1500}{230.1}=6.52\ metres\ per\ \sec.$$

For 3000 m, Average speed in women's category= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 3000}{586.1}=6.17\ metres\ per\ \sec.$$

$$\alpha\ =1-\ \frac{\ 6.17}{6.52}=1-0.9463=0.0536$$

In the third option, we need to check the speeds for 5000 m and 10000 m.

For 5000 m, Average speed in women's category= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 5000}{851.2}=5.874\ metres\ per\ \sec.$$

For 10000 m, Average speed in women's category= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 10000}{1757.5}=5.6912\ metres\ per\ \sec.$$

$$\alpha\ =1-\ \frac{\ 5.6912}{5.874}=1-0.9688=0.0311$$

In the fourth option, we need to check the speeds for 400 m and 800 m.

For 400 m, Average speed in women's category= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 400}{47.6}=8.40336\ metres\ per\ \sec.$$

For 800 m, Average speed in women's category= $$\frac{\ Total\ dis\tan ce}{Total\ time}=\ \frac{\ 800}{113.3}=7.0609\ metres\ per\ \sec.$$

$$\alpha\ =1-\ \frac{\ 7.0609}{8.40336}=1-0.84024=0.16$$

Therefore, lowest $$\alpha\ $$ is in the first option

Students who play all the three games are 4, and with the rest information given in the question, we can at best draw the venn diagram as:

Now, total students=40.

So, Chess+ 5+8+15-x=40

=>18+28-x=40

=> 46-x=40

.'. x=6

Let S= $$\left(0\times\ ^{20}C_0\right)+\left(1\times^{20}C_1\ \right)+\left(2\times^{20}C_2\ \right)+\left(3\times^{20}C_3\ \right)+...\left(19\times^{20}C_{19}\ \right)+\left(20\times^{20}C_{20}\ \right)$$.---------Eqn 1.

When we reverse the RHS and write again, we get

S= $$\left(20\times\ ^{20}C_20\right)+\left(19\times^{20}C_19\ \right)+\left(18\times^{20}C_18\ \right)+\left(17\times^{20}C_17\ \right)+...\left(1\times^{20}C_{1}\ \right)+\left(0\times^{20}C_{0}\ \right)$$.--------- Eqn 2.

When we use $$^nC_r=^nC_{n-r}$$, we can re-write the second equation as:

S= $$\left(20\times\ ^{20}C_0\right)+\left(19\times^{20}C_1\ \right)+\left(18\times^{20}C_2\ \right)+\left(17\times^{20}C_3\ \right)+...\left(1\times^{20}C_{19}\ \right)+\left(0\times^{20}C_{20}\ \right)$$.-----------Eqn 3.

Adding Eqn 1 and Eqn 3, we will get,

2S= 20($$^{20}C_0+^{20}C_1+^{20}C_2+^{20}C_3.....+^{20}C_{19}+^{20}C_{20}$$

=> 2S=20$$\times\ 2^{20}$$

=>S= $$\ \frac{\ 20\times\ 2^{20\ }}{2}$$

.'. S= 20$$\times2^{19}\ $$

First, we will try and find the point of intersection of the two curves.

We do so by equating the two curves.

So, $$x^2+\ y^2-8=y^2-2x$$

=> $$x^2+2x\ -8=0$$

=> (x+4)(x-2)=0

=> x= -4 or 2.

But since, the point of intersection lies on the first quadrant, x=2.

By putting the value of x in any of the two curves, we can find the value of y. 

'.' $$y^2=2x$$

=> $$y^2=4$$

=> y= 2 or -2. But the point of intersection is in first quadrant, so y=2.

So, P= (2,2)

Now, for the parabola $$y^2=2x$$, tangent can be found by differentiating the curve at (2,2).

=> $$2y\ \frac{\ dy}{dx}=2$$

=>$$\frac{\ dy}{dx}=\ \frac{\ 1}{y}$$

=> $$\frac{\ dy}{dx}=\ \frac{\ 1}{2}$$

.'. Angle made by the tangent to parabola at P, tan$$\alpha\ $$= $$\ \frac{\ 1}{2}$$.

Similarly, tangent at the circle $$x^2+y^2=8$$ can be found by differentiating the curve at (2,2)

=> $$2x+2y\ \frac{\ dy}{dx}=0$$

=> $$2y\ \frac{\ dy}{dx}=-2x$$ 

=> $$\frac{\ dy}{dx}=\ \frac{\ -x}{y}$$

=> $$\frac{\ dy}{dx}=\ -1$$.

.'. Angle made by the tangent to circle at P, tan$$\beta\ $$= -1

Now, the acute angle between the two tangents is represented by $$\alpha\ -\beta\ $$.

We will use the expression $$\tan\left(\alpha-\beta\right)\ =\ \ \frac{\ \tan\ \alpha\ -\tan\ \beta\ }{1+\tan\ \alpha\ \tan\ \beta\ }$$

=> $$\tan\left(\alpha\ -\beta\ \right)=\ \ \frac{\ \ \frac{\ 1}{2}-\left(-1\right)}{1-\frac{\ 1}{2}}$$

=>$$\tan\left(\alpha\ -\beta\ \right)=\ \ \frac{\ \ \ \frac{\ 3}{2}}{\ \ \frac{\ 1}{2}}$$

=>$$\tan\left(\alpha\ -\beta\ \right)=\ \ 3$$

.'. $$\alpha\ -\beta\ =\ \ \tan^{-1}\left(3\right)$$. therefore, option A

Let the isosceles sides be of the length 3x units each. Then, we can draw the triangle with given point as:

Angle SPT= B-A.

tanA= $$\ \frac{\ RT}{PR}=\ \frac{\ x}{3x}=\ \frac{\ 1}{3}\ $$

tanB= $$\ \frac{\ RS}{PR}=\ \frac{\ 2x}{3x}=\ \frac{\ 2}{3}\ $$

So, $$\tan\left(B-A\right)=\ \ \frac{\ \tan B-\tan\ A}{1+\tan\ A\tan\ B}\ $$.

=> $$\ \tan\left(B-A\right)=\ \ \frac{\ \ \frac{\ 2}{3}-\ \frac{\ 1}{3}}{1+\ \frac{\ 2}{9}}$$

=>$$\ \tan\left(B-A\right)=\ \ \frac{\ \ \frac{\ 1}{3}}{\ \ \frac{\ 11}{9}}=\ \frac{\ 3}{11}$$

$$\ h\left(x\right)=\ \ \frac{\ f\left(\ \frac{\ x}{1+x}\right)}{1+f\left(\ \frac{\ x}{1+x}\right)}=\ \frac{1+\ f\left(\ \frac{\ x}{1+x}\right)-1}{1+f\left(\ \frac{\ x}{1+x}\right)}=1-\ \frac{\ 1}{1+f\left(\ \frac{\ x}{1+x}\right)}=1-\ \frac{\ 1}{1+f\left(\ \frac{\ 1+x-1}{1+x}\right)}=1-\ \frac{\ 1}{1+f\left(\ 1-\frac{\ 1}{1+x}\right)}$$.

Now, Since f is a strictly decreasing function and its domain is positive, then, $$f\left(\ \frac{\ 1}{x+1}\right)$$ will be a strictly increasing function. Again, $$f\left(\ 1-\frac{\ 1}{x+1}\right)$$ will be a strictly decreasing function.

Let us assume $$1+f\left(\ 1-\frac{\ 1}{x+1}\right)$$= g(x).

So, g(x) is a strictly decreasing function.

h(x)= $$1-\ \frac{\ 1}{g\left(x\right)}$$. 

 $$\frac{\ 1}{g\left(x\right)}$$ becomes strictly increasing function, but $$1-\frac{\ 1}{g\left(x\right)}$$ will finally be strictly decreasing function. 

Therefore, h(x) becomes strictly decreasing function- Option A.