Properties of inequalities

It is given that $$5^{n-1} < 3^{n + 1}$$, where n is a natural number. By inspection, we can say that the inequality holds when n = 1, 2, 3 4, and 5.

Now, we need to find the least integer value of m that satisfies $$3^{n+1} < 2^{n+m}$$

For, n =1, the least integer value of m is 3.

For, n = 2, the least integer value of m is 3

For, n = 3, the least integer value of m is 4.

For, n = 4, the least integer value of m is 4.

For, n= 5, the least integer value of m is 5.

Hence, the least integer value of m such that for all the values of n, the equation holds is 5.

It is given that $$\frac{x}{y}<\ \frac{\ x+3}{y-3}$$, which can be written as $$\frac{x}{y}-\frac{\ x+3}{y-3}<0$$

=> $$\ \frac{\ x\left(y-3\right)-y\left(x+3\right)}{y\left(y-3\right)}<0$$

=> $$\ \frac{\ xy-3x-xy-3y}{y\left(y-3\right)}<0$$

=> $$\ \frac{\ -3\left(x+y\right)}{y\left(y-3\right)}<0$$

=> $$\ \frac{\ 3\left(x+y\right)}{y\left(y-3\right)}>0$$

From this inequality, we can say that, when $$y<0=>y(y-3)>0$$. Now to satisfy the given equation $$\ \frac{\ 3\left(x+y\right)}{y\left(y-3\right)}>0$$, 

$$(x+y)$$ must be greater than zero Hence, $$x>0$$ and $$\left|x\right|\ >\left|y\right|$$

Therefore, the magnitude of $$x$$ is greater than the magnitude of $$y$$. 

Hence, $$x>y$$, and $$\left|x\right|\ >\ \left|y\right|$$ =>$$-x\ <\ y$$ (Since the magnitude of $$x$$ is greater than the magnitude of $$y$$.)

The correct option is B.

It is given that if a certain amount of money is divided equally among n persons, each one receives Rs 352. Hence, the total amount of money is (352*n) = 352n ... Eq(1)

It is also known that if two persons receive Rs 506 each and the remaining amount is divided equally among the other persons, each of them receives less than or equal to Rs 330

Hence, the maximum amount of money with them = 506*2 + (n-2)*330 = 1012+330n-660 = 352+330n

Now, $$352+330n\ \ge\ 352n\ $$

=> $$22n\ \le\ 352$$ 

=> $$n\ \le\ 16$$

Hence, the maximum value is 16