JEE Novice ( new 11)

Let $n=111111$ . Then $x=\frac{n-1}{n}$ , $y=\frac{2n-1}{2n+1}$ , and $z = \frac{3n-2}{3n+1}$ .

Then, $x=1-\frac{1}{n}$ .

Let $a = 2n+1$ , then $y = \frac{a-2}{a} = 1 - \frac{2}{a} = 1 - \frac{2}{2n+1} = 1 - \frac{1}{n+\frac{1}{2}}$ .

Finally, let $b = 3n+1$ , then $z = \frac{b-3}{b} = 1 - \frac{3}{b} = 1 - \frac{3}{3n+1} = 1 - \frac{1}{n+\frac{1}{3}}$

It is evident that $\frac{1}{n} > \frac{1}{n+\frac{1}{3}} > \frac{1}{n+\frac{1}{2}}$ , so therefore $1 - \frac{1}{n+\frac{1}{2}} > 1 - \frac{1}{n+\frac{1}{3}} > 1-\frac{1}{n}$ , or $\boxed{y > z > x}$ .

x = 1 - 1/111111 = 1 - 2/222222 = 1 - 3/333333

y = 1 - 2/222223 = 1 - 6/666669

z = 1 - 3/333334 = 1 - 6/666668

1 - m/n < 1 - m/(n+1) when m and n are both positive

1 - 2/222222 < 1 - 2/222223 so x < y

1 - 3/333333 < 1 - 3/333334 so x < z

1 - 6/666668 < 1 - 6/666669 so z < y

Result: x < z < y ( <=> y > z > x)

My technique was not as sophisticated as the other solutions, but since it is a multiple choice question, it still works and is simpler. Multiplying the numerator and denominator of x by 2 gives $x=\frac{222220}{222222}$ , which means y>x and multiplying the numerator and denominator of x by 3 gives $x=\frac{333330}{333333}$ which means z>x. There is only one solution where x is the least so $\boxed{y>z>x}$ .

Its very simple. 111110/111111 = 1-(1/111110) 222221/222223= 1-(2/222223)=1-(1/111111.5) 333331/333334= 1-(3/333334)=1-(1/111111.3) Look at the denominators. Larger the denominator larger the value.

Let a = 111110, b = 111111 then, x = a/b, y = (2a + 1)/(2b + 1) = (a + (1/2))/(b + (1/2)) z = (3a + 1)/(3b + 1) = (a + (1/3))/(b + (1/3)) It is obvious that (x + p)/(y + p) > (x + q)/(y + q), provided p > q. Hence, (a + (1/2))/(b + (1/2)) > (a + (1/3))/(b + (1/3)) > (a + 0)/(b +0) Thus, y > z > x