IAS 2022 — CSAT Maths

Correct Option

The problem provides two relationships between three numbers x, y, and z.

From the first statement: "When 70% of a number x is added to another number y, the sum becomes 165% of the value of y."

This can be expressed as an equation:

$$0.7x + y = 1.65y$$0.7x+y=1.65y

Subtracting y from both sides yields:

$$0.7x = 0.65y$$0.7x=0.65y

From this, we can express y in terms of x:

$$y = \frac{0.7}{0.65}x = \frac{70}{65}x = \frac{14}{13}x$$y=0.650.7​x=6570​x=1314​x

Since $\frac{14}{13} > 1$1314​>1, it implies that $y > x$y>x.

From the second statement: "When 60% of the number x is added to another number Z, then the sum becomes 165% of the value of z."

This can be expressed as an equation:

$$0.6x + z = 1.65z$$0.6x+z=1.65z

Subtracting z from both sides yields:

$$0.6x = 0.65z$$0.6x=0.65z

From this, we can express z in terms of x:

$$z = \frac{0.6}{0.65}x = \frac{60}{65}x = \frac{12}{13}x$$z=0.650.6​x=6560​x=1312​x

Since $\frac{12}{13} < 1$1312​<1, it implies that $z < x$z<x.

Now, comparing y and z relative to x:

We have $y = \frac{14}{13}x$y=1314​x and $z = \frac{12}{13}x$z=1312​x.

Since $\frac{14}{13} > \frac{12}{13}$1314​>1312​, it follows that $y > z$y>z.

Combining the inequalities $y > x$y>x, $z < x$z<x, and $y > z$y>z, the correct order is $z < x < y$z<x<y.

Incorrect Options

Options 2, 3, and 4 are incorrect as they do not align with the derived order of the numbers $z < x < y$z<x<y.