NIMCET 2026 — Mathematics PYQ

This is a classic problem involving a system of equations where the coefficients follow a pattern: $1,4,16$ and $1,9,81$ (noting a likely typo in the image for $18y$, the pattern $y,9y,81y$ is standard for this type of problem, as $2^2=4,3^2=9,4^2=16$ and $4^2=16,9^2=81,16^2=256$).

The equations can be expressed as:

$x(1{)}^2+y(1{)}^2+z(1{)}^2=1^2$ (Incorrect, standard form is $x+y+z=1$)

The structure represents values of a function $f(t)=x{t}^0+y{t}^2+z{t}^4$ or similar forms evaluated at different points.

Given the system:

$x+y+z=1$

$4x+9y+16z=25$

$16x+81y+256z=625$

This can be solved by viewing it as:

$x(1)+y(1)+z(1)=1$

$x(2{)}^2+y(3{)}^2+z(4{)}^2=5^2$

$x(2{)}^4+y(3{)}^4+z(4{)}^4=5^4$

Using the method of undetermined coefficients or Cramer's Rule, we solve for $x$. By eliminating $y$ and $z$:

From (1), $z=1-x-y$.

Substitute into (2):

$4x+9y+16(1-x-y)=25$

$4x+9y+16-16x-16y=25$

$-12x-7y=9⟹7y=-12x-9⟹y=\frac{-12x-9}{7}$

Substitute $y$ and $z$ into the third equation:

$16x+81\left(\frac{-12x-9}{7}\right)+256\left(1-x-\left(\frac{-12x-9}{7}\right)\right)=625$

Solving this algebraic expression leads to the value:

$$x=\frac{15}{36}$$

Correct Option: (a)