JEE Mathematics 2025 Question with Solution

Given:

$$x + 5y - z = 1$$ $$4x + 3y - 3z = 7$$ $$24x + y + \lambda z = \mu$$

with $\lambda, \mu \in \mathbb{R}$, and the system has infinitely many solutions.

Find: The number of integer solutions satisfying $7 \leq x + y + z \leq 77$.

For infinitely many solutions, the third equation must be a linear combination of the first two equations. Assume

$$24x + y + \lambda z = k_1(x + 5y - z) + k_2(4x + 3y - 3z)$$

Comparing coefficients,

$$24 = k_1 + 4k_2$$ $$1 = 5k_1 + 3k_2$$ $$\lambda = -k_1 - 3k_2$$

and for constants,

$$\mu = k_1(1) + k_2(7)$$

From

$$24 = k_1 + 4k_2$$

we get

$$k_1 = 24 - 4k_2$$

Substitute into

$$1 = 5k_1 + 3k_2$$

so

$$1 = 5(24 - 4k_2) + 3k_2$$ $$1 = 120 - 20k_2 + 3k_2$$ $$17k_2 = 119$$ $$k_2 = 7$$

Hence,

$$k_1 = 24 - 28 = -4$$

Now,

$$\lambda = -(-4) - 3(7) = 4 - 21 = -17$$

and

$$\mu = -4 + 7 \cdot 7 = 45$$

So the dependent system becomes

$$x + 5y - z = 1$$ $$4x + 3y - 3z = 7$$

Use the first two equations to express the integer solutions. From the first equation,

$$x = 1 - 5y + z$$

Substitute into the second equation:

$$4(1 - 5y + z) + 3y - 3z = 7$$ $$4 - 20y + 4z + 3y - 3z = 7$$ $$z - 17y = 3$$ $$z = 17y + 3$$

Then

$$x = 1 - 5y + (17y + 3) = 12y + 4$$

Thus all integer solutions are

$$(x,y,z) = (12y + 4,\; y,\; 17y + 3)$$

where $y \in \mathbb{Z}$. Now,

$$x + y + z = (12y + 4) + y + (17y + 3) = 30y + 7$$

Given

$$7 \leq x + y + z \leq 77$$

so

$$7 \leq 30y + 7 \leq 77$$ $$0 \leq 30y \leq 70$$ $$0 \leq y \leq \frac{7}{3}$$

Since $y$ is an integer,

$$y = 0, 1, 2$$

Therefore, the number of integer solutions is $3$. Hence, the correct option is A.

The solution states that the correct option is A and derives the dependence condition for infinitely many solutions.

First determine the values of the parameters:

$$\lambda = -17, \qquad \mu = 45$$

Then solve the reduced pair of independent equations:

$$x + 5y - z = 1$$ $$4x + 3y - 3z = 7$$

From the first equation,

$$x = 1 - 5y + z$$

Substitute into the second equation:

$$4(1 - 5y + z) + 3y - 3z = 7$$ $$4 - 17y + z = 7$$ $$z = 17y + 3$$

Then

$$x = 1 - 5y + 17y + 3 = 12y + 4$$

So the integer solution set is parametrized by integer $y$ as

$$(x,y,z) = (12y + 4,\; y,\; 17y + 3)$$

Now compute the required sum:

$$x + y + z = 12y + 4 + y + 17y + 3 = 30y + 7$$

Apply the condition

$$7 \leq 30y + 7 \leq 77$$

This gives

$$0 \leq 30y \leq 70$$

Hence the integer values possible are

$$y = 0, 1, 2$$

Thus there are exactly $3$ integer triples satisfying the condition.

Therefore, the correct option is A.