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CAT 2024 Slot 1 — QA Question 7

Number TheoryEasy

Let x, y, and z be real numbers satisfying

4(x2 + y2 + z2) = a,

4(x - y - z) = 3 + a

Then a equals

Answer & solution

Correct answer: 3

  • A

    1

  • B

    4

  • 3

  • D

    113

Solution

Medium

Substitute aa from the second equation into the first, move everything to one side, and complete the square in each variable. A sum of squares equal to zero forces each square to vanish, fixing x,y,zx,y,z and hence aa.

1

Eliminate aa. From the second equation a=4(x−y−z)−3a=4(x-y-z)-3. Substitute into the first.

4(x2+y2+z2)=4(x−y−z)−3⇒ 4x2+4y2+4z2−4x+4y+4z+3=0(bring all to one side)\begin{aligned} &4(x^2+y^2+z^2)=4(x-y-z)-3\\ &\Rightarrow\ 4x^2+4y^2+4z^2-4x+4y+4z+3=0 \quad\text{(bring all to one side)} \end{aligned}
2

Complete the square. Split the constant 3=1+1+13=1+1+1 to package each variable.

(4x2−4x+1)+(4y2+4y+1)+(4z2+4z+1)=0⇒ (2x−1)2+(2y+1)2+(2z+1)2=0(perfect squares)\begin{aligned} &(4x^2-4x+1)+(4y^2+4y+1)+(4z^2+4z+1)=0\\ &\Rightarrow\ (2x-1)^2+(2y+1)^2+(2z+1)^2=0 \quad\text{(perfect squares)} \end{aligned}
3

Force each square to zero (real squares are non-negative), then read off the values.

x=12,y=−12,z=−12\begin{aligned} &x=\tfrac12,\quad y=-\tfrac12,\quad z=-\tfrac12 \end{aligned}
4

Compute aa from the first equation using step 3.

a=4(14+14+14)⇒ a=4⋅34(sum the squares)⇒ a=3\begin{aligned} &a=4\left(\tfrac14+\tfrac14+\tfrac14\right)\\ &\Rightarrow\ a=4\cdot\tfrac34 \quad\text{(sum the squares)}\\ &\Rightarrow\ a=3 \end{aligned}
a=3a=3

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CAT 2024 Slot 1 QA Q7: Let x, y, and z be real numbers satisfying 4(x 2 + y 2 + z 2 ) = a, 4(x - y - z) = 3 + a Then a equals — Solution | TheCATExam