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SSC Selection Post 2020 PYQ — Question 55

SSC Selection Post - Matriculation Level - 9 Nov 2020 - Shift 3 • Quantitative Aptitude • Q55

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If 4x2 + 9y2 + z2 + 49 = 12(x + y + z), then what is the value of (4x + 9y −z)?

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Correct answer: C

Explanation

The correct answer is option 3 i.e. 6.

 

4x2+9y2+z2+49=12(x+y+z)4x^2 + 9y^2 + z^2 + 49 = 12(x + y + z)

4x2+9y2+z2−12x−12y−12z+49=0 4x^2 + 9y^2 + z^2 - 12x - 12y - 12z + 49 = 0

For x:

4x2−12x=4(x2−3x)=4(x−32)2−94x^2 - 12x = 4(x^2 - 3x) = 4(x - \tfrac{3}{2})^2 - 9

For y:

9y2−12y=9(y2−43y)=9(y−23)2−49y^2 - 12y = 9(y^2 - \tfrac{4}{3}y) = 9(y - \tfrac{2}{3})^2 - 4

For z:

z2−12z=(z−6)2−36z^2 - 12z = (z - 6)^2 - 36

[4(x−32)2−9]+[9(y−23)2−4]+[(z−6)2−36]+49=0[4(x - \tfrac{3}{2})^2 - 9] + [9(y - \tfrac{2}{3})^2 - 4] + [(z - 6)^2 - 36] + 49 = 0

4(x−32)2+9(y−23)2+(z−6)2−9−4−36+49=04(x - \tfrac{3}{2})^2 + 9(y - \tfrac{2}{3})^2 + (z - 6)^2 - 9 - 4 - 36 + 49 = 0

⇒4(x−32)2+9(y−23)2+(z−6)2=0 \Rightarrow 4(x - \tfrac{3}{2})^2 + 9(y - \tfrac{2}{3})^2 + (z - 6)^2 = 0

x=32,y=23,z=6x = \tfrac{3}{2}, \quad y = \tfrac{2}{3}, \quad z = 6

4x+9y−z=4(32)+9(23)−6=6+6−6=6 4x + 9y - z = 4(\tfrac{3}{2}) + 9(\tfrac{2}{3}) - 6 = 6 + 6 - 6 = {6}