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SSC Selection Post 2020 PYQ — Question 51

SSC Selection Post - Matriculation Level 2020 - 10 Nov 2020 - Shift 1 • Quantitative Aptitude • Q51

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If x2 + 3y2 + 4z2 + 19 = 43\sqrt3(x + y + 4z), then the value of (x - y + 4z) is:

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Correct answer: C

Explanation

The correct answer is option 3  i.e.  2833\frac{28\sqrt{3}}{3}.

​​​​​​​x2+3y2+4z2+19=43(x+y+4z)​​​​​​​x^2 + 3y^2 + 4z^2 + 19 = 4\sqrt{3}(x + y + 4z)

⇒ x2+3y2+4z2−43(x+y+4z)+19=0x^2 + 3y^2 + 4z^2 - 4\sqrt{3}(x + y + 4z) + 19 = 0

⇒ (x−2(x−23)2−12+3(y−233)2−4+4(z−23)2−48+19=0 (x−2(x - 2\sqrt{3})^2 - 12 + 3\left(y - \frac{2\sqrt{3}}{3}\right)^2 - 4 + 4(z - 2\sqrt{3})^2 - 48 + 19 = 0

⇒ (x−23)2+3(y−233)2+4(z−23)2−45=0(x - 2\sqrt{3})^2 + 3\left(y - \frac{2\sqrt{3}}{3}\right)^2 + 4(z - 2\sqrt{3})^2 - 45 = 0

⇒ (x−23)2+3(y−233)2+4(z−23)2=45 (x - 2\sqrt{3})^2 + 3\left(y - \frac{2\sqrt{3}}{3}\right)^2 + 4(z - 2\sqrt{3})^2 = 45

⇒ (23−233+83)=(6−2+243)3=2833\left(2\sqrt{3} - \frac{2\sqrt{3}}{3} + 8\sqrt{3}\right) = \left(\frac{6 - 2 + 24}{3}\right)\sqrt{3} = \frac{28}{3}\sqrt{3}