SSC CHSL 5 Aug 2021 Shift 3 Paper

Section: Quantitative Aptitude

Question : 52 of 100

Marks: +1, -0

x+y+z=13, x^{2}+y^{2}+z^{2}=91

and

x z=y^{2}

, then the difference between

z

and

x

Solution:

Given,

x+y+z=13

x^{2}+y^{2}+z^{2}=91 \;\text " and "\; x z=y^{2}

We know that,

(a+b+c)^{2}

=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a

∴ (x+y+z)^{2}

=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x

⇒ \;\; (13)^{2}=91+2 x y+2 y z+2 y^{2}

⇒ \;\; (13)^{2}=91+2 y(x+y+z)

⇒ \;\; 169=91+2 y × 13

∴ \;\; y=3

∴ \;\; x z=(3)^{2}=9

and

x^{2}+z^{2}=91-9=82

∴ \;\; (x-z)^{2}=x^{2}+z^{2}-2 x z

=82-2 × 9=64

∴ \;\; x-z=√{64}=8

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