SarkariPYQ

SSC Selection Post 2020 PYQ — Question 74

SSC Selection Post - Matriculation Level 2020 - 10 Nov 2020 - Shift 1 • Quantitative Aptitude • Q74

Back to paper

If x4 + x2y2 + y4 = 91, and x2 − xy + y2 = 13, then what is the value of |x − y|?
 

Show answer & explanation

Correct answer: D

Explanation

The correct answer is option 4 i.e. 4.

  1. x4 + 1x2y2 + y4  = 91x4 + x2y2 + y4 = 91
  2. x2 − xy + y2 = 13x2 - xy + y2 = 13

⇒ x4+x2y2+y4=(x2+y2)2−x2y2 x^4 + x^2y^2 + y^4 = (x^2 + y^2)^2 - x^2y^2

So, from (1):

⇒ (x^2 + y^2)^2 - x^2y^2 = 91 \tag{A}

Also, from (2):

⇒ x^2 - xy + y^2 = 13 \Rightarrow x^2 + y^2 = 13 + xy \tag{B}

S=x2+y2,P=xyS = x^2 + y^2 , P = xy

Then equation (B) gives:

⇒ S = 13 + P

Substitute into equation (A):

⇒(13+P)2−P2=91 (13 + P)^2 - P^2 = 91

⇒ 169+26P+P2−P2=91⇒26P=91−169=−78⇒P=−3 169 + 26P + P^2 - P^2 = 91 \Rightarrow 26P = 91 - 169 = -78 \Rightarrow P = -3

xy=−3andx2+y2=13+(−3)=10xy = -3 \quad\text{and}\quad x^2 + y^2 = 13 + (-3) = 10

⇒(x−y)2=x2+y2−2xy=10−2(−3)=10+6=16⇒∣x−y∣=16 (x - y)^2 = x^2 + y^2 - 2xy = 10 - 2(-3) = 10 + 6 = 16 \Rightarrow |x - y| = \sqrt{16} = 4