JEE Main 2026 · 2 Apr · Shift 1 · Mathematics Q2

← JEE Main 2026 · 2 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 2 of 75

Let $x$ and $y$ be real numbers such that $50\left(\dfrac{2x}{1+3i}-\dfrac{y}{1-2i}\right)=31+17i$ , $i=\sqrt{-1}$ . Then the value of $10(x-3y)$ is :

Answer and worked solution

Official answer: D

Rationalise. $\dfrac{2x}{1+3i}=\dfrac{x(1-3i)}5$ and $\dfrac y{1-2i}=\dfrac{y(1+2i)}5$ , so
$50(\cdots)=10(x-y)+10i(-3x-2y)=31+17i.$

Solve. $x-y=3.1$ and $3x+2y=-1.7$ give $y=-2.2$ , $x=0.9$ .

Value. $10(x-3y)=10(0.9+6.6)=75$ .

(D)

Topics:Algebra of complex numbers

← Previous question Next question →

JEE Main 2026 · 2 Apr · Shift 1, Mathematics question 2