Complex NumbersmediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Complex Equation for 10(x - 3y): Value = 75 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let xx and yy be real numbers such that 50(2x1+3iy12i)=31+17i50\left(\dfrac{2x}{1+3i}-\dfrac{y}{1-2i}\right)=31+17i, i=1i=\sqrt{-1}. Then the value of 10(x3y)10(x-3y) is
A2020
B3131
C3535
D7575correct
Solution
Step 1: Rationalise each denominator:
2x1+3i=2x(13i)10,y12i=y(1+2i)5.\frac{2x}{1+3i}=\frac{2x(1-3i)}{10},\qquad \frac{y}{1-2i}=\frac{y(1+2i)}{5}.
So
50(2x(13i)10y(1+2i)5)=31+17i.50\left(\frac{2x(1-3i)}{10}-\frac{y(1+2i)}{5}\right)=31+17i.
Step 2: Simplify: 502x(13i)10=10x(13i)50\cdot\dfrac{2x(1-3i)}{10}=10x(1-3i) and 50y(1+2i)5=10y(1+2i)50\cdot\dfrac{y(1+2i)}{5}=10y(1+2i):
10x(13i)10y(1+2i)=31+17i,10x(1-3i)-10y(1+2i)=31+17i,
(10x10y)+(30x20y)i=31+17i.\big(10x-10y\big)+\big(-30x-20y\big)i=31+17i.
Step 3: Compare real and imaginary parts:
10(xy)=31,30x20y=17.10(x-y)=31,\qquad -30x-20y=17.
Step 4: From the first, xy=3.1x-y=3.1. Solving the two equations gives
x=0.9,y=2.2.x=0.9,\qquad y=-2.2.
Step 5:
10(x3y)=10(0.93(2.2))=10(0.9+6.6)=75.10(x-3y)=10\big(0.9-3(-2.2)\big)=10(0.9+6.6)=75.
Correct answer: (4)
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