The sum of the squares of two numbers is $3341$ and the difference of their squares is $891$. The numbers are

### Concept & Formula This problem simplifies to a basic system of linear equations if you treat the squares of the numbers as single variables (e.g., let $A = x^2$ and $B = y^2$). $$x^2 + y^2 = \text{Sum}$$ $$x^2 - y^2 = \text{Difference}$$ ### Step-by-Step Solution * **Given:** Let the two numbers be $x$ and $y$. * Equation 1: $x^2 + y^2 = 3341$ * Equation 2: $x^2 - y^2 = 891$ * To find $x^2$, add Equation 1 and Equation 2: $$(x^2 + y^2) + (x^2 - y^2) = 3341 + 891$$ $$2x^2 = 4232$$ * Divide by $2$: $$x^2 = 2116$$ * Take the square root to find $x$: $$x = \sqrt{2116} = 46$$ * To find $y^2$, subtract Equation 2 from Equation 1: $$(x^2 + y^2) - (x^2 - y^2) = 3341 - 891$$ $$2y^2 = 2450$$ * Divide by $2$: $$y^2 = 1225$$ * Take the square root to find $y$: $$y = \sqrt{1225} = 35$$ * The numbers are $46$ and $35$. ### Exam Strategy & Shortcut Instead of full square root extraction, use **Unit Digits**. We know $2x^2 = 4232$, so $x^2$ ends in $6$. The square roots of numbers ending in $6$ must end in $4$ or $6$. Option (c) has $46$, which fits. Similarly, $2y^2 = 2450 \Rightarrow y^2 = 1225$. A square ending in $25$ guarantees the root ends in $5$. Option (c) contains $35$ and $46$, seamlessly matching the unit digit logic without doing the heavy lifting of calculating $\sqrt{2116}$. ### Common Pitfall Students often get intimidated by large numbers like $3341$ and $2116$ and assume they made an arithmetic error. Do not panic; trust the linear elimination method. Keep track of which variable is the larger one to match the options correctly (though options here are pairs). ### Final Answer **Therefore, the correct answer is 35, 46.**