Subject: General Aptitude · Chapter: Area · Exam: 2010 · Marks: · Difficulty:
Twenty-nine times the area of a square is one square metre less than six times the area of the second square and nine times its side exceeds the perimeter of other square by 1 metre. The difference in the sides of these squares is
(a)5 m
(b)$\frac{54}{11} \mathrm{m}$
(c)6 m
(d)11 m
Answer
Answer (as printed): C
Explanation
Let the sides of the two squares be $x$ metres and $y$ metres respectively. Then, $29 x^{2}=6 y^{2}-1$ $$\text { And, } 9 x-4 y=1 \Rightarrow 4 y=9 x-1 \Rightarrow y=\frac{9 x-1}{4}$$ From (i) and (ii), we get: $$\begin{aligned} & 29 x^{2}=6\left(\frac{9 x-1}{4}\right)^{2}-1 \Rightarrow 29 x^{2}=6\left(\frac{81 x^{2}+1-18 x}{16}\right)-1 \\ & \Rightarrow 243 x^{2}+3-54 x-8=232 x^{2} \\ & \Rightarrow 11 x^{2}-54 x-5=0 \Rightarrow(x-5)(11 x+1)=0 \Rightarrow x=5 \mathrm{m} . \\ & \therefore y=\frac{9 x-1}{4}=\frac{9 \times 5-1}{4}=11 \mathrm{m} . \end{aligned}$$ Required difference $=(11-5) \mathrm{m}=6 \mathrm{m}$.