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CAT Exam Model Paper 2 with solutions for free online practice

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Question : 84 of 100

Marks: +1, -0

Inspired by the ‘Golden quadrilateral project’ UP Government recently accomplished a diamond triangular project. Under this project the State Government laid down 6 lane roads connecting three cities Ayodhya, Banaras and Chitrakoot, which are equally separated from each other i.e., in terms of geometry they form an equilateral triangle. Angad and Bajrang start simultaneously from Ayodhya and Banaras respectively, towards Chitrakoot. When Angad covers 100 kms, Bajrang covers such a distance that the distance between Angad and Bajrang makes 90° angle with the road joining Banaras and Chitrakoot. When Bajrang reaches Chitrakoot, Angad is still 150 km away from Chitrakoot. What is the distance between Ayodhya and Banaras?

Your Answer:

Solution:

Let the distance between Q and C = x km

\text{ then, } \cos 60^{\circ} = \frac{CM}{QC} \Rightarrow CM = x \times \frac{1}{2} \Rightarrow CM = \frac{x}{2} \text{ km}

\frac{100 + \frac{x}{2}}{100} = \frac{100 + x}{100 + x - 150} = \frac{\text{Speed of Bajrang}}{\text{Speed of Angad}}

\Rightarrow \frac{200+x}{200} = \frac{100+x}{x-50}

\Rightarrow (200+x)(x-50) \& = 200(100+x)

x^{2} + 150x - 10000 = 20000 + 200x

\Rightarrow x^{2} - 50x - 30000 = 0

\Rightarrow (x-200)(x+150) = 0 \Rightarrow x = 200 \text{ km}

Therefore, distance between Ayodhya and Banaras is 300 km since AB = BC = AC.

(With the help of trigonometry we can find the value of MC in terms of x i.e., cos 60° =

\frac{MC}{QC} = \frac{1}{2}

. Hence MC

\frac{X}{2}

)

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