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Quadratic Equations
CBSE 2026 Board Exam Set 3 (Code 30/1/3)
CBSE_BOARD_PYQ_2026_30_1_3
Grade 10
Question:
[Section D]
A faster train takes $1$ hour less than a slower train for a journey of $200$ km. If the average speed of the slower train is $10$ km/h less than that of the faster train, find the speeds of the two trains.
OR
Sum of the areas of two squares is $640$ m$^2$. If the difference of their perimeters is $64$ m, find the sides of the two squares.
Step-by-Step Solution
Key Concept: Speed equation $\dfrac{200}{x-10} - \dfrac{200}{x} = 1$ OR Square side equations $x^2 + y^2 = 640$ and $4x - 4y = 64$.
Main: Let speed of faster train $= x$ km/h. $\dfrac{200}{x-10} - \dfrac{200}{x} = 1 \Rightarrow 2000 = x(x-10) \Rightarrow x^2 - 10x - 2000 = 0 \Rightarrow (x-50)(x+40) = 0 \Rightarrow x = 50$ km/h. Faster train $= 50$ km/h, Slower train $= 40$ km/h. [5.0 Marks]
OR: Let sides be $x$ and $y$ ($x > y$). $4x - 4y = 64 \Rightarrow x - y = 16 \Rightarrow x = y + 16$. $x^2 + y^2 = 640 \Rightarrow (y+16)^2 + y^2 = 640 \Rightarrow 2y^2 + 32y - 384 = 0 \Rightarrow y^2 + 16y - 192 = 0 \Rightarrow (y+24)(y-8) = 0 \Rightarrow y = 8$ m, $x = 24$ m. Sides are $24$ m and $8$ m. [5.0 Marks]
Correct Answer:Faster = 50 km/h, Slower = 40 km/h OR Sides = 24 m and 8 m
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