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Mathematics · 2024 · 5 marks
CBSE 2024 · Region 4 · Set 1 · Q34
(a) A train travels a distance of 90 km at a constant speed. Had the speed been 15 km/h more, it would have taken 30 minutes less for the journey.…
(a)
A train travels a distance of $\displaystyle 90$ km at a constant speed. Had the speed been $\displaystyle 15$ km/h more, it would have taken $\displaystyle 30$ minutes less for the journey. Find the original speed of the train.
OR
(b)
Find the value of 'c' for which the quadratic equation \[(c+1) x^{2}-6(c+1) x+3(c+9)=0 ; c \neq-1 \] has real and equal roots.
Marking-scheme solution
Let the original speed be x km/h
New speed \(\displaystyle =(\mathrm{x}+15) \mathrm{km} / \mathrm{h}\)
A.T.Q.
\[\begin{aligned}
& \frac{90}{x}-\frac{90}{x+15}=\frac{1}{2} \\
& \Rightarrow x^{2}+15 x-2700=0 \\
& \Rightarrow(x+60)(x-45)=0 \\
& x \neq-60, x=45
\end{aligned}
\]
The original speed of the train \(\displaystyle =45 \mathrm{~km} / \mathrm{h}\)
For real and equal roots,
\[\begin{aligned}
& \{-6(c+1)\}^{2}-4(c+1) \times 3(c+9)=0 \\
& \Rightarrow 12(c+1)(2 c-6)=0 \\
& c \neq-1 \text { So, } c=3
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.