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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 1 · Set 1 · Q33
(A) A faster train takes one hour less than a slower train for a journey of 200 km. If the speed of the slower train is 10 km/hr less than that of…
(A)
A faster train takes one hour less than a slower train for a journey of $\displaystyle 200$ km. If the speed of the slower train is $\displaystyle 10$ km/hr less than that of the faster train, find the speeds of the two trains.
OR
(B)
The sum of the areas of two squares is $\displaystyle 640 \mathrm{~m}^{2}$. If the difference in their perimeters is $\displaystyle 64$ m, find the sides of the two squares.
Marking-scheme solution
Let the speed of faster train be \(\displaystyle x \mathrm{~km} / \mathrm{h}\)
∴ speed of slower train \(\displaystyle =(x-10) \mathrm{km} / \mathrm{h}\)
According to the question,
\(\displaystyle \frac{200}{x-10}-\frac{200}{x}=1\)
\[\begin{aligned}
& \Rightarrow x^{2}-10 x-2000=0 \\
& \Rightarrow(x-50)(x+40)=0 \\
& \therefore x=50 \\
& x=-40 \text { (Rejected) }
\end{aligned}
\]
Hence, speed of faster train \(\displaystyle =50 \mathrm{~km} / \mathrm{h}\) and speed of slower train \(\displaystyle =40 \mathrm{~km} / \mathrm{h}\)
Let the sides of the two squares be \(\displaystyle x \mathrm{~m}\) and \(\displaystyle y \mathrm{~m}(x>y)\)
\[x^{2}+y^{2}=640
\]
and \(\displaystyle 4 x-4 y=64 \Rightarrow y=x-16\)
\[\therefore x^{2}+(x-16)^{2}=640
\]
\(\displaystyle \Rightarrow x^{2}-16 x-192=0\)
\[\Rightarrow(x-24)(x+8)=0
\]
\(\displaystyle \therefore x=24\)
\(\displaystyle x=-8\) (Rejected)
\[\Rightarrow y=24-16=8
\]
Hence the sides of the two squares are $\displaystyle 24$ m and $\displaystyle 8$ m
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CBSE Class 10 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.