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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 2 · Set 1 · Q33
(a) A person on a tour has ₹ 4,200 for expenses. If he extends his tour for 3 days, he has to cut down his daily expenses by ₹ 70. Find the original…
(a)
A person on a tour has ₹ $\displaystyle 4{,}200$ for expenses. If he extends his tour for $\displaystyle 3$ days, he has to cut down his daily expenses by ₹ 70. Find the original duration of the tour.
OR
(b)
The area of a right-angled triangle is $\displaystyle 600 \mathrm{~cm}^{2}$. If the base of the triangle exceeds the altitude by $\displaystyle 10$ cm, find all the three dimensions of the triangle.
Marking-scheme solution
Let the original duration of the tour be x days
Original daily expenses \(\displaystyle =₹ \frac{4200}{\mathrm{x}}\)
New daily expenses \(\displaystyle =₹ \frac{4200}{\mathrm{x}+3}\)
\(\displaystyle \frac{4200}{x}-\frac{4200}{x+3}=70\)
\(\displaystyle \Rightarrow \quad \mathrm{x}^{2}+3 \mathrm{x}-180=0\)
\(\displaystyle \Rightarrow \quad(\mathrm{x}+15)(\mathrm{x}-12)=0\)
⇒ \(\displaystyle \mathrm{x}=-15, \mathrm{x}=12\)
\(\displaystyle \mathrm{x}=-15\) (rejected)
\(\displaystyle \mathrm{x}=12\)
∴ Original duration of the tour = $\displaystyle 12$ days
Let the altitude of triangle be x cm
then Base of triangle \(\displaystyle =(\mathrm{x}+10) \mathrm{cm}\)
Area of triangle \(\displaystyle =600 \mathrm{~cm}^{2}\)
\(\displaystyle \frac{1}{2} \times x \times(x+10)=600\)
\(\displaystyle \Rightarrow \quad \mathrm{x}^{2}+10 \mathrm{x}-1200=0\)
\(\displaystyle \Rightarrow \quad(\mathrm{x}+40)(\mathrm{x}-30)=0\)
\(\displaystyle \Rightarrow \quad \mathrm{x}=-40, \mathrm{x}=30\)
\(\displaystyle \mathrm{x}=-40\) (rejected)
\(\displaystyle \mathrm{x}=30\)
∴ Altitude \(\displaystyle =30 \mathrm{~cm}\)
Base \(\displaystyle =40 \mathrm{~cm}\)
Hypotenuse \(\displaystyle =\sqrt{(30)^{2}+(40)^{2}}=50 \mathrm{~cm}\)
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