CBSE 2025 · Region 5 · Set 1 · Q32 · 5 marks
A $\displaystyle 2$-digit number is seven times the sum of its digits and two ($\displaystyle 2$) more than $\displaystyle 5$ times the product of its digits. Find the number.Find the value(s) of p for which the quadratic equation given as $\displaystyle (\mathrm{p}+4) x^{2}-(\mathrm{p}+1) x+1=0$ has real and equal roots. Also, find the roots of the equation(s) so obtained.
A $\displaystyle 2$-digit number is seven times the sum of its digits and two ($\displaystyle 2$) more than $\displaystyle 5$ times the product of its digits. Find the number.
Find the value(s) of p for which the quadratic equation given as $\displaystyle (\mathrm{p}+4) x^{2}-(\mathrm{p}+1) x+1=0$ has real and equal roots. Also, find the roots of the equation(s) so obtained.
Marking-scheme solution
Let digit at unit place be \(\displaystyle x\) and digit at tens place be \(\displaystyle y\)
\[\begin{aligned}
& \therefore \text { number }=10 y+x \\
& \text { ATQ } \\
& \quad 10 y+x=7(x+y) \\
& \Rightarrow \quad 3 y=6 x \text { or } y=2 x \quad \ldots(1)
\end{aligned}
\]
Also, \(\displaystyle 10 \mathrm{y}+\mathrm{x}=5 \mathrm{xy}+2 \quad \ldots(2)\)
from ($\displaystyle 1$) and ($\displaystyle 2$), we get
\[\begin{aligned}
& 10 x^{2}-21 x+2=0 \\
\Rightarrow \; & (x-2)(10 x-1)=0
\end{aligned}
\]
\(\displaystyle \therefore \mathrm{x}=2\)
So, \(\displaystyle \mathrm{y}=4\)
∴ Required number is 42.
For real and equal roots, \(\displaystyle \mathrm{D}=0\)
\[\begin{aligned}
& \therefore[-(\mathrm{p}+1)]^{2}-4(\mathrm{p}+4)=0 \\
& \Rightarrow \mathrm{p}^{2}-2 \mathrm{p}-15=0 \\
& \Rightarrow(\mathrm{p}-5)(\mathrm{p}+3)=0 \\
& \therefore \mathrm{p}=5,-3
\end{aligned}
\]
For \(\displaystyle \mathrm{p}=5\),
\[\begin{aligned}
& 9 x^{2}-6 x+1=0 \\
\Rightarrow & (3 x-1)(3 x-1)=0 \\
\therefore x= & \frac{1}{3}, \frac{1}{3}
\end{aligned}
\]
For \(\displaystyle \mathrm{p}=-3\),
\[\begin{aligned}
& x^{2}+2 x+1=0 \\
\Rightarrow & (x+1)(x+1)=0 \\
\therefore x= & -1,-1
\end{aligned}
\]
Hence roots are \(\displaystyle \frac{1}{3}, \frac{1}{3}\) and \(\displaystyle -1,-1\) for \(\displaystyle \mathrm{p}=5\) and \(\displaystyle \mathrm{p}=-3\) respectively.
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CBSE Class 10 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.