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Mathematics · 2024 · 5 marks
CBSE 2024 · Region 5 · Set 2 · Q35
A $\displaystyle 2$-digit number is such that the product of the digits is 14. When $\displaystyle 45$ is added to the number, the digits are reversed. Find the number.The side of a square exceeds the side of another square by $\displaystyle 4$ cm and the sum of the areas of the two squares is $\displaystyle 400 \mathrm{~cm}^{2}$. Find the sides of the squares.
A $\displaystyle 2$-digit number is such that the product of the digits is 14. When $\displaystyle 45$ is added to the number, the digits are reversed. Find the number.
The side of a square exceeds the side of another square by $\displaystyle 4$ cm and the sum of the areas of the two squares is $\displaystyle 400 \mathrm{~cm}^{2}$. Find the sides of the squares.
Marking-scheme solution
Let the two digit number be \(\displaystyle 10 x+y\)
\[\begin{aligned}
& x y=14 \ldots \ldots \text { (i) } \\
& 10 x+y+45=10 y+x \\
& y-x=5 \ldots . . \text { (ii) }
\end{aligned}
\]
from (i) and (ii)
\[\begin{aligned}
& \frac{14}{x}-x=5 \\
& x^{2}+5 x-14=0 \\
& (x+7)(x-2)=0 \\
& x \neq-7 \\
& x=2 \\
& \text { Number }=27
\end{aligned}
\]
Let the side of first square \(\displaystyle =x \mathrm{~cm}\)
\[\begin{aligned}
& \therefore \text { Side of second square }=(x+4) \mathrm{cm} \\
& \quad x^{2}+(x+4)^{2}=400 \\
& x^{2}+4 x+192=0 \\
& (\mathrm{x}+16)(\mathrm{x}-12)=0 \\
& \mathrm{x} \neq-16 \\
& \mathrm{x}=12
\end{aligned}
\]
Side of square \(\displaystyle =12 \mathrm{~cm}\) and $\displaystyle 16$ cm
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CBSE Class 10 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.