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Mathematics · 2023 · 3 marks
CBSE 2023 · Region 6 · Set 1 · Q29
(A) The sum of two numbers is 15. If the sum of their reciprocals is 3/10, find the two numbers.
(A)
The sum of two numbers is $\displaystyle 15$ . If the sum of their reciprocals is $\displaystyle \frac{3}{10}$, find the two numbers.
OR
(B)
If $\displaystyle \alpha$ and $\displaystyle \beta$ are roots of the quadratic equation $\displaystyle x^{2}-7 x+10=0$, find the quadratic equation whose roots are $\displaystyle \alpha^{2}$ and $\displaystyle \beta^{2}$.
Marking-scheme solution
Let one number be \(\displaystyle \mathrm{x} \Rightarrow\) another number \(\displaystyle =15-\mathrm{x}\)
Therefore, \(\displaystyle \frac{1}{x}+\frac{1}{15-\mathrm{x}}=\frac{3}{10}\)
\[\frac{15-x+x}{x(15-x)}=\frac{3}{10} \Rightarrow 150=3 x(15-x)
\]
\[3 x^{2}-45 x+150=0
\]
\[x^{2}-15 x+50=0 \Rightarrow(x-10)(x-5)=0
\]
\[\Rightarrow \mathrm{x}=10,5
\]
Numbers are $\displaystyle 10$, $\displaystyle 5$ or $\displaystyle 5$, $\displaystyle 10$
\[\begin{aligned}
& x^{2}-7 x+10=0 \\
& \alpha+\beta=7, \alpha \beta=10 \\
& \alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=49-20=29 \\
& \alpha^{2} \beta^{2}=(10)^{2}=100
\end{aligned}
\]
Quadratic Equation with roots \(\displaystyle \alpha^{2}, \beta^{2}\) is
\[\begin{aligned}
& \therefore x^{2}-\left(\alpha^{2}+\beta^{2}\right) x+\alpha^{2} \beta^{2}=0 \\
& \text { i.e. } x^{2}-29 x+100=0
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.