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Mathematics · 2023 · 4 marks
CBSE 2023 · Region 5 · Set 1 · Q38
In a pool at an aquarium, a dolphin jumps out of the water travelling at $\displaystyle 20$ cm per second. Its height above water level after t seconds is given by $\displaystyle \mathrm{h}=20 \mathrm{t}-16 \mathrm{t}^{2}$.
Based on the above, answer the following questions :(i)Find zeroes of polynomial $\displaystyle \mathrm{p}(\mathrm{t})=20 \mathrm{t}-16 \mathrm{t}^{2}$.(ii)Which of the following types of graph represents p(t) ?(b)
(d)
(iii)What would be the value of h at $\displaystyle \mathrm{t}=\frac{3}{2}$ ? Interpret the result.How much distance has the dolphin covered before hitting the water level again ?
In a pool at an aquarium, a dolphin jumps out of the water travelling at $\displaystyle 20$ cm per second. Its height above water level after t seconds is given by $\displaystyle \mathrm{h}=20 \mathrm{t}-16 \mathrm{t}^{2}$.
Based on the above, answer the following questions :
(i)
Find zeroes of polynomial $\displaystyle \mathrm{p}(\mathrm{t})=20 \mathrm{t}-16 \mathrm{t}^{2}$.
(ii)
Which of the following types of graph represents p(t) ?
(b)
(d)
(iii)
What would be the value of h at $\displaystyle \mathrm{t}=\frac{3}{2}$ ? Interpret the result.
How much distance has the dolphin covered before hitting the water level again ?
Marking-scheme solution
(i)
\(\displaystyle -\mathbf{1 6} \mathbf{t}^{\mathbf{2}}+\mathbf{2 0 t}=\mathbf{0} \Rightarrow \mathbf{4 t}(-\mathbf{4 t}+\mathbf{5})=\mathbf{0}\)
\(\displaystyle \mathrm{t}=\mathrm{0}, \mathrm{t}=\frac{5}{4}\)
(a)
(iii)
At \(\displaystyle \mathbf{t}=\frac{\mathbf{3}}{\mathbf{2}}, \mathbf{h}=-\mathbf{1 6} \times \frac{\mathbf{9}}{\mathbf{4}}+\mathbf{2 0} \times \frac{\mathbf{3}}{\mathbf{2}}=-\mathbf{3 6}+\mathbf{3 0}=-\mathbf{6}\)
It means after \(\displaystyle \frac{\mathbf{3}}{\mathbf{2}}\) seconds, dolphin has reached $\displaystyle 6$ cm below water level.
Speed of dolphin \(\displaystyle =\mathbf{2 0 ~ c m}\) per second.
In one second, distance covered = $\displaystyle 20$ cm
In \(\displaystyle \frac{\mathbf{5}}{\mathbf{4}}\) seconds, distance covered \(\displaystyle \boldsymbol{=} \mathbf{2 0} \boldsymbol{\times} \frac{\mathbf{5}}{\mathbf{4}} \boldsymbol{=} \mathbf{2 5 ~ c m}\)
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CBSE Class 10 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.