Case Study – 3 In a pool at an aquarium, a dolphin jumps out of the water travelling at 20 cm per second. Its height above water level after t seconds is given by h = 20t - 16t². Based on the above, answer the following questions : (i) Find zeroes of polynomial p(t) = 20t - 16t². (ii) Which of the following types of graph represents p(t) ? (a) Parabola opening downward from the origin, crossing the time axis at 5/4, with the curve above the time axis between 0 and 5/4. (b) Parabola with vertex above the time axis at t = 5/4, crossing the time axis at 0 and at a point beyond 5/4. (c) Parabola opening upward, with vertex below the time axis, crossing the axis at 0 and 5/4. (d) Downward-opening parabola with vertex at t = 5/4 on the time axis, starting above the origin on the height axis. (iii) (a) What would be the value of h at t = 3/2 ? Interpret the result. (iii) (b) How much distance has the dolphin covered before hitting the water level again ?

MathsCBSE Class 10 · Standard · Board exam 2023, set 30/5/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2023: question 38

Set in the CBSE Class 10 board exam of 2023, paper 30/5/1, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedPolynomials

Case Study – 3

In a pool at an aquarium, a dolphin jumps out of the water travelling at 20 cm per second. Its height above water level after t seconds is given by $h = 20 t - 16 t^{2}$ .

Based on the above, answer the following questions :

(i)
Find zeroes of polynomial $p (t) = 20 t - 16 t^{2}$ .
1 mark
(ii)
Which of the following types of graph represents $p (t)$ ?
(a) Parabola opening downward from the origin, crossing the time axis at $\frac{5}{4}$ , with the curve above the time axis between $0$ and $\frac{5}{4}$ .
(b) Parabola with vertex above the time axis at $t = \frac{5}{4}$ , crossing the time axis at $0$ and at a point beyond $\frac{5}{4}$ .
(c) Parabola opening upward, with vertex below the time axis, crossing the axis at $0$ and $\frac{5}{4}$ .
(d) Downward-opening parabola with vertex at $t = \frac{5}{4}$ on the time axis, starting above the origin on the height axis.
1 mark
(iii)
(a) What would be the value of h at $t = \frac{3}{2}$ ? Interpret the result.
2 marks
OR
(iii)
(b) How much distance has the dolphin covered before hitting the water level again ?
2 marks

Figure in the paper: Four graph options labelled (a) to (d), each with axes x′, x (time) and y′, y (height). Option (a): upward-opening arc from origin to a point at 5/4 on the time axis. Option (b): arc peaking at t = 5/4 with dashed vertical line, returning to the time axis. Option (c): upward parabola dipping below the axis between 0 and 5/4. Option (d): downward-opening curve with vertex at 5/4 on the time axis. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 30/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Take t common from 20t − 16t² to find where the height is zero.

Hint 2

The sign of the t² coefficient tells you whether the parabola opens up or down; match the zeroes to the graph.

Hint 3

A negative h means below the surface. For (iii)(b), the dolphin is in the air between the two zeroes — combine that time with its speed.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Put $p (t) = 0$ : $20 t - 16 t^{2} = 0 \Rightarrow 4 t (5 - 4 t) = 0$
$\Rightarrow t = 0$ or $t = \frac{5}{4}$ . Zeroes are $0$ and $\frac{5}{4}$ .
1 mark
(ii) The coefficient of $t^{2}$ is $- 16 < 0$ , so the graph is a downward-opening parabola, and it cuts the time axis at $t = 0$ and $t = \frac{5}{4}$ , staying above the axis in between. This matches graph (a).
1 mark
(iii)(a) At $t = \frac{3}{2}$ : $h = 20 \times \frac{3}{2} - 16 \times \frac{9}{4} = 30 - 36 = - 6$ cm
1 mark
(iii)(a) Interpretation: a negative height means the dolphin is below the water surface; after $\frac{3}{2}$ s it is 6 cm below water level (it re-entered the water at $t = \frac{5}{4}$ s).
1 mark
(iii)(b) The dolphin is out of the water from $t = 0$ to $t = \frac{5}{4}$ s (the zeroes of $p (t)$ ). Travelling at 20 cm per second, distance covered $= 20 \times \frac{5}{4} = 25$ cm.
2 marks

Answer: (i) 0 and 5/4; (ii) graph (a); (iii)(a) h = −6 cm, i.e. 6 cm below water level; (iii)(b) 25 cm

Where the marks are

(i) Factorises and finds zeroes 0 and 5/4 (1 mark)
(ii) Chooses graph (a) (1 mark)
(iii)(a) Computes h = −6 (1 mark)
(iii)(a) Interprets as 6 cm below water level (1 mark)
(iii)(b) Uses time 5/4 s with speed 20 cm/s to get 25 cm (2 marks)

Revise the chapter

Chapter: Polynomials (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Polynomials, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 30/5/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2023: question 38

Hints

Step-by-step solution

Revise the chapter

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