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Scientific notation

Scientific Notation — Exam-Style Problem Set

These questions combine scientific notation, modelling with spheres, and percentage error.
All solutions show full reasoning and correct significant figures (3 s.f. unless stated).

Instructions

  • Write answers in the form a\times10^k, where 1\le a<10.
  • Use 3 significant figures unless otherwise stated.
  • Show full working.
  • Use: V=\frac{4}{3}\pi r^3.

Question 1

A rare metallic asteroid contains platinum deposits valued at 9875 quadrillion euros, where 1 quadrillion =10^{15}.

(a) Express the value in scientific notation.
(b) Model the asteroid as a sphere of radius 128 km. Find its volume.
(c) The actual volume is 7.385\times10^{6} km^3. Calculate the percentage error.

Solution

(a)

    \[9875\times10^{15}\]

    \[= (9.875\times10^3)\times10^{15}\]

    \[= \boxed{9.875\times10^{18}\text{ EUR}}\]

(b)

    \[r=128\]

    \[r^3=128^3=2\,097\,152\]

    \[V=\frac{4}{3}\pi(2\,097\,152)\]

    \[V\approx 8.79\times10^6\text{ km}^3\]

(c)

    \[\varepsilon=\left|\frac{8.79-7.385}{7.385}\right|\times100\]

    \[\varepsilon\approx 19.0\%\]

Answer: 9.875\times10^{18} EUR, 8.79\times10^6 km^3, 19.0\%

Question 2

An exoplanet fragment is valued at 12,345 trillion euros, where 1 trillion =10^{12}.

(a) Express this in scientific notation.
(b) Model it as a sphere of radius 2100 km and find the volume.
(c) The actual volume is 4.71\times10^{10} km^3. Find the percentage error.

Solution

(a)

    \[12\,345\times10^{12}\]

    \[= 1.2345\times10^{16}\]

(b)

    \[V=\frac{4}{3}\pi(2100)^3\]

    \[V\approx 3.88\times10^{10}\text{ km}^3\]

(c)

    \[\varepsilon=\left|\frac{3.88-4.71}{4.71}\right|\times100\]

    \[\varepsilon\approx 17.6\%\]

Answer: 1.23\times10^{16} EUR, 3.88\times10^{10} km^3, 17.6\%

Self-Practice Questions

  1. Write 4,250 quadrillion in scientific notation.
  2. Find the volume of a sphere of radius 85 km (3 s.f.).
  3. Find the percentage error if estimated volume is 2.57\times10^6 km^3 and exact value is 2.20\times10^6 km^3.
  4. Write 678 trillion in scientific notation.
  5. Write 9,600 billion in scientific notation.

Click to Reveal Full Solutions

Solution 1

    \[1 \text{ quadrillion} = 10^{15}\]

    \[4\,250 \text{ quadrillion} = 4\,250\times10^{15}\]

Write 4,250 in standard form:

    \[4\,250 = 4.25\times10^3\]

    \[= (4.25\times10^3)\times10^{15}\]

    \[= 4.25\times10^{18}\]

Answer: \boxed{4.25\times10^{18}}

Solution 2

Volume formula:

    \[V=\frac{4}{3}\pi r^3\]

    \[r=85\]

First compute:

    \[85^3 = 614\,125\]

Now substitute:

    \[V=\frac{4}{3}\pi(614\,125)\]

    \[V\approx 4.18879\times614\,125\]

    \[V\approx 2\,571\,818\]

Convert to scientific notation:

    \[V\approx 2.57\times10^6 \text{ km}^3\]

Answer: \boxed{2.57\times10^6\text{ km}^3}

Solution 3

Percentage error formula:

    \[\varepsilon=\left|\frac{v_A-v_E}{v_E}\right|\times100\]

Substitute:

    \[\varepsilon=\left|\frac{2.57\times10^6-2.20\times10^6}{2.20\times10^6}\right|\times100\]

Factor out 10^6:

    \[=\left|\frac{2.57-2.20}{2.20}\right|\times100\]

    \[=\frac{0.37}{2.20}\times100\]

    \[=0.168\times100\]

    \[=16.8\%\]

Answer: \boxed{16.8\%}

Solution 4

    \[1 \text{ trillion} = 10^{12}\]

    \[678 \text{ trillion} = 678\times10^{12}\]

Write 678 in standard form:

    \[678 = 6.78\times10^2\]

    \[= (6.78\times10^2)\times10^{12}\]

    \[= 6.78\times10^{14}\]

Answer: \boxed{6.78\times10^{14}}

Solution 5

    \[1 \text{ billion} = 10^9\]

    \[9\,600 \text{ billion} = 9\,600\times10^9\]

Write 9,600 in standard form:

    \[9\,600 = 9.6\times10^3\]

    \[= (9.6\times10^3)\times10^9\]

    \[= 9.6\times10^{12}\]

To 3 significant figures:

    \[= 9.60\times10^{12}\]

Answer: \boxed{9.60\times10^{12}}

———————————–

Self-Practice Questions

These questions are designed in the same style as the solved and practice examples above.
Try each one carefully, show your working, and check your percentage error at the end.

Practice Question 1

A newly identified asteroid belt fragment is estimated to contain iron worth
4,250 quadrillion euros, where 1 quadrillion =10^{15}.

(a) Express this value in the form a \times 10^k, where 1 \leq a < 10 and k \in \mathbb{Z}.

(b) Astronomers approximate the fragment as a sphere of radius 85 km. Find its volume (km^3).

(c) If new data shows the actual volume is 2.57 \times 10^6 km^3, calculate the percentage error of the estimate.

(HL extension — density modelling)
(d) Assume the fragment is solid iron with density 7.80\times10^3 kg m^{-3}.
Estimate its mass in kg.

Hint: 1 km^3 = 10^9 m^3.

Practice Question 2

A comet core is estimated to hold frozen gases valued at
678 trillion euros, where 1 trillion =10^{12}.

(a) Write this in scientific notation.

(b) Model the core as a sphere with radius 12.5 km and compute its volume (km^3).

(c) Later, refined measurements give the actual volume as 8.00 \times 10^3 km^3.
Find the percentage error of the original estimate.

(HL extension — density modelling)
(d) Comet material is very porous. Assume the average density is 6.00\times10^2 kg m^{-3}.
Estimate the comet’s mass in kg.

Practice Question 3

A small exoplanet fragment is believed to contain precious minerals worth
9,600 billion euros, where 1 billion =10^{9}.

(a) Express this amount in standard scientific notation.

(b) The fragment is modelled as a sphere of radius 6.4 km. Calculate its approximate volume (km^3).

(c) If the true volume is 1.15 \times 10^3 km^3, determine the percentage error in the estimate.

(HL extension — density modelling)
(d) Assume the fragment has average density 3.20\times10^3 kg m^{-3}.
Estimate its mass in kg.

Click to Reveal Full Worked Solutions (Self-Practice Answers)



Practice Question 1 — Solutions

(a)

    \[4250\ \text{quadrillion} = 4250\times10^{15}\]

Write 4250 in scientific notation:

    \[4250 = 4.25\times10^3\]

So,

    \[4250\times10^{15}=(4.25\times10^3)\times10^{15}=4.25\times10^{18}\]

Answer (a): \boxed{4.25\times10^{18}\ \text{EUR}}

(b)

    \[V=\frac{4}{3}\pi r^3,\quad r=85\ \text{km}\]

    \[85^3 = 614125\]

    \[V=\frac{4}{3}\pi(614125)\approx 2.572\times10^6\ \text{km}^3\]

(to 3 s.f.)

Answer (b): \boxed{2.57\times10^6\ \text{km}^3}

(c) Percentage error:

    \[\varepsilon=\left|\frac{v_A-v_E}{v_E}\right|\times100\%\]

Here, estimated v_A\approx 2.572\times10^6 km^3 and exact v_E=2.57\times10^6 km^3:

    \[\varepsilon=\left|\frac{2.572\times10^6-2.57\times10^6}{2.57\times10^6}\right|\times100\%\]

Factor 10^6:

    \[=\left|\frac{2.572-2.57}{2.57}\right|\times100\% =\frac{0.002}{2.57}\times100\%\]

    \[\varepsilon\approx 0.0950\%\]

Answer (c): \boxed{0.0950\%}

(d) (HL) Convert volume to m^3:

    \[1\ \text{km}^3=10^9\ \text{m}^3\]

Using V\approx 2.572\times10^6 km^3:

    \[V\approx 2.572\times10^6\times10^9=2.572\times10^{15}\ \text{m}^3\]

Mass:

    \[m=\rho V=(7.80\times10^3)(2.572\times10^{15})\]

    \[m\approx 2.01\times10^{19}\ \text{kg}\]

Answer (d): \boxed{2.01\times10^{19}\ \text{kg}}



Practice Question 2 — Solutions

(a)

    \[678\ \text{trillion}=678\times10^{12}\]

    \[678=6.78\times10^2\]

    \[678\times10^{12}=(6.78\times10^2)\times10^{12}=6.78\times10^{14}\]

Answer (a): \boxed{6.78\times10^{14}\ \text{EUR}}

(b)

    \[V=\frac{4}{3}\pi r^3,\quad r=12.5\ \text{km}\]

    \[12.5^3=1953.125\]

    \[V=\frac{4}{3}\pi(1953.125)\approx 8.18\times10^3\ \text{km}^3\]

Answer (b): \boxed{8.18\times10^3\ \text{km}^3}

(c)
Estimated v_A\approx 8.18\times10^3, exact v_E=8.00\times10^3:

    \[\varepsilon=\left|\frac{8.18\times10^3-8.00\times10^3}{8.00\times10^3}\right|\times100\%\]

Factor 10^3:

    \[=\left|\frac{8.18-8.00}{8.00}\right|\times100\% =\frac{0.18}{8.00}\times100\%\]

    \[\varepsilon\approx 2.27\%\]

Answer (c): \boxed{2.27\%}

(d) (HL)
Convert volume:

    \[V\approx 8.18\times10^3\times10^9=8.18\times10^{12}\ \text{m}^3\]

Mass:

    \[m=\rho V=(6.00\times10^2)(8.18\times10^{12})\]

    \[m\approx 4.91\times10^{15}\ \text{kg}\]

Answer (d): \boxed{4.91\times10^{15}\ \text{kg}}



Practice Question 3 — Solutions

(a)

    \[9600\ \text{billion}=9600\times10^9\]

    \[9600=9.60\times10^3\]

    \[9600\times10^9=(9.60\times10^3)\times10^9=9.60\times10^{12}\]

Answer (a): \boxed{9.60\times10^{12}\ \text{EUR}}

(b)

    \[V=\frac{4}{3}\pi r^3,\quad r=6.4\ \text{km}\]

    \[6.4^3=262.144\]

    \[V=\frac{4}{3}\pi(262.144)\approx 1.10\times10^3\ \text{km}^3\]

Answer (b): \boxed{1.10\times10^3\ \text{km}^3}

(c)
Estimated v_A\approx 1.10\times10^3, exact v_E=1.15\times10^3:

    \[\varepsilon=\left|\frac{1.10\times10^3-1.15\times10^3}{1.15\times10^3}\right|\times100\%\]

Factor 10^3:

    \[=\left|\frac{1.10-1.15}{1.15}\right|\times100\% =\frac{0.05}{1.15}\times100\%\]

    \[\varepsilon\approx 4.52\%\]

Answer (c): \boxed{4.52\%}

(d) (HL)
Convert volume:

    \[V\approx 1.098\times10^3\times10^9=1.098\times10^{12}\ \text{m}^3\]

Mass:

    \[m=\rho V=(3.20\times10^3)(1.098\times10^{12})\]

    \[m\approx 3.51\times10^{15}\ \text{kg}\]

Answer (d): \boxed{3.51\times10^{15}\ \text{kg}}

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