IBDP Maths Applications and Interpretation Topic : SL 1.1 Operations with numbers SL Paper 2

Question

The diameter of a spherical planet is 6 × 104 km .
(a) Write down the radius of the planet. [1]
The volume of the planet can be expressed in the form π(a × 10k) km3 where 1 ≤ a <10 and k ∈  \(\mathbb{Z}\)
(b) Find the value of a and the value of k .

Answer/Explanation

Ans

(a)  \(3 \times 10^4\)
(b) \(\frac{4}{3}\pi(3 \times 10^4)^3\)
\(=\frac{4}{3}\pi\times 27 \times 10^{12}\)
\(=\pi(3.6 \times 10^{13})(km)^3\)
Hence
\(a= 3.6 \) ,  \(k =13\)

Question

Nickel in the asteroid 16 Psyche is said to be valued at 8973 quadrillion euros (EUR), where one quadrillion = 1015.

    1. Write down the value of the Nicke in the form a × 10k where 1 a < 10 , k∈Z . [2]

      Charlie believes the asteroid is approximately spherical with radius 113 km. He uses this information to estimate its volume.

    2. Calculate Charlie’s estimate of its volume, in km3. [2]

      The actual volume of the asteroid is found to be 6.074 × 106 km3 .

    3. Find the percentage error in Charlie’s estimate of the volume. [2]

Answer/Explanation

Ans: 

(a)

8.97 × 1018 (EUR)(8.973 × 1018)

(b)

\(\frac{4\times \pi 113^{3}}{3}\)

\(6040000(km^{3})(6.04\times 10^{6},\frac{5771588\pi }{3}, 6043992.82)\)

(c)

\(|\frac{6043992.82-6.074\times 10^{6}}{6.074 \times 10^{6}}|\) x 100

0.494(%)   (0.494026…(%))

Question

Write down the following numbers in increasing order.

\(3.5\), \(1.6 \times 10^{−19}\), \(60730\), \(6.073 \times 10^{5}\), \(0.006073 \times 10^6\), \(\pi\), \(9.8 \times 10^{−18}\).[3]

a.

Write down the median of the numbers in part (a).[1]

b.

State which of the numbers in part (a) is irrational.[1]

c.
Answer/Explanation

Markscheme

\(1.6 \times 10^{−19}\), \(9.8 \times 10^{−18}\), \(\pi\), \(3.5\), \(0.006073 \times 10^6\), \(60730\), \(6.073 \times 10^{5}\)     (A4)

Award (A1) for \(\pi\) before 3.5

Award (A1) for \(1.6 \times 10^{−19}\) before \(9.8 \times 10^{−18}\)

Award (A1) for the three numbers containing 6073 in the correct order.

Award (A1) for the pair with negative indices placed before 3.5 and \(\pi\) and the remaining three numbers placed after (independently of the other three marks).

Award (A3) for numbers given in correct decreasing order.

Award (A2) for decreasing order with at most 1 error     (C4)[3 marks]

a.

The median is 3.5.     (A1)(ft)

Follow through from candidate’s list.     (C1)[1 mark]

b.

\(\pi\) is irrational.     (A1)     (C1)[1 mark]

c.

Question

Calculate exactly \(\frac{{{{(3 \times 2.1)}^3}}}{{7 \times 1.2}}\).[1]

a.

Write the answer to part (a) correct to 2 significant figures.[1]

b.

Calculate the percentage error when the answer to part (a) is written correct to 2 significant figures.[2]

c.

Write your answer to part (c) in the form \(a \times {10^k}\) where \(1 \leqslant a < 10{\text{ and }}k \in \mathbb{Z}\).[2]

d.
Answer/Explanation

Markscheme

\(29.7675\)     (A1)     (C1)

Note: Accept exact answer only.[1 mark]

a.

\(30\)     (A1)(ft)     (C1)[1 mark]

b.

\(\frac{{30 – 29.7675}}{{29.7675}} \times 100\% \)     (M1)

For correct formula with correct substitution.

\( = 0.781\% \)     accept \(0.78\% \) only if formula seen with \(29.7675\) as denominator     (A1)(ft)     (C2)[2 marks]

c.

\(7.81 \times {10^{ – 1}}\% \) (\(7.81 \times {10^{ – 3}}\) with no percentage sign)     (A1)(ft)(A1)(ft)     (C2)[2 marks]

d.

Question

Given that \(h = \sqrt {{\ell ^2} – \frac{{{d^2}}}{4}} \) ,

Calculate the exact value of \(h\) when \(\ell  = 0.03625\) and \(d = 0.05\) .[2]

a.

Write down the answer to part (a) correct to three decimal places.[1]

b.

Write down the answer to part (a) correct to three significant figures.[1]

c.

Write down the answer to part (a) in the form \(a \times {10^k}\) , where \(1 \leqslant a < 10{\text{, }}k \in \mathbb{Z}\).[2]

d.
Answer/Explanation

Markscheme

\(h = \sqrt {{{0.03625}^2} – \frac{{{{0.05}^2}}}{4}} \)     (M1)
\( = 0.02625\)     (A1)     (C2)

Note: Award (A1) only for \(0.0263\) seen without working[2 marks]

a.

\(0.026\)     (A1)(ft)     (C1)[1 mark]

b.

\(0.0263\)     (A1)(ft)     (C1)[1 mark]

c.

\(2.625 \times {10^{ – 2}}\)

for \(2.625\) (ft) from unrounded (a) only     (A1)(ft)

for \( \times {10^{ – 2}}\)     (A1)(ft)     (C2)[2 marks]

d.
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