# IBDP Maths AA: Topic : SL 1.1: Operations with numbers: IB style Questions SL Paper 1

### 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 . 

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. 

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. 

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

Five pipes labelled, “6 metres in length”, were delivered to a building site. The contractor measured each pipe to check its length (in metres) and recorded the following;

5.96, 5.95, 6.02, 5.95, 5.99.

(i) Find the mean of the contractor’s measurements.

(ii) Calculate the percentage error between the mean and the stated, approximate length of 6 metres.

a.

Calculate $$\sqrt {{{3.87}^5} – {{8.73}^{ – 0.5}}}$$, giving your answer

(i) correct to the nearest integer,

(ii) in the form $$a \times 10^k$$, where 1 ≤ a < 10, $$k \in {\mathbb{Z}}$$ .

b.

## Markscheme

(i) Mean = (5.96 + 5.95 + 6.02 + 5.95 + 5.99) / 5 = 5.974 (5.97)     (A1)

(ii) $${\text{% error}} = \frac{{error}}{{actualvalue}} \times 100\%$$

$$= \frac{{6 – 5.974}}{{5.974}} \times 100\% = 0.435\%$$     (M1)(A1)(ft)

(M1) for correctly substituted formula.

Allow 0.503% as follow through from 5.97

Note: An answer of 0.433% is incorrect.     (C3)[3 marks]

a.

number is 29.45728613

(i) Nearest integer = 29     (A1)

(ii) Standard form = 2.95 × 101 (accept 2.9 × 101)     (A1)(ft)(A1)

Award (A1) for each correct term

Award (A1)(A0) for 2.95 × 10     (C3)[3 marks]

b.

## Question

Write down the following numbers in increasing order.

$$3.5$$, $$1.6 \times 10^{−19}$$, $$60730$$, $$6.073 \times 10^{5}$$, $$0.006073 \times 10^6$$, $$\pi$$, $$9.8 \times 10^{−18}$$.

a.

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

b.

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

c.

## 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 $$\frac{{77.2 \times {3^3}}}{{3.60 \times {2^2}}}$$.

a.

Express your answer to part (a) in the form $$a \times 10^k$$, where $$1 \leqslant a < 10$$ and $$k \in {\mathbb{Z}}$$.

b.

Juan estimates the length of a carpet to be 12 metres and the width to be 8 metres. He then estimates the area of the carpet.

(i) Write down his estimated area of the carpet.

When the carpet is accurately measured it is found to have an area of 90 square metres.

(ii) Calculate the percentage error made by Juan.

c.

## Markscheme

$$144.75\left( { = \frac{{579}}{4}} \right)$$     (A1)

accept 145     (C1)[1 mark]

a.

$$1.4475 \times 10^2$$     (A1)(ft)(A1)(ft)

accept $$1.45 \times 10^2$$     (C2)[2 marks]

b.

Unit penalty (UP) is applicable in question part (c)(i) only.

(UP) (i) Area = 96 m2     (A1)

(ii) $$\% {\text{ error}} = \frac{{(96 – 90)}}{{90}} \times 100$$     (M1)

$$= \frac{{6 \times 100}}{{90}}$$

$$\frac{{20}}{3}\%$$ or 6.67 %     (A1)(ft)     (C3)[3 marks]

c.

## Question

Calculate exactly $$\frac{{{{(3 \times 2.1)}^3}}}{{7 \times 1.2}}$$.

a.

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

b.

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

c.

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

d.

## 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 $$d = 0.05$$ .

a.

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

b.

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

c.

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

d.

## 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.

## Question

A rectangle is 2680 cm long and 1970 cm wide.

Find the perimeter of the rectangle, giving your answer in the form $$a \times {10^k}$$, where $$1 \leqslant a \leqslant 10$$ and $$k \in \mathbb{Z}$$.

a.

Find the area of the rectangle, giving your answer correct to the nearest thousand square centimetres.

b.

## Markscheme

Note: Unit penalty (UP) applies in this part

(2680 + 1970) × 2     (M1)

(UP)     = 9.30 × 103 cm     (A1)(A1)     (C3)

Notes: Award (M1) for correct formula.

(A1) for 9.30 (Accept 9.3).

(A1) for 103.[3 marks]

a.

2680 × 1970     (M1)

= 5279600     (A1)

= 5,280,000 (5280 thousand)     (A1)(ft)     (C3)

Note: Award (M1) for correctly substituted formula.

Accept 5.280 × 106.

Note: The last (A1) is for specified accuracy, (ft) from their answer.

The (AP) for the paper is not applied here.[3 marks]

b.

## Question

The following diagram shows a rectangle with sides of length 9.5 ×102 m and 1.6 ×103 m. Write down the area of the rectangle in the form a × 10k, where 1 ≤ a < 10, k ∈ $$\mathbb{Z}$$.

a.

Helen’s estimate of the area of the rectangle is $$1\,600\,000{\text{ }}{{\text{m}}^2}$$.

Find the percentage error in Helen’s estimate.

b.

## Markscheme

UP applies in part (a).

$$9.5 \times 10^2 \times 1.6 \times 10^3$$     (M1)
(UP)     $$= 1.52 \times {10^6}{\text{ }}{{\text{m}}^2}$$     (A1)(A1)     (C3)

Notes: Award (M1) for multiplication of the two numbers.

Award (A1) for $$1.52$$, (A1) for $$10^6$$.[3 marks]

a.

$$\frac{{1600000 – 1520000}}{{1520000}} \times 100$$     (M1)(A1)(ft)

Note: Award (M1) for substitution in formula, (A1)(ft) for their correct substitution.

= 5.26 % (percent sign not required).     (A1)(ft)     (C3)

Note: Accept positive or negative answer.[3 marks]

b.

## Question

The volume of a sphere is $$V{\text{ = }}\sqrt {\frac{{{S^3}}}{{36\pi }}}$$, where $$S$$ is its surface area.

The surface area of a sphere is 500 cm2 .

Calculate the volume of the sphere. Give your answer correct to two decimal places.

a.

b.

Write down your answer to (b) in the form $$a \times {10^n}$$, where $$1 \leqslant a < 10$$ and $$n \in \mathbb{Z}$$.

c.

## Markscheme

$$V{\text{ = }}\sqrt {\frac{{{500^3}}}{{36\pi }}}$$     (M1)
Note: Award (M1) correct substitution into formula.

V = 1051.305 …     (A1)
V = 1051.31 cm3     (A1)(ft)    (C3)

Note: Award last (A1)(ft) for correct rounding to 2 decimal places of their answer. Unrounded answer must be seen so that the follow through can be awarded.[3 marks]

a.

1051     (A1)(ft)[1 mark]

b.

$$1.051 \times {10^3}$$     (A1)(ft)(A1)(ft)     (C2)

Note: Award (A1) for 1.051 (accept 1.05) (A1) for $$\times {10^3}$$.[2 marks]

c.

## Question

Consider the following four numbers.

$$p = 0.00314{\text{ ; }}q = 0.00314 \times {10^2}{\text{ ; }}r = \frac{\pi }{{1000}}{\text{ ; }}s = 3.14 \times {10^{ – 2}}$$

One of these numbers is written in the form $$a \times {10^k}$$ where $$1 \leqslant a < 10$$ and $$k \in \mathbb{Z}$$. Write down this number.

a.

Write down the smallest of these numbers.

b.

Write down the value of q + s.

c.

Give your answer to part (c) in the form $$a \times {10^k}$$ where $$1 \leqslant a < 10$$ and $$k \in \mathbb{Z}$$.

d.

## Markscheme

3.14 × 10–2 or s     (A1)     (C1)[1 mark]

a.

0.00314 or 3.14 × 10–3 or p     (M1)(A1)     (C2)

Note: Award (M1) for indication of comparing numbers where at least one of them is converted. The converted number does not have to be correct. A single converted number is sufficient for (M1) to be awarded.[2 marks]

b.

0.3454 (0.345)     (A1)     (C1)[1 mark]

c.

3.454 × 10–1 (3.45 × 10–1)     (A1)(A1)(ft)     (C2)

Notes: Follow through from their (c).

Award (A1) for 3.454 (3.45) (A1) for 10–1.[2 marks]

d.

## Question

A satellite travels around the Earth in a circular orbit $$500$$ kilometres above the Earth’s surface. The radius of the Earth is taken as $$6400$$ kilometres.

Write down the radius of the satellite’s orbit.

a.

Calculate the distance travelled by the satellite in one orbit of the Earth. Give your answer correct to the nearest km.

b.

Write down your answer to (b) in the form $$a \times {10^k}$$ , where $$1 \leqslant a < 10{\text{, }}k \in \mathbb{Z}$$ .

c.

## Markscheme

$$6900$$ km     (A1)     (C1)[1 mark]

a.

$$2\pi (6900)$$     (M1)(A1)(ft)

Notes: Award (M1) for substitution into circumference formula, (A1)(ft) for correct substitution. Follow through from part (a).

$$= 43354$$     (A1)(ft)     (C3)

Notes: Follow through from part (a). The final (A1) is awarded for rounding their answer correct to the nearest km. Award (A2) for $$43 400$$ shown with no working.[3 marks]

b.

$$4.3354 \times {10^4}$$     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for $$4.3354$$, (A1)(ft) for $$\times {10^4}$$ . Follow through from part (b). Accept $$4.34 \times {10^4}$$ .[2 marks]

c.

## Question

The planet Earth takes one year to revolve around the Sun. Assume that a year is 365 days and the path of the Earth around the Sun is the circumference of a circle of radius $$150000000{\text{ km}}$$. Calculate the distance travelled by the Earth in one day.

a.

Give your answer to part (a) in the form $$a \times {10^k}$$ where $$1 \leqslant a \leqslant 10$$ and $$k \in \mathbb{Z}$$ .

b.

## Markscheme

$$2\pi \frac{{150000000}}{{365}}$$     (M1)(A1)(M1)

Notes: Award (M1) for substitution in correct formula for circumference of circle.

Award (A1) for correct substitution.

Award (M1) for dividing their perimeter by $$365$$.

Award (M0)(A0)(M1) for $$\frac{{150000000}}{{365}}$$ .

$$2580000{\text{ km}}$$     (A1)     (C4)[4 marks]

a.

$$2.58 \times {10^6}$$     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for $$2.58$$, (A1)(ft) for $${10^6}$$ . Follow through from their answer to part (a). The follow through for the index should be dependent not only on the answer to part (a), but also on the value of their mantissa. No (AP) penalty for first (A1) provided their value is to 3 sf or is all their digits from part (a).[2 marks]

b.

## Question

Given that $$z = \frac{{12\cos (A)}}{{4q + r}}$$ and that $$A = {60^ \circ }$$, $$q = 8$$ and $$r = 32$$;

Find the exact value of $$z$$.

a.

b.i.

b.ii.

Write your answer to part (a) in the form $$a \times {10^k}$$, where 1 ≤ a < 10, $$k \in {\mathbb{Z}}$$ .

b.iii.

## Markscheme

$$z = \frac{{12\cos ({{60}^ \circ })}}{{(4(8) + 32)}}$$     (M1)

Note: Award (M1) for correct substituted formula seen.

$$= 0.09375\left( {\frac{3}{{32}}} \right)$$     (A1)(C2)[2 marks]

a.

$$0.09$$     (A1)(ft)     (C1)[1 mark]

b.i.

$$0.0938$$     (A1)(ft)     (C1)[1 mark]

b.ii.

$$9.375 \times {10^{ – 2}}$$ ($$9.38 \times {10^{ – 2}}$$)     (A1)(ft)(A1)(ft)     (C2)

Note: Award (A1)(ft) for $$9.375$$, (A1)(ft) for $$\times {10^{ – 2}}$$. Follow through from their part (a).[2 marks]

b.iii.

## Question

Ross is a star that is 82 414 080 000 000 km away from Earth. A spacecraft, launched from Earth, travels at 48 000 kmh–1 towards Ross.

Calculate the exact time, in hours, for the spacecraft to reach the star Ross.

a.

Give your answer to part (a) in years. (Assume 1 year = 365 days)

b.

Give your answer to part (b) in the form a×10k, where 1 ≤ a < 10 and $$k \in \mathbb{Z}$$.

c.

## Markscheme

$$\frac{{{\text{82 414 080 000 000}}}}{{48\;000}}$$     (M1)

Note: Award (M1) for correct substitution in correct formula.

1 716 960 000 (hours)     (A1)     (C2)[2 marks]

a.

$$\frac{{{\text{their (a)}}}}{{24 \times 365}}$$     (M1)

196 000 (years)     (A1)(ft)     (C2)

Note: Award (A1)(ft) from their part (a).[2 marks]

b.

1.96×105     (A1)(ft)(A1)(ft)     (C2)

Note: Award (A1)(ft) for 1.96 (accept 1.96000), (A1)(ft) for 105 . Follow through from their answer to part (b).[2 marks]

c.

## Question

Consider c = 5200 and d = 0.0000037.

Write down the value of r = c × d.

a.

Write down your value of r in the form a × 10k, where 1 ≤ a < 10 and $$k \in \mathbb{Z}$$.

b.

Consider the following statements about c, d and r. Only three of these statements are true.

Circle the true statements. c.

## Markscheme

r = 0.01924     (A1)     (C1)

Note: Accept 0.0192 and 0.019[1 mark]

a.

r = 1.924 × 10−2     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1) for 1.924, (A1) for 10−2. Accept 1.92 and 1.9. Follow through from their part (a).[2 marks]

b. (A1)(A1)(A1)     (C3)

Notes: Award (A1) for each true statement circled. Do not follow through from part (a). Award (A1)(A1)(A0) if 1 extra term seen. Award (A1)(A0)(A0) if 2 extra terms seen. Award (A0)(A0)(A0) if all terms circled. Accept other indications of the correct statements i.e. highlighted or ticks.[3 marks]

c.

## Question

The length, in cm, of six baseball bats was measured. The lengths are given below.

104.5, 105.1, 104.8, 105.2, 104.9, 104.9

Calculate the exact value of the mean length.

a.

Write your answer to part (a) in the form a × 10k where 1 ≤ a < 10 and $$k \in \mathbb{Z}$$.

b.

Marian calculates the mean length and finds it to be 105 cm.

Calculate the percentage error made by Marian.

c.

## Markscheme

$$\left( {\frac{{104.5 + 105.1 + …}}{6}} \right)$$     (M1)

Note: Award (M1) for use of mean formula.

= 104.9 (cm)     (A1)     (C2)[2 marks]

a.

1.049 × 102     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for 1.049, (A1)(ft) for 102. Follow through from their part (a).[2 marks]

b.

$$\frac{{105 – 104.9}}{{104.9}} \times 100$$  (%)     (M1)

Notes: Award (M1) for their correctly substituted % error formula.

% error = 0.0953  (%)     (0.0953288…)     (A1)(ft)     (C2)

Notes: A 2sf answer of 0.095 following $$\frac{{105 – 104.9}}{{105}} \times 100$$ working is awarded no marks. Follow through from their part (a), provided it is not 105. Do not accept a negative answer. % sign not required.[2 marks]

c.

## Question

$$z = \frac{{17{x^2}}}{{a – b}}$$.

Find the value of z when x = 12.5, a = 0.572 and b = 0.447. Write down your full calculator display.

a.

(i) correct to the nearest 1000 ;

(ii) correct to three significant figures.

b.

Write your answer to part (b)(ii) in the form a × 10k, where 1 ≤ a < 10, $$k \in \mathbb{Z}$$.

c.

## Markscheme

$$z = \frac{{17{{(12.5)}^2}}}{{(0.572 – 0.447)}}$$     (M1)

Note: Award (M1) for correct substitution into formula.

= 21250     (A1)     (C2)[2 marks]

a.

(i) 21000     (A1)(ft)

(ii) 21300     (A1)(ft)     (C2)

Note: Follow through from part (a).[2 marks]

b.

$$2.13 \times 10^4$$     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for 2.13, (A1)(ft) for $$\times 10^4$$. Follow through from part (b)(ii).[ 2 marks]

c.

## Question

A cuboid has the following dimensions: length = 8.7 cm, width = 5.6 cm and height = 3.4 cm.

Calculate the exact value of the volume of the cuboid, in cm3.

a.

(i) one decimal place;

(ii) three significant figures.

b.

Write your answer to part (b)(ii) in the form $$a \times 10^k$$, where $$1 \leqslant a < 10 , k \in \mathbb{Z}$$.

c.

## Markscheme

$${\text{V}} = 8.7 \times 5.6 \times 3.4$$     (M1)

Note: Award (M1) for multiplication of the 3 given values.

$$=165.648$$     (A1)     (C2)

a.

(i) 165.6     (A1)(ft)

(ii) 166     (A1)(ft)     (C2)

b.

$$1.66 \times 10^2$$     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for 1.66, (A1)(ft) for $$10^2$$. Follow through from their answer to part (b)(ii) only. The follow through for the index should be dependent on the value of the mantissa in part (c) and their answer to part (b)(ii).

c.

## Question

Let $$p = \frac{{2\cos x – \tan x}}{{\sqrt y – z}}.$$

Calculate the value of $$p$$ when $$x = 45^\circ$$, $$y = 8192$$ and $$z = 64$$. Write down your full calculator display.

a.

(i)     correct to two decimal places;

(ii)     correct to four significant figures;

(iii)     in the form $$a \times {10^k}$$, where $$1 \leqslant a < 10,{\text{ }}k \in \mathbb{Z}$$.

b.

## Markscheme

$$\frac{{2\cos 45^\circ – \tan 45^\circ }}{{\sqrt {8192} – 64}}$$     (M1)

$$= 0.015625$$     (A1)     (C2)

Notes:     Accept $$\frac{1}{{64}}$$ and also $$1.5625 \times {10^{ – 2}}$$.[2 marks]

a.

(i)     0.02     (A1)(ft)

(ii)     0.01563     (A1)(ft)

Notes:     For parts (i) and (ii), accept equivalent standard form representations.

(iii)     $$1.5625 \times {10^{ – 2}}$$     (A2)(ft)     (C4)

Notes:     Award (A1)(A0) for correct mantissa, between 1 and 10, with incorrect index.

Where the candidate has correctly rounded their mantissa from part (a) and has the correct exponent, award (A0)(A1)

Award (A0)(A0) for answers of the type: $$15.625 \times {10^{ – 3}}$$.[4 marks]

b.

## Question

The average radius of the orbit of the Earth around the Sun is 150 million kilometres. The average radius of the orbit of the Earth around the Sun is 150 million kilometres. Write down this radius, in kilometres, in the form $$a \times {10^k}$$, where $$1 \leqslant a < 10,{\text{ }}k \in \mathbb{Z}$$.

a.

The Earth travels around the Sun in one orbit. It takes one year for the Earth to complete one orbit.

Calculate the distance, in kilometres, the Earth travels around the Sun in one orbit, assuming that the orbit is a circle.

b.

Today is Anna’s 17th birthday.

Calculate the total distance that Anna has travelled around the Sun, since she was born.

c.

## Markscheme

$$1.5 \times {10^8}{\text{ (km)}}$$     (A2)     (C2)

Notes: Award (A2) for the correct answer.

Award (A1)(A0) for 1.5 and an incorrect index.

Award (A0)(A0) for answers of the form $$15 \times {10^7}$$.[2 marks]

a.

$$2\pi 1.5 \times {10^8}$$     (M1)

$$= 942\,000\,000{\text{ (km) (942}}\,{\text{477}}\,{\text{796.1}} \ldots {\text{, }}3 \times {10^8}\pi ,{\text{ }}9.42 \times {10^8})$$     (A1)(ft)     (C2)

Notes: Award (M1) for correct substitution into correct formula. Follow through from part (a).

Do not accept calculator notation $$9.42{\text{E}}8$$.

Do not accept use of $$\frac{{22}}{7}$$ or$$3.14$$ for $$\pi$$.[2 marks]

b.

$$17 \times 942\,000\,000$$     (M1)

$$= 1.60 \times {10^{10}}{\text{ (km) }}\left( {{\text{1.60221}} \ldots \times {{10}^{10}}{\text{, 1.6014}} \times {{10}^{10}},{\text{ 16}}\,{\text{022}}\,{\text{122}}\,{\text{530, }}(5.1 \times {{10}^9})\pi } \right)$$     (A1)(ft)     (C2)

Note: Follow through from part (b).[2 marks]

c.

## Question

Mandzur, a farmer, takes out a loan to buy a buffalo. He borrows 900 000 Cambodian riels (KHR) for 2 years. The nominal annual interest rate is 15%, compounded monthly.

Find the amount of the interest that Mandzur must pay. Give your answer correct to the nearest 100 KHR.

a.

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

b.

## Markscheme

$$FV = 900000{\left( {1 + \frac{{15}}{{12 \times 100}}} \right)^{24}}$$     (M1)(A1)

Note: Award (M1) for substitution in the compound interest formula (either $$FV$$ or interest), do not penalize if $$–PV$$ not seen.

Award (A1) for correct substitution.

OR

$${\text{N}} = 2$$

$${\text{I% = 15}}$$

$${\text{PV = 900}}\,{\text{000}}$$

$${\text{P/Y}} = 1$$

$${\text{C/Y}} = 12$$     (A1)(M1)

Note: Award (A1) for $${\text{C/Y}} = 12$$ seen, (M1) for other correct entries.

OR

$${\text{N}} = 24$$

$${\text{I% = 15}}$$

$${\text{PV = 900}}\,{\text{000}}$$

$${\text{P/Y}} = 12$$

$${\text{C/Y}} = 12$$     (A1)(M1)

Note: Award (A1) for $${\text{C/Y}} = 12$$ seen, (M1) for other correct entries.

$${\text{interest = 321615.945}}$$     (A1)

$$= 312\,600\;\;\;{\text{(KHR)}}$$     (A1)(ft)     (C4)

Notes: Award the final (A1) for the correct rounding of their unrounded answer.

If final amount is $$1 212 600$$ and working is shown award (M1)(A1)(A0)(A1)(ft).

Award (A2) for $$(FV = )\ 1212600$$ if no working is seen.

a.

$$3.126 \times {10^5}$$     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for their $$3.126$$ $$(3.13)$$, (A1)(ft) for $$\times {10^5}$$.

b.

## Question

Assume the Earth is a perfect sphere with radius 6371 km.

Calculate the volume of the Earth in $${\text{k}}{{\text{m}}^3}$$. Give your answer in the form $$a \times {10^k}$$, where $$1 \leqslant a < 10$$ and $$k \in \mathbb{Z}$$.

a.

The volume of the Moon is $$2.1958 \times {10^{10}}\;{\text{k}}{{\text{m}}^3}$$.

Calculate how many times greater in volume the Earth is compared to the Moon.

b.

## Markscheme

$$\frac{4}{3}\pi {(6371)^3}$$     (M1)

Note: Award (M1) for correct substitution into volume formula.

$$= 1.08 \times {10^{12}}\;\;\;(1.08320 \ldots \times {10^{12}})$$     (A2)     (C3)

Notes: Award (A1)(A0) for correct mantissa between 1 and 10, with incorrect index.

Award (A1)(A0) for $$1.08\rm{E}12$$

Award (A0)(A0) for answers of the type: $$108 \times {10^{10}}$$.

a.

$$\frac{{1.08320 \ldots \times {{10}^{12}}}}{{2.1958 \times {{10}^{10}}}}$$     (M1)

Note: Award (M1) for dividing their answer to part (a) by $$2.1958 \times {10^{10}}$$.

$$= 49.3308 \ldots$$     (A1)(ft)

Note: Accept $$49.1848…$$ from use of 3 sf answer to part (a).

$$= 49$$     (A1)     (C3)

Notes: Follow through from part (a).

The final (A1) is awarded for their unrounded non-integer answer seen and given correct to the nearest integer.

Do not award the final (A1) for a rounded answer of 0 or if it is incorrect by a large order of magnitude.

b.

## Question

The distance $$d$$ from a point $${\text{P}}(x,{\text{ }}y)$$ to the point $${\text{A}}(1,{\text{ }} – 2)$$ is given by $$d = \sqrt {{{(x – 1)}^2} + {{(y + 2)}^2}}$$

Find the distance from $${\text{P}}(100,{\text{ }}200)$$ to $${\text{A}}$$. Give your answer correct to two decimal places.

a.

Write down your answer to part (a) correct to three significant figures.

b.

Write down your answer to part (b) in the form $$a \times {10^k}$$, where $$1 \leqslant a < 10$$ and $$k \in \mathbb{Z}$$.

c.

## Markscheme

$$\sqrt {{{(100 – 1)}^2} + {{(200 + 2)}^2}}$$     (M1)

$$\sqrt {50605} \;\;\;( = 224.955 \ldots )$$     (A1)

Note: Award (M1)(A1) if $$\sqrt {50605}$$ seen.

$${\text{224.96}}$$     (A1)     (C3)

Note: Award (A1) for their answer given correct to 2 decimal places.

a.

$$225$$     (A1)(ft)     (C1)

Note: Follow through from their part (a).

b.

$$2.25 \times {10^2}$$     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(A0) for $$2.25$$ and an incorrect index value.

Award (A0)(A0) for answers such as $$22.5 \times {10^1}$$.

c.

## Question

Assume that the Earth is a sphere with a radius, $$r$$ , of $$6.38 \times {10^3}\,{\text{km}}$$ . i)     Calculate the surface area of the Earth in $${\text{k}}{{\text{m}}^2}$$.

ii)    Write down your answer to part (a)(i) in the form $$a \times {10^k}$$ , where $$1 \leqslant a < 10$$ and $$k \in \mathbb{Z}$$ .

a.

The surface area of the Earth that is covered by water is approximately $$3.61 \times {10^8}{\text{k}}{{\text{m}}^2}$$ .

Calculate the percentage of the surface area of the Earth that is covered by water.

b.

## Markscheme

i)     $$4\pi {(6.38 \times {10^3})^2}$$       (M1)

Note: Award (M1) for correct substitution into the surface area of a sphere formula.

$$= 512\,000\,000\,\,\,(511506576,\,\,162\,817\,600\pi )$$       (A1)    (C2)

Note: Award at most (M1)(A0) for use of $$3.14$$ for $$\pi$$, which will give an answer of $$511\,247\,264$$.

ii)    $$5.12 \times {10^8}\,\,\,(5.11506… \times {10^8},\,\,1.628176\pi \times {10^8})$$       (A1)(ft)(A1)(ft)    (C2)

Note: Award (A1) for $$5.12$$ and (A1) for $$\times {10^8}$$.
Award (A0)(A0) for answers of the type: $$5.12 \times {10^7}$$.

a.

$$\frac{{3.61 \times {{10}^8}}}{{5.11506…\,\, \times {{10}^8}}} \times 100$$  OR $$\frac{{3.61}}{{5.11506…\,}} \times 100$$  OR $$0.705758… \times 100$$        (M1)

Note: Award (M1) for correct substitution. Multiplication by $$100$$ must be seen.

$$= 70.6\,(\% )\,\,\,\,(70.5758…\,(\% ))$$        (A1)(ft)   (C2)

Note: Follow through from part (a). Accept the use of $$3$$ sf answers, which gives a final answer of $$70.5\,(\% )\,\,\,\,(70.5758…\,(\% ))$$ .

b.

## Question

Let $$p = \frac{{\cos x + \sin y}}{{\sqrt {{w^2} – z} }}$$,

where $$x = 36^\circ ,{\text{ }}y = 18^\circ ,{\text{ }}w = 29$$ and $$z = 21.8$$.

Calculate the value of $$p$$. Write down your full calculator display.

a.

(i)     correct to two decimal places;

(ii)     correct to three significant figures.

b.

Write your answer to part (b)(ii) in the form $$a \times {10^k}$$, where $$1 \leqslant a < 10,{\text{ }}k \in \mathbb{Z}$$.

c.

## Markscheme

$$\frac{{\cos 36^\circ + \sin 18^\circ }}{{\sqrt {{{29}^2} – 21.8} }}$$    (M1)

Note:     Award (M1) for correct substitution into formula.

$$= 0.0390625$$    (A1)     (C2)

Note:     Accept $$\frac{5}{{128}}$$.[2 marks]

a.

(i)    0.04     (A1)(ft)

(ii)     0.0391     (A1)(ft)     (C2)

Note:     Follow through from part (a).[2 marks]

b.

$$3.91 \times {10^{ – 2}}$$    (A1)(ft)(A1)(ft)     (C2)

Note:     Answer should be consistent with their answer to part (b)(ii). Award (A1)(ft) for 3.91, and (A1)(ft) for $${10^{ – 2}}$$. Follow through from part (b)(ii).[2 marks]

c.

## Question

In the Canadian city of Ottawa:

$\begin{array}{*{20}{l}} {{\text{97% of the population speak English,}}} \\ {{\text{38% of the population speak French,}}} \\ {{\text{36% of the population speak both English and French.}}} \end{array}$

The total population of Ottawa is $$985\,000$$.

Calculate the percentage of the population of Ottawa that speak English but not French.

a.

Calculate the number of people in Ottawa that speak both English and French.

b.

Write down your answer to part (b) in the form $$a \times {10^k}$$ where $$1 \leqslant a < 10$$ and k $$\in \mathbb{Z}$$.

c.

## Markscheme

$$97 – 36$$     (M1)

Note:     Award (M1) for subtracting 36 from 97.

OR (M1)

Note:     Award (M1) for 61 and 36 seen in the correct places in the Venn diagram.

$$= 61{\text{ }}(\% )$$     (A1)     (C2)

Note:     Accept 61.0 (%).[2 marks]

a.

$$\frac{{36}}{{100}} \times 985\,000$$     (M1)

Note:     Award (M1) for multiplying 0.36 (or equivalent) by $$985\,000$$.

$$= 355\,000{\text{ }}(354\,600)$$     (A1)     (C2)[2 marks]

b.

$$3.55 \times {10^5}{\text{ }}(3.546 \times {10^5})$$     (A1)(ft)(A1)(ft)     (C2)

Note:     Award (A1)(ft) for 3.55 (3.546) must match part (b), and (A1)(ft) $$\times {10^5}$$.

Award (A0)(A0) for answers of the type: $$35.5 \times {10^4}$$. Follow through from part (b).[2 marks]

c.

## Question

Consider the numbers $$p = 2.78 \times {10^{11}}$$ and $$q = 3.12 \times {10^{ – 3}}$$.

Calculate $$\sqrt{{\frac{p}{q}}}$$. Give your full calculator display.

a.

Write down your answer to part (a) correct to two decimal places;


b.i.

Write down your answer to part (a) correct to three significant figures.

b.ii.

Write your answer to part (b)(ii) in the form $$a \times {10^k}$$, where $$1 \leqslant a < 10,{\text{ }}k \in \mathbb{Z}$$.

c.

## Markscheme

$$\sqrt{{\frac{{2.78 \times {{10}^{11}}}}{{3.12 \times {{10}^{ – 3}}}}}}$$$$\,\,\,$$OR$$\,\,\,$$$$\sqrt{{8.91025 \ldots \times {{10}^{13}}}}$$     (M1)

Note:     Award (M1) for correct substitution into given expression.

44664.59503     (A1)     (C2)

Note:     Award (A1) for a correct answer with at least 8 digits.

Accept 44664.5950301.[2 marks]

a.

44664.60     (A1)(ft)     (C1)

Note:     For a follow through mark, the answer to part (a) must be to at least 3 decimal places.[1 mark]

b.i.

44700     (A1)(ft)     (C1)

Notes:     Answer to part (a) must be to at least 4 significant figures.

Accept any equivalent notation which is correct to 3 significant figures.

For example $$447 \times {10^2}$$ or $$44.7 \times {10^3}$$.

Follow through from part (a).[1 mark]

b.ii.

$$4.47 \times {10^4}$$     (A1)(ft)(A1)(ft)     (C2)

Notes:     Award (A1)(ft) for 4.47 and (A1)(ft) for $${10^4}$$.

Award (A0)(A0) for answers such as $$44.7 \times {10^3}$$.

Follow through from part (b)(ii) only.[2 marks]

c.

## Question

The speed of light is $${\text{300}}\,{\text{000}}$$ kilometres per second. The average distance from the Sun to the Earth is 149.6 million km.

A light-year is the distance light travels in one year and is equal to $${\text{9}}\,{\text{467}}\,{\text{280}}$$ million km. Polaris is a bright star, visible from the Northern Hemisphere. The distance from the Earth to Polaris is 323 light-years.

Calculate the time, in minutes, it takes for light from the Sun to reach the Earth.

a.

Find the distance from the Earth to Polaris in millions of km. Give your answer in the form $$a \times {10^k}$$ with $$1 \leqslant a < 10$$ and $$k \in \mathbb{Z}$$.

b.

## Markscheme

$$\frac{{149600000}}{{300000 \times 60}}$$     (M1)(M1)

Note:     Award (M1) for dividing the correct numerator (which can be presented in a different form such as $$149.6 \times {10^6}$$ or $$1.496 \times {10^8}$$) by $${\text{300}}\,{\text{000}}$$ and (M1) for dividing by 60.

$$= 8.31{\text{ }}({\text{minutes}}){\text{ }}(8.31111 \ldots {\text{, 8 minutes 19 seconds}})$$     (A1)     (C3)[3 marks]

a.

$$323 \times 9467\,280$$     (M1)

Note:     Award (M1) for multiplying 323 by $$9\,467\,280$$, seen with any power of 10; therefore only penalizing incorrect power of 10 once.

$$= 3.06 \times {10^9}{\text{ ( = }}3.05793 \ldots \times {10^9})$$     (A1)(A1)     (C3)

Note:     Award (A1) for 3.06.

Award (A1) for $$\times {10^9}$$

Award (A0)(A0) for answers of the type: $$30.6 \times {10^8}$$[3 marks]

b.

## Question

Each year the soccer team, Peterson United, plays 25 games at their home stadium. The owner of Peterson United claimed that last year the mean attendance per game at their home stadium was 24500.

The actual total attendance last year was 617700.

Based on the owner’s claim, calculate the total attendance for the games at Peterson United’s home stadium last year.

a.

Calculate the percentage error in the owner’s claim.

b.

Write down your answer to part (b) in the form a × 10k where 1 ≤ a < 10, $$k \in \mathbb{Z}$$.

c.

## Markscheme

24500 × 25   (M1)

Note: Award (M1) for multiplying 24500 by 25.

= 613000  (612500)   (A1) (C2)[2 marks]

a.

$$\left| {\frac{{612500 – 617700}}{{617700}}} \right| \times 100$$    (M1)

Note: Award (M1) for correct substitution into the percentage error formula.

= 0.842  (0.841832)    (A1)(ft)  (C2)

Note: Follow through from part (a).[2 marks]

b.

8.42 × 10−1    (A1)(ft)(A1)(ft) (C2)

Note: Award (A0)(A0) for answers of the type 84.2 × 10−2. Follow through from part (b). Ignore ‘%’ sign.[2 marks]

c.

## Question

Last year a South American candy factory sold 4.8 × 108 spherical sweets. Each sweet has a diameter of 2.5 cm.

The factory is producing an advertisement showing all of these sweets placed in a straight line. The advertisement claims that the length of this line is x times the length of the Amazon River. The length of the Amazon River is 6400 km.

Find the length, in cm, of this line. Give your answer in the form a × 10k , where 1 ≤ a < 10 and k ∈ $$\mathbb{Z}$$.

a.

Write down the length of the Amazon River in cm.

b.i.

Find the value of x.

b.ii.

## Markscheme

4.8 × 108 × 2.5     (M1)

Note: Award (M1) for multiplying by 2.5.

1.2 × 109 (cm)     (A1)(ft)(A1)(ft) (C3)

Note: Award (A0)(A0) for answers of the type 12 × 108.[3 marks]

a.

640 000 000 (cm)  (6.4 × 108 (cm))     (A1)[1 mark]

b.i.

$$\frac{{1.2 \times {{10}^9}}}{{6.4 \times {{10}^8}}}$$     (M1)

Note: Award (M1) for division by 640 000 000.

= 1.88 (1.875)     (A1)(ft) (C3)

Note: Follow through from part (a) and part (b)(i).[2 marks]

b.ii.