# IBDP Maths AI: Topic : SL 1.6: Approximation: decimal places, significant figures: IB style Questions SL Paper 1

## 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.

[3]
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, $$k \in {\mathbb{Z}}$$ .

[3]
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

1 Brazilian Real (BRL) = 2.607 South African Rand (ZAR). Giving answers correct to two decimal places,

(i) convert 300 BRL to ZAR,

(ii) find how many Real it costs to purchase 300 Rand.

## Markscheme

Financial accuracy penalty (FP) is applicable where indicated in the left hand column.

1 BRL = 2.607 ZAR

(FP) (i) $$300 \times 2.607 = 782.10 {\text{ ZAR}}$$     (A1)

Note: 782.1 is (A0)(FP)

(FP) (ii) $$300 \times \frac{1}{{2.607}} = 115.07{\text{ BRL}}$$     (A1)(ft)

Note: Follow through only if processes are reversed.     (C2)

[2 marks]

[3 marks]

## 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.

## Markscheme

$$29.7675$$     (A1)     (C1)

[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

The table below shows some exchange rates for the Japanese Yen ($${\text{JPY}}$$).

Minbin has $$1250$$ Japanese Yen which she wishes to exchange for Chinese Yuan.

[2]
a.

Rupert has $$855$$ Canadian Dollars which he wishes to exchange for Japanese Yen.

[2]
b.

Find how many Norwegian Kroner there are to the Euro. Give your answer correct to 2 decimal places.

[2]
c.

## Markscheme

Financial accuracy penalty (FP) is applicable where indicated in the left hand column.

Multiplying $$1250$$ by $$0.07127$$ or $$0.7127$$     (M1)

(FP)     $$89$$     (A1)     (C2)

[2 marks]

a.

Financial accuracy penalty (FP) is applicable where indicated in the left hand column.

Dividing by $$0.010406$$ or $$0.10406$$     (M1)

(FP)     $$82164$$     (A1)     (C2)

Note: If candidate has divided in (a) and multiplied in (b) award (M1)(A1)(ft) for $$9$$ in (b).

[2 marks]

b.

Financial accuracy penalty (FP) is applicable where indicated in the left hand column.

(FP)     $$\frac{{0.057319}}{{0.0072591}}$$ allow $$0.57319$$ and/or $$0.072591$$     (M1)

$$7.90$$      (A1)     (C2)

Note: The (M1) is being allowed for misreading values from the table but do not (ft) to candidate’s answers.

[2 marks]

c.

## 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.

## 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

The exchange rate between Indian rupees (INR) and Singapore dollars (S\$) is $$100{\text{ INR}} = {\text{S\ }}3.684$$

Kwai Fan changes $${\text{S\ }}500$$ to Indian rupees.

Calculate the number of Indian rupees she will receive using this exchange rate. Give your answer correct to the nearest rupee.

[2]
a.

On her return to Singapore, Kwai Fan has $$2500$$ Indian rupees left from her trip. She wishes to exchange these rupees back to Singapore dollars. There is a $$3\%$$ commission charge for this transaction and the exchange rate is $$100{\text{ INR}} = {\text{S\}}3.672$$.

Calculate the commission in Indian rupees that she is charged for this exchange.

[2]
b.

On her return to Singapore, Kwai Fan has $$2500$$ Indian rupees left from her trip. She wishes to exchange these rupees back to Singapore dollars. There is a $$3\%$$ commission charge for this transaction and the exchange rate is $$100{\text{ INR}} = {\text{S\}}3.672$$.

Calculate the amount of money she receives in Singapore dollars, correct to two decimal places.

[2]
c.

## Markscheme

Financial penalty (FP) applies in this question.

$$500 \times \frac{{100}}{{3.684}}$$     (M1)

FP     $$= 13572$$     (A1)     (C2)

Note: (M1) for multiplication by $$\frac{{100}}{{3.684}}$$

[2 marks]

a.

$$2500 \times 0.03$$     (M1)

$$= 75{\text{ }}(75.0{\text{, }}75.00)$$     (A1)     (C2)

If $$2500 \times 0.03 \times \frac{{3.672}}{{100}}$$

$$= 2.75$$

Award (M1)(A0)

[2 marks]

b.

Financial penalty (FP) applies in this question.

$$2425 \times \frac{{3.672}}{{100}}$$     (M1)(ft)

FP     $$= 89.05$$     (A1)(ft)

OR

$$\frac{{3.672}}{{100}} \times 0.97 \times 2500$$     (M1)(ft)

FP     $$= 89.05$$     (A1)(ft)

OR

$$3\% {\text{ of }}91.8 = 2.754$$

$$91.8 – 2.754$$     (M1)(ft)

FP     $$= 89.05$$     (A1)(ft)     (C2)

Note: (ft) in (c) if the conversion process is reversed consistently through the question, i.e. multiplication in (a) followed by division in (c).

[2 marks]

c.

## 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 $$k \in \mathbb{Z}$$.

[3]
a.

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

[3]
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 exchange rates between the British pound (GBP) and the United States dollar (USD) and between the USD and the Euro (EUR) are given below.

Find the exchange rate between GBP and EUR in the form 1 GBP = k EUR, where k is a constant. Give your answer correct to two decimal places.

[2]
a.

Isabella changes 400 USD into Euros and is charged 2 % commission.

[4]
b.

## Markscheme

k = 2.034 × 0.632 (M1)

= 1.29 (1 GBP = 1.29 EUR) (A1) (C2)

Note: Accept 1.29 only

[2 marks]

a.

Financial penalty (FP) applies in part (b).

400 × 0.632 (M1)

= 252.80 EUR (A1)

2 % of 252.80 = 5.06 EUR (A1)

(FP) She receives 247.74 EUR (A1)

OR

(FP) 0.98 × 252.80 = 247.74 EUR (A1)(A1) (C4)

Note: Accept (A1) for 0.98 seen.

[4 marks]

b.

## Question

Mr Tan invested 5000 Swiss Francs (CHF) in Bank A at an annual simple interest rate of r %, for four years. The total interest he received was 568 CHF.

Mr Black invested 5000 CHF in Bank B at a nominal annual interest rate of 3.6 %, compounded quarterly for four years.

Calculate the total interest he received at the end of the four years. Give your answer correct to two decimal places.

## Markscheme

Financial penalty (FP) applies in part (b).

$$I = 5000(1.009)^{16} – 5000$$     (M1)(A1)Note: Award (M1) for substitution into the compound interest formula, (A1) for correct values.

(FP)     I = 770.70 CHF     (A1)     (C3)

[3 marks]

## 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.

[3]
a.

[1]
b.

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

[2]
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

Astrid invests 1200 Euros for five years at a nominal annual interest rate of 7.2 %, compounded monthly.

Find the interest Astrid has earned during the five years of her investment. Give your answer correct to two decimal places.

## Markscheme

$$I = 1200{\left( {1 + \frac{{7.2}}{{600}}} \right)^{5 \times 12}} – 1200$$     (M1)(A1)
I = 518.15 Euros     (A1)     (C3)

Notes: Award (M1) for substitution in the compound interest formula, (A1) for correct substitutions, (A1) for correct answer.

If final amount found is 1718.15 and working shown award (M1) (A1)(A0).

[3 marks]

## Question

A rumour spreads through a group of teenagers according to the exponential model

$$N = 2 \times {(1.81)^{0.7t}}$$

where N is the number of teenagers who have heard the rumour t hours after it is first started.

Find the number of teenagers who started the rumour.

[2]
a.

Write down the number of teenagers who have heard the rumour five hours after it is first started.

[1]
b.

Determine the length of time it would take for 150 teenagers to have heard the rumour. Give your answer correct to the nearest minute.

[3]
c.

## Markscheme

N = 2 × (1.81)0.7×0     (M1)
N = 2     (A1)     (C2)

Notes: Award (M1) for correct substitution of t = 0.

[2 marks]

a.

16.0 (3 s.f)     (A1)     (C1)

Note: Accept 16 and 15.

[1 mark]

b.

150 = 2 × (1.81)0.7t     (M1)

t = 10.39… h     (A1)

t = 624 minutes     (A1)(ft)     (C3)

Notes: Accept 10 hours 24 minutes. Accept alternative methods.

Award last (A1)(ft) for correct rounding to the nearest minute of their answer.

Unrounded answer must be seen so that the follow through can be awarded.

[3 marks]

c.

## Question

Given $$p = x – \frac{{\sqrt y }}{z}$$ , $$x = 1.775$$ , $$y = 1.44$$ and $$z = 48$$ .

Calculate the value of $$p$$.

[2]
a.

Barry first writes $$x$$ , $$y$$ and $$z$$ correct to one significant figure and then uses these values to estimate the value of $$p$$ .
(i)     Write down $$x$$ , $$y$$ and $$z$$ each correct to one significant figure.
(ii)    Write down Barry’s estimate of the value of $$p$$ .

[2]
b.

Calculate the percentage error in Barry’s estimate of the value of $$p$$ .

[2]
c.

## Markscheme

$$p = 1.775 – \frac{{\sqrt {1.44} }}{{48}}$$     (M1)

Note: Award (M1) for correctly substituted equation for $$p$$.

$$= 1.75$$ $$\left( {1.750{\text{, }}\frac{7}{4}} \right)$$     (A1)(C2)

[2 marks]

a.

(i)     $$x = 2$$, $$y =1$$, $$z = 50$$     (A1)

(ii)    $$p =1.98$$ $$\left( {\frac{{99}}{{50}}} \right)$$     (A1)(ft)     (C2)

Note: Follow through from part (b)(i), irrespective of whether working is shown.

Note: If 2 s.f. used throughout part (b)(i) award (A1)(ft) for $$1.78$$ or $$1.8$$.

[2 marks]

b.

$$\frac{{1.98 – 1.75}}{{1.75}} \times 100$$     (M1)

Note: Award (M1) for correctly substituted $$\%$$ error formula.

Note: Follow through from parts (a) and (b).

$$= 13.1\%$$     (A1)(ft)     (C2)

Notes: $$\%$$ sign not required. Do not accept $$– 13.1\%$$. If 100 missing and incorrect answer, award (M0)(A0). If 100 missing and answer incorrectly rounded, award (M1)(A1).

[2 marks]

c.

## 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.

[1]
a.

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

[3]
b.

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

[2]
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

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

Find the exact value of $$z$$.

[2]
a.

[1]
b.i.

[1]
b.ii.

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

[2]
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.

[2]
a.

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

[2]
b.

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

[2]
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

$$z = \frac{{17{x^2}}}{{a – b}}$$.

Find the value of z when x = 12.5, a = 0.572 and b = 0.447. Write down your full calculator display.[2]

a.

(i) correct to the nearest 1000 ;

(ii) correct to three significant figures.[2]

b.

Write your answer to part (b)(ii) in the form a × 10k, where 1 ≤ a < 10, $$k \in \mathbb{Z}$$.[2]

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

Neung is going home to Vietnam after working in Singapore.

She has 5000 Singapore dollars (SGD) and changes these into American dollars (USD)
to take home.

The exchange rate between Singapore dollars (SGD) and American dollars (USD) is

1 USD = 1.2945 SGD.

There is also a 2.4 % commission on the exchange.

Calculate the amount of commission on the exchange in SGD.[2]

a.

Calculate the number of American dollars (USD) Neung takes home. Give your answer correct to 2 decimal places.[2]

b.

At the airport in Vietnam, Neung changes 150 USD into Vietnamese dong (VND) to pay for her transport home.

The exchange rate between American dollars (USD) and Vietnamese dong (VND) is

1 USD = 19 495 VND.

There is no commission.

Calculate the number of Vietnamese dong that Neung receives. Give your answer correct to the nearest thousand dong.[2]

c.

## Markscheme

5000 × 0.024     (M1)

Note: Award (M1) for multiplication by 0.024.

=120     (A1)     (C2)

a.

$$4880 \times \frac{1}{{1.2945}}$$     (M1)

Note: Award (M1) for multiplication by $$\frac{1}{{1.2945}}$$.

= 3769.80     (A1)(ft)     (C2)

Note: Correct answer to 2 dp only. Follow through from their part (a).

b.

$$150 \times 19495$$     (M1)

Note: Award (M1) for $$\times 19495$$.

$$= 2924000$$     (A1)     (C2)

Notes: Correct answer to nearest 1000 only. Do not penalize incorrect accuracy in (c) if this has already been penalized in part (b).

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.[2]

a.

(i) one decimal place;

(ii) three significant figures.[2]

b.

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

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

Yoshi is spending a year travelling from Japan to Italy and then to the United States of America.

Before Yoshi leaves Japan he changes 100 000 Japanese Yen (JPY) into euro (EUR). The exchange rate is 1 JPY = 0.006 EUR.

Calculate the amount Yoshi receives, in EUR.[2]

a.

Yoshi spends 426.70 EUR in Italy. In an American bank he changes the remaining amount, into US dollars (USD), at an exchange rate of 1 USD = 0.673 EUR.

The bank charges 1.5 % commission.

Calculate the amount, in USD, Yoshi receives after commission. Give your answer correct to the nearest USD.[4]

b.

## Markscheme

$$0.006 \times 100000$$     (M1)

Note: Award (M1) for multiplication by 0.006.

$$= 600$$     (A1)     (C2)

a.

$$\frac{{(600 – 426.70)}}{{0.673}} \times 0.985$$     (M1)(M1)(M1)

Note: Award (M1) for subtracting 426.70 from their answer to part (a), (M1) for division by 0.673, (M1) for multiplication by 0.985 (or equivalent).

OR

$$\frac{{173.30 – (600 – 426.70) \times 0.015}}{{0.673}}$$     (M1)(M1)(M1)

Note: Award (M1) for subtracting 426.70 from their answer to part (a), (M1) for division by 0.673, (M1) for multiplication by 0.015 (or equivalent) and subtraction from their 173.30.

254     (A1)(ft)     (C4)

Notes: Follow through from their part (a). In order to award the final (A1)(ft) the answer must be given correct to the nearest dollar. If division used in part (a) and multiplication in part (b) award at most (M1)(M1)(M1)(A0).

b.

## Question

Marcus has been given 500 Australian dollars (AUD) by his grandmother for his 18th birthday.

He plans to deposit it in a bank which offers a nominal annual interest rate of 6.0 %, compounded quarterly, for three years.

Calculate the total amount of interest Marcus would earn, in AUD, over the three years. Give your answer correct to two decimal places.[3]

a.

Marcus would earn the same amount of interest, compounded annually, for three years if he deposits the 500 AUD in a second bank.

Calculate the interest rate the second bank offers.[3]

b.

## Markscheme

$$500{\left( {1 + \frac{6}{{100 \times 4}}} \right)^{4 \times 3}} – 500$$     (M1)(A1)

Note: Award (M1) for substitution in correct formula (A1) for correct substitutions.

$$= 97.81$$     (A1)     (C3)

Note: The answer must be given to 2 dp or the final (A1) is not awarded.

a.

$$97.8090… = 500{\left( {1 + \frac{r}{{100}}} \right)^3} – 500$$     (M1)(A1)(ft)

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

$$= 6.14$$ (6.13635…)     (A1)(ft)     (C3)

b.

## 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.[2]

a.

(i)     correct to two decimal places;

(ii)     correct to four significant figures;

(iii)     in the form $$a \times {10^k}$$, where $$1 \leqslant a < 10,{\text{ }}k \in \mathbb{Z}$$.[4]

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

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.[4]

a.

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

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

$$T = \frac{{\left( {\tan (2z) + 1} \right)\left( {2\cos (z) – 1} \right)}}{{{y^2} – {x^2}}}$$, where $$x = 9$$, $$y = 41$$ and $$z = 30^\circ$$.

Calculate the exact value of $$T$$.[2]

a.

Give your answer to $$T$$ correct to

(i)     two significant figures;

(ii)     three decimal places.[2]

b.

Pyotr estimates the value of $$T$$ to be $$0.002$$.

Calculate the percentage error in Pyotr’s estimate.[2]

c.

## Markscheme

$$\frac{{\left( {\tan (2 \times 30) + 1} \right)\left( {2\cos (30) – 1} \right)}}{{{{41}^2} – {9^2}}}$$     (M1)

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

$$= 0.00125\;\;\;\left( {\frac{1}{{800}}} \right)$$     (A1)     (C2)

Note: Using radians the answer is $$– 0.000570502$$, award at most (M1)(A0).

a.

(i)     $$0.0013$$    (A1)(ft)

Note: Follow through from part (a).

(ii)     $$0.001$$     (A1)(ft)     (C2)

Note: Follow through from part (a).

b.

$$\left| {\frac{{0.002 – 0.00125}}{{0.00125}}} \right| \times 100$$     (M1)

Notes: Award (M1) for their correct substitution into the percentage error formula. Absolute value signs are not required.

Their unrounded answer from part (a) must be used.

Do not accept use of answers from part (b).

$$= 60{\text{ (%)}}$$     (A1)(ft)     (C2)

Notes: The $${\text{%}}$$ sign is not required.

The answer from radians is $$450.568{\text{%}}$$, award (M1)(A1)(ft).

c.

## 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 $$k \in \mathbb{Z}$$.[3]

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

Pierre invests 5000 euros in a fixed deposit that pays a nominal annual interest rate of 4.5%, compounded monthly, for seven years.

Calculate the value of Pierre’s investment at the end of this time. Give your answer correct to two decimal places.[3]

a.

Carla has 7000 dollars to invest in a fixed deposit which is compounded annually.

She aims to double her money after 10 years.

Calculate the minimum annual interest rate needed for Carla to achieve her aim.[3]

b.

## Markscheme

$$5000{\left( {1 + \frac{{4.5}}{{12 \times 100}}} \right)^{12 \times 7}}$$     (M1)(A1)

Note: Award (M1) for substitution into compound interest formula, (A1) for correct substitutions.

OR

$$N = 7$$

$$I\% = 4.5$$

$$PV = ( \pm )5000$$

$$P/Y = 1$$

$$C/Y = 12$$     (A1)(M1)

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

OR

$$N = 84$$

$$I\% = 4.5$$

$$PV = ( \pm )5000$$

$$P/Y = 12$$

$$C/Y = 12$$     (A1)(M1)

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

$$= 6847.26{\text{ (euros)}}$$     (A1)     (C3)

Note: Answer must be correct to 2 decimal places for the final (A1) to be awarded.

a.

$$14000 = 7000{\left( {1 + \frac{r}{{100}}} \right)^{10}}$$     (M1)(A1)

Notes: Award (M1) for substitution into compound interest formula equated to 14000 or equivalent.

Award (A1) for correct substitutions.

OR

$$N = 10$$

$$PV = \pm 7000$$

$$FV \mp 14000$$

$$P/Y = 1$$

$$C/Y = 1$$     (A1)(M1)

Note: Award (A1) for $$C/Y = 1$$ seen, (M1) for other correct entries. $$PV$$ and $$FV$$ must have opposite signs.

$$r = 7.18\% \;\;\;(7.17734 \ldots \% ,{\text{ }}0.0718)$$     (A1)     (C3)

Note: Do not penalize if $$\%$$ sign is missing. Do not accept $$0.0718\%$$.

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.[3]

a.

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

b.

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

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

Minta deposits 1000 euros in a bank account. The bank pays a nominal annual interest rate of 5%, compounded quarterly.

Find the amount of money that Minta will have in the bank after 3 years. Give your answer correct to two decimal places.[3]

a.

Minta will withdraw the money from her bank account when the interest earned is 300 euros.

Find the time, in years, until Minta withdraws the money from her bank account.[3]

b.

## Markscheme

$$1000{\left( {1 + \frac{5}{{4 \times 100}}} \right)^{4 \times 3}}$$     (M1)(A1)

Note: Award (M1) for substitution into compound interest formula, (A1) for correct substitution.

OR

$${\text{N}} = 3$$

$${\text{I}}\% = 5$$

$${\text{PV}} = – 1000$$

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

$${\text{C/Y}} = 4$$     (A1)(M1)

Note: Award (A1) for $${\text{C/Y}} = 4$$ seen, (M1) for other correct entries.

OR

$${\text{N}} = 12$$

$${\text{I}}\% = 5$$

$${\text{PV}} = – 1000$$

$${\text{P/Y}} = 4$$

$${\text{C/Y}} = 4$$     (A1)(M1)

Note: Award (A1) for $${\text{C/Y}} = 4$$ seen, (M1) for other correct entries.

$$= 1160.75$$ (€)    (A1)     (C3)

a.

$$1000{\left( {1 + \frac{5}{{4 \times 100}}} \right)^{4 \times t}} = 1300$$     (M1)(A1)

Note: Award (M1) for using the compound interest formula with a variable for time, (A1) for substituting correct values and equating to $$1300$$.

OR

$${\text{I}}\% = 5$$

$${\text{PV}} = \pm 1000$$

$${\text{FV}} = \mp 1300$$

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

$${\text{C/Y}} = 4$$     (A1)(M1)

Note: Award (A1) for 1300 seen, (M1) for the other correct entries.

OR

$${\text{I}}\% = 5$$

$${\text{PV}} = \pm 1000$$

$${\text{FV}} = \mp 1300$$

$${\text{P/Y}} = 4$$

$${\text{C/Y}} = 4$$     (A1)(M1)

Note: Award (A1) for 1300 seen, (M1) for the other correct entries.

OR

Sketch drawn of two appropriate lines which intersect at a point

Note: Award (M1) for a sketch with a straight line intercepted by appropriate curve, (A1) for a numerical answer in the range $$5.2-5.6$$.

$$t = 5.28{\text{ (years)}}\;\;\;(5.28001 \ldots )$$     (A1) (C3)

b.

## Question

Yun Bin invests $$5000{\text{ euros}}$$ in an account which pays a nominal annual interest rate of $$6.25\%$$ , compounded monthly.

Give all answers correct to two decimal places.

Find the value of the investment after 3 years.[3]

a.

Find the difference in the final value of the investment if the interest was compounded quarterly at the same nominal rate.[3]

b.

## Markscheme

$$FV = 5000{\left( {1 + \frac{{6.25}}{{1200}}} \right)^{3 \times 12}}$$   (M1)(A1)

Note: Award (M1) for substituted compound interest formula, (A1) for correct substitutions.

OR

$$N = 3$$
$$I\% = 6.25$$
$$PV = – 5000$$
$$P/Y = 1$$
$$C/Y = 12$$     (M1)(A1)

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

OR

$$N = 36$$
$$I\% = 6.25$$
$$PV = – 5000$$
$$P/Y = 12$$
$$C/Y = 12$$     (M1)(A1)

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

$$= 6028.22$$     (A1)     (C3)

Note: The answer should be given correct to two decimal places or the final (A1) is not awarded.

a.

$$FV = 5000{\left( {1 + \frac{{6.25}}{{400}}} \right)^{3 \times 4}}$$     (M1)

Note: Award (M1) for correctly substituted compound interest formula.

OR

$$N = 3$$
$$I\% = 6.25$$
$$PV = – 5000$$
$$P/Y = 1$$
$$C/Y = 4$$     (M1)

Note: Award (M1) for all correct entries seen.

OR

$$N = 12$$
$$I\% = 6.25$$
$$PV = – 5000$$
$$P/Y = 4$$
$$C/Y = 4$$     (M1)

Note: Award (M1) for all correct entries seen.

$$FV = 6022.41$$     (A1)
$${\text{Difference}} = 5.80$$     (A1)(ft)     (C3)

Notes: Accept $$5.81$$. This answer should be given correct to two decimal places or the final (A1) is not awarded unless this has already been penalized in part (a). Follow through from part (a).
Notes: Illustrating use of GDC notation acceptable in this case only. However on P2 an answer given with no working would receive G2.

b.

## Question

Calculate $$\frac{{77.2 \times {3^3}}}{{3.60 \times {2^2}}}$$.[1]

a.

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

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.[3]

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

The diagram shows triangle ABC in which angle BAC $$= 30^\circ$$, BC $$= 6.7$$ cm and AC $$= 13.4$$ cm.

Calculate the size of angle ACB.[4]

a.

Nadia makes an accurate drawing of triangle ABC. She measures angle BAC and finds it to be 29°.

Calculate the percentage error in Nadia’s measurement of angle BAC.[2]

b.

## Markscheme

$$\frac{{\sin {\text{A}}{\operatorname{\hat B}}{\text{C}}}}{{13.4}} = \frac{{\sin 30^\circ }}{{6.7}}$$     (M1)(A1)

Note: Award (M1) for correct substituted formula, (A1) for correct substitution.

$${\text{A}}{\operatorname{\hat B}}{\text{C}}$$ = 90°     (A1)

$${\text{A}}{\operatorname{\hat C}}{\text{B}}$$ = 60°     (A1)(ft)     (C4)

Note: Radians give no solution, award maximum (M1)(A1)(A0).

[4 marks]

a.

$$\frac{{29 – 30}}{{30}} \times 100$$     (M1)

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

% error = −33.3 %     (A1)     (C2)

Notes: Percentage symbol not required. Accept positive answer.

[2 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 ∈ $$\mathbb{Z}$$.[3]

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.[3]

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

A shipping container is a cuboid with dimensions $${\text{16 m}}$$, $${\text{1}}\frac{{\text{3}}}{{\text{4}}}{\text{ m}}$$ and $${\text{2}}\frac{{\text{2}}}{{\text{3}}}{\text{ m}}$$.

Calculate the exact volume of the container. Give your answer as a fraction.[3]

a.

Jim estimates the dimensions of the container as 15 m, 2 m and 3 m and uses these to estimate the volume of the container.

Calculate the percentage error in Jim’s estimated volume of the container.[3]

b.

## Markscheme

$$V = 16 \times 1\frac{3}{4} \times 2\frac{2}{3}$$     (M1)

Note: Award (M1) for correct substitution in volume formula. Accept decimal substitution of $$2.66$$ or better.

$$= 74.6666{\text{ }} \ldots$$     (A1)
$$= {\text{74}}\frac{{\text{2}}}{{\text{3}}}{\text{ }}{{\text{m}}^{\text{3}}}{\text{ }}\left( {\frac{{{\text{224}}}}{{\text{3}}}{\text{ }}{{\text{m}}^{\text{3}}}} \right)$$     (A1)    (C3)

[3 marks]

a.

$${\text{% error}} = \frac{{\left( {90 – 74\frac{2}{3}} \right) \times 100}}{{74\frac{2}{3}}}$$     (A1)(M1)

Note: Award (A1) for $$90$$ seen, or inferred in numerator, (M1) for correct substitution into percentage error formula.

$$= 20.5$$     (A1)(ft)     (C3)

Note: Accept $$– 20.5$$.

[3 marks]

b.

## Question

José stands 1.38 kilometres from a vertical cliff.

Express this distance in metres.[1]

a.

José estimates the angle between the horizontal and the top of the cliff as 28.3° and uses it to find the height of the cliff.

Find the height of the cliff according to José’s calculation. Express your answer in metres, to the nearest whole metre.[3]

b.

José estimates the angle between the horizontal and the top of the cliff as 28.3° and uses it to find the height of the cliff.

The actual height of the cliff is 718 metres. Calculate the percentage error made by José when calculating the height of the cliff.[2]

c.

## Markscheme

1380 (m)     (A1)     (C1)[1 mark]

a.

$$1380\tan 28.3$$     (M1)

$$= 743.05 \ldots$$ .     (A1)(ft)

$$= 743$$ (m)     (A1)(ft)     (C3)

Notes: Award (M1) for correct substitution in tan formula or equivalent, (A1)(ft) for their 743.05 seen, (A1)(ft) for their answer correct to the nearest m.[3 marks]

b.

$${\text{percentage error}} = \frac{{743.05 \ldots – 718}}{{718}} \times 100$$     (M1)

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

= 3.49 % (% symbol not required)     (A1)(ft)     (C2)

Notes: Accept 3.48 % for use of 743.

c.

## Question

In a television show there is a transparent box completely filled with identical cubes. Participants have to estimate the number of cubes in the box. The box is 50 cm wide, 100 cm long and 40 cm tall.

Find the volume of the box.[2]

a.

Joaquin estimates the volume of one cube to be 500 cm3. He uses this value to estimate the number of cubes in the box.

Find Joaquin’s estimated number of cubes in the box.[2]

b.

The actual number of cubes in the box is 350.

Find the percentage error in Joaquin’s estimated number of cubes in the box.[2]

c.

## Markscheme

$$50 \times 100 \times 40 = 200\,000{\text{ c}}{{\text{m}}^3}$$     (M1)(A1)     (C2)

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

[2 marks]

a.

$$\frac{{200\,000}}{{500}} = 400$$     (M1)(A1)(ft)     (C2)

Note: Award (M1) for dividing their answer to part (a) by 500.

[2 marks]

b.

$$\frac{{400 – 350}}{{350}} \times 100 = 14.3{\text{ }}\%$$     (M1)(A1)(ft)     (C2)

Notes: Award (M1) for correct substitution in the percentage error formula.

Accept –14.3 %.

% sign not necessary.

[2 marks]

c.

## Question

Given $$p = x – \frac{{\sqrt y }}{z}$$ , $$x = 1.775$$ , $$y = 1.44$$ and $$z = 48$$ .

Calculate the value of $$p$$.[2]

a.

Barry first writes $$x$$ , $$y$$ and $$z$$ correct to one significant figure and then uses these values to estimate the value of $$p$$ .
(i)     Write down $$x$$ , $$y$$ and $$z$$ each correct to one significant figure.
(ii)    Write down Barry’s estimate of the value of $$p$$ .[2]

b.

Calculate the percentage error in Barry’s estimate of the value of $$p$$ .[2]

c.

## Markscheme

$$p = 1.775 – \frac{{\sqrt {1.44} }}{{48}}$$     (M1)

Note: Award (M1) for correctly substituted equation for $$p$$.

$$= 1.75$$ $$\left( {1.750{\text{, }}\frac{7}{4}} \right)$$     (A1)(C2)

[2 marks]

a.

(i)     $$x = 2$$, $$y =1$$, $$z = 50$$     (A1)

(ii)    $$p =1.98$$ $$\left( {\frac{{99}}{{50}}} \right)$$     (A1)(ft)     (C2)

Note: Follow through from part (b)(i), irrespective of whether working is shown.

Note: If 2 s.f. used throughout part (b)(i) award (A1)(ft) for $$1.78$$ or $$1.8$$.

[2 marks]

b.

$$\frac{{1.98 – 1.75}}{{1.75}} \times 100$$     (M1)

Note: Award (M1) for correctly substituted $$\%$$ error formula.

Note: Follow through from parts (a) and (b).

$$= 13.1\%$$     (A1)(ft)     (C2)

Notes: $$\%$$ sign not required. Do not accept $$– 13.1\%$$. If 100 missing and incorrect answer, award (M0)(A0). If 100 missing and answer incorrectly rounded, award (M1)(A1).

[2 marks]

c.

## Question

$$80$$ matches were played in a football tournament. The following table shows the number of goals scored in all matches.

Find the mean number of goals scored per match.[2]

a.

Find the median number of goals scored per match.[2]

b.

A local newspaper claims that the mean number of goals scored per match is two. Calculate the percentage error in the local newspaper’s claim.[2]

c.

## Markscheme

$$\frac{{0 \times 16 + 1 \times 22 + 2 \times 19 \ldots }}{{80}}$$     (M1)

Note: Award (M1) for substituting correct values into mean formula.

1.75     (A1)     (C2)[2 marks]

a.

An attempt to enumerate the number of goals scored.     (M1)

$$2$$     (A1)     (C2)[2 marks]

b.

$$\frac{{2 – 1.75}}{{1.75}} \times 100$$     (M1)
$$14.3 \%$$     (A1)(ft)     (C2)

Notes: Award (M1) for correctly substituted $$\%$$ error formula. $$\%$$ sign not required. Follow through from their answer to part (a). If $$100$$ is missing and answer incorrect award (M0)(A0). If $$100$$ is missing and answer incorrectly rounded award (M1)(A1)(ft)(AP).[2 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.[2]

a.

Write your answer to part (a) in the form a × 10k where 1 ≤ a < 10 and $$k \in \mathbb{Z}$$.[2]

b.

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

Calculate the percentage error made by Marian.[2]

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

Ludmila takes a loan of 320 000 Brazilian Real (BRL) from a bank for two years at a nominal annual interest rate of 10%, compounded half yearly.

Write down the number of times interest is added to the loan in the two years.[1]

a.

Calculate the exact amount of money that Ludmila must repay at the end of the two years.[3]

b.

Ludmila estimates that she will have to repay $${\text{360}}\,{\text{000}}$$ BRL at the end of the two years.

Calculate the percentage error in her estimate.[2]

c.

## Markscheme

4     (A1)     (C1)[1 mark]

a.

$$320\,000{\left( {1 + \frac{{10}}{{2 \times 100}}} \right)^{2 \times 2}}$$     (M1)(A1)

Note: Award (M1) for substituted compound interest formula, (A1) for correct substitutions.

OR

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

$${\text{I}}\% = 10$$

$${\text{PV}} = – 320000$$

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

$${\text{C }}/{\text{ Y}} = 2$$     (A1)(M1)

Note: Award (A1) for $${\text{C }}/{\text{ Y}} = 2$$ seen, (M1) for correctly substituted values from the question into the finance application.

OR

$${\text{N}} = 4$$

$${\text{I}}\% = 10$$

$${\text{PV}} = – 320000$$

$${\text{P }}/{\text{ Y}} = 2$$

$${\text{C }}/{\text{ Y}} = 2$$     (A1)(M1)

Note: Award (A1) for $${\text{C }}/{\text{ Y}} = 2$$ seen, (M1) for correctly substituted values from the question into the finance application.

amount to repay $$= 388962$$     (A1)     (C3)

Note: Award (C2) for final answer $$389000$$ if $$388962$$ not seen previously.[3 marks]

b.

$$\left| {\frac{{360\,000 – 388\,962}}{{388\,962}}} \right| \times 100$$     (M1)

Note: Award (M1) for correctly substituted percentage error formula.

$$= 7.45{\text{ (% ) }}(7.44597 \ldots )$$     (A1)(ft)     (C2)

c.

## Question

$$T = \frac{{\left( {\tan (2z) + 1} \right)\left( {2\cos (z) – 1} \right)}}{{{y^2} – {x^2}}}$$, where $$x = 9$$, $$y = 41$$ and $$z = 30^\circ$$.

Calculate the exact value of $$T$$.[2]

a.

Give your answer to $$T$$ correct to

(i)     two significant figures;

(ii)     three decimal places.[2]

b.

Pyotr estimates the value of $$T$$ to be $$0.002$$.

Calculate the percentage error in Pyotr’s estimate.[2]

c.

## Markscheme

$$\frac{{\left( {\tan (2 \times 30) + 1} \right)\left( {2\cos (30) – 1} \right)}}{{{{41}^2} – {9^2}}}$$     (M1)

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

$$= 0.00125\;\;\;\left( {\frac{1}{{800}}} \right)$$     (A1)     (C2)

Note: Using radians the answer is $$– 0.000570502$$, award at most (M1)(A0).

a.

(i)     $$0.0013$$    (A1)(ft)

Note: Follow through from part (a).

(ii)     $$0.001$$     (A1)(ft)     (C2)

Note: Follow through from part (a).

b.

$$\left| {\frac{{0.002 – 0.00125}}{{0.00125}}} \right| \times 100$$     (M1)

Notes: Award (M1) for their correct substitution into the percentage error formula. Absolute value signs are not required.

Their unrounded answer from part (a) must be used.

Do not accept use of answers from part (b).

$$= 60{\text{ (%)}}$$     (A1)(ft)     (C2)

Notes: The $${\text{%}}$$ sign is not required.

The answer from radians is $$450.568{\text{%}}$$, award (M1)(A1)(ft).

c.

## Question

Fabián stands on top of a building, T, which is on a horizontal street.

He observes a car, C, on the street, at an angle of depression of 30°. The base of the building is at B. The height of the building is 80 metres.

The following diagram indicates the positions of T, B and C.

Show, in the appropriate place on the diagram, the values of

(i)     the height of the building;

(ii)     the angle of depression.[2]

a.

Find the distance, BC, from the base of the building to the car.[2]

b.

Fabián estimates that the distance from the base of the building to the car is 150 metres. Calculate the percentage error of Fabián’s estimate.[2]

c.

## Markscheme

(A1)(A1)     (C2)

Notes: Award (A1) for 80 m in the correct position on diagram.

Award (A1) for 30° in a correct position on diagram.

a.

$$\tan 30^\circ = \frac{{80}}{{{\text{BC}}}}\;\;\;$$OR$$\;\;\;\tan 60^\circ = \frac{{{\text{BC}}}}{{80}}\;\;\;$$OR$$\;\;\;\frac{{80}}{{\sin 30^\circ }} = \frac{{{\text{BC}}}}{{\sin 60^\circ }}$$     (M1)

Note: Award (M1) for a correct trigonometric or Pythagorean equation for BC with correctly substituted values.

$$({\text{BC}} = ){\text{ 139 (m)}}\;\;\;\left( {138.564 \ldots {\text{ (m)}}} \right)$$     (A1)(ft)     (C2)

Notes: Accept an answer of $$80\sqrt 3$$ which is the exact answer.

b.

$$\left| {\frac{{150 – 138.564 \ldots }}{{138.564 \ldots }}} \right| \times 100$$     (M1)

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

$$= 8.25(\% )\;\;\;(8.25317 \ldots \% )$$     (A1)(ft)     (C2)

Notes: Accept $$7.91(\%)$$ ($$7.91366…$$ if $$139$$ is used.

Accept $$8.23(\%)$$ ($$8.22510…$$ if $$138.6$$ is used.

If answer to part (b) is $$46.2$$, answer to part (c) is $$225\%$$, award (M1)(A1)(ft) with or without working seen. If answer to part (b) is negative, award at most (M1)(A0).