N5 Applications of Maths 2026 Paper 2 — Questions & Video Solutions

1

Question 1

Finance

4 Marks

2026 P2 Q1

A new housing development is being built in a village.
The village currently has a population of 750.
The developers expect the population to increase by 36% for each of the next 3 years.
Calculate the expected population after 3 years.
Give your answer rounded to 2 significant figures.

Show answer

A 36% increase gives a multiplier of 1.36.

$\displaystyle 750\times1.36^{3}=1886.592$

Rounded to 2 significant figures: 1900

2

Question 2

Geometry

2 Marks

2026 P2 Q2

Rachel's front door is in the shape of a rectangle with a semi-circular window.
The dimensions are shown.

Rectangular front door 0.76 m wide and 1.98 m high, with a semi-circular window of diameter 0.54 m near the top

Rachel wants to paint the front of the door excluding the window.
Calculate the area of the door to be painted.

Show answer

Area of door $\displaystyle =0.76\times1.98=1.5048$ m²

Window has radius 0.27 m, so area $\displaystyle =\frac{1}{2}\times\pi\times0.27^{2}=0.1145...$ m²

Area to paint $\displaystyle =1.5048-0.1145=$ 1.39 m² (to 2 d.p.)

3

Question 3

Finance

(2, 2)4 Marks

2026 P2 Q3

Lee earns a gross annual salary of £42,000.
National Insurance is calculated on a person's pay before deductions such as pension contributions.

National Insurance rates per year
Up to £12,584	0%
From £12,584 to £50,284	8%
Over £50,284	2%

(a) Calculate Lee's annual National Insurance payment.

Lee pays 11.2% of their gross annual salary into their pension.
Lee's annual income tax is £4787.12.

(b) Calculate Lee's annual net pay.

Show answer

(a) Pay taxed at 8% $\displaystyle =42000-12584=£29\,416$
NI $\displaystyle =0.08\times29416=$ £2353.28

(b) Pension $\displaystyle =0.112\times42000=£4704$
Total deductions $\displaystyle =4704+4787.12+2353.28=£11\,844.40$
Net pay $\displaystyle =42000-11844.40=$ £30,155.60

4(a)

Question 4(a)

Finance

3 Marks

2026 P2 Q4(a)

Shehbaz is travelling from Glasgow to Barcelona and then Istanbul.

Rates of Exchange
Pounds Sterling (£)	Other Currencies
1	1.15 euros
1	44 Turkish lira

Shehbaz converted £840 into euros.
He stayed in Barcelona for 4 days.
He spent 205 euros each day that he was in Barcelona.

He converted his remaining euros into Turkish lira.

(a) Calculate how many Turkish lira he received.

Show answer

$\displaystyle 840\times1.15=966$ euros, and he spent $\displaystyle 205\times4=820$ euros.

Remaining $\displaystyle =966-820=146$ euros $\displaystyle =146\div1.15=£126.9565...$

$\displaystyle 126.9565...\times44=$ 5586.09 Turkish lira

4(b)

Question 4(b)

Measurement

3 Marks

2026 P2 Q4(b)

Shehbaz flies from Glasgow to Barcelona.
The plane departs at 1:30 pm local time.
The time in Barcelona is 1 hour ahead of Glasgow.
The plane flew 1680 kilometres at an average speed of 600 kilometres per hour.

(b) Calculate the local time the plane landed in Barcelona.

Show answer

Flight time $\displaystyle =\frac{1680}{600}=2.8$ hours = 2 hours 48 minutes.

$\displaystyle 13\!:\!30+2\text{ h }48\text{ min}=16\!:\!18$ UK time, and Barcelona is 1 hour ahead.

17:18 (5:18 pm)

4(c)

Question 4(c)

Measurement

(3, 2)5 Marks

2026 P2 Q4(c)

Shehbaz flies from Glasgow to Barcelona and then from Barcelona to Istanbul.

The flight from Glasgow to Barcelona:

distance: 1680 km
bearing: 161°.

The flight from Barcelona to Istanbul:

distance: 2250 km
bearing: 082°.

(c) (i) Construct a scale drawing to illustrate this journey.
Use a scale of 1 cm : 300 km.

Starting point for the scale drawing: a north arrow above a point labelled Glasgow

The plane then returns directly to Glasgow from Istanbul.

(ii) Use your scale drawing to determine the distance and bearing of Glasgow from Istanbul.

Show answer

(i) Glasgow to Barcelona: $\displaystyle 1680\div300=5.6$ cm on a bearing of 161°.
Barcelona to Istanbul: $\displaystyle 2250\div300=7.5$ cm on a bearing of 082°.

(ii) Measuring Istanbul back to Glasgow gives about 10.2 cm, so the distance is approximately 3060 km on a bearing of approximately 295°.

5(a)

Question 5(a)

Finance

3 Marks

2026 P2 Q5(a)

Alan works for a company that lays patios.
His current hourly rate is £14 per hour.
He has a contract for 35 hours of work per week but regularly works extra hours.
Alan has been offered two pay options:

Option 1: Earn time and a half for any extra hours.
Option 2: Receive a 10% increase in his hourly rate. All hours worked will be paid at this rate.

(a) Determine which option gives Alan a higher gross pay for working 40 hours in a week.

Show answer

Option 1: $\displaystyle 35\times14=£490$ , plus $\displaystyle 5\times(14\times1.5)=5\times21=£105$ , giving £595.

Option 2: $\displaystyle 14\times1.10=£15.40$ per hour, so $\displaystyle 40\times15.40=£616$ .

Option 2 gives the higher gross pay (£616 > £595).

5(b)

Question 5(b)

Finance

3 Marks

2026 P2 Q5(b)

Alan needs to buy 120 paving slabs for a patio. He is comparing prices from three different shops, each with its own pricing.

Shop A: The cost per slab is £10, and there is a 'Buy 2, Get 1 Free' deal.
Shop B: The cost per slab is £9, with a 15% discount applied to the total cost of the slabs.
Shop C: The cost per slab is £7.50.

(b) Determine the cheapest option for buying 120 slabs.

Show answer

Shop A: he pays for 2 slabs in every 3, so $\displaystyle 120\div3\times2=80$ slabs, $\displaystyle 80\times10=£800$ .

Shop B: $\displaystyle 120\times9=£1080$ , then $\displaystyle 1080\times0.85=£918$ .

Shop C: $\displaystyle 120\times7.50=£900$ .

Shop A is the cheapest at £800.

5(c)

Question 5(c)

Numeracy

2 Marks

2026 P2 Q5(c)

It takes 4 workers 15 hours to lay a patio.
The patio company are able to provide 2 extra workers.
All workers work at the same rate.

(c) Calculate how long it will take to lay the patio.

Show answer

The job takes $\displaystyle 4\times15=60$ worker-hours.

With $\displaystyle 4+2=6$ workers: $\displaystyle 60\div6=$ 10 hours

5(d)

Question 5(d)

Numeracy

2 Marks

2026 P2 Q5(d)

The patio company completed a job for a customer.

It cost the company £2300 for materials and £672 for wages to complete the job.
The customer paid £3800 for the job.

(d) Calculate the company's percentage profit.

Show answer

Costs $\displaystyle =2300+672=£2972$ , so profit $\displaystyle =3800-2972=£828$ .

$\displaystyle \frac{828}{2972}\times100=27.86...\%=$ 27.9% (to 1 d.p.)

6(a)

Question 6(a)

Measurement

3 Marks

2026 P2 Q6(a)

Katy is conducting quality control for a garden centre.
Bags of fertiliser must weigh $\displaystyle 25\text{ kg}\pm4.8\%$
Below are the weights of ten bags of fertiliser, in kilograms:

24.7 25.8 29.9 24.6 23.7
26.5 26.0 24.4 20.1 25.9

(a) Identify which of these weights are not suitable.

Show answer

$\displaystyle 4.8\%$ of 25 kg $\displaystyle =1.2$ kg, so the acceptable range is 23.8 kg to 26.2 kg.

Outside this range: 29.9, 23.7, 26.5 and 20.1

6(b)-(c)

Question 6(b)-(c)

Statistics

(4, 2)6 Marks

2026 P2 Q6(b)-(c)

The garden centre recorded how many products it sells.
A sample of the number of spades sold per month is shown.

35 18 28 30 32 40 20

(b) Calculate the mean and standard deviation of the number of spades sold per month.

The garden centre also sells hose pipes.
In the same months the mean number of hose pipes sold each month was 18 and the standard deviation was 13.

(c) Make two valid comments comparing the number of spades sold and the number of hose pipes sold.

Show answer

(b) Mean $\displaystyle =\frac{203}{7}=29$ .
$\displaystyle \sum(x-\bar{x})^{2}=36+121+1+1+9+121+81=370$ , so $\displaystyle s=\sqrt{\frac{370}{6}}=$ 7.85 (to 2 d.p.)

(c) On average, more spades were sold per month than hose pipes, because the mean number of spades sold is higher (29 > 18).
The number of spades sold each month was more consistent than the number of hose pipes sold, because the standard deviation for spades is lower (7.85 < 13).

6(d)

Question 6(d)

Finance

3 Marks

2026 P2 Q6(d)

Margaret buys a greenhouse from the garden centre.
The advertised price was £900.
Margaret used a payment plan to purchase the greenhouse.
The total price of the payment plan was 18% more than the advertised price.
The payments are calculated as follows:

the deposit is $\displaystyle \frac{1}{5}$ of the total price.
there are 10 equal monthly instalments.
followed by a final payment of £160.

(d) Calculate the cost of each monthly instalment.

Show answer

Total price $\displaystyle =900\times1.18=£1062$

Deposit $\displaystyle =1062\div5=£212.40$ , leaving $\displaystyle 1062-212.40=£849.60$ .

After the final payment: $\displaystyle 849.60-160=£689.60$

$\displaystyle 689.60\div10=$ £68.96

7(a)

Question 7(a)

Numeracy

2 Marks

2026 P2 Q7(a)

A golf club has adult, senior and junior members.
The ratio of members is 4:3:1 respectively.
There are 102 senior members.

(a) Calculate the total number of members in the club.

Show answer

3 parts = 102, so 1 part $\displaystyle =102\div3=34$ .

Total $\displaystyle =8\times34=$ 272 members

7(b)

Question 7(b)

Geometry

3 Marks

2026 P2 Q7(b)

The club holds an annual Ladies' Championship. The winner receives a trophy.
The trophy has a metal top, consisting of a cone and sphere.
The cone and sphere both have a radius of 3.5 cm.
The total height of the metal top is 11 cm.

Trophy top made of a sphere resting on a downward-pointing cone, with a total height of 11 cm

(b) Calculate the volume of metal in the trophy top.

Show answer

The sphere has height $\displaystyle 2\times3.5=7$ cm, so the cone has height $\displaystyle 11-7=4$ cm.

$\displaystyle V_{\text{sphere}}=\frac{4}{3}\pi\times3.5^{3}=179.59...$ cm³

$\displaystyle V_{\text{cone}}=\frac{1}{3}\pi\times3.5^{2}\times4=51.31...$ cm³

Total $\displaystyle =179.59+51.31=$ 230.9 cm³

7(c)

Question 7(c)

Geometry

4 Marks

2026 P2 Q7(c)

The length of a golf hole is the distance between the tee and the flag.
Laura plays a golf hole that is 430 metres long.
Laura stands at the tee and hits her ball. Her ball stops 274 metres from the tee, as shown in the diagrams.
The distance between Laura's ball and the flag is shown by the dashed line in Diagram 2.

Diagram 1 shows the golf hole with the tee, the flag 430 m away and Laura's ball off to the left. Diagram 2 is a right-angled diagram with the ball 274 m from the tee, 270 m of that measured along the line from tee to flag, and a dashed line from the ball to the flag

(c) Calculate the distance between Laura's ball and the flag.
Do not use a scale drawing.

Show answer

Horizontal offset: $\displaystyle x^{2}=274^{2}-270^{2}=75076-72900=2176$

Remaining distance towards the flag $\displaystyle =430-270=160$ m.

$\displaystyle d^{2}=160^{2}+2176=25600+2176=27776$

$\displaystyle d=\sqrt{27776}=$ 166.66 m

7(d)

Question 7(d)

Measurement

3 Marks

2026 P2 Q7(d)

The club shop sells boxes of golf balls with dimensions as shown.

Box of golf balls measuring 17.2 cm by 12.9 cm by 4.3 cm, marked THIS WAY UP

The boxes are shipped in a container with internal dimensions as shown.
The boxes must be aligned in the same direction.

Shipping container measuring 65 cm by 52 cm by 40 cm, marked THIS WAY UP

(d) Calculate the maximum number of golf ball boxes than can fit in the container.

Show answer

THIS WAY UP means the 4.3 cm height stands upright: $\displaystyle 40\div4.3=9.30...$ , so 9 layers.

Base layer, first orientation: $\displaystyle 65\div17.2=3.77...\to3$ and $\displaystyle 52\div12.9=4.03...\to4$ , giving 12.

Base layer, second orientation: $\displaystyle 65\div12.9=5.03...\to5$ and $\displaystyle 52\div17.2=3.02...\to3$ , giving 15.

$\displaystyle 15\times9=$ 135 boxes