Maths Challenge 6 # Maths Challenge 6

## 1: Upgrade CostPercentages

You are playing a game which involves building a city.
You need to upgrade your Headquarters building immediately.
However, the timer says it will take 3 days, 16 hours and 48 minutes.
The last 15 minutes of the upgrade are completely free to rush, however you can speed it up with payments of 1 gold per additional minute of rush.
Thankfully, you have some free speed-ups available to use, these are:
12 x 1 hour construction rush
25% reduction in construction time
15 x 5 minute construction rush
50% reduction in construction time
What is the least amount of gold you will need to spend to complete the upgrade immediately?

You will need to apply the percentage discounts first to maximise their effect

One approach:

Convert time to hours = (3x24) + 16 + (48/60) = 88.8hours

50% of 88.8 = 44.4
25% of 44.4 = 11.1
So, after percentage removal, we have 88.8 - 44.4 - 11.1 = 33.3 hours remaining

Now to deduct time boosts:
33.3hours - (12 x 1 hour boosts) - (15 x 5 minute boosts) - 15 minutes free boosts
= 33.3 - 12 - 1.25 - 0.25 = 19.8 hours = 19 hours 48 minutes

Amount of gold for boost = (19 x 60) + 48 = 1188

## 2: Space LaunchForces (Higher GCSE)

A rocket is powered by three engines, each of which produces a thrust of 2000 Newtons.
The mass of the rocket and its fuel is 500kg.
1. When the engines are fired:
1. Calculate the total thrust on the rocket
2. Explain why the rocket moves upwards
3. Calculate the resultant force on the rocket
4. Calculate the take-off acceleration of the rocket

2. After 2 seconds, the rocket engines have used up 20kg of fuel. Assuming that the thrust of the engines is constant, calculate:
1. the mass of the rocket and fuel after 2 seconds,
2. the resultant force on the rocket after 2 seconds,
3. the acceleration of the rocket after 2 seconds.

3. Assuming that the thrust of the engines is constant, explain why the acceleration of the rocket will continue to increase for as long as the engines are fired.

From WJEC GCSE Higher Tier Physics Exam (Paper 2, Jan 2010)

Acceleration = Resultant Force ÷ Mass

1. 3 x 2000 = 6000N
2. Resultant Force is Upwards,
or,
Upwards force is greater than the downwards force.
3. Using gravity = 10m/s2
Resultant Force = 6000N - 5000N
4. Using Acceleration = Resultant Force ÷ Mass:
Acceleration = 1000/500 = 2m/s2

1. 500kg - 20kg = 480kg
2. 6000N - 4800N = 1200N
3. Acceleration = 1200N/480kg = 2.5m/s2

1. Resultant Force increases because...
• mass decreases
• weight decreases
• air resistance decreases with height
• gravity decreases with height

## 3: Snack ShopBasic Operations and Tables

This chart shows the number of packets of different flavours of crisps sold by a shop.

Monday Tuesday Wednesday Thursday Friday
Ready Salted 3 1 2 4 0
Salt 'n' Vinegar 4 2 5 3 1
Cheese 'n' Onion 5 1 3 1 4
Roast Beef 3 2 6 4 1
Prawn 1 1 2 4 4

1. Which flavour(s) was the best seller on each of the days?

2. The shop makes the below profit on each flavour:
Flavour Profit per Packet
Ready Salted 2p
Salt 'n' Vinegar 4p
Cheese 'n' Onion 1p
Roast Beef 3p
Prawn 5p

During the week which flavour(s) gave the shopkeeper the most profit?

It may help for part 2 to add a totals column to the original table

1. Monday: Cheese 'n' Onion
Tuesday: Salt 'n' Vinegar and Roast Beef
Wednesday: Roast Beef
Thursday: Ready Salted, Roast Beef and Prawn
Friday: Cheese 'n' Onion and Prawn

2. Ready Salted Total Profit = 2p x (3 + 1 + 2 + 4 + 0) = 2p x 10 = 20p
Salt 'n' Vinegar Total Profit = 4p x (4 + 2 + 5 + 3 + 1) = 4p x 15 = 60p
Cheese 'n' Onion Total Profit = 1p x (5 + 1 + 3 + 1 + 4) = 1p x 14 = 14p
Roast Beef Total Profit = 3p x (3 + 2 + 6 + 4 + 1) = 3p x 16 = 48p
Prawn Total Profit = 5p x (1 + 1 + 2 + 4 + 4) = 5p x 12 = 60p

## 4: Tarsia PuzzleBasic Algebra

This puzzle involves joining triangles together to form a hexagon where the value of x matches on touching sides.
You will need to print/copy and cut out the triangles from the tarsia puzzle pdf (there are 3 pages).
Text version of puzzle for screen readers (with solution)

Created using software from Hermitech Laboratory

The blank sides go around the outer edge of the final hexagon

## 5: Playing CricketMechanics (A-Level)

A batsman strikes a ball at a height of 1.5m above the ground, giving it an initial speed of 29 ms-1 at an angle of 30° to the horizontal.

1. What is the maximum distance of the boundary from the batsman if he scores a 'six' (i.e. the ball passes over the boundary line without first bouncing)?

A fielder, who is capable of catching a ball at a height of 2.75m or below, goes to the boundary line.

2. What speed must the batsman give the ball if he is to hit it at the same height and elevation as before and ensure that he will score a six and not be caught by the fielder?

Use the 'suvat' equations. As a reminder, see the below table:

Value Horizontal Motion Vertical Motion
s x = (u cos ∝)t y = (u sin ∝)t - ½gt2
u ux = u cos ∝ uy = u sin ∝
v vx = u cos ∝ vy = u sin ∝ - gt
a ẍ = 0 ÿ = -g

1. 75.3m
2. 29.9ms-1