Ian Roblin Maths Resources

Ian Roblin Maths Resources Retail of Maths resources for Key Stages 2/3, GCSE and 'A' level in the Cardiff area. www.ianroblinm See my website www.ianroblinmathstuition.co.uk.

Examples of GCSE & 'AS/A' worksheets I prepared can be found on my main Facebook photos page (click on the business card).

Many non-mathematically minded people question the utility of Trigonometric equations etc. Firstly, this question contai...
19/11/2023

Many non-mathematically minded people question the utility of Trigonometric equations etc. Firstly, this question contains a lot of mathematical & logical content and, secondly, it underpins scientific, engineering & technological specialisms (Rocket scientists & chemists etc were trained on this) :

15/10/2023

Modelling vertical downward motion with a resistive force directly
-------------------------------------------------------------------------
proportional to the cube of the velocity
----------------------------------------------

A body of mass m kg falls vertically from rest, from a fixed point O, which is at a height of h metres above horizontal ground. The resistive force to motion is directly proportional to the cube of the velocity at any moment.
a) Find an expression for the time taken of the particle in terms of velocity.
b) “ “ “ “ “ displacement from point O.

Solution : Let R = resistive force. => R α v³ and R = kv³
where k = constant of proportionality

dv dt
From F = ma, mdv = mg - kv³ => ∫ --------- = ∫ ----
dt mg - kv³ m

dv dv
1 ∫dt = ∫ -------- => t = ∫ ------- + c (1)
-- mg - kv³ --- mg - kv³
m m

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20/09/2023

In the FA Cup Preliminary round, there are
n teams from both North & South. Find the probability of n all North v South ties.

Solution :
-----------
n teams from north, n teams from south, => n ties
Draw could go :
N S N S N S S N

Draw 1 Draw 2


S N S N S N N S

Draw 3 Draw 4

P(Draw 1) = P(Draw 2) =P(Draw 3) = P(Draw 4)
= n x n x (n - 1) x (n - 1) x ...
-- ------- -------- --------
2n (2n - 1) (2n - 2) (2n - 3)

= n²!
-----
(2n)!

Let X = n ties of N-S matches. => P(X) = 4(n²!)
-------
(2n!)

If n = 2, P(X) = 4(4) = 16/24 = 2/3
-----
4!

eg NNSS, NSNS, NSSN => P(X) = 4/6 = 2/3
SSNN, SNSN, SNNS

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24/08/2023

Prove that log xⁿ = nlog x
a a
------------------------------

Let y = log x => aʸ = x
a

Raising both sides to power n, (aʸ)ⁿ = xⁿ

aⁿʸ = xⁿ

Taking logs to base a on both sides, log aⁿʸ = log xⁿ
a a

=> ny = log xⁿ
a

nlog x = log xⁿ QED
a a

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23/08/2023

Calculating log values to various bases
--------------------------------------------

a) Let log 32 = t => 2ᵗ = 32 = 2⁵ and t = 5
2

b) Let log 27 = t => 3ᵗ = 27 = 3³ and t = 3
3

c) log 64² = 2log 64 = 2log 4³
4 4 4

= 3 x 2 x log 4
4

= 6log 4
4

= 6(1) = 6 since 4¹ = 4

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10/08/2023

The creative mathematical muse has smothered me overnight.

---------------------------------------------------------------------
Angela has placed :
3 red balls & 1 white ball in Box A,
1 red ball & 2 white balls in Box B,
2 of each in Box C

Angela selects two boxes only, at random, and then
randomly selects a ball from each box.
Calculate the probability of her obtaining 2 red balls.

Solution :
-----------
Angela could choose combinations (A, B), (A, C), (B, C).
Let R = red ball.
P(RR) = P(R in A) x P(A) x P(R in B) x P(B)

+ P(R in A) x P(A) x P(R in C) x P(C)

+ P(R in B) x P(B) x P(R in C) x P(C)

= ¾ x 1/3 x 1/3 x ½ + ¾ x 1/3 x 2/4 x ½
+ 1/3 x 1/3 x 2/4 x ½

= ¾ x 1/3 x 1/3 x ½ + ¾ x 1/3 x 1/2 x ½
+ 1/3 x 1/3 x 1/2 x ½

= 3/72 + 3/48 + 1/36

= (6 + 9 + 4) = 19
------------- -----
144 144

----------------------------------------------------------------

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10/08/2023

JoJo’s snooker balls
-----------------------
JoJo places various snooker balls in Box 1 & Box 2
(3 of the 8 balls in Box 1 are red ; 4 of the 12 balls in Box 2 are red).
She draws a ball, at random, from both boxes (she calls this a trial), notes the colour and then replaces the balls in their respective boxes. What is the probability that JoJo draws two reds, for the first time only, on trial 3 ?

Solution :
------------
Let n = the number of trials.
Let R(1) = red is drawn in Box 1, R(2) = red is drawn in Box 2.

P[R(1)] = 3/8, P[R(2)] = 4/12 = 1/3

P{R(1) & R(2)}’ = P[R(1) & R(2)’] + P[R(1)’ & R(2)] + P[R(1)’ & R(2)’]
= 1 - P[R(1) & R(2)]


P[R(1) & R(2) in n = 3]
= P{R(1) & R(2)}’ x P{R(1) & R(2)}’ x P[R(1) & R(2)]

= {1 - P[R(1) & R(2)] } x {1 - P[R(1) & R(2)]} x P[R(1) & R(2)]

= {1 - P[R(1) & R(2)]}² x P[R(1) & R(2)]
= {1 - 3/8 x 1/3}² x 3/8 x 1/3
= {1 - 1/8}² x 1/8 = [7/8]² x 1/8 = 49/512 = 0.0957 (4 d.p)

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05/08/2023

Implicit differentiation (tangents to a circle)
-------------------------------------------------
x² + y² = r² where r = constant
a) find dy/dx where :
1) x = 0 2) x = r
b) Interpret your calculations in (a) geometrically

Solution :
------------
a) d (x² + y²) = d (r²) => 2x + d (y²).dy = 0
--- --- --- ---
dx dx dy dx

2x + 2y.dy = 0 => dy = -2x = -x (1)
--- --- --- ---
dx dx 2y y

1) At x = 0, 0² + y² = r² => y² = r² and y = ±r

In (1) :
at (0, r), dy = 0 = 0 ; at (0, -r), dy = 0 = 0
--- --- --- ---
dx r dx -r

2) At x = r, r² + y² = r² => y² = r² - r² = 0 and y = 0

In (1) :
at (r,0), dy = -r => dy → -∞
--- --- ---
dx 0 dx

b) x² + y² = r² is the equation of a circle
[centre (0, 0) and radius = r].
dy/dx = 0 (horizontal line at x = 0) while dy/dx is a vertical line at x = r and so dy/dx → -∞.
Sketch this scenario showing all relevant points.

--------------------------------------------------------------

28/07/2023

Theodore Wright's Cost law (Learning curve)
-------------------------------------------------
As production doubles, the unit cost decreases by 15%.
Let T = initial total costs, Q(1) = initial production level,
C(1) = initial cost per unit.
=> C(1) = T (1)
----
Q(1)

Let Q(2) = 2Q(1) => Q(1) = Q(2)
-----
2

Let C(2) = unit cost at Q(2) => C(2) = 0.85C(1)
= 0.85 x T
----
Q(1)

1
 C(2) = 0.85 x T ÷ Q(2) = 0.85 x 2T (2)
----- -----
2 Q(2)

Let Q(3) = 2Q(2) = 2 .2Q(1) = 4Q(1)
=> Q(2) = Q(3)
-----
2

Let C(3) = unit cost at Q(3)
=> C(3) = 0.85C(2)
= 0.85 x 0.85 x 2T
-----
Q(2)

2 2
=> C(3) = 0.85² x 2T ÷ Q(3) = 0.85 x 2 x T (3)
----- ----
2 Q(3)

In general :
n n
C(n+1) = 0.85 x 2 x T where n = number of periods where
-----
Q(n+1) output doubling occurs

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19/07/2023

Lemmy's Aces in a Poker hand
----------------------------------
Calculate the probability of Lemmy getting the Ace of Spades
& the Ace of Clubs in a Poker hand.

Solution :
-----------
Consider the two Aces as a certain entity and we must also discount the other 2 Aces from appearing in the remaining 3 cards.

P(Ace of S & Ace of C) = 1 x 48C3/52C5
= 17,296/2,598,960
= 0.00665 (5 d.p)

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