Maths Olympiad - Permutations 1

Question:

The letters of the words MATHEMATICS are arranged in such a way that the first four letters are all vowels. Find the total no. of distinct arrangements that can be formed this way.


Answer:

There are 4 vowels - AAEI with the letter A appearing twice
There are 7 consonants - MTHMTCS with the letters M and T appearing twice

Using factorials, the no. of distinct combinations of vowels are:
4!/2! = 4 x 3 x 2 x 1 / 2 x 1 = 12 (divide by 2! because the letter A appears twice)

*Note: The formula is the factorial of the total no. of letters divided by the the multiplication of the factorials of the total no. of repeated letters.

Using factorials, the no. of distinct combinations of consonants are:
7!/(2! x 2!) = 7 x 5 x 4 x 3 x 2 x 1 / (2 x 1)(2 x 1) = 210
(divide by 2! twice because the letters M and T appear twice)

Therefore, the total no. of distinct arrangements that can be formed where the first four letters are all vowels = 12 x 210 = 2520

Maths Olympiad - Algebra 3

Question:




In the above table each letter of the alphabet is given a value. The algebraic expression 4x − 3 is used as a key to convert the letters P S R X B O E into the word H A R M O N Y. Find the algebraic expression used to convert S R X B into G O L D ?



Answer:


For S R X B to become G O L D ,

S -> G where 4 -> 14
R -> O where 1 -> 5
X -> L where 3 -> 11
B -> D where 2 -> 8


Using the graphical y = mx + c expression, we will take the above numbers as x, y coordinates.

So, we have the following coordinates - (4,14), (1,5), (3,11) and (2,8).


Therefore,
y = mx + c
14 = 4m + c ----- sub. (4,14)
c = 14 - 4m ----- Eq. 1


y = mx + c
8 = 2m + c ---- sub. (2,8)
c = 8 - 2m ---- Eq. 2


Combine Eq.1 and Eq.2,
14 - 4m = 8 - 2m
6 = 2m
m = 3


Sub. m = 3 into Eq.2,
c = 8 - 2(3)
c = 8 - 6
c = 2


Sub. m = 3 and c = 2 into y = mx + c:
y = 3x + 2


Therefore, the algebraic expression used to convert S R X B into G O L D is:
3x + 2

Maths Olympiad - Algebra 2

Question:

If 30% of p is q, and 20% of q is 12, what is 50% of p + q?

Answer:

20% q = 12
100% q = 12 x 5
q = 60

30% p = q
30% p = 60
100% p = 100/30 x 60
p = 200

Therefore, 50% of p + q = 50% (200 + 60)
= 50% (260)
= 130

Maths Olympiad - Algebra 1

Question:

where a and b are real numbers.

Find the maximum possible value of (a - 5b).

Answer:

Since the square of any number cannot be negative, the minimum value of

is ZERO.

(*Since its minimum value would make the resulting number the maximum possible value because it is multiplied by a negative number -4)

Therefore, the maximum possible value of (a - 5b)

= - 4 (0 - 125)

= 500

Maths Olympiad - Prime Factors 1

Question:

When 2007 bars of soap are packed into N boxes, there is a remainder of 5. How many possible values of N are there?

Answer:

No. of bars of soap to be divided equally among the boxes = 2007 - 5 = 2002



2002 = 2 x 11 x 7 x 13

Therefore, N, the no. of boxes could be a combination of these prime factors
= 2, 7, 11, 13, 14 (2x7), 22 (2x11), 26 (2x13), 77 (7x11), 91 (7x13), 143 (13x11), 154 (2x7x11), 182 (2x7x13), 286 (2x11x13), 1001 (7x11x13)

*Note: You do not actually need to multiply these factors because it's not required for this question.

Counting the no. of boxes that could divide 2002 bars of soaps equally = 14
Therefore, N = 14

Maths Olympiad - Triangles 1

Question:

PQ = 40cm, RS = 20cm, PS = 60cm and Angle QPS = Angle RSP = 60 degrees. Find Angle QRS.

Answer:

Draw line RX parallel to PS. Since Angle QPS = Angle RSP = 60 degrees, then Angle QXR = 60 degrees and line QX = XP = 20cm.

Angle PYS = 180 - (QPS + RSP) = 60 degrees

Therefore, Triangle YPS is equilateral because all of its angles are 60 degrees.

Since, the lengths of all sides are equal in an equilateral triangle, PY = SY = PS = 60cm.

Therefore, QY = 20cm (Diagram not drawn to scale) and Q is the midpoint of line XY.

Therefore, line QR is a bisector of Angle XRY = 60 degrees which makes Angle QRX = 60 / 2 = 30 degrees.

Angle XRS = 180 - PSR (internal angles) = 120 degrees.

Therefore, Angle QRS = QRX + XRS = 150 degrees.

Maths Olympiad - Probability 1

Question:

A bag contains x green and y red sweets. A sweet is selected at random and its colour noted. It is then replaced into the bag together with 10 additional sweets of the same colour. A second sweet is next randomly drawn. Find the probability that the second sweet is red.

Answer:

Total no. of sweets in the bag at the start = (x + y)

Probability that first sweet is GREEN = x / (x +y)

Probability that first sweet is RED = y / (x +y)

If the first sweet was GREEN:

Then, 10 more GREEN sweets will be added.

Therefore, total no. of GREEN sweets = (x + 10)

Therefore, total no. of sweets in the bag is now = (x + 10) + y

Therefore, the probability that the second sweet is RED = y / (x + 10 + y)

If the first sweet was RED:

Then, 10 more RED sweets will be added.
Therefore, total no. of RED sweets = (y + 10)
Therefore, total no. of sweets in the bag is now = (y + 10) + x
Therefore, the probability that the second sweet is RED = (y + 10) / (y + 10 + x)

Therefore, the probability that the second sweet is RED

= [probability of first GREEN, then RED] + [probability of first RED, then RED]

= [x / (x +y)] * [y / (x + 10 + y)] + [y / (x +y)] * [(y + 10) / (y + 10 + x)]

= y / (x + y)

*You may also use a probability tree to illustrate the solution.