IGCSE at Mathematics Realm

Vectors on Triangles (Part 1)

noreply@blogger.com (Mathrealm) — Wed, 02 Dec 2015 12:55:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 2 of October/November 2007.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Vectors
(b) Specific topic: Addition/Subtraction of Vectors, Position Vectors, Naming a Vector in Terms of other Two Vectors
(c) Concepts Needed: Triangle Law for Addition/Subtraction of Vectors

To answer (i), take note that vector OA = a and vector OB = b. Let us divide each vector into three parts as shown:

Since we need 1/3 of a, we will take 1 part of a. On the other hand, we need 2/3 of b, meaning 2 parts of b.

Here are the vectors 1/3 a and 2/3 b:

Following the triangle law of addition for vectors, the end of the first vector should be the start of the second vector. This means that the the tip of 1/3 a should be the start of 2/3 b.

From this, we can now draw the resulting vector 1/3 a + 2/3 b.

However, we are finding point C, which is the tip/end of this resulting vector.

Therefore, point C is located at

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Speed - Time Graph

noreply@blogger.com (Mathrealm) — Mon, 30 Nov 2015 13:05:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 43 of October/November 2013.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Algebra
(b) Specific topic: Speed-Time Graphs
(c) Formulas/Concepts Needed: gradient, area of a trapezium, average speed

For (i), let us determine which part of the graph is needed. Since the question is about deceleration, which means decreasing speed, it means that the graph should be downward. That could be seen on the last part of the graph as shown here (in red).

To find the deceleration, let us recall that the deceleration is the change of speed over time. We can achieve this by getting the gradient of the line. We will use two different methods here.

First, let us use the common notion for gradient - that it is rise/run. In the graph, the rise is the vertical distance and the run is the horizontal distance.

We can see that the rise is equal to 16 and the run is from 20 to 25. That is

Second, let us use the formula for gradient - that is change in y over the change in x. To start with, let us determine the coordinates of the endpoints of the line.

Now we can use the formula to find the gradient.

For both methods, notice that the answer is negative. This suggests that it is decreasing at that constant speed. The negative sign may be disregarded in the final answer.

To answer (ii), let us recall that average speed requires total distance and total time. To find the total distance covered by the car, let us find the area of the figure under the graph. In this graph, it is a trapezium. To find the area of the trapezium, let us identify the parts needed.

Using the formula for the area of a trapezium, we can now find the area under the graph.

Further, we can now determine the average speed by getting the quotient of the total distance and the total time taken.

ANSWERS:

(i) 16/5 m/s2

(ii) 56/5 or 11.2 m/s

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Product of Matrices

noreply@blogger.com (Mathrealm) — Sat, 28 Nov 2015 03:21:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 42 of May/June 2013.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Matrices
(b) Specific topic: Operations on Matrices

The common mistake students do with this item is to get directly the square of each element of the matrix. Here is the common wrong solution and answer:

Take note that:

It means that it is the product of matrix D multiplied to itself.

Before you start multiplying, determine first the order (dimension) of each matrix and the order of the resulting matrix.

It is possible to multiply the matrices since the number of columns of the first is equal to the number of rows of the second matrix. Follow the method of multiplying matrices - multiply each row of the first to each column of the second. That is

Simplify.

Therefore,

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Graphical Solutions

noreply@blogger.com (Mathrealm) — Sat, 09 May 2015 06:15:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 43 of May/June 2011.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Algebra
(b) Specific topic: Graphical Solutions

For (i), we should take note that y = f(x).

Since the given is f(2), then x = 2. This suggest that we should draw the line x = 2 then find the value of y in the intersection of the line x = 2 and the given graph.

For (ii), since y = f(x) and the given is f(x) = 0, then y = 0. This means that the line y = 0 must be drawn. The line y = 0 is also the x-axis. Find the value of x in the intersection between the x-axis and the given graph.

For (iii), we are looking for the value of k where there are three solutions. This means that we are looking for a horizontal line that passes through the given graph at exactly three points. We should also take note that the value of k is an integer. The possible values of k are 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5 (based on the range of the grid given). Let us draw the horizontal lines and mark the intersection(s) of each.

Let us summarize the number of intersections in the following table then determine the highest value of k where there are three intersections.

For (iv), to solve for the equation f(x) = x, we should take note that f(x) = y. Hence, f(x) = x is the same as y = x. By now, you should be familiar with the graph of this equation. If not, you may use a table to assign values of x then get the corresponding values of y. The graph is a straight line passing through the origin and slanting to the right.

Let us determine the coordinates of the intersection between the line y=x and the given graph.

Therefore,

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Vectors and a Hexagon

noreply@blogger.com (Mathrealm) — Thu, 07 May 2015 15:00:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 22 of October/November 2013.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Vectors
(b) Specific topic: Vectors, Position Vectors, Addition/Subtraction of Vectors
(c) Concept Needed:

Let us start with the relations of the sides of a regular hexagon.

Before we answer (a), let us start with the relations of the vectors given in the figure.

We cannot get the direct value of BE using the given vectors. Instead, we can find the value of OB first, then use the relation of the vectors BE = 2BO.

Considering vectors OC and CB, the resultant is vector OB. This uses the concept of triangle law of vector addition, since vector OC and CB are in a nose to tail position.

However, we are finding BO. Hence, we can reverse the directions and find the resultant BO in terms of a and c.

We can find BE by using vector BO.

For (b), let us start with the relations of the vectors needed.

Using the triangle law of subtraction of vectors.

For (iii), we should take note that it is a position vector. It means that the vector starts with O then ends with E. Hence, we are finding vector OE.

Using the relation of vector OE and vector BE,

Therefore,

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Similar Triangles and Bearings

noreply@blogger.com (Mathrealm) — Wed, 06 May 2015 10:21:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 41 of May/June 2011.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Trigonometry
(b) Specific topic: Bearings, Similar Triangles
(c) Concept Needed:

Let us start with the path/route taken by the senior students. Let us determine first the length of all sides of their triangular path. Let us use the given that they walked a total of 18.9 km.

For the path taken by the junior students, let us start to find first the measure of the angles of the triangle formed.

From the given, it is stated that the junior students follow a similar path to that of the senior students. This suggests that the triangles formed by their paths are similar. Here are the corresponding measurements.

Let us use triangle similarity to find the missing measurements.

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Transforming a Formula

noreply@blogger.com (Mathrealm) — Tue, 05 May 2015 14:18:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 33 of October/November 2010.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Algebra
(b) Specific topic: Transforming a Formula

We should take note that r, alone, should be placed on one side of the equation. Using the properties of multiplication/division and radicals, we can make r as the subject of the formula.

Vectors and Triangles

noreply@blogger.com (Mathrealm) — Sat, 02 May 2015 12:46:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 41 of October/November 2010.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Vectors
(b) Specific topic: Addition and Subtraction of Vectors
(c) Concept needed:

To answer (i), let us use the given statement and name vector QT in terms of t.

To answer (ii), let us also use the given then determine the relation of the sides.

Let us take note that vector OT and OR starts at the same vertex. This suggest that the resultant is the difference of these two vectors. That is

To answer (iii), let us use the answers from (i) and (ii), and the triangle that they formed.

Therefore,

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Simplifying Rational Expressions

noreply@blogger.com (Mathrealm) — Sat, 02 May 2015 00:42:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 43 of October/November 2010.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Algebra
(b) Specific topic: Simplifying Rational Expressions

We will try to answer this item in different methods.
For the first method, we need to find the LCD (least common denominator) of the fractional expressions. Here are the steps needed:

Another method is the following:

Therefore,

Simplifying Expressions

noreply@blogger.com (Mathrealm) — Fri, 01 May 2015 14:22:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 43 of October/November 2010.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Algebra
(b) Specific topic: Simplifying Algebraic Expressions

Take note that there are terms enclosed by parentheses and a term outside each group. This suggests the use of distributive property.

Therefore, x(x+3) + 4x(x-1) = 5x^2 - x.
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Reflection of a Shape

noreply@blogger.com (Mathrealm) — Wed, 29 Apr 2015 17:39:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 43 of May/June 2011.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Transformations
(b) Specific topic: Reflection

Let us start by drawing the mirror line.

We should take note that the mirror line is exactly halfway between the object (shape) and its image. This means that distance of each point in the object from the mirror line is the same as the distance of the corresponding image from the mirror line. Let us find each of the corresponding image of each vertex of the shape given.

Therefore, shape Y is a trapezium with vertices (6,-4),(4,-4),(2,-6) and (6,-6).

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Trigonometrical Ratios

noreply@blogger.com (Mathrealm) — Sat, 18 Apr 2015 16:58:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 12 of May/June 2014.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Trigonometry
(b) Specific topic: Trigonometrical Ratios
(c) Formula Needed:
To easily remember the trigonometrical ratios, take note of SOH - CAH - TOA.

Let us start to identify the given angle and the given sides.Furthermore, identify the relation of the given sides to the given angle.

Now, let us identify which of the trigonometrical ratios is applicable. First, side AB is adjacent to the given angle and AC is the hypotenuse. This leads us to use cosine.

Simplifying and finding the length of AB.

Therefore, the length of AB is 7.06 cm.

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Speed Upper Bound

noreply@blogger.com (Mathrealm) — Tue, 07 Apr 2015 14:49:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 4 of October/November 2014.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Numbers / Algebra
(b) Specific topic: Limits of Accuracy
(c) Formula Needed:

Let us start with the first quantity, which is the distance. We should take note that the distance is given correct to the nearest kilometre. In other words, it is correct to the nearest whole number. It is then expected that the bounds would consist of one (1)
decimal place, which round off to the whole number.

The second quantity is the time. We take note that it is given correct to the nearest half hour. To get the lower and upper bounds of the time, let us first illustrate how to round off to the nearest half hour, given the 2.5 in the middle.

We should take note that 2 to 2.5 is half an hour; the same case for 2.5 to 3. If we look at the number line, starting from 2, before it round off to the nearest half hour, it should reach first halfway between 2 and 2.5. It means that if it reaches 2.25, it can be rounded off to 2.5 already. The same case for the other side. At 2.75, it will round off to 3 already. These numbers then are the lower and upper bounds of 2.5.

The upper bound then for the speed will be

Therefore, the upper bound of the speed is 10 km/h.

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Age Problem

noreply@blogger.com (Mathrealm) — Sun, 29 Mar 2015 05:57:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 31 of May/June 2013.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Algebra
(b) Specific topic: Age Problems
Let us first complete the table for (a)(i). The difference between 2013 and 2001 is 12. In other way, if 1 is added to 2001, the result if 2013. Hence, if 2001 is represented by x, then 2013 is represented by x + 12.

For Ken's age on 2001, let us go back to the given in the problem: "Ken is 34 years younger than Arnold". In other words, the difference of their ages is 34. The age of Ken on 2013 can also be calculated by adding 12 to his age on 2001.

For (a)(ii), let us use the expressions for their ages on 2013. Arnold is (x + 12) and Ken is (x - 22) years old. By that time, Arnold is "three times" as old as Ken. It means that to get the age of Arnold, multiply the age of Ken by three.

Solving for the value of x.

Therefore, the value of x is 39, which is the age of Arnold on 2001.

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Area of Similar Figures

noreply@blogger.com (Mathrealm) — Tue, 24 Mar 2015 15:12:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 41 of May/June 2013.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Mensuration
(b) Specific topic: Area of Similar Figures
(c) Formula Needed:

Let us first determine the given in the problem.

Using the relation of the areas and corresponding lengths of similar figures,

Cross multiply:

Simplify then calculate the value of l by getting the square root:

Another way is to directly get the square root of both sides to eliminate the exponent/power.

Then manipulate the equation to get the value of l.

Therefore, the length (l) of the base of the first figure is 5.75 cm.

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Arithmetic Sequence

noreply@blogger.com (Mathrealm) — Mon, 23 Mar 2015 13:00:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 32 of May/June 2013.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Algebra
(b) Specific topic: Arithmetic Sequence
(c) Formula Needed:

Let us identify first the given and details of the sequence. Let us start with the first term and the common difference of the sequence. The common difference is the result of subtracting any consecutive term of the sequence.

Using the first term and the common difference,

Another method (shortcut method) could be used to check at an instant the nth term of the sequence.

Therefore, the nth term of the arithmetic sequence is 4n + 3.

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Lower Bound and Upper Bound of Perimeter

noreply@blogger.com (Mathrealm) — Sun, 22 Mar 2015 12:55:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 23 of May/June 2014.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Geometry
(b) Specific topic: Perimeter of Plane Figures,Limits of Accuracy
(c) Formula Needed:

Let us start with the lower bound and upper bound of the length of the rectangle.

Now, let us find the lower and upper bounds of the width of the rectangle.

To find the upper bound of the perimeter of the rectangle, calculate using the upper bounds of its length and width.

To find the lower bound of the perimeter of the rectangle, calculate using the lower bounds of its length and width.

Therefore, the upper and lower bounds of the perimeter of the rectangle are 427.8 cm and 427.4 cm, respectively.

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Quadratic Formula

noreply@blogger.com (Mathrealm) — Mon, 09 Mar 2015 14:36:00 +0000

Let us take note that the given is a quadratic equation but it is not factorable. We can then find the solution(s) of the equation by using the quadratic formula.

Let us first identify the values of a, b and c for the quadratic formula.

We can now substitute the values of a, b and c in the formula:

At this point, we should not get the decimal value first of the square root of 17. Leave it as is.

We can now separate the solution into two, separating the "+"and "-" signs, then simplify.

Therefore, the solutions for the equation is x = -0.22 or x = -2.28.

*We should take note also that the complete working is required in this item. The most important part of it is the correct quadratic formula with correct values of a, b and c.

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The Cosine Rule

noreply@blogger.com (Mathrealm) — Tue, 03 Mar 2015 14:22:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 43 of October/November 2010.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Trigonometry
(b) Specific topic: The Cosine Rule
(c) Formula Needed:

Let us first identify the parts needed:

Since the unknown is angle PQR, let us write the cosine rule using PQR as the subject.

Let us now substitute the given values in the formula.

Simplify

We should take note that the problem requires us to show that angle PQR round to 110.7 degrees. This means that we should show the complete calculation for the angle and the full calculator display, which is 110.74238 degrees. This will verify that the rounded value of the angle is 110.7 degrees.

Therefore, the measure of angle PQR is 110.7 degrees.

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Probability Tree Diagram

noreply@blogger.com (Mathrealm) — Sat, 28 Feb 2015 16:34:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 43 of May/June 2013.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Probability
(b) Specific topic: Tree Diagrams and Dependent Events

Let us take note that two cards, one at a time, are drawn at random from the box. Let us also take note that the first card chosen is not replaced in the box, which means that it is a dependent event. It means that the result on the first card chosen affects the result on the second card.

There are 3 blue cards out of 10 cards in the box so the probability that the blue card is chosen on the first pick/draw is 3/10.

Since the first card chosen is not replaced, the number of cards left in the box is decreased by one(1) for the second draw. This means that if the first card chosen is red, there will be 6 red cards and 3 blue cards left in the box. The corresponding probabilities will be

If the first card chosen is blue, there will be 7 red cards and 2 blue cards left in the box.

Therefore the complete tree diagram is

For the second question, let us take note that the two cards chosen must have different colors. The possible combinations of the cards will be

Let us look at the tree diagram for the two possible combinations of the cards.

Using the tree diagram, we can now multiply the corresponding probabilities.

Since there are two possible combinations (cases), let us add the probabilities.

Therefore, the probability that the two card have different colours is 42/90 or 7/15.

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Average Speed

noreply@blogger.com (Mathrealm) — Fri, 27 Feb 2015 13:19:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 23 of October/November 2010.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Algebra
(b) Specific topic: Average Speed

(c) Formula Needed:

Let us identify the given in the problem:

Take note that the given time is in minutes and seconds. The problem requires us to convert it into hours. Using the conversion factors,

We can now find the average speed by dividing the total distance taken and the total time spent.

Therefore, the average speed of Priyantha is 10.8 km/h or 10 70/83 km/h.

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Area of a Triangle using Trigonometry

noreply@blogger.com (Mathrealm) — Thu, 26 Feb 2015 15:55:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 43 of May/June 2014.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Trigonometry
(b) Specific topic: Area of a Triangle

(c) Formula Needed:

Let us take note that the area of the triangle is already given and the measure of one of the angles is missing. Using the formula above let us assign the parts of the triangle.

Substitute the given measurements in the formula.

Let us leave sin PQR on one side of the equation by dividing both sides by (1/2)(12)(17).

Simplify, then find the measure of angle PQR.

Therefore, the measure of angle PQR is 72.0 degrees.

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Bearing

noreply@blogger.com (Mathrealm) — Sun, 22 Feb 2015 04:48:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 32 of October/November 2012.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Trigonometry
(b) Specific topic: Bearing

Since the bearing is from B, let us draw/highlight the north line on B.

The north line is the initial (starting) side of the angle for the bearing. Now, let us also identify the terminal (end) side of the angle. Just simply connect points A and B.

With the north line on B as the initial side and the line connecting A and B as the terminal side, we can now determine the angle of the bearing.

To solve for this. Let us take note that the given angle is on point A. Let us also take note that the north lines drawn from each point are parallel.

With the north lines parallel, the line connecting the two points A and B will serve as a transversal.

On parallel lines cut by a transversal, the interior angles on the same side of the transversal are supplementary.

It means that the measure of the other interior angle is 180 - 72 = 108 degrees.

Take note that the angle measuring 108 degrees and the bearing of A from B forms one whole revolution, which means that their sum is 360 degrees.

We can now find the bearing of A from B by subtracting 108 degrees from 360 degrees. That is 360 - 108 = 252 degrees.

Therefore, the bearing of A from B is 252 degrees.

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Completing the Square

noreply@blogger.com (Mathrealm) — Sat, 21 Feb 2015 03:45:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 41 of October/November 2014.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Algebra
(b) Specific topic: Quadratic Equations

Our goal is to make the left side of the equation to be of the form (x - p)^2 - q. We cannot use factoring for this quadratic since it is not a perfect square trinomial.

Let us then use completing the square:

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Volume of a Cylinder

noreply@blogger.com (Mathrealm) — Thu, 19 Feb 2015 12:27:00 +0000

This item is taken from IGCSE Mathematics (0580) Paper 31 of May/June 2012.

To deal with this item, let us summarize the details of the problem:
(a) Syllabus area: Geometry/Mensuration
(b) Specific topic: Volume of a Cynlinder
(c) Formula Needed:

Using the given in the problem and the formula for the volume of a cylinder:

Therefore, the volume of the cylinder is 226 cubic centimetre.

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