Categories

# ID#:080600100003

COMPLETING THE SQUARE

• A quadratic expression is one of the form aX² + bX + c, where a, b & c are constants.
• We may not always be able to factorize a quadratic expression, thus, in order to simplify it, we write it in the form, a(X + h) ² + k or k- a(X +h) ²
• (X + h) ² = X² + 2hX + h², this expression is a perfect square.

STRATEGY FOR SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE.

1. Make the coefficient of X² unity or 1.
2. Add half the coefficient of X all squared.
3. Because we have added this amount in step 2, we will have to subtract it at the end of the equation.
4. We can now solve for X since the equation had been simplified.

EXAMPLE 1): X² + 6X + 5 = 0, solve this equation by completing the square.

Step 1): The coefficient of X² is already 1.

Step 2): The coefficient of X is 6, we need to all half of 6 all squared, that is, (3) ².

The equation now becomes: X² + 6X + (3) ² + 5 = 0

Step 3): Because we have added (3) ², we need to subtract it back at the end of the equation. The equation now becomes:

X² + 6X + (3) ² + 5 – (3) ² = 0

• (X + 3) ² + 5 – 9 = 0
• (X + 3) ² – 4 = 0

Step 4): Solving for X, from the above equation,

(X + 3) ² – 4 = 0

• (X + 3) ² = 4
• (X + 3) = ± 4
• (X + 3) = ±2
• X = ±2 – 3
• X = 2- 3 = -1 OR  X = -2 -3 = -5

EXAMPLE 2): X² – 3X – 6 = 0, solve this equation by completing the square.

Step 1): The coefficient of X² is already 1.

Step 2): The coefficient of X is -3. We need to add half of -3 all squared, that is, (-3/2) ².

The equation now becomes: X² – 3X + (-3/2) ² -6 = 0

Step 3): Because we have added (-3/2) ², we need to subtract it back at the end of the equation. The equation now becomes:

X² – 3X + (-3/2) ² – 6 – (-3/2) ² = 0

=>(X – 3/2) ² – 6 – (9/4) = 0

=> (X – 3/2) ² – 8¼ = 0

Step 4): Solving for X from the above equation:

(X – 3/2) ² – 8¼ = 0

=>(X – 3/2) ² = 8¼

=> (X – 3/2) = ±

=> X = ±8¼ + 3/2

=> X = +8¼ + 3/2 OR X= –8¼ + 3/2

EXAMPLE 3): 2X² + 3X – 2 = 0, solve this equation by completing the square.

NOTE: In this equation, the coefficient of X² 1

Step 1):  2X² + 3X – 2 = 0, becomes, [2X² + 3X] – 2 = 0

We now have to make the coefficient on X², 1. Thus, the equation becomes:

2[X² + 3/2X] -2 = 0

Step 2): The coefficient of X = 3/2.

We are required to add half of 3/2, squared = (3/2 x ½)²

= (¾)²

Therefore, the equation now becomes: 2[X² + 3/2X + (¾)²] -2 = 0

=> 2[X + ¾]² -2 = 0

Step 3): Because we have added (¾)², we now have 2 subtract this back at the end of the

equation, but one must however keep in mind that (¾)² has to be multiplied by 2,

since when it was added, it is being multiplied by 2. The equation now becomes:

2[X + ¾]² -2 – 2(¾)² = 0

=>2[X + ¾]² – 25/8 = 0

Step 4): Solving the equation: 2[X + ¾]² – 25/8 = 0

=>2[X + ¾]² = 25/8

=> [X + ¾]² = 25/8 x ½

=> [X + ¾]² = 25/16

=> [X + ¾] = ± 25/16

=> [X + ¾] = ±5/4

=> X = ±5/4 – ¾

=> X = 5/4 – ¾ = ½ OR X = -5/4 – ¾ = -2

PRACTICE QUESTIONS

Express each of the following in the form a(X +h) ² + k OR k-a(X +h) ²

1. X² + 6X + 5
2. X² – 8X – 3
3. 3X² + 4X -5

4) X² – 9X -1

1. -2X² +5X +7

Solve the following equations, using the method of completion of squares.

1. X² + 9X + 10 = 0
2. 4X² + 7X -5 = 0
3. 3X² + 2X -2 = 0

Source: Caribbean Secondary Education Certificate Mathematics

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# ID#:080601000004

SOLVING  SIMULTANEOUS  EQUATIONS

• A linear equation is one of the form: Y = aX + b, where a and b are constants and X and Y are variables.
• Two linear equations can be solved to find values of X and Y, which satisfy both equations simultaneously or at the same time.

• There are 3 basic methods of solving simultaneous equations (excluding matrix method):

1. The Method Of Elimination
2. The Method Of Substitution
3. Graphically.

THE METHOD OF ELIMINATION

In this method, we use the two given equations to produce a new equation containing one variable only.

EXAMPLE 1)

Suppose we wanted to find the two numbers X and Y whose sum is 12 and whose difference is 2. We may form the equations:

X + Y = 12 ………….. (1)

X – Y = 2 …………… (2)

We want to solve (1) and (2) simultaneously.

From these two equations, we want to form one equation in one variable only, either X or Y. If we add equation (1) and equation (2), we will get:

2X = 14 (one equation in only one variable)

Therefore, X = 7.

Now, having found the value of one variable, X, we may now substitute this is either equation (1) or equation (2) and obtain the other variable.

Putting X = 7 in equation (1):

7 + Y = 12

Therefore, Y = 12 – 7

= 5

Solution: X = 7 and Y = 5

• In the above example the coefficient of Y in equation (1) was 1 and in equation (2) was -1. Therefore when we added these two equations the term in y was eliminated. We could have also eliminated the term in X by subtracting the two equations.

• In the elimination method, therefore we perform operations on the two equations so that the coefficient of the term to be eliminated will be the same in both equations or one coefficient will be opposite of the other, that is, if one is 2, the other will b -2.

EXAMPLE 2)

Solve the pairs of equations:

2X + Y = 1 ………….(1)

3X + 4Y = -6 ………..(2)

Let’s try to eliminate X. The coefficient of X in equation (1) is 2 and the coefficient of X in equation (2) is 3.

If we multiply equation (1) by 3 and equation (2) by 2, then the coefficient of X in both equations will be 6. We therefore have two new equivalent equations.

Equation (1) becomes: 6X + 3Y = 3 ……….. (3)

Equation (2) becomes: 6X + 8Y = -12 ……… (4)

Subtracting equation (3) from equation (4) will give:

-5Y = 15

Therefore Y = -3

Substituting for y in equation (1), we get:

2X – 3 =1

2X = 4

X = 2.

Solution: X = 2 and Y = -3

METHOD OF SUBSTITUTION

Step 1) Make one of the variables the subject of the linear equation.

Step 2) Substitute this into the second equation.

Step 3) The result is an equation in either X or Y, solve this equation.

Step 4) Substitute this solution into the first linear equation to find the value of the second                                                                                                                variable.

EXAMPLE 3)

Using the method of substitution, solve the following equations simultaneously:

2X + Y = 1 …………… (1)

3X + 4Y = -6 ………….. (2)

From Equation (1), making Y the subject of the formula, Y = 1 – 2X ………….(3)

By substituting equation (3) into equation (2), we get an equation in X only, that is:

3X + 4(1 – 2X) = -6

3X + 4 – 8X = -6

-5X = -10

Hence, X = 2

We will now substitute X = 2 into equation (3), this gives:

Y = 1 – 2(2)

= 1 – 4

= -3.

Solution: X = 2, Y = -3

• Observe that in example 2) and example 3) we are given the same equation to solve using two different methods and we get the same result using either method.
• Note that in example 3), both equations are of a linear form but example 4) we will examine a question where one equation is of a linear form and the other is a quadratic.

EXAMPLE 4)

Solve the simultaneous equations:

3X – Y = 1 …………….. (1)

X² + 2XY = Y² – 1 …….. (2)

Since equation (1) is the linear equation, we will transpose for Y in equation (1).

Y = 3X – 1 …………… (3)

Now, substituting equation (3) into equation (2):

X² + 2X (3X – 1) = (3X – 1) ² – 1

X² + 6X² – 2X = 9X² – 6X + 1 – 1

7X² – 2X = 9X² – 6X

2X² – 4X = 0

2X (X – 2) = 0

Hence, X = 0 or X = 2

We are now required to substitute each of these values of X into equation (3) to solve for Y.

Substituting X = 0, Y = 3(0) – 1

Y = -1

Substituting X = 2, Y = 3(2) – 1

= 6 – 1

= 5

Hence, when X = 0, Y = -1 and when X = 2, Y = 5.

Solution: X = 0, Y = -1, or X = 2, Y = 5.

GRAPHICAL METHOD

Step 1) For each equation given, using the domain given, find corresponding values for the range, this can be done by drawing a simple table and will be explained further in the examples.

Step 2) Using graph paper, plot each point in the domain with its’ corresponding value in the range, (this is again done for each equation given.)

Step 3) From the graph write down the point or points of intersection. This is the solution to the simultaneous equations.

EXAMPLE 5)

Using the graphical method, solve the following simultaneous equations, (-2 ≤ X ≤ 4) :

2X + Y = 1 …………… (1)

3X + 4Y = -6 ………… (2)

Making Y the subject of the formula in both equations (1) and (2):

Equation (1) becomes: Y = 1 – 2X …………. (3)

Equation (2) becomes: 4Y = -6 – 3X

Y = ¼ (-6 – 3X) ………….. (4)

Using Equation (3), Y = 1 – 2X

 X -2 -1 0 1 2 3 4 -2X -2 x -2 = 4 -2 x -1 = 2 -2 x 0 = 0 -2 x 1 = -2 -2 x 2 = -4 -2 x 3 = -6 -2 x 4 = -8 1 – 2X 1 + 4 = 5 1 + 2 =   3 1 – 0 =    1 1 – 2 =   -1 1 – 4 =       -3 1 – 6 = -5 1 – 8 = -7 Y 5 3 1 -1 -3 -5 -7

Therefore from equation (3) we obtain the points:

(-2, 5), (-1, 3), (0, 1), (1,-1), (2, -3), (3, -5), (4, -7)

Now, using equation (4), Y = ¼ (-6 – 3X)

 X -2 -1 0 1 2 3 4 -3X -3 x -2 = 6 -3 x -1 = 3 -3 x 0 = 0 -3 x 1 = -3 -3 x 2 = -6 -3 x 3 = -9 -3 x 4 = -12 -6 – 3X -6 + 6 = 0 -6 +3 = -3 -6 – 0 = -6 -6 – 3 = -9 -6 – 6 = -12 -6 – 9 = -15 -6 – 12 = -18 ¼  x (-6– 3X) ¼ x  0 = 0 ¼ x -3 = -¾ ¼ x -6 = – 1½ ¼ x -9 = -2¼ ¼ x -12= -3 ¼ x -15 =  –3¾ ¼ x -18 = –4½ Y 0 -¾ – 1½ -2¼ -3 -3¾ -4½

Therefore from equation (4), we obtain the following points:

(-2, 0), (-1, -¾), (0, – 1½), (1, -2¼), (2, -3), (3, -3¾), (4, -4½)

We now need to plot the two sets of points obtained from equation (3) and equation (4) on one graph, using graph paper!

From the graph, we see that the two lines intersect at the point (2, -3). This is the solutions to these simultaneous equations.

QUESTIONS

1. X + 2Y = 8                                              2)               6X + 5Y -7 = 0

3X + Y = 9                                                                4X – 3Y + 35 = 0

3)               X – Y = 2                                                4)               3X – Y = 7

X² – Y² = 8                                                               3XY + Y² = 4

USING SIMULTANEOUS EQUATIONS TO SOLVE WORD PROBLEMS.

EXAMPLE 6)

The sum of two numbers is 6 and the difference of their squares is 12. Find the       numbers.

• Let: The first number be X and the second number be Y

• If the sum of the two numbers is 6, this would imply that: X + Y = 6 …….. (1)

• The square of X is X² and similarly, the square of Y is Y².

• If the difference of their squares is 12, this would imply that X² – Y² = 12….. (2)

We have therefore derived two equations, equation (1) and equation (2) which we can solve simultaneously using the method of substitution, since equation (1) is of a linear form and equation (2) is a quadratic.

QUESTIONS

1. Find the dimensions of a rectangle if its’ perimeter is 22cm and its area is 24cm².
2. Two squares have a total area of 244cm² and the sum of their perimeters is 88cm. Calculate the lengths of their sides.

HOMEWORK

1. Using the method of elimination, solve the following simultaneous equations:

3X + 4Y = -5 ………… (1)

2X – 3Y = 8 …………. (2)

1. Using the graphical method for solving simultaneous equations, solve the following pair of equations:

X + Y = 5 …………….. (1)

2X + Y = 9 …………… (2)

Sources: 1) Additional Mathematics, Pure and Applied

Authors: JF TALBERT & HH HENG

2) Caribbean Examinations Council, Caribbean Secondary Education Certificate

Categories

## Ratio (CXC CSEC Math Study Guide)

RATIO

• Ratios compare quantities of the same kind, for example, 4kg : 7kg.
• The given units may be different, for example, 1cm and 5km but the quantities are of the same kind, for it is possible to express km in cm.

EXAMPLE: Suppose the prices of two chairs are \$60 and \$80. The ratio of the prices is 60 : 80, that is , “sixty to eighty”.

• Ratios behave in the same way as fractions, for example:

60/80 = 30/40 = 120/160 = 3/4, and 60 : 80 = 30 : 40 = 120 : 160 = 3 : 4

Therefore, both parts of a ratio may be multiplied or divided by the same number.

It is usual to express ratios as fractions in their lowest terms.

EXAMPLE: Express the ratio 8cm to 3.5cm as simply as possible.

8cm : 3.5cm = 8 : 3.5

= 2 x 8 : 2 x 3.5

= 16 : 7

• Notice, we do not give units in a ratio.

EXAMPLE: Express the following ratio, 96¢ : \$1.20 = 96¢  = 96  = 96 ÷ 24 =   4

—-     —–    ———-   —-

120¢    120    120 ÷ 24    5

Therefore, the ratio is 4:5.

• Notice, quantities must be the same units before they can be given in a ratio.

FINDING TWO RATIOS THAT ARE EQUAL.

EXAMPLE:  2 : 7 = ⁯ : 28, what is ⁯?

Let: ⁯ = a ==> 2 :7 = a: 28    OR    2/7 = a/28 = a /(7 x 4)

Therefore, a = 2 x 4 = 8.

⁯ = 8.

SHARING

To divide a quantity into 2 parts, which are in the ratio a : b, first divide it into (a + b) equal shares. The required parts will then be “a” shares and “b” shares.

EXAMPLE: Two boys share 35 oranges in the ratio 2:3. How many oranges does each boy get?

2 : 3 means 1 boy gets 2 shares and the other gets 3 shares. Therefore, there are 5 shares altogether.

Thus, the number of oranges in 5 shares = 35

The number of oranges in 1 share = 35 /5 = 7

The number of oranges in 2 shares = 2 x 7 = 14

The number of oranges in 3 shares = 3 x 7 = 21

è 1 boy gets 14 oranges & the other gets 21 oranges.

Source:

Mr. C. Layne & Mr. M. Caine & Ms. M. Maxwell& Ms. M. Griffith (2004)
Caribbean Secondary Education Certificate (CSEC)

Categories

# ID#:080700100001

Using the method of completing the square, the following quadratic was derived and this can be used to solve any quadratic equation. Let us recall that a quadratic form is of the form:

aX² + bX + c = 0

The solution of the equation is:

X = – b ± √ (b² – 4ac)

———————

2a

Example 1): Using the equation, solve: 2X² – 5X + 3 = 0

In this example, a = the coefficient of X² = 2

b = the coefficient of X = -5

c = the constant = 3

Using the quadratic formula: X = – (-5) ± √ [(-5) ² – 4(2) (3)]

———————————

2(2)

= + 5 ± √ [25 – 24]

——————–

4

= +5 ± √ 1

———–

4

= +5 ± 1

———

4

Therefore, X = (5+1)/4 = 6/4 =1½

OR X = (5-1)/4 = 4/4 = 1

Example 2): Solve: 3X² + 5X – 4 = 0

In this example, a = 3, b = 5, c = – 4

Using the quadratic formula: X = -5 ± √ [(5)² -4(3) (-4)]

—————————-

2(3)

= -5 ± √ [25 + 48]

——————–

2(3)

= -5 ± √73

————-

6

•        -5 + √73                                                       -5 – √73

X = ———— = 0.0591         OR             X =     ———- = -2.26

6                                                                   6

QUESTIONS

1. 2X² – 4X = 5
2. 3X² – 6X + 2 = 0
3. 12 = 5X + 2X²
4. 5X( X + 1) – 2X(2X – 1) = 20

HOMEWORK

1. 2X² – 7X -12 = 3
2. 5X² – 4X – 1= 0
3. 10X² + 1 = -10X
Categories

# ID#:080700100002

CONSUMER ARITHMETIC

PERCENTAGES

Let us recall:

1. The meaning of percent – Percent means per hundred (or out of 100)

EXAMPLE:  30% = 30/100

123% = 123/100

1.25% = 1.25/100

1. Fractions with a denominator of 100 are called percentages –

EXAMPLE: 25/100 = 25%

10/100 = 10%

1. To convert a fraction to a percentage, multiply the fraction by 100% –

EXAMPLE: 4/5 = 4/5 × 100% = 80%

11/20 = 11/20 × 100% = 55%

1. To convert a decimal to a percentage, multiply the decimal by 100%-

EXAMPLE: 0.45 = 0.45 × 100% = 45%

0.059 = 0.059 × 100% = 5.9%

1. To find the percentage of a quantity, express the percentage as a fraction and

multiply by the number-

EXAMPLE: 30% of 42 = 30/100 × 42 = 12.6

12.5% of 120 = 12.5/100 × 120 = 15

1. To express one number as a percentage of another, write the number as a fraction

and then convert the fraction to a percentage:

EXAMPLE: 14 as a percentage of 35 = 14/35 × 100 = 40%

90 as a percentage of 180 = 90/180 × 100 = 50%

PROFIT AND LOSS

• When a buyer buys goods, the price he pays for the article is called the BUYING PRICE.
• The price for which the article is sold is called the SELLING PRICE.
• If the selling price is greater than the cost price, then a profit is made.

PROFIT = SELLING PRICE – COST PRICE

• The percentage profit is always calculated on the cost price,

% PROFIT = PROFIT / COST PRICE × 100

• If the selling price is less than the cost price then a loss is made.

LOSS = COST PRICE – SELLING PRICE

• The loss percent is always calculated on the cost price,

% LOSS = LOSS / COST PRICE × 100

EXAMPLE 1):

An article which cost \$45.00 was later sold for \$48.50. Express the profit made as a percentage.

Profit = \$48.50 – \$45.00

= \$ 8.50

Recall: % PROFIT = PROFIT / COST PRICE × 100

% Profit = 3.50 / 45.00 × 100

= 7.5%

EXAMPLE 2):

A table which originally cost \$245.73 was sold for \$168.25. Express the loss made as a percentage of the cost price.

Loss = \$245.73 – \$ 168.25

= \$77.48

Recall: % LOSS = LOSS / COST PRICE × 100

% Loss = 77.48 / 245.73 × 100

= 31.5%

EXAMPLE 3):

An article which was bought for \$225.00 was sold at a profit of 9%. For how much was the article sold?

In this question, Cost Price = \$225.00 and Profit = 9%

Recall: % PROFIT = PROFIT / COST PRICE × 100

Therefore: 9 = profit / 225.00 × 100

9/100 = profit / 225

Profit = 9/100 × 225

= \$20.25

EXAMPLE 4):

By selling an article for \$95.34, a profit of 15% was made on the cost price. What is the cost of the article?

Remember: the selling price = the cost price + profit

We have to let the cost price = 100% which would mean that the selling price = 115%

In other words, \$95.34 represents 115% of the cost price

• \$95.34 = 115/100 × cost price
• cost price = \$95.34 × 100/115
• cost price = \$82.90

REVIEW QUESTIONS

1. A local village shop-keeper buys cans of baked beans from the wholesaler at \$1.50 per can. He sells them at \$1.75 per can. Calculate his profit percent.
2. A table costs \$750.00 to make but because of a flaw it was sold for \$675.00. Calculate the loss percentage.
3. A supermarket advertises month end specials, milk which cost \$1.50 is sold for \$1.45, drink mix which cost \$2.45 is sold for \$2.30 and tea bags which cost \$2.80 is sold for \$2.59.Calculate total loss as a percentage of total cost.
4. Soya milk was sold for \$4.69, representing a loss of 5%. What is the cost price of the milk?

DISCOUNT

Discount means the “money off” goods or services. Sometimes you are offered discounts during sales on the purchase of a large quantity of articles or as an incentive for customers who pay cash for certain articles. The discount is the amount deducted from the cost price.

SALES TAX

In the tax structure of many countries, a tax is added to goods and services at the point of purchase. This tax is called that sales tax or value added tax (VAT). Most stores include the sales tax as part of the marked price on the article.

MARKED PRICE = SELLING PRICE + THE TAX ADDED

EXAMPLE 5):

Max Furniture Store has a store-wide sale on the last Saturday of the month. There is a 20% discount on all furniture and a 15% discount on all large appliances, Calculate:

1. The selling price of a living room suite which is marked at \$2389.00
2. The selling price of a refrigerator which has a marked price of \$1253.79

SOLUTION

1. Amount of discount = 20% of \$2389.00

= 20/100 × \$2389.00

= \$477.80

Selling Price = \$2389 – \$477.80

= \$1911.20

1. Amount of discount on refrigerator = 15% of \$1253.79

= 15/100 × \$1253.79

= \$188.07

Selling Price = \$1253.79 – \$188.07

= \$1065.72

EXAMPLE 6): The marked price on a colour T.V set was \$783.20 less 5% discount for cash. How much money would you save by buying the T.V set for cash?

Discount = 5% of \$783.20

= 5/100 × \$783.20

= \$39.16

EXAMPLE 7): A table has a marked price of \$800.00 including the sales tax of 15%. A cash discount of 12% is given on the price of the table. How much will be paid for the table if it is bought for cash?

The marked price of the table = the selling price + the sales tax

Therefore \$800.00 = 115% of the selling price,

Actual selling price = 800 × 100/115

= \$695.65

Discount of 12% of \$695.65 = \$695.65 × 12/100

= \$83.48

Discounted price of table = \$695.65 – \$83.48

= \$ 612.17

New price of table = \$ 612.17 + sales tax

Sales Tax = 15% of \$ 612.17

= \$91.81

New price of table = \$ 612.17 + \$91.81

= \$703.98

REVIEW QUESTIONS

1) What is the marked price on an article which costs \$234.56 after the sales tax of 15% added?

2) The regular price on a Munch bar is \$1.25. At a sale, it was marked at \$1.10. Calculate the discount as a percentage of the regular price?

3) Edgar’s Hardware store offers a 5% discount on purchases over \$250.00. What is the discount on a bill totaling \$956.72?

HIRE PURCHASES AND MORTGAGES

The method of buying an article by paying a deposit and then paying the rest of the money by equal monthly or weekly payments or installments over some fixed period is called Hire Purchase.

EXAMPLE 8): Morty wants to buy a new stove. The stove costs \$1199.00 cash. Morty does not have this amount of money. At the Affordable Appliance Store, Morty can buy the stove using two methods –

1. Pay the \$1199.00 cash, or
2. Pay a deposit of \$200.00 and then \$54.00 every month for two years.

Since Morty does not have the money he will have to pay using the second method.

Morty pays a deposit of \$200.00 plus \$54.00 every month for two years.

The total payment will be: \$200 + (\$54.00 × 24)

= \$200 + \$1296

= \$1496.00

Thus, Morty pays an additional \$297.00 for the stove.

EXAMPLE 9): A CD player costs \$1079.85 cash. Using hire purchase, the player may be obtained by making a deposit of 20% of the cash price and then paying an interest of 15% on the outstanding balance. How much money is paid for the CD player? If the CD player is to be paid for over 12 months with equal payments, how much is each payment?

Deposit paid = 20% × \$1079.85

= \$215.97

Amount outstanding = \$1079.85 – \$215.97

= \$ 863.88

Interest = 15% of \$ 863.88

= \$129.58

Total hire purchase = \$215.97 + \$ 863.88 + \$129.58

= \$1209.43

Monthly payments = \$(863.88 + 129.58)/12

= \$82.79

Mortgage: This is the amount of money borrowed or lent against the security of a property. It is usually used in terms of buying houses, land or other property in which payments are made over a long period of time. A person may borrow a certain amount of money to buy or build a house from a bank or other lending agency. A deposit is required on the loan; the balance is repaid by equal payments over a fixed period. The payments usually include interest and principal repayments.

EXAMPLE 10): Mr. Davies wants to buy a house which cost \$175 000. The bank will lend him 90% of this amount to be repaid over 20 years with an interest of 9.5%. Calculate the amount of the loan repaid at the end of three years. The yearly mortgage payments are \$19 858.42.

The payment of \$19 858.42 includes the interest on the loan. Since a deposit of 10% of the value of the house was made, the loan was 90% of \$175 000 = \$157 500

Payments for year 1:

Amount outstanding = \$157 500

Interest owed = 9.5% of \$157 500

= \$14 962.50

Payment made = \$19 858.42

Amount paid to loan = \$19 858.42 – \$14 962.50

= \$4895.92

Amount outstanding = \$157 500 – \$4895.92

= \$152 604.08

Payments for year 2:

Amount outstanding = \$152 604.08

Interest owed = 9.5% of \$152 604.08

= \$14 497.39

Payment made = \$19 858.42

Amount paid to loan = \$19 858.42 – \$14 497.39

= \$5361.03

Amount outstanding = \$152 604.08 – \$5361.03

= \$ 147 243.05

Payments for year 3:

Amount outstanding = \$ 147 243.05

Interest owed = 9.5% of \$147 243.05

= \$13 988.09

Payment made = \$19 858.42

Amount paid to loan = \$19 858.42 – \$13 988.09

= \$5870.33

Amount outstanding = \$ 147 243.05 – \$5870.33

= \$141 372.72

Tabulating the information we get:

 Amt. Outstanding Payment made Interest paid Principal repaid \$ 157 500 \$ 19 858.42 \$14 962.50 \$4895.92 \$ 152 604.08 \$ 19 858.42 \$ 14 497.39 \$5361.03 \$ 147 243.05 \$ 19 858.42 \$ 13 988.09 \$5870.33 \$ 141 372.72

At the end of three years, \$5870.33 has been paid towards the loan. The remaining amount of money paid has been paid as interest.

WAGES AND SALARIES

• Payment by the hour: Many people who work in factories, in the building and construction industry are paid by the hour. Often there is a basis wage for a certain number of hours, called a basic week.
• Overtime: Any number of hours worked beyond the basic week is paid at a different rate than the basic rate. These extra hours are referred to over-time. Overtime rate are calculated as:

Time and a quarter = 1.25 times the basic rate

Time and a half = 1.5 times the basic rate

Double time = twice the basic rate.

The total wages for the week are then calculated from the basic wages plus the over time wages.

• Commission: Most sales people are paid a commission in addition to their basic wage. A commission is a percentage of their total value of the goods they have sold, or the sales of the company. Sometimes, this commission is paid at intervals and is called a bonus.

EXAMPLE 11): A labourer is paid \$297.50 for a basic 35 hour work week. What is the hourly rate paid?

Hourly rate paid = \$297.50 / 35

= \$8.50

EXAMPLE 12): A road cleaner works a 40 hour week for a basic wage of \$270. He also works 4 hours overtime at time and a half and 3 hours overtime at double time. Calculate his total wages for the week.

Basic wage = \$270

Basic week = 40 hours

Basic rate = 270/40

= \$6.75

Time and a half = \$6.75 × 1.5

= \$10.125

4 hours at time and a half = 4 × \$10.125

= \$40.50

Double time = \$6.75 × 2

= \$13.50

3 hours at double time = 3 × \$13.50

= \$40.50

Total wages = \$270 + \$40.50 + \$40.50

= \$ 351.00

EXAMPLE 13): A salesman is paid a basic wage of \$250 per week plus 1½% of the total sales of computers that he made during the week. If his total sales for last week were \$16 550 what was his gross wages?

Gross wages = basic wage + commission

= \$250 + (\$16 550 × 0.015)

= \$250 + \$248.25

= \$498.25

REVIEW QUESTIONS

1. A man’s basic wage for a 40 hour work week is \$225. Calculate his total wages for the week that he works an additional 5 hours overtime at time and a half.
2. A taxi company pays its drivers a basic wage of \$150 per week plus a 25% commission on all jobs done for the week. Calculate the total wages for a driver whose jobs total \$2765 in a given week.
3. A man is paid a basic hourly rate of \$5.60 for a 40 hour week. On weekends, he is paid overtime at time and a half. If his wages for a given week was \$308, how many hours of overtime did he work?

Mr. C. Layne & Mr. M. Caine & Ms. M. Maxwell& Ms. M. Griffith (2004)
Caribbean Secondary Education Certificate (CSEC)

Categories

## Arithmetic Laws (CXC CSEC Math Study Guide)

ID#: 080400100005

NUMBER SYSTEMS

• Numbers are the building blocks of mathematics.
• Without them we will not have a means of representing the interesting “stuff” that mathematicians discover, for example that the earth’s gravity is 9.8N.
• Clearly numbers are important. As such they are categorized for easier usage. The basic groups are provided below. The initial letter is used to represent the group. Following is a short example of the numbers in the specific groups:

• The natural numbers are the counting numbers:
• N = { 1, 2, 3, 4, . .  .}

• This means that the set of numbers 1, 2, 3… is the set of numbers called the Natural numbers which is referred to as just N.

• The whole numbers are the natural numbers including 0.
• W= { 0, 1, 2, 3, . . . }

• The integers consist of the whole numbers and their negatives:
• Z = {. . . , -3, -2, -1, 0, 1, 2, 3, . . . }

• These include the numbers smaller than zero (0). The number before the zero can also be called “negative whole numbers,” but it is more accurate to say negative integers.

Table showing the different groups of numbers.

 Numbers Name and Symbol 0, 1, 2, 3, 4, 5, … Whole Numbers    W 1, 2, 3, 4, 5, … Natural Numbers  N … -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, … Integers              Z

http://www.mathsisfun.com/whole-numbers.html

http://www.ttpti.com (Navigate to “Study Guides”)

ARITHMETIC LAWS

• Commutative Law: states that the order of the elements when added or multiplied does not matter since the result will be the same.

• Illustration:               (5 + 2) = (2 + 5) = 7
• (5 × 2) = (2 × 5) = 10

• Subtraction is NOT commutative since 4 – 2 2 – 4.
• Division is NOT commutative since 4 ÷ 2 2 ÷ 4.

• Associative Law: states that the order in which the brackets are placed does NOT matter.

• Illustration:    (2 + 3) + 4 = 5 + 4 = 9
• 2 + (3 + 4) = 2 + 7 = 9

• (2 × 3) × 4 = 6 × 4 = 24
• 2 × (3 × 4) = 2 ×12 = 24

• Distributive Law: This law works with multiplication and addition as follows:
• For example;               2 × (3 + 4)

• 2 × (3 + 4) = 2 × (7) = 14
• However, (2 × 3) + (2 × 4) = 6 + 8 = 14.

• We can distribute the terms. (i.e. multiplication distributes over addition).

• Another example.

• Evaluate 2 + (5 × 4).

• We work out brackets first. This gives 2 + 20 = 22

• Let’s use the distributive law. We get (2 + 5) × (2 + 4) = 7 × 6 = 42

• Therefore, the law only holds in cases with the following array a × (b + c) where a, b and c can be any number.
Categories

## Matrices (CXC CSEC Math Study Guide)

ID#: 080400100007

Matrices

• MATRIX – An array of elements in rows and columns is called a matrix.

• A matrix looks like:               a  b  c            where all the letter are scalars or numbers. This is a 3×3 matrix since                d  e  f            there are three rows and three columns.

g  h  i

Determinant

Consider the following demonstration.

• Given a 2×2 matrix               a   b              the determinant is found by [ad – bc]

c   d

• eg. Find the determinant of                4  6

1  5

SOLUTION

• We call the matrix A. the determinant of A written as “det A” or |A| is equal to (4×5 – 6×1) = 20 – 6 = 14.
• Hence, |A| = 14.

• Note:
• We can only find the determinant of a square matrix (one which has equal number of rows and columns).
• Determinant of a matrix with a row of zeroes is zero.
• If two columns or rows are equal then the determinant is zero.
• For a triangular matrix, the determinant is the product of the diagonal     elements.

3  6  5

Eg of 3. The determinant of              0  2  3              is 3×2×5 = 30.

0  0  5

Inverse

A-1 = adj A

|A|

• INVERSE MATRIX – found by finding the adjoint of a matrix A and dividing that by the determinant of the matrix.

• To calculate adjoint of matrix follow this procedure:

1. Calculate Minor for each element of the matrix.
2. Form Cofactor matrix from the minors calculated. (+_+)

Form Adjoint from cofactor matrix. (by transposing.)

Categories

## Functions, Relations, Graphs (CXC CSEC Math Study Guide)

ID#: 080400100008

RELATIONS

In everyday life, many quantities depend on one or more changing variables eg:

• plant growth depends on sunlight and rainfall
• speed depends on distance traveled and time taken
• voltage depends on current and resistance
• test marks depend on attitude, listening in lectures and doing tutorials (among many other variables)

 Relation S : X thinks that Y likes Z Relation S : X thinks that Y likes Z Relation S : X thinks that Y likes Z Person X Person Y Person Z Alice Bob Denise Charles Alice Bob Charles Charles Alice Denise Denise Denise

Each row of the Table records a fact or makes an assertion of the form “X thinks that Y likes Z “. For instance, the first row says, in effect, “Alice thinks that Bob likes Denise”. The Table represents a relation S over the set P of people under discussion:

P = {Alice, Bob, Charles, Denise}

The data of the Table are equivalent to the following set of ordered triples:

S = {(Alice, Bob, Denise), (Charles, Alice, Bob), (Charles, Charles, Alice), (Denise, Denise, Denise)}

[LINE GRAPHS]

Line graphs compare two variables. Each variable is plotted along an axis .

A line graph has a vertical axis and a horizontal axis. So, for example, if you wanted to graph the height of a ball after you have thrown it, you could put time along the horizontal, or x-axis, and height along the vertical, or y-axis.

As mentioned before, each type of graph has characteristics that make it useful in certain situations. Some of the strengths of line graphs are that:

1. They are good at showing specific values of data, meaning that given one variable the other can easily be determined.
2. They show trends in data clearly, meaning that they visibly show how one variable is affected by the other as it increases or decreases.

They enable the viewer to make predictions about the results of data not yet recorded.

# [BAR GRAPHS]

Bar graphs are used to display data in a similar way to line graphs. However, rather than using a point on a plane to define a value, a bar graph uses a horizontal or vertical rectangular bar that levels off at the appropriate level.

There are many characteristics of bar graphs that make them useful. Some of these are that:

1. They make comparisons between different variables very easy to see.
2. They clearly show trends in data, meaning that they show how one variable is affected as the other rises or falls.
3. Given one variable, the value of the other can be easily determined.

# [HISTOGRAMS]

When drawing a histogram, the y-axis is labeled ‘relative frequency’ or ‘frequency density’.

Area of bar = frequency x standard width.

Draw a histogram for the following information.

 Height (feet): Frequency Relative Frequency 0-2 0 0 2-4 1 1 4-5 4 8 5-6 8 16 6-8 2 2

[Frequency Polygon]

[FUNCTIONS & RELATIONS]

• A Relation is a set of ordered pairs. It is just a relationship between sets of information.
• A function is a rule that relates how one quantity depends on other quantities. A function is a 1-1 or many-one relation. This means that there is one or a unique output/answer for any given input.

Notation: f : x       x2 ; f(x) = x2. “function f of x”. e.g. f(x) = 2x2 + 5x + 3.

Demonstrated in Arrow Diagrams as;

 This is a function. You can tell by tracing from each x to each y. There is only one y for each x; there is only one arrow coming from each x. This is a function! There is only one arrow coming from each x; there is only one y for each x. It just so happens that it’s always the same y, but it is only one y. So this is a function; just a very boring function! This one is not a function. If you’ll notice, there are two arrows coming from the number 1; the number 1 is associated with two different range elements. So this is a relation, but it is not a function. Okay, this one’s a trick question. Each element of the domain that has a pair in the range is nicely well-behaved. But what about that 16? It is in the domain, but it has no range element that corresponds to it! This is not a function.

Given the relation {(2, 3), (2, –2)}? We already know that this is not a function, since x = 2 goes to both y = 3 and y = –2.

 If we graph this relation, it looks like: Notice that you can draw a vertical line through the two points, like this:

This characteristic of non-functions was codified in The Vertical Line Test: Given the graph of a relation, if you can draw a vertical line that crosses the graph in more than one place, then the relation is not a function

Categories

## Factorization (CXC CSEC Math Study Guide)

ID#: 080400100006

Factorization

[PROCESS]

1. The first term of the trinomial is the product of the first terms of the binomials.
2. The last term of the trinomial is the product of the last terms of the binomials.
3. The coefficient of the middle term of the trinomial is the sum of the last terms of the binomials.

Note: If all the signs in the trinomial are positive, all signs in both binomials are positive.

Example

Factorize:   x2 – 14x – 15

Solution:

1. First, write down two sets of parentheses to indicate the product.

( )    )

1. Since the first term in the trinomial is the product of the first terms of the binomials, enter x as the first term of each binomial

(x     ) (x     )

1. The product of the last terms of the binomials must equal -15, and their sum must equal -14One of the binomials’ terms has to be negative. Four different pairs of factors have a product that equals to -15.

(3)(-5) = -15     (-15)(1) = -15

(-3)(5) = -15     (15)(-1) = -15

However, only one of those pairs has a sum of -14

(-15) + (1) = -14

Therefore, the second terms in the binomial are -15  and 1 because these are the only two factors whose product is -15 (the last term of the trinomial) and whose sum is -14 (the coefficient of the middle term in the trinomial).

(x – 15)(x + 1) is the answer.

Quadratic polynomials can sometimes be factored into two binomials with simple integer coefficients, without the need to use the quadratic formula. In a quadratic equation, this will expose its two roots. The formula

would be factored into:

where

Categories

## Consumer Arithmetic (CXC CSEC Math Study Guide)

CONSUMER ARITHMETIC

PROFIT AND LOSS

• When a buyer buys goods, the price he pays for the article is called the BUYING PRICE.
• The price for which the article is sold is called the SELLING PRICE.
• If the selling price is greater than the cost price, then a profit is made.

PROFIT = SELLING PRICE – COST PRICE

• The percentage profit is always calculated on the cost price,

% PROFIT = PROFIT / COST PRICE × 100

• If the selling price is less than the cost price then a loss is made.

LOSS = COST PRICE – SELLING PRICE

• The loss percent is always calculated on the cost price,

% LOSS = LOSS / COST PRICE × 100

EXAMPLE 1):

An article which cost \$45.00 was later sold for \$48.50. Express the profit made as a percentage.

Profit = \$48.50 – \$45.00

= \$ 8.50

Recall: % PROFIT = PROFIT / COST PRICE × 100

% Profit = 3.50 / 45.00 × 100

= 7.5%

EXAMPLE 2):

A table which originally cost \$245.73 was sold for \$168.25. Express the loss made as a percentage of the cost price.

Loss = \$245.73 – \$ 168.25

= \$77.48

Recall: % LOSS = LOSS / COST PRICE × 100

% Loss = 77.48 / 245.73 × 100

= 31.5%

EXAMPLE 3):

An article which was bought for \$225.00 was sold at a profit of 9%. For how much was the article sold?

In this question, Cost Price = \$225.00 and Profit = 9%

Recall: % PROFIT = PROFIT / COST PRICE × 100

Therefore: 9 = profit / 225.00 × 100

9/100 = profit / 225

Profit = 9/100 × 225

= \$20.25

EXAMPLE 4):

By selling an article for \$95.34, a profit of 15% was made on the cost price. What is the cost of the article?

Remember: the selling price = the cost price + profit

We have to let the cost price = 100% which would mean that the selling price = 115%

In other words, \$95.34 represents 115% of the cost price

• \$95.34 = 115/100 × cost price
• cost price = \$95.34 × 100/115
• cost price = \$82.90

REVIEW QUESTIONS

1. A local village shop-keeper buys cans of baked beans from the wholesaler at \$1.50 per can. He sells them at \$1.75 per can. Calculate his profit percent.
2. A table costs \$750.00 to make but because of a flaw it was sold for \$675.00. Calculate the loss percentage.
3. A supermarket advertises month end specials, milk which cost \$1.50 is sold for \$1.45, drink mix which cost \$2.45 is sold for \$2.30 and tea bags which cost \$2.80 is sold for \$2.59.Calculate total loss as a percentage of total cost.
4. Soya milk was sold for \$4.69, representing a loss of 5%. What is the cost price of the milk?

DISCOUNT

Discount means the “money off” goods or services. Sometimes you are offered discounts during sales on the purchase of a large quantity of articles or as an incentive for customers who pay cash for certain articles. The discount is the amount deducted from the cost price.

SALES TAX

In the tax structure of many countries, a tax is added to goods and services at the point of purchase. This tax is called that sales tax or value added tax (VAT). Most stores include the sales tax as part of the marked price on the article.

MARKED PRICE = SELLING PRICE + THE TAX ADDED

EXAMPLE 5):

Max Furniture Store has a store-wide sale on the last Saturday of the month. There is a 20% discount on all furniture and a 15% discount on all large appliances, Calculate:

1. The selling price of a living room suite which is marked at \$2389.00
2. The selling price of a refrigerator which has a marked price of \$1253.79

SOLUTION

1. Amount of discount = 20% of \$2389.00

= 20/100 × \$2389.00

= \$477.80

Selling Price = \$2389 – \$477.80

= \$1911.20

1. Amount of discount on refrigerator = 15% of \$1253.79

= 15/100 × \$1253.79

= \$188.07

Selling Price = \$1253.79 – \$188.07

= \$1065.72

EXAMPLE 6): The marked price on a colour T.V set was \$783.20 less 5% discount for cash. How much money would you save by buying the T.V set for cash?

Discount = 5% of \$783.20

= 5/100 × \$783.20

= \$39.16

EXAMPLE 7): A table has a marked price of \$800.00 including the sales tax of 15%. A cash discount of 12% is given on the price of the table. How much will be paid for the table if it is bought for cash?

The marked price of the table = the selling price + the sales tax

Therefore \$800.00 = 115% of the selling price,

Actual selling price = 800 × 100/115

= \$695.65

Discount of 12% of \$695.65 = \$695.65 × 12/100

= \$83.48

Discounted price of table = \$695.65 – \$83.48

= \$ 612.17

New price of table = \$ 612.17 + sales tax

Sales Tax = 15% of \$ 612.17

= \$91.81

New price of table = \$ 612.17 + \$91.81

= \$703.98

REVIEW QUESTIONS

1) What is the marked price on an article which costs \$234.56 after the sales tax of 15% added?

2) The regular price on a Munch bar is \$1.25. At a sale, it was marked at \$1.10. Calculate the discount as a percentage of the regular price?

3) Edgar’s Hardware store offers a 5% discount on purchases over \$250.00. What is the discount on a bill totaling \$956.72?

HIRE PURCHASES AND MORTGAGES

The method of buying an article by paying a deposit and then paying the rest of the money by equal monthly or weekly payments or installments over some fixed period is called Hire Purchase.

EXAMPLE 8): Morty wants to buy a new stove. The stove costs \$1199.00 cash. Morty does not have this amount of money. At the Affordable Appliance Store, Morty can buy the stove using two methods –

1. Pay the \$1199.00 cash, or
2. Pay a deposit of \$200.00 and then \$54.00 every month for two years.

Since Morty does not have the money he will have to pay using the second method.

Morty pays a deposit of \$200.00 plus \$54.00 every month for two years.

The total payment will be: \$200 + (\$54.00 × 24)

= \$200 + \$1296

= \$1496.00

Thus, Morty pays an additional \$297.00 for the stove.

EXAMPLE 9): A CD player costs \$1079.85 cash. Using hire purchase, the player may be obtained by making a deposit of 20% of the cash price and then paying an interest of 15% on the outstanding balance. How much money is paid for the CD player? If the CD player is to be paid for over 12 months with equal payments, how much is each payment?

Deposit paid = 20% × \$1079.85

= \$215.97

Amount outstanding = \$1079.85 – \$215.97

= \$ 863.88

Interest = 15% of \$ 863.88

= \$129.58

Total hire purchase = \$215.97 + \$ 863.88 + \$129.58

= \$1209.43

Monthly payments = \$(863.88 + 129.58)/12

= \$82.79

Mortgage: This is the amount of money borrowed or lent against the security of a property. It is usually used in terms of buying houses, land or other property in which payments are made over a long period of time. A person may borrow a certain amount of money to buy or build a house from a bank or other lending agency. A deposit is required on the loan; the balance is repaid by equal payments over a fixed period. The payments usually include interest and principal repayments.

EXAMPLE 10): Mr. Davies wants to buy a house which cost \$175 000. The bank will lend him 90% of this amount to be repaid over 20 years with an interest of 9.5%. Calculate the amount of the loan repaid at the end of three years. The yearly mortgage payments are \$19 858.42.

The payment of \$19 858.42 includes the interest on the loan. Since a deposit of 10% of the value of the house was made, the loan was 90% of \$175 000 = \$157 500

Payments for year 1:

Amount outstanding = \$157 500

Interest owed = 9.5% of \$157 500

= \$14 962.50

Payment made = \$19 858.42

Amount paid to loan = \$19 858.42 – \$14 962.50

= \$4895.92

Amount outstanding = \$157 500 – \$4895.92

= \$152 604.08

Payments for year 2:

Amount outstanding = \$152 604.08

Interest owed = 9.5% of \$152 604.08

= \$14 497.39

Payment made = \$19 858.42

Amount paid to loan = \$19 858.42 – \$14 497.39

= \$5361.03

Amount outstanding = \$152 604.08 – \$5361.03

= \$ 147 243.05

Payments for year 3:

Amount outstanding = \$ 147 243.05

Interest owed = 9.5% of \$147 243.05

= \$13 988.09

Payment made = \$19 858.42

Amount paid to loan = \$19 858.42 – \$13 988.09

= \$5870.33

Amount outstanding = \$ 147 243.05 – \$5870.33

= \$141 372.72

Tabulating the information we get:

 Amt. Outstanding Payment made Interest paid Principal repaid \$ 157 500 \$ 19 858.42 \$14 962.50 \$4895.92 \$ 152 604.08 \$ 19 858.42 \$ 14 497.39 \$5361.03 \$ 147 243.05 \$ 19 858.42 \$ 13 988.09 \$5870.33 \$ 141 372.72

At the end of three years, \$5870.33 has been paid towards the loan. The remaining amount of money paid has been paid as interest.

WAGES AND SALARIES

• Payment by the hour: Many people who work in factories, in the building and construction industry are paid by the hour. Often there is a basis wage for a certain number of hours, called a basic week.
• Overtime: Any number of hours worked beyond the basic week is paid at a different rate than the basic rate. These extra hours are referred to over-time. Overtime rate are calculated as:

Time and a quarter = 1.25 times the basic rate

Time and a half = 1.5 times the basic rate

Double time = twice the basic rate.

The total wages for the week are then calculated from the basic wages plus the over time wages.

• Commission: Most sales people are paid a commission in addition to their basic wage. A commission is a percentage of their total value of the goods they have sold, or the sales of the company. Sometimes, this commission is paid at intervals and is called a bonus.

EXAMPLE 11): A labourer is paid \$297.50 for a basic 35 hour work week. What is the hourly rate paid?

Hourly rate paid = \$297.50 / 35

= \$8.50

EXAMPLE 12): A road cleaner works a 40 hour week for a basic wage of \$270. He also works 4 hours overtime at time and a half and 3 hours overtime at double time. Calculate his total wages for the week.

Basic wage = \$270

Basic week = 40 hours

Basic rate = 270/40

= \$6.75

Time and a half = \$6.75 × 1.5

= \$10.125

4 hours at time and a half = 4 × \$10.125

= \$40.50

Double time = \$6.75 × 2

= \$13.50

3 hours at double time = 3 × \$13.50

= \$40.50

Total wages = \$270 + \$40.50 + \$40.50

= \$ 351.00

EXAMPLE 13): A salesman is paid a basic wage of \$250 per week plus 1½% of the total sales of computers that he made during the week. If his total sales for last week were \$16 550 what was his gross wages?

Gross wages = basic wage + commission

= \$250 + (\$16 550 × 0.015)

= \$250 + \$248.25

= \$498.25

REVIEW QUESTIONS

1. A man’s basic wage for a 40 hour work week is \$225. Calculate his total wages for the week that he works an additional 5 hours overtime at time and a half.
2. A taxi company pays its drivers a basic wage of \$150 per week plus a 25% commission on all jobs done for the week. Calculate the total wages for a driver whose jobs total \$2765 in a given week.
3. A man is paid a basic hourly rate of \$5.60 for a 40 hour week. On weekends, he is paid overtime at time and a half. If his wages for a given week was \$308, how many hours of overtime did he work?

Mr. C. Layne & Mr. M. Caine & Ms. M. Maxwell& Ms. M. Griffith (2004)
Caribbean Secondary Education Certificate (CSEC)