LET_Math

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APPENDIX

MATHEMATICS

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Factors and Multiples

• Factors and Multiples – Given: 5 X 7 = 35; the numbers 5 and 7 are the factors of 35. Since 35 is divisible by 5 and 7, 35 is called a multiple of both 5 and 7. Examples: a. List four factors of 25 Four factors of 25 are: 1 X 25 and 5 X 5 b. List two multiples of 7 Two multiples of 7 are: 2 X 7 = 14 and 8 X 7 = 56 • Prime Number – number whose factors are 1 and itself. Examples: 2, 3, 7, 11, 13, 17, 19, 23, 29, 31, 37, 43, 61, 79, 83, 91... • Composite Number – number with many factors including 1 and itself. Examples: 4, 6, 9, 10, 12, 14, 15, 24, 28, 34, 45, 54, 62, 68, 76, 77 ... • Twin Primes – are pairs of prime numbers whose difference is 2 such as 7 and 5; 13 and 11; 19 and 17. • Divisibility of Numbers – is the property of a whole number that can be divided by another whole number, the quotient of which is a whole number with 0 as a remainder. 7 5 35 5 35 7 35 1 35 35 35 35 0 0 0 35 is divisible by 1, 5, 7 2 3 4 5 6 7 8 9 10 11 Even Sum of the digits is divisible by 3 Last two digits divisible by 4 Numbers end in 0 or 5 Even numbers divisible by 3 Difference between twice the last digit and the other digits divisible by 7 Last 3 digits divisible by 8 Sum of the digits divisible by 9 Numbers end in 0 Difference between sums of alternate digits is 0 or multiples of 11 DIVISIBLE BY CHARACTERISTICS OF THE NUMBERS DIVISIBILITY TESTS

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Copyright 1999 • Prime Factors – a number can be reduced to its prime factors by factoring. Example: 45 = 5 X 9 45 = 5 X 3 X 3 • Greatest Common Factor (GCF) – is the biggest whole number from the sets of factors of two or more numbers. It can divide exactly all the given numbers. Examples: a. Find the GCF of 18 and 24. ÞFactors of 18 are: 1, 2, 3, 6, 9, 18 ÞFactors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24 • The common factors are:1, 2, 3, 6 and the GCF is 6. To find the GCF, express every number as a product of prime factors, then get the prime factor common to all numbers and obtain their product. b. Find the GCF of 12, 24, 36 12 24 36 2 X 6 4 X 6 4 X 9 2 X 3 2 X 2 X 2 X 3 2 X 2 X 3 X 3 12 = 2 X 2 X 3 24 = 2 X 2 X 3 X 2 36 = 2 X 2 X 3 X 3 GCF = 2 X 2 X 3 = 12 • Least Common Multiple (LCM) – the smallest number which is the multiple of 2 or more given numbers. It is the smallest number that can be divided by all the given numbers. To determine the LCM, express each given number as a product of prime fac tors. Then list each factor according to the number of times it appears and find their products. Example: Find the LCM of 24, 36, 48 24 = 2 X 3 X 2 X 2 36 = 2 X 3 X 2 X 3 48 = 2 X 3 X 2 X 2 X 2 LCM = 2 X 3 X 2 X 2 X 3 X 2 = 144 13 3

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Place Value

Place Value means that the value or the number of things for which a digit stands, depends on the place it occupies. The decimal numeral system has only nine digits (nu merals) and zero (0). However, the numbers larger than nine can be expressed using these numerals. Take the number twenty five in using the system of place value, the digits 2 and 5 are needed to represent the number. The first place on the right stands for single things and the second place to the left for groups of ten things. Thus, the first on the right is called the ones place or units place. It is for the numerals from 1 to 9. The second place to the left is called the tens place. It is for the groups of 10 to 90. When you write large numerals like 8, 504, 971, 351, set off every three digits with a comma starting at the right and counting to the left. These groups of three digits set off by commas are called periods. a. Whole Numbers hundredmillions tenmillions millions hundredthousands tenthousands thousands hundreds tens ones 805, 694, 357 b.Decimals .9753246 ten millionths millionths hundredthousandths tenthousands thousandths hundredths tenths

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Copyright 1999 • Roundingoff numbers Rules for roundingoff numbers: 1. If the digit to be dropped is more than 5, add 1 to the significant digit. 2. If the digit to be dropped is less than 5, leave the significant digit unchanged. 3. If the digit to be dropped is 5: a. Add 1 to the significant digit if it is an odd number. b. Leave the significant digit unchanged if it is an even number. 3. Add 1 to the significant digit if there is a digit (except 0) after 5. Examples: 1. Round 7,881 to the nearest hundreds dropping digit significant digit Ans: = 7,900 c. Monetary (Peso) thousandpeso hundredpeso tenpeso onepeso tencentavo centavo tenth of a centavo hundredth of a centavo P 2,467.9853 Examples: 1. 53,841 – Fifty three thousand, eight hundred forty one 2. 0.730 – Seventy three hundredths or seven hundred thirty thousandths 3. 578.09 – Five hundred seventy eight and nine hundredths 4. .10945 – Ten thousand nine hundred fortyfive hundredthousandths 5. .78 – Seventy eight hundredth of a centavo 6. P.2783 – Twentyseven and eightythree hundredth of a centavo 7. P .05 M – Fifty thousand pesos 8. P 3.9 M – Three million, nine hundred thousand pesos 9. P 2.009 M – Two million, nine thousand pesos 10. .35 ctv. – Thirty five hundredth of a centavo 13 5

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2. Round 0.636 to the nearest tenths dropping digit significant digit Ans: = 0.6 3. a. Round 835 to the nearest tens dropping digit significant digit Ans: = 840 b. Round 0.9745 to the nearest thousandths dropping digit significant digit Ans: = 0.974 c. Round P724.53 to the nearest peso droping digit significant digit Ans: = P725.00 • Operations on Decimals 1. Addition Examples: .75 18.3 .478 405.06 .935 8.456 .00356 72.38 .8 .008 2.96656 504.204 2. Subtraction Examples: .8 478.3 .358 56.06 .442 422.24

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Fractions

• Conversions Involving Fractions: 1. Changing Inproper Fractions to Mixed Numbers Rule: Divide the numerator by the denominator and write the remainder as a frac tion. Examples: 28 3 39 3 1 5 5 6 6 2 2. Changing Mixed Numbers to Improper Fractions Rule: Multiply the denominator by the whole number and add the product to the numerator. Write the sum over the same denominator. Examples: 3 59 3 91 7 7 4 4 .21 4 .84 8 4 4 X X X + + 13 7 = 5 ; = 6 = 6 8 = ; 22 =

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3. Reducing Fractions to Lowest Terms Rule: Divide the numerator and denominator of the given fraction by their greatest common factor (GCF). Examples: 21 7 220 44 39 13 365 73 GCF = 3 GCF = 5 4. Raising Fractions to Higher Terms Rule: Multiply the numerator and denominator of the given fraction by the same number.

Examples: Raise to the 24ths ; =

• Operations with Fractions 1. Addition a. Like Fractions Rule: Add the numerators and copy the common denominator. Example: 6 8 5 19 23 23 23 23 b. Unlike Fractions Rule: First, find the least common denominator (LCD), then divide the LCD by the denominator of each fraction and multiply the result by its nu merator, apply (a). Example: 1 2 3 3 + 16 + 18 37 13 8 3 4 24 24 24 2. Subtraction a. Like Fractions Rule: Subtract the numerators and copy the common denominator. Example: 20 9 11 23 23 23 + + = – = = ; = 3 8 5 9 N 54 3 8 9 24 5 9 30 54 = = + + = = or 1

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Copyright 1999 b. Unlike Fractions Rule: Find the least common denominator (LCD). Divide the LCD by the denominator of each fraction and multiply the result by its numerator, then apply (a). Example: 4 1 16 – 15 1 15 4 60 60 3. Multiplication Rule: a. Multiply the numerators of the given fractions to get the numerator of the product and multiply their denominators to get the denominator of the prod uct. Then simplify the result. Example: 3 4 12 4 5 9 45 15 Rule: b. Another method is by cancellation. That is, a numerator may be cancelled with any denominator in the given fractions dividing each by their GCF. Then apply (a). Example: 14 49 7 21 56 12 4. Division Rule: Multiply the dividend by the reciprocal, or multiplicative inverse, of the divi sor. Example: 16 4 16 7 4 35 7 35 4 5

Percentages, Ratio and Proportion

Per cent is an expression indicating the number of parts taken from a hundred. Literally, "per cent" means by, or, on the hundred. Instead of saying "hundredths" we often use the term per cent. For example 4/100 or 0.04 may be written as 4 %. Ratio is a relationship between two numbers or like quantities. It may be expressed in the form of a fraction. The ratio 36:27 of the number of girls to that of boys (read as 36 is to 27), may be written as 4:3 or in the simplest form 4/3. X = or 13 9 – = = : = X = 4 5 1 1 X = 1 3 7 4

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Proportion is an expression of two equivalent ratios. hence 3:4 = 9:12 is a proportion. In the example, 3 and 12, or the first and fourt terms are called extremes, while 4 and 9, which are the second and third terms, are called means. Therefore, if a sentence is a proportion, the product of the extremes is equal to the product of the means. • Conversion Techniques A. Changing Decimal to Per cent To change decimal to per cent, multiply the decimal number by 100 which is equivalent to moving the decimal point two places to the right, and then affix the per cent sign. Examples: .18 = 18 = 18% .135 = 135 = 13.5% .05 = 05 = 5% B. Changing Per cent to Decimal

To change per cent to decimal, divide the given number by 100, which is equivalent to moving the decimal point two places to the left, and then drop the percent sign. Examples: 15% = .15 12.5% = .125 8% = .08 4 1/2% = .045 C. Changing Fractions to Per cent To change fraction to per cent, change the fraction first to its decimal equiva lent. Then move the decimal point two places to the right and affix the per cent sign. Examples: 5/8 = .625 = 62.5% 4/5 = .8 = 80% 1/7 = .14 2/7 = 14 2/7% 5/6 = .83 1/3 = 83 1/3% D. Changing Per cent to Fraction To change per cent to fraction, drop the per cent sign first, then write the per cent as a decimal; change the decimal to a fraction and change to simplest form.

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Copyright 1999 Examples: 20% = .20 = 20/100 or 1/5 180% = 180/100 = 9/5 or 1 4/5 33 1/3% = 33 1/3 / 100 = 100/3 X 1/100 = 1/3 .05% = .05/100 = .05/100 X 100/100 = 5/1000 or 1/3000 • The Three Types of Percentage Problems A. Finding a per cent of a number: Given the base and the rate, to find the percentage use the formula P = B X R, P = Percentage, B = Base, R = Rate Examples: 1. Find 25 % of P500. B = P500 R = .25 P = B X R = P500 X .25 = P125.00 2. 40 % of P500 is what number? B = P500 R = .40 P = B X R = P500 X .40 = P200.00 B. Finding what per cent one number is of another. Given the base and the percentage, to find the rate, use the formula R = P/B 1. 140 is what per cent of 280 P = 140 B = 280 R = P/B = 140/280 X 100 = 1/2 X 100 = 50% 14 1

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2. P600 is what per cent of P4,000? P = P600 B = P4,000 R = P/B = P600/PP4,000 X 100 = .15 X 100 = 15 % C. Finding a number when a per cent of that number is known. Given the rate and the percentage, to find the base, use the formula B = P/R 1. 8 is 2.5% of what number? P = 8 R = .025 B = P/R = 8/.025 = 300 2. 2.5 % of what number equals 75? P = 75 R = .025 B = P/R = 75/.025 = 3,000 D. Other types of Problems Involving Fractional and Percent Relationship 1. 20 is what per cent greater than 16? 20 – 16 = 4 4/16 X 100 = 1/4 X 100 = 25 % 2. 36 is what per cent smaller than 40? 40 – 36 = 4 4/40 X 100 = 10% 3. 20 increased by 15 % of itself equals what number? 20 + 15% of 20 = the number 20 = known number 15% of 20 is 3 = the increase 23 = the number 4. 30 decreased by 25 % of itself equals what number? 30 – 25% of 30 = the number 30 = known number 25% of 30 is 7.5 = the decrease 22.5 = the number

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Copyright 1999 5. What number increased by 10% of itself equals 22? 100% X the number = the number 10% X the number = the increase 110% X the number = the nmber + the increase or = 22 the number = 22/110% = 20 Check: 20 + 10% of 20 = 20 + 2 = 22 • Finding Percent of Increase or Decrease 1. Divide the amount of the increase by the initial amount. Formula for the percent of increase. Percent of Increase Amount of Increase Initial Amount Example: After Lawrence graduated from college, his earnings on his first job were P75,000 a year. At present his yearly salary is P105,000. What percent of increase has he received? Difference = P105,000 – P75,000 = P30,000 Percent of Increase = P30,000/P75,000 = .40 or 40% 2. Divide the amount of the decrease by the initial amount. Formula for the percent of decrease Percent of Decrease Amount of Decrease Initial Amount Example: There were 140 students enrolled in a judo class during the first semester at U.E., P.E. Department. In the second semester, only 110 were enrolled in the class. What was the percent of decrease in the enrollment? Difference = 140 – 110 = 30 Percent of Decrease = 30/140 = .2143 = 21.43% 14 3 = =

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• Ratio and Proportion A ratio is the relation between two similar quantities. It is consist of two numbers, the antecedent which is the number mentioned and the consequent, the second number. Just like a fraction, a ratio is written in fractional form like 2/3 or in this form, 2:3. Reduce ratios to their lowest terms in the same manner that fractions are reduced. Examples: 1. What is the ratio of 16 ft. to 6 ft.? 16 ft./6 ft = 8/3 2. What is the ratio of 10 yd. to 2 yd.? 10 yd./2 yd = 5/1 3. What is the ratio of 2 lb. to 4 oz.? 2 lb/4 oz = 32 oz/ 4 oz = 8/1 Note: If the units are different, apply conversion. A proportion is a statement that two ratios are equal. The ratios 3/4 and 9/12 are two equal ratios and therefore, constitute a proportion. A proportion is written as: 3/4 = 9/12 or 3:4 :: 9:12. Rule of Proportion: The product of the mean equals the product of the extremes: In 3:4 :: 9:12, the means are 4 and 9 while 3 and 12 are called extremes. The double :: is read equals. • Finding an unknown term in a proportion 1. Assuming the second term is unknown, the proportion may be written as 6:n = 12:14. Using the rule, the product of the means equals the product of the extremes, 6 X n = 12 X 14 12n = 84 n = 84/12 n = 7 2. A motorist drove 140 miles in 5 hours. At the same rate of speed, how far can he drive in 7 hours?

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Copyright 1999 Solution: N represents the missing distance N/140 is the ratio of the missing distance to the known time 7/5 is the ratio of the missing distance to the length of the known time Equal ratios: N/140 = 7/5 5 X N = 140 X 7 N = 980/5 N = 196 miles

Interest

Interest is usually refered to as the sum paid for the use of one's money. Persons in need of cash or financial credit avail themselves of loans from banks or individuals with an agreement to pay a certain amount for the use of the borrowed amount for a given time. Interest is com puted on an agreed rate of interest and the formula used is: I = P r t where I is the interest P is the principal or face value r is the rate t is the number of years, months or days for which the money will be used • Ordinary and Exact Interest Ordinary interest is interest computed for a given number of days, divided by 360, assuming that there are only 360 days in a year. Exact interest is interest computed for a given number of days, divided by 365, which is the actual or exact number of days in a year. In computing both the ordinary and exact interests, the formula is I = Prt where the time varies depending on which one is being calculated.:

I _O = Pr (given in the problem)

I _E = Pr (given in the problem)

Final amount is the sum of the principal and the interest as computed. It is also called maturity value. The formula is F = P + I, where P is the principal and I is the interest. days 360 days 360 14 5

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Example: 1. Find the interest and the final amount on P4,200 for 3 years at 10 %. Given: P = P4,200 r = 10% t = 3 years I = ? F = ? Solution: I = Prt = P4,200 X .10 X 3 = P1260 F = P + I = P4,200 + P1260 = P5,460 2. How much interest is due on P3,500 at 6% for 10 months? Given: P = P3,500 r = 6% t = 10/12 Solution: I = Prt = P3,500 X .06 X = P175 3. Solve for the ordinary interest on P2,960 for 110 days at 5%. Given: P = P2960 r = 5% t = 110/360 Solution: I _O = Prt

= P2960 X .05 X = P45.22 Find the exact interest on P2,400 for 90 days at 4%. Given: P = P2400 r = 4% t = 90/365 10 12 110 360

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= P2400 X .04 X = P23.67 • Rate of Interest To solve for the rate of interest, if principal, time and interest are given in the problem: r = I/Pt Example: 1. The interest on a loan of P2,500 is P60. If the loan is to be paid after 180 days, what is the rate of interest charged? Given: P = P2,500 I = P60 t = Solution: r = I/Pt = = = = .048 or 4.8% 2. What is the rate of interest charged on a loan of P3,400 if the interest paid is P150 at the end of 8 months? Given: P = P3,400 I = P150 t = Solution: r = I/Pt = 90 365 180 360 P60 P2500 180 360 X P60 P2500 1 2 X P150 P3400 8 12 X P60 X 2 P2500 14 7 8 12

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= = = .066 or 6.6% • Finding the Time To solve for time if the principal, interest, and rate are given, the formula is t = I/Pr Example: 1. The interest on a loan of P1,800 is P20. If the rate of interest is 5%, when is the laon due? Given: P = P1,800 I = P20 r = 5% Solution: t = I/Pr = = 0.22 years Multiply quotient by 12 if answer is desired to be expressed in months and multiply by 360 if answer is desired to be expressed in days. t = 0.22 years X 12 = 2.64 months = 0.22 years X 360 = 79.2 days Actual Time and Approximate Time Two methods of determining with in two dates: 1. Approximate number of days 2. Actual number of days Example: 1. Solve for the actual and approximate number of days from March 18, 1992 to No vember 30,1992. P150 P3400 2 3 X P150 X 3 P3400 X 2 P20 P1800 X .05

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Year Month Day

1992 11 30 1992 3 18 8 12 since all months are assumed to have only 30 days then 8 X 30 240 + 12 252 approximate no. of days 2. Actual no. of days March (31, 18) 13 April 30 May 31 June 30 July 31 August 31 September 30 October 31 November 30 257 actual no. of days

Integers

Integer is a whole number or any number which is not a fraction. 1. The set of integers consists of positive, negative numbers and zero. 2. The absolute value of a whole number is the distance of that number from zero. Using the notation for absolute values: [+5] = 5 [–5] = 5 3. Addition of integers: a. If two positive integers are added, the sum is positive • 5 + 9 = 14 • 19 + 17 = 36 b. If two negative integers are added, the sum is negative • (15) + (14) = 29 • (21) + (12) = 33 14 9 0 1 2 3 4 5 1 2 3 4 5

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c. If a positive and a negative integer are added the sum is obtained by subtracting their absolute values. The sign of the sum follows that of the addend with the greater absolute value. • (8) + (4) = 4 • (10) + (2) = 8 4. Subtraction of integers: Subtracting an integer is the same as adding the opposite of that integer. a. 5 – (4) = 9 c. (8) – (2) = 6 5. Multiplication of integers: a. The product of two positive or two negative integer is a positive integer. • (4) X (5) = 20 • (4) X (5) = 20 b. The product of a negative and a positive integers is always a negative integer. • (4) X (5) = 20 • (4) X (5) = 20 6. Division of integers: a. The quotient of two integers with the same sign is always a positive integer. • (8) (2) = 4 • (90) (10) = 20 b. The quotient of two integers with unlike signs is a negative integer. • (16) (4) = 4 • (24) (3) = 8

First Degree Equation Problems

1. Translations • • • _• • • • _• the sum of x and y x increased by y x added to y x more than y x exceeds y by c x exceeds y by 6 the difference bet. x and y x decreased by y x subtracted to y x less than y two quantities whose sum is c x + y x + y y + x y + x x = y + c x = y + 6 x – y x – y y – x y – x x and (c – x)

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Examples:

a. There are three numbers such that the second is 3 times the first, and the third is 2 less than the first. If their sum is 18, find the three numbers.

Solution: Let X = the first number then 3X = the second number and X – 2 = the third number X + 3X + X – 2 = 18 5X = 20 X = 4 first number 3X = 3 X 4 = 12 second number X – 2 = 4 – 2 = 2 third number b. Separate 22 into two parts such that the larger divided by the smaller gives a quotient of 3 and a remainder of 2.

Solution: Let X = the smaller part then 22 – X = the larger part Division Law: Dividend = Quotient X Divisor + Remainder Larger = 3 (smaller) + 2 22 – X = 3X + 2 4X = 20 X = 5 smaller part 22 – 5 = 17 larger part 3. Consecutive Integer (Whole Number) Problems Example: a. Find three consecutive odd integers such that the sum of of the first two is 25 more than the third.

Solution: Let X = the first consecutive odd integer then X + 2 = the second X + 4 = the third to form the equation, we use First + second = third + 25 X + X + 2 = X + 4 + 25 X = 27 (first) X + 2 = 29 (second) X + 4 = 31 (third) • • • 15 1 Consecutive Integers Consecutive Even Integers Consecutive Odd Integers x, x + 1, x + 2, x + 3.... x, x + 2, x + 4, x + 6.... x, x + 2, x + 4, x + 6.... E: 5, 6, 7, 8, .... E: 4, 6, 8, 10, .... E: 3, 5, 7, 9, ....

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4. Triangle Problems Examples:

a. The second angle of a triangle is twice the first. The third angle exceeds the sum of the first two angles by 12 0 _{. Find the three angles:}

Solution: Let X = degrees in first angle then 2X = degrees in second angle 3X + 12 = degrees in third angle the sum of the three angles of any triangle is 180 0 _. X + 2X + 3X + 12 = 180 0 6X = 168 0 X = 28 degrees in first angle 2X = 56 degrees in second angle 3X + 12 = 96 degrees in third angle b. The second side of a triangle is 5'' more than the first. The third side is 3'' less than twice the first side. The perimeter of the triangle is 38''. Find the sides.

Solution: Let X = first side in inches then X + 5 = second side in inches 2X – 3 = third side in inches Perimeter is sum of the three sides X + X + 5 + 2X – 3 = 38'' 4X = 36'' X = 9'' (first side) X + 5 = 14'' (second side) 2X – 3 = 15'' (third side) 5. Rectangle Problems Example: a. The length of a rectangle exceeds 3 times the width by 5. If the perimeter equals 58, find the length and the width.

Solution: Let X = width

then 3X + 5 = length The perimeter equals the sum of all four sides 3X + 5 + X + 3X + 5 + X = 58 8X = 48 X = 6 (width) 3X + 5 = 23 (length)

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Example:

a. Mrs. Tuazon is 24 years older than her daughter. In 3 years, she will be 4 times as old as her daughter. Find their present ages.

Solution: Let X = daughter's age in years and X + 24 = Mrs. Tuazon's age in years in 3 yrs, X + 3 = daughter's age and X + 27 = Mrs. Tuazon's age Mrs. Tuazon's age in 3 yrs = 4(daughter's age in 3 yrs) X + 27 = 4(X + 3) X + 27 = 4X + 12 3X = 15 X = 5 yrs. (daughter's age) X + 24 = 29 yrs. (Mrs. Tuazon's age) 7. Coin Problems Example: a. A purse contains 5 cents, 10 cents and 25 cents. The number of 5 cents is 1/3 the number of 10 cents and there are 7 more 25 cents than 10 cents. The total value of all the coins is P8.35. How many of each kind are there? Solution: To avoid fractions, let 3X = number of 10 cents Total value of all coins = 835 cents 5X + 30X + 25(3X + 7) = 835 5X + 30X + 75X + 175 = 835 110X = 660 X = 6 (5 cents) 3X = 18 (10 cents) 3X + 7 = 25 (25 cents) 8. Investment Problems Example: a. P6000 is to be invested, part at 6% and the rest at 3%, to give a total annual income of P294. How much should be invested at each rate? Solution: If interest is computed annually, the formula I = Prt becomes I = Pr(1), or I = Pr. 15 3 5 cents 10 cents 25 cents X 3X 3X + 7 5X 30X 25(3X + 7) 5 10 25 No. of cents per coins X coin = Value in cents

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Income at 6% + Income at 3% = 294 .06X + .03(6000 – X) = 294 Multiplied by LCD: 100 6X + 3(6000 – X) = 29400 6X + 18000 – 3X = 29400 3X = 11400 X = P3800 at 6% 6000 – X = P2200 at 3% 9. Mixture Problems Example: a. How many pounds each of nuts worth 72 cents a lb. and nuts worth 84 cents a lb. should be used to obtain a 40 lb. mixture worth 75 cents a lb.? Solution: Value of less expensive nuts + value of more expensive nuts = value of mixture 72X + 84 (40 – X) = 75(40) 72X + 3360 – 84X = 3000 12X = 360 X = 30 lb. of the 72 cents nuts 40 – X = 10 lb. of the 84 cents nuts 10. Motion Problems Examples: a. A bus goes from San Teodoro to Roxas at the rate of 80 mph; a second bus goes from Roxas to San Teodoro at 60 mph. If both start at 11 a. m. and the two towns are 455 miles apart, at what time should they pass each other? The total distance traveled by both bus is 455 miles Principal in Peso X Rate of Interest = Interest or Income in Peso at 6% at 3% X 6000 – X .06 .03 .06X .03(6000 – X) San Teodoro to Roxas Roxas to San Teodoro 80 60 X X 80X 60X R(mph) X T(hr) = D(mi) No. of lb. X cents per lb. = value in cents Less Expensive More Expensive Mixture X 40 – X 40 72 84 75 72X 84(40 – X) 75(40)

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Copyright 1999 80X + 60X = 455 140X = 455 X = 3 1/4 hr. Ans: 2:15 p.m. b. Illissa starts from home on her bicycle at 15 mph. One hour later, her father starts after her in his car going 40 mph. How soon will he overtake her? Solution: (Since Illissa had a 1 hr. headstart her traveling time is 1 hr. more than her father's) Father's distance = Illissa's distance 40X = 15(X + 1) 40X = 15X + 15 25X = 15 X = 3/5 hr. Ans: 36 minutes c. A patrol pilot can go east in his plane at 150 mph, and can return at 100 mph. If the plane has 5 hrs of flying time, how far east can he go? Solution: (Distance going = distance returning) (T column obtained from T = D/R) Time Going + Time Returning = 5 hrs. X/150 + X/100 = 5 X = 300 mi d. To find the rate of the wind, a pilot whose plane has an airspeed of 220 mph goes with the wind for 10 min. and returns over the same distance in 12 min. What is the rate of the wind?

Solution: (Rate against wind = airspeed – wind rate: Rate with wind = airspeed + wind rate) Illissa Father 15 40 X + 1 X 15(X + 1) 40X R(mph) X T(hr) = D(mi) Going Returning 150 100 X/150 X/100 X X R(mph) X T(hr) = D(mi) 15 5

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(Convert 12 min. and 10 min. to hr) Distances are equal 1/5(220 – X) = 1/6(220 + X) X = 20 mph. 11. Work Problems Example: a. Lawrence can mow a lawn in 36 min.; Rondon can in 30 min. If both boys work together, how long will the job take? Part of Job done by Lawrence + part of job done by Rondon = whole Job (we represent the whole job by 1) X/36 + X/30 = 1 (Multiply by 180; etc.) Ans: 16 4/11 min.

Geometric Figures

• Angle(Ð) – plane figure formed by two rays with a common endpoint and do not lie on a straight line. The two rays are the sides and the common endpoint is the vertex. 1. Interior and Exterior of an Angle: A point X is in the interior of an angle if it is in the region between the sides of an angle. A point Y is at the exterior of an angle if it is in the region not enclosed by an angle. Against Wind With Wind 220 – X 220 + X 1/5 1/6 1/5(220 – X) 1/6(220 + X) R(mph) X T(hr) = D(mi) No. of min. to do job alone Part of job done in 1 min. No. of min. actually worked Part of Job done X ₌ 36 30 1/36 1/30 X X X/36 X/30 Lawrence Rondon A 2 ÐA Ð2 D ÐBDG B G

X is the interior of ÐA Y is the exterior of ÐB

A X B

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Copyright 1999 2. Measure of an angle – the unit of measure of an angle is called the angle degree ( 0 _). Þ 75 0 _, 900 _, 1800 3. Classification of Angles: a. Acute angle – its measure is greater than 0 0 _{but less than 90}0 _. b. Right Angle – its measure is 90 0 _. c. Obtuse Angle – its measure is greater than 90 0 _{but less than 180}0 _. • Polygons – are closed figures formed by joining three or more line segments. 1. Equilateral Polygon – a polygon whose sides are of the same length. 2. Equiangular Polygon – a polygon whose angles are all equal. 3. Regular Polygon – a polygon which is both equilateral and equiangular. 4. Congruent Figure – a figure like a line or polygon with the same size and shape. Angles with the same measure are congruent. • Triangle(D) – polygon with three sides, three vertices and three angles. The sum of the measures of its three angles is equal to 180 0 _. 15 7 Polygon Triangle Quadrilateral Pentagon Hexagon Heptagon No. of Sides 3 4 5 6 7 Polygon Octagon Nonagon Decagon Undecagon Dodecagon No. of Sides 8 9 10 11 12 CLASSIFICATION OF POLYGONS 36 0 90 0 110 0

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Parts of Triangle 1. Altitude(h) – is the perpendicular ( ) line from a vertex to the opposite side. 2. Median –line segment which connects the vertex with the midpoint of the opposite side. 3. Angle Bisector – divides an angle into two equal measures. Kinds of Triangle 1. According to angle size: a. acute – if each angle is less than 90 0 _. b. Right – triangle with 90 0 _angle. c. Obtuse – triangle with an angle more than 90 0 _. According to Sides a. Equilateral – all three sides are equal or congruent. JV is the angle bisector of DJHN A P F V Y G D PF is the altitude of DAVP TG is the median of DDTY T J H N V

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Copyright 1999 b. Isosceles – has two congruent sides. c. Scalene – has no equal sides. • Quadrilaterals – polygon with four sides, four angles and four vertices. 1. Kinds of Quadrilaterals a. General Quadrilaterals – has no pair of opposite sides parallel. b. parallelogram – has both pairs of opposite sides parallel. Parallel lines are those that lie on the same plane but does not intersect each other. c. Trapezoid – has only one pair of opposite sides parallel. 2. Kinds of Parallelograms: a. Square – a parallelogram with all sides equal and all its angles are right angles. b. Rectangle – a parallelogram whose angles are all right angles. c. Rhombus – a parallelogram whose sides are all equal in length and whose an gles are all oblique (acute or obtuse). 3. Parts of a Trapezoid: a. Bases – the sides that are parallel. b. Legs – are the two nonparallel sides. c. Median – segment connecting the midpoints of the two legs. d. altitude – segment drawn from the lower base to the upper base. • Circle – the set of all points in plane m whose distance from point 0 is r units. Point 0 is the center of the circle. Parts of a Circle 1. Radius – line segment from the center to any point on the circle 2. Chord – line segment joining any two points on the circle. 3. Diameter – a chord that passes through the center of the circle. 4. Circumference – the distance around the circle. 5.Semicircle – an arc whose endpoints are the endpoints of the diameter of a circle. 6. Arc – a part of the circumference of acircle. 7. Minor Arc – an arc less than a semicircle. 8. Major Arc – an arc greater than a semicircle. 9. Central Angle – an angle whose vertex is the center of the circle and whose sides are the radii. 10. Intercepted Arc of an Angle – an arc whose endpoints are on each side of the angle and all other points on the arc are in the interior of the angle. 15 9 r units 0 m

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English 12 ins = 1 ft 3 ft = 1 yd 5,280 ft = 1 mile 16 oz = 1 lb 2000 lbs. = 1 ton ( t ) Counting 1 doz = 12 units 1 quire = 24 units 1 gross = 12 dozens 1 gross = 144 units 1 ream = 550 sheets

Units of Measure

Metric Units of Length 100 cm = 1 m 1,000 m = 1 km Units of Weight 100 g = 1 kg 10 hg = 1 kg 1000 kg = 1 metric ton ( mt ) Units of Capacity 1000 cc = 1 liter 1000 L = 1 kilometer = 1 cu m 1 gal = 4 qt 1 gal = 231 cu in Useful Equivalent 1 kg = 2.2 lbs 1 lb = 454 g 1 liter = 1057 quarts 1 gal = 3785 L. 1 hectare = 10,000 sq m

Useful Equivalents

1 in = 2.54 cm 1 ft = .3048 m 1 yd = .9144 m 1 m = 38.37 in 1 mi = 1.609 km 1 not = 1. 152 mi 1 sq in = 6.452 sq cm 1 sq in = .0929 sq m 1 sq yd = .8361 sq m 1 sq mi = 2.590 sq km 1 sq mi = 640 acres 1 cu in = 16.39 cu cm 1 cu ft = .02832 cu cm 1 cu yd = 76.46 cu cm 1 cu ft = 62.32 liters 1 cu ft = 2.205 lbs 1 kg = 15 grains 1 gram = .0002902 1 gram = .0002909 Sine 1 = .01745 Rad. 1 in mercury = 1.133 ft water 1 in mercury = .4912 lbs per sq. in

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Copyright 1999 1 U.S. gal = 3.785 liters 1 U.S. gal = 231 cu in 1 U.S. gal = 8.345 lbs water 1 ft water = .4335 lbs per sq in 1 atmosphere = 14.7 lbs per sq in 1 B.T.U. = 778 ft lbs 1 H.P. = .707 B.T.U. per sec 1 H.P. = .746 kw 1 mile per hour = 1.467 ft per sec side of Square = .707 Diagonal of sq 16 1 Log e x ₌ _{2.3026 Log x} Circumference of a circle = pD Area of circle = p r 2 Surface area of a Sphere = p d 2 Vol. of a Sphere = .5236 d 3 1 milligram ( mg ) = 1000 micrograms 29 = 5.39 1 Radian = 180 0 _{/p = 57.3}0