Math 118 Review Review objectives The review includes: 1. Basic algebraic operations i. Bed math (order of operation): Addition, subtraction, multiplication, division and brackets ii. Laws of exponents 2. Simple equations i. Solve for the indicated letters (in terms of numbers or other letters) ii. Equations involving fractions iii. System of two linear equations 3. Formulas 4. Calculator problems i. Evaluate complex calculation with careful use of brackets ii. Percent conversion 5. Ratio and percent i. Reduce the given ratio to its lowest term i. Allocation based on proportion Page 1
Basic Algebra Four operations: addition, subtraction, multiplication and division Multiplication and Division 1. + + + When signs are the same, we get positive 3x4=12 + (-3)x(-4)=12 2. + When signs are different, we get negative 3x(-4)=-12 + (-3)x4=-12 Addition 1. + + + When signs are same, we get the number with same sign 3+4=7-3+(-4)=-(3+4)=-7 2. + to be determined Comparing two numbers, the result= sign of the bigger one X ( bigger smaller) For 3+(-4), we know that 4 is bigger than 3, so the result should have negative sign with value of 4-3, so 3+(-4)=-(4-3)=-1 For 4+(-3), we know that 4 is bigger than 3, so the result should have positive sign with value of 4-3, so 4+(-3)=+(4-3)=1 3. + (same as the previous one, since we can change the order for addition) Subtraction (the strategy is to change it to addition) 1. + + When signs are different, we follow the first one + 3-(-4)=3+4=7 minus & minus becomes positive -3-4=-3+(-4)=-(3+4)=-7 2. + + to be determined Comparing the two numbers, the result= sign of the bigger one X ( bigger smaller) 3-4=3+(-4)=-(4-3)=-1 Page 2
3. to be determined Comparing a and b, the result= sign of the bigger one X ( bigger smaller) -3-(-4)=-3+4=4+(-3)=+(4-3)=1 About -a b = -axaxaxa.xa (b of a) -3 4 =-3x3x3x3=-81 (-a) b =(-a)(-a)(-a) (-a) (b of a) = -a b if b is odd (-4) 3 =-64 a b if b is even (-3) 4 =81 In general a(-b)(-c)d=adcd If the number of is even, we get positive (-7)(-2)(-3)(-2)= 84 (-a)(-b)(-c)d=-abcd If the number of is odd, we get negative (-2)(5)(3)=-30 About 0 Any number + 0 or 0 = same number Any number X 0 = 0 Any number 0 does not exist (because 0 cannot be denominator) 0 any number(not equal to 0) = 0 About () d+(a+b-c) = d+a+b+c If there is positive before bracket, we don t change the sign in the bracket x+(x 3 +x 2-1)=x+x 3 +x 2-1 d-(a+b-c) = d-a-b+c If there is negative before bracket, we change the sign in the bracket x-(x 3 +x 2-1)=x-x 3 -x 2 +1 a(b+c)=ab+ac -a(b+c)=(-a)(b)+(-a)(c)=-ab-ac only used in simplifying or solving equations Page 3
Order of operations priority bracket multiplication & division addition & subtraction ( ) X + - priority [ ] { } 3+(-2 3 )(-4 2 ) {2 2 [(-3 2 +5)-4]} work out the exponent =3+(-8)(-16) { 4[(-9+5)-4]} =3+128 {4[-4-4]} First step, remove ( ) =3+128 {4 [-8]} =3+128 {-32} Second step, remove [ ] =3+{-4} = 3-4 Last we remove { } = -1 Percent Conversion Percent to decimal (be careful about the rounding) Correct to 3 decimal places: 1.68763-> 1.688 3 places Given decimal (or integer) %, decimal (or integer) 100 = 18%=0.18 Given fraction %, integer part ab/c numerator ab/c denominator 100 = ab/c 17⅓=0.1733 Page 4
Percent to lowest terms Given decimal (or integer) %, decimal (or integer) 100 = ab/c 55%= Given fraction %, integer part ab/c numerator ab/c denominator 100 = 4 %= Decimal or fraction to percent(exact form) Given decimal (or integer), decimal (or integer) x 100 = Remember to put the % after the calculation 0.7543=75.43% Given fraction, integer part ab/c numerator ab/c denominator x 100 = ab/c Remember to put the % after the calculation 2 = 283.333%(after rounding up to 3 decimal places) Laws of exponents a 0 = 1 where a is any number (-5) 0 =1 -(5) 0 =-1 a -b = 3-4 = = (1+x) - 2 = (ab) c =a c b c (2a) 3 =2 3 a 3 =8a 3 ( ) c = ( ) 3 = = Page 5
To simplify equations, we usually use a c a d = a c+d = a c a -c = a c-d If the question requires that with positive exponents only we compare the exponents on both top and bottom. We keep the position of big exponent, and reduce that exponent by the small exponent. = x 6-2 =x 4 = = (a c ) d = a cd the sign for (-3) -4 is -, do you know why? Common factoring ab+ac= (ab+ac)=a( )=a( + )=a(b+c) Looking for the common part of each of the terms, put it outside of the bracket and leave the terms (after dividing by the common part ) in the bracket For 5 +5 z, the common factor is 5, so after taking it outside of bracket we have 5 ( = 5 (a Formulas that are good to know (a+b) 2 =a 2 +2ab+b 2 (1+x) 2 =1 2 +2(1)(x)+x 2 =1+2x+x 2 (a-b) 2 =a 2-2ab+b 2 (1-x) 2 =1 2-2(1)(x)+x 2 =1-2x+x 2 (a+b)(a-b)=a 2 -b 2 (1+x)(1-x)=1 2 -x 2 =1-x 2 Solving equations for x We always want x on the left side and number on the right side If there is number on the left, we subtract that number on both sides x+1=5 x+1-1=5-1 x=4 If there is x on the left, we subtract x on both sides 2x+1=5+x 2x+1-x=5+x-x x+1=5 If x is multiplied by a number, we divide both sides by that number 3x=6 = x=2 If x is divided by a number, we multiply both sides by that number = 2 * 3 = 2*3 x=6 Page 6
Calculator problems To type equations into calculator in one step, we must be very careful about how to use brackets correctly. If we have a fraction, we have to put a bracket around it Ex. To evaluate, we type ( 0.345 7 ) ^ ( 6 7 ) ( 8 7) If there is more than one number in the denominator or numerator, we need to add a bracket Ex. To evaluate, we type (45 + 23.1 ^ 4) (1.23 ^ 4 1 ) Solving equations involving fractions First step is to find the least common denominator (LCD) Next, multiplying both sides (or each of the terms) by that LCD Try to cancel each denominator with the LCD Finally we always get an equation without fraction and solve for x from that For example, the LCD of 12,8,6 is found by 12,8,6 divided by 2 so the LCD is 2 X 3 X 2 X 2 =24 6,4,3 divided by 3 multiplying both sides by 24 gives 24 X 24X or 24 cancel numerator with dominator gives 2x-3=4(x+2) solving gives x= 2,4,1 divided by 2 1,2,1 divided by 2 1,1,1 we stop here Page 7
Formula Rearrangement The whole process is to get closer to the unknown parameter we want to solve for Ex. For c= a + b Solve for a: Minus both sides by b, and we get a= c - b Solve for b: Minus both sides by a, and we get c-a= b Divide both sides by, and we get b= Solve for e: Minus both sides by a, and we get c-a= b Divide both sides by b, and we get = Take power of on both sides, we get d= Ratio Problems Reduce to lowest form 1. Case 1 (basic) integer : integer or integer : integer : integer We need to find common factor for each integer given and reduce each integer by that common factor using division, then we do these steps all over again until the common factor is only 1 Ex. 45 : 30 : 75 common factor 5 Dividing by 5 gives 9: 6: 15 common factor 3 Dividing by 3 gives 3: 2: 5 common factor only 1, so we stop here 2. Case 2 (decimal) decimal : decimal or decimal : decimal : decimal First, we need to multiply by the same number to change all of the decimals to integers, and the number we used to multiply is determined by the longest length after the decimal place Ex: 2.3:1.2 multiplied by 10 for both numbers gives 23:12 2.34:1.23 multiplied by 100 for both numbers gives 234:123 2.3:1.23 multiplied by 100 for both numbers gives 230:123 After changing to integers, we use the techniques talked about in case1 Page 8
3. Case 3 (fraction) fraction : fraction or fraction : fraction : fraction First we need to find least common denominator and multiply each term by that LCD to change fractions into integers. Ex. : : multiply by 6 for every number gives 3:1:2 After changing to integers, we use the techniques talked about in Case 1 Allocation First step is always reducing the given ratio to its lowest form. Next step is adding the ratio to get the total ratio Ex. If we have a:b:c in integers and total amount is given as Q, then the total ratio is a+b+c and each allocation is calculated by a ; b ; c Proportion problems Given ratio in forms of a:b=c:d, we can change it into fraction form by =. Applying cross-multiply = gives us a d=c b from where we can solve for any unknown parameter. Note: When doing application problems involving proportions, we have to make sure that the units in numerators and units in denominators are same respectively. Percent problems Whenever we have a decimal, we can change it into percent by multiplying by 100 and adding % Whenever we have a percent, we can change it into decimal by dividing by 100 and deleting % 1. Percentage(amount) = rate base A=BR or rate= New number=rate original number or rate= Page 9
Ex. (1) 33 % of 3 is calculated by 33 ab/c 1 ab/c 3 100 3, which gives 1 (2) What is 30% of 40? 12 is 30% of what amount? X = 0.3 * 40 12 = 30% * X Note: the word of in question always means multiplied by in mathematical expression. 2. Original number + increase (or decrease) = new number Or we can instead use Base + amount=total B+A=T or B+BR=T or B(1+R)=T Before the change After the change Amount of change Original number + original number rate of increase = new number Or base + increase amount=total Ex. 20 increased by 30% = 20+30%(20)=26 Original number - original number rate of decrease = new number Or base decrease amount=total Ex. 30% less than 20=20 30%(20)=14 Note: base is changed (decreased or increased) to total by amount 3. Rate of increase = Rate of decrease = Note: The words more than or less than are always followed by original number. The word amount to is always followed by new number. Page 10
Solving system of linear equations We always try to manipulate the two equations to eliminate one of the unknowns and solve for another parameter, then we substitute back the solved parameter into one of the original equations to get the eliminated parameter Case 1: coefficients are numerically equal or opposite ax+by=c❶ when the sign is same, we use subtraction ax+dy=e❷ ❶ ❷ to eliminate x and gives by-dy=c-e or y= Then we substitute y= into either❶or❷to solve for x Ex. x-y=-2❶ x+2y=10❷ ❶ ❷to eliminate x and gives y-2y=-2-10 or -3y=-12 or y=4 Then we substitute y=4 into ❶ and get x-4=-2 or x=2 ax+by=c❶ -ax+dy=e❷ when the sign is opposite, we use addition ❶ ❷ to eliminate x and gives by+dy=c+e or y= Then we substitute y= into either❶or❷to solve for x Ex. -x-y=-2❶ x+2y=10❷ ❶ ❷to eliminate x and gives y+2y=-2+10 or y=8 Then we substitute y=8 into ❶ and get -x-8=-2 or -x=6 or x=-6 bx+ay=c❶ dx+ay=e❷ when the sign is same, we use subtraction ❶ ❷ to eliminate y and gives bx-dx=c-e or x= Then we substitute x= into either❶or❷to solve for y Ex. 2x+2y=-4❶ x+2y=10❷ ❶ ❷to eliminate y and gives 2x-x=-4-10 or x=-14 Then we substitute x=-14 into ❷ and get -14+2y=10 or 2y=24 or y=12 Page 11
bx+ay=c❶ dx-ay=e❷ when the sign is opposite, we use addition ❶ ❷ to eliminate y and gives bx+dx=c+e or x= Then we substitute x= into either❶or❷to solve for y Ex. x-2y=-2❶ x+2y=10❷ ❶ ❷to eliminate y and gives x+x=-2+10 or 2x=8 or x=4 Then we substitute x=4 into ❶ and get 4-2y=-2 or -2y=-6 or y=3 Case 2: coefficients are not numerically equal or opposite ax+by=c❶ dx+ey=f❷ Option 1: To eliminate x, we need ❶ d and ❷ a, then we get adx+bdy=cd❸ adx+aey=af❹ ❸ ❹ gives bdy-aey=cd-af or y= Option 2: To eliminate y, we need ❶ e and ❷ b aex+bey=ce❸ bdx+bey=bf❹ ❸ ❹ gives aex-bdx=ce-bf or x= Ex. 4x+y=-13❶ x-5y=-19❷ Option 1: To eliminate x, we need ❶ 1 and ❷ 4, then we get 4x+y=-13❸ 4x-20y=-76❹ ❸ ❹ gives y-(-20y)=-13-(-76) or 21y=63 or y=3 Then we substitute y=-3 into ❶ and get 4x+3=-13 or 4x=-16 or x=- Option 2: To eliminate y, we need ❶ 5 and ❷ 1, then we get 20x+5y=-65❸ x-5y=-19❹ ❸ ❹ gives 20x+x=-65+(-19) or 21x=-84 or x=-4 Then we substitute x=-4 into ❷ and get -4-5y=-19 or -5y=-15 or y=3 Page 12
Case 3: If we are given decimal or fraction as coefficients, we need to change those coefficients into integers and solve the changed equations as indicated in first two cases 1. For decimal, we multiply both sides by 10 to the power of the # of decimal Ex. 0.4x+1.5y=16.8❶ # of decimal for both equations is 1, so multiplied by 10 1.1x-0.9y=6.0❷ To eliminate decimal, ❶ 10 and ❷ 10 give 4x+15y=168 11x-9y=6 2. For fraction, we multiply both sides by the LCD for each equation Ex. ❶ the LCD is 12 ❷ the LCD is 20 To eliminate fractions, ❶ 12 and ❷ 20 give which is 9x-8y=-26 which is 16x+15y=246 Page 13
Math 119 Review Review objectives This review includes: 1. Business pricing system i. Trade discount ii. Cash discount with term iii. Retail price Mark-up Mark-down 2. The time value of money and financial mathematics i. Simple interest rate ii. iii. Present value and future value Equivalent payments Compound interest rate Present value and future value Equivalent payments and rates Annuities simple annuity and general annuity ordinary annuity and annuity due Page 14
Trade Discount Formula Discount=List price*rate of discount Net price= List price*(1-rate of discount) D=Ld N=L(1-d) Price after the discount reduced price Price before the discount list price: price before the reduction Serial discount Net price = List price*(1-rate of discount 1)*(1-rate of discount 2)* *(rate of discount n) N=L(1-d1)(1-d2) (1-dn) Single equivalent rate Since we have 1-d = (1-d1)*(1-d2)*..*(1-dn) So solving for d gives d= (1-d1)*(1-d2)*..*(1-dn)-1 Ex. Match the competitor s price means that Net price of store 1 = Net price of store 2 Note: Some of the question requires rearranging the formula For example, from D=Ld we can get L= and d= from N=L(1-d) we can get L= and d= Cash discount with term Invoice always has a form of cash discount/length of discount period, n/length of credit period Formula Payment = Credit (1- rate of cash discount) Payment=Cr(1-d) Where payment is the actual cash outflow and credit is the amount by which the outstanding balance is reduced Page 15
Interpretation of the term invoice issued date last date of discount= invoice issued date + length of discount discount period non-discounted period Outstanding balance Old outstanding balance Credit (balance is reduced by ) = New outstanding balance (balance is reduced to ) End-of-month dating(e.o.m) Invoice starts at the end-of-month of the original stated date Receipt-of-goods dating(r.o.g) Invoice starts at the date of receipt of the shipment Mark-up Formula(requires skills in solving for different parameters) Regular selling price= Cost of buying +Expense (overhead) + Profit S=C+E+P Sometimes, we need to take the serial discount to get this price Expense (overhead) + Profit = Mark-up Regular selling price = Cost of buying + Mark-up Ex. If M=30%ofC, then S=0.3C+C or S=1.3C If M=30%ofS, then S=0.3S+c or 0.7S=C E+P=M S=C+M Rate of mark-up based on cost= Rate of mark-up based on selling price= Ex. If M=60%S, then C=S-M=40%S and rate of mark-up based on cost= = Page 16
Mark-down Formula Mark-down = Regular selling price*rate of mark-down Md=Sd Sales price= regular selling price mark-down sales price=s-md = regular selling price*(1-rate of mark-down) sales price=s(1-d) Note: for all formulas above, we can replace all terms of regular selling price by marked price, or MP Total cost=cost of buying + expense TC=C+E Profit realized during promotion = sales price total cost P=sales price-tc = sales price cost of buying expense P=sales price-c-e Terminology Break-even-point: Profit = 0 Operating cost: the sales price is insufficient to cover the total cost Absolute cost: the sales price is insufficient to cover the cost of buying S<TC=C+E S<C Note: Usually we use the mark-up to find the regular selling price and mark-down to find the sales price In diagram, Mark-up Mark-down Cost of buying Regular selling price= mark-up + cost of buying Sales price= Mark-up + cost of buying Discount Marked price= regular selling + discount Sales price= marked price-mark-down Mark-down Page 17
Simple Interest Rate Formula p.a Interest= Principal * Rate * Time I=Prt In years, or if given in days, divided by 365; if given in weeks, divided by 52; if given in months, divided by 12; if given in quarters, divided by 3. After rearranging, we can also get P= r= t= Future value(or Maturity value) = Principal + Interest S=P+I or S=P(1+rt) After rearranging, we can also get P= Ex.Repayment of a loan is the future value of the money borrowed Compound Interest Rate Formula S=P* or FV =PV* where i= and n=mt Note: 1. j and t is always given in the question, but first we need to convert the time t to number of years. 2. For interest compounded annually, m=1 For interest compounded semi-annually, m=2 For interest compounded quarterly, m=3 For interest compounded monthly, m=12 For interest compounded weekly, m=52 Rearranging gives us PV= or PV =FV* Page 18
Single Equivalent Payment years t1 t2(focal date) t3 From t1 to t2, we are given the Present value, so using FV =PV* to get the equivalent value at focal date Where n=t2-t1 From t3 to t2, we are given the Future value, so using PV= to get the equivalent value at focal date Where n=t3-t2 Note: For questions having borrow Q with a term of n years, we need to first find out the future value of Q after n years and use it as due amount. More about Compound Interest Rate Effective rate of interest FV=PV Rearranging gives f=( -1 Note: the effective rate is also equal to nominal rate compounded annually Equivalent rates 1+f= = where i1= n1=m1 t i2= n2=m2 t 1. Given j1 and m1, calculate f nominal to effective f= -1 2. Given j1, m1 and m2, calculate j2 nominal to nominal j2= m2 [ ] 3. Given f and m1, calculate j1 effective to nominal j1= m1 [ ] Page 19
Simple annuity (a special case of general annuity: # of payments per year= # of interest per year) Future value FVn = PMT[ ] where i= and n=m t Key words for future values: accumulated to, amount to Note: for deposit (cash inflow), Interest = FV contributions for loan or mortgage (cash outflow), Interest = contributions FV Present value PVn = PMT[ ] where i= and n=m t Key words for present values: discounted value, purchase value, immediate payment, value of contract Ex. PV= given instalment down payment= given instalment (1 rate of down) Ex. At any point of time on the time line, deposit=withdraw deposit withdraw t1 t2 t3 t4 FVdeposit = deposit[ ] FV =PV PVwithdraw = withdraw[ ] Ex. Change of interest rate( or payment) To apply the formula for annuity, we must have fixed interest rate, so we need to separate the whole time interval into two intervals with fixed interest rate. annuity 1 annuity 2 t1 t2 t3 For annuity 1, from t1 to t2 we have FVn = PMT[ ] From t2 to t3 we have FV =PV* For annuity 2, from t2 to t3 we have FVn = PMT[ ] Page 20
General annuity First, we have to calculate C = = And p = where i = Then we have Future value FVg = PMT[ ] where n=m1 t Present value PVg = PMT[ ] where n=m1 t Combo questions: general annuity + compound interest Case1 : PVg FVg or PV FV(we are solving for) n1 n2 n3 unit: year annuity with period( n2-n1) compound interest with period (n3-n2) From the time line above, we can get two equations (1) FVg = PMT[ ] where n=(n2-n1) m1 (2) FV=PV where i= and n=(n3-n2) m1 Case2 : PV(we are solving for) FV or PVg FVg n1 n2 n3 compound interest with period( n2-n1) annuity with period (n3-n2) From the time line above, we can get two equations (1) (1) PVg = PMT[ ] where n=(n2-n1) m1 (2) PV=FV where i= and n=(n3-n2) m1 Page 21
Simple annuities due (payments are made at the beginning of each period) Future value FVn = PMT[ ](1+i) where i= and n=m t Present value PVn = PMT[ ](1+i) where i= and n=m t General annuities due First, we have to calculate C = = And p = where i = Then we have Future value FVg = PMT[ ](1+p) where i= and n=m1 t Present value PVg = PMT[ ](1+p) where i= and n=m1 t Page 22