Download Algebra Chapter 7 Review Section 7.1 Write each ratio in simplest

Algebra Chapter 7 Review Section 7.1 Write each ratio in simplest form 1) 18:24 2) 6Y:9Y 3) 72RS5/12R2S2 4) 2kg:90g 3:4 2:3 6S3/R 200:9 2) What is the ratio of $2.00 to 60¢ 10:3 3) The measure of the angles of a triangle are in the ratio of 3:4:5. Find the measurement of the smallest angle of the triangle. Let 3X = small angle; 4X = 2nd angle; 5X = largest angle 3X + 4X + 5X = 180 12X = 180 X = 15 Smallest angle measurement is 45˚. 4) The perimeter of a rectangle is 68 ft. Find the dimensions of the rectangle if the ratio of the length to the width is 9:8 Let 9X = length in ft; 8X = Width in ft 2(9X) + 2(8X) = 68 18X + 16X = 68 34X = 68 X = 2 The rectangle is 18 ft long and 16 ft wide 5) Find the ratio of X:Y in the equation cx – ay = aby –bcx cx = aby + ay – bcx cx + bcx = aby + ay x(c + bc) = y(ab + a) x = y(ab +a)/(c + bc)_ x/y = (ab + a)/(c + bc) x/y = a(b+ 1)/c(1+ b) x:y = a:c NOTICE THE WAY THE FINAL ANSWER IS WITTEN. 6) What is a ratio? A comparison of two quantities of the same kind; 7) The ratio of freshmen to sophomores is 4:5 and there are 120 freshman. How many sophomores are there? Let X = # of sophomores 4/5 = 120/X 4X = 600 X = 150 There are 150 sophopmores Section 7.2 1) What are the means of the proportion 4/X = 7/Y Extremes are 4 and Y Means are X and 7 2) What are the extremes of the proportion 5:T = 8:M Extremes are 5 and m Means are T and 8 3) What is a proportion? An EQUATION that states two ratios are equal 4) For the proportion give the equation that states the product of the extremes equals the product of the means: (do not solve) a) 4X/5 = 9/2 b) 4/9 = (a-­‐2)/3 2 • 4X = 5 • 9 4 • 3 = 9 (a – 2) You could simplify these if you want 5) Solve: a) 5 = (4 + 3Y)/2 b) (3 + 2X)/(3 -­‐ 2X) = -­‐3 c) (2n-­‐ 9)/7 = (3-­‐n)/4 10 = 4 + 3Y 3 + 2X = -­‐3(3 – 2X) 4(2n – 9) = 7(3 – n) 6 = 3Y 3 + 2X = -­‐ 9 + 6X 8n – 36 = 21 – 7n Y = 2 12 = 4X 15n = 57 X = 3 n = 57/15 = 19/5 6) Find the ratio of X:Y in the equation (b -­‐ a)/aY = (b2 -­‐ a2)/bX (b – a)bx = (b2 – a2)ay (b – a)bx/y = (b2 – a2)a x/y = [(b2 – a2)a]/[ (b – a)b] x/y = (b + a)a/b 7) Sam drove 72km on 6L of gas. How far can he drive on 50 L of gas? Let X = distance in miles 72/6 = X/50 6X = (72)(50) 6X = 3600 X = 600 He can drive 600 miles on 50 L of gasoline. 8) A recipe for 2 1/2 dozen muffins requires 600g of flour. How many muffins can be made with 900 g of flour? Let X = # of muffins 2 1/2/600 = X/900 (5/2)/600 = X/900 5/2 (900) = 600X 2250 = 600X X = 3 ¾ You can make 3 and ¾ dozen muffins or 45 muffins. Section 7.3 1) Write the LCD for the fractions in the equation (X -­‐1)/5 + X/3 = 7/10 LCD = 30 2) Solve the above equation for X 30[(X -­‐1)/5 + X/3) = (7/10)(30) 6(X – 1) + 10X = 21 6X – 6 + 10X = 21 16X= 27 X = 27/16 3) Solve: a) (X + 4)/3 – X/7 = (X + 7)/5 b) 2/3(X – 1) – 1/5(X – 2) = (X +2)/3 105[(X + 4)/3 – X/7) = 105[(X + 7) /5] 35(X + 4) – 15X = 21(X + 7) 35X + 140 – 15X = 21X + 147 20X + 140 = 21X + 147 -­‐7 = X X = -­‐7 4) Two numbers are in a ratio of 5:2. ½ of their sum is 10 ½. Find the numbers. Let 5X = first number and 2X = 2nd number ½(5X + 2X) = 10 ½ ½(7X) = 21/2 7X = 21 X = 3 The two numbers are 15 and 6. 5) The width of a rectangle is 4 cm less than the length. If each dimension were increased by 3 cm, the width of the new rectangle formed would be 2/3 the length of the new rectangle. Find the dimensions of the original rectangle. Let L = length of the rectangle in cm L + 3 = length of new rectangle L – 4 = width of the rectangle in cm L -­‐ 1 = width of new rectangle L -­‐ 1 = 2/3(L + 3)_ 3(L -­‐ 1) = 2(L + 3) 3L -­‐ 3 = 2L + 6 L = 9 The length of the original rectangle is 9 cm and its width is 5 cm 6) The lengths of the sides of a triangle are consecutive integers. Half of the perimeter is 14 more than the length of the longest side. Find the perimeter. Let X = 1st side; X +1 = 2nd side; X + 2 = 3rd side 1/2[( X + (X +1) +(X + 2)] = (X + 2) + 14 ½(3X + 3) =X + 16 3X + 3 = 2(X + 16) 3X + 3= 2X + 32 X = 29 The perimeter is 90 units. Section 7.4 Solve: a) 2/(X – 1) = X b) 1/(X – 1) + X/3 = X/(X – 1 X(X – 1) = 2 3(X -­‐1)[1/(X -­‐1) + X/3] = 3(X -­‐1)[[X/(X -­‐1)] 2
X – X = 2 3 + X(X – 1) = 3X X2 – X – 2 = 0 3 + X2 – X = 3X (X – 2)(X + 1) = 0 X2 -­‐ 4X + 3 = 0 X = 2 or -­‐1 (X + 3)(X -­‐ 1) = 0 X = 3 or 1, but 1 is a restricted value So the solution is X = 3 c) (X2 + 1)/(X2 -­‐1) = 2/(X -­‐1) + X/(X + 1) (X2 – 1)[ (X2 + 1)/(X2 -­‐1)] = [2/(X -­‐1) + X/(X + 1)] (X2 – 1) X2 + 1 = 2(X + 1) + X(X – 1) X2 + 1 = 2X + 2 + X2 – X X = -­‐1 But -­‐1 is a restricted value and so there is no solution. d) (N – 2)/N -­‐ (N – 3)/(N – 6) = 1/N N(N – 6)[ (N – 2)/N -­‐ (N – 3)/(N – 6)] = N(N -­‐ 6)(1/N) (N – 6)(N – 2) – N(N – 3) = N – 6 N2 – 8N + 12 – N2 + 3N = N – 6 -­‐ 5N + 12 = N – 6 -­‐6N = -­‐ 18 N = 3 Section 7.5 1) Find 16% of 50 2) 36% of what number is 180? 3) What percent of 80 is 260? .16 • 50 = 8 .36X = 180 X/100 • 80 = 260 X = 50 80X/100 = 260 80X = 26000 X = 325 325/100 = 325% 2) Solve: .08X + .06(1000 – X ) = 72 8X + 6(1000 – X) = 7200 8X + 6000 – 6X = 7200 2X + 6000 = 7200 2X = 1200 X = 600 3) A $60 coat is marked down 15%. What is the new price? .15(60) = 9 Coat was marked down $9 which brings the new price to 60 – 9 or $51. 4) You invested a certain amount of money at 7.5% annual interest and earned $30. How much did you invest? Interest = principal • interest rate Let X = amount invested (principal) 30 = X (.075) X = 30/.075 X = 400 You invested $400. 5) You invested $2500 and earned $275 annual interest. What was the interest rate? Interest = principal • interest rate Let X = interest rate 275 = 2500 • X/100 275 = 2500X/100 27500 = 2500X X = 11 11/100 = 11% The interest rate was 11% Section 7.6 1) The price of a dress was $75. During a sale the price of the dress was marked down to $60. What was the percent decrease? Let X = % decrease X/100 = (75 – 60)/75 X/100= 15/75 75X = 1500 X = 20 20/100 = 20% The percent decrease was 20% 2) Suppose you paid $24 for a pair of shoes that were discounted 20%. What was the original price of the shoes? Let X = original shoe price 20/100 = (X – 24)/X 20X = 100(X – 24) 20X = 100X – 2400 -­‐80X = -­‐2400 X = 30 The shoes originally cost #30. 3) Sam Murray bought a care for $15,000. A year later it was worth $12,000. What was the percent decrease in value? Let X = percent decrease X/100 = (15,000 -­‐12,000)/15,000 X/100 = 1/5 5X = 100 X = 20 The percent decrease was 20% 4) Frank invests part of $8000 at 6% interest and the rest at 12% interest. Her total income from the investments is $570. How much is invested at the 12% rate? Let X = investment at 6% rate 8000 – X = invstement at 12% rate .06X + .12(8000-­‐X) = 570 6X + 12(8000-­‐X) = 57000 6X + 96000 – 12X = 57000 -­‐6X = -­‐39000 X =6500 He invested $1500 at the 12% rate. Section 7.7 1) How many liters of pure acid must be added to 6L of a 75% acid solution to make an 85% solution? Let X = # liters of pure acid to add Amount mixture % Acid Total Acid Acid X 100 1X Acid solution 6 75 .75(6) Mixture X + 6 85 .85(X + 6) X + .75(6) = .85(X + 6) X + 4.5 = .85X + 5.1 .15X = .6 X = 4 4 liters must be added. 2) How many liters of water must be evaporated from 20L of a 30% salt solution to produce a 50% salt solution? Let X = amount of water to be evaporated Amount mixture % salt Total Salt Salt solution 20 30 .2(30) Water X 0 0 Salt Sol -­‐ Water 20 – X 50 .5(20 – X) .2(30) = .5(20 – X) 2(30) = 5(20 – X) 60 = 100 -­‐5X -­‐40 = -­‐5X X = 8 8 liters of water must be evaporated. 3) A collection of 50 coins is worth $5.20. There are 12 more nickels than dimes, and the rest are quarters. How many coins of each type are there in the collection? Let D = # of dimes D + 12 = # of nickels 50 –(D + 12) = # of quarters # of coins Value of coins Total value of coins Dimes D 10 10D Nickels D + 12 5 5(D + 12) Quarters 50 – [D + (D + 12)] 25 25{50 – [D + (D + 12)} 10D +5(D + 12) + 25{50 – [D + (D + 12)}= 520 10D + 5D + 60 + 25(38 – 2D) = 520 10D + 5D + 60 + 950 – 50 D = 520 -­‐35D + 1010 = 520 -­‐ 35D = -­‐ 490 D = 14 The collection had 14 dimes, 26 nickels, and 10 Quarters. 4) A grocer mixes two kinds of cereal. One kind costs $5/kg and the other $5.80/kg. How many kg of each type are needed to produce a mixture of 40kg worth $5.50/kg? Let X = amount in kg of the $5/kg cereal – cereal A # kg Price per kg Total cost Cereal A X 5 5X Cereal B 40 – X 5.8 5.8(40 – X) Total 40 5.5 5.5(40) 5X + 5.8(40 – X) = 5.5(40) 5X + 232 – 5.8X = 220 -­‐.8X + 232 = 220 -­‐.8X = -­‐12 X = 15 You will need 15kg of the $5 cereal and 25 of the $5.80 cereal. Section 7.8 1) One computer can process a bank’s statements in 30 min and a second can process the same number of statements in 20 min. How long would it take to process the statements if both computers were used at the same time? Let X = time in minutes for both computers Work rate Time Work done First computer 1/30 X X/30 nd
2 computer 1/20 X X/20 X/30 + X/ 20 = 1 600(X/30 + X/20) = 1(600) 20X + 30X = 600 50X = 600 X = 12 It would take 12 minutes when both computers work together. 2) Bill can shovel the walk in 45 min. When carol helps him they both can shovel the walk together in 18 min. How long would it take Carol to shovel the walk if she worked alone? Let X = Carol’s time alone Work Rate Time Work Done Carol 1/X 18 18/X Bill 1/45 18 18/45 18/X + 18/45 = 1 45X(18/X + 18/45) = 45X 810 + 18X = 45X 810 = 27X X = 30 It would take Carol 30 minutes to shovel the walk by herself. Section 7.9 1) Simplify: a) (X-­‐4/X)-­‐2 b) (4/5) -­‐1 c) (3X)-­‐2 d) (4X-­‐2Y3)-­‐3 5/4 1/(3X)2 1/9X2 4-­‐3(X6)(Y-­‐9) X6/64Y9 b) 3,009,000,000 c) (4.5 X 10-­‐4)(3.1 X 10-­‐6) (X-­‐4)-­‐2/X-­‐2 X8/X-­‐2 10 X
Section 7.10 Write in scientific notation: a) 345,000,000,000 c) 37 X 105 Mulitply or divide and express in scientific notation: 2.1X10 9
a) (3.5 X 105)(2.8 X 107) b) 3X10 4
Complete each statement: €
a) 30 L = ____ ml b) 5 km = ____ m d) 4dam = ___m e) 2m = ____ cm The answers to this section are on the next page. c) 3mg = ___ g f) 30 cm =____ m a) 3.45 X 1011 c) 3.7 X 106 b) 3.009 X 109 b) .7 X 105 = 7 X 104 c) 13.95 X 10-­‐10 = 1.395 X 10-­‐9
b) 5000 m d) .003 g f) 200 cm f) .3 m a) 9.8 X 1012 a) 30,000 ml e) .4 m