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1.05 Word Problems Intermediate Algebra: One Step at a Time
Page 51 - 76: #3, 4, 7, 8, 17, 18, 21, 25, 27, 29, 30, 31, 32, 35, 36, 40, 45, 47, Extra Problems!
Dr. Robert J. Rapalje Seminole State College of Florida Sanford, FL 32773
To see Section 1.05, with explanations, examples, exercises, and answers, click here!
p.55. 3. Three numbers are such that the second number is 4 more than three times the first, and the third number is 12 less than the sum of the first two numbers. The sum of the three numbers is 44. Find the numbers. Solution: STEP 1: Let x =_____. Let x = First number (since it is at the end of the first phrase!) 3x + 4 = Second number x + 3x + 4 – 12 = Third number (sum of the first two numbers, less 12) 4 x – 8 = Third number STEP 2: Write the equation. The sum of the three numbers is 44. x + 3x + 4 + 4 x – 8 = 44
STEP 3: Solve the equation. 8x – 4 = 44 8x = 48 x = 6
STEP 4: Answer the question. x = 6 First number 3x +4 = 3(6) + 4 = 22 Second number 4x – 8 = 4(6) – 8 = 16 First number
STEP 5: Check. Check: The sum of the numbers is 44. 6+22 + 16 = 44 44 = 44 It checks!!
4. Three numbers are such that the first number is 10 less than twice the second, and the second number is 4 more than three times the third. Twice the second number is equal to the sum of the first and third numbers. Find the numbers. Solution: STEP 1: Let x =_____. Let x = Third number (since it is at the end of the first sentence!) 3x + 4 = Second number 2(3x + 4 ) – 10 = First number 6x + 8 – 10 = First number 6x – 2 = First number STEP 2: Write the equation. Twice the second number is equal to the sum of the first and third. 2 (3x + 4 ) = 6x – 2 + x STEP 3: Solve the equation. 6x + 8 = 6x – 2 + x 6x + 8 = 7x – 2 10 = x
STEP 4: Answer the question. x = 10 Third number 3x +4 = 3(10) + 4 = 34 Second number 6x – 2 = 6(10) – 2 = 58 First number
STEP 5: Check. Check: Twice the second equals sum of the first and third. 2(34) = 58 + 10 68 = 68 It checks!!
7. Find two consecutive odd integers such that twice the second plus the first is 121. Solution: STEP 1: Let x =_____. Let x = First number x + 2 = Second number
STEP 2: Write the equation.
Twice the second number plus the first is 121. 2 (x + 2 ) + x = 121 STEP 3: Solve the equation. 2x + 4 + x = 121 3x + 4 = 121 3x =117 x = 39
STEP 4: Answer the question. x = 39 First odd number x + 2 = 41 Second odd number STEP 5: Check. Check: Twice the second plus the first is 121. 2 (41) + 39 = 121 82 + 39 = 121 It checks!!
8. Three consecutive integers are such that the first, plus twice the second, plus three times the third is equal to 200. Find the numbers. SOLUTION: STEP 1: Let x =_____. Let x = first number x+1 = second number (since these are consecutive integers!) x+2 = third number STEP 2: Write the equation. First + 2 * Second + 3 * Third = 200 ( x ) + 2 * (x+1) + 3 * (x+2) = 200 STEP 3: Solve the equation. x + 2x +2 + 3x+6 = 200 6x + 8 = 200 Subtract 8 from each side: 6x + 8 – 8 = 200 – 8 6x = 192 Divide both sides by 6: x = 192/6 = 32 STEP 4: Answer the question. x = 32 First number x + 1 = 33 Second number x + 2 = 34 Third number STEP 5: Check. Check: ( x ) + 2 * (x+1) + 3 * (x+2) = 200 32 + 2 * 33 + 3 * 34 32 + 66 + 102 = 200 It checks!!
17. The length of a rectangle is three less than five times the width. The perimeter is ten times the width. Find the dimensions and perimeter of the rectangle. SOLUTION: STEP 1: Let x =_____. Let x = width of the rectangle 5x-3 = length of the rectangle 10x = Perimeter
STEP 2: Write the equation. 2( ) + 2( ) = Perimeter 2( x ) + 2( 5x-3) = 10x
STEP 3: Solve the equation. 2x + 10x – 6 = 10x 12x – 6 = 10x Subtract 12x from each side: 12x – 12x – 6 = 10x -12x – 6 = –2x Divide both sides by -2: x = 3
STEP 4: Answer the question. x = 3 Width 5x – 3 = 5(3) – 3 = 12 Length 10x = 30 Perimeter
STEP 5: Check. Check: 2W + 2 L = P 2(3) + 2(12) = 30 6 + 24 = 30 It checks!!
18. The perimeter of a rectangle is 46. Twice the length is 4 more than five times the width. Find the length and width of the rectangle. SOLUTION: STEP 1: Let x =_____. Let x = width of the rectangle 5x+4 = two lengths of the rectangle STEP 2: Write the equation. 2( W ) + 2( L ) = Perimeter 2( x ) + 5x+4 = 46 STEP 3: Solve the equation. 2x + 5x + 4 = 46 7x + 4 = 46 Subtract 4 from each side: 7x +4 – 4 = 46 – 4 7x = 42 Divide both sides by 7: x = 6
STEP 4: Answer the question. x = 6 Width 5x + 4 = 5(6) + 4 = 34 = 2•Length 17 = Length
STEP 5: Check. Check: 2W + 2 L = P 2(6) + 2(17) = 46 12 + 34 = 46 It checks!!
21. A certain number of quarters, four times as many pennies as quarters, and 6 more dimes than pennies are worth $3.36. How many of each coin are there? SOLUTION: STEP 1: Let x =_____. Let x = number of quarters 4x = number of pennies 4x+6 = number of dimes
STEP 2: Write the equation from the last column of the chart above. 25(x) + 1(4x) + 10(4x+6) = 336 STEP 3: Solve the equation.
25 x + 4x + 40x + 60 = 336 69x + 60 = 336 69x = 276 x = 4 STEP 4: Answer the question. x = 4 Quarters 4x =4(4) = 16 Pennies 4x+6 = 16+6= 22 Dimes STEP 5: Check. Check: 4 Quarters $1.00 16 Pennies .16 22 Dimes 2.20 TOTAL: $3.36 It checks!!
25. A box contains $6.60 in nickels, dimes, and quarters. There are three times as many nickels as quarters, and the number of dimes is 4 less than the number of nickels. How many of each coin are there? SOLUTION: STEP 1: Let x =_____. Let x = number of quarters 3x = number of nickels 3x-4 = number of dimes
STEP 2: Write the equation from the last column of the chart above. 25(x) + 5(3x) + 10(3x-4) = 660 STEP 3: Solve the equation. 25 x + 15x + 30x - 40 = 660 70x - 40 = 660 70x = 700 x = 10 STEP 4: Answer the question. x = 10 Quarters 3x =3(10) = 30 Nickels 3x-4 = 30-4= 26 Dimes
STEP 5: Check. Check: 10 Quarters $2.50 30 Nickels 1.50 26 Dimes 2.60 TOTAL: $6.60 It checks!!
27. A certain number of pennies, four times as many dimes as pennies, and a number of quarters which is 16 less than twice the number of dimes, are worth $24.92. How many of each coin are there? SOLUTION: STEP 1: Let x =_____. Let x = number of pennies 4x = number of dimes 2(4x)-16 = number of quarters 8x - 16 = number of quarters
STEP 2: Write the equation from the last column of the chart above. 1(x) + 10(4x) + 25(8x-16) = 2492 STEP 3: Solve the equation. 1 x + 40x + 200x - 400 = 2492 241x - 400 = 2492
x = 12 STEP 4: Answer the question. x = 12 Pennies 4x =4(12) = 48 Dimes 8x-16 = 96-16 = 80 Quarters
STEP 5: Check. Check: 12 Pennies $ 0.12 48 Dimes 4.80 80 Quarters 20.00 TOTAL: $24.92 It checks!!
29. A box contains nickels, dimes, and quarters worth $12.60. The number of dimes is 2 less than three times the number of nickels, and the number of quarters is 4 less than twice the number of dimes. How many of each coin are there? SOLUTION: STEP 1: Let x =_____. Let x = number of nickels 3x – 2 = number of dimes 2(3x– 2) –4 = number of quarters 6x – 4 – 4 or 6x – 8 = number of quarters
STEP 2: Write the equation from the last column of the chart above. 5(x) + 10(3x – 2) + 25(6x – 8) = 1260
STEP 3: Solve the equation. 5 x + 30x - 20 + 150x - 200 = 1260 185 x - 220 = 1260 185 x = 1480
x = 8 STEP 4: Answer the question. x = 8 Nickels 3x – 2 = 3(8) – 2 = 22 Dimes 6x – 8 = 6(8) – 8 = 40 Quarters
STEP 5: Check. Check: 8 Nickels $ 0.40 22 Dimes 2.20 40 Quarters 10.00 TOTAL: $ 12.60 It checks!!
30. A box contains nickels, dimes, and quarters worth $69.50. The number of nickels is 10 more than twice the number of dimes. There are as many quarters as nickels and dimes combined. How many of each coin are there? SOLUTION: STEP 1: Let x =_____. Let x = number of dimes 2x+10 = number of nickels x + 2x+10 = number of quarters 3x + 10 = number of quarters
STEP 2: Write the equation from the last column of the chart above. 10(x) + 5(2x+10) + 25(3x+10) = 6950 STEP 3: Solve the equation. 10 x + 10x + 50 + 75x + 250 = 6950 95x + 300 = 6950 95x = 6650
x = 70 STEP 4: Answer the question. x = 70 Dimes 2x+10 =2(70) +10 = 150 Nickels x+2x+10 = 70 + 150 = 220 Quarters
STEP 5: Check. Check: 70 Dimes $ 7.00 150 Nickels 7.50 220 Quarters 55.00 TOTAL: $69.50 It checks!!
31. A sum of money was invested at 8% simple interest, and three times this much at 10%. The total interest earned for the year was $190. How much was invested at each rate. SOLUTION: STEP 1: Let x =_____. Let x = principle invested 8% 3x = principle invested at 10%
STEP 2: Write the equation from the last column of the chart above. .08(x) + .10(3x) = 190 STEP 3: Solve the equation. .08 x + .30x = 190 .38x = 190
STEP 4: Answer the question. x = $ 500 at 8% 3x = $1500 at 10% STEP 5: Check. Check: $ 500 @ 8% = $ 40 $1500 @ 10% = $150 TOTAL: $190 It checks!!
32. A sum of money was invested at 12% simple interest, and $1000 less than this at 10%. The total interest earned for the year was $1000. How much was invested at each rate. SOLUTION: STEP 1: Let x =_____. Let x = principle invested 12% x−1000 = principle invested at 10%
STEP 2: Write the equation from the last column of the chart above. .12(x) + .10(x−1000) = 1000 STEP 3: Solve the equation. .12 x + .10x − 100 = 1000 .22x − 100 = 1000 .22x = 1100
STEP 4: Answer the question. x = $5000 at 12% x−1000 = $5000-$1000 = $4000 at 10% STEP 5: Check. Check: $5000 @ 12% = $ 600 $4000 @ 10% = $ 400 TOTAL: $1000 It checks!!
35. A man has $10,000 to invest, some in a relatively safe account earning 5% interest per year, and the rest in more speculative investments earning 12% per year. If the total interest earned for the year was $955, how much was invested at each rate? SOLUTION: STEP 1: Let x =_____. Let x = principle invested 5% 10000− x = principle invested at 12%
STEP 2: Write the equation from the last column of the chart above. .05(x) + .12(10000−x) = 955 STEP 3: Solve the equation. .05 x + 1200 −.12x = 955 −.07x + 1200 = 955 −.07x = −245
STEP 4: Answer the question. x = $3500 at 5% 10000−x = 10000-3500 = $6500 at 12% STEP 5: Check. Check: $3500 @ 5% = $ 175 $6500 @ 12% = $ 780 TOTAL: $ 955 It checks!!
36. A sum of money was invested at 5% annual interest, and $500 less than twice this amount was invested at 12%. If the total interest earned for the year was $375, how much was invested at each rate? SOLUTION: STEP 1: Let x =_____. Let x = principle invested 5% 2x−500 = principle invested at 12%
STEP 2: Write the equation from the last column of the chart above. .05(x) + .12(2x−500) = 375 STEP 3: Solve the equation. .05 x + .24x − 60 = 375 .29x − 60 = 375 .29x = 435
STEP 4: Answer the question. x = $1500 at 5% 2x−500 = $3000-$500 = $2500 at 12% STEP 5: Check. Check: $1500 @ 5% = $ 75 $2500 @ 12% = $ 300 TOTAL: $ 375 It checks!!
40. How much water must be added to 50% solution to obtain 100 liters of 10% solution? SOLUTION: STEP 1: Let x =_____. Let x = number of liters of water (which is 0% alcohol!)
STEP 2: Write the equation from the last column of the chart above. 0 + .50(100-x) = .10 (100) STEP 3: Solve the equation. 50 - .50 x = 10 -.50 x = - 40
STEP 4: Answer the question. x = 80 liters of water 100 – x = 20 liters of 50% alcohol STEP 5: Check. Check: 20 liters @ 50% solution = 10 liters 100 liters @ 10% solution = 10 liters . It checks!!
45. A merchant mixes some candy worth $3.50 per pound with cheap stuff worth $1.00 per pound. There are 10 more pounds of the cheap stuff than the more expensive candy. If the total value of the mixture is $28, how many pounds of each are there? SOLUTION: STEP 1: Let x =_____. Let x = number of pounds of expensive candy x+10 = number of cheap stuff
STEP 2: Write the equation from the last column of the chart above. 3.50(x) + 1(x+10) = 28 STEP 3: Solve the equation. 3.5 x + 1.0x + 10 = 28 4.5x + 10 = 28 4.5x = 18
x = 4
STEP 4: Answer the question. x = 4 pounds Expensive Candy x+10 = 4 +10 = 14 pounds Cheap Stuff STEP 5: Check. Check: 4 Expensive @ $3.50 = $ 14.00 14 Cheap Stuff @ $1.00 = 14.00 TOTAL: $28.00 It checks!!
47. Fifty tickets were sold to a chicken barbeque for a total of $219. Children’s tickets sold for $2.50, youth tickets sold for $3.50, and adult tickets sold for $5.00. There were 10 more youth tickets than children’s tickets. How many of each ticket were sold? SOLUTION: STEP 1: Let x =_____. Let x = number of children’s tickets (at the end of the sentence!!) x+10 = number of youth tickets 2x+10 = number of children and youth tickets combined 50-(2x+10) = number of adult tickets 50 - 2x - 10= number of adult tickets 40 - 2x = number of adult tickets
STEP 2: Write the equation from the last column of the chart above. 2.50(x) + 3.50(x+10) + 5.00(40 - 2x) = 219 STEP 3: Solve the equation. 2.5 x + 3.5x + 35 + 200 - 10 x = 219 6.0 x + 235 - 10 x = 219 - 4 x + 235 = 219 - 4 x + 235 - 235 = 219 – 235 - 4 x = -16 x = 4 STEP 4: Answer the question. x = 4 Children’s Tickets x+10 = 4 +10 = 14 Youth Tickets 40 - 2x = 40 - 8 = 32 Adult Tickets STEP 5: Check. Check: 4 Children’s @ $2.50 = $ 10.00 14 Youth @ $3.50 = 49.00 32 Adult @ $5.00 = 160.00 TOTAL = 50 Tickets TOTAL: $219.00 It checks!!
Extra Problem, from Bernard in Chicago. If there is a stock 95% solution of ethanol, how much of this stock solution is needed to produce 700 ml. of a 5% solution? A. how much pure water is needed, B. how much 95% ethanol is needed? We are dealing with a slightly diluted solute, literally making a weaker solution from a stronger solution. SOLUTION: It doesn’t matter whether the mixture is liquid or solid, it is still a mixture problem, and the methods of this section still apply!! STEP 1: Let x =_____. Let x = number of liters of pure water (which is 0% ethanol!)
STEP 2: Write the equation from the last column of the chart above. 0 + .95(700-x) = .05 (700) STEP 3: Solve the equation. 665 - .95 x = 35 -.95 x = - 630
STEP 4: Answer the question. x ≈ 663.16 ml of water 700 – x ≈ 36.84 ml of 95% ethanol STEP 5: Check. Check: 36.84 ml @ 95% solution = 34.998 ml. (note round off error!) 700 ml @ 5% solution = 35 ml . It checks!! NOTE: Life doesn’t always come out even, does it!!
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