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50 challenging linear word problems with solutions

Fifty challenging real-world problems solvable with first-degree equations. Solutions are hidden and, when useful, include a method without an equation.

Articles /50-challenging-linear-word-problems-with-solutions

75 min

This collection contains fifty challenging first-degree word problems. Percentages, ages, rates, mixtures, work, geometry, and allocations all conceal the same structure: a linear relationship with one unknown.

How to use the collection. First try to build the model yourself. Then open that problem’s solution to compare the equation, steps, and check. Whenever a natural arithmetic argument exists, a method without an equation is included as well.

A solving strategy

Choose the unknown and state its unit and constraints; express every other quantity in terms of it; translate one relationship from the text into an equation; solve it; and finally check the result in context. A numerically correct value may still be invalid if it makes a quantity negative, exceeds a capacity, or is not an integer when objects are counted.

The 50 problems

  1. The manuscript in three days

    An editor proofreads a manuscript in three days. On the first day she reads 3/8 of its pages. On the second day she reads 40% of the pages left after the first day, plus another 12 pages. On the third day she finishes by reading the final 69 pages. How many pages are in the manuscript?

    Show the solution to problem 1

    Unknown. Let x be the total number of pages.

    After the first day, 5x/8 pages remain; 40% of that amount is (2/5)(5x/8) = x/4. Thus, on the second day she reads x/4 + 12 pages.

    Equation. x − 3x/8 − (x/4 + 12) = 69.

    Steps. 3x/8 − 12 = 69, so 3x/8 = 81 and x = 216.

    Check. She reads 81 pages on the first day, leaving 135; on the second she reads 40% of 135 + 12 = 54 + 12 = 66. The number left is 216 − 81 − 66 = 69, as required.

    Answer. 216 pages

    Solution without an equation

    Forty percent of the remaining 5/8 is 1/4 of the manuscript. Without the 12 extra pages, the share for the third day would therefore be 1 − 3/8 − 1/4 = 3/8 of the total. Put those 12 pages back with the final pages: 69 + 12 = 81 represents 3/8. One eighth is 27 pages, so the whole manuscript has 216 pages.

  2. The tank after two operations

    A tank can hold 200 litres. From an unknown initial amount, one fifth of the contents is drained, followed immediately by another 12 litres. Then 30 litres are added. At the end, the tank is 3/4 full. How many litres did it contain initially?

    Show the solution to problem 2

    Unknown. Let x be the initial number of litres.

    The total amount drained is x/5 + 12; the final level is 3/4 · 200 = 150 litres.

    Equation. x − (x/5 + 12) + 30 = 150.

    Steps. 4x/5 + 18 = 150, so 4x/5 = 132 and x = 165.

    Check. One fifth of 165 is 33; draining that and another 12 litres leaves 165 − 33 − 12 = 120 litres. Adding 30 gives 150, which is three quarters of 200.

    Answer. 165 litres

    Solution without an equation

    There are 150 litres at the end, so before the 30-litre addition there were 120. Put back the fixed 12 litres mentally: 132 litres represents 4/5 of the initial contents. One fifth is therefore 33 litres, and the initial amount was 165 litres.

  3. Luca's and his mother's ages

    Today Luca's mother is 4 years older than three times Luca's age. Eight years ago, the mother was exactly five times as old as Luca was then. How old are they today?

    Show the solution to problem 3

    Unknown. Let x be Luca's current age; his mother's age is 3x + 4.

    Equation. Eight years ago their ages were x − 8 and 3x + 4 − 8, so 3x − 4 = 5(x − 8).

    Steps. 3x − 4 = 5x − 40, hence 36 = 2x and x = 18. His mother is 3 · 18 + 4 = 58.

    Check. Eight years ago Luca was 10 and his mother was 50; indeed, 50 = 5 · 10. Today, 58 = 3 · 18 + 4.

    Answer. Luca is 18; his mother is 58

  4. Full-price and reduced tickets

    A total of 340 tickets are sold for a show. A full-price ticket costs €12.50 and a reduced ticket costs €8. The total revenue is €3,512. How many tickets of each type were sold?

    Show the solution to problem 4

    Unknown. Let x be the number of full-price tickets; the number of reduced tickets is 340 − x.

    Equation. 12.50x + 8(340 − x) = 3512.

    Steps. 12.50x + 2720 − 8x = 3512, so 4.50x = 792 and x = 176. There are 340 − 176 = 164 reduced tickets.

    Check. 176 · 12.50 = 2200 euros and 164 · 8 = 1312 euros; 2200 + 1312 = 3512 euros, and 176 + 164 = 340 tickets.

    Answer. 176 full-price and 164 reduced tickets

    Solution without an equation

    If all 340 tickets were reduced, the revenue would be 340 · 8 = 2720 euros, leaving a difference of 792 euros. Replacing one reduced ticket with a full-price ticket adds 12.50 − 8 = 4.50 euros, so 792 ÷ 4.50 = 176 replacements are needed. The other 164 tickets are reduced.

  5. A mixture at the required concentration

    How many kilograms of an 18% saline solution must be mixed with 5 kg of a 42% solution to obtain a 24% solution?

    Show the solution to problem 5

    Unknown. Let x be the kilograms of 18% solution.

    Equation. The mass of salt before and after mixing is the same: 0.18x + 0.42 · 5 = 0.24(x + 5).

    Steps. 0.18x + 2.10 = 0.24x + 1.20, so 0.90 = 0.06x and x = 15.

    Check. The first solution contains 15 · 0.18 = 2.7 kg of salt and the second contains 2.1 kg. That is 4.8 kg of salt in 20 kg of mixture: 4.8/20 = 0.24 = 24%.

    Answer. 15 kg of the 18% solution

    Solution without an equation

    Using alligation, 18% is 6 percentage points below 24%, while 42% is 18 points above it. The amounts must be in the inverse ratio 18 : 6 = 3 : 1. Since the 42% part weighs 5 kg, the 18% part must weigh 3 · 5 = 15 kg.

  6. The average after a correction

    A student has eight recorded scores with an average of 74. One score, mistakenly entered as 52, must be replaced by the correct score. After the correction, the student takes a ninth test and scores 86; the average of all nine scores becomes 78. What was the correct score?

    Show the solution to problem 6

    Unknown. Let x be the correct score.

    The sum of the eight recorded scores is 8 · 74 = 592. Remove 52, insert x, and add 86.

    Equation. 592 − 52 + x + 86 = 9 · 78.

    Steps. 626 + x = 702, so x = 76.

    Check. The corrected total for the first eight scores is 592 − 52 + 76 = 616; with the ninth score it is 616 + 86 = 702, and 702/9 = 78.

    Answer. 76

    Solution without an equation

    An average of 78 over nine tests requires 702 points in total. The seven unchanged scores contribute 592 − 52 = 540 points and the ninth test adds 86, so 626 points are already known. The missing score is 702 − 626 = 76.

  7. Two speeds and a stop

    A coach travels a 180 km route. It travels at 60 km/h on the first section and at 90 km/h for the rest. It stops for 20 minutes during the journey; the total time from departure to arrival is 2 hours 50 minutes. How long is the first section?

    Show the solution to problem 7

    Unknown. Let x be the kilometres travelled at 60 km/h; the second section is 180 − x km.

    The stop lasts 1/3 hour and the total time is 17/6 hours, so the driving time is 5/2 hours.

    Equation. x/60 + (180 − x)/90 = 5/2.

    Steps. Multiplying by 180 gives 3x + 2(180 − x) = 450; hence x + 360 = 450 and x = 90.

    Check. The first 90 km take 1.5 hours and the other 90 km take 1 hour. Adding the 20-minute stop gives 2 hours 50 minutes.

    Answer. 90 km

    Solution without an equation

    If all 180 km were travelled at 90 km/h, the driving time would be 2 hours; it is actually 2 hours 30 minutes, half an hour longer. Each kilometre moved from 90 km/h to 60 km/h adds 1/60 − 1/90 = 1/180 hour. Adding half an hour therefore requires (1/2) · 180 = 90 km at the slower speed.

  8. Pay at different rates

    Two craftspeople work on the same project. Ada works 32 hours at an unknown hourly rate. Bruno works 28 hours at a rate 20% higher than Ada's and also receives a fixed €60 bonus. Their combined pay is €1,700. What is each person's hourly rate?

    Show the solution to problem 8

    Unknown. Let x be Ada's hourly rate; Bruno's is 1.20x.

    Equation. 32x + 28(1.20x) + 60 = 1700.

    Steps. 32x + 33.6x = 1640, so 65.6x = 1640 and x = 25. Bruno earns 1.20 · 25 = 30 euros per hour.

    Check. Ada receives 32 · 25 = 800 euros; Bruno receives 28 · 30 + 60 = 900 euros. Their total is 800 + 900 = 1700 euros.

    Answer. Ada €25/h; Bruno €30/h

  9. Fencing with a gate

    A rectangular courtyard's length is 8 m more than 3/2 of its width. Fencing it requires 102 m of mesh, with a 4 m wide gate left unfenced. What are the courtyard's dimensions?

    Show the solution to problem 9

    Unknown. Let x be the width; the length is 3x/2 + 8.

    Equation. The gate is excluded from the fenced perimeter: 2[x + (3x/2 + 8)] − 4 = 102.

    Steps. 5x + 12 = 102, so 5x = 90 and x = 18. The length is 3 · 18/2 + 8 = 35 m.

    Check. The perimeter is 2(18 + 35) = 106 m; after excluding the 4 m gate, exactly 102 m of mesh is needed.

    Answer. 18 m × 35 m

    Solution without an equation

    The mesh and gate make a perimeter of 102 + 4 = 106 m, so width plus length is 53 m. Removing the extra 8 m in the length leaves 45 m split into 1 + 3/2 = 5/2 width-parts. The width is 45 ÷ (5/2) = 18 m and the length is 35 m.

  10. The number with reversed digits

    A two-digit number has digits whose sum is 11. Reversing the digits makes the number 27 smaller. What is the original number?

    Show the solution to problem 10

    Unknown. Let x be the tens digit; the units digit is 11 − x.

    The original number is 10x + (11 − x); the reversed number is 10(11 − x) + x.

    Equation. [10x + (11 − x)] − [10(11 − x) + x] = 27.

    Steps. 18x − 99 = 27, so 18x = 126 and x = 7. The units digit is 4.

    Check. The number is 74: 7 + 4 = 11 and 74 − 47 = 27.

    Answer. 74

    Solution without an equation

    The difference between a two-digit number and its reversal is nine times the difference between its digits. Since 27 ÷ 9 = 3, the tens digit is 3 greater than the units digit. The two digits with sum 11 and difference 3 are 7 and 4, so the number is 74.

  11. Successive discount and tax

    An item is sold at a 15% discount on its list price. A 22% tax is then applied to the discounted price. The customer pays €124.44. What was the list price?

    Show the solution to problem 11

    Unknown. Let x be the list price in euros.

    The discount leaves 85% of the price, and the tax multiplies that result by 1.22.

    Equation. 1.22 · 0.85x = 124.44.

    Steps. 1.037x = 124.44, so x = 124.44/1.037 = 120.

    Check. Fifteen percent of 120 is 18, so the discounted price is 102 euros. Twenty-two percent of 102 is 22.44 euros, and 102 + 22.44 = 124.44 euros.

    Answer. €120

    Solution without an equation

    Undo the operations in reverse order. Remove the tax: 124.44 ÷ 1.22 = 102 euros. Those 102 euros are 85% of the list price, so the original price is 102 ÷ 0.85 = 120 euros.

  12. Capital split between two returns

    A capital sum of €15,000 is split between two one-year investments: the first earns 3.2% and the second 4.7%. After one year, the total interest is €592.50. How much was placed in each investment?

    Show the solution to problem 12

    Unknown. Let x be the amount invested at 3.2%; the amount at 4.7% is 15,000 − x euros.

    Equation. 0.032x + 0.047(15,000 − x) = 592.50.

    Steps. 0.032x + 705 − 0.047x = 592.50, hence 0.015x = 112.50 and x = 7500.

    Check. 7500 · 0.032 = 240 euros and 7500 · 0.047 = 352.50 euros; their sum is 592.50 euros.

    Answer. €7,500 at 3.2% and €7,500 at 4.7%

    Solution without an equation

    The average return is 592.50/15,000 = 3.95%. Since 3.95% lies exactly halfway between 3.2% and 4.7%, the two rates must have equal weights. The capital is therefore divided into two equal amounts of €7,500.

  13. Downstream and back

    A boat travels the same stretch of river first downstream at 18 km/h and then upstream at 12 km/h. Its total travelling time is 2 hours 30 minutes. How long is the stretch in one direction?

    Show the solution to problem 13

    Unknown. Let x be the length in kilometres travelled in each direction.

    Equation. The sum of the times is x/18 + x/12 = 5/2.

    Steps. Multiplying by 36 gives 2x + 3x = 90, so 5x = 90 and x = 18.

    Check. Downstream, the boat takes 18/18 = 1 hour; upstream, it takes 18/12 = 1.5 hours. The total is 2.5 hours.

    Answer. 18 km

  14. The break-even point between two plans

    For someone using more than 50 minutes per month, plan A costs €24 and includes the first 50 minutes; each additional minute costs €0.08. Plan B costs €8 plus €0.14 for every minute, starting with the first. At how many monthly minutes do the two plans cost the same?

    Show the solution to problem 14

    Unknown. Let x be the monthly minutes, with x > 50.

    Equation. 24 + 0.08(x − 50) = 8 + 0.14x.

    Steps. 20 + 0.08x = 8 + 0.14x, so 12 = 0.06x and x = 200.

    Check. Plan A costs 24 + 0.08 · 150 = 36 euros; plan B costs 8 + 0.14 · 200 = 36 euros.

    Answer. 200 minutes; both cost €36

    Solution without an equation

    At 50 minutes, plan A costs €24 and plan B costs 8 + 0.14 · 50 = 15 euros, so A must make up a €9 difference. After minute 50, A is €0.06 cheaper per minute. It takes 9/0.06 = 150 additional minutes to close the gap, so break-even occurs at 200 minutes.

  15. An inheritance with linked shares

    An inheritance of €108,000 is divided among Anna, Bruno, and Carla. Bruno receives €3,000 plus 3/4 of Anna's share. Carla receives €4,000 plus 2/3 of Bruno's share. How much does each person receive?

    Show the solution to problem 15

    Unknown. Let x be Anna's share. Bruno receives 3x/4 + 3000; Carla receives 2/3(3x/4 + 3000) + 4000.

    Equation. x + (3x/4 + 3000) + [2/3(3x/4 + 3000) + 4000] = 108,000.

    Steps. Carla's share simplifies to x/2 + 6000; therefore 9x/4 + 9000 = 108,000, 9x/4 = 99,000, and x = 44,000. Bruno receives €36,000 and Carla €28,000.

    Check. 3/4 · 44,000 + 3000 = 36,000; 2/3 · 36,000 + 4000 = 28,000; finally, 44,000 + 36,000 + 28,000 = 108,000.

    Answer. Anna €44,000; Bruno €36,000; Carla €28,000

  16. Output from three production lines

    Three lines together produce a batch of components. Line A makes 40% of the total; line B makes 120 more components than line A. Line C is left with 520 components. How many components are in the batch?

    Show the solution to problem 16

    Unknown. Let x be the total number of components. A makes 0.40x and B makes 0.40x + 120.

    Equation. x − 0.40x − (0.40x + 120) = 520.

    Steps. 0.20x − 120 = 520, so 0.20x = 640 and x = 3200.

    Check. A makes 0.40 · 3200 = 1280 components; B makes 1,400 and C makes 520. Their sum is 1280 + 1400 + 520 = 3200.

    Answer. 3,200 components

    Solution without an equation

    Without B's additional 120 components, A and B together would cover 80% of the batch, leaving 20% for C. Those extra 120 components are taken from C's share: adding them back to the actual 520 gives 640 components, which is 20% of the total. The batch therefore contains 640 · 5 = 3200 components.

  17. Books on three shelves

    Three shelves hold 249 books in total. The second holds 6 books more than 75% of the number on the first. The third holds 18 books more than half the number on the first. How many books are on each shelf?

    Show the solution to problem 17

    Unknown. Let x be the number of books on the first shelf. The second holds 0.75x + 6 and the third x/2 + 18.

    Equation. x + (0.75x + 6) + (x/2 + 18) = 249.

    Steps. 2.25x + 24 = 249, so 2.25x = 225 and x = 100. The second shelf has 81 books and the third 68.

    Check. 75% of 100 + 6 = 81; 100/2 + 18 = 68; finally, 100 + 81 + 68 = 249.

    Answer. 100, 81, and 68 books

    Solution without an equation

    Removing the additional 6 and 18 books from the total leaves 225 books. These equal 1 + 3/4 + 1/2 = 9/4 times the first shelf's contents. One quarter of that contents is 225 ÷ 9 = 25 books, so the first shelf holds 100.

  18. The reserve on a map

    On a 1:25,000 map, a rectangular nature reserve's length is 3.5 cm greater than its width. Its actual perimeter is 6.75 km. What are its dimensions on the map and in reality?

    Show the solution to problem 18

    Unknown. Let x be the map width in centimetres; the length is x + 3.5. At a scale of 1:25,000, 1 cm represents 250 m.

    Equation. 2[x + (x + 3.5)] · 250 = 6750.

    Steps. 500(2x + 3.5) = 6750, so 2x + 3.5 = 13.5 and x = 5. The map length is 8.5 cm.

    Check. The actual dimensions are 5 · 250 = 1250 m and 8.5 · 250 = 2125 m. Their perimeter is 2(1250 + 2125) = 6750 m, or 6.75 km.

    Answer. Map: 5 cm × 8.5 cm; actual: 1.25 km × 2.125 km

    Solution without an equation

    The 6,750 m actual perimeter corresponds to 6750 ÷ 250 = 27 cm on the map. The semiperimeter is 13.5 cm: width and length sum to 13.5 cm and differ by 3.5 cm. The smaller measure is (13.5 − 3.5) ÷ 2 = 5 cm and the larger is 8.5 cm.

  19. Available seats in the hall

    A hall has 18 rows with the same number of seats in each. Twelve seats cannot be used. Of the remaining seats, one sixth is reserved for guests and 260 are available to the public. How many seats are in each row?

    Show the solution to problem 19

    Unknown. Let x be the seats in each row. There are 18x seats in total and 18x − 12 usable seats.

    Since one sixth is reserved, the 260 public seats represent 5/6 of the usable seats.

    Equation. 5/6(18x − 12) = 260.

    Steps. 18x − 12 = 312, so 18x = 324 and x = 18.

    Check. There are 18 · 18 = 324 seats; removing the 12 unusable ones leaves 312. One sixth, or 52, is reserved, leaving 312 − 52 = 260 public seats.

    Answer. 18 seats per row

    Solution without an equation

    The 260 public seats are five sixths of the usable seats. One sixth is 260 ÷ 5 = 52, so there are 52 · 6 = 312 usable seats. Adding the 12 unusable ones gives 324 seats in total; divided among 18 rows, that is 18 seats per row.

  20. The clock hands meet

    Between 3:00 and 4:00, at exactly what time does the minute hand first catch the hour hand? Include the seconds.

    Show the solution to problem 20

    Unknown. Let x be the minutes elapsed after 3:00.

    The minute hand is 6x degrees clockwise from 12; the hour hand starts at 90 degrees and moves another 0.5x degrees.

    Equation. 6x = 90 + 0.5x.

    Steps. 5.5x = 90, so x = 180/11 = 16 4/11 minutes. The remaining 4/11 minute is 240/11 = 21 9/11 seconds.

    Check. At x = 180/11, both angular positions equal 6 · 180/11 = 1080/11 degrees and 90 + 0.5 · 180/11 = 1080/11 degrees.

    Answer. approximately 3:16:21.82 (exactly 3:16:21 9/11)

    Solution without an equation

    At 3:00 the minute hand must make up a 90-degree gap. Its speed relative to the hour hand is 6 − 0.5 = 5.5 degrees per minute. Catching up therefore takes 90 ÷ 5.5 = 180/11 minutes, or 16 minutes and 21 9/11 seconds.

  21. Work begun together

    Ada can complete a job alone in 12 days; Bruno would take 24 days alone. They work together for some days, then Bruno leaves the project and Ada finishes the job in another 3 days. For how many days did they work together?

    Show the solution to problem 21

    Unknown. Let x be the days worked together. In one day Ada completes 1/12 of the job and Bruno 1/24.

    Equation. x(1/12 + 1/24) + 3/12 = 1.

    Steps. x/8 + 1/4 = 1, so x/8 = 3/4 and x = 6.

    Check. Together they complete 6 · (1/8) = 3/4 of the job; Ada completes 3 · (1/12) = 1/4 during the last three days. The two parts make the whole job.

    Answer. 6 days

    Solution without an equation

    Imagine dividing the job into 24 units. Ada completes 2 units per day and Bruno 1. In the final three days Ada completes 6 units, so 18 units had already been done. Together they produce 3 units per day, and 18 ÷ 3 = 6 days are needed.

  22. Reconstructing income from a budget

    Twenty-eight percent is withheld for tax from a gross monthly income. A rent of €720 is then paid from the net income. Of the money left after rent, 3/8 is used for other expenses, leaving €990. What was the gross income?

    Show the solution to problem 22

    Unknown. Let x be the gross monthly income. After tax, 0.72x euros remain, and after rent the amount is 0.72x − 720.

    Since 3/8 is spent, the €990 represents 5/8 of the amount available after rent.

    Equation. 5/8(0.72x − 720) = 990.

    Steps. 0.72x − 720 = 1584, so 0.72x = 2304 and x = 3200.

    Check. Twenty-eight percent of 3,200 is 896, leaving a net income of €2,304. After €720 rent, €1,584 remains; 3/8 is €594 and 1584 − 594 = 990.

    Answer. €3,200 gross per month

    Solution without an equation

    Work backwards. If €990 is 5/8, the amount after rent was 990 · 8/5 = 1584 euros. Before rent, net income was 1584 + 720 = 2304 euros. Net income is 72% of gross income, so gross income was 2304 ÷ 0.72 = 3200 euros.

  23. The angles of a triangle

    In a triangle, the second angle is 18° greater than twice the first; the third angle is 12° less than three times the first. Find all three angles.

    Show the solution to problem 23

    Unknown. Let x be the first angle. The second is 2x + 18 degrees and the third is 3x − 12 degrees.

    Equation. Interior angles sum to 180°: x + (2x + 18) + (3x − 12) = 180.

    Steps. 6x + 6 = 180, so 6x = 174 and x = 29. The other angles are 2 · 29 + 18 = 76 degrees and 3 · 29 − 12 = 75 degrees.

    Check. 29 + 76 + 75 = 180; moreover, 76 = 2 · 29 + 18 and 75 = 3 · 29 − 12. All angles are positive.

    Answer. 29°, 76°, and 75°

    Solution without an equation

    Temporarily ignore the adjustments of +18° and −12°. The three angles then contain 1 + 2 + 3 = 6 equal base-parts. The adjustments add 6° overall, so the six base-parts total 180° − 6° = 174°. Each part is 174° ÷ 6 = 29°, from which 76° and 75° follow.

  24. The population before the changes

    A town's population increases by 8% during one year. Immediately afterwards, 520 residents move away and the population falls to 19,352. How many residents were there at the beginning of the year?

    Show the solution to problem 24

    Unknown. Let x be the initial population. After the 8% increase, the population is 1.08x.

    Equation. 1.08x − 520 = 19,352.

    Steps. 1.08x = 19,872, so x = 19,872/1.08 = 18,400.

    Check. Eight percent of 18,400 is 1,472; after the increase there are 18,400 + 1,472 = 19,872 residents. Subtracting 520 gives 19,352.

    Answer. 18,400 residents

    Solution without an equation

    Work backwards. Before the residents moved away, the population was 19,352 + 520 = 19,872. This is 108% of the initial population; dividing by 1.08 gives 19,872 ÷ 1.08 = 18,400.

  25. Compression, encryption, and metadata

    A digital archive is compressed, reducing its size by 25%. Encryption then increases the compressed file's size by 8%. Finally, 12 MB of metadata is added; the resulting package occupies 660 MB. What was the original size?

    Show the solution to problem 25

    Unknown. Let x be the original size in megabytes. After compression, 0.75x remains; after encryption, it becomes 1.08 · 0.75x.

    Equation. 1.08 · 0.75x + 12 = 660.

    Steps. 0.81x + 12 = 660, so 0.81x = 648 and x = 800.

    Check. Compression takes 800 MB to 800 · 0.75 = 600 MB. Encryption takes it to 600 · 1.08 = 648 MB; adding 12 MB gives 660 MB.

    Answer. 800 MB

    Solution without an equation

    Undo the transformations backwards. Removing the metadata leaves 660 − 12 = 648 MB. Before the 8% increase, the file occupied 648 ÷ 1.08 = 600 MB. That is 75% of the original, so the initial size was 600 ÷ 0.75 = 800 MB.

  26. The notebook warehouse

    A warehouse receives an unknown number of boxes, each containing 24 notebooks. Quality control discards 5% of all the notebooks. Another 180 loose notebooks then arrive. The available notebooks fill 74 new cartons of 30, with 12 notebooks left over. How many boxes arrived originally?

    Show the solution to problem 26

    Unknown: let x be the original number of boxes; they contain 24x notebooks.

    Equation: 95% remain after inspection, and 180 are then added: 0.95 · 24x + 180 = 74 · 30 + 12.

    Steps: 22.8x + 180 = 2232, so 22.8x = 2052 and x = 90.

    Check: 90 boxes contain 90 · 24 = 2160 notebooks; 5% is 108, leaving 2,052. Adding 180 gives 2,232 notebooks, exactly 74 · 30 + 12.

    Answer. Originally there were 90 boxes, or 2,160 notebooks.

    Solution without an equation

    At the end there are 74 · 30 + 12 = 2232 notebooks. Removing the 180 added after inspection leaves 2,052, which is 95% of the original quantity. Dividing by 0.95 gives 2,160 notebooks, and dividing by 24 gives 90 boxes.

  27. The tournament table

    A team plays 38 matches. It has 4 fewer draws than twice its number of losses; all other matches are wins. A win is worth 3 points, a draw 1 and a loss 0. After a 5-point penalty, the team finishes with 82 points. How many wins, draws and losses did it record?

    Show the solution to problem 27

    Unknown: let x be the number of losses. Draws number 2x − 4, and wins number 38 − x − (2x − 4) = 42 − 3x.

    Equation: including the penalty, 3(42 − 3x) + (2x − 4) − 5 = 82.

    Steps: 126 − 9x + 2x − 4 − 5 = 82, so 117 − 7x = 82, 7x = 35, and x = 5. There are 6 draws and 27 wins.

    Check: 27 + 6 + 5 = 38 matches, and the final score is 3 · 27 + 6 − 5 = 81 + 6 − 5 = 82.

    Answer. The team recorded 27 wins, 6 draws and 5 losses.

  28. Currency exchange with two commissions

    A customer brings a sum in euros to a currency exchange. The exchange keeps 2% of the sum and a fixed €6 fee, then converts the remainder at 1.08 dollars per euro. The correspondent bank finally deducts $9. The customer receives $1,042.92. How many euros were originally handed over?

    Show the solution to problem 28

    Unknown: let x be the initial sum in euros. After both euro fees, 0.98x − 6 euros remain.

    Equation: after conversion and the dollar fee, 1.08(0.98x − 6) − 9 = 1042.92.

    Steps: 1.08(0.98x − 6) = 1051.92, so 0.98x − 6 = 974, 0.98x = 980, and x = 1000.

    Check: 2% of €1,000 is €20; after the additional €6 fee, €974 remain. Conversion gives 974 · 1.08 = 1051.92 dollars, and the final $9 deduction leaves $1,042.92.

    Answer. The customer originally handed over €1,000.

    Solution without an equation

    Undo the operations: add $9 to get $1,051.92, divide by 1.08 to get €974, and add back the fixed €6 to get €980. This is 98% of the original sum, so 980 ÷ 0.98 = 1000 euros.

  29. The train that catches up

    A train leaves at 8:15 a.m. travelling at 72 km/h. A second train leaves the same station on the same route at 9:05 a.m. at 96 km/h, but makes a 10-minute stop before catching the first train. At what time does it catch up?

    Show the solution to problem 29

    Unknown: let t be the hours from 9:05 a.m. to the meeting. The first train travels for t + 5/6 hours; the second is moving for t − 1/6 hours.

    Equation: at the catch-up point their distances are equal: 72(t + 5/6) = 96(t − 1/6).

    Steps: 72t + 60 = 96t − 16, so 76 = 24t and t = 19/6 hours, or 3 hours 10 minutes. Adding this to 9:05 a.m. gives 12:15 p.m.

    Check: the first train has travelled for 4 hours and covers 72 · 4 = 288 km. Excluding its stop, the second has travelled for 3 hours and covers 96 · 3 = 288 km.

    Answer. The second train catches the first at 12:15 p.m., 288 km from the station.

  30. The transformed rectangle

    A rectangle’s length is 12 cm less than three times its width. If the length is reduced by 4 cm and the width increased by 5 cm, the new perimeter is 130 cm. Find the original dimensions.

    Show the solution to problem 30

    Unknown: let x be the original width in centimetres; the length is 3x − 12.

    Equation: the new dimensions are x + 5 and 3x − 16, so 2[(x + 5) + (3x − 16)] = 130.

    Steps: 2(4x − 11) = 130, hence 8x − 22 = 130, 8x = 152, and x = 19. The original length is 3 · 19 − 12 = 45 cm.

    Check: the new dimensions are 24 cm and 41 cm; their perimeter is 2(24 + 41) = 130 cm.

    Answer. The original rectangle measures 19 cm by 45 cm.

    Solution without an equation

    The new semiperimeter is 65 cm. Reducing one side by 4 cm and increasing the other by 5 cm increases their sum by 1 cm, so the original sum was 64 cm. The length is 12 cm less than three widths: add 12 to 64 and divide the resulting four widths by 4. This gives a width of 19 cm and then a length of 45 cm.

  31. Dosage in a simulation

    In a simulated veterinary exercise, the theoretical dose is 5 mg per kilogram of mass plus a fixed 30 mg. Preparation requires loading 20% more than the theoretical dose. At the end, 18 mg remain in the line and 462 mg have actually been delivered. What animal mass does the simulation represent?

    Show the solution to problem 31

    Unknown: let x be the simulated animal mass in kilograms. The theoretical dose is 5x + 30 mg, and the loaded amount is 1.20(5x + 30) mg.

    Equation: subtracting what remains in the line gives 1.20(5x + 30) − 18 = 462.

    Steps: 1.20(5x + 30) = 480, so 5x + 30 = 400, 5x = 370, and x = 74.

    Check: at 74 kg the theoretical dose is 5 · 74 + 30 = 400 mg. Adding 20% makes the loaded amount 480 mg; subtracting the 18 mg left in the line gives 462 mg.

    Answer. The simulated mass is 74 kg.

    Solution without an equation

    Work backwards: add the 18 mg left in the line to the 462 mg delivered, giving 480 mg loaded. Divide by 1.20 to remove the extra 20%, obtaining a 400 mg theoretical dose. After removing the fixed 30 mg, the remaining 370 mg represent 5 mg/kg: 370 ÷ 5 = 74 kg.

  32. The three-digit number

    In a three-digit number, the hundreds digit is twice the units digit, and the tens digit is 3 less than the hundreds digit. Reversing the digits decreases the number by 396. What is the number?

    Show the solution to problem 32

    Unknown: let x be the units digit. The hundreds and tens digits are 2x and 2x − 3.

    Equation: the number is 100(2x) + 10(2x − 3) + x, and its reverse is 100x + 10(2x − 3) + 2x. Hence [100(2x) + 10(2x − 3) + x] − [100x + 10(2x − 3) + 2x] = 396.

    Steps: the tens terms cancel, leaving 99x = 396, so x = 4. The digits are 8, 5 and 4.

    Check: the number is 854, its reverse is 458, and 854 − 458 = 396; also 8 = 2 · 4 and 5 = 8 − 3.

    Answer. The number is 854.

    Solution without an equation

    When the hundreds and units are reversed, the tens digit does not affect the difference. Each unit of difference between the outside digits contributes 99; since 396 ÷ 99 = 4, the outside digits differ by 4. The hundreds digit is twice the units digit, so that difference is the units digit itself: 4. The other digits are then 8 and 5.

  33. Perpendicular clock hands

    Between 3:30 and 4:00, when are a clock’s hands perpendicular for the first time? Treat the hour hand as moving continuously.

    Show the solution to problem 33

    Unknown: let t be the minutes after 3:00. In the stated interval the minute hand is ahead of the hour hand.

    Equation: their angles from 12 are 6t degrees for the minute hand and 90 + 0.5t for the hour hand. Their difference must be 90°: 6t − (90 + 0.5t) = 90.

    Steps: 5.5t − 90 = 90, hence 5.5t = 180 and t = 360/11 = 32 8/11 minutes. The 8/11 minute equals 480/11 = 43 7/11 seconds.

    Check: at that instant the minute hand is at 2160/11° and the hour hand at 1170/11°; their difference is 990/11° = 90°.

    Answer. At about 3:32:43.64 (exactly 3:32:43 and 7/11 seconds).

  34. The boat and the current

    A boat travels at 18 km/h in still water. It goes upstream for 2 hours 30 minutes and then downstream for 1 hour 30 minutes, covering 69 km altogether. Find the constant speed of the current.

    Show the solution to problem 34

    Unknown: let x be the current’s speed in km/h. The effective speeds are 18 − x upstream and 18 + x downstream.

    Equation: the total distance gives 2.5(18 − x) + 1.5(18 + x) = 69.

    Steps: 45 − 2.5x + 27 + 1.5x = 69, so 72 − x = 69 and x = 3.

    Check: upstream the boat travels at 15 km/h and covers 2.5 · 15 = 37.5 km; downstream it travels at 21 km/h and covers 1.5 · 21 = 31.5 km. The total is 69 km.

    Answer. The current flows at 3 km/h.

    Solution without an equation

    With no current, the boat would cover 72 km in 4 hours. The current subtracts its speed for 2.5 hours and adds it for 1.5 hours, so its net effect is to subtract the current’s speed once. The 3 km shortfall from 72 therefore means a 3 km/h current.

  35. Tickets for the show

    A show has 180 attendees. There are 10 more students than twice the number of adults; everyone else is a child. Tickets cost €18 for adults, €12 for students and €7 for children. Total receipts are €2,150. How many attendees are in each category?

    Show the solution to problem 35

    Unknown: let x be the number of adults. There are 2x + 10 students and 180 − x − (2x + 10) = 170 − 3x children.

    Equation: the receipts give 18x + 12(2x + 10) + 7(170 − 3x) = 2150.

    Steps: 18x + 24x + 120 + 1190 − 21x = 2150, hence 21x + 1310 = 2150, 21x = 840, and x = 40. There are 90 students and 50 children.

    Check: 40 + 90 + 50 = 180, and receipts are 40 · 18 + 90 · 12 + 50 · 7 = 720 + 1080 + 350 = 2150 euros.

    Answer. There are 40 adults, 90 students and 50 children.

    Solution without an equation

    If everyone were a child, receipts would be €1,260. Each adult in place of a child adds €11 and each student adds €5. The 10 extra students add €50; each adult also brings two students, adding 11 + 2 · 5 = 21 euros per such group. The unexplained amount is 2150 − 1260 − 50 = 840 euros, or 40 groups of €21. Hence 40 adults, 90 students and 50 children.

  36. The battery energy balance

    An 80 kWh battery starts with an unknown amount of energy. During a journey, traction consumes 30% of the initial energy and onboard systems consume another 9 kWh. Regenerative braking returns 40% of the energy used for traction. On arrival, 20 kWh are drawn from the grid, but charging efficiency is 85%. The battery finally contains 49 kWh. How much energy, and what percentage charge, did it start with?

    Show the solution to problem 36

    Unknown: let x be the initial energy in kWh. Traction uses 0.30x, regeneration returns 0.40 · 0.30x, and charging adds 0.85 · 20 kWh.

    Equation: x − 0.30x − 9 + 0.40(0.30x) + 0.85 · 20 = 49.

    Steps: x − 0.30x + 0.12x − 9 + 17 = 49, so 0.82x + 8 = 49, 0.82x = 41, and x = 50.

    Check: traction uses 15 kWh and onboard systems 9, leaving 26 kWh; regeneration returns 6 and charging adds 17. Thus 26 + 6 + 17 = 49 kWh. The initial charge was 50/80 = 62.5%.

    Answer. The battery started with 50 kWh, or 62.5% of its capacity.

  37. The trip contribution

    A trip is booked by 48 people, each paying the same amount. Eight withdraw; the organiser obtains a €320 reduction in the total cost, and each of the remaining 40 people pays €4 more than the original amount. What was the original contribution?

    Show the solution to problem 37

    Unknown: let x be the original contribution in euros. The initial cost was 48x.

    Equation: after the reduction the cost is 48x − 320, covered by 40 payments of x + 4: 40(x + 4) = 48x − 320.

    Steps: 40x + 160 = 48x − 320, so 480 = 8x and x = 60.

    Check: the initial cost was 48 · 60 = 2880 euros. The reduced cost is €2,560, and the remaining people pay 40 · 64 = 2560 euros.

    Answer. The original contribution was €60 per person.

    Solution without an equation

    The eight withdrawals remove eight original contributions. The remaining people provide an extra €160, and the discount covers another €320; together these replace €480, equal to the eight missing contributions. Each was therefore 480 ÷ 8 = 60 euros.

  38. Added salt and evaporation

    A container holds 72 kg of a 25% salt solution. Six kilograms of pure salt are added, and then only water is allowed to evaporate until the concentration reaches 40%. How many kilograms of water evaporate?

    Show the solution to problem 38

    Unknown: let x be the kilograms of water evaporated. Initially there are 0.25 · 72 = 18 kg of salt; after adding salt there are 24 kg, while total mass before evaporation is 78 kg.

    Equation: after evaporation the total mass is 78 − x, of which salt is 40%: 24 = 0.40(78 − x).

    Steps: 24 = 31.2 − 0.40x, so 0.40x = 7.2 and x = 18.

    Check: 78 − 18 = 60 kg of solution remain, containing 24 kg of salt; 24/60 = 0.40, or 40%.

    Answer. 18 kg of water evaporate.

    Solution without an equation

    After the addition there are 24 kg of salt, which does not evaporate. If this is to be 40% of the final solution, the final mass must be 24 ÷ 0.40 = 60 kg. There were 78 kg before evaporation, so 18 kg of water evaporate.

  39. The two printers

    One printer produces 40 pages per minute. A second printer, producing 55 pages per minute, starts 6 minutes after the first. They stop at the same time, after producing 2,520 pages altogether. How long does each printer run?

    Show the solution to problem 39

    Unknown: let t be the first printer’s running time in minutes. The second runs for t − 6 minutes.

    Equation: total output gives 40t + 55(t − 6) = 2520.

    Steps: 40t + 55t − 330 = 2520, hence 95t = 2850 and t = 30. The second runs for 24 minutes.

    Check: the first produces 40 · 30 = 1200 pages and the second 55 · 24 = 1320; together they produce 2,520 pages.

    Answer. The first printer runs for 30 minutes and the second for 24 minutes.

    Solution without an equation

    In the first 6 minutes, the first printer produces 240 pages. The remaining 2,280 pages are then produced together at 40 + 55 = 95 pages per minute, requiring 2280 ÷ 95 = 24 minutes. Thus the first runs for 30 minutes and the second for 24.

  40. Angles in a trapezoid

    In an isosceles trapezoid, each acute angle is 8° less than three fifths of each obtuse angle. Find all four angles.

    Show the solution to problem 40

    Unknown: let x be an obtuse angle in degrees. An adjacent acute angle is 180 − x, because the angles along each leg are supplementary.

    Equation: the stated relation becomes 180 − x = (3/5)x − 8.

    Steps: 188 = x + (3/5)x = (8/5)x, so x = 188 · 5/8 = 117.5. The acute angle is 180 − 117.5 = 62.5 degrees.

    Check: three fifths of 117.5° is 70.5°, and 70.5° − 8° = 62.5°; also 117.5° + 62.5° = 180°.

    Answer. The angles are 62.5°, 62.5°, 117.5° and 117.5°.

  41. Banknotes in the cashbox

    A cashbox contains only €5, €10 and €20 notes, 87 notes in total. There are 12 more €5 notes than €10 notes, while the number of €20 notes is half the number of €10 notes. How many notes of each kind are there, and what is their total value?

    Show the solution to problem 41

    Unknown: let x be the number of €10 notes. There are x + 12 €5 notes and x/2 €20 notes.

    Equation: counting all notes gives x + (x + 12) + x/2 = 87.

    Steps: (5/2)x + 12 = 87, so (5/2)x = 75 and x = 30. There are 42 €5 notes and 15 €20 notes.

    Check: 42 + 30 + 15 = 87. Their value is 42 · 5 + 30 · 10 + 15 · 20 = 210 + 300 + 300 = 810 euros.

    Answer. There are 42 €5 notes, 30 €10 notes and 15 €20 notes, worth €810 in total.

    Solution without an equation

    After removing the 12 extra €5 notes, 75 notes remain. Group them into packets containing two €5 notes, two €10 notes and one €20 note: five notes per packet, in the required ratio. There are 15 packets, giving base counts of 30, 30 and 15; adding back the 12 €5 notes gives 42, 30 and 15.

  42. A family’s ages

    In a family of four, the father is 4 years older than three times the son’s age. The mother is 2 years younger than the father, and the sister is 3 years younger than the son. Their four ages total 123 years. Find each person’s age.

    Show the solution to problem 42

    Unknown: let x be the son’s age. The father is 3x + 4, the mother 3x + 2, and the sister x − 3.

    Equation: the age total gives x + (3x + 4) + (3x + 2) + (x − 3) = 123.

    Steps: 8x + 3 = 123, so 8x = 120 and x = 15. The father is 49, the mother 47, and the sister 12.

    Check: 15 + 49 + 47 + 12 = 123; also 49 = 3 · 15 + 4, 47 = 49 − 2, and 12 = 15 − 3.

    Answer. The son is 15, the sister 12, the mother 47 and the father 49.

    Solution without an equation

    Eight copies of the son’s age account for the variable parts, while the fixed adjustments total 4 + 2 − 3 = 3 years. Removing those 3 years from 123 leaves 120, equal to eight son-ages: 120 ÷ 8 = 15. The other ages then follow.

  43. A three-stage journey

    A tourist walks at 5 km/h for some time. The tourist then cycles at 18 km/h for half an hour less than the walking time. Finally, the tourist rides a bus for 45 minutes at 40 km/h. The total distance is 67 km. How long is the whole journey?

    Show the solution to problem 43

    Unknown: let x be the walking time in hours. Cycling lasts x − 0.5 hours and the bus ride 0.75 hours.

    Equation: adding the distances gives 5x + 18(x − 0.5) + 40 · 0.75 = 67.

    Steps: 5x + 18x − 9 + 30 = 67, so 23x + 21 = 67, 23x = 46, and x = 2. Cycling lasts 1.5 hours.

    Check: the distances are 5 · 2 = 10 km, 18 · 1.5 = 27 km, and 40 · 0.75 = 30 km, totalling 67 km. Total time is 2 + 1.5 + 0.75 = 4.25 hours.

    Answer. The journey lasts 4 hours 15 minutes.

  44. The ballot review

    In a preliminary count of 1,200 ballots, all treated as valid, candidate A receives an unknown number of votes and B receives the rest. A review invalidates 5% of the ballots initially assigned to A; in addition, 60 valid ballots mistakenly assigned to B are transferred to A. After the corrections, A holds 7/13 of the valid votes. How many votes were initially assigned to each candidate?

    Show the solution to problem 44

    Unknown: let x be A’s initial votes; B had 1200 − x. After invalidation and transfer, A has 19x/20 + 60 votes, and the valid total is 1200 − x/20.

    Equation: 19x/20 + 60 = (7/13)(1200 − x/20).

    Steps: multiplying by 260 gives 247x + 15600 = 168000 − 7x; hence 254x = 152400 and x = 600. B also initially had 600 votes.

    Check: 5% of 600 is 30: after invalidation A falls to 570 votes and after the transfer rises to 630; B falls to 540. There are 1,170 valid votes, and 630/1170 = 7/13.

    Answer. In the preliminary count, A and B each had 600 votes.

  45. The pallet load

    An empty lorry with its driver weighs 9,080 kg and carries a 420 kg refrigeration unit. It is loaded with identical pallets: each pallet has a tare mass of 30 kg and holds 45 packages, each consisting of 5.5 kg of product and 0.5 kg of packaging. The weighbridge reads 17,300 kg. How many pallets and packages is the lorry carrying?

    Show the solution to problem 45

    Unknown: let x be the number of pallets. Each package weighs 5.5 + 0.5 = 6 kg, so a loaded pallet weighs 30 + 45 · 6 = 300 kg.

    Equation: 9080 + 420 + x[30 + 45(5.5 + 0.5)] = 17300.

    Steps: 9500 + 300x = 17300, so 300x = 7800 and x = 26. The package count is 26 · 45 = 1170.

    Check: the loaded pallets weigh 26 · 300 = 7800 kg; adding the 9,080 kg lorry and the 420 kg unit gives 17,300 kg.

    Answer. The lorry carries 26 pallets and 1,170 packages.

    Solution without an equation

    Subtract the fixed 9,500 kg of lorry and refrigeration unit from the gross mass, leaving 7,800 kg of cargo. Each pallet unit weighs 30 kg plus 270 kg of packages, or 300 kg. Thus there are 7800 ÷ 300 = 26 pallets and 26 · 45 = 1170 packages.

  46. The uncalibrated sensor

    The true temperature is C degrees Celsius, but an uncalibrated sensor sends the software the value S = 0.95C + 2. The software converts the received value S to Fahrenheit using F = (9/5)S + 32. At one instant, the displayed Fahrenheit number is 24 greater than twice the true Celsius temperature. What is the true temperature, and what does the display show?

    Show the solution to problem 46

    Unknown: let x be the true temperature in degrees Celsius. The sensor sends 0.95x + 2, and the display shows (9/5)(0.95x + 2) + 32.

    Equation: (9/5)(0.95x + 2) + 32 = 2x + 24.

    Steps: 1.71x + 3.6 + 32 = 2x + 24, so 11.6 = 0.29x and x = 40.

    Check: at 40 °C the sensor sends 0.95 · 40 + 2 = 40. The software displays (9/5) · 40 + 32 = 104 °F, and 2 · 40 + 24 = 104.

    Answer. The true temperature is 40 °C and the display shows 104 °F.

  47. A tank with inlet and drain

    A tank contains 240 litres of water. An inlet supplies 18 litres per minute while a drain removes 25 litres per minute. Both run together for some time; the drain is then closed and the inlet alone runs for another 6 minutes. There are 264 litres at the end. After how many minutes was the drain closed?

    Show the solution to problem 47

    Unknown: let x be the minutes for which inlet and drain run together. During this phase the volume falls by 25 − 18 = 7 litres per minute.

    Equation: subtract the first phase’s loss from the initial volume and add the final 18 · 6 = 108 litres: 240 − 7x + 108 = 264.

    Steps: 348 − 7x = 264, so 7x = 84 and x = 12.

    Check: after 12 minutes with both flows, 240 − 7 · 12 = 156 litres remain; the following 6 minutes add 108 litres, and 156 + 108 = 264.

    Answer. The drain was closed after 12 minutes; the whole process lasts 18 minutes.

    Solution without an equation

    Before the final 6 minutes, the tank must have held 264 − 108 = 156 litres. It had therefore lost 84 litres from the initial 240. Since the two open flows cause a loss of 7 litres per minute, the drain was closed after 84 ÷ 7 = 12 minutes.

  48. The balanced beam

    A uniform 4 m beam with a mass of 18 kg rests on a pivot 1.5 m from its left end. A crate of unknown mass is placed at the left end; a 30 kg load is 0.8 m from that end, and a 24 kg load is at the right end. The beam is horizontally balanced. What is the crate’s mass? Ignore the thickness of the loads.

    Show the solution to problem 48

    Unknown: let x be the crate’s mass in kilograms. Relative to the pivot, the crate’s lever arm is 1.5 m and the 30 kg load’s left lever arm is 1.5 − 0.8 = 0.7 m. On the right, the 24 kg load has a 2.5 m arm, and the beam’s weight, acting at its centre, has a 0.5 m arm.

    Equation: equating moments and cancelling gravitational acceleration gives 1.5x + 0.7 · 30 = 2.5 · 24 + 0.5 · 18.

    Steps: 1.5x + 21 = 60 + 9, so 1.5x = 48 and x = 32.

    Check: the total left moment is 1.5 · 32 + 0.7 · 30 = 48 + 21 = 69; the right moment is 2.5 · 24 + 0.5 · 18 = 60 + 9 = 69.

    Answer. The crate has a mass of 32 kg.

    Solution without an equation

    The known right moments total 69 units, while the known left load supplies 21. The crate must therefore supply 48 units of moment. Since it is 1.5 m from the pivot, its mass is 48 ÷ 1.5 = 32 kg.

  49. Production after quality control

    Line A makes an unknown number of components; line B makes 400 more. Quality control rejects 3% of A’s output and 8% of B’s. Of the components that pass, 120 are then assigned to destructive testing. This leaves 3,272 saleable components. How many components did each line originally make?

    Show the solution to problem 49

    Unknown: let x be line A’s output. Line B makes x + 400; after inspection, 0.97x and 0.92(x + 400) conforming components remain.

    Equation: 0.97x + 0.92(x + 400) − 120 = 3272.

    Steps: 0.97x + 0.92x + 368 − 120 = 3272, so 1.89x + 248 = 3272, 1.89x = 3024, and x = 1600. Line B makes 2,000.

    Check: A rejects 3% of 1600 = 48 components and keeps 1,552; B rejects 8% of 2000 = 160 and keeps 1,840. There are 3,392 conforming components, and removing 120 for testing leaves 3,272.

    Answer. Line A made 1,600 components and line B made 2,000.

  50. The weighted average

    Four tests contribute 20%, 25%, 25% and 30% to a final score. The scores are 78, 84, an unknown score and 92. After calculating the weighted average, 2 bonus points are added, giving a final result of 85.2. What is the unknown score?

    Show the solution to problem 50

    Unknown: let x be the third test score.

    Equation: 0.20 · 78 + 0.25 · 84 + 0.25x + 0.30 · 92 + 2 = 85.2.

    Steps: the known contributions, including the bonus, total 15.6 + 21 + 27.6 + 2 = 66.2. Thus 66.2 + 0.25x = 85.2, 0.25x = 19, and x = 76.

    Check: the average before the bonus is 15.6 + 21 + 0.25 · 76 + 27.6 = 15.6 + 21 + 19 + 27.6 = 83.2; adding 2 gives 85.2.

    Answer. The third test score is 76.

    Solution without an equation

    Subtract the bonus and the three known contributions from the final result: 85.2 − 2 − 15.6 − 21 − 27.6 = 19. Those 19 points are 25% of the missing score; dividing by 0.25 gives 76.