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Lesson 12 of 13

Finding the Unknown · Lesson 12 of 13

Connected Word Problems and Challenges

“Combine relationships, pattern counts, and equation reasoning in richer problems.”

Learning Objectives

• Express related digits, prices, shares, and measurements using one unknown. • Derive and test rules for two stick sequences. • Find related expression values without unnecessary solving. • Solve bracket equations and use solutions to navigate a maze. • Check both numerical and contextual conditions in a mixed problem.

Turn several clues into one relationship

A word problem may contain several unknown-looking quantities, but its clues often link them. Choose one quantity as the unknown and express the others in terms of it. Then use a total or equality to write one equation. After solving, return to every clue: finding a number is only part of solving a story.

Three connected digits

Problem
A three-digit number has a hundreds digit 3 less than its tens digit, and a tens digit 3 less than its units digit. Its digit sum is 15. Find it.

  1. 1.Let the tens digit be t. The hundreds digit is t − 3 and the units digit is t + 3.
  2. 2.Write (t − 3) + t + (t + 3) = 15. The −3 and +3 cancel: 3t = 15.
  3. 3.Divide by 3: t = 5. The digits are 2, 5, and 8, so the number is 258.
  4. 4.Check: 2 is 3 less than 5, 5 is 3 less than 8, and 2 + 5 + 8 = 15. All are valid digits.
A brick contains half its own weight in the statement

Problem
A brick weighs 1 kg more than half its weight. Find its weight.

  1. 1.Let w be the weight in kilograms. The relationship is w = 1 + w/2.
  2. 2.Subtract w/2 from both sides: w/2 = 1. Multiply by 2: w = 2 kg.
  3. 3.Check: half of 2 kg is 1 kg, and 1 kg more is 2 kg. The 1 kg was not the whole brick.
A quarter returns to the whole

Problem
One quarter of a number, increased by 9, equals that number. Find it.

  1. 1.Let the number be a. Write a/4 + 9 = a.
  2. 2.Subtract a/4: 9 = 3a/4. Multiply by 4: 36 = 3a. Divide by 3: a = 12.
  3. 3.Check: 12/4 + 9 = 3 + 9 = 12.
Two restaurant prices

Problem
A fruit juice costs ₹15 less than a milkshake. Four juices and seven milkshakes cost ₹600. Find each price.

  1. 1.Let j be the juice price in rupees. The milkshake price is j + 15.
  2. 2.Write 4j + 7(j + 15) = 600. Expand: 4j + 7j + 105 = 600.
  3. 3.Combine and subtract 105: 11j = 495. Divide by 11: j = ₹45.
  4. 4.The milkshake costs ₹60. Check: 4 × 45 + 7 × 60 = 180 + 420 = ₹600, and 60 − 45 = ₹15.
A triple and a total

Problem
Two numbers sum to 76, and one is three times the other. Find them.

  1. 1.Let the smaller number be a; the larger is 3a. Write a + 3a = 76.
  2. 2.Combine: 4a = 76. Divide by 4: a = 19. The larger is 3 × 19 = 57.
  3. 3.Check: 19 + 57 = 76 and 57 = 3 × 19.

Watch what each comparison refers to

Successive comparisons must be read in sequence. “Three times the second share” is different from “three times the first share”. A measurement described as a fixed amount plus half itself also needs the whole measurement on the other side. Careful reading prevents an equation that is easy to solve but describes the wrong situation.

Successive shares in a manuscript problem

Problem
The second person receives twice the first share, the third receives three times the second, and the fourth four times the third. The total is 132. Find each share.

  1. 1.Let the first share be a. The second is 2a; the third is 3(2a) = 6a; the fourth is 4(6a) = 24a.
  2. 2.The total is a + 2a + 6a + 24a = 132, so 33a = 132.
  3. 3.Divide by 33: a = 4. The shares are 4, 8, 24, and 96.
  4. 4.Check the total: 4 + 8 + 24 + 96 = 132. Check each multiplier: 8 = 2 × 4, 24 = 3 × 8, 96 = 4 × 24.
  5. 5.This problem appears in the Bakhshali Manuscript tradition. Its successive relationships, rather than its historical notation, determine the equation.
A giraffe’s height

Problem
A giraffe is 2.5 m taller than half its height. Find the height.

  1. 1.Let h be its height in metres. Write h = 2.5 + h/2.
  2. 2.Subtract h/2: h/2 = 2.5. Multiply by 2: h = 5 m.
  3. 3.Check: half of 5 m is 2.5 m; adding 2.5 m gives 5 m.
Increased by a fixed amount

Problem
A number increased by 36 equals ten times itself. Find it.

  1. 1.Let the number be a. Write a + 36 = 10a.
  2. 2.Subtract a: 36 = 9a. Divide by 9: a = 4.
  3. 3.Check: 4 + 36 = 40 and 10 × 4 = 40.

Count sticks, not just squares

The first new sequence is a row of n joined squares with a triangular cap at the right end. The cap shares the row’s final vertical edge. The row alone needs 4 sticks for the first square and 3 for each later square, so it uses 3n + 1. The cap contributes only two extra sticks. Thus the whole arrangement uses 3n + 3 sticks.

Position 11 square6 sticksPosition 22 squares9 sticksPosition 33 squares12 sticksPosition 44 squares15 sticks
Sequence A: squares with a shared cap— The triangle contributes two new sticks because its vertical side already belongs to the row.
Use sequence A in both directions

Problem
Find squares and sticks at position 11. Decide whether 85 or 150 sticks can make an arrangement.

  1. 1.At position 11 there are 11 squares. Sticks = 3 × 11 + 3 = 36.
  2. 2.For 85 sticks, solve 3n + 3 = 85: 3n = 82, so n = 82/3. This is not a whole position; the arrangement is impossible.
  3. 3.For 150 sticks, 3n + 3 = 150 gives 3n = 147 and n = 49, which is valid.
  4. 4.Check: 3 × 49 + 3 = 150. The triangle is not another square, so it does not raise the square count.

Sequence B forms a staircase of squares. At position 1 there are four squares, in two rows of two. Each new position extends the staircase by three squares. Thus position n has 3n + 1 squares. The first arrangement needs 13 sticks. Each extension adds three squares but shares three new edges with existing or newly added squares, so it adds 12 − 3 = 9 sticks. The total is 13 + 9(n − 1) = 9n + 4.

Position 14 squares13 sticksPosition 27 squares22 sticksPosition 310 squares31 sticksPosition 413 squares40 sticks
Sequence B: a staircase of shared squares— Each extension adds three squares and nine sticks. Count each shared edge once.
Use sequence B and compare with A

Problem
Find the square and stick counts at position 11; test totals 85 and 150.

  1. 1.Squares at position 11 = 3 × 11 + 1 = 34. Sticks = 9 × 11 + 4 = 103.
  2. 2.For 85 sticks, 9n + 4 = 85 gives 9n = 81, so n = 9. Check: 9 × 9 + 4 = 85.
  3. 3.For 150 sticks, 9n + 4 = 150 gives n = 146/9, not a whole position, so it is impossible.
  4. 4.The answers differ from sequence A because the shapes have different growth and edge-sharing relationships.
Common mistake

Do not multiply a square count by 4 when adjoining squares share edges. Do not reuse one sequence’s rule for another shape. Derive the rule from the structure and check the first few positions.

Use a given equality directly

Sometimes the question asks for a related expression rather than the unknown. Transform the given equality to produce that expression. This can avoid finding a fractional unknown and then substituting it. The same equality operations used for solving can also reveal a quantity directly.

Related expressions from 4k + 1 = 13

Problem
Find 8k + 2, 4k, k, 4k − 1, and −k − 2.

  1. 1.Double the whole equality: 8k + 2 = 26. Subtract 1 from the original: 4k = 12.
  2. 2.Divide 4k = 12 by 4: k = 3.
  3. 3.For 4k − 1, subtract 2 from 4k + 1 = 13: 4k − 1 = 11.
  4. 4.Using k = 3, −k − 2 = −3 − 2 = −5. Each requested expression is a different quantity.
Find expressions without finding p

Problem
Given 28p − 36 = 98, find 14p − 19 and 28p − 38.

  1. 1.Divide the given equality by 2: 14p − 18 = 49. Subtract 1 from both sides: 14p − 19 = 48.
  2. 2.Subtract 2 from the original: 28p − 38 = 96.
  3. 3.Neither result required computing p. If checked separately, 28p = 134, so p = 67/14, which produces both values.

Brackets and signs in a mixed set

The same few principles handle equations that look quite different. Expand every term when it helps; otherwise undo an outer multiplication first. Subtracting a whole bracket reverses all its term signs. After collecting unknown terms, divide by the remaining non-zero coefficient and verify the original brackets.

An unknown appears in two brackets

Problem
Solve −3(u + 2) = 2(u − 1).

  1. 1.Expand both sides: −3u − 6 = 2u − 2.
  2. 2.Subtract 2u: −5u − 6 = −2. Add 6: −5u = 4.
  3. 3.Divide by −5: u = −4/5.
  4. 4.Check: left = −3(−4/5 + 2) = −18/5; right = 2(−4/5 − 1) = −18/5.
Two brackets on the right

Problem
Solve 10 − 5x = 3(x − 4) − 2(x − 7).

  1. 1.Expand the right: 3x − 12 − 2x + 14 = x + 2.
  2. 2.The equation is 10 − 5x = x + 2. Subtract x: 10 − 6x = 2.
  3. 3.Subtract 10: −6x = −8. Divide by −6: x = 4/3.
  4. 4.Check: left = 10 − 20/3 = 10/3; right = 3(−8/3) − 2(−17/3) = −8 + 34/3 = 10/3.
EquationUseful intermediate equalitySolution
5(r + 2) = 10r + 2 = 2r = 0
2(7 − 2n) = −67 − 2n = −3; −2n = −10n = 5
2(x − 4) = −16x − 4 = −8x = −4
6(x − 1) = 2(x − 1) − 46x − 6 = 2x − 6; 4x = 0x = 0
3 − 7s = 7 − 3s−4s = 4s = −1
2x + 1 = 6 − (2x − 3)2x + 1 = 9 − 2x; 4x = 8x = 2

Check each table result in its original equation, including the bracket signs. For example, x = 0 in 6(x − 1) = 2(x − 1) − 4 gives −6 on both sides. The presence of x − 1 in both places does not justify cancelling it by division: it could be zero, and a subtraction of 4 is also present.

Follow an equation maze

At each box in the maze, solve that box’s equation. Choose the connecting path whose label equals your solution, and move to its next box. Continue until you reach the End box. Letters can change between boxes because each box is a separate problem. You may solve the End equation too, but reaching that box completes the route.

−4−13−2138+444−441652−10−5−45218−8−10−4A8x = 20 + 3xB−7 = 11 − 3xC15 = 19 − 4xD2x − 9 = −3E−2x = −42F2x + 3 = x + 5G8m + 8 = −72H2(x + 1) − 10 = 18I2x + 5 = 3(x − 1)J−4 = 16 − 5kK2x − 9 = 3 − xL30 = 4 − 50nStartEnd
Equation maze with every connecting path— Solve a box, then follow a connector labelled with that solution. Start at A and finish on arrival at L.
Solve the maze route

Problem
Find a path from Start A to End L, recording the solution at each visited box.

  1. 1.A: 8x = 20 + 3x gives 5x = 20, x = 4; follow 4 to D.
  2. 2.D: 2x − 9 = −3 gives 2x = 6, x = 3; follow 3 to E.
  3. 3.E: −2x = −42 gives x = 21; follow 21 to C. C: 15 = 19 − 4x gives −4 = −4x, x = 1; follow 1 to F.
  4. 4.F: 2x + 3 = x + 5 gives x = 2; follow 2 to I. I: 2x + 5 = 3x − 3 gives 8 = x; follow 8 to H.
  5. 5.H: 2x + 2 − 10 = 18 gives 2x = 26, x = 13; follow 13 to G. G: 8m + 8 = −72 gives 8m = −80, m = −10; follow −10 to J.
  6. 6.J: −4 = 16 − 5k gives −20 = −5k, k = 4; follow +4 to K. K: 2x − 9 = 3 − x gives 3x = 12, x = 4; follow 4 to L, the End.
  7. 7.The route is A → D → E → C → F → I → H → G → J → K → L. Solving the End equation separately gives −50n = 26, n = −13/25.

Heads and feet describe the same group

A total number of heads counts individuals, while a total number of feet weights those individuals differently. Each child contributes two feet and each donkey four. Let one count be the unknown; the head total determines the other count. Then the foot total supplies an equation. Check both totals after finding the answer.

Children and donkeys

Problem
Children and donkeys together have 28 heads and 80 feet. Find their numbers.

  1. 1.Let d be the number of donkeys. There are 28 − d children.
  2. 2.Feet give 4d + 2(28 − d) = 80. Expand: 4d + 56 − 2d = 80.
  3. 3.Combine: 2d + 56 = 80. Subtract 56: 2d = 24. Divide by 2: d = 12.
  4. 4.There are 28 − 12 = 16 children. Check: 12 + 16 = 28 heads and 4 × 12 + 2 × 16 = 48 + 32 = 80 feet.

Quiz

Quick check

If the tens digit is t and the hundreds digit is 3 less, which represents the hundreds digit?

Quick check

If successive shares are a, twice the first, three times the second, and four times the third, what is the fourth?

Quick check

How many sticks are needed at position 11 of sequence A, with rule 3n + 3?

Quick check

Which sequence B position uses 85 sticks under 9n + 4?

Quick check

Given 28p − 36 = 98, what is 14p − 19?

Quick check

What solves 2x + 1 = 6 − (2x − 3)?

Quick check

With d donkeys among 28 children and donkeys, which counts the feet?

Practice Problems

Practice Problems
  1. Solve and check 5(r + 2) = 10, 2(7 − 2n) = −6, and 2(x − 4) = −16.
  2. Solve and check 6(x − 1) = 2(x − 1) − 4 and 3 − 7s = 7 − 3s. Explain any zero or negative answer.
  3. Solve 2x + 1 = 6 − (2x − 3) and 10 − 5x = 3(x − 4) − 2(x − 7), showing every bracket expansion.
  4. For sequences A and B, verify the square and stick counts in their first four positions. Then compare the counts at position 11.
  5. Use both pattern rules to test 85 and 150 sticks, and explain why a fractional position is rejected.
  6. Redraw the maze and record a solution at every box you visit. Solve the unused box B separately and check it.
  7. Given 4k + 1 = 13, find 8k + 2 and 4k − 1 by direct equality operations rather than first finding k.
  8. A juice costs ₹10 less than a shake; three juices and five shakes cost ₹370. Find both prices and check.
  9. Explain why the successive shares a, 2a, 6a, 24a cannot be replaced by a, 2a, 3a, 4a.
  10. Children and donkeys have 20 heads and 56 feet. Find both counts and verify both totals.
  11. Create a three-digit-number problem in which the digits are related by equal differences. Give a valid solution and verify the digits.

Key Takeaways

Key Takeaways

• Use relationships to express several quantities with one unknown. • Read successive comparisons in their stated order. • Shared edges determine stick counts, and positions must be whole numbers. • Transform a given equality to find related expressions directly. • Check bracket signs, every story condition, and the meaning of the final answer.

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Equations from Diagrams and Machines

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Chapter Summary and Practice