Arithmetic Expressions · Lesson 6 of 7
Applying and Creating Expressions
“Combine the chapter’s ideas to model real situations, check repeated patterns and create expressions under clear rules.”
• Construct expressions for repeated deliveries, savings and reading schedules. • Check units and identify when a final step differs from a repeated cycle. • Group alternating sums and compare equivalent expressions structurally. • Create many expressions for one value. • Solve and design number puzzles while checking every operation and digit constraint.
Expressions for Real Situations
A correct calculation must first represent the situation you intend to describe. Identify what repeats, what is added or removed once, and which unit the answer should have. The same symbols can produce a correct numerical value for the wrong quantity, so keep the story beside the expression while constructing it.
Problem
Rahim delivers 9 kg of mangoes daily and Shyam delivers 11 kg daily. How much do they deliver together in seven days?
- 1.One combined day contributes 9 + 11 = 20 kg.
- 2.There are seven such days: 7 × (9 + 11) = 7 × 20 = 140 kg.
- 3.Distribution gives another route: 7 × 9 + 7 × 11 = 63 + 77 = 140 kg. Each person’s daily delivery is repeated seven times.
Problem
Binu earns ₹20000 per month. Monthly rent, food and other expenses are ₹5000, ₹5000 and ₹2000. Assuming these amounts stay the same, find one year’s savings.
- 1.First combine monthly expenses: 5000 + 5000 + 2000 = ₹12000.
- 2.Monthly savings are 20000 − (5000 + 5000 + 2000) = ₹8000.
- 3.Twelve months give 12 × [20000 − (5000 + 5000 + 2000)] = 12 × 8000 = ₹96000.
- 4.You can also subtract annual expenses from annual income: 12 × 20000 − 12 × (5000 + 5000 + 2000). Both terms then describe yearly amounts.
| Situation | Repeated unit | Expression and final unit |
|---|---|---|
| Mango delivery | One day: 9 + 11 kg | 7 × (9 + 11) = 140 kg |
| Savings | One month: ₹20000 − ₹12000 | 12 × (20000 − 12000) = ₹96000 |
| Reading schedule | One week: 7 − 2 reading days | 8 × (7 − 2) = 40 stories |
A Repeated Pattern and Its Final Step
A pattern may repeat only while the task remains unfinished. Before multiplying a complete cycle by a number of cycles, ask whether every cycle really occurs in full. Reaching a destination, filling a container or finishing a reading task may change the last step.
Problem
A snail climbs 3 cm up a 10 cm post each day and slips 2 cm each night. Starting at the bottom, when does it first reach the top?
- 1.A completed day-and-night cycle gains 3 − 2 = 1 cm. After seven full cycles, the snail is at 7 cm.
- 2.On day 8 it climbs 3 cm from 7 cm to 10 cm and reaches the top. The task is complete before another nightly slip.
- 3.Check the previous day: day 7 starts at 6 cm and reaches only 9 cm before slipping to 7 cm. It has not arrived earlier.
- 4.The correct expression for height at the end of the final climb is 7 × (3 − 2) + 3 = 10. Therefore the answer is 8 days, not 10.
Problem
Melvin reads one two-page short story each day except Tuesdays and Saturdays. How many stories does he read in eight full weeks?
- 1.Two days each week have no story, so each week has 7 − 2 = 5 stories.
- 2.For eight weeks, (7 − 2) × 8 = 40 stories. Equivalently, start with all 56 days and remove 16 skipped days: 7 × 8 − 2 × 8 = 40.
- 3.The story length is not needed to count stories. Multiplying 40 by 2 gives 80 pages, which answers a different question.
| Day | Monday | Tuesday | Wednesday | Thursday | Friday | Saturday | Sunday |
|---|---|---|---|---|---|---|---|
| Stories in one week | 1 | 0 | 1 | 1 | 1 | 0 | 1 |
For Melvin, 5 × 2 × 8 counts pages, not stories. For the snail, 3 − 2 describes a completed day-and-night cycle, not the final daytime climb. Check the meaning of a repeated unit before multiplying it.
Different Methods, Comparisons and Equivalent Expressions
The chapter’s properties let you choose an efficient route without changing the quantity. Group complete signed terms, expand or collect repeated groups, and compare common parts. Equivalent expressions need not look alike, but an apparent rearrangement fails if it changes signs or the scope of a factor.
Problem
Evaluate 1 − 2 + 3 − 4 + 5 − 6 + 7 − 8 + 9 − 10, and a sum of five consecutive 1 − 1 pairs.
- 1.In the first sum, group adjacent pairs: (1 − 2) + (3 − 4) + (5 − 6) + (7 − 8) + (9 − 10). Each pair is −1, so the sum is −5.
- 2.Alternatively, the positive odd terms total 25 and the negative even terms total −30, giving −5. These are the same signed terms regrouped.
- 3.In the second sum, every pair is 1 − 1 = 0. Five zero pairs total 0. Grouping all five positive ones and all five negative ones also gives 5 − 5 = 0.
| Compare | Reason | Sign |
|---|---|---|
| 49 − 7 + 8 and 49 − 7 + 8 | The expressions are identical. | = |
| 83 × 42 − 18 and 83 × 40 − 18 | The first has two extra groups of 83. | > |
| 145 − 17 × 8 and 145 − 17 × 6 | The first removes two more groups of 17. | < |
| 23 × 48 − 35 and 23 × (48 − 35) | The second removes 23 × 35, not just 35. | > |
| (16 − 11) × 12 and −11 × 12 + 16 × 12 | Distribution and swapping complete signed terms give equality. | = |
| (76 − 53) × 88 and 88 × (53 − 76) | The first is positive and the second negative. | > |
| 25 × (42 + 16) and 25 × (43 + 15) | Both inner sums are 58. | = |
| 36 × (28 − 16) and 35 × (27 − 15) | Both differences are 12; the first has one more group. | > |
For 83 − 37 − 12, the signed terms are 83, −37 and −12. The expression 84 − 38 − 12 is equal because its first two terms compensate; −37 + 83 − 12 is equal because it swaps the same signed terms. But 84 − (37 + 12) increases only the starting quantity, and 83 − 38 − 13 removes two more overall. Do not mistake a similar appearance for equivalence.
For 93 + 37 × 44 + 76, the product 37 × 44 is one complete term. It can move to the beginning: 37 × 44 + 93 + 76. Changing the bracket or attaching 44 to a different sum would change that term. Check the three outer terms before deciding whether two complicated expressions agree.
Many Expressions for One Value
You can work backwards from a desired value to create a calculation that reaches it. Begin with an easy sum or difference, then replace one part with an equivalent product or grouped expression. Check each new expression independently instead of assuming a small change preserves the value.
| Ten expressions for 26 | Check |
|---|---|
| 10 + 16 | 26 |
| 30 − 4 | 26 |
| 13 × 2 | 26 |
| 52 ÷ 2 | 26 |
| 5 + 3 × 7 | 5 + 21 = 26 |
| 5 × 5 + 1 | 25 + 1 = 26 |
| 20 + 6 | 26 |
| 2 × (15 − 2) | 2 × 13 = 26 |
| 3 × 10 − 4 | 30 − 4 = 26 |
| 2 + 6 × 4 | 2 + 24 = 26 |
Expression Engineer
Number puzzles ask you to construct expressions under restrictions, rather than simply evaluate expressions already given. State the allowed numbers and operations before starting, and count every use of a number after you have checked the value. A successful construction must satisfy both the arithmetic and the rules.
With three 3s, the expressions (3 + 3) ÷ 3, 3 + 3 − 3 and 3 × 3 + 3 give 2, 3 and 12 respectively. They all use exactly three separate 3s. Brackets let the first expression combine two 3s before dividing by the third; without that grouping, it would describe a different calculation.
Use exactly four separate 4s, the operations +, −, ×, ÷ and brackets. Do not join digits into 44 or add decimal points, roots or factorials. Explore targets from 1 to 20 and record valid constructions. Under these basic rules, 10, 11, 13, 14, 18 and 19 cannot be made; failure to find them is not a calculation error. A puzzle allowing extra notation would need to state and explain that notation.
(4 + 4) ÷ (4 + 4) = 1. Count the four digits and evaluate the brackets and products first.
Another challenge uses 1, 2, 3, 4 and 5 exactly once each, in any order, with the same four operations and brackets. Try to make as many integers from −10 to 10 as possible. You can start with 1 + 2 + 3 + 4 − 5 = 5, or (5 − 4) × (3 − 2) − 1 = 0. The second construction makes two differences of 1 and then subtracts the remaining 1.
Problem
Make 100 with each digit 0–9 exactly once using ordinary operations and brackets.
- 1.Choose two products with a total above 100: 9 × 8 = 72 and 7 × 6 = 42; their sum is 114.
- 2.The unused digits 5, 4, 3 and 2 sum to 14. Subtracting them brings the value to 100.
- 3.Use the remaining 1 and 0 as the zero term 1 × 0. One valid expression is 9 × 8 + 7 × 6 − 5 − 4 − 3 − 2 + 1 × 0 = 100.
- 4.Check the digit list: 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 each occurs once. The zero product changes no value.
Design a small puzzle with specified numbers, allowed operations and a target. First write at least one valid solution yourself, then give only the rules and target to a partner. For example, use 2, 3 and 5 once each with +, −, ×, ÷ and brackets to make 11; 2 × 3 + 5 verifies that the challenge is feasible.
Check Your Understanding
Try each question before opening the explanations. A useful answer includes the reason for your choice, not just its letter.
Quiz
Two farmers deliver 6 kg and 8 kg daily for five days. Which expression gives their combined delivery?
A snail climbs 4 m daily and slips 3 m nightly on a 9 m post. When does it first reach the top?
A reader reads one three-page story on four days each week for six weeks. How many stories are read?
Which expression is equivalent to 70 − 24 − 8?
Which construction uses exactly four separate 4s and has value 9?
Each day includes both deliveries, so repeat their sum five times.
Work these problems on paper and explain any rewrite you use. Open the matching solution after you have made a first attempt; more than one valid expression or method may be possible.
Practice Problems
- Write two expressions for seven days of deliveries of 9 kg and 11 kg each day.
- Explain why 12 × [20000 − (5000 + 5000 + 2000)] models annual savings, whereas 12 × 20000 − (5000 + 5000 + 2000) does not.
- List the snail’s daytime height for days 1–8 on the 10 cm post. Why is the net daily gain insufficient by itself?
- For Melvin’s eight-week schedule, decide which count stories: (a) 5 × 2 × 8; (b) (7 − 2) × 8; (c) 8 × 7; (d) 7 × 2 × 8; (e) 7 × 5 − 2; (f) (7 + 2) × 8; (g) 7 × 8 − 2 × 8; (h) (7 − 5) × 8.
- Find two methods for each: 1 − 2 + 3 − 4 + 5 − 6 + 7 − 8 + 9 − 10; 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1.
- Explain the signs in all eight comparisons in the comparison table without finding every complete value.
- For 83 − 37 − 12, select all equivalent choices: (i) 84 − 38 − 12; (ii) 84 − (37 + 12); (iii) 83 − 38 − 13; (iv) −37 + 83 − 12.
- For 93 + 37 × 44 + 76, select all equivalent choices: (i) 37 + 93 × 44 + 76; (ii) 93 + 37 × 76 + 44; (iii) (93 + 37) × (44 + 76); (iv) 37 × 44 + 93 + 76.
- Choose a number and write ten distinct expressions for it. Include both grouped and ungrouped expressions.
- Using four separate 4s and only +, −, ×, ÷ and brackets, make 5, 7, 12 and 20.
- Using 1, 2, 3, 4 and 5 exactly once, make a negative value, zero and a positive value between −10 and 10. Then explore more targets.
- Verify the all-digits expression for 100 and invent a different feasible puzzle.
7 × (9 + 11) and 7 × 9 + 7 × 11 both give 140 kg. The first groups by day; the second by person.
Key Takeaways
• Construct an expression for the requested quantity before calculating. • Repeat all parts of a daily, weekly or monthly unit consistently. • Check whether the final stage follows a different rule from earlier cycles. • The same signed terms and valid compensation support equivalent expressions. • Alternating sums can be evaluated by pairs or by grouping positive and negative terms. • A puzzle solution must satisfy the digit and operation rules as well as the target value.