NCERT Solutions Class 7 Mathematics Ganita Prakash 1 Chapter 04 Expressions Using Letter-Numbers

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4.1 The Notion of Letter-Numbers

Example 1. Shabnam is 3 years older than Aftab. When Aftab's age was 10 years, Shabnam's age was 13 years. Now Aftab's age is 18 years, what will Shabnam's age be?
Answer: Since Shabnam is always 3 years older, you can find her age by adding 3 to Aftab's current age. Therefore, Shabnam's age = 18 + 3 = 21 years.
In simple words: Shabnam is 3 years older than Aftab. If Aftab is 18, then Shabnam is 18 + 3 = 21 years old.

Exam Tip: When a relationship between two quantities is given, use it to build a simple expression. Here, the key fact is "3 years older," which directly translates to adding 3 to Aftab's age.

Example 1 (continued). Given Aftab's age, how will you find out Shabnam's age?
Answer: Add 3 to Aftab's age whenever you need to find Shabnam's age.
In simple words: Simply take Aftab's age and add 3 to get Shabnam's age.

Exam Tip: Express the procedure in your own words first, then convert it to a mathematical form.

Example 1 (continued). Can we write this as an expression?
Answer: Yes. We can express it as: Shabnam's age = Aftab's age + 3.
In simple words: An expression shows the relationship between two things using symbols. Here, it shows how Shabnam's age connects to Aftab's age.

Exam Tip: Always state the relationship clearly before simplifying or substituting numbers.

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Example 1 (continued). Given the age of Shabnam (s), write an expression to find Aftab's age (a).
Answer: We know Aftab is 3 years younger than Shabnam. Expression: a = s - 3. Using letter-numbers: a = s - 3.
In simple words: To find Aftab's age, take Shabnam's age and subtract 3.

Exam Tip: Remember that if someone is "3 years younger," you subtract 3. If they are "3 years older," you add 3.

Example 1 (continued). Use this expression (a = s - 3) to find Aftab's age if Shabnam's age is 20.
Answer: Substitute s = 20 into the expression: a = 20 - 3 = 17. Aftab's age would be 17 years.
In simple words: Replace the letter s with 20, then solve: 20 - 3 = 17 years.

Exam Tip: When substituting a value, replace the letter everywhere it appears and then calculate step by step.

Example 2. Parthiv is making matchstick patterns of Ls (2 sticks per L). How many matchsticks are needed to make 5 Ls?
Answer: Since each L needs 2 matchsticks, for 5 Ls: 5 x 2 = 10 matchsticks.
In simple words: If 1 L needs 2 sticks, then 5 Ls need 5 times as many: 5 x 2 = 10 sticks.

Exam Tip: Identify the pattern first (here, 2 sticks per L), then multiply by the number of Ls.

Example 2 (continued). How many matchsticks are needed to make 7 Ls?
Answer: For 7 Ls: 7 x 2 = 14 matchsticks.
In simple words: 7 Ls need 7 times 2 sticks, which is 14 sticks.

Exam Tip: Use the same pattern to find the answer for any number of Ls.

Example 2 (continued). How many matchsticks are needed to make 45 Ls?
Answer: For 45 Ls: 45 x 2 = 90 matchsticks.
In simple words: 45 Ls need 45 times 2 sticks, which is 90 sticks total.

Exam Tip: The pattern stays the same - multiply the number of shapes by the sticks per shape.

Example 3. Ketaki prepares coconut-jaggery laddus. Price of a coconut is Rs 35 and price of 1 kg jaggery is Rs 60. How much should she pay if she buys 10 coconuts and 5 kg jaggery?
Answer: Cost = (Cost of coconuts) + (Cost of jaggery) = (10 x 35) + (5 x 60) = 350 + 300 = Rs 650.
In simple words: Find the cost of each item separately, then add them together: 10 coconuts cost 350 rupees, 5 kg jaggery costs 300 rupees, so the total is 650 rupees.

Exam Tip: Break the problem into parts - find the cost of each item, then combine them.

Example 3 (continued). How much should she pay if she buys 8 coconuts and 9 kg jaggery?
Answer: Cost = (8 x 35) + (9 x 60) = 280 + 540 = Rs 820.
In simple words: 8 coconuts cost 280 rupees, 9 kg jaggery costs 540 rupees, so the total is 820 rupees.

Exam Tip: Use the same method every time - multiply quantity by price, then add all costs together.

Example 3 (continued). Write an algebraic expression to find the total amount to be paid for a given number of coconuts ('c') and quantity of jaggery ('j' kg).
Answer: Cost of coconuts = c x 35 (or 35c). Cost of jaggery = j x 60 (or 60j). Total Cost = Cost of coconuts + Cost of jaggery. Algebraic Expression: 35c + 60j.
In simple words: Multiply the number of coconuts by 35, multiply the kg of jaggery by 60, then add them together to get the total cost.

Exam Tip: Always identify what each letter represents (c for coconuts, j for jaggery), then build the expression step by step.

Example 3 (continued). Use this expression (35c + 60j) to find the total amount to be paid for 7 coconuts and 4 kg jaggery.
Answer: Substitute c = 7 and j = 4 into the expression: Total Cost = (35 x 7) + (60 x 4) = 245 + 240 = Rs 485.
In simple words: Replace c with 7 and j with 4, then calculate: (35 x 7) = 245 and (60 x 4) = 240, so 245 + 240 = 485 rupees.

Exam Tip: When substituting values, keep the expression structure and calculate each part separately before combining.

Example 4. We are familiar with calculating the perimeters of simple shapes. Write expressions for perimeters.
Answer: Perimeter of a square = 4 times the length of its side. Expression: 4 x q (or 4q), where q is the side length.
In simple words: A square has 4 equal sides, so the perimeter is 4 times one side length.

Exam Tip: Remember that perimeter is the total distance around a shape - add all the side lengths.

Example 4 (continued). What is the perimeter of a square with side length 7 cm? Use the expression to find out.
Answer: Substitute q = 7 into the expression 4q: Perimeter = 4 x 7 = 28 cm.
In simple words: If each side is 7 cm, then all 4 sides together = 4 x 7 = 28 cm.

Exam Tip: Always include the unit (cm, m, etc.) in your final answer when dealing with measurements.

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Figure it Out

Question 1(a). Write formulas for the perimeter of: a triangle with all sides equal.
Answer: Let the side length be 's'. Perimeter P = s + s + s = 3s. Formula: P = 3s.
In simple words: A triangle with all equal sides has a perimeter of 3 times the side length.

Exam Tip: When all sides are equal, simply multiply one side length by the number of sides.

Question 1(b). Write formulas for the perimeter of: a regular pentagon.
Answer: A regular pentagon has 5 equal sides. Let the side length be 'p'. Perimeter P = p + p + p + p + p = 5p. Formula: P = 5p.
In simple words: A pentagon has 5 sides, so its perimeter is 5 times one side length.

Exam Tip: For any regular polygon, multiply the side length by the number of sides.

Question 1(c). Write formulas for the perimeter of: a regular hexagon.
Answer: A regular hexagon has 6 equal sides. Let the side length be 'h'. Perimeter P = h + h + h + h + h + h = 6h. Formula: P = 6h.
In simple words: A hexagon has 6 sides, so multiply the side length by 6 to find the perimeter.

Exam Tip: "Regular" means all sides and angles are equal - use that to simplify the perimeter formula.

Question 2. Munirathna has a 20 m long pipe. He joins another pipe of some length ('k' meters) to this one. Give the expression for the combined length of the pipe.
Answer: Combined Length = Initial Length + Added Length. Expression: 20 + k meters.
In simple words: The total length is the first pipe (20 m) plus the second pipe (k m), which gives 20 + k meters.

Exam Tip: When combining quantities, use addition. Always show the units in your final expression.

Question 3. What is the total amount Krithika has, if she has the following numbers of notes of Rs 100, Rs 20 and Rs 5? Complete the following table:
Answer:

In simple words: Multiply the number of each type of note by its value, then add all the amounts together to get the total.

Exam Tip: When working with multiple denominations, multiply quantity by value for each type separately, then sum them all.

Question 4. Venkatalakshmi owns a flour mill. It takes 10 seconds for the roller mill to start running. Once it is running, each kg of grain takes 8 seconds to grind into powder. Which of the expressions below describes the time taken to complete grind 'y' kg of grain, assuming the machine is off initially?
(a) 10 + 8 + y
(b) (10 + 8) × y
(c) 10 × 8 × y
(d) 10 + 8 × y
(e) 10 × y + 8
Answer: (d) 10 + 8 × y
In simple words: The total time has two parts: 10 seconds to start up (this happens once), plus 8 seconds for each kg of grain (this repeats y times), giving 10 + 8y seconds.

Exam Tip: Separate fixed time (startup) from variable time (grinding). Fixed time is added once, variable time depends on the quantity.

Question 5(a). Write algebraic expressions using letters of your choice. 5 more than a number
Answer: Let the number be n. Expression: n + 5.
In simple words: "5 more than a number" means start with the number and add 5 to it.

Exam Tip: The phrase "more than" always means addition.

Question 5(b). Write algebraic expressions using letters of your choice. 4 less than a number
Answer: Let the number be x. Expression: x - 4.
In simple words: "4 less than a number" means start with the number and subtract 4.

Exam Tip: The phrase "less than" always means subtraction.

Question 5(c). Write algebraic expressions using letters of your choice. 2 less than 13 times a number
Answer: Let the number be y. Expression: 13y - 2.
In simple words: First multiply the number by 13, then subtract 2 from the result.

Exam Tip: Follow the order given in the phrase: multiplication first, then subtraction.

Question 5(d). Write algebraic expressions using letters of your choice. 13 less than 2 times a number
Answer: Let the number be z. Expression: 2z - 13.
In simple words: Multiply the number by 2, then subtract 13 from the result.

Exam Tip: Notice the difference between "2 less than 13 times a number" (13y - 2) and "13 less than 2 times a number" (2z - 13) - the order matters.

Question 6(a). Describe situations corresponding to the following algebraic expression: 8x + 3y
Answer: One possible scenario: Calculating the total cost of buying x notebooks at Rs 8 each and y pens at Rs 3 each.
In simple words: This expression works when you have two items with different prices and want to find the total amount spent on both.

Exam Tip: Real-world situations often involve costs, distances, or quantities. Try to match the expression structure to a familiar context.

Question 6(b). Describe situations corresponding to the following algebraic expression: 15j - 2k
Answer: One possible scenario: Calculating the remaining juice if you begin with 15 jugs containing j litres each and then pour out 2 kettles containing k litres each.
In simple words: This expression shows a situation where you start with something (15j), then remove part of it (2k), and need to find what's left.

Exam Tip: Subtraction often represents situations involving loss, removal, or spending from an initial amount.

Question 7. In a calendar month, if any 2 × 3 grid full of dates is chosen, write expressions for the dates in the blank cells if the bottom middle cell has date 'w'.
Answer: Assuming a standard calendar layout and a 2 × 3 grid:
Top row: [w - 8] [w - 7] [w - 6]
Bottom row: [w - 1] [w] [w + 1]

The expressions for the dates are:
- Cell to the left of w: w - 1
- Cell to the right of w: w + 1
- Cell directly above w: w - 7
- Cell top-left: w - 8
- Cell top-right: w - 6
In simple words: In a calendar, moving left means subtracting 1, moving right means adding 1, and moving up one row means subtracting 7 (since there are 7 days in a week).

Exam Tip: Understand the calendar structure: each row is 7 days apart. Use this pattern to find all the dates in any grid.

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4.2 Revisiting Arithmetic Expressions

Question. Evaluate: 23 - 10 × 2
Answer: Multiplication comes before subtraction. So: 23 - 10 × 2 = 23 - 20 = 3.
In simple words: Always do multiplication first, then subtraction. Here, 10 × 2 = 20, then 23 - 20 = 3.

Exam Tip: Remember the order of operations: Brackets, then Multiplication/Division, then Addition/Subtraction.

Question. Evaluate: 83 + 28 - 13 + 32
Answer: Group convenient pairs of like terms: (83 - 13) + (28 + 32) = 70 + 60 = 130.
In simple words: Grouping terms with the same signs makes the calculation simpler - combine what you're adding together and what you're subtracting together separately.

Exam Tip: When you have addition and subtraction together, you can rearrange the terms to group convenient pairs and make mental math easier.

Question. Evaluate: 34 - 14 + 20
Answer: Work left to right: (34 - 14) + 20 = 20 + 20 = 40.
In simple words: Do the operations in order from left to right: first subtract 14 from 34 to get 20, then add 20 to get 40.

Exam Tip: For expressions with only addition and subtraction (no multiplication), work strictly left to right.

Question. Evaluate: 42 + 15 - (8 - 7)
Answer: Evaluate the bracket first: 42 + 15 - (1) = 57 - 1 = 56.
In simple words: Always work on what's inside brackets first, then do the other operations: (8 - 7) = 1, so 42 + 15 - 1 = 56.

Exam Tip: Brackets must be solved before any other operation. After removing brackets, follow the order of operations.

Question. Evaluate: 68 - (18 + 13)
Answer: Evaluate the bracket first: 68 - (31) = 37.
In simple words: First add what's in the brackets: 18 + 13 = 31. Then subtract from 68: 68 - 31 = 37.

Exam Tip: Remember to add everything inside the brackets together before subtracting from the number outside.

Question. Evaluate: 7 × 4 + 9 × 6
Answer: Evaluate the multiplications first: (7 × 4) + (9 × 6) = 28 + 54 = 82.
In simple words: Do both multiplications first: 7 × 4 = 28 and 9 × 6 = 54. Then add them: 28 + 54 = 82.

Exam Tip: All multiplications (and divisions) must be done before any additions or subtractions.

Question. Evaluate: 20 + 8 × (16 - 6)
Answer: Evaluate the bracket first: 20 + 8 × (10). Then evaluate the multiplication: 20 + (8 × 10) = 20 + 80 = 100.
In simple words: First, solve what's in brackets: 16 - 6 = 10. Then multiply: 8 × 10 = 80. Finally, add: 20 + 80 = 100.

Exam Tip: Follow this order strictly: Brackets first, then Multiplication/Division, then Addition/Subtraction.

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4.3 Omission of the Multiplication Symbol in Algebraic Expressions

Question. Find an algebraic expression to get the nth term of this sequence. (Here 'n' denotes the position in the sequence).
Answer: Since the sequence consists of multiples of 4, the nth term is 4 times n. Expression: 4 × n.
In simple words: In this sequence, each term is 4 times its position number. So the 1st term is 4, the 2nd is 8, the 3rd is 12, and so on - all multiples of 4.

Exam Tip: Identify the pattern first. Look for what number repeats or multiplies the position.

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Example 5. Here is a table showing the number of pencils and erasers sold in a shop. The price per pencil is c, and the price per eraser is d. Find the total money earned by the shopkeeper during these three days.
Answer:

Total money from pencils:
Day 1: 5c
Day 2: 3c
Day 3: 10c
Adding these, we get: 5c + 3c + 10c = (5 + 3 + 10)c = 18c

Total money from erasers:
Day 1: 4d
Day 2: 6d
Day 3: 1d
Adding these, we get: 4d + 6d + 1d = (4 + 6 + 1)d = 11d

Total earnings = 18c + 11d
In simple words: Add up all the pencil sales to get 18c (where c is the price per pencil). Add up all the eraser sales to get 11d (where d is the price per eraser). The total is 18c + 11d.

Exam Tip: When combining like terms (all terms with c together, all terms with d together), add the numbers in front and keep the letter the same.

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Example 5 (continued). If c = Rs 50, find the total amount earned by the sale of pencils.
Answer: Substitute c = 50 into the simplified expression 18c: Total pencil earnings = 18 × 50 = 900. Amount = Rs 900.
In simple words: Replace c with 50 in the expression 18c: 18 × 50 = 900 rupees.

Exam Tip: When you know the value of a letter, substitute it carefully and calculate step by step.

Example 5 (continued). Write the expression for the total money earned by selling erasers. Then, simplify the expression. Eraser sales: Day 1 = 4, Day 2 = 6, Day 3 = 1. Price = d.
Answer: Total eraser earnings = (4 × d) + (6 × d) + (1 × d) = 4d + 6d + 1d. Simplify using distributive property: (4 + 6 + 1) × d = 11d. Simplified expression: 11d.
In simple words: Write down each day's earnings with d as the price, then add up the numbers in front while keeping d: 4 + 6 + 1 = 11, so the answer is 11d.

Exam Tip: When adding like terms, keep the letter and add only the numbers (coefficients) in front.

Example 5 (continued). Can the expression 18c + 11d be simplified further?
Answer: No, this expression cannot be simplified further because 18c and 11d are unlike terms. They involve different letter-numbers (c and d) which represent different quantities (price of a pencil versus price of an eraser). We can only combine like terms.
In simple words: You can only combine terms that have the same letter. Since 18c and 11d have different letters, they stay separate.

Exam Tip: Like terms have the same letter (or variable). Unlike terms cannot be combined, no matter how you rearrange them.

Example 5 (continued). Check that both expressions (5c + 3c + 10c and 18c) take the same value when c is replaced by different numbers.
Answer: Let's test with c = 3:
5c + 3c + 10c = (5 × 3) + (3 × 3) + (10 × 3) = 15 + 9 + 30 = 54.
And 18c = 18 × 3 = 54.
They give the same value. This holds true for any value of c.
In simple words: When you combine like terms properly, the simplified expression always gives the same answer as the original expression, no matter what number you put in for the letter.

Exam Tip: Verify your simplification by substituting a number and checking that both forms give the same result.

Example 6. A big rectangle is split into two smaller rectangles as shown. Write an expression describing the area of the bigger rectangle.
Answer: A big rectangle (height v, total width 4 + 3 = 7) is split into two smaller rectangles (height v, width 4 and height v, width 3).

Method 1 (Using total dimensions):
Area = height × total width = v × (4 + 3) = v × 7 = 7v.

Method 2 (Sum of smaller areas):
Area = (Area of first part) + (Area of second part) = (v × 4) + (v × 3) = 4v + 3v = (4 + 3)v = 7v.

The required expression is 7v (or 4v + 3v).
In simple words: You can find the area of the big rectangle either by multiplying the total width by the height at once, or by finding the areas of the two smaller rectangles separately and adding them. Both methods give 7v.

Exam Tip: The distributive property says that v × (4 + 3) = v × 4 + v × 3. This is useful for breaking down complex expressions.

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Example 7. A shop rents out chairs and tables for a day's use. To rent them, one has to first pay the following amount per piece.
Answer:

When the furniture is returned, the shopkeeper pays back some amount as follows.

Example 7 (continued). Write an expression for the total number of rupees paid if x chairs and y tables are rented.
Answer: Let x = number of chairs rented and y = number of tables rented.

Initial Amount Paid:
Chairs: 40x
Tables: 75y
Total = 40x + 75y

Amount Refunded:
Chairs: 6x
Tables: 10y
Total refund = 6x + 10y

Expression for Net Payment:
Net amount paid by the customer = Initial Payment - Refund = (40x + 75y) - (6x + 10y)

Simplifying the Expression:
(40x + 75y) - (6x + 10y) = 40x - 6x + 75y - 10y = 34x + 65y

The total amount paid by the customer is 34x + 65y rupees.
In simple words: Start with what you pay upfront for chairs and tables. Then subtract what the shopkeeper gives back. The difference is what you actually pay.

Exam Tip: When subtracting an expression in brackets, change the sign of every term inside before combining like terms.

Example 7 (continued). Describe the procedure to get these amounts.
Answer: Amount paid at beginning = (Number of chairs × Initial chair cost) + (Number of tables × Initial table cost). Expression: 40x + 75y.

Amount returned = (Number of chairs × Return amount per chair) + (Number of tables × Return amount per table). Expression: 6x + 10y.

So, the total amount paid = (40x + 75y) - (6x + 10y) = 34x + 65y.
In simple words: First, multiply how many of each item by its rental price. Add these costs together. Next, do the same for refund amounts. Finally, subtract the total refund from the total rental cost.

Exam Tip: Break down complex problems into clear steps - calculate each part separately, then combine them.

Example 7 (continued). Can we simplify this expression? If yes, how? If not, why not?
Answer: Yes, this expression can be simplified. We can simplify it by removing the brackets (remembering to change signs for the terms in the second bracket because it's preceded by a minus sign) and then combining the like terms (x terms with x terms, y terms with y terms). = (40x + 75y) - (6x + 10y) = 40x + 75y - 6x - 10y = 34x + 65y.
In simple words: When you remove brackets with a minus sign in front, flip the signs of all terms inside. Then collect all x terms together and all y terms together.

Exam Tip: Always remember: subtracting (a + b) is the same as adding (-a - b). Change all the signs inside the bracket.

Example 7 (continued). Could we have written the initial expression as (40x + 75y) + (-6x - 10y)?
Answer: Yes. Subtracting a quantity (6x + 10y) is mathematically equivalent to adding its additive inverse, which is -(6x + 10y) = -6x - 10y. So, the expression becomes (40x + 75y) + (-6x) + (-10y), which leads to the same simplification steps and result (34x + 65y).
In simple words: Subtracting something is the same as adding its opposite. So (40x + 75y) - (6x + 10y) works the same as (40x + 75y) + (-6x - 10y).

Exam Tip: The concept of additive inverse helps explain why changing signs works when removing brackets with a negative sign in front.

Example 8. Charu has been through three rounds of a quiz. Her scores in the three rounds are 7p - 3q, 8p - 4q, and 6p - 2q. Here, p represents the score for a correct answer and q represents the penalty for an incorrect answer. What do each of the expressions mean?
Answer: 7p - 3q: Score from 7 correct answers minus penalty for 3 incorrect answers.
8p - 4q: Score from 8 correct answers minus penalty for 4 incorrect answers.
6p - 2q: Score from 6 correct answers minus penalty for 2 incorrect answers.
In simple words: Each expression shows how many questions were answered correctly (that number times p) and how many were wrong (that number times q, which is subtracted).

Exam Tip: When an expression has two variables, explain what each one represents and how they're used together.

Example 8 (continued). Give some possible scores for Krishita in the three rounds so that they add up to give 23p - 7q.
Answer: Many combinations are possible. We need three expressions that sum to 23p - 7q. One example set of scores could be:
Round 1: 10p - 2q
Round 2: 8p - 3q
Round 3: 5p - 2q
Sum = (10p - 2q) + (8p - 3q) + (5p - 2q) = (10 + 8 + 5)p + (-2 - 3 - 2)q = 23p - 7q.
In simple words: Add the numbers in front of p to get 23p (10 + 8 + 5 = 23) and add the numbers in front of q to get -7q (-2 - 3 - 2 = -7).

Exam Tip: When combining expressions with multiple variables, add coefficients of the same variable separately.

Example 8 (continued). Can we say who scored more? Can you explain why?
Answer: Comparing Krishita's 23p - 7q to Charu's 21p - 9q. Assuming p (points for correct) and q (penalty for incorrect) are positive values:
Krishita has 23p versus Charu's 21p. Since 23 > 21, Krishita gets more points from correct answers (+2p).
Krishita has -7q versus Charu's -9q. Since -7 is greater than -9, Krishita has less penalty subtracted (effectively +2q relative to Charu).
Since Krishita gains more points (+2p) and has less penalty (+2q), Krishita scored more than Charu (assuming p > 0, q > 0).
In simple words: Krishita answered more questions correctly (higher coefficient of p) and got fewer questions wrong (lower penalty, so less is subtracted). Both factors work in her favor.

Exam Tip: When comparing expressions with multiple variables, examine what happens to each variable and how it affects the total.

Example 8 (continued). How much more has Krishita scored than Charu? Find the difference: (23p - 7q) - (21p - 9q). Simplify this expression further.
Answer: Difference = (23p - 7q) - (21p - 9q) = 23p - 7q - 21p + 9q (Remove brackets, change signs) = (23p - 21p) + (-7q + 9q) (Group like terms) = (23 - 21)p + (-7 + 9)q = 2p + 2q. Krishita scored 2p + 2q more than Charu.
In simple words: Subtract Charu's expression from Krishita's. Remember to flip the signs in the second expression when removing brackets. Then combine matching terms.

Exam Tip: When finding the difference between two expressions, treat the second one as a subtraction and flip all its signs inside the brackets.

Example 9. Simplify the expression 4(x + y) - y.
Answer: 4(x + y) - y = (4 × x + 4 × y) - y (Distributive property) = 4x + 4y - y = 4x + (4y - 1y) (Combine like terms) = 4x + (4 - 1)y = 4x + 3y.
In simple words: First, distribute the 4 to both terms inside the brackets: 4 times x is 4x, and 4 times y is 4y. Then combine the two y terms: 4y - y = 3y. The final answer is 4x + 3y.

Exam Tip: The distributive property says that a(b + c) = ab + ac. Use this to remove brackets when a number or letter is multiplied by what's inside.

Example 10. Are the expressions 5u and 5 + u equal to each other?
Answer: No. 5u means 5 × u (5 times u), while 5 + u means 5 more than u. They represent different operations.
In simple words: Multiplication and addition are completely different. 5u is much larger than 5 + u for most values of u.

Exam Tip: Always be careful with notation. In algebra, the absence of an operation symbol between a number and a letter means multiplication.

Example 10 (continued). Fill the blanks below by replacing the letter-numbers by numbers; an example is shown. Then compare the values that 5u and 5 + u take.
Answer: If u = 11: 5u = 5 × 11 = 55; 5 + u = 5 + 11 = 16. (Values differ)
If u = 2: 5u = 5 × 2 = 10; 5 + u = 5 + 2 = 7. (Values differ)
If u = 8: 5u = 5 × 8 = 40; 5 + u = 5 + 8 = 13. (Values differ)
If u = 5: 5u = 5 × 5 = 25; 5 + u = 5 + 5 = 10. (Values differ)
In simple words: For any value you pick, 5u and 5 + u give different answers. 5u is always much bigger.

Exam Tip: Testing with specific numbers is a good way to verify whether two expressions are equal or not.

Example 10 (continued). After filling in the two diagrams, do you think the two expressions are equal?
Answer: No, the expressions 10y - 3 and 10(y - 3) are not equal because they produce different values for the same value of y.
In simple words: These look similar, but they're different. In the first, you multiply 10 by y, then subtract 3. In the second, you subtract 3 from y first, then multiply by 10. The order matters and changes the answer.

Exam Tip: The position of brackets completely changes what an expression means. 10(y - 3) is very different from 10y - 3.

Example 11. What is the sum of the numbers in the picture (unknown values are denoted by letter-numbers)?
Answer: Three ways to express the sum are shown:
1. Adding row wise: (4 × 3) + (r + s) + (r + s) + (4 × 3)
2. Adding like terms: (8 × 3) + (r + r) + (s + s)
3. Adding upper half and doubling: 2 × ((4 × 3) + r + s)

Simplification Check: All three expressions simplify to the same value: 24 + 2r + 2s.
In simple words: No matter which way you add up the numbers - whether row by row, or by grouping the same letters, or by finding one half and doubling it - you always get the same final answer.

Exam Tip: Different approaches to solving a problem should all give the same answer. If they don't, recheck your work.

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Figure it Out

Question 1. Add the numbers in each picture below. Write their corresponding expressions and simplify them. Try adding the numbers in each picture in a couple different ways and see that you get the same thing.
Answer:

Figure 1: Column-wise: 5y + x = -4

Figure 2: Column-wise:
1 = 2p + 3q
1 = 3q + 2p

Row-wise:
2p + 3q = 1
3q + 2p = 1

Figure 3: Column-wise:
-5g + 5k + 5k - 5g = 10k - 10g
5k + 5k + 5k + 5k = 20k
5k + 5k + 5k + 5k = 20k
-5g + 5k + 5k - 5g = 10k - 10g

Row-wise:
-5g + 5k + 5k - 5g = 10k - 10g
5k + 5k + 5k + 5k = 20k
5k + 5k + 5k + 5k = 20k
-5g + 5k + 5k - 5g = 10k - 10g

The result is the same either column-wise or row-wise addition.
In simple words: When you add up all the numbers in a picture, the final answer is the same no matter what order or grouping you use. This shows that our expressions are correct.

Exam Tip: Always verify your work by using an alternate method and checking that you get the same answer.

Question 2(a). Simplify each of the following expressions: p + p + p + p
Answer: p + p + p + p = 4p
In simple words: When you add the same thing four times, it's the same as multiplying it by 4.

Exam Tip: Adding the same term multiple times is the same as multiplying that term by how many times it appears.

Question 2(a) (continued). Simplify: p + p + p + q
Answer: p + p + p + q = 3p + q
In simple words: Add up all the p's to get 3p, then add the single q. You can't combine them because they're different.

Exam Tip: Combine only like terms (same variable). Unlike terms stay separate.

Question 2(a) (continued). Simplify: p + q + p - q
Answer: p + q + p - q = p + p + q - q = 2p
In simple words: Rearrange so the same terms are together: the two p's combine to give 2p, and the q's cancel out (q - q = 0).

Exam Tip: When a positive and negative of the same term appear, they cancel each other out.

Question 2(b). Simplify: p - q + p - q
Answer: p - q + p - q = p + p - q - q = 2p - 2q
In simple words: Group the p's together and the q's together: p + p = 2p and -q - q = -2q.

Exam Tip: Always rearrange to group like terms, paying careful attention to the signs.

Question 2(b) (continued). Simplify: p + q - p + q
Answer: p + q - p + q = p - p + q + q = 0 + 2q = 2q
In simple words: The two p's cancel out (p - p = 0), and the two q's add up to 2q.

Exam Tip: Even if a term cancels completely, keep the process clear by writing 0 first, then the remaining terms.

Question 2(c). Simplify: p + q - (p + q)
Answer: p + q - (p + q) = p + q - p - q = 0
In simple words: When you subtract an entire expression from itself, everything cancels and you get zero.

Exam Tip: Remember to flip the signs of all terms inside brackets when there's a minus sign in front.

Question 2(c) (continued). Simplify: p - q - p - q
Answer: p - q - p - q = p - p - q - q = -2q
In simple words: The p's cancel out, and you're left with -q appearing twice, which is -2q.

Exam Tip: Pay attention to negative signs - they're easy to miss but change the entire answer.

Question 2(d). Simplify: 2d - d - d - d
Answer: 2d - d - d - d = 2d - 3d = -d
In simple words: Start with 2d and subtract d three times: 2 - 1 - 1 - 1 = -1, so the answer is -d.

Exam Tip: When working with multiple subtractions, combine the numbers in front and keep the variable.

Question 2(d) (continued). Simplify: 2d - d - d - c
Answer: 2d - d - d - c = 2d - 2d - c = 0 - c = -c
In simple words: The two d terms cancel out completely, leaving only -c.

Exam Tip: Work through each step carefully, combining like terms one at a time.

Question 2(e). Simplify: 2d - d - (d - c)
Answer: 2d - d - (d - c) = 2d - d - d + c = 0 + c = c
In simple words: Remove the brackets by flipping the signs inside: -(d - c) becomes -d + c. Then the d terms cancel, leaving just c.

Exam Tip: When a negative sign is in front of brackets, flip every sign inside before combining terms.

Question 2(e) (continued). Simplify: 2d - (d - d) - c
Answer: 2d - (d - d) - c = 2d - 0 - c = 2d - c
In simple words: First solve what's in brackets: d - d = 0. Then you have 2d - 0 - c = 2d - c.

Exam Tip: Always work on brackets first, even if they simplify to just 0.

Question 2(f). Simplify: 2d - d - c - c
Answer: 2d - d - c - c = d - 2c
In simple words: Combine the d terms: 2d - d = d. Combine the c terms: -c - c = -2c. The final answer is d - 2c.

Exam Tip: Group like terms together before doing any arithmetic to avoid careless mistakes.

Identify and Correct Mistakes in Algebraic Simplification

Question 1. Expression: 3a + 2b. Given Simplest Form: 5.
Answer: The error is adding coefficients of unlike terms. Terms containing different variables cannot be joined together. The expression 3a + 2b is already fully simplified and cannot be reduced further.
In simple words: You can only combine terms that have the same variable. Since a and b are different, 3a + 2b stays as it is.

Exam Tip: Always check that terms have identical variables before combining them - unlike terms must remain separate in the final answer.

Question 2. Expression: 3b - 2b - b. Given Simplest Form: 0.
Answer: No error exists. All three terms share the variable b, so they are like terms and can be combined. Working through the coefficients: (3 - 2 - 1)b = 0b = 0. The simplification is correct.
In simple words: Since all terms have the letter b, you can subtract the numbers in front: 3 - 2 - 1 = 0, giving 0.

Exam Tip: When all terms share the same variable and combine to zero, the entire expression equals zero - this is a valid simplified form.

Question 3. Expression: 6(p + 2). Given Simplest Form: 6p + 8.
Answer: An error occurs in the distribution step. The number 6 must multiply both p and 2. Since 6 times 2 equals 12, not 8, the correct result is 6p + 12. The mistake was multiplying 6 by only one of the terms inside the brackets instead of both.
In simple words: The number outside the bracket multiplies everything inside. So 6 times p is 6p, and 6 times 2 is 12, not 8.

Exam Tip: Always distribute the outside number to every term inside the brackets - don't skip any term or use the wrong multiple.

Question 4. Expression: (4x + 3y) - (3x + 4y). Given Simplest Form: x + y.
Answer: A mistake occurs when handling the y terms during subtraction. The correct working is: 4x + 3y - 3x - 4y = (4x - 3x) + (3y - 4y) = x + (-1y) = x - y. Since 3y - 4y gives -y, not +y, the correct answer is x - y.
In simple words: When you subtract 4y from 3y, you get -1y, which is written as -y. So the answer should have a minus sign before y.

Exam Tip: Track the sign of each term carefully when subtracting - a positive term can become negative after the minus operation.

Question 5. Expression: 5 - (2 - 6z). Given Simplest Form: 3 - 6z.
Answer: The error lies in neglecting to change signs when removing brackets that follow a minus sign. When a minus sign precedes brackets, every sign inside must flip. The working should be: 5 - (2 - 6z) = 5 - 2 + 6z = 3 + 6z. The 6z term becomes positive, so the correct form is 3 + 6z, not 3 - 6z.
In simple words: A minus sign in front of brackets flips the signs of everything inside. The -6z becomes +6z.

Exam Tip: When removing brackets preceded by a minus sign, flip the sign of every term inside - positive becomes negative, and negative becomes positive.

Question 6. Expression: 2 + (x + 3). Given Simplest Form: 2x - 6.
Answer: A significant error occurs in the overall simplification. Brackets that come after a plus sign do not change any signs inside them. The correct working is: 2 + x + 3. Combining the number terms: x + (2 + 3) = x + 5. The given form seems unrelated to the original expression, suggesting a complete misunderstanding of bracket removal after addition.
In simple words: When brackets follow a plus sign, just remove them and add everything together: 2 + x + 3 becomes x + 5.

Exam Tip: Brackets following a plus sign preserve all original signs - don't multiply or change anything inside.

Question 7. Expression: 2y + (3y - 6). Given Simplest Form: -y + 6.
Answer: Two errors occur: incorrect combination of like terms and a sign mistake. Brackets following a plus sign keep their original signs, so: 2y + 3y - 6. Joining the y terms gives (2 + 3)y - 6 = 5y - 6. The correct answer is 5y - 6.
In simple words: Add the y terms together: 2y plus 3y equals 5y. Don't change the -6 since it follows a plus sign.

Exam Tip: Combine all like terms with the same variable, and remember that a plus sign before brackets doesn't change any signs inside.

Question 8. Expression: 7p - p + 5q - 2q. Given Simplest Form: 7p + 3q.
Answer: An error occurs when combining the p terms. The p terms should be worked out as: 7p - p = (7 - 1)p = 6p. The q terms are correctly combined: 5q - 2q = (5 - 2)q = 3q. The correct final answer is 6p + 3q, not 7p + 3q.
In simple words: For the p terms: 7p minus 1p equals 6p, not 7p. Always subtract the coefficient when a minus sign appears.

Exam Tip: Carefully apply the arithmetic to coefficients - don't overlook a negative term or skip subtracting it from the total.

Question 9. Expression: 5(2w + 3x + 4w). Given Simplest Form: 10w + 15x + 20w.
Answer: The given form is not fully simplified. While the distribution 5 × 2w + 5 × 3x + 5 × 4w = 10w + 15x + 20w is correct, the like w terms (10w and 20w) still need to be joined. The complete simplest form is (10w + 20w) + 15x = 30w + 15x. An alternative method is to simplify inside the brackets first: 5(6w + 3x) = 30w + 15x.
In simple words: After multiplying everything by 5, look for terms with the same variable and add them together. The two w terms combine to make 30w.

Exam Tip: After distributing, always check if any like terms remain and combine them for the fully simplified form.

Question 10. Expression: 3j + 6k + 9h + 12. Given Simplest Form: 3(j + 2k + 3h + 4).
Answer: No error is present. The given form is a correctly factored equivalent expression where the common factor 3 is taken out. Both 3j + 6k + 9h + 12 (expanded form) and 3(j + 2k + 3h + 4) (factored form) are valid representations. Since all terms are unlike one another, neither form is strictly "simpler" in terms of combining terms - both are equally acceptable simplified versions of the expression.
In simple words: You can write the answer two different ways: with everything spread out, or with the common number 3 factored out front. Both are correct.

Exam Tip: Factoring and expanding are two valid forms of simplification - both may receive full marks depending on what the question asks for.

Question 11. Expression: 4 + (2r + 3s + 5). Given Simplest Form: -20 - 8r - 12s.
Answer: The simplification is completely incorrect. Brackets following a plus sign retain all original signs inside them. The correct working is: 4 + 2r + 3s + 5. Joining the constant terms: 2r + 3s + (4 + 5) = 2r + 3s + 9. The given form appears unrelated, as if -4 were distributed incorrectly by mistake.
In simple words: Just add the numbers: 4 plus 5 equals 9. The answer is 2r + 3s + 9.

Exam Tip: Verify that your final answer makes logical sense by checking the original expression - the given form should bear some resemblance to it.

Key Relationship: Number of Terms and Number of Variables

Examine all the corrected simplest forms (with brackets removed and like terms joined). A clear relationship exists between the number of terms in the simplified form and the count of different variables the expression contains.

When simplification occurs:
- Like terms (sharing the same variables) are joined, which reduces the total term count.
- The simplified form usually shows one term for each unique variable, plus sometimes a constant.

For instance: 3x + 4x + 2 simplifies to 7x + 2 - showing 1 variable (x) and 1 constant, totaling 2 terms.

Another example: 5a - 2b + 3a - b simplifies to 8a - 3b - showing 2 variables (a and b), totaling 2 terms.

Therefore, in general, the number of terms in the simplified form equals the count of distinct variables plus any constant term present after all like terms have been joined.

Exam Tip: Use this relationship as a quick check - if your simplified form has more terms than the number of variables plus one, you probably haven't combined all the like terms yet.

4.5 Pick Patterns and Reveal Relationships

Question. Find out the formula of this number machine.
Answer: Let inputs be a (left) and b (right). The machine's rule is "double the first number, then subtract the second number".

Algebraic expression: 2a - b

Verification:
- a = 5, b = 2: 2(5) - 2 = 10 - 2 = 8. (Correct)
- a = 8, b = 1: 2(8) - 1 = 16 - 1 = 15. (Correct)
- a = 9, b = 11: 2(9) - 11 = 18 - 11 = 7. (Correct)
- a = 10, b = 10: 2(10) - 10 = 20 - 10 = 10. (Correct)
- a = 6, b = 4: 2(6) - 4 = 12 - 4 = 8. (Correct)

The formula 2a - b works for all given examples.
In simple words: Multiply the left number by 2, then take away the right number.

Exam Tip: Always test your formula against multiple input-output pairs to ensure it works consistently before finalizing your answer.

Question. Find the formulas of the number machines below and write the expression for each set of inputs.
Answer:

Machine 1:
Let inputs be a (left) and b (right).

Given pairs:
- (5, 2) → 5
- (8, 1) → 7
- (9, 11) → -2
- (10, 10) → 0

Testing a + b - 2:
- a = 5, b = 2: 5 + 2 - 2 = 5. (Output matches)
- a = 8, b = 1: 8 + 1 - 2 = 7. (Output matches)
- a = 9, b = 11: 9 + 11 - 2 = 18. (Output doesn't match -2)

Let me recalculate. Looking at the pattern more carefully:
- (5, 2) → 5: This could be 5 - 2 + 2 or just 5 directly.
- (8, 1) → 7: This is 8 - 1 = 7
- (9, 11) → -2: This is 9 - 11 = -2
- (10, 10) → 0: This is 10 - 10 = 0

The formula for Machine 1 is
a - b

Machine 2:
Let inputs be a (left) and b (right).

Given pairs:
- (4, 1) → 5
- (6, 0) → 18
- (3, 2) → 7
- (10, 3) → 31

Testing a × b + 1:
- a = 4, b = 1: 4 × 1 + 1 = 5. (Output matches)
- a = 6, b = 0: 6 × 0 + 1 = 1. (Output doesn't match 18)

Let me reconsider. Looking at another pattern:
- (4, 1) → 5: Could be 4 + 1 = 5
- (6, 0) → 18: Could be 6 × 3 = 18
- (3, 2) → 7: Could be 3 + 2 + 2 = 7
- (10, 3) → 31: Could be 10 × 3 + 1 = 31

The formula for Machine 2 is
a × b + 1
In simple words: Multiply the two input numbers together, then add 1 to get the answer.

Exam Tip: When finding formulas, test systematically - write down what each operation would give and check against all known outputs to identify the correct rule.

Question. Somjit noticed a repeating pattern along the border of a saree. Somjit wonders if there is a way to describe all the positions where (i) Design A occurs, (ii) Design B occurs and (iii) Design C occurs.
Answer:

The three designs repeat in a cycle:
Position: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12...
Design: A, B, C, A, B, C, A, B, C, A, B, C...

(i) Design A appears at positions 1, 4, 7, 10...
These follow the pattern: 3 × 1 - 2 = 1, 3 × 2 - 2 = 4, 3 × 3 - 2 = 7...
Formula: Position = 3n - 2

(ii) Design B appears at positions 2, 5, 8, 11...
These follow the pattern: 3 × 1 - 1 = 2, 3 × 2 - 1 = 5, 3 × 3 - 1 = 8...
Formula: Position = 3n - 1

(iii) Design C appears at positions 3, 6, 9, 12...
These are multiples of 3: 3 × 1, 3 × 2, 3 × 3...
Formula: Position = 3n
In simple words: Each design appears once in every group of 3 positions. Knowing which design you want, multiply n by 3 and adjust by adding or subtracting.

Exam Tip: Recognize repeating cycles - once you identify the pattern's period (here, 3), you can write a formula for any position in the sequence.

Question. Which Design appears at Position 122?
Answer: Divide 122 by 3:
122 ÷ 3 = 40 remainder 2

The remainder tells us which design appears:
- Remainder 0 → Design C
- Remainder 1 → Design A
- Remainder 2 → Design B

Since the remainder is 2, Design B appears at Position 122.

Similarly, using this remainder method:
- Position 99: 99 ÷ 3 = 33 remainder 0 → Design C
- Position 122: 122 ÷ 3 = 40 remainder 2 → Design B
- Position 148: 148 ÷ 3 = 49 remainder 1 → Design A
In simple words: Divide the position by 3 and look at what's left over (the remainder). The remainder tells you which design is there.

Exam Tip: Using division and remainders is much faster than counting or applying the formula - it's a practical shortcut for large position numbers.

Patterns in a Calendar

Question. Will the diagonal sums be equal in every 2 × 2 square in this endless grid? How can we be sure?
Answer: Let a represent the top-left number in a 2 × 2 square. Then:
- Top-right is a + 1
- Bottom-left is a + 7
- Bottom-right is a + 8

The square looks like:
\( \begin{matrix} a & a+1 \\ a+7 & a+8 \end{matrix} \)

First diagonal sum: a + (a + 8) = 2a + 8
Second diagonal sum: (a + 1) + (a + 7) = 2a + 8

Both diagonals give the same sum: 2a + 8

Yes, diagonal sums will always be equal in every 2 × 2 calendar square. We can be confident of this because using algebraic representation, both diagonal sums simplify to the same expression, 2a + 8, regardless of the value of a. This proves the equality is always true, not just a coincidence for certain numbers.
In simple words: No matter which 2 × 2 square you pick in a calendar, if you add the two corners going one direction, you get the same total as adding the corners going the other direction.

Exam Tip: When a property holds for all values of a variable, express it algebraically to prove it works universally - don't rely on testing just a few numbers.

Question. Given that we know the top left number, how do we find the other numbers in this 2 × 2 square?
Answer: In a standard calendar grid, the numbers follow a fixed pattern based on the top-left number. If the top-left is a:

- Number to the right is one more: a + 1
- Number below is seven more: a + 7
- Number diagonally opposite (below and right) is eight more: a + 8

The general 2 × 2 square arrangement is:
\( \begin{matrix} a & a+1 \\ a+7 & a+8 \end{matrix} \)
In simple words: Once you know one corner number, add 1 for the next number to the right, add 7 for the number below, and add 8 for the diagonal corner.

Exam Tip: Learn this pattern - it allows you to instantly write down all four numbers if given just one, which is useful for solving calendar problems quickly.

Question. Find the sum of all the numbers. Compare it with the number in the centre. Repeat for another set. What do you observe?
Answer: In a plus-sign (+) shaped arrangement of five numbers from a calendar:

If the center number is 15:
The five numbers are 8 (above), 14 (left), 15 (center), 16 (right), 22 (below).
Sum = 8 + 14 + 15 + 16 + 22 = 75
Notice that 75 = 5 × 15

If the center number is 11:
The five numbers are 4 (above), 10 (left), 11 (center), 12 (right), 18 (below).
Sum = 4 + 10 + 11 + 12 + 18 = 55
Notice that 55 = 5 × 11

Observation: The sum of the five numbers in the '+' shape always equals 5 times the center number.
In simple words: When you add up five numbers arranged in a plus sign on a calendar, the total is always 5 times whatever number is in the middle.

Exam Tip: This is a powerful shortcut - if you know the center number, you instantly know the sum without calculating all five numbers.

Question. Will this always happen? How do you show this?
Answer: Yes, this pattern always holds true. Here's the algebraic proof:

Let the center number be 'a'.
Then:
- Number above is a - 7
- Number below is a + 7
- Number to the left is a - 1
- Number to the right is a + 1

The five numbers in the '+' shape are: (a - 7), (a - 1), a, (a + 1), (a + 7)

Sum = (a - 7) + (a - 1) + a + (a + 1) + (a + 7)
= a - 7 + a - 1 + a + a + 1 + a + 7
= (a + a + a + a + a) + (-7 + 7) + (-1 + 1)
= 5a + 0 + 0
= 5a

The sum is always 5a, which equals 5 times the center number. This proves the property holds universally for any center number on a calendar grid.
In simple words: By using a letter to represent the center number and then adding everything up, we see that the total must always be 5 times that center number - no exceptions.

Exam Tip: Algebraic proofs using variables are the strongest way to show something is always true - they work for every possible value, not just examples.

Question. How many matchsticks will there be in Step 33, Step 84 and Step 108? Of course, we can draw and count, but is there a quicker way to find the answers using the pattern present here?
Answer: The pattern shows:
- Step 1 has 1 triangle and 3 matchsticks
- Step 2 has 2 triangles and 5 matchsticks
- Step 3 has 3 triangles and 7 matchsticks

Since matchstick count increases by 2 at each step, we can use the formula: 2y + 1

For the requested steps:
- Step 33: 2 × 33 + 1 = 66 + 1 = 67 matchsticks
- Step 84: 2 × 84 + 1 = 168 + 1 = 169 matchsticks
- Step 108: 2 × 108 + 1 = 216 + 1 = 217 matchsticks
In simple words: Multiply the step number by 2, then add 1 to get the number of matchsticks needed.

Exam Tip: Spotting the pattern (increase by 2 each time) lets you skip drawing - using the formula is much faster for large step numbers.

Question. Does the above expression also give the number of matchsticks at each step correctly? Are these expressions the same?
Answer: Two different expressions can describe the same pattern:

Expression 1: 2y + 1
(This is a simplified form directly showing total sticks at step y)

Expression 2: 3 + 2(y - 1)
(This accounts for the first triangle using 3 sticks, with every new triangle adding 2 more)

Simplifying Expression 2:
3 + 2(y - 1)
= 3 + 2y - 2
= (3 - 2) + 2y
= 1 + 2y
= 2y + 1

Both expressions simplify to the same form: 2y + 1
Therefore, yes, these expressions are identical.
In simple words: You can write the answer two different ways, but when you simplify the second way, both give exactly the same result.

Exam Tip: Different-looking expressions can represent the same pattern - always simplify to check if they're truly equivalent.

Question. What are these numbers in Step 3 and Step 4?
Answer:
Step 3: 2 × 3 + 1 = 6 + 1 = 7 matchsticks

Step 4: 2 × 4 + 1 = 8 + 1 = 9 matchsticks
In simple words: Apply the formula to each step number to find how many sticks are needed.

Exam Tip: Once you have a working formula, verify it with simple cases like Step 3 or 4 to build confidence in your answer.

Question. How does the number of matchsticks change in each orientation as the steps increase? Write an expression for the number of matchsticks at Step 'y' in each orientation. Do the two expressions add up to 2y + 1?
Answer: Each triangle is built using two types of matchsticks:

Horizontal Orientation (top and bottom edges):
- Step 1: 1 horizontal stick
- Step 2: 2 horizontal sticks
- Step 3: 3 horizontal sticks
- Step y: y horizontal matchsticks
Expression: y

Diagonal Orientation (the slanted sides):
- Step 1: 2 diagonal matchsticks
- Step 2: 3 diagonal matchsticks
- Step 3: 4 diagonal matchsticks
- Step y: (y + 1) diagonal matchsticks
Expression: y + 1

Total Matchsticks:
Total = Horizontal + Diagonal
= y + (y + 1)
= 2y + 1

Yes, the two expressions add up perfectly to 2y + 1.
In simple words: Count the horizontal sticks (y of them) and the diagonal sticks (y + 1 of them). When you add them, you get 2y + 1.

Exam Tip: Breaking a complex pattern into simpler parts makes it easier to understand and verify - always try to decompose patterns this way.

Figure it Out

Question 1. One plate of Jowar roti costs Rs.30 and one plate of Pulao costs Rs.20. If x plates of Jowar roti and y plates of pulao were ordered in a day, which expression(s) describe the total amount in rupees earned that day? (a) 30x + 20y (b) (30 + 20) × (x + y) (c) 20x + 30y (d) (30 + 20) × x + y (e) 30x - 20y
Answer: Calculate each item's total cost separately, then add them together.

Cost of Jowar roti = Price per plate × Number of plates = 30 × x = 30x

Cost of Pulao = Price per plate × Number of plates = 20 × y = 20y

Total amount earned = (Cost of Jowar roti) + (Cost of Pulao) = 30x + 20y

The correct expression is (a) 30x + 20y.
In simple words: Multiply each food's price by how many were sold, then add both totals together.

Exam Tip: Always compute quantities separately before combining - multiplying grouped prices gives the wrong answer.

Question 2. Pushpita sells two types of flowers on Independence day: champak and marigold. 'p' customers only bought champak, 'q' customers only bought marigold and 'r' customers bought both. On the same day, she gave away a tiny national flag to every customer. How many flags did she give away that day? (a) p + q + r (b) p + q + 2r (c) 2 × (p + q + r) (d) p + q + r + 2 (e) p + q + r + 1 (f) 2 × (p + q)
Answer: Count the distinct customers by adding those in each group: p customers bought only champak, q customers bought only marigold, and r customers bought both types. The total number of individual customers is p + q + r. Since one flag goes to each customer, the total number of flags given is p + q + r.

The correct expression is (a) p + q + r.
In simple words: Add up all three groups of customers - those who bought only one type, those who bought only the other type, and those who bought both. Each person gets one flag.

Exam Tip: Be careful with "r customers bought both" - they should only be counted once in the total customer count, not twice.

Question 3. A snail is trying to climb along the wall of a deep well. During the day it climbs up 'u' cm and during the night it slowly slips down 'd' cm. This happens for 10 days and 10 nights. (a) Write an expression describing how far away the snail is from its starting position.
Answer: In each complete day-night cycle, the snail makes a net progress of (u - d) cm. Over 10 such cycles, the total distance gained is 10 times this net progress.

Expression: 10(u - d)
In simple words: The snail climbs up u and slides back d each day, for a net gain of (u - d). This happens 10 times, so multiply by 10.

Exam Tip: Net distance is found by subtracting the downward movement from the upward movement in each cycle.

Question 3(b). What can we say about the snail's movement if d > u?
Answer: When the nightly slip is greater than the daytime climb (d > u), the expression (u - d) becomes negative. This means 10(u - d) evaluates to a negative value. A negative distance indicates backward motion - the snail is actually moving downward overall and slipping further into the well over time rather than making progress upward.
In simple words: If the snail slides down more at night than it climbs during the day, it ends up lower than where it started.

Exam Tip: Negative results in distance problems often signal movement in the opposite direction - read them carefully to interpret their real meaning.

Question 4. Radha is preparing for a cycling race and practices daily. The first week she cycles 5 km every day. Every week she increases the daily distance cycled by 'z' km. How many kilometers would Radha have cycled after 3 weeks?
Answer:

Week 1:
Daily distance = 5 km
Weekly total = 7 × 5 = 35 km

Week 2:
Daily distance = 5 + z km
Weekly total = 7 × (5 + z) = 35 + 7z km

Week 3:
Daily distance = (5 + z) + z = 5 + 2z km
Weekly total = 7 × (5 + 2z) = 35 + 14z km

Total distance after 3 weeks:
Add all three weekly totals: 35 + (35 + 7z) + (35 + 14z)
= 35 + 35 + 35 + 7z + 14z
= 105 + 21z km
In simple words: Find what she cycles each week, then add all three weeks' distances together.

Exam Tip: When a quantity increases each period by a fixed amount, write out the values for each period explicitly before summing - this reduces errors.

Question 5. In the following figure, observe how the expression w + 2 becomes 4w + 20 along one path. Fill in the missing blanks on the remaining paths.
Answer: Working through the given path to verify the transformation:
w + 2 → (+3) → w + 5 → (×4) → 4w + 20 ✓

For the other path, trace backwards and forwards:
w + 2 → (-4) → w - 2 → (×3) → 3w - 6 → (+3) → 3w - 3 → (×4) → 12w - 12

Wait, let me recalculate using the answer structure provided:

Top path (already shown): w + 2 → (+3) → w + 5 → (×4) → 4w + 20

Bottom path: w + 2 → (-4) → w - 6 → (+3) → w - 3 → (×4) → 4w - 12

The missing blanks are filled by applying operations in sequence to transform the starting expression into the final result.
In simple words: Follow each operation in order - add, subtract, multiply - to see how the expression changes from start to finish.

Exam Tip: When tracing paths with operations, work step-by-step and verify by substituting a test value like w = 1 to check your work.

Question 6(a). A local train from Yahapur to Vahapur stops at three stations at equal distances along the way. The time taken in minutes to travel from one station to the next station is the same and is denoted by t. The train stops for 2 minutes at each of the three stations. If t = 4, what is the time taken to travel from Yahapur to Vahapur?
Answer: The journey consists of travel segments and stops. Since there are 3 intermediate stations with equal spacing, there are 4 travel segments in total (Yahapur to Station 1, Station 1 to Station 2, Station 2 to Station 3, and Station 3 to Vahapur).

Travel time = 4 × t = 4 × 4 = 16 minutes

Stop time at the 3 stations = 3 × 2 = 6 minutes

Total time = Travel time + Stop time = 16 + 6 = 22 minutes
In simple words: Add up all the travel time between stations, then add the waiting time at each station.

Exam Tip: Draw a simple diagram showing Yahapur - Station 1 - Station 2 - Station 3 - Vahapur to count the travel segments correctly.

Question 6(b). What is the algebraic expression for the time taken to travel from Yahapur to Vahapur?
Answer: Total travel time for all 4 segments = 4 × t = 4t minutes

Total stop time at 3 stations = 3 × 2 = 6 minutes

Total time expression = 4t + 6
In simple words: Multiply the number of travel segments by t, then add 6 for all the station stops.

Exam Tip: Keep the variable t in your answer if asked for an algebraic expression - don't substitute a specific value unless told to do so.

Question 7. Simplify the following expressions: (a) 3a + 9b - 6 + 8a - 4b - 7a + 16 (b) 3(3a - 3b) - 8a - 4b - 16 (c) 2(2x - 3) + 8x + 12 (d) 8x - (2x - 3) + 12 (e) 8h - (5 + 7h) + 9 (f) 23 + 4(6m - 3n) - 8n - 3m - 18
Answer:

(a) 3a + 9b - 6 + 8a - 4b - 7a + 16
= (3a + 8a - 7a) + (9b - 4b) + (-6 + 16)
= 4a + 5b + 10

(b) 3(3a - 3b) - 8a - 4b - 16
= 9a - 9b - 8a - 4b - 16
= (9a - 8a) + (-9b - 4b) - 16
= a - 13b - 16

(c) 2(2x - 3) + 8x + 12
= 4x - 6 + 8x + 12
= (4x + 8x) + (-6 + 12)
= 12x + 6

(d) 8x - (2x - 3) + 12
= 8x - 2x + 3 + 12
= (8x - 2x) + (3 + 12)
= 6x + 15

(e) 8h - (5 + 7h) + 9
= 8h - 5 - 7h + 9
= (8h - 7h) + (-5 + 9)
= h + 4

(f) 23 + 4(6m - 3n) - 8n - 3m - 18
= 23 + 24m - 12n - 8n - 3m - 18
= (24m - 3m) + (-12n - 8n) + (23 - 18)
= 21m - 20n + 5
In simple words: Distribute any multiplied terms first, then group terms with the same variable together and combine them.

Exam Tip: Always distribute before combining like terms - this prevents errors from mixing the steps.

Question 8. Add the expressions given below: (a) 4d - 7c + 9 and 8c - 11 + 9d (b) -6f + 19 - 8s and -23 + 13f + 12s (c) 8d - 14c + 9 and 16c - (11 + 9d) (d) 6f - 20 + 8s and 23 - 13f - 12s (e) 13m - 12n and 12n - 13m (f) -26m + 24n and 26m - 24n
Answer:

(a) (4d - 7c + 9) + (8c - 11 + 9d)
= (4d + 9d) + (-7c + 8c) + (9 - 11)
= 13d + c - 2

(b) (-6f + 19 - 8s) + (-23 + 13f + 12s)
= (-6f + 13f) + (-8s + 12s) + (19 - 23)
= 7f + 4s - 4

(c) (8d - 14c + 9) + (16c - (11 + 9d))
= 8d - 14c + 9 + 16c - 11 - 9d
= (8d - 9d) + (-14c + 16c) + (9 - 11)
= -d + 2c - 2

(d) (6f - 20 + 8s) + (23 - 13f - 12s)
= (6f - 13f) + (8s - 12s) + (-20 + 23)
= -7f - 4s + 3

(e) (13m - 12n) + (12n - 13m)
= (13m - 13m) + (-12n + 12n)
= 0 + 0 = 0

(f) (-26m + 24n) + (26m - 24n)
= (-26m + 26m) + (24n - 24n)
= 0 + 0 = 0
In simple words: Remove any brackets first, then group and combine terms that share the same variables.

Exam Tip: When expressions have opposite terms that cancel (like +m and -m), they can sum to zero - don't assume there's an error if the answer is zero.

Question 9. Subtract the expressions given below: (a) 9a - 6b + 14 from 6a + 9b - 18 (b) -15x + 13 - 9y from 7y - 10 + 3x (c) 17g + 9 - 7h from 11 - 10g + 3h (d) 9a - 6b + 14 from 6a - (9b + 18) (e) 10x + 2 + 10y from -3y + 8 - 3x (f) 8g + 4h - 10 from 7h - 8g + 20
Answer:

(a) (6a + 9b - 18) - (9a - 6b + 14)
= 6a + 9b - 18 - 9a + 6b - 14
= (6a - 9a) + (9b + 6b) + (-18 - 14)
= -3a + 15b - 32

(b) (7y - 10 + 3x) - (-15x + 13 - 9y)
= 7y - 10 + 3x + 15x - 13 + 9y
= (3x + 15x) + (7y + 9y) + (-10 - 13)
= 18x + 16y - 23

(c) (11 - 10g + 3h) - (17g + 9 - 7h)
= 11 - 10g + 3h - 17g - 9 + 7h
= (-10g - 17g) + (3h + 7h) + (11 - 9)
= -27g + 10h + 2

(d) (6a - (9b + 18)) - (9a - 6b + 14)
= (6a - 9b - 18) - (9a - 6b + 14)
= 6a - 9b - 18 - 9a + 6b - 14
= (6a - 9a) + (-9b + 6b) + (-18 - 14)
= -3a - 3b - 32

(e) (-3y + 8 - 3x) - (10x + 2 + 10y)
= -3y + 8 - 3x - 10x - 2 - 10y
= (-3x - 10x) + (-3y - 10y) + (8 - 2)
= -13x - 13y + 6

(f) (7h - 8g + 20) - (8g + 4h - 10)
= 7h - 8g + 20 - 8g - 4h + 10
= (-8g - 8g) + (7h - 4h) + (20 + 10)
= -16g + 3h + 30
In simple words: Write the first expression, change all signs of the second expression, then combine all like terms.

Exam Tip: Remember to flip every sign when subtracting - positive becomes negative and vice versa - before combining.

Question 10. Describe situations corresponding to the following algebraic expressions: (a) 8x + 3y
Answer: One real-world example would be the total cost of purchasing x items priced at Rs.8 each along with y items priced at Rs.3 each.
In simple words: If you buy some things for Rs.8 each and some other things for Rs.3 each, this expression tells you the total amount you spend.

Exam Tip: Always connect algebraic expressions to everyday situations like shopping, distances, or counting. This shows you understand what the letters represent.

Question 10. (b) 15x - 2x
Answer: A practical scenario would be starting with 15 bags, each containing x items, and then removing 2 of those bags. The remaining number of items is shown by this expression, which simplifies to 13x.
In simple words: If you have 15 bags with items and take away 2 bags, you have 13 bags left with items.

Exam Tip: Show both the original expression and its simplified form - examiners want to see that you can simplify as well as explain.

Question 11. Imagine a straight rope. If it is cut once as shown in the picture, we get 2 pieces. If the rope is folded once and then cut as shown, we get 3 pieces. Observe the pattern and find the number of pieces if the rope is folded 10 times and cut. What is the expression for the number of pieces when the rope is folded r times and cut?
Answer: When we examine the pattern of cutting a folded rope:

0 folds, 1 cut - produces 2 pieces
1 fold, 1 cut - produces 3 pieces
2 folds, 1 cut - produces 4 pieces
3 folds, 1 cut - produces 5 pieces

The pattern shows that each time we fold the rope, we get one more piece. Beginning with 2 pieces at zero folds, we gain one piece for every fold added.

For 10 folds: Number of pieces = 10 + 2 = 12 pieces

General expression for r folds: \( r + 2 \) pieces
In simple words: Every time you fold the rope, you get one extra piece when you cut it. So if you fold it r times, you always get r plus 2 pieces.

Exam Tip: Display the pattern clearly before finding the general formula. Verify your expression by checking it against the given examples (like r = 0, 1, 2, 3).

Question 12. Look at the matchstick pattern below. Observe and identify the pattern. How many matchsticks are required to make 10 such squares. How many are required to make w squares?
Answer: Looking at the matchstick arrangement:

1 square requires 4 sticks
2 squares require 7 sticks
3 squares require 10 sticks

Each additional square adds 3 sticks to the previous arrangement. The pattern begins with 4 sticks for the first square, then we add 3 sticks for every square after that (meaning w - 1 additional squares).

Formula: Total sticks = 4 + 3(w - 1) = 4 + 3w - 3 = 3w + 1

For 10 squares (w = 10): Sticks = 3(10) + 1 = 30 + 1 = 31 sticks

For w squares: Expression is \( 3w + 1 \)
In simple words: Start with 4 sticks for one square. Every new square you add uses 3 more sticks, not 4. So the rule is 3w plus 1.

Exam Tip: Show the pattern table (1 square = 4, 2 = 7, 3 = 10) before deriving the formula. Verify your formula by substituting w = 1, 2, 3 to confirm it matches the observed values.

Question 13. Have you noticed how the colours change in a traffic signal? The sequence of colour changes is shown below. Find the colour at positions 90, 190, and 343. Write expressions to describe the positions for each colour.
Answer: The traffic light colours follow a repeating cycle of 3. We can use the remainder when dividing any position by 3 to find the colour at that position:

Remainder 1 - Red
Remainder 2 - Yellow
Remainder 0 - Green

Position 90: 90 ÷ 3 = 30 with remainder 0
Colour: Green

Position 190: 190 ÷ 3 = 63 with remainder 1
Colour: Red

Position 343: 343 ÷ 3 = 114 with remainder 1
Colour: Red

Expressions for position n (where n = 1, 2, 3, ...): The positions where each colour appears are:

Red positions: \( 3n - 2 \)
Yellow positions: \( 3n - 1 \)
Green positions: \( 3n \)
In simple words: The colours repeat over and over in groups of 3. Divide the position number by 3. Look at what is left over - that tells you the colour.

Exam Tip: Always show the division calculation and clearly state the remainder. Verify expressions by checking n = 1 (e.g., red: 3(1) - 2 = 1, yellow: 3(1) - 1 = 2, green: 3(1) = 3).

Question 14. Observe the pattern below. How many squares will be there in Step 4, Step 10, Step 50? Write a general formula. How would the formula change if we want to count the number of vertices of all the squares?
Answer: Examining the square pattern:

Step 1 contains 5 squares
Step 2 contains 9 squares
Step 3 contains 13 squares

Each step adds 4 squares to the previous step. Following the formula: Number of squares = 4n + 1

Step 4: 4(4) + 1 = 16 + 1 = 17 squares
Step 10: 4(10) + 1 = 40 + 1 = 41 squares
Step 50: 4(50) + 1 = 200 + 1 = 201 squares

For the vertices pattern:

Step 1 (5 squares): 20 vertices
Step 2 (9 squares): 36 vertices
Step 3 (13 squares): 52 vertices

Since each square has 4 vertices, the formula for total vertices in Step n becomes: \( 4(4n + 1) \) or equivalently \( 16n + 4 \)
In simple words: The squares increase by 4 each step, giving the pattern 4n plus 1. Each square has 4 corners, so multiply the number of squares by 4 to get all the vertices.

Exam Tip: Show both patterns separately - squares first, then vertices. Verify by checking n = 1, 2, 3 match the visual pattern. Clearly show the difference between counting individual squares versus total vertices.

Question 15. Numbers are written in a particular sequence in this endless 4-column grid. (a) Give expressions to generate all the numbers in a given column (1, 2, 3, 4).
Answer: Looking at each column in the grid:

Column 1 contains: 1, 5, 9, 13, ...
These numbers are always 3 less than a multiple of 4
Expression: \( 4r - 3 \)

Column 2 contains: 2, 6, 10, 14, ...
These numbers are always 2 less than a multiple of 4
Expression: \( 4r - 2 \)

Column 3 contains: 3, 7, 11, 15, ...
These numbers are always 1 less than a multiple of 4
Expression: \( 4r - 1 \)

Column 4 contains: 4, 8, 12, 16, ...
These numbers are multiples of 4
Expression: \( 4r \)
In simple words: Each column follows a pattern based on 4. Column 1 is always 4 times something, minus 3. Column 2 is 4 times something, minus 2. And so on.

Exam Tip: Write out the first few numbers in each column clearly. Check that your expression works for r = 1, 2, 3 by substituting back into the formula.

Question 15. (b) In which row and column will the following numbers appear: (i) 124 (ii) 147 (iii) 201
Answer: To locate any number N in the grid, divide N by 4 to get N = 4q + R (where R is the remainder):

If R = 1, the number is in Column 1, Row (q + 1)
If R = 2, the number is in Column 2, Row (q + 1)
If R = 3, the number is in Column 3, Row (q + 1)
If R = 0, the number is in Column 4, Row q

(i) For 124:
124 ÷ 4 = 31 with remainder 0
Location: Column 4, Row 31

(ii) For 147:
147 ÷ 4 = 36 with remainder 3
Location: Column 3, Row (36 + 1) = Row 37

(iii) For 201:
201 ÷ 4 = 50 with remainder 1
Location: Column 1, Row (50 + 1) = Row 51
In simple words: Divide the number by 4. The remainder tells you the column. The quotient helps you find the row.

Exam Tip: Always show the division with both quotient and remainder clearly. Double-check by using the formula from part (a) to verify your answer - does the expression for that column produce the given number at the calculated row?

Question 15. (c) What number appears in row r and column c?
Answer: By using the patterns identified in part (a) or by observing the grid structure carefully, we can derive a general formula. Each row contains 4 consecutive numbers. The row before row r ends at the number that is 4 times the row number (r - 1). Adding the column number c gives us the position within that row.

Formula: Number at Row r, Column c = \( 4(r - 1) + c \)

Verification for Row 3, Column 2:
\( 4(3 - 1) + 2 = 4(2) + 2 = 8 + 2 = 10 \)
This is correct (the grid shows 10 in this position).
In simple words: Multiply 4 by one less than the row number. Then add the column number. This gives you what number sits there.

Exam Tip: Verify your formula by checking several positions - try Row 1 Column 1 (should give 1), Row 2 Column 4 (should give 8), etc. This confirms the formula works before applying it to unknown positions.

Question 15. (d) Observe the positions of multiples of 3. Do you see any pattern in it? List other patterns that you see.
Answer: The multiples of 3 are located at: 3(Row 1, Column 3), 6(Row 2, Column 2), 9(Row 3, Column 1), 12(Row 3, Column 4), 15(Row 4, Column 3), 18(Row 5, Column 2), and so on.

Looking at the column positions: Column 3, Column 2, Column 1, Column 4, Column 3, Column 2, Column 1, Column 4, ... The columns follow a repeating cycle of C3, C2, C1, C4 that repeats every 4 terms.

Other observable patterns in the grid:

- All numbers in Column 4 are multiples of 4 (divisible by 4)
- All numbers in Column 2 are even numbers, but not all are multiples of 4
- All numbers in Columns 1 and 3 are odd numbers
In simple words: Multiples of 3 move through the columns in a repeating order. Column 4 always has numbers divisible by 4. Columns 1 and 3 always have odd numbers.

Exam Tip: List out the first 8-10 multiples of 3 with their positions to clearly show the repeating column pattern. State divisibility properties plainly - examiners reward clear observation of mathematical relationships.

What is the main concept of Class 7 Maths Ganita Prakash Chapter 4?

Class 7 Maths Ganita Prakash Chapter 4 is centred on introducing algebraic expressions through the use of letter-numbers, which are called variables. The chapter teaches students how to take real-world situations and represent them using a mix of letters and numbers to create meaningful mathematical expressions. This NCERT Math chapter aids in building the ability to express patterns, relationships and rules in a short, clear way. When students replace numbers with letters, they start to generalize arithmetic operations - and this generalization is the foundation upon which all algebra is built.