Class 7 : Maths β Lesson 4. Expressions using Letter-Numbers
EXPLANATION AND ANALYSIS
π΅ Introduction: When Numbers Are Not Fixed
π§ In earlier lessons, we worked with numbers whose values were known and fixed. However, in many situations, we do not know the exact value of a quantity. For example, the cost of one pen may vary, the number of students in a class may change, or the length of a side of a shape may not be given.
πΏ To represent such changing or unknown quantities, we use letters along with numbers. These combinations are called expressions using letter-numbers. This chapter introduces us to the idea of using letters to represent numbers and forming meaningful mathematical expressions.
π’ Meaning of Letter-Numbers
π§ A letter-number is a letter used to represent a number whose value is not fixed or is unknown.
πΉ Letters like x, y, a, b, p, q are commonly used
πΉ These letters stand for numbers
π Example
If x represents a number, then x + 5 is an expression using a letter-number.
π‘ Concept:
Letter-numbers help us write rules and relationships in a general form.
π΅ Expressions Using Letter-Numbers
π§ An expression using letter-numbers is formed when letters and numbers are combined using arithmetic operations.
πΉ It does not contain an equal sign
πΉ It represents a value that depends on the value of the letter
π Examples
πΉ x + 7
πΉ 3a
πΉ 5y β 2
βοΈ Note:
The value of such an expression changes when the value of the letter changes.
π’ Why Do We Use Letters in Mathematics?
π§ Letters make mathematics more powerful and flexible.
πΉ They help represent unknown values
πΉ They help write general rules
πΉ They make formulas easy to remember
πΉ They reduce lengthy calculations
π Example
The perimeter of a square with side length a is written as
4a
π‘ Concept:
Using letters allows us to describe many cases using one expression.
π΅ Forming Simple Algebraic Expressions
π§ We can form expressions using letter-numbers by combining letters with numbers and operations.
πΉ Addition: x + 4
πΉ Subtraction: y β 3
πΉ Multiplication: 5a
πΉ Division: b Γ· 2
π Example
If the cost of one book is p rupees, then the cost of 3 books is
3p
βοΈ Note:
Multiplication of a number and a letter is written without using the multiplication sign.
π’ Terms in an Expression
π§ An expression using letter-numbers is made up of terms.
πΉ A term may be a number
πΉ A term may be a letter
πΉ A term may be a product of numbers and letters
π Example
In the expression 3x + 5
πΉ 3x is one term
πΉ 5 is another term
π‘ Concept:
Terms are separated by addition or subtraction signs.
π΅ Like and Unlike Terms
π§ Terms that have the same letters raised to the same power are called like terms.
πΉ Like terms can be added or subtracted
πΉ Unlike terms cannot be combined
π Examples
πΉ 3x and 7x are like terms
πΉ 4a and 4b are unlike terms
βοΈ Note:
Only like terms can be combined in an expression.
π’ Evaluating Expressions Using Letter-Numbers
π§ Evaluating an expression means finding its value when the value of the letter is given.
πΉ Substitute the given value of the letter
πΉ Perform the operations step by step
π Example
Evaluate 2x + 5 when x = 3
πΉ Substitute x = 3
πΉ 2 Γ 3 + 5 = 6 + 5
πΉ Value = 11
π‘ Concept:
Evaluation helps us find numerical results from expressions.
π‘ Using Expressions in Daily Life
π§ Expressions using letter-numbers are useful in many real-life situations.
πΉ Calculating total cost when price per item is unknown
πΉ Finding perimeter and area of shapes
πΉ Writing rules in science and commerce
πΉ Representing patterns
π Example
If the length of a rectangle is l and breadth is b, then
Perimeter = 2(l + b)
π΄ Common Mistakes to Avoid
π΄ Using multiplication sign between number and letter
π΄ Adding unlike terms
π΄ Forgetting to substitute correct value while evaluating
π΄ Mixing up letters
βοΈ Note:
Always check whether terms are like or unlike before combining them.
π’ Importance of Expressions Using Letter-Numbers
π§ Learning this chapter helps students to:
πΉ Understand the basics of algebra
πΉ Solve problems with unknown quantities
πΉ Write mathematical rules clearly
πΉ Prepare for higher classes
This chapter is the first step towards algebra, which is a very important branch of mathematics.
π Summary
π΅ Letters are used to represent unknown numbers
π’ Expressions using letter-numbers combine letters and numbers
π‘ Such expressions do not have an equal sign
π΄ Letters help write general rules
π΅ Like terms can be combined
π’ Expressions can be evaluated by substitution
π‘ Letter-number expressions are widely used in daily life
π Quick Recap
π Quick Recap
π΅ Letters represent unknown values
π’ Letter-numbers form algebraic expressions
π‘ Expressions change with the value of letters
π΄ Like terms can be added or subtracted
π΅ These expressions are the foundation of algebra
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TEXTBOOK QUESTIONS
π΅ THE NOTION OF LETTER-NUMBERS
π β 1. Write formulas for the perimeter of:
π β (a) a triangle with all sides equal
π β Answer:
πΉ Let the length of one side be a
πΉ All three sides are equal
βοΈ Final:
Perimeter = 3 Γ a
π β (b) a regular pentagon
π β Answer:
πΉ A regular pentagon has 5 equal sides
πΉ Let the length of one side be a
βοΈ Final:
Perimeter = 5 Γ a
π β (c) a regular hexagon
π β Answer:
πΉ A regular hexagon has 6 equal sides
πΉ Let the length of one side be a
βοΈ Final:
Perimeter = 6 Γ a
π β 2. Munirathna has a 20 m long pipe. He joins another pipe of length βkβ. Write an expression for the combined length.
π β Answer (Teacher-like explanation):
πΉ First pipe length = 20 m
πΉ Second pipe length = k m
βοΈ Final:
Combined length = 20 + k meters
π β 3. Find the total amount Krithika has. Complete the table.
π β Answer (row-wise):
πΉ Row 1:
3 βΉ100 notes, 5 βΉ20 notes, 6 βΉ5 notes
πΈ Expression:
3 Γ 100 + 5 Γ 20 + 6 Γ 5
= 300 + 100 + 30
βοΈ Final:
βΉ430
πΉ Row 2 (given example):
βοΈ βΉ695 (already correct)
πΉ Row 3:
8 βΉ100 notes, 4 βΉ20 notes, z βΉ5 notes
πΈ Expression:
8 Γ 100 + 4 Γ 20 + z Γ 5
βοΈ Final:
800 + 80 + 5z = 880 + 5z
πΉ Row 4:
x βΉ100 notes, y βΉ20 notes, z βΉ5 notes
βοΈ Final Expression:
100x + 20y + 5z
π β 4. Which expression shows the time to grind βyβ kg of grain?
π β Answer (Teacher-like explanation):
πΉ Time to start machine = 10 seconds
πΉ Time to grind 1 kg = 8 seconds
πΉ Time for y kg = 8 Γ y seconds
πΉ Total time = starting time + grinding time
βοΈ Final:
π β
10 + 8 Γ y
(correct option d)
π β 5. Write algebraic expressions using letters of your choice:
π β (a) 5 more than a number
π β Answer:
πΉ Let the number be x
βοΈ Final:
x + 5
π β (b) 4 less than a number
π β Answer:
βοΈ Final:
x β 4
π β (c) 2 less than 13 times a number
π β Answer:
πΉ 13 times a number = 13x
πΉ 2 less than that = subtract 2
βοΈ Final:
13x β 2
π β (d) 13 less than 2 times a number
π β Answer:
πΉ 2 times a number = 2x
πΉ 13 less than that = subtract 13
βοΈ Final:
2x β 13
π β 6. Describe situations for the expressions:
π β (a) 8 Γ x + 3 Γ y
π β Answer:
πΉ Suppose one notebook costs βΉ8 and one pen costs βΉ3
πΉ x notebooks and y pens are bought
βοΈ Final:
Total cost = 8x + 3y
π β (b) 15 Γ j β 2 Γ k
π β Answer:
πΉ Suppose j tickets cost βΉ15 each
πΉ A discount of βΉ2 is given on each of k tickets
βοΈ Final:
Total cost after discount = 15j β 2k
π β 7. In a 2 Γ 3 calendar grid, the bottom middle cell has date βwβ. Write expressions for the other dates.
π β Answer (Teacher-like explanation):
πΉ Dates in a calendar increase by 1 each day
πΉ The bottom row is consecutive dates
βοΈ Final:
πΉ Bottom left cell = w β 1
πΉ Bottom middle cell = w
πΉ Bottom right cell = w + 1
πΉ Top left cell = w β 8
πΉ Top middle cell = w β 7
πΉ Top right cell = w β 6
π΅ SIMPLIFICATION OF ALGEBRAIC EXPRESSIONS
π β 1. Add the numbers in each picture below. Write their corresponding expressions and simplify them.
π β Answer (Picture-wise):
πΉ Picture 1
Numbers shown:
5y, β6, x, x, 2, 5y
Corresponding expression:
5y + (β6) + x + x + 2 + 5y
Simplifying:
πΉ 5y + 5y = 10y
πΉ x + x = 2x
πΉ β6 + 2 = β4
βοΈ Final:
π β
10y + 2x β 4
πΉ Picture 2
Numbers shown:
2p, 3q, β2, 3,
3q, 2p, 3, β2,
2p, 3q,
3q, 2p
Corresponding expression:
(2p + 2p + 2p + 2p) + (3q + 3q + 3q + 3q) + (β2 + 3 + 3 β 2)
Simplifying:
πΉ 2p Γ 4 = 8p
πΉ 3q Γ 4 = 12q
πΉ β2 + 3 + 3 β 2 = 2
βοΈ Final:
π β
8p + 12q + 2
πΉ Picture 3
Numbers shown:
Four β5g circles at the corners
Twelve 5k circles inside
Corresponding expression:
(β5g β 5g β 5g β 5g) + (5k Γ 12)
Simplifying:
πΉ β5g Γ 4 = β20g
πΉ 5k Γ 12 = 60k
βοΈ Final:
π β
60k β 20g
π β 2. Simplify each of the following expressions:
π β (a)
p + p + p + p,
p + p + p + q,
p + q + p β q
π β
Answers:
πΉ p + p + p + p = 4p
πΉ p + p + p + q = 3p + q
πΉ p + q + p β q = 2p
π β (b)
p β q + p β q,
p + q β p + q
π β
Answers:
πΉ p β q + p β q = 2p β 2q
πΉ p + q β p + q = 2q
π β (c)
p + q β (p + q),
p β q β p β q
π β
Answers:
πΉ p + q β (p + q) = 0
πΉ p β q β p β q = β2q
π β (d)
2d β d β d β d,
2d β d β d β c
π β
Answers:
πΉ 2d β d β d β d = βd
πΉ 2d β d β d β c = βc
π β (e)
2d β d β (d β c),
2d β (d β d) β c
π β
Answers:
πΉ 2d β d β (d β c) = c
πΉ 2d β (d β d) β c = 2d β c
π β (f)
2d β d β c β c
π β
Answer:
πΉ 2d β d β c β c = d β 2c
π΅ FIGURE IT OUT ?
π β 1. One plate of Jowar roti costs βΉ30 and one plate of Pulao costs βΉ20. 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?
π’1οΈβ£ 30x + 20y
π΅2οΈβ£ (30 + 20) Γ (x + y)
π‘3οΈβ£ 20x + 30y
π£4οΈβ£ (30 + 20) Γ x + y
πΉ (Option (e) is also printed, but MCQ format here uses 4 slots; we answer by correct expression.)
π β
Answer (Teacher-like explanation):
πΉ Amount from x plates of Jowar roti = 30 Γ x = 30x
πΉ Amount from y plates of pulao = 20 Γ y = 20y
πΉ Total amount earned = 30x + 20y
βοΈ Answer: π’1οΈβ£
π β 2. Pushpita sells two types of flowers on Independence day: champak and marigold. βpβ customers only bought champak, βqβ customers only bought marigold, and βrβ 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?
π’1οΈβ£ p + q + r
π΅2οΈβ£ p + q + 2r
π‘3οΈβ£ 2 Γ (p + q + r)
π£4οΈβ£ p + q + r + 2
πΉ (Options (e) and (f) are also printed in the book line, but the correct expression is clear.)
π β
Answer (Teacher-like explanation):
πΉ Customers are counted person-wise, not flower-wise
πΉ p customers (only champak) get 1 flag each
πΉ q customers (only marigold) get 1 flag each
πΉ r customers (bought both) are still r people, so they also get 1 flag each
πΉ Total flags = total customers = p + q + r
βοΈ Answer: π’1οΈβ£
π β 3. A snail is trying to climb along the wall of a deep well. During the day it climbs up βuβ cm and during the night it slowly slips down βdβ cm. This happens for 10 days and 10 nights.
π β (a) Write an expression describing how far away the snail is from its starting position.
π β
Answer (Teacher-like explanation):
πΉ In 1 day and 1 night, net movement = u β d
πΉ This repeats for 10 days and 10 nights
πΉ Net distance from start after 10 cycles = 10(u β d)
π β Final: 10(u β d)
π β (b) What can we say about the snailβs movement if d > u?
π β
Answer (Teacher-like explanation):
πΉ If d > u, then u β d is negative
πΉ Negative net movement means slipping down is more than climbing up
π β
Final: The snail moves downward overall and ends up below its starting position after 10 days and 10 nights.
π β 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 βzβ km. How many kilometers would Radha have cycled after 3 weeks?
π β
Answer (Teacher-like explanation):
πΉ Week 1 daily distance = 5
πΉ Week 2 daily distance = 5 + z
πΉ Week 3 daily distance = 5 + 2z
πΉ Each week has 7 days
πΉ Total distance in 3 weeks
πΈ Week 1 total = 7 Γ 5
πΈ Week 2 total = 7 Γ (5 + z)
πΈ Week 3 total = 7 Γ (5 + 2z)
πΉ Add them
πΈ = 7Γ5 + 7(5 + z) + 7(5 + 2z)
πΈ = 35 + (35 + 7z) + (35 + 14z)
πΈ = 105 + 21z
π β Final: 105 + 21z km
π β 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. (The ovals contain expressions and the boxes contain operations.)
π β
Answer (Teacher-like explanation):
πΉ Start from the center oval: w + 2
πΉ Follow arrows exactly and apply the operation in each box
πΉ Top-left path (ends at a blank oval)
πΈ w + 2, then β5 gives: (w + 2) β 5 = w β 3 (this oval is already shown)
πΈ Then Γ3 gives: 3(w β 3) = 3w β 9
π β
Blank top-left oval: 3w β 9
πΉ Bottom-left path (two blank ovals)
πΈ w + 2, then β8 gives: (w + 2) β 8 = w β 6
π β
Blank middle bottom oval: w β 6
πΈ Then β4 gives: (w β 6) β 4 = w β 10
π β
Blank left bottom oval: w β 10
πΉ Bottom-right path (one blank oval before Γ4)
πΈ After Γ4, result is 3w β 6
πΈ So the oval just before Γ4 must be: (3w β 6)/4
π β
Blank bottom-right oval: (3w β 6)/4
π β Question 6
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.
(a) If t = 4, what is the time taken to travel from Yahapur to Vahapur?
(b) What is the algebraic expression for the time taken to travel from Yahapur to Vahapur?
π β Answer:
πΉ First understand the journey structure
πΈ Yahapur β Station 1 β Station 2 β Station 3 β Vahapur
πΈ Total 4 travel segments, each taking t minutes
πΉ Travel time
πΈ Total travel time = 4 Γ t = 4t minutes
πΉ Stoppage time
πΈ Train stops at 3 stations
πΈ Each stop = 2 minutes
πΈ Total stoppage time = 3 Γ 2 = 6 minutes
πΉ (a) When t = 4
πΈ Travel time = 4 Γ 4 = 16 minutes
πΈ Total time = 16 + 6 = 22 minutes
πΉ (b) Algebraic expression
π Total time = 4t + 6
π β Question 7
Simplify the following expressions:
(a) 3a + 9b β 6 + 8a β 4b β 7a + 16
π β
Answer:
πΉ Combine like terms
πΈ (3a + 8a β 7a) + (9b β 4b) + (β6 + 16)
πΈ = 4a + 5b + 10
(b) 3(3a β 3b) β 8a β 4b β 16
π β
Answer:
πΉ First multiply
πΈ 9a β 9b β 8a β 4b β 16
πΈ = a β 13b β 16
(c) 2(2x β 3) + 8x + 12
π β
Answer:
πΉ Multiply first
πΈ 4x β 6 + 8x + 12
πΈ = 12x + 6
(d) 8x β (2x β 3) + 12
π β
Answer:
πΉ Remove bracket carefully
πΈ 8x β 2x + 3 + 12
πΈ = 6x + 15
(e) 8h β (5 + 7h) + 9
π β
Answer:
πΉ Distribute minus sign
πΈ 8h β 5 β 7h + 9
πΈ = h + 4
(f) 23 + 4(6m β 3n) β 8n β 3m β 18
π β
Answer:
πΉ Multiply first
πΈ 23 + 24m β 12n β 8n β 3m β 18
πΈ = 21m β 20n + 5
π β Question 8
Add the expressions given below:
(a) 4d β 7c + 9 and 8c β 11 + 9d
π β
Answer:
πΉ Add like terms
πΈ (4d + 9d) + (β7c + 8c) + (9 β 11)
πΈ = 13d + c β 2
(b) β6f + 19 β 8s and β23 + 13f + 12s
π β
Answer:
πΉ = 7f + 4s β 4
(c) 8d β 14c + 9 and 16c β (11 + 9d)
π β
Answer:
πΉ Simplify second expression
πΈ 16c β 11 β 9d
πΈ = βd + 2c β 2
(d) 6f β 20 + 8s and 23 β 13f β 12s
π β
Answer:
πΉ = β7f β 4s + 3
(e) 13m β 12n and 12n β 13m
π β
Answer:
πΉ = 0
(f) β26m + 24n and 26m β 24n
π β
Answer:
πΉ = 0
π β Question 9
Subtract the expressions given below:
(a) 9a β 6b + 14 from 6a + 9b β 18
π β
Answer:
πΉ (6a + 9b β 18) β (9a β 6b + 14)
πΈ = β3a + 15b β 32
(b) β15x + 13 β 9y from 7y β 10 + 3x
π β
Answer:
πΉ = 18x + 16y β 23
(c) 17g + 9 β 7h from 11 β 10g + 3h
π β
Answer:
πΉ = β27g + 10h + 2
(d) 9a β 6b + 14 from 6a β (9b + 18)
π β
Answer:
πΉ = β3a β 32
(e) 10x + 2 + 10y from β3y + 8 β 3x
π β
Answer:
πΉ = β13x β 13y + 6
(f) 8g + 4h β 10 from 7h β 8g + 20
π β
Answer:
πΉ = β16g + 3h + 30
π β Question 10
Describe situations corresponding to:
(a) 8x + 3y
π β
Answer:
πΉ Cost of x items at βΉ8 each and y items at βΉ3 each
(b) 15x β 2x
π β
Answer:
πΉ Total of 15 groups reduced by 2 groups of x items
π β Question 11
A rope is cut once β 2 pieces
Fold once and cut β 3 pieces
πΉ Observe the pattern
πΈ Each fold increases pieces by 1
π β
Answer:
πΉ For 10 folds β 11 pieces
πΉ For r folds β r + 1 pieces
π β Q12. 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?
π β Answer:
πΉ Step 1: Observe the pattern
- 1 square needs 4 matchsticks
- 2 squares share one side β 7 matchsticks
- 3 squares β 10 matchsticks
πΉ Each new square adds 3 matchsticks.
πΉ General pattern
- Matchsticks = 4 + 3 Γ (number of extra squares)
πΉ For w squares
- Matchsticks = 4 + 3(w β 1)
- = 3w + 1
πΉ For 10 squares
- Matchsticks = 3 Γ 10 + 1
- = 31 matchsticks
βοΈ Final Answer:
πΉ 10 squares β 31 matchsticks
πΉ w squares β 3w + 1
π β Q13. Traffic signal colour pattern is shown.
Find the colour at positions 90, 190, and 343.
Write expressions for the positions of each colour.
π β Answer:
πΉ Observed colour sequence (repeats every 4):
1 β Red
2 β Yellow
3 β Green
4 β Yellow
πΉ Cycle length = 4
πΉ Position 90
- 90 Γ· 4 = remainder 2
- Position 2 β Yellow
πΉ Position 190
- 190 Γ· 4 = remainder 2
- Position 2 β Yellow
πΉ Position 343
- 343 Γ· 4 = remainder 3
- Position 3 β Green
πΉ General expressions
- Red positions β 4n β 3
- Yellow positions β 4n β 2 and 4n
- Green positions β 4n β 1
βοΈ Final Answer:
πΉ 90 β Yellow
πΉ 190 β Yellow
πΉ 343 β Green
π β Q14. 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?
π β Answer:
πΉ Understanding the pattern
πΈ Step 1 has 5 squares
πΈ Step 2 has 9 squares
πΈ Step 3 has 13 squares
πΉ Observation
πΈ Each step increases by 4 squares
πΉ General formula
πΈ Number of squares in Step n = 4n + 1
πΉ Applying the formula
πΈ Step 4: 4 Γ 4 + 1 = 17 squares
πΈ Step 10: 4 Γ 10 + 1 = 41 squares
πΈ Step 50: 4 Γ 50 + 1 = 201 squares
πΉ Counting vertices
πΈ Each square has 4 vertices
πΈ Total vertices = 4 Γ (number of squares)
πΉ General formula for vertices
πΈ Vertices = 4 Γ (4n + 1)
πΈ = 16n + 4
βοΈ Final:
πΉ Squares in Step n = 4n + 1
πΉ Vertices in Step n = 16n + 4
π β Q15. Numbers are written in a particular sequence in this endless 4-column grid.
π β Answer:
πΉ (a) Expressions for each column
πΈ Column 1 numbers: 1, 5, 9, 13, β¦
πΈ Expression β 4n β 3
πΈ Column 2 numbers: 2, 6, 10, 14, β¦
πΈ Expression β 4n β 2
πΈ Column 3 numbers: 3, 7, 11, 15, β¦
πΈ Expression β 4n β 1
πΈ Column 4 numbers: 4, 8, 12, 16, β¦
πΈ Expression β 4n
πΉ (b) Row and column of given numbers
πΈ 124
πΉ 124 Γ· 4 = 31 remainder 0
πΈ Row = 31, Column = 4
πΈ 147
πΉ 147 Γ· 4 = 36 remainder 3
πΈ Row = 37, Column = 3
πΈ 201
πΉ 201 Γ· 4 = 50 remainder 1
πΈ Row = 51, Column = 1
πΉ (c) Number in row r and column c
πΈ Number = 4(r β 1) + c
πΉ (d) Pattern of multiples of 3
πΈ Multiples of 3 occur at regular intervals
πΈ Their positions repeat because numbers increase by 4 in each new row
πΈ Column positions follow a repeating cycle
βοΈ Final:
πΉ General number at row r, column c = 4(r β 1) + c
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OTHER IMPORTANT QUESTIONS
(MODEL QUESTION PAPER)
ESPECIALLY MADE FOR THIS LESSON ONLY
π΅ Section A β Very Short Answer (1 Γ 6 = 6 marks)
π β Question 1
What is meant by a letter-number?
π β
Answer:
πΉ A letter-number is a letter used to represent an unknown or variable number
π β Question 2
Write an expression for βfive more than xβ.
π β
Answer:
πΉ The expression is x + 5
π β Question 3
Is 3a an expression or an equation?
π β
Answer:
πΉ 3a is an expression
πΉ It has no equal sign
π β Question 4
How many terms are there in the expression 4x + 7?
π β
Answer:
πΉ There are two terms: 4x and 7
π β Question 5
True or False: 5x and 3x are like terms.
π β
Answer:
πΉ True
π β Question 6
What is the value of 2x when x = 4?
π β
Answer:
πΉ 2 Γ 4 = 8
π’ Section B β Short Answer I (2 Γ 6 = 12 marks)
π β Question 7
Write two expressions using the letter y.
π β
Answer:
πΉ y + 6
πΉ 3y β 2
π β Question 8
Write the expression for βthe cost of 4 pencils if the cost of one pencil is p rupeesβ.
π β
Answer:
πΉ Cost of 4 pencils = 4p
π β Question 9
State whether the following are like or unlike terms: 7a and 7b.
π β
Answer:
πΉ 7a and 7b are unlike terms
πΉ They have different letters
π β Question 10
Evaluate 3x + 5 when x = 2.
π β
Answer:
πΉ Substitute x = 2
πΉ 3 Γ 2 + 5 = 6 + 5
πΉ Value = 11
π β Question 11
Why is multiplication written as 5a and not 5 Γ a?
π β
Answer:
πΉ In algebra, multiplication sign is omitted
πΉ 5a makes expressions simpler
π β Question 12
Write the terms of the expression 6y β 4.
π β
Answer:
πΉ The terms are 6y and β4
π‘ Section C β Short Answer II (3 Γ 10 = 30 marks)
π β Question 13
Form an expression for βtwice a number increased by 7β.
π β
Answer:
πΉ Let the number be x
πΉ Twice the number = 2x
πΉ Expression = 2x + 7
π β Question 14
Identify the terms in the expression 5a + 3b β 9.
π β
Answer:
πΉ The terms are 5a, 3b, and β9
π β Question 15
Evaluate 4x β 3 when x = 5.
π β
Answer:
πΉ Substitute x = 5
πΉ 4 Γ 5 β 3 = 20 β 3
πΉ Value = 17
π β Question 16
Explain what like terms are with an example.
π β
Answer:
πΉ Like terms have the same letter with the same power
πΉ Example: 3x and 7x
π β Question 17
Write an expression for the perimeter of a rectangle with length l and breadth b.
π β
Answer:
πΉ Perimeter of rectangle = 2(l + b)
π β Question 18
State whether 4x and 4xΒ² are like or unlike terms. Give reason.
π β
Answer:
πΉ They are unlike terms
πΉ The powers of x are different
π β Question 19
Evaluate 2a + 3 when a = 6.
π β
Answer:
πΉ Substitute a = 6
πΉ 2 Γ 6 + 3 = 12 + 3
πΉ Value = 15
π β Question 20
Write two expressions using different letters.
π β
Answer:
πΉ 5p + 2
πΉ 3q β 4
π β Question 21
Explain why only like terms can be added.
π β
Answer:
πΉ Like terms represent the same quantity
πΉ Unlike terms represent different quantities and cannot be combined
π β Question 22
Form an expression for βthe sum of a number x and its doubleβ.
π β
Answer:
πΉ Double of x = 2x
πΉ Expression = x + 2x
π΄ Section D β Long Answer (4 Γ 8 = 32 marks)
π β Question 23
Explain the use of letters in mathematics with suitable examples.
π β
Answer:
πΉ Letters represent unknown or variable quantities
πΉ They help write general rules
πΉ Example: Cost of n books at p rupees each = np
πΉ Letters make mathematics flexible and powerful
π β Question 24
Evaluate 3x + 2y when x = 2 and y = 5.
π β
Answer:
πΉ Substitute x = 2 and y = 5
πΉ 3 Γ 2 + 2 Γ 5 = 6 + 10
πΉ Value = 16
π β Question 25
Explain like and unlike terms with examples.
π β
Answer:
πΉ Like terms have the same letter and power
πΉ Example: 4a and 9a
πΉ Unlike terms have different letters or powers
πΉ Example: 3x and 3y
π β Question 26
Form an expression for the following and find its value when x = 4.
βThree times a number decreased by 5β.
π β
Answer:
πΉ Let the number be x
πΉ Expression = 3x β 5
πΉ Substitute x = 4
πΉ 3 Γ 4 β 5 = 12 β 5
πΉ Value = 7
π β Question 27
Write four common mistakes students make while working with expressions using letter-numbers.
π β
Answer:
πΉ Using multiplication sign between number and letter
πΉ Adding unlike terms
πΉ Forgetting to substitute correct values
πΉ Mixing different letters
π β Question 28
Write an expression for the area of a square with side s and explain it.
π β
Answer:
πΉ Area of square = s Γ s
πΉ Expression = sΒ²
πΉ The area depends on the value of s
π β Question 29
Explain how expressions using letter-numbers are useful in daily life.
π β
Answer:
πΉ Used to calculate cost when price is unknown
πΉ Used to find perimeter and area
πΉ Used to write formulas
πΉ Used in patterns and rules
π β Question 30
Explain why expressions using letter-numbers are important for learning algebra.
π β
Answer:
πΉ They introduce variables
πΉ They help understand algebraic rules
πΉ They prepare students for higher mathematics
πΉ They simplify complex calculations
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