School Assignments for Class 7 Mathematics: Chapter 04 Expressions using Letter Numbers
Access comprehensive school assignments for Chapter 04 Expressions using Letter Numbers using the CBSE Class 7 Mathematics Expressions using Letter Numbers Practice Paper Assignment Set 02 with Answers. Designed to align with the 2026-27 CBSE academic guidelines, these practice sets help Class 7 Mathematics students reinforce core concepts and improve their problem-solving accuracy.
Practice Class 7 Mathematics Assignments: Chapter 04 Expressions using Letter Numbers
View or download the dedicated CBSE Class 7 Mathematics Expressions using Letter Numbers Practice Paper Assignment Set 02 with Answers resource below. Engaging with these assignments under focused study conditions ensures continuous academic progress and mastery of the 2026-27 curriculum.
Class 7 Maths Ganita Prakash Chapter 4 Expressions using Letter-Numbers Practice Paper 10 is a 40-mark test with 20 questions in Sections A to E, and every question here comes with a step-by-step answer. It checks writing expressions for perimeters, ropes and shopping bills, simplifying by grouping like terms and opening brackets, substituting values (including negative numbers), a number machine, and the diagonal pattern in a 2×2 block of calendar dates.
Section A: Multiple Choice Questions (1 Mark Each)
(a) 5p + q
(b) 5p + 4q
(c) p + q + 9
(d) 5p + 4
Show Answer & Explanation
Answer: (b) 5p + 4q
Explanation:
1. A regular pentagon has 5 equal sides, so its perimeter is 5 × p = 5p.
2. A square has 4 equal sides, so its perimeter is 4 × q = 4q.
3. Total perimeter = 5p + 4q.
4. 5p and 4q are unlike terms, so they cannot be combined further. The answer is 5p + 4q.
Teacher's Note:
1. Perimeter of a regular polygon = number of sides × side length.
2. Do not add the numbers 5 and 4 when the letters are different.
(a) (50 − x) × 2
(b) \( \frac{50}{2} \) − x
(c) \( \frac{50 - x}{2} \)
(d) 50 − \( \frac{x}{2} \)
Show Answer & Explanation
Answer: (c) \( \frac{50 - x}{2} \)
Explanation:
1. After cutting x m from 50 m, the rope left is (50 − x) m.
2. This remaining piece is divided into 2 equal parts, so we divide the whole of (50 − x) by 2.
3. Each part = \( \frac{50 - x}{2} \) m.
4. The bracket matters: we halve the remaining length, not just the 50 or just the x. The answer is (c).
Teacher's Note:
1. First write the remaining length, then divide it.
2. Check with a number: if x = 10, each part should be 40 ÷ 2 = 20 m.
(a) 3n + 5p − 10
(b) 3n × 5p − 10
(c) 3(n − 10) + 5p
(d) 10 − (3n + 5p)
Show Answer & Explanation
Answer: (a) 3n + 5p − 10
Explanation:
1. Cost of 3 notebooks = 3 × n = 3n rupees.
2. Cost of 5 pens = 5 × p = 5p rupees.
3. Total before discount = 3n + 5p.
4. A discount of ₹10 is taken off the whole bill, so the amount to pay is 3n + 5p − 10.
Teacher's Note:
1. A discount is subtracted from the total bill.
2. Costs of different items are added, not multiplied.
(a) 9a − 8b
(b) 9a + 2b
(c) 5a + 8b
(d) 5a − 2b
Show Answer & Explanation
Answer: (b) 9a + 2b
Explanation:
1. Group the like terms: (7a + 2a) + (−3b + 5b).
2. 7a + 2a = 9a.
3. −3b + 5b = 2b.
4. So the expression simplifies to 9a + 2b.
Teacher's Note:
1. Each term carries the sign written just before it.
2. Only terms with the same letter-number can be combined.
(a) 23
(b) 7
(c) 11
(d) −23
Show Answer & Explanation
Answer: (a) 23
Explanation:
1. Replace x by −4: 15 − 2 × (−4).
2. 2 × (−4) = −8.
3. 15 − (−8) = 15 + 8 = 23.
4. So the value is 23.
Teacher's Note:
1. Put a bracket around a negative value when substituting.
2. Subtracting a negative number is the same as adding it.
(a) l × b
(b) 2(l × b)
(c) l + b + l + b
(d) 2l × 2b
Show Answer & Explanation
Answer: (c) l + b + l + b
Explanation:
1. The perimeter is the total length of all four sides of the rectangle.
2. The sides are l, b, l and b, so the perimeter = l + b + l + b.
3. Grouping like terms, l + l + b + b = 2l + 2b, the same expression.
4. The other options multiply l and b, which gives area-type expressions, not the perimeter. The answer is l + b + l + b.
Teacher's Note:
1. Perimeter is found by adding sides, never by multiplying them.
2. Test with numbers: l = 5, b = 3 gives 16 for both 2l + 2b and l + b + l + b.
(a) 3 − 10y
(b) 10(y − 3)
(c) 10y − 3
(d) 3y − 10
Show Answer & Explanation
Answer: (c) 10y − 3
Explanation:
1. "10 times a number y" is 10 × y = 10y.
2. "3 less than" this quantity means we subtract 3 from 10y.
3. So the expression is 10y − 3.
Teacher's Note:
1. "3 less than A" means A − 3, not 3 − A.
2. Build the expression in the order the words describe the operations.
(a) 7x and 3x
(b) 2ab and 5ba
(c) 10p and 11q
(d) 4y² and 6y²
Show Answer & Explanation
Answer: (c) 10p and 11q
Explanation:
1. Like terms have exactly the same letter-number part; only the number in front may differ.
2. 7x and 3x both have x, so they are like terms.
3. 2ab and 5ba both have a × b (order of multiplication does not matter), so they are like terms.
4. 4y² and 6y² both have y², so they are like terms.
5. 10p and 11q have different letter-numbers, so they are unlike terms.
Teacher's Note:
1. ab and ba are the same product, so they are like terms.
2. Compare only the letter-number parts, not the numbers in front.
Assertion (A): The simplified form of the expression 5(2w + 3x + 4w) is 30w + 15x.
Reason (R): The distributive property is used to multiply the number outside the bracket with each term inside the bracket.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Inside the bracket, combine like terms: 2w + 4w = 6w, so 5(2w + 3x + 4w) = 5(6w + 3x).
2. Multiply 5 with each term: 5 × 6w + 5 × 3x = 30w + 15x. So A is true.
3. R correctly states the distributive property, so R is true.
4. The distributive property is exactly the step that turns 5(6w + 3x) into 30w + 15x, so R explains A.
5. The answer is (a).
Teacher's Note:
1. Simplify inside the bracket first, then distribute.
2. Multiply the outside number with every term in the bracket.
Assertion (A): The expressions 10y − 3 and 10(y − 3) are equal.
Reason (R): Algebraic expressions with different operations on the same variables can have the same values for some specific values of the variable.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. 10(y − 3) = 10y − 30, which is different from 10y − 3.
2. Check with y = 5: 10y − 3 = 47, but 10(y − 3) = 20. So A is false.
3. R is a true general fact: for example, 2y and y + 2 are different expressions but both equal 4 when y = 2.
4. So A is false but R is true. The answer is (d).
Teacher's Note:
1. Two expressions are equal only if they agree for every value of the letter-number.
2. One counter-example is enough to show two expressions are not equal.
Section B: Short Answer Questions (2 Marks Each)
Question 11. Find the value of the expression 2p − 3q + 5 when p = 4 and q = −1.
Answer:
Substitute p = 4 and q = −1: 2 × 4 − 3 × (−1) + 5.
2 × 4 = 8 and 3 × (−1) = −3.
So the value = 8 − (−3) + 5 = 8 + 3 + 5 = 16.
In simple words: Put p = 4 and q = −1 in the expression to get 8 + 3 + 5 = 16.
Teacher's Note:
1. Use brackets for the negative value of q.
2. Subtracting −3 means adding 3.
Question 12. The perimeter of a regular hexagon is given by the expression 6s. If the side of the hexagon is increased by 2 cm, write a new expression for its perimeter.
Answer:
A regular hexagon has 6 equal sides, each of length s, so its perimeter is 6s.
New side length = (s + 2) cm.
New perimeter = 6 × (s + 2) = 6(s + 2) cm.
Using the distributive property, 6(s + 2) = 6s + 12 cm.
In simple words: Each of the 6 sides grows by 2 cm, so the perimeter becomes 6s + 12 cm.
Teacher's Note:
1. Increase the side first, then multiply by 6.
2. The perimeter grows by 6 × 2 = 12 cm, not just 2 cm.
Question 13. Simplify the expression 12m − (3m + 4n) + 2n.
Answer:
Remove the bracket. The minus sign before it changes the sign of every term inside: 12m − 3m − 4n + 2n.
Group like terms: (12m − 3m) + (−4n + 2n).
12m − 3m = 9m and −4n + 2n = −2n.
So the simplified expression is 9m − 2n.
In simple words: Open the bracket carefully and combine: 9m − 2n.
Teacher's Note:
1. A minus sign before a bracket flips the sign of each term inside.
2. Check with m = 1, n = 1: 12 − 7 + 2 = 7 and 9 − 2 = 7.
Question 14. A shopkeeper sells x kg of sugar at ₹50 per kg and y kg of rice at ₹40 per kg. He spent ₹20 on transport. Write an algebraic expression for his total cost.
Answer:
Cost of x kg of sugar = 50 × x = ₹50x.
Cost of y kg of rice = 40 × y = ₹40y.
Transport cost = ₹20, which is added to the bill.
Total cost = ₹(50x + 40y + 20).
In simple words: Add the cost of sugar, the cost of rice and the transport: 50x + 40y + 20.
Teacher's Note:
1. Price per kg × number of kg gives the cost of each item.
2. The fixed transport charge is a number term without a letter.
Section C: Short Answer Questions (3 Marks Each)
Question 15. Simplify the expression (40x + 75y) − (6x + 10y) by grouping like terms and then find the value of the simplified expression if x = 2 and y = 3.
Answer:
Remove the brackets (the minus sign changes the signs inside the second bracket): 40x + 75y − 6x − 10y.
Group like terms: (40x − 6x) + (75y − 10y) = 34x + 65y.
Simplified expression: 34x + 65y.
Now put x = 2 and y = 3: 34 × 2 + 65 × 3 = 68 + 195 = 263.
Check with the original: (80 + 225) − (12 + 30) = 305 − 42 = 263. ✓
In simple words: The expression simplifies to 34x + 65y, which is 263 when x = 2 and y = 3.
Teacher's Note:
1. Both terms in the second bracket are subtracted.
2. Checking with the original expression catches sign mistakes.
Question 16. A number machine takes in two numbers, a and b, as inputs. The output is given by the formula 3a + 5b. If the output is 31 and the value of b is 4, find the value of a.
Answer:
Output = 3a + 5b, with b = 4: 3a + 5 × 4 = 3a + 20.
The output is 31, so 3a + 20 = 31.
3a = 31 − 20 = 11.
a = 11 ÷ 3 = \( \frac{11}{3} \), that is, 3\( \frac{2}{3} \).
Check: 3 × \( \frac{11}{3} \) + 20 = 11 + 20 = 31. ✓
In simple words: 5 × 4 = 20 uses up 20 of the 31, so 3a = 11 and a = 11/3.
Teacher's Note:
1. Undo the operations in reverse order: first take away 20, then divide by 3.
2. The input need not be a whole number.
Question 17. A school has p classrooms, each with 20 students. On a particular day, q students are absent from each classroom. Write an expression for the total number of students present on that day.
Answer:
Students in each classroom = 20; absent from each classroom = q.
Students present in each classroom = 20 − q.
There are p classrooms, so total present = p × (20 − q) = p(20 − q).
By the distributive property, this is also 20p − pq.
(Another way: total students 20p, total absent pq, so present = 20p − pq.)
In simple words: Each class has 20 − q present, and there are p classes, so p(20 − q) students are present.
Teacher's Note:
1. Find the number for one classroom first, then multiply.
2. Both p(20 − q) and 20p − pq are correct answers.
Section D: Long Answer Question (5 Marks)
Question 18. Simplify the following expression and find its value if x = −2 and y = 5.
5(2x − 3y) + 3(4y − 6x) − (x + y)
Answer:
Step 1: Remove the brackets. 5(2x − 3y) = 10x − 15y; 3(4y − 6x) = 12y − 18x; −(x + y) = −x − y.
So the expression = 10x − 15y + 12y − 18x − x − y.
Step 2: Group like terms. x-terms: 10x − 18x − x = (10 − 18 − 1)x = −9x.
y-terms: −15y + 12y − y = (−15 + 12 − 1)y = −4y.
Simplified expression: −9x − 4y.
Step 3: Substitute x = −2, y = 5. −9 × (−2) − 4 × 5 = 18 − 20 = −2.
Check with the original: 5(−4 − 15) + 3(20 + 12) − (−2 + 5) = 5(−19) + 3(32) − 3 = −95 + 96 − 3 = −2. ✓
In simple words: The expression simplifies to −9x − 4y, whose value at x = −2, y = 5 is −2.
Teacher's Note:
1. Multiply the outside number with every term in its bracket.
2. −(x + y) means subtract both x and y.
3. Substituting into both forms is a good check.
Section E: Case Study Based Questions (4 Marks Each)
Question 19. Case Study 1: A travel agent offers different packages for a trip. The cost of one person's travel is ₹c, and the cost of one night's stay is ₹n. The agent charges an additional service fee of ₹100 for each booking.
(a) Write an expression for the total cost for a group of 5 people staying for 3 nights. (2 marks)
(b) If the travel cost per person is ₹2000 and the stay cost per night is ₹1500, what is the total cost for a group of 4 people staying for d nights? (2 marks)
Answer:
Here the stay cost ₹n is for one person for one night, and the group makes one booking, so ₹100 is added once.
(a) Travel for 5 people = 5 × c = 5c. Stay for 5 people for 3 nights = 5 × 3 × n = 15n. Service fee = 100.
Total cost = ₹(5c + 15n + 100).
(b) For 4 people and d nights, total cost = 4c + 4 × d × n + 100 = 4c + 4dn + 100.
Put c = 2000 and n = 1500: 4 × 2000 + 4 × d × 1500 + 100 = 8000 + 6000d + 100.
Total cost = ₹(6000d + 8100).
In simple words: Total cost = travel for everyone + stay for everyone for every night + ₹100 fee.
Teacher's Note:
1. The stay cost is multiplied by both the number of people and the number of nights.
2. The service fee is added only once per booking.
| a | a + 1 |
| a + 7 | a + 8 |
(b) If the sum of the two diagonal pairs together (all four dates) in such a 2×2 grid is 56, find the dates in the grid. (2 marks)
Answer:
The date below any date is 7 more (one week later), and the date to its right is 1 more.
(a) Diagonal 1 (top-left and bottom-right): a + (a + 8) = 2a + 8.
Diagonal 2 (top-right and bottom-left): (a + 1) + (a + 7) = 2a + 8.
Both sums are 2a + 8, so the diagonal sums are always equal, whatever the value of a.
(b) Sum of both diagonal pairs = (2a + 8) + (2a + 8) = 4a + 16.
4a + 16 = 56, so 4a = 40 and a = 10.
The dates are 10, 11, 17 and 18.
| 10 | 11 |
| 17 | 18 |
Check: 10 + 18 = 28, 11 + 17 = 28, and 28 + 28 = 56. ✓
In simple words: Each diagonal adds to 2a + 8; all four dates add to 4a + 16 = 56, so a = 10.
Teacher's Note:
1. Moving down one row in a calendar adds 7.
2. Always check that the dates found can exist in a month (none above 31).
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Chapter 04 Expressions using Letter Numbers Printable Assignments & Solutions for Class 7 Mathematics
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