Class 7 Mathematics Practice Sheet: CBSE Class 7 Mathematics Expressions using Letter Numbers Practice Paper Worksheet Set 02 with Answers
Explore structured practice materials through the CBSE Class 7 Mathematics Expressions using Letter Numbers Practice Paper Worksheet Set 02 with Answers. Tailored for Class 7 learners, utilizing these Mathematics worksheets ensures thorough preparation and strengthens problem-solving accuracy before final school evaluations.
Download Chapter 04 Expressions using Letter Numbers Worksheet PDF with Answers
Access the complete worksheet PDF for Class 7 Mathematics below. Regular practice with these targeted academic tasks builds familiarity with standard question patterns and helps secure higher marks in final school examinations.
Practice Paper 11 for Class 7 Maths Ganita Prakash Chapter 4 Expressions using Letter-Numbers has 20 questions worth 40 marks, split into Sections A to E, with clear step-by-step solutions for each one. Its questions cover matchstick patterns, taking out a common factor, like and unlike terms, expressions with negatives such as 10 − (−x), number machines, distance with speed and time, a flour mill's grinding time and adding up quiz scores in p and q.
Section A: Multiple Choice Questions (1 Mark Each)
(a) 14
(b) 20
(c) 17
(d) 18
Show Answer & Explanation
Answer: (b) 20
Explanation:
1. "2 less than 5" means y = 5 − 2 = 3.
2. Substitute y = 3: 3 × 3 + 11 = 9 + 11 = 20.
3. So the value is 20.
Teacher's Note:
1. First find the value of y from the words.
2. Then multiply before adding.
(a) length 12, width 4
(b) length 8, width n
(c) length 12 − 4, width n
(d) (b) and (c)
Show Answer & Explanation
Answer: (d) (b) and (c)
Explanation:
1. 12n − 4n = (12 − 4)n = 8n.
2. Area = length × width. Length 8 and width n give 8 × n = 8n, so (b) works.
3. Length 12 − 4 is also 8, and with width n the area is (12 − 4) × n = 8n, so (c) works too.
4. Length 12 and width 4 give 48, which has no n, so (a) is wrong.
5. The answer is (d) (b) and (c).
Teacher's Note:
1. 12n − 4n = (12 − 4) × n by the distributive property.
2. Read every option before choosing; here two options are correct.
(a) 3(j + 6k + 9h + 12)
(b) 3(j + 2k + 3h + 4)
(c) 3(j + 2k + 3h + 12)
(d) 3j + 6k + 9h + 4
Show Answer & Explanation
Answer: (b) 3(j + 2k + 3h + 4)
Explanation:
1. 3, 6, 9 and 12 are all multiples of 3.
2. 3j = 3 × j, 6k = 3 × 2k, 9h = 3 × 3h and 12 = 3 × 4.
3. Taking out 3: 3j + 6k + 9h + 12 = 3(j + 2k + 3h + 4).
4. Check by multiplying back: 3 × j + 3 × 2k + 3 × 3h + 3 × 4 gives the original expression.
Teacher's Note:
1. Divide every term by the common factor.
2. Multiply back to check the answer.
(a) 8c + 10d
(b) 10c + 8d
(c) 5c + 4d + 3c + 6d
(d) (a) and (c)
Show Answer & Explanation
Answer: (d) (a) and (c)
Explanation:
1. Monday: 5c + 4d. Tuesday: 3c + 6d.
2. Total = 5c + 4d + 3c + 6d, which is option (c).
3. Grouping like terms: (5c + 3c) + (4d + 6d) = 8c + 10d, which is option (a).
4. Both expressions are equal, so the answer is (d) (a) and (c).
Teacher's Note:
1. An expression and its simplified form are equivalent.
2. In option (b) the coefficients of c and d are swapped.
(a) 3 + s
(b) s³
(c) 3s
(d) s + s
Show Answer & Explanation
Answer: (c) 3s
Explanation:
1. An equilateral triangle has three equal sides, each of length s.
2. Perimeter = s + s + s = 3 × s = 3s.
3. 3 + s adds instead of multiplying, s³ = s × s × s multiplies the sides, and s + s counts only two sides.
4. So the answer is 3s.
Teacher's Note:
1. Perimeter adds the lengths of all the sides.
2. 3s means 3 × s, not 3 + s.
(a) 3 + n
(b) 3n
(c) n − 3
(d) n ÷ 3
Show Answer & Explanation
Answer: (b) 3n
Explanation:
1. Each 'C' shape uses 3 matchsticks.
2. For n separate 'C' shapes we need 3 + 3 + 3 + … (n times) = 3 × n matchsticks.
3. So the expression is 3n.
4. Check: 2 'C' shapes need 6 sticks, and 3 × 2 = 6.
Teacher's Note:
1. Repeated addition of the same number is multiplication.
2. Test the expression with a small value of n.
(a) 10 + x
(b) 10 − (−1)x
(c) 10 + (−x)
(d) 10 − (0 − x)
Show Answer & Explanation
Answer: (c) 10 + (−x)
Explanation:
1. 10 − (−x) = 10 + x, because subtracting a negative is adding.
2. (b) (−1)x = −x, so 10 − (−1)x = 10 − (−x) = 10 + x. Equivalent.
3. (d) 0 − x = −x, so 10 − (0 − x) = 10 + x. Equivalent.
4. (c) 10 + (−x) = 10 − x, which is different from 10 + x.
5. So the expression that is NOT equivalent is 10 + (−x).
Teacher's Note:
1. Subtracting −x is the same as adding x.
2. Adding −x is the same as subtracting x.
(a) 5 × 4 − 2
(b) 10 + 20
(c) 2a + 3b
(d) 40
Show Answer & Explanation
Answer: (c) 2a + 3b
Explanation:
1. An algebraic expression contains at least one letter-number (variable).
2. 5 × 4 − 2, 10 + 20 and 40 contain only numbers; they are numerical expressions.
3. 2a + 3b contains the letter-numbers a and b.
4. So the answer is 2a + 3b.
Teacher's Note:
1. Look for letters standing for numbers.
2. Numerical expressions have a fixed value; algebraic ones change with the letters.
Assertion (A): The total number of matchsticks for n 'L' shapes is 2n.
Reason (R): Each 'L' shape requires 2 matchsticks, and we multiply the number of shapes by the number of matchsticks per shape.
(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. An 'L' is made of 2 matchsticks: one standing and one lying.
2. For n separate 'L' shapes, the number of sticks is 2 × n = 2n. So A is true.
3. R gives the number of sticks per shape and the rule of multiplying, so R is true.
4. R is exactly how the expression 2n is formed, so R explains A. The answer is (a).
Teacher's Note:
1. Sticks per shape × number of shapes gives the total.
2. Check: 3 'L' shapes need 6 sticks.
Assertion (A): The expression 2d − d − d − c simplifies to −c.
Reason (R): Like terms can be added or subtracted to get a single term.
(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. Combine the like terms in d: 2d − d − d = (2 − 1 − 1)d = 0d = 0.
2. So 2d − d − d − c = 0 − c = −c. A is true.
3. R is a true rule: like terms can be combined into one term.
4. Combining the like terms in d is exactly what gives −c, so R explains A. The answer is (a).
Teacher's Note:
1. 0 × d = 0, so the d-terms cancel out.
2. The term −c has no like term to combine with.
Section B: Short Answer Questions (2 Marks Each)
Question 11. Simplify the expression 2d − d − (d − c) and find its value if c = −3.
Answer:
Open the bracket (the minus sign changes both signs inside): 2d − d − d + c.
Group the d-terms: (2 − 1 − 1)d = 0d = 0.
So the simplified expression is c.
When c = −3, the value is −3 (whatever the value of d).
In simple words: All the d's cancel, leaving just c, which is −3.
Teacher's Note:
1. −(d − c) = −d + c.
2. The value does not depend on d at all.
Question 12. The length of a rectangle is 5 cm more than its breadth. If the breadth is b cm, write an expression for its perimeter.
Answer:
Breadth = b cm, so length = (b + 5) cm.
Perimeter = 2 × (length + breadth) = 2 × ((b + 5) + b) = 2 × (2b + 5).
Perimeter = (4b + 10) cm.
Check with b = 3: length 8, perimeter 2 × (8 + 3) = 22, and 4 × 3 + 10 = 22. ✓
In simple words: Length is b + 5, so the perimeter is 2(b + 5 + b) = 4b + 10.
Teacher's Note:
1. "5 more than b" is b + 5.
2. Perimeter of a rectangle = 2 × (length + breadth).
Question 13. A number machine takes in one number, m, and outputs the value of the expression 5m + 3. If the output is 28, what is the input number, m?
Answer:
The machine works out 5m + 3, so 5m + 3 = 28.
Undo the "+ 3": 5m = 28 − 3 = 25.
Undo the "× 5": m = 25 ÷ 5 = 5.
Check: 5 × 5 + 3 = 28. ✓
In simple words: Take away 3 and divide by 5: (28 − 3) ÷ 5 = 5.
Teacher's Note:
1. Undo the steps in reverse order.
2. Always put the answer back into the machine to check.
Question 14. Write an expression for the area of a square whose side is given by 3x + 2.
Answer:
Area of a square = side × side.
Side = 3x + 2, so area = (3x + 2) × (3x + 2).
Area = (3x + 2)² square units.
The brackets are needed: the whole side 3x + 2 is multiplied by itself.
In simple words: Multiply the side 3x + 2 by itself: (3x + 2)².
Teacher's Note:
1. Keep the side inside brackets when squaring.
2. 3x + 2² would be a different expression.
Section C: Short Answer Questions (3 Marks Each)
Question 15. Simplify the expression 7p − p + 5q − 2q and check if its value is the same as the value of 6p + 3q when p = 4 and q = 1.
Answer:
Group like terms: (7p − p) + (5q − 2q) = (7 − 1)p + (5 − 2)q = 6p + 3q.
Value of the original expression at p = 4, q = 1: 7 × 4 − 4 + 5 × 1 − 2 × 1 = 28 − 4 + 5 − 2 = 27.
Value of 6p + 3q at p = 4, q = 1: 6 × 4 + 3 × 1 = 24 + 3 = 27.
Both values are 27, so they are the same, as expected, because the two expressions are equivalent.
In simple words: 7p − p + 5q − 2q simplifies to 6p + 3q, and both give 27 when p = 4, q = 1.
Teacher's Note:
1. p on its own means 1 × p.
2. Equivalent expressions give the same value for any values of the letters.
Question 16. A train travels at a speed of s km/hr for 2 hours and then at a speed of t km/hr for 3 hours. Write an expression for the total distance covered. If s = 80 km/hr and t = 60 km/hr, find the total distance.
Answer:
Distance = speed × time.
First part: s × 2 = 2s km. Second part: t × 3 = 3t km.
Total distance = (2s + 3t) km.
With s = 80 and t = 60: 2 × 80 + 3 × 60 = 160 + 180 = 340 km.
In simple words: Add the distances of the two parts: 2s + 3t, which is 340 km.
Teacher's Note:
1. Find the distance for each part of the journey separately.
2. Write the unit (km) with the final answer.
Question 17. Describe a situation that can be represented by the algebraic expression 15 × j − 2 × k.
Answer:
15 × j − 2 × k means "15 times j, minus 2 times k".
Situation: Ravi sells j glasses of lemonade at ₹15 each, so he collects 15j rupees. He used k paper cups that cost ₹2 each, so he spent 2k rupees.
His profit is ₹(15j − 2k).
For example, if he sells 20 glasses using 20 cups, profit = 15 × 20 − 2 × 20 = 300 − 40 = ₹260.
Another situation: in a game you get 15 points for each of j correct answers and lose 2 points for each of k wrong answers; your score is 15j − 2k.
In simple words: 15j − 2k can be money earned at ₹15 each minus costs of ₹2 each.
Teacher's Note:
1. Each term needs a meaning: 15 per j and 2 per k.
2. The minus sign should mean something taken away, such as a cost or a penalty.
Section D: Long Answer Question (5 Marks)
Question 18. Simplify the expression 4(2r + 3s + 5) − 20 − 8r − 12s and show that the simplified expression takes the same value for any value of r and s.
Answer:
Step 1: Use the distributive property: 4(2r + 3s + 5) = 8r + 12s + 20.
So the expression = 8r + 12s + 20 − 20 − 8r − 12s.
Step 2: Group like terms: (8r − 8r) + (12s − 12s) + (20 − 20).
= 0r + 0s + 0 = 0.
Step 3: The simplified expression is 0 and contains no r or s, so its value is always 0, whatever values r and s take.
Check with r = 1, s = 2: 4(2 + 6 + 5) − 20 − 8 − 24 = 52 − 52 = 0. Check with r = 7, s = −3: 4(14 − 9 + 5) − 20 − 56 + 36 = 40 − 40 = 0. ✓
In simple words: Everything cancels out, so the expression is always 0.
Teacher's Note:
1. Multiply 4 with all three terms in the bracket, including 5.
2. Terms such as 8r and −8r cancel each other.
Section E: Case Study Based Questions (4 Marks Each)
Question 19. Case Study 1: Venkatlakshmi, a dedicated flour mill owner, ensures her mill is running efficiently for her customers. When a customer brings in a new batch of grain, she powers on the machine, which takes 10 seconds to get to full speed. Once it's operational, she can grind the grain at a consistent rate of 8 seconds per kilogram. This two-step process allows her to accurately estimate the time each order will take.
(a) Write an expression for the total time taken to grind y kg of grain. (2 marks)
(b) How much time will it take to grind 50 kg of grain? (2 marks)
Answer:
(a) Starting time = 10 seconds (once for each batch).
Grinding y kg at 8 seconds per kg takes 8 × y = 8y seconds.
Total time = (10 + 8y) seconds.
(b) Put y = 50: 10 + 8 × 50 = 10 + 400 = 410 seconds.
That is 6 minutes 50 seconds (since 410 = 6 × 60 + 50).
In simple words: Time = 10 seconds to start + 8 seconds for each kg, so 50 kg take 410 seconds.
Teacher's Note:
1. The fixed starting time is added only once.
2. Multiply the rate by the quantity for the grinding time.
Question 20. Case Study 2: Charu, an enthusiastic participant in a quiz competition, aims to maximize her total score across three challenging rounds. The scoring is determined by two variables: p, the points awarded for each correct answer, and q, the points deducted for each incorrect answer. Her scores for the three rounds are expressed in terms of these variables as 7p − 3q, 8p − 4q, and 6p − 2q. Charu's goal is to track her performance and calculate her cumulative score after all three rounds.
(a) What is Charu's total score after three rounds? Simplify the expression. (2 marks)
(b) Her friend Krishita's total score is 23p − 7q. How much more has Krishita scored than Charu? Simplify the expression for the difference. (2 marks)
Answer:
(a) Total = (7p − 3q) + (8p − 4q) + (6p − 2q).
p-terms: 7p + 8p + 6p = 21p. q-terms: −3q − 4q − 2q = −9q.
Charu's total score = 21p − 9q.
(b) Difference = (23p − 7q) − (21p − 9q) = 23p − 7q − 21p + 9q.
= (23 − 21)p + (−7 + 9)q = 2p + 2q.
So Krishita scored 2p + 2q points more than Charu.
In simple words: Charu has 21p − 9q; Krishita's extra is 23p − 7q − (21p − 9q) = 2p + 2q.
Teacher's Note:
1. Subtracting −9q means adding 9q.
2. Keep the p-terms and q-terms separate while adding.
Free study material for Mathematics
Free CBSE Practice Worksheets: Class 7 Mathematics Chapter 04 Expressions using Letter Numbers
Practice Exercises for Class 7 Mathematics Chapter 04 Expressions using Letter Numbers
Review targeted practice exercises for Class 7 Mathematics Chapter 04 Expressions using Letter Numbers. Curated to match official CBSE guidelines, these printable problem sets support daily revision and improve overall test readiness.
Step-by-Step Solutions and Practice Guidelines
Each worksheet draws directly from authorized standard textbooks to maintain academic accuracy. Evaluating your finished exercises against expert-verified solutions helps master the formal presentation standards expected in school evaluations.
Enhance Speed with Online Practice
Consistent engagement with these exercises builds familiarity with recurring exam themes. If specific areas within Chapter 04 Expressions using Letter Numbers cause trouble, utilize our dedicated NCERT solutions for Class 7 Mathematics to clear up doubts immediately.
FAQs
You can download the latest chapter-wise printable worksheets for Class 7 Mathematics Chapter 04 Expressions using Letter Numbers for free from StudiesToday.com. These have been made as per the latest CBSE curriculum for this academic year.
Yes, Class 7 Mathematics worksheets for Chapter 04 Expressions using Letter Numbers focus on activity-based learning and also competency-style questions. This helps students to apply theoretical knowledge to practical scenarios.
Yes, we have provided solved worksheets for Class 7 Mathematics Chapter 04 Expressions using Letter Numbers to help students verify their answers instantly.
Yes, our Class 7 Mathematics test sheets are mobile-friendly PDFs and can be printed by teachers for classroom.
For Chapter 04 Expressions using Letter Numbers, regular practice with our worksheets will improve question-handling speed and help students understand all technical terms and diagrams.