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CHAPTER 2 — ARITHMETIC EXPRESSIONS
Class VII Mathematics — Challenge Paper Set 4
Preeti Kushwah Classes
General Instructions:
1. All questions are compulsory.
2. Section A: 6 questions of 1 mark each (6 marks).
3. Section B: 5 questions of 2 marks each (10 marks).
4. Section C: 4 questions of 3 marks each (12 marks).
5. Section D: 2 questions of 5 marks each (10 marks).
6. Bonus: 1 question of 2 marks (optional).
7. This is an expert-level paper. Show all steps clearly!
1. All questions are compulsory.
2. Section A: 6 questions of 1 mark each (6 marks).
3. Section B: 5 questions of 2 marks each (10 marks).
4. Section C: 4 questions of 3 marks each (12 marks).
5. Section D: 2 questions of 5 marks each (10 marks).
6. Bonus: 1 question of 2 marks (optional).
7. This is an expert-level paper. Show all steps clearly!
Section A — MCQ / Short Answer (1 Mark Each) [6 × 1 = 6]
Q1.1
Evaluate: 50 − {3 × (10 − 4)}
Choose the correct answer:
Choose the correct answer:
- (a) 20
- (b) 32
- (c) 14
- (d) 44
Q2.1
What happens to the signs inside a bracket when a minus sign in front of it is removed?
Choose the correct answer:
Choose the correct answer:
- (a) All signs stay the same
- (b) All signs flip (+ becomes − and − becomes +)
- (c) Only the first sign changes
- (d) Signs are multiplied by 2
Q3.1
How many terms are there in the expression: 5a + 3b − 2c + 7?
Q4.1
Evaluate using BODMAS: 20 − 4 × 3 + 6 ÷ 2
Choose the correct answer:
Choose the correct answer:
- (a) 14
- (b) 11
- (c) 23
- (d) 5
Q5.1
Which of the following correctly shows the distributive property?
- (a) 5 × (3 + 4) = 5 × 3 + 5 × 4
- (b) 5 × (3 + 4) = (5 × 3) + 4
- (c) 5 × (3 + 4) = 5 + 3 × 4
- (d) 5 × (3 + 4) = 5 + 3 + 4 × 5
Q6.1
Which pair of expressions gives the same result?
- (a) 3 × (6 + 4) and 3 × 6 + 4
- (b) 2 × (5 − 3) and 2 × 5 − 2 × 3
- (c) 4 × (7 − 2) and 4 × 7 − 2
- (d) 6 × (3 + 1) and 6 × 3 + 1
Section B — (2 Marks Each) [5 × 2 = 10]
Q7.2
Evaluate step by step, showing all working:
100 − [5 × {4 + (18 ÷ 6)}]
100 − [5 × {4 + (18 ÷ 6)}]
Q8.2
Simplify by removing all brackets:
a − (b − c + d) + (e − f)
Show which signs change and why.
a − (b − c + d) + (e − f)
Show which signs change and why.
Q9.2
Use the distributive property to calculate quickly:
(a) 7 × 98 (b) 12 × 103
(a) 7 × 98 (b) 12 × 103
Q10.2
A shopkeeper bought 15 pens at ₹8 each and 15 notebooks at ₹12 each. Write a mathematical expression for the total cost and use the distributive property to calculate it in one step.
Q11.2
Verify whether LHS = RHS:
4 × (15 − 7) = 4 × 15 − 4 × 7
Show full calculation on both sides.
4 × (15 − 7) = 4 × 15 − 4 × 7
Show full calculation on both sides.
Section C — (3 Marks Each) [4 × 3 = 12]
Q12.3
(a) Remove brackets from: 3x − (2x − 5y + z)
Explain clearly: why does the sign of 5y change? [2 marks]
(b) Remove brackets from: 2a + (3b − 4c) − (a − 2b + c)
Write the simplified expression. [1 mark]
Explain clearly: why does the sign of 5y change? [2 marks]
(b) Remove brackets from: 2a + (3b − 4c) − (a − 2b + c)
Write the simplified expression. [1 mark]
Q13.3
At a school mela, Class 7A collected ₹(500 + 3 × 80) from ticket sales and ₹(2 × 150 − 50) from food stalls.
(a) Calculate the amount collected from ticket sales. [1 mark]
(b) Calculate the amount collected from food stalls. [1 mark]
(c) They need ₹1200 for the school charity fund. Do they have enough? If not, how much more do they need? [1 mark]
(a) Calculate the amount collected from ticket sales. [1 mark]
(b) Calculate the amount collected from food stalls. [1 mark]
(c) They need ₹1200 for the school charity fund. Do they have enough? If not, how much more do they need? [1 mark]
Q14.3
Consider the expression: 5 + [8 − {3 × (2 + 1)}] + 4 × 2
(a) Find the value of the expression. Show all steps. [1 mark]
(b) How many terms are there in the expression? List them. [1 mark]
(c) If the innermost bracket changes from (2 + 1) to (2 + 3), find the new value. [1 mark]
(a) Find the value of the expression. Show all steps. [1 mark]
(b) How many terms are there in the expression? List them. [1 mark]
(c) If the innermost bracket changes from (2 + 1) to (2 + 3), find the new value. [1 mark]
Q15.3
Study the number pattern:
1 × (1 + 1) = 1 × 2 = 2
2 × (2 + 1) = 2 × 3 = 6
3 × (3 + 1) = 3 × 4 = 12
(a) Write the next two values (for n = 4 and n = 5) without using a calculator. [1 mark]
(b) Write the general expression for any value of n. [1 mark]
(c) Verify that 8 × (8 + 1) = 72 using the pattern. Also check: is 9 × 8 = 72? What property does this show? [1 mark]
1 × (1 + 1) = 1 × 2 = 2
2 × (2 + 1) = 2 × 3 = 6
3 × (3 + 1) = 3 × 4 = 12
(a) Write the next two values (for n = 4 and n = 5) without using a calculator. [1 mark]
(b) Write the general expression for any value of n. [1 mark]
(c) Verify that 8 × (8 + 1) = 72 using the pattern. Also check: is 9 × 8 = 72? What property does this show? [1 mark]
Section D — (5 Marks Each) [2 × 5 = 10]
Q16.5
A school has 4 sections in Class 7. The number of students in each section is given as an expression:
(a) Evaluate each expression to find the number of students in each section. [1 mark]
(b) Find the total number of students in Class 7. [1 mark]
(c) For the annual school trip, each student pays ₹(50 + 15 × 2 + 30 ÷ 3). Evaluate this expression for the cost per student. [1 mark]
(d) Find the total amount collected from Class 7. [1 mark]
(e) 5 more students join Class 7 from a new school. Write the new total as an expression and calculate the new total collection at the same price per student. [1 mark]
| Section | Students |
|---|---|
| 7A | (8 + 12) |
| 7B | (25 − 3) |
| 7C | (6 × 5) |
| 7D | (40 ÷ 2 + 5) |
(b) Find the total number of students in Class 7. [1 mark]
(c) For the annual school trip, each student pays ₹(50 + 15 × 2 + 30 ÷ 3). Evaluate this expression for the cost per student. [1 mark]
(d) Find the total amount collected from Class 7. [1 mark]
(e) 5 more students join Class 7 from a new school. Write the new total as an expression and calculate the new total collection at the same price per student. [1 mark]
Q17.5
Evaluate the following expression and verify your answer:
250 − {3 × (20 − 8) + 4 × (15 − 6)} + 2 × (10 + 5)
(a) First evaluate all the innermost brackets. [1 mark]
(b) Simplify inside the curly brackets { }. [1 mark]
(c) Complete the final calculation. [1 mark]
(d) Verify your answer: does the result equal 208? Show this clearly. [1 mark]
(e) If the last term changes from +2 × (10 + 5) to −2 × (10 + 5), what is the new answer? [1 mark]
250 − {3 × (20 − 8) + 4 × (15 − 6)} + 2 × (10 + 5)
(a) First evaluate all the innermost brackets. [1 mark]
(b) Simplify inside the curly brackets { }. [1 mark]
(c) Complete the final calculation. [1 mark]
(d) Verify your answer: does the result equal 208? Show this clearly. [1 mark]
(e) If the last term changes from +2 × (10 + 5) to −2 × (10 + 5), what is the new answer? [1 mark]
Bonus Question (Optional) [2 Marks]
Q18.2CHALLENGE
★ A mystery number puzzle using the distributive property:
It is given that: a × (b + c) = 120 and a × b = 72
(i) Find the value of a × c. [1 mark]
(ii) If a = 8, find the individual values of b and c. [1 mark]
It is given that: a × (b + c) = 120 and a × b = 72
(i) Find the value of a × c. [1 mark]
(ii) If a = 8, find the individual values of b and c. [1 mark]
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Answer Key & Detailed Solutions
Q1. [1 Mark]
50 − {3 × (10 − 4)}
Step 1: Innermost bracket: (10 − 4) = 6
Step 2: Curly bracket: {3 × 6} = 18
Step 3: 50 − 18 = 32
Answer: (b) 32
Step 1: Innermost bracket: (10 − 4) = 6
Step 2: Curly bracket: {3 × 6} = 18
Step 3: 50 − 18 = 32
Answer: (b) 32
Q2. [1 Mark]
When a minus sign precedes a bracket, removing the bracket flips all the signs inside.
Example: −(a − b + c) = −a + b − c
Answer: (b) All signs flip (+ becomes − and − becomes +)
Example: −(a − b + c) = −a + b − c
Answer: (b) All signs flip (+ becomes − and − becomes +)
Q3. [1 Mark]
In the expression 5a + 3b − 2c + 7, the terms are separated by + and − signs:
Term 1: 5a Term 2: 3b Term 3: −2c Term 4: 7
Answer: 4 terms
Term 1: 5a Term 2: 3b Term 3: −2c Term 4: 7
Answer: 4 terms
Q4. [1 Mark]
20 − 4 × 3 + 6 ÷ 2
BODMAS: Multiplication and Division first (left to right), then Addition and Subtraction:
= 20 − 12 + 3
= 8 + 3 = 11
Answer: (b) 11
BODMAS: Multiplication and Division first (left to right), then Addition and Subtraction:
= 20 − 12 + 3
= 8 + 3 = 11
Answer: (b) 11
Q5. [1 Mark]
The distributive property states: a × (b + c) = a × b + a × c
5 × (3 + 4) = 5 × 3 + 5 × 4 = 15 + 20 = 35 ✓
Answer: (a) 5 × (3 + 4) = 5 × 3 + 5 × 4
5 × (3 + 4) = 5 × 3 + 5 × 4 = 15 + 20 = 35 ✓
Answer: (a) 5 × (3 + 4) = 5 × 3 + 5 × 4
Q6. [1 Mark]
Check option (b): 2 × (5 − 3) = 2 × 2 = 4
2 × 5 − 2 × 3 = 10 − 6 = 4 ✓
Both give 4. This is exactly the distributive property!
Answer: (b)
2 × 5 − 2 × 3 = 10 − 6 = 4 ✓
Both give 4. This is exactly the distributive property!
Answer: (b)
Q7. [2 Marks]
100 − [5 × {4 + (18 ÷ 6)}] [1 mark for correct order of brackets]
Step 1: Innermost (): 18 ÷ 6 = 3
Step 2: {4 + 3} = 7
Step 3: [5 × 7] = 35
Step 4: 100 − 35 = 65 [1 mark]
Step 1: Innermost (): 18 ÷ 6 = 3
Step 2: {4 + 3} = 7
Step 3: [5 × 7] = 35
Step 4: 100 − 35 = 65 [1 mark]
Q8. [2 Marks]
a − (b − c + d) + (e − f) [1 mark for correct rule application]
The first bracket has a minus sign in front → all signs inside flip:
−(b − c + d) = −b + c − d
The second bracket has a plus sign in front → signs stay the same:
+(e − f) = e − f
Result: a − b + c − d + e − f [1 mark]
The first bracket has a minus sign in front → all signs inside flip:
−(b − c + d) = −b + c − d
The second bracket has a plus sign in front → signs stay the same:
+(e − f) = e − f
Result: a − b + c − d + e − f [1 mark]
Q9. [2 Marks]
(a) 7 × 98 [1 mark]
= 7 × (100 − 2) = 7 × 100 − 7 × 2 = 700 − 14 = 686
(b) 12 × 103 [1 mark]
= 12 × (100 + 3) = 12 × 100 + 12 × 3 = 1200 + 36 = 1236
= 7 × (100 − 2) = 7 × 100 − 7 × 2 = 700 − 14 = 686
(b) 12 × 103 [1 mark]
= 12 × (100 + 3) = 12 × 100 + 12 × 3 = 1200 + 36 = 1236
Q10. [2 Marks]
Total cost = 15 × 8 + 15 × 12 [1 mark for expression]
Using distributive property:
= 15 × (8 + 12)
= 15 × 20
= ₹300 [1 mark]
Using distributive property:
= 15 × (8 + 12)
= 15 × 20
= ₹300 [1 mark]
Q11. [2 Marks]
LHS = 4 × (15 − 7) = 4 × 8 = 32 [1 mark]
RHS = 4 × 15 − 4 × 7 = 60 − 28 = 32 [1 mark]
LHS = RHS = 32 ✓
This verifies the distributive property of multiplication over subtraction.
RHS = 4 × 15 − 4 × 7 = 60 − 28 = 32 [1 mark]
LHS = RHS = 32 ✓
This verifies the distributive property of multiplication over subtraction.
Q12. [3 Marks]
(a) 3x − (2x − 5y + z) [2 marks]
The bracket has a minus sign in front, so ALL signs inside flip:
−(2x − 5y + z) = −2x + 5y − z
Result: 3x − 2x + 5y − z = x + 5y − z
Why does 5y become +5y? Because the original sign of 5y inside the bracket was −. When the − outside removes the bracket, − × (−5y) = +5y.
(b) 2a + (3b − 4c) − (a − 2b + c) [1 mark]
First bracket (+ in front): signs stay → +3b − 4c
Second bracket (− in front): signs flip → −a + 2b − c
Result: 2a + 3b − 4c − a + 2b − c = a + 5b − 5c
The bracket has a minus sign in front, so ALL signs inside flip:
−(2x − 5y + z) = −2x + 5y − z
Result: 3x − 2x + 5y − z = x + 5y − z
Why does 5y become +5y? Because the original sign of 5y inside the bracket was −. When the − outside removes the bracket, − × (−5y) = +5y.
(b) 2a + (3b − 4c) − (a − 2b + c) [1 mark]
First bracket (+ in front): signs stay → +3b − 4c
Second bracket (− in front): signs flip → −a + 2b − c
Result: 2a + 3b − 4c − a + 2b − c = a + 5b − 5c
Q13. [3 Marks]
(a) Ticket sales: ₹(500 + 3 × 80) [1 mark]
= 500 + 240 = ₹740
(b) Food stalls: ₹(2 × 150 − 50) [1 mark]
= 300 − 50 = ₹250
(c) Total collected = 740 + 250 = ₹990 [1 mark]
They need ₹1200. They have only ₹990.
No, they do not have enough. They are short by ₹1200 − ₹990 = ₹210.
= 500 + 240 = ₹740
(b) Food stalls: ₹(2 × 150 − 50) [1 mark]
= 300 − 50 = ₹250
(c) Total collected = 740 + 250 = ₹990 [1 mark]
They need ₹1200. They have only ₹990.
No, they do not have enough. They are short by ₹1200 − ₹990 = ₹210.
Q14. [3 Marks]
(a) Evaluate: 5 + [8 − {3 × (2 + 1)}] + 4 × 2 [1 mark]
Step 1: (2 + 1) = 3
Step 2: {3 × 3} = 9
Step 3: [8 − 9] = −1
Step 4: 4 × 2 = 8
Step 5: 5 + (−1) + 8 = 12
(b) Terms in the expression: [1 mark]
Term 1: 5 Term 2: [8 − {3 × (2 + 1)}] Term 3: 4 × 2
There are 3 terms.
(c) If (2 + 1) changes to (2 + 3): [1 mark]
Step 1: (2 + 3) = 5
Step 2: {3 × 5} = 15
Step 3: [8 − 15] = −7
Step 4: 5 + (−7) + 8 = 6
Step 1: (2 + 1) = 3
Step 2: {3 × 3} = 9
Step 3: [8 − 9] = −1
Step 4: 4 × 2 = 8
Step 5: 5 + (−1) + 8 = 12
(b) Terms in the expression: [1 mark]
Term 1: 5 Term 2: [8 − {3 × (2 + 1)}] Term 3: 4 × 2
There are 3 terms.
(c) If (2 + 1) changes to (2 + 3): [1 mark]
Step 1: (2 + 3) = 5
Step 2: {3 × 5} = 15
Step 3: [8 − 15] = −7
Step 4: 5 + (−7) + 8 = 6
Q15. [3 Marks]
(a) Next two values: [1 mark]
n = 4: 4 × (4 + 1) = 4 × 5 = 20
n = 5: 5 × (5 + 1) = 5 × 6 = 30
(b) General expression for n: [1 mark]
n × (n + 1)
(c) Verification for n = 8: [1 mark]
8 × (8 + 1) = 8 × 9 = 72 ✓
Also: 9 × 8 = 72 ✓
This shows the commutative property of multiplication (a × b = b × a).
n = 4: 4 × (4 + 1) = 4 × 5 = 20
n = 5: 5 × (5 + 1) = 5 × 6 = 30
(b) General expression for n: [1 mark]
n × (n + 1)
(c) Verification for n = 8: [1 mark]
8 × (8 + 1) = 8 × 9 = 72 ✓
Also: 9 × 8 = 72 ✓
This shows the commutative property of multiplication (a × b = b × a).
Q16. [5 Marks]
(a) Students per section: [1 mark]
7A: (8 + 12) = 20 students
7B: (25 − 3) = 22 students
7C: (6 × 5) = 30 students
7D: (40 ÷ 2 + 5) = 20 + 5 = 25 students
(b) Total students: [1 mark]
20 + 22 + 30 + 25 = 97 students
(c) Cost per student: [1 mark]
₹(50 + 15 × 2 + 30 ÷ 3)
= 50 + 30 + 10 = ₹90
(d) Total collection: [1 mark]
97 × 90 = ₹8,730
(e) 5 more students join: [1 mark]
New total students = 97 + 5 = 102
New expression for collection: (97 + 5) × 90 = 102 × 90
Using distributive property: 97 × 90 + 5 × 90 = 8730 + 450 = ₹9,180
7A: (8 + 12) = 20 students
7B: (25 − 3) = 22 students
7C: (6 × 5) = 30 students
7D: (40 ÷ 2 + 5) = 20 + 5 = 25 students
(b) Total students: [1 mark]
20 + 22 + 30 + 25 = 97 students
(c) Cost per student: [1 mark]
₹(50 + 15 × 2 + 30 ÷ 3)
= 50 + 30 + 10 = ₹90
(d) Total collection: [1 mark]
97 × 90 = ₹8,730
(e) 5 more students join: [1 mark]
New total students = 97 + 5 = 102
New expression for collection: (97 + 5) × 90 = 102 × 90
Using distributive property: 97 × 90 + 5 × 90 = 8730 + 450 = ₹9,180
Q17. [5 Marks]
Expression: 250 − {3 × (20 − 8) + 4 × (15 − 6)} + 2 × (10 + 5)
(a) Innermost brackets: [1 mark]
(20 − 8) = 12 (15 − 6) = 9 (10 + 5) = 15
(b) Curly brackets: [1 mark]
{3 × 12 + 4 × 9}
= {36 + 36} = 72
(c) Final calculation: [1 mark]
250 − 72 + 2 × 15
= 250 − 72 + 30
= 178 + 30 = 208
(d) Verification: [1 mark]
208 = 208 ✓
Yes, the result equals 208, confirming our calculation is correct.
(e) Changed to −2 × (10 + 5): [1 mark]
250 − 72 − 2 × 15
= 250 − 72 − 30
= 178 − 30 = 148
(a) Innermost brackets: [1 mark]
(20 − 8) = 12 (15 − 6) = 9 (10 + 5) = 15
(b) Curly brackets: [1 mark]
{3 × 12 + 4 × 9}
= {36 + 36} = 72
(c) Final calculation: [1 mark]
250 − 72 + 2 × 15
= 250 − 72 + 30
= 178 + 30 = 208
(d) Verification: [1 mark]
208 = 208 ✓
Yes, the result equals 208, confirming our calculation is correct.
(e) Changed to −2 × (10 + 5): [1 mark]
250 − 72 − 2 × 15
= 250 − 72 − 30
= 178 − 30 = 148
Q18. Bonus [2 Marks]
Given: a × (b + c) = 120 and a × b = 72
(i) Find a × c: [1 mark]
By the distributive property: a × (b + c) = a × b + a × c
120 = 72 + a × c
a × c = 120 − 72 = 48
(ii) Find b and c when a = 8: [1 mark]
a × b = 72 → 8 × b = 72 → b = 72 ÷ 8 = 9
a × c = 48 → 8 × c = 48 → c = 48 ÷ 8 = 6
Verification: 8 × (9 + 6) = 8 × 15 = 120 ✓
(i) Find a × c: [1 mark]
By the distributive property: a × (b + c) = a × b + a × c
120 = 72 + a × c
a × c = 120 − 72 = 48
(ii) Find b and c when a = 8: [1 mark]
a × b = 72 → 8 × b = 72 → b = 72 ÷ 8 = 9
a × c = 48 → 8 × c = 48 → c = 48 ÷ 8 = 6
Verification: 8 × (9 + 6) = 8 × 15 = 120 ✓