CBSE Class 7 Mathematics Arithmetic Expressions Practice Paper Assignment Set 02 with Answers

This is Class 7 Maths Ganita Prakash Chapter 2 Arithmetic Expressions Practice Paper 04, a 40-mark test with 20 questions arranged in Sections A to E, and every question comes with a step-by-step answer. It checks the order of operations (brackets, then × and ÷, then + and −), squares and cubes inside expressions, removing brackets, balancing equalities and comparing values. The word problems on a shop bill, packing fruit by weight, a birthday order with a discount and class stationery show how to turn a real situation into one arithmetic expression.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Simplify: 108 ÷ 9 + 12 × (7 − 4)
Answer:
Brackets first: 7 − 4 = 3.
The expression becomes 108 ÷ 9 + 12 × 3.
Division and multiplication next: 108 ÷ 9 = 12 and 12 × 3 = 36.
Add: 12 + 36 = 48.
In simple words: Work out the bracket, then the ÷ and ×, and finally add: 12 + 36 = 48.

Teacher's Note:
1. Write each step on a new line so no term is lost.
2. Check: 108 ÷ 9 = 12 because 9 × 12 = 108.

Question 12. Solve the following by inserting suitable values:
(a) 35 + 9 = x + 6 (b) y + 15 = 67 (c) 14 × z = 98 (d) w − 20 = 35
Answer:
(a) 35 + 9 = 44, so 44 = x + 6. The number that gives 44 when 6 is added is 44 − 6 = 38. So x = 38.
(b) y + 15 = 67, so y = 67 − 15. So y = 52.
(c) 14 × z = 98, so z = 98 ÷ 14. So z = 7 (since 14 × 7 = 98).
(d) w − 20 = 35, so w = 35 + 20. So w = 55.
Check: 38 + 6 = 44, 52 + 15 = 67, 14 × 7 = 98, 55 − 20 = 35.
In simple words: Undo the operation: x = 38, y = 52, z = 7 and w = 55.

Teacher's Note:
1. An equals sign means both sides have the same value, like a balanced scale.
2. Always put your value back in to check that both sides match.

Question 13. Rewrite the expressions without brackets and compute:
(a) 40 + (16 + 14) (b) 50 − (20 + 15) (c) 70 + (30 − 18) (d) 60 − (25 − 5)
Answer:
(a) A '+' before the bracket keeps the signs: 40 + 16 + 14 = 70.
(b) A '−' before the bracket changes the signs inside: 50 − 20 − 15 = 15.
(c) 70 + 30 − 18 = 100 − 18 = 82.
(d) 60 − 25 + 5 = 35 + 5 = 40.
Check with brackets: 40 + 30 = 70, 50 − 35 = 15, 70 + 12 = 82, 60 − 20 = 40.
In simple words: Plus before a bracket keeps the signs; minus before a bracket flips them.

Teacher's Note:
1. Subtracting (25 − 5) means taking away 25 but giving back 5.
2. Check by computing the bracket first; the answers must agree.

Question 14. Arrange the values in increasing order: 120 ÷ 6, 5 × 5, 13 + 7, 3³, 90 − 30
Answer:
Find each value:
120 ÷ 6 = 20; 5 × 5 = 25; 13 + 7 = 20; 3³ = 3 × 3 × 3 = 27; 90 − 30 = 60.
Values in increasing order: 20, 20, 25, 27, 60.
So the order is 120 ÷ 6 = 13 + 7 < 5 × 5 < 3³ < 90 − 30.
In simple words: Work out each value first, then list them from smallest to largest.

Teacher's Note:
1. 3³ means 3 × 3 × 3, not 3 × 3.
2. Two expressions can have the same value; show them as equal.

Section C: Short Answer Questions (3 Marks Each)

Question 15. (a) Simplify and calculate: (60 − 4 × 9) + (27 ÷ 3)²
(b) Simplify and calculate: (90 − 6 × 7) + (30 ÷ 5)³
Answer:
(a) First bracket: 4 × 9 = 36, so 60 − 36 = 24.
Second bracket: 27 ÷ 3 = 9, and 9² = 81.
Total: 24 + 81 = 105.
(b) First bracket: 6 × 7 = 42, so 90 − 42 = 48.
Second bracket: 30 ÷ 5 = 6, and 6³ = 6 × 6 × 6 = 216.
Total: 48 + 216 = 264.
In simple words: Finish each bracket, then the power, then add: 105 and 264.

Teacher's Note:
1. Inside a bracket the usual order (× and ÷ before + and −) still applies.
2. Powers act on the value of the bracket after it is simplified.

Question 16. (a) A carton holds 25 oranges. Each orange weighs 200 grams. If a box can hold 10 kg, how many oranges fit in a box? Write expression and find answer.
(b) A box holds 30 mangoes, each weighing 150 grams. How many mangoes can be packed in a 15 kg crate? Write expression and solve.
Answer:
(a) Change kilograms to grams: 10 kg = 10 × 1000 g = 10000 g.
Number of oranges = (10 × 1000) ÷ 200 = 10000 ÷ 200 = 50 oranges.
(The carton size of 25 is extra information; it is not needed to answer.)
(b) 15 kg = 15 × 1000 g = 15000 g.
Number of mangoes = (15 × 1000) ÷ 150 = 15000 ÷ 150 = 100 mangoes.
Check: 50 × 200 g = 10000 g = 10 kg and 100 × 150 g = 15000 g = 15 kg.
In simple words: Total weight ÷ weight of one fruit: 50 oranges and 100 mangoes.

Teacher's Note:
1. Use the same unit (grams) for both weights before dividing.
2. Real problems often contain extra numbers; pick only the ones you need.

Question 17. Evaluate:
(a) \( \frac{(72 \div 9) + (7 \times 4) - 20}{5} \)
(b) \( \frac{(81 \div 9) + (8 \times 5) - 30}{7} \)
Answer:
(a) Numerator: 72 ÷ 9 = 8 and 7 × 4 = 28, so 8 + 28 − 20 = 16.
Value = 16 ÷ 5 = \( \frac{16}{5} \) = 3.2.
(b) Numerator: 81 ÷ 9 = 9 and 8 × 5 = 40, so 9 + 40 − 30 = 19.
Value = 19 ÷ 7 = \( \frac{19}{7} \) = \( 2\frac{5}{7} \).
In simple words: Simplify the top completely, then divide by the bottom number.

Teacher's Note:
1. A fraction bar works like a bracket around the whole numerator.
2. When the division is not exact, leave the answer as a fraction or mixed number.

Section D: Long Answer Question (5 Marks)

Question 18. Mr. Arjun, a friendly shopkeeper in the neighborhood, often helps customers find exactly what they need for school and office work. One afternoon, a customer visited his shop to buy supplies for her children. She selected 6 books priced at ₹120 each, 4 pens costing ₹15 each, and 2 sturdy school bags priced at ₹350 each. Mr. Arjun quickly calculated the total but also allowed the customer to pay in cash immediately. She handed him ₹2000, and he checked whether the amount was enough or if more was needed.
(a) Write expression for total cost.
(b) Calculate total cost.
(c) If he receives ₹2000, find balance or shortage.
Answer:
(a) Cost of each kind of item = number × price. Total cost = 6 × 120 + 4 × 15 + 2 × 350.
(b) Books: 6 × 120 = ₹720. Pens: 4 × 15 = ₹60. Bags: 2 × 350 = ₹700.
Total cost = 720 + 60 + 700 = ₹1480.
(c) ₹2000 is more than ₹1480, so the amount is enough and there is no shortage.
Balance returned = 2000 − 1480 = ₹520.
In simple words: The bill is ₹1480, so from ₹2000 the customer gets back ₹520.

Teacher's Note:
1. Each product term (number × price) is one term of the expression.
2. Compare the amount paid with the bill before deciding between balance and shortage.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. A family is excitedly preparing for their daughter Anaya's birthday celebration at home. They have invited friends, relatives, and neighbors, ensuring everyone will enjoy delicious food. After visiting the local bakery, they decided to place an order for 10 large pizzas priced at ₹250 each, 12 bottles of fresh fruit juice at ₹80 each, and 25 beautifully decorated cupcakes at ₹45 each. The bakery owner, happy to receive such a big order, also mentioned a special discount for bulk purchases.
(a) What will be the total cost before discount?
(b) If the bakery gives a 12% discount on the total, what will be the final bill?
Answer:
(a) Total cost = 10 × 250 + 12 × 80 + 25 × 45.
Pizzas: 10 × 250 = ₹2500. Juice: 12 × 80 = ₹960. Cupcakes: 25 × 45 = ₹1125.
Total before discount = 2500 + 960 + 1125 = ₹4585.
(b) 12% means 12 out of every 100. Discount = 4585 × 12 ÷ 100 = 55020 ÷ 100 = ₹550.20.
Final bill = 4585 − 550.20 = ₹4034.80.
In simple words: The order costs ₹4585; after taking off 12% (₹550.20) the family pays ₹4034.80.

Teacher's Note:
1. Find the full total first; the discount is taken on the whole bill.
2. ₹550.20 means 550 rupees and 20 paise.

Question 20. At the start of the new academic session, Mrs. Rao, the dedicated class teacher of Class 7, wanted every student to have the basic stationery needed for a smooth learning experience. She purchased 2 notebooks for each student at ₹25 each, 3 pens for each at ₹10 each, and 1 eraser for each at ₹5 each, ensuring no one lacked essential items. She distributed them among all 30 students during the first day of class. The total expense was compared to the amount collected from students.
(a) Find the total cost of all the items.
(b) If ₹3000 was collected from students, how much amount is left or needed?
Answer:
(a) Cost for one student = 2 × 25 + 3 × 10 + 1 × 5 = 50 + 30 + 5 = ₹85.
Total cost = 30 × (2 × 25 + 3 × 10 + 1 × 5) = 30 × 85 = ₹2550.
(Check item-wise: 60 notebooks × ₹25 = ₹1500, 90 pens × ₹10 = ₹900, 30 erasers × ₹5 = ₹150; 1500 + 900 + 150 = 2550.)
(b) ₹3000 is more than ₹2550, so money is left over.
Amount left = 3000 − 2550 = ₹450.
In simple words: Each student's kit costs ₹85, so 30 kits cost ₹2550 and ₹450 of the ₹3000 is left.

Teacher's Note:
1. Finding the cost for one student first makes the calculation shorter.
2. Both methods must give the same total; use one to check the other.