CBSE Class 7 Mathematics Arithmetic Expressions Practice Paper Assignment Set 03 with Answers

Here is Class 7 Maths Ganita Prakash Chapter 2 Arithmetic Expressions Practice Paper 06: 20 questions for 40 marks across Sections A to E, each followed by a step-by-step solution. This paper focuses on the terms of an expression, removing and inserting brackets, the distributive property, filling blanks to balance two expressions, comparing products without calculating and turning phrases such as 'the sum of 10 and y, divided by 2' into expressions. Real-life questions include a fruit vendor's earnings, an Annual Day budget, a birthday shopping list and a monthly water bill.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Add brackets to make the statements true:
(a) 34 − 9 + 12 = 13
(b) 56 − 14 − 8 = 50
Answer:
(a) Without brackets, 34 − 9 + 12 = 37. To get 13 we must subtract the whole of 9 + 12:
34 − (9 + 12) = 13, since 34 − 21 = 13.
(b) Without brackets, 56 − 14 − 8 = 34. To get 50 we must subtract only the difference 14 − 8:
56 − (14 − 8) = 50, since 56 − 6 = 50.
In simple words: Group the numbers that must be worked out first: 34 − (9 + 12) = 13 and 56 − (14 − 8) = 50.

Teacher's Note:
1. Try each possible place for the bracket and check the value.
2. A bracket after a minus sign changes the signs of the terms inside.

Question 12. Fill in the blanks to balance the equations:
(a) 13 + 4 = ____ + 6
(b) 22 + ____ = 6 × 5
Answer:
(a) Left side: 13 + 4 = 17. We need a number which gives 17 when 6 is added: 17 − 6 = 11.
So 13 + 4 = 11 + 6.
(b) Right side: 6 × 5 = 30. We need a number which gives 30 when added to 22: 30 − 22 = 8.
So 22 + 8 = 6 × 5.
In simple words: Make both sides equal: the blanks are 11 and 8.

Teacher's Note:
1. Work out the complete side first, then find the missing number.
2. Check: 11 + 6 = 17 and 22 + 8 = 30.

Question 13. Evaluate:
(a) 8 × 4 − 2 + 6 ÷ 3
(b) 45 ÷ 5 + 6 − 3 × 2
Answer:
(a) Multiply and divide first: 8 × 4 = 32 and 6 ÷ 3 = 2.
Then 32 − 2 + 2 = 30 + 2 = 32.
(b) Divide and multiply first: 45 ÷ 5 = 9 and 3 × 2 = 6.
Then 9 + 6 − 6 = 15 − 6 = 9.
In simple words: Do × and ÷ first, then + and − from left to right: 32 and 9.

Teacher's Note:
1. Underline the × and ÷ terms before you start.
2. After that, add and subtract from left to right.

Question 14. Write expressions for these and find the value:
(a) Multiply 5 by sum of 7 and 2
(b) Subtract the product of 4 and 3 from 30
Answer:
(a) The sum of 7 and 2 is (7 + 2). Expression: 5 × (7 + 2) = 5 × 9 = 45.
(b) The product of 4 and 3 is 4 × 3, and it is subtracted from 30.
Expression: 30 − 4 × 3 = 30 − 12 = 18.
In simple words: 5 × (7 + 2) = 45 and 30 − 4 × 3 = 18.

Teacher's Note:
1. 'Subtract A from B' means B − A.
2. The sum needs brackets; the product does not, because × is done first.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Write the expression for: 3 times the sum of 45 and 28, then subtract 72 from it. Find its value.
Answer:
The sum of 45 and 28 is (45 + 28); 3 times it is 3 × (45 + 28); then 72 is subtracted.
Expression: 3 × (45 + 28) − 72.
45 + 28 = 73, so 3 × 73 = 219.
219 − 72 = 147.
In simple words: The expression 3 × (45 + 28) − 72 has the value 147.

Teacher's Note:
1. Keep the sum in brackets because the whole sum is multiplied by 3.
2. Check: 3 × 45 + 3 × 28 − 72 = 135 + 84 − 72 = 147.

Question 16. A vendor sells 15 apples at ₹12 each and 10 oranges at ₹8 each. He spends ₹50 on transport. Write an expression for his earnings after paying for transport and find it.
Answer:
Money from sales (revenue) = 15 × 12 + 10 × 8 = 180 + 80 = ₹260.
The ₹50 spent on transport is taken away.
Expression: 15 × 12 + 10 × 8 − 50.
Value = 180 + 80 − 50 = ₹210.
In simple words: He takes in ₹260 from sales and keeps ₹210 after paying ₹50 for transport.

Teacher's Note:
1. Money received is added and money spent is subtracted.
2. Revenue is the money from sales (₹260); after costs he is left with ₹210.

Question 17. Evaluate: 120 − 3 × {15 − (28 ÷ 4)}
Answer:
Innermost bracket first: 28 ÷ 4 = 7.
Curly bracket: 15 − 7 = 8.
Multiply: 3 × 8 = 24.
Subtract: 120 − 24 = 96.
In simple words: Work from the inside out: 28 ÷ 4 = 7, 15 − 7 = 8, 3 × 8 = 24, and 120 − 24 = 96.

Teacher's Note:
1. Order of brackets: ( ) first, then { }.
2. Do not subtract 120 − 3 first; multiplication comes before subtraction.

Section D: Long Answer Question (5 Marks)

Question 18. For Annual Day, the school orders:
120 chairs at ₹150 each
6 loudspeakers at ₹2500 each
15 decorative lights at ₹320 each
(a) Form an expression for total cost.
(b) Find total cost.
(c) If ₹40,000 was budgeted, find the remaining amount or shortage.
Answer:
(a) Total cost = 120 × 150 + 6 × 2500 + 15 × 320.
(b) Chairs: 120 × 150 = ₹18,000. Loudspeakers: 6 × 2500 = ₹15,000. Lights: 15 × 320 = ₹4,800.
Total cost = 18,000 + 15,000 + 4,800 = ₹37,800.
(c) The budget ₹40,000 is more than ₹37,800, so there is no shortage.
Remaining amount = 40,000 − 37,800 = ₹2,200.
In simple words: The order costs ₹37,800, so ₹2,200 of the ₹40,000 budget is left.

Teacher's Note:
1. 120 × 150 can be found as 12 × 15 × 100 = 18,000.
2. Compare the bill with the budget before subtracting.

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

Question 19. Case Study 1: Birthday Budget
Meera, with her birthday approaching, was buzzing with excitement as she planned her party. She wanted to make it special for all her friends. With a budget in mind, she visited a local store to buy everything she needed. She carefully picked out six gift items for the party guests, each costing ₹35. Next, she selected five packets of colorful balloons, with each packet priced at ₹18. The most important part, the birthday cake, was a beautiful creation that cost ₹250. With all her purchases in hand, she headed home to prepare for the celebration.
(a) Write an arithmetic expression for the total cost
(b) Find total expenditure
(c) If she had ₹600, how much is left?
(d) If she adds 2 return gifts at ₹30 each, what will be the new total?
Answer:
(a) Total cost = 6 × 35 + 5 × 18 + 250.
(b) 6 × 35 = 210 and 5 × 18 = 90, so total = 210 + 90 + 250 = ₹550.
(c) Money left = 600 − 550 = ₹50.
(d) Extra cost = 2 × 30 = ₹60. New total = 550 + 60 = ₹610.
(With the return gifts the total, ₹610, is ₹10 more than the ₹600 she has.)
In simple words: Meera spends ₹550 and keeps ₹50; with 2 more gifts the total becomes ₹610.

Teacher's Note:
1. The cake is a single item, so it is just the term 250.
2. In (d) add only the cost of the new items to the old total.

Question 20. Case Study 2: Water Bill Calculation
Every month, the family braced for the dreaded utility bill. This month, they were particularly careful with their water usage, knowing each unit was priced at ₹7. Despite their best efforts, the bill showed they had consumed 30 units of water. Along with their standard ₹250 maintenance charge, a small oversight meant a ₹50 late fine was added to their total, turning a simple bill into a frustrating calculation of their monthly household expenses.
(a) Write a numerical expression for the total water bill
(b) Calculate total
(c) If the household paid ₹600, find the balance
(d) What would be the bill if they used 40 units instead?
Answer:
(a) Total bill = 30 × 7 + 250 + 50.
(b) 30 × 7 = 210, so total = 210 + 250 + 50 = ₹510.
(c) Balance returned = 600 − 510 = ₹90.
(d) With 40 units: 40 × 7 + 250 + 50 = 280 + 250 + 50 = ₹580.
(Only the water term changes: 10 more units add 10 × 7 = ₹70, and 510 + 70 = 580.)
In simple words: The bill is ₹510, the family gets ₹90 back from ₹600, and 40 units would cost ₹580.

Teacher's Note:
1. Fixed charges (maintenance and fine) stay the same when usage changes.
2. Spotting which term changes saves recalculating everything.