CBSE Class 7 Mathematics Ganita Prakash Term 1 Sample Paper with Answers

This is the CBSE Class 7 Mathematics Ganita Prakash Term 1 Sample Paper 2025-26, a 60-mark paper of 29 questions in Sections A to E, given here with step-by-step answers. It covers the first three chapters: Large Numbers Around Us (Indian and International place value, lakhs, crores and millions, rounding and estimation), Arithmetic Expressions (order of operations, brackets and writing expressions for shopping bills) and A Peek Beyond the Point (comparing, adding and subtracting decimals and converting cm, m, g and kg).

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 16. Write the number name of 760840120 in both Indian and International number systems.
Answer:
Indian system: place commas as 76,08,40,120 (groups: crore | lakh | thousand | ones).
Number name: Seventy-six crore eight lakh forty thousand one hundred twenty.
International system: place commas as 760,840,120 (groups: million | thousand | ones).
Number name: Seven hundred sixty million eight hundred forty thousand one hundred twenty.
In simple words: 76,08,40,120 is 76 crore 8 lakh 40 thousand 120, and 760,840,120 is 760 million 840 thousand 120.

Teacher's Note:
1. Put the commas first; then read each group with its period name.
2. Indian: 3 digits, then groups of 2. International: groups of 3.

Question 17. Find the difference between the greatest 8-digit number and smallest 7-digit number.
Answer:
Greatest 8-digit number = 9,99,99,999.
Smallest 7-digit number = 10,00,000.
Difference = 9,99,99,999 − 10,00,000 = 9,89,99,999.
Check: 9,89,99,999 + 10,00,000 = 9,99,99,999.
In simple words: 99,999,999 − 1,000,000 = 98,999,999, which is 9,89,99,999 in Indian commas.

Teacher's Note:
1. Greatest n-digit number: all nines. Smallest n-digit number: 1 followed by zeros.
2. Check a subtraction by adding the answer back.

Question 18. Arrange the values in increasing order: 120 ÷ 6, 5 × 5, 13 + 7, 3³, 90 − 30
Answer:
120 ÷ 6 = 20
5 × 5 = 25
13 + 7 = 20
3³ = 3 × 3 × 3 = 27
90 − 30 = 60
Increasing order: 120 ÷ 6 = 13 + 7 < 5 × 5 < 3³ < 90 − 30, that is 20, 20, 25, 27, 60.
In simple words: Work out each value (20, 25, 20, 27, 60) and then put 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; write them with an equals sign.

Question 19. A notebook is 2.45 cm thick. If 10 such notebooks are stacked, what will be the total height in meters?
Answer:
Total height = 10 × 2.45 cm = 24.5 cm (multiplying by 10 moves the decimal point one place right).
100 cm = 1 m, so 24.5 cm = 24.5 ÷ 100 m = 0.245 m.
In simple words: 10 notebooks are 24.5 cm tall, which is 0.245 m.

Teacher's Note:
1. × 10 moves the decimal point one place to the right.
2. ÷ 100 (cm to m) moves it two places to the left.

Question 20. (a) Evaluate: (90 ÷ 9) + 8 × (6 − 3)
(b) Find the value: (20 × 5) − (80 ÷ 8) + 15
Answer:
(a) Brackets first: 90 ÷ 9 = 10 and 6 − 3 = 3.
Then multiply: 8 × 3 = 24. Then add: 10 + 24 = 34.
(b) Brackets first: 20 × 5 = 100 and 80 ÷ 8 = 10.
Then work left to right: 100 − 10 + 15 = 90 + 15 = 105.
In simple words: (a) 10 + 24 = 34; (b) 100 − 10 + 15 = 105.

Teacher's Note:
1. Solve all brackets before anything else.
2. For + and − together, work from left to right.

Section C: Short Answer Questions (3 Marks Each)

Question 21. A company had a turnover of ₹4,68,75,999 in 2021 and ₹5,84,23,071 in 2022.
(a) Find the increase in turnover.
(b) Write the increase in International number system.
(c) Estimate the total turnover in both years.
Answer:
(a) Increase = ₹5,84,23,071 − ₹4,68,75,999 = ₹1,15,47,072.
Check: 4,68,75,999 + 1,15,47,072 = 5,84,23,071.
(b) In International commas: ₹11,547,072, read as eleven million five hundred forty-seven thousand seventy-two rupees.
(c) Round each turnover to the nearest lakh: ₹4,68,75,999 ≈ ₹4,69,00,000 and ₹5,84,23,071 ≈ ₹5,84,00,000.
Estimated total ≈ ₹4,69,00,000 + ₹5,84,00,000 = ₹10,53,00,000, that is about ₹10.5 crore.
(Exact total = ₹10,52,99,070, which is very close to the estimate.)
In simple words: The turnover grew by ₹1,15,47,072 (11,547,072), and the two years together come to about ₹10.5 crore.

Teacher's Note:
1. Check a subtraction by adding the answer to the smaller number.
2. An estimate uses rounded numbers; say which place you rounded to.

Question 22. A shopkeeper sells 3.75 m of cloth on Monday, 4.5 m on Tuesday, and 2.85 m on Wednesday.
(a) Find the total cloth sold in three days.
(b) How much more cloth was sold on Tuesday than Wednesday?
Answer:
(a) Line up the decimal points: 3.75 + 4.50 + 2.85.
3.75 + 4.50 = 8.25; 8.25 + 2.85 = 11.10.
Total cloth sold = 11.10 m (= 11.1 m).
(b) Tuesday − Wednesday = 4.50 − 2.85 = 1.65 m.
So 1.65 m more cloth was sold on Tuesday.
In simple words: The shop sold 11.1 m in all, and 1.65 m more on Tuesday than on Wednesday.

Teacher's Note:
1. Write 4.5 as 4.50 so all numbers have two decimal places.
2. 'How much more' means subtract.

Question 23. (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) The box can hold 10 kg = 10,000 g; each orange weighs 200 g.
Expression: 10000 ÷ 200 (or 10 ÷ 0.2 in kg).
10000 ÷ 200 = 100 ÷ 2 = 50 oranges. (The 25 oranges in a carton is extra information not needed here.)
(b) The crate can hold 15 kg = 15,000 g; each mango weighs 150 g.
Expression: 15000 ÷ 150 (or 15 ÷ 0.15 in kg).
15000 ÷ 150 = 1500 ÷ 15 = 100 mangoes. (The 30 mangoes in a box is not needed for the crate.)
In simple words: Divide the total weight the container can hold by the weight of one fruit: 50 oranges and 100 mangoes.

Teacher's Note:
1. Change kg to g first so both weights are in the same unit.
2. Not every number in a word problem is needed; pick the useful ones.

Teacher's Note:
1. Add zeros so every number has the same number of decimal places.
2. × 1000 moves the decimal point three places to the right.

Section D: Long Answer Questions (5 Marks Each)

Question 25. The population of City A is 37,45,280 and City B is 1,02,95,800. A new survey predicts that the population of City A will increase by 8,45,000 in the next 5 years, and City B’s population will rise by 5,00,000 in the same period.
(a) Estimate and find the future population of both cities.
(b) Compare their future populations using <, >, or =.
(c) Round off the future population of both cities to the nearest lakh.
Answer:
(a) Estimate (round to the nearest lakh): City A ≈ 37 lakh + 8 lakh = 45 lakh; City B ≈ 103 lakh + 5 lakh = 108 lakh.
Exact future population:
City A = 37,45,280 + 8,45,000 = 45,90,280.
City B = 1,02,95,800 + 5,00,000 = 1,07,95,800.
Both exact answers are close to the estimates, so they are reasonable.
(b) 45,90,280 has 7 digits and 1,07,95,800 has 8 digits, so 45,90,280 < 1,07,95,800. City B will still have more people.
(c) City A: 45,90,280 lies between 45 lakh and 46 lakh; the thousands part 90,280 is more than 50,000, so it rounds up to 46,00,000 (46 lakh).
City B: 1,07,95,800 lies between 107 lakh and 108 lakh; 95,800 is more than 50,000, so it rounds up to 1,08,00,000 (108 lakh = 1.08 crore).
In simple words: City A will have 45,90,280 people (about 46 lakh) and City B 1,07,95,800 (about 1.08 crore), so City A < City B.

Teacher's Note:
1. When comparing, a number with more digits is greater.
2. For rounding to the nearest lakh, look at the part after the lakh place: 50,000 or more rounds up.

Question 26. 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) Total cost = 6 × 120 + 4 × 15 + 2 × 350 (in ₹).
(b) 6 × 120 = 720 (books), 4 × 15 = 60 (pens), 2 × 350 = 700 (bags).
Total cost = 720 + 60 + 700 = ₹1480.
(c) ₹2000 is more than ₹1480, so there is no shortage; Mr. Arjun must return a balance.
Balance = 2000 − 1480 = ₹520 returned to the customer.
In simple words: The supplies cost ₹1480, so from ₹2000 the customer gets ₹520 back.

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

Question 27. A tailor bought two rolls of fabric:
Roll 1: 15.275 m
Roll 2: 9.65 m
(a) Find the total fabric length.
(b) If 12.5 m was used, how much fabric was left?
(c) Convert the leftover length into centimetres.
(d) If 80 cm of fabric is needed for one shirt, how many shirts can be made?
(e) How much fabric will remain unused after making the shirts?
Answer:
(a) Total = 15.275 + 9.650 = 24.925 m.
(b) Left = 24.925 − 12.500 = 12.425 m.
(c) 1 m = 100 cm, so 12.425 m = 12.425 × 100 cm = 1242.5 cm.
(d) 1242.5 ÷ 80: 80 × 15 = 1200 and 80 × 16 = 1280, which is more than 1242.5.
So the tailor can make 15 shirts (a shirt cannot be made from a part piece).
(e) Fabric used for shirts = 15 × 80 = 1200 cm. Unused = 1242.5 − 1200 = 42.5 cm.
In simple words: Out of 24.925 m, 12.425 m (1242.5 cm) is left; it makes 15 shirts with 42.5 cm to spare.

Teacher's Note:
1. Write 9.65 as 9.650 and 12.5 as 12.500 to line up decimals.
2. Only whole shirts count, so take the whole-number part of the division.

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

Question 28. A mobile phone factory currently produces 18,000 phones daily. Management plans to expand production to 25,000 phones daily starting next year. Based on daily output, they need an annual production estimate to prepare raw materials, staffing, and storage facilities. Without calculating large numbers efficiently, the company risks running short of resources or overstocking products.
Based on this, answer the following questions:
(i) Calculate current annual production (365 days). (1)
(ii) Calculate future projected annual production. (1)
(iii) Express both in lakhs and crores. (2)
Answer:
(i) Current annual production = 18,000 × 365.
18 × 365 = 6570, so 18,000 × 365 = 65,70,000 phones.
(ii) Future annual production = 25,000 × 365.
25 × 365 = 9125, so 25,000 × 365 = 91,25,000 phones.
(iii) 1 lakh = 1,00,000 and 1 crore = 100 lakh.
Current: 65,70,000 = 65.7 lakh = 65.7 ÷ 100 crore = 0.657 crore.
Future: 91,25,000 = 91.25 lakh = 91.25 ÷ 100 crore = 0.9125 crore.
In simple words: The factory makes 65.7 lakh phones a year now and will make 91.25 lakh (just under 1 crore) after expanding.

Teacher's Note:
1. Multiply 18 × 365 first, then attach the three zeros.
2. To change lakhs to crores, divide by 100.

Question 29. Arjun is organizing a weekend picnic for his friends and has taken charge of purchasing the snacks and refreshments. He decides to buy 6 packets of freshly made sandwiches priced at ₹50 each, 8 bottles of chilled juice costing ₹30 each, and 12 packets of crispy chips priced at ₹20 each. He visits a nearby store that offers a special flat discount of ₹100 on the total bill, regardless of the purchase amount. Happy to save some money, Arjun calculates the total cost of all the items, subtracts the discount, and prepares the final payment for the shopkeeper.
(a) Write the arithmetic expression for the total cost after discount and calculate amount to pay.
(b) If Arjun buys 3 more packets of chips and the price per chip packet increases by 10%, find new total after discount.
Answer:
(a) Expression: 6 × 50 + 8 × 30 + 12 × 20 − 100.
= 300 + 240 + 240 − 100 = 780 − 100 = ₹680 to pay.
(b) New price of one chip packet: 10% of ₹20 = ₹2, so the price becomes 20 + 2 = ₹22.
New number of chip packets = 12 + 3 = 15, costing 15 × 22 = ₹330.
New total after discount = 6 × 50 + 8 × 30 + 15 × 22 − 100 = 300 + 240 + 330 − 100 = ₹770.
In simple words: Arjun pays ₹680 at first; with 15 chip packets at ₹22 each he would pay ₹770.

Teacher's Note:
1. The flat discount is subtracted once from the whole bill.
2. 10% of a number is one-tenth of it: 10% of 20 is 2.