Chapter 1 of 15
Chapter 1: We the Travellers – I
हम यात्री – १ (बड़ी संख्याएँ एवं यात्रा गणना) | CBSE Session 2026–27
Core Pedagogical Competencies & Concepts
- Reading, writing and representing 4-digit, 5-digit, and 6-digit numbers using the Indian Place Value System.
- Indian Numbering System comma placement: Ones, Tens, Hundreds, Thousands (Th, T-Th), Lakhs (L, T-L).
- Place Value and Face Value: Place value of a digit = Face value × Value of place position.
- Expanded notation: e.g., 58,406 = 50,000 + 8,000 + 400 + 0 + 6.
- Forming smallest and greatest numbers from given digits and solving digit-swap puzzles.
- Travel arithmetic: Distance, Speed, Time relationship: Distance = Speed × Time.
All Textbook Exercises & In-Text Tasks (24 Questions)
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Q1
Indian Numeration • Periods & Commas
Read the following numbers, insert commas according to the Indian System of Numeration, and write their number names:
(a) 8045
(b) 20869
(c) 75034
(d) 10599
[3 Marks]
(a) 8045
(b) 20869
(c) 75034
(d) 10599
✅ Step-by-Step Solution & Working
Indian System Commas & Number Names:
• (a) 8045: Comma insertion: 8,045. Number Name: Eight thousand forty-five.
• (b) 20869: Comma insertion: 20,869. Number Name: Twenty thousand eight hundred sixty-nine.
• (c) 75034: Comma insertion: 75,034. Number Name: Seventy-five thousand thirty-four.
• (d) 10599: Comma insertion: 10,599. Number Name: Ten thousand five hundred ninety-nine.
• (a) 8045: Comma insertion: 8,045. Number Name: Eight thousand forty-five.
• (b) 20869: Comma insertion: 20,869. Number Name: Twenty thousand eight hundred sixty-nine.
• (c) 75034: Comma insertion: 75,034. Number Name: Seventy-five thousand thirty-four.
• (d) 10599: Comma insertion: 10,599. Number Name: Ten thousand five hundred ninety-nine.
CBSE Step-by-Step Marking Scheme
Part (a): 8,045 & Number name
0.75 M
Part (b): 20,869 & Number name
0.75 M
Part (c): 75,034 & Number name
0.75 M
Part (d): 10,599 & Number name
0.75 M
Q2
Number Sequences • Pattern Finding
Find the missing terms in the following number patterns by identifying the sequence rule:
(a) 456, 567, 678, _____, _____
(b) 10,100, 10,200, 10,300, _____, _____, 10,600
(c) 73,005, 72,004, 71,003, _____, _____
(d) 2,500, 5,000, 7,500, _____, _____
[3 Marks]
(a) 456, 567, 678, _____, _____
(b) 10,100, 10,200, 10,300, _____, _____, 10,600
(c) 73,005, 72,004, 71,003, _____, _____
(d) 2,500, 5,000, 7,500, _____, _____
✅ Step-by-Step Solution & Working
Pattern Rules & Solutions:
• (a) 456, 567, 678: Rule is add 111 (+111). Next: 678 + 111 = 789; then 789 + 111 = 900.
• (b) 10,100, 10,200, 10,300: Rule is add 100 (+100). Next: 10,400, then 10,500.
• (c) 73,005, 72,004, 71,003: Rule is subtract 1,001 (-1,001). Next: 70,002, then 69,001.
• (d) 2,500, 5,000, 7,500: Rule is add 2,500 (+2,500). Next: 10,000, then 12,500.
• (a) 456, 567, 678: Rule is add 111 (+111). Next: 678 + 111 = 789; then 789 + 111 = 900.
• (b) 10,100, 10,200, 10,300: Rule is add 100 (+100). Next: 10,400, then 10,500.
• (c) 73,005, 72,004, 71,003: Rule is subtract 1,001 (-1,001). Next: 70,002, then 69,001.
• (d) 2,500, 5,000, 7,500: Rule is add 2,500 (+2,500). Next: 10,000, then 12,500.
CBSE Step-by-Step Marking Scheme
Part (a): 789, 900
0.75 M
Part (b): 10,400, 10,500
0.75 M
Part (c): 70,002, 69,001
0.75 M
Part (d): 10,000, 12,500
0.75 M
Q3
Comparing Numbers • Ordering
(a) Compare using > or < signs: 68,950 and 68,905.
(b) Arrange in ascending order: 23,516; 52,823; 30,512; 30,125; 42,222.
(c) Riya took the number 26,053 and swapped two digits to make a new number between 50,000 and 60,000. Which digits did she swap, and what is the new number?
[3 Marks]
(b) Arrange in ascending order: 23,516; 52,823; 30,512; 30,125; 42,222.
(c) Riya took the number 26,053 and swapped two digits to make a new number between 50,000 and 60,000. Which digits did she swap, and what is the new number?
✅ Step-by-Step Solution & Working
Step-by-Step Working:
• (a) Comparison: In ten-thousands, thousands, and hundreds places, both numbers are equal (6, 8, 9). Looking at tens place, 5 > 0. Therefore: 68,950 > 68,905.
• (b) Ascending Order: Smallest to largest: 23,516 < 30,125 < 30,512 < 42,222 < 52,823.
• (c) Digit Swap: To get a number in the 50,000s, the ten-thousands digit (2) must be replaced by 5. Swapping digits 2 and 5 gives: 56,023 (which lies between 50,000 and 60,000).
• (a) Comparison: In ten-thousands, thousands, and hundreds places, both numbers are equal (6, 8, 9). Looking at tens place, 5 > 0. Therefore: 68,950 > 68,905.
• (b) Ascending Order: Smallest to largest: 23,516 < 30,125 < 30,512 < 42,222 < 52,823.
• (c) Digit Swap: To get a number in the 50,000s, the ten-thousands digit (2) must be replaced by 5. Swapping digits 2 and 5 gives: 56,023 (which lies between 50,000 and 60,000).
CBSE Step-by-Step Marking Scheme
Part (a): 68,950 > 68,905
1.0 M
Part (b): Correct ascending order
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Part (c): Swapped 2 & 5 to get 56,023
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Q4
Travel Arithmetic • Distance & Speed
A family travels from New Delhi to Jaipur covering 280 km in 5 hours, then from Jaipur to Udaipur covering 395 km in 7 hours.
(a) Find the total distance traveled.
(b) Find their average speed during the New Delhi to Jaipur journey.
(c) Find the total travel time for the entire journey.
[3 Marks]
(a) Find the total distance traveled.
(b) Find their average speed during the New Delhi to Jaipur journey.
(c) Find the total travel time for the entire journey.
✅ Step-by-Step Solution & Working
Calculations:
• (a) Total Distance: 280 km + 395 km = 675 km.
• (b) Average Speed (Delhi to Jaipur): Speed = Distance ÷ Time = 280 ÷ 5 = 56 km/h.
• (c) Total Travel Time: 5 hours + 7 hours = 12 hours.
• (a) Total Distance: 280 km + 395 km = 675 km.
• (b) Average Speed (Delhi to Jaipur): Speed = Distance ÷ Time = 280 ÷ 5 = 56 km/h.
• (c) Total Travel Time: 5 hours + 7 hours = 12 hours.
CBSE Step-by-Step Marking Scheme
Total distance = 675 km
1.0 M
Speed calculation = 56 km/h
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Total travel time = 12 hours
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Q5
Forming Numbers • Place Value Rules
(a) Using the digits 3, 0, 8, 5, 2 only once each, form the greatest 5-digit number and the smallest 5-digit number.
(b) Calculate the difference between the greatest and smallest numbers formed.
[3 Marks]
(b) Calculate the difference between the greatest and smallest numbers formed.
✅ Step-by-Step Solution & Working
Formation & Difference:
• Greatest 5-digit number: Arrange digits in descending order: 85,320.
• Smallest 5-digit number: 0 cannot be the leading digit. Smallest non-zero digit is 2, followed by 0, then ascending digits: 20,358.
• (b) Difference: 85,320 - 20,358 = 64,962.
• Greatest 5-digit number: Arrange digits in descending order: 85,320.
• Smallest 5-digit number: 0 cannot be the leading digit. Smallest non-zero digit is 2, followed by 0, then ascending digits: 20,358.
• (b) Difference: 85,320 - 20,358 = 64,962.
CBSE Step-by-Step Marking Scheme
Greatest number = 85,320
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Smallest number = 20,358
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Difference = 64,962
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Q6
Place Value Expansion • 6-Digit Numbers
Write the number 8,45,219 in words according to the Indian Numbering System and show its place value expansion table.
[3 Marks]
✅ Step-by-Step Solution & Working
1. Number in Words: Eight Lakh Forty-Five Thousand Two Hundred Nineteen.
2. Place Value Expansion Table:
• Lakhs (L): 8 × 1,00,000 = 8,00,000
• Ten Thousands (TTh): 4 × 10,000 = 40,000
• Thousands (Th): 5 × 1,000 = 5,000
• Hundreds (H): 2 × 100 = 200
• Tens (T): 1 × 10 = 10
• Ones (O): 9 × 1 = 9
Expanded Notation: 8,00,000 + 40,000 + 5,000 + 200 + 10 + 9.
2. Place Value Expansion Table:
• Lakhs (L): 8 × 1,00,000 = 8,00,000
• Ten Thousands (TTh): 4 × 10,000 = 40,000
• Thousands (Th): 5 × 1,000 = 5,000
• Hundreds (H): 2 × 100 = 200
• Tens (T): 1 × 10 = 10
• Ones (O): 9 × 1 = 9
Expanded Notation: 8,00,000 + 40,000 + 5,000 + 200 + 10 + 9.
CBSE Step-by-Step Marking Scheme
Correct word form
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Accurate place value table and expanded form
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Q7
Place Value Comparison • City Populations
The population of City A is 7,85,420 and City B is 7,82,990.
(a) Which city has a greater population?
(b) By how much is it greater?
(c) Round both populations to the nearest thousand.
[3 Marks]
(a) Which city has a greater population?
(b) By how much is it greater?
(c) Round both populations to the nearest thousand.
✅ Step-by-Step Solution & Working
Analysis:
• (a) Comparison: Lakhs (7) and Ten-Thousands (8) are equal. At Thousands place, City A has 5 and City B has 2. Since 5 > 2, City A has a greater population (7,85,420 > 7,82,990).
• (b) Difference: 7,85,420 - 7,82,990 = 2,430 people.
• (c) Rounding to Nearest Thousand:
- City A (7,85,420): Hundreds digit is 4 (< 5) → 7,85,000.
- City B (7,82,990): Hundreds digit is 9 (≥ 5) → 7,83,000.
• (a) Comparison: Lakhs (7) and Ten-Thousands (8) are equal. At Thousands place, City A has 5 and City B has 2. Since 5 > 2, City A has a greater population (7,85,420 > 7,82,990).
• (b) Difference: 7,85,420 - 7,82,990 = 2,430 people.
• (c) Rounding to Nearest Thousand:
- City A (7,85,420): Hundreds digit is 4 (< 5) → 7,85,000.
- City B (7,82,990): Hundreds digit is 9 (≥ 5) → 7,83,000.
CBSE Step-by-Step Marking Scheme
Correct comparison
1.0 M
Difference = 2,430
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Accurate rounding (7,85,000 & 7,83,000)
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Q8
Number Operations • Predecessor & Successor
Find the predecessor and successor of the largest 6-digit number. What is their difference?
[3 Marks]
✅ Step-by-Step Solution & Working
Step-by-Step Solution:
1. Largest 6-digit number: 9,99,999.
• Predecessor (Number - 1): 9,99,999 - 1 = 9,99,998.
• Successor (Number + 1): 9,99,999 + 1 = 10,00,000 (Ten Lakhs, smallest 7-digit number).
2. Difference: 10,00,000 - 9,99,998 = 2.
Mathematical property: The difference between the successor and predecessor of any whole number is always 2.
1. Largest 6-digit number: 9,99,999.
• Predecessor (Number - 1): 9,99,999 - 1 = 9,99,998.
• Successor (Number + 1): 9,99,999 + 1 = 10,00,000 (Ten Lakhs, smallest 7-digit number).
2. Difference: 10,00,000 - 9,99,998 = 2.
Mathematical property: The difference between the successor and predecessor of any whole number is always 2.
CBSE Step-by-Step Marking Scheme
Correct predecessor and successor
2.0 M
Accurate difference = 2
1.0 M
Q9
Rounding Off • Real-World Context
Round off the following vehicle odometer readings to the nearest hundred and nearest thousand:
(a) 43,267 km
(b) 89,849 km
(c) 15,092 km
[3 Marks]
(a) 43,267 km
(b) 89,849 km
(c) 15,092 km
✅ Step-by-Step Solution & Working
Rounding Calculations:
• (a) 43,267 km:
- Nearest 100: Tens digit is 6 (≥ 5) → 43,300 km.
- Nearest 1,000: Hundreds digit is 2 (< 5) → 43,000 km.
• (b) 89,849 km:
- Nearest 100: Tens digit is 4 (< 5) → 89,800 km.
- Nearest 1,000: Hundreds digit is 8 (≥ 5) → 90,000 km.
• (c) 15,092 km:
- Nearest 100: Tens digit is 9 (≥ 5) → 15,100 km.
- Nearest 1,000: Hundreds digit is 0 (< 5) → 15,000 km.
• (a) 43,267 km:
- Nearest 100: Tens digit is 6 (≥ 5) → 43,300 km.
- Nearest 1,000: Hundreds digit is 2 (< 5) → 43,000 km.
• (b) 89,849 km:
- Nearest 100: Tens digit is 4 (< 5) → 89,800 km.
- Nearest 1,000: Hundreds digit is 8 (≥ 5) → 90,000 km.
• (c) 15,092 km:
- Nearest 100: Tens digit is 9 (≥ 5) → 15,100 km.
- Nearest 1,000: Hundreds digit is 0 (< 5) → 15,000 km.
CBSE Step-by-Step Marking Scheme
Part (a) rounding
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Part (b) rounding
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Part (c) rounding
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Q10
Travel Budget • Railway Timetable
The train distance from Delhi to Mumbai is 1,384 km. The Rajdhani Express covers this distance in 16 hours.
(a) Find the average speed of the train in km per hour.
(b) If a passenger ticket costs ₹2,150, what is the total fare collected from a coach of 54 passengers?
[3 Marks]
(a) Find the average speed of the train in km per hour.
(b) If a passenger ticket costs ₹2,150, what is the total fare collected from a coach of 54 passengers?
✅ Step-by-Step Solution & Working
Calculations:
• (a) Average Speed: Speed = Distance ÷ Time = 1,384 ÷ 16 = 86.5 km/h.
• (b) Total Fare: Number of passengers × Fare per ticket = 54 × ₹2,150 = ₹1,16,100.
• (a) Average Speed: Speed = Distance ÷ Time = 1,384 ÷ 16 = 86.5 km/h.
• (b) Total Fare: Number of passengers × Fare per ticket = 54 × ₹2,150 = ₹1,16,100.
CBSE Step-by-Step Marking Scheme
Speed calculation: 1384 ÷ 16 = 86.5 km/h
1.5 M
Total fare: 54 × 2150 = ₹1,16,100
1.5 M
Q11
Number Line • Interval Estimation
On a number line marked from 50,000 to 60,000 with major ticks at intervals of 2,000:
(a) How many equal subdivisions lie between 50,000 and 60,000?
(b) Between which two markings does 57,400 lie?
(c) Which major marking is 57,400 closest to?
[3 Marks]
(a) How many equal subdivisions lie between 50,000 and 60,000?
(b) Between which two markings does 57,400 lie?
(c) Which major marking is 57,400 closest to?
✅ Step-by-Step Solution & Working
Number Line Analysis:
• (a) Subdivisions: Total interval = 60,000 - 50,000 = 10,000. Subdivisions = 10,000 ÷ 2,000 = 5 equal intervals (50,000; 52,000; 54,000; 56,000; 58,000; 60,000).
• (b) Position: 57,400 lies between 56,000 and 58,000.
• (c) Proximity: Distance to 56,000 = 57,400 - 56,000 = 1,400. Distance to 58,000 = 58,000 - 57,400 = 600. Since 600 < 1,400, it is closest to 58,000.
• (a) Subdivisions: Total interval = 60,000 - 50,000 = 10,000. Subdivisions = 10,000 ÷ 2,000 = 5 equal intervals (50,000; 52,000; 54,000; 56,000; 58,000; 60,000).
• (b) Position: 57,400 lies between 56,000 and 58,000.
• (c) Proximity: Distance to 56,000 = 57,400 - 56,000 = 1,400. Distance to 58,000 = 58,000 - 57,400 = 600. Since 600 < 1,400, it is closest to 58,000.
CBSE Step-by-Step Marking Scheme
Part (a): 5 intervals identified
1.0 M
Part (b): Between 56,000 and 58,000
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Part (c): Closest to 58,000
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Q12
Mental Arithmetic • Number Puzzles
I am a 5-digit number. My ten-thousands digit is 7. My thousands digit is 3 less than ten-thousands digit. My hundreds digit is 0. My tens digit is double my thousands digit, and my ones digit is 1. What number am I?
[3 Marks]
✅ Step-by-Step Solution & Working
Deduction Step-by-Step:
1. Ten-thousands digit = 7.
2. Thousands digit = 7 - 3 = 4.
3. Hundreds digit = 0.
4. Tens digit = 2 × 4 = 8.
5. Ones digit = 1.
The secret number is 74,081 (Seventy-four thousand eighty-one).
1. Ten-thousands digit = 7.
2. Thousands digit = 7 - 3 = 4.
3. Hundreds digit = 0.
4. Tens digit = 2 × 4 = 8.
5. Ones digit = 1.
The secret number is 74,081 (Seventy-four thousand eighty-one).
CBSE Step-by-Step Marking Scheme
Correct deduction of individual place values
1.5 M
Final number: 74,081
1.5 M
Q13
Distance Comparison • Highway Markers
A highway distance stone shows: Chandigarh 145 km, Shimla 260 km, Manali 495 km from Delhi.
(a) What is the distance between Chandigarh and Shimla?
(b) How much further is Manali from Chandigarh than Shimla is from Chandigarh?
(c) If an electric bus travels 45 km per hour, how long will it take to reach Shimla from Chandigarh?
[3 Marks]
(a) What is the distance between Chandigarh and Shimla?
(b) How much further is Manali from Chandigarh than Shimla is from Chandigarh?
(c) If an electric bus travels 45 km per hour, how long will it take to reach Shimla from Chandigarh?
✅ Step-by-Step Solution & Working
Calculations:
• (a) Distance Chandigarh to Shimla: 260 km - 145 km = 115 km.
• (b) Manali to Chandigarh vs Shimla to Chandigarh:
- Manali to Chandigarh = 495 km - 145 km = 350 km.
- Extra distance = 350 km - 115 km = 235 km.
• (c) Travel Time: Time = Distance ÷ Speed = 115 ÷ 45 = 2.55 hours = 2 hours 33 minutes (approx 2.56 h).
• (a) Distance Chandigarh to Shimla: 260 km - 145 km = 115 km.
• (b) Manali to Chandigarh vs Shimla to Chandigarh:
- Manali to Chandigarh = 495 km - 145 km = 350 km.
- Extra distance = 350 km - 115 km = 235 km.
• (c) Travel Time: Time = Distance ÷ Speed = 115 ÷ 45 = 2.55 hours = 2 hours 33 minutes (approx 2.56 h).
CBSE Step-by-Step Marking Scheme
Part (a): 115 km
1.0 M
Part (b): 235 km extra
1.0 M
Part (c): Time calculation (approx 2 h 33 min)
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Q14
Word Problem • Flight Fuel & Passengers
An airliner carries 180 passengers from New Delhi to Bengaluru (distance 1,740 km).
(a) If each ticket is priced at ₹4,500, find the total revenue collected.
(b) The flight consumes 4 litres of aviation fuel per kilometre. How many litres of fuel are needed for the one-way flight?
[3 Marks]
(a) If each ticket is priced at ₹4,500, find the total revenue collected.
(b) The flight consumes 4 litres of aviation fuel per kilometre. How many litres of fuel are needed for the one-way flight?
✅ Step-by-Step Solution & Working
Working:
• (a) Total Ticket Revenue: 180 × ₹4,500 = ₹8,10,000.
• (b) Fuel Required: 1,740 km × 4 L/km = 6,960 litres.
• (a) Total Ticket Revenue: 180 × ₹4,500 = ₹8,10,000.
• (b) Fuel Required: 1,740 km × 4 L/km = 6,960 litres.
CBSE Step-by-Step Marking Scheme
Ticket revenue: 180 × 4,500 = ₹8,10,000
1.5 M
Fuel: 1740 × 4 = 6,960 L
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Q15
Indian vs International System • Comparison
Write the number 450320 in:
(a) Indian System with commas and number name.
(b) International System with commas and number name.
(c) How many thousands make 1 Lakh?
[3 Marks]
(a) Indian System with commas and number name.
(b) International System with commas and number name.
(c) How many thousands make 1 Lakh?
✅ Step-by-Step Solution & Working
System Comparison:
• (a) Indian System: 4,50,320 → Four Lakh Fifty Thousand Three Hundred Twenty.
• (b) International System: 450,320 → Four Hundred Fifty Thousand Three Hundred Twenty.
• (c) Equivalence: 1 Lakh = 1,00,000 = 100 × 1,000 = 100 thousands.
• (a) Indian System: 4,50,320 → Four Lakh Fifty Thousand Three Hundred Twenty.
• (b) International System: 450,320 → Four Hundred Fifty Thousand Three Hundred Twenty.
• (c) Equivalence: 1 Lakh = 1,00,000 = 100 × 1,000 = 100 thousands.
CBSE Step-by-Step Marking Scheme
Indian system: 4,50,320 and name
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International system: 450,320 and name
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1 Lakh = 100 thousands
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Q16
HOTS • Digit Value Puzzle
In the number 5,75,425, what is the sum of the place values of the digit '5' appearing in all three positions?
[3 Marks]
✅ Step-by-Step Solution & Working
Finding Place Values of 5:
• First 5 (Lakhs place): 5 × 1,00,000 = 5,00,000.
• Second 5 (Thousands place): 5 × 1,000 = 5,000.
• Third 5 (Ones place): 5 × 1 = 5.
• Sum of place values: 5,00,000 + 5,000 + 5 = 5,05,005.
• First 5 (Lakhs place): 5 × 1,00,000 = 5,00,000.
• Second 5 (Thousands place): 5 × 1,000 = 5,000.
• Third 5 (Ones place): 5 × 1 = 5.
• Sum of place values: 5,00,000 + 5,000 + 5 = 5,05,005.
CBSE Step-by-Step Marking Scheme
Identifying 500,000; 5,000; and 5
1.5 M
Sum = 5,05,005
1.5 M
Q17
In-Text Activity • Travel Log
A family maintained a travel odometer log:
• Start at Home: 12,450 km
• Lunch stop at Highway Dhaba: 12,680 km
• Evening Arrival at Hill Resort: 12,910 km
(a) Find distance covered before lunch and after lunch.
(b) Was the travel distance split equally before and after lunch?
[3 Marks]
• Start at Home: 12,450 km
• Lunch stop at Highway Dhaba: 12,680 km
• Evening Arrival at Hill Resort: 12,910 km
(a) Find distance covered before lunch and after lunch.
(b) Was the travel distance split equally before and after lunch?
✅ Step-by-Step Solution & Working
Analysis:
• Distance before lunch = 12,680 - 12,450 = 230 km.
• Distance after lunch = 12,910 - 12,680 = 230 km.
• (b) Conclusion: Yes, both legs of the journey covered exactly 230 km each, so the travel distance was split perfectly equally.
• Distance before lunch = 12,680 - 12,450 = 230 km.
• Distance after lunch = 12,910 - 12,680 = 230 km.
• (b) Conclusion: Yes, both legs of the journey covered exactly 230 km each, so the travel distance was split perfectly equally.
CBSE Step-by-Step Marking Scheme
Before lunch: 230 km
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After lunch: 230 km
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Valid affirmative conclusion
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Q18
Place Value • Digit Interchange
A cashier entered ₹45,290 into the computer by mistake instead of ₹42,590.
(a) Which two digits were interchanged?
(b) How much extra amount was recorded?
(c) In which place values did this error occur?
[3 Marks]
(a) Which two digits were interchanged?
(b) How much extra amount was recorded?
(c) In which place values did this error occur?
✅ Step-by-Step Solution & Working
Error Analysis:
• (a) Interchanged Digits: Digits 5 and 2 were interchanged.
• (b) Excess Amount: ₹45,290 - ₹42,590 = ₹2,700.
• (c) Place Values: The error occurred between the Thousands place and Hundreds place.
• (a) Interchanged Digits: Digits 5 and 2 were interchanged.
• (b) Excess Amount: ₹45,290 - ₹42,590 = ₹2,700.
• (c) Place Values: The error occurred between the Thousands place and Hundreds place.
CBSE Step-by-Step Marking Scheme
Part (a): Digits 5 and 2
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Part (b): Difference ₹2,700
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Part (c): Thousands & Hundreds
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Q19
Real-Life Application • Toll Plaza Data
At a national highway toll plaza, 3,450 cars passed on Monday, 4,120 cars on Tuesday, and 2,980 cars on Wednesday.
(a) Find the total number of cars that passed over the three days.
(b) If each car paid ₹85 as toll, calculate the total collection on Tuesday.
[3 Marks]
(a) Find the total number of cars that passed over the three days.
(b) If each car paid ₹85 as toll, calculate the total collection on Tuesday.
✅ Step-by-Step Solution & Working
Calculations:
• (a) Total Cars: 3,450 + 4,120 + 2,980 = 10,550 cars.
• (b) Tuesday Toll Collection: 4,120 cars × ₹85 = ₹3,50,200.
• (a) Total Cars: 3,450 + 4,120 + 2,980 = 10,550 cars.
• (b) Tuesday Toll Collection: 4,120 cars × ₹85 = ₹3,50,200.
CBSE Step-by-Step Marking Scheme
Total cars = 10,550
1.5 M
Tuesday revenue = ₹3,50,200
1.5 M
Q20
Number Logic • Predecessor Riddle
The predecessor of a number $X$ is 99,999. (a) What is the number $X$? (b) How many digits does $X$ have? (c) What is the successor of $X$?
[3 Marks]
✅ Step-by-Step Solution & Working
Working:
• (a) Finding X: Predecessor = X - 1 = 99,999 → X = 99,999 + 1 = 1,00,000 (One Lakh).
• (b) Number of Digits: 1,00,000 has 6 digits (it is the smallest 6-digit number).
• (c) Successor of X: 1,00,000 + 1 = 1,00,001 (One Lakh One).
• (a) Finding X: Predecessor = X - 1 = 99,999 → X = 99,999 + 1 = 1,00,000 (One Lakh).
• (b) Number of Digits: 1,00,000 has 6 digits (it is the smallest 6-digit number).
• (c) Successor of X: 1,00,000 + 1 = 1,00,001 (One Lakh One).
CBSE Step-by-Step Marking Scheme
Part (a): X = 1,00,000
1.0 M
Part (b): 6 digits
1.0 M
Part (c): Successor = 1,00,001
1.0 M
Q21
Travel Estimation • Walking vs Cycling
Rohan's school is 3,600 metres from his house.
(a) Express this distance in kilometres.
(b) Rohan walks at 60 metres per minute. How many minutes does he take to walk to school?
(c) If he rides his bicycle at 180 metres per minute, how much time does he save?
[3 Marks]
(a) Express this distance in kilometres.
(b) Rohan walks at 60 metres per minute. How many minutes does he take to walk to school?
(c) If he rides his bicycle at 180 metres per minute, how much time does he save?
✅ Step-by-Step Solution & Working
Steps & Solutions:
• (a) In Kilometres: 3,600 m ÷ 1,000 = 3.6 km.
• (b) Walking Time: 3,600 ÷ 60 = 60 minutes (1 hour).
• (c) Cycling Time & Time Saved:
- Cycling time = 3,600 ÷ 180 = 20 minutes.
- Time saved = 60 min - 20 min = 40 minutes saved.
• (a) In Kilometres: 3,600 m ÷ 1,000 = 3.6 km.
• (b) Walking Time: 3,600 ÷ 60 = 60 minutes (1 hour).
• (c) Cycling Time & Time Saved:
- Cycling time = 3,600 ÷ 180 = 20 minutes.
- Time saved = 60 min - 20 min = 40 minutes saved.
CBSE Step-by-Step Marking Scheme
Distance: 3.6 km
1.0 M
Walking: 60 minutes
1.0 M
Time saved: 40 minutes
1.0 M
Q22
Mental Math • Quick Multiplications
Evaluate using place value breakdown:
(a) 250 × 40
(b) 1,200 × 5
(c) 99 × 35 (using 100 × 35 - 35)
[3 Marks]
(a) 250 × 40
(b) 1,200 × 5
(c) 99 × 35 (using 100 × 35 - 35)
✅ Step-by-Step Solution & Working
Mental Math Strategies:
• (a) 250 × 40: 25 × 4 = 100, append two zeros → 10,000.
• (b) 1,200 × 5: 12 × 5 = 60, append two zeros → 6,000.
• (c) 99 × 35: (100 - 1) × 35 = (100 × 35) - (1 × 35) = 3,500 - 35 = 3,465.
• (a) 250 × 40: 25 × 4 = 100, append two zeros → 10,000.
• (b) 1,200 × 5: 12 × 5 = 60, append two zeros → 6,000.
• (c) 99 × 35: (100 - 1) × 35 = (100 × 35) - (1 × 35) = 3,500 - 35 = 3,465.
CBSE Step-by-Step Marking Scheme
Part (a): 10,000
1.0 M
Part (b): 6,000
1.0 M
Part (c): 3,465 with distributive working
1.0 M
Q23
Practical Application • Bus Capacity
A state transport bus has 52 passenger seats. For an inter-city tour, 380 people register.
(a) How many buses are required to seat all passengers comfortably?
(b) How many vacant seats will remain in the last bus?
[3 Marks]
(a) How many buses are required to seat all passengers comfortably?
(b) How many vacant seats will remain in the last bus?
✅ Step-by-Step Solution & Working
Division with Remainders:
• 380 ÷ 52: 52 × 7 = 364 passengers.
• Remaining passengers = 380 - 364 = 16 passengers.
• (a) Buses Required: Since 16 passengers cannot be left behind, 7 + 1 = 8 buses are needed.
• (b) Vacant Seats in 8th Bus: 52 - 16 = 36 vacant seats.
• 380 ÷ 52: 52 × 7 = 364 passengers.
• Remaining passengers = 380 - 364 = 16 passengers.
• (a) Buses Required: Since 16 passengers cannot be left behind, 7 + 1 = 8 buses are needed.
• (b) Vacant Seats in 8th Bus: 52 - 16 = 36 vacant seats.
CBSE Step-by-Step Marking Scheme
Division: 380 ÷ 52 gives quotient 7, remainder 16
1.5 M
Total 8 buses & 36 vacant seats
1.5 M
Q24
CBSE Revision • Comprehensive Problem
A book publisher prints 25,000 copies of a Class 5 textbook.
(a) If 18,450 copies are dispatched to government schools and 4,820 copies to private bookstores, how many copies remain in the warehouse?
(b) Write the number of remaining copies in words.
[3 Marks]
(a) If 18,450 copies are dispatched to government schools and 4,820 copies to private bookstores, how many copies remain in the warehouse?
(b) Write the number of remaining copies in words.
✅ Step-by-Step Solution & Working
Calculations:
• Total books dispatched = 18,450 + 4,820 = 23,270 copies.
• (a) Remaining Copies: 25,000 - 23,270 = 1,730 copies.
• (b) In Words: One thousand seven hundred thirty.
• Total books dispatched = 18,450 + 4,820 = 23,270 copies.
• (a) Remaining Copies: 25,000 - 23,270 = 1,730 copies.
• (b) In Words: One thousand seven hundred thirty.
CBSE Step-by-Step Marking Scheme
Total dispatched = 23,270
1.0 M
Remaining copies = 1,730
1.0 M
In words: One thousand seven hundred thirty
1.0 M
🌟 NEP 2020 Competency-Based Case Study
Competency-Based Question: Highway Toll & Fuel Economy
A tour bus travels across 3 states covering 1,840 km in 4 days. The bus has a fuel tank of 150 litres and gives an average mileage of 8 km per litre of diesel. Diesel costs ₹90 per litre.
Analytical Task: (a) How many litres of diesel were consumed during the 1,840 km trip?
(b) What was the total cost of diesel consumed?
(c) If 40 tourists shared the diesel expense equally, how much did each passenger pay?
(b) What was the total cost of diesel consumed?
(c) If 40 tourists shared the diesel expense equally, how much did each passenger pay?
Evaluation & Model Answer:
Suggested Model Solution:
(a) Diesel Consumption: 1840/8 = 230 litres.
(b) Total Fuel Cost: 230 × ₹90 = ₹20,700.
(c) Share per Passenger: ₹20,700/40 = ₹517.50.
Rubric: 1M concept identification + 1M step execution + 1M conclusion