Chapter 6 of 15
Chapter 6: The Dairy Farm
दुग्धशाला (धारिता, लीटर-मिलीलीटर एवं व्यापारिक गणना) | CBSE Session 2026–27
Core Pedagogical Competencies & Concepts
- Standard units of liquid volume: 1 Litre (L) = 1,000 millilitres (mL).
- Common fractional benchmarks: 1/2 L = 500 mL, 1/4 L = 250 mL, 3/4 L = 750 mL.
- Adding and subtracting compound units of Litres and Millilitres.
- Packaging milk into pouches and cartons; commercial billing at given rates per litre.
- (a) True: Since $6 = 2 \times 3$, any number divisible by 6 is divisible by both of its factors (2 and 3).
- (b) False: 1 has only one factor (itself). A prime number must have exactly two distinct factors (1 and the number itself).
- (c) True: 1 is a factor of every integer, hence it is always the smallest common factor.
All Textbook Exercises & In-Text Tasks (24 Questions)
100% Complete Ingestion
Q1
Dairy Farm Operations • Daily Milk Production
A dairy cooperative farm has 12 cows. Each cow yields an average of 8 litres of milk per day.
(a) How many litres of milk does the farm produce in one day?
(b) How many litres are produced in the month of June (30 days)?
(c) If milk is sold at ₹52 per litre, calculate the total revenue in June.
[3 Marks]
(a) How many litres of milk does the farm produce in one day?
(b) How many litres are produced in the month of June (30 days)?
(c) If milk is sold at ₹52 per litre, calculate the total revenue in June.
✅ Step-by-Step Solution & Working
Calculations:
• (a) Daily Milk: 12 cows × 8 L/cow = 96 litres per day.
• (b) June Production (30 days): 96 L × 30 = 2,880 litres.
• (c) June Revenue: 2,880 L × ₹52 = ₹1,49,760.
• (a) Daily Milk: 12 cows × 8 L/cow = 96 litres per day.
• (b) June Production (30 days): 96 L × 30 = 2,880 litres.
• (c) June Revenue: 2,880 L × ₹52 = ₹1,49,760.
CBSE Step-by-Step Marking Scheme
Daily milk = 96 L
1.0 M
June milk = 2,880 L
1.0 M
Revenue = ₹1,49,760
1.0 M
Q2
Capacity Distribution • Can Sizes
A milk booth receives 65 litres of milk in the morning. They fill:
• 40 packets of 500 mL each
• 30 bottles of 1 Litre each
(a) What is the total volume of milk packaged in packets and bottles?
(b) How much loose milk remains in the can?
[3 Marks]
• 40 packets of 500 mL each
• 30 bottles of 1 Litre each
(a) What is the total volume of milk packaged in packets and bottles?
(b) How much loose milk remains in the can?
✅ Step-by-Step Solution & Working
Volume Calculations:
• (a) Packaged Milk:
- 40 packets × 500 mL = 20,000 mL = 20 Litres.
- 30 bottles × 1 L = 30 Litres.
- Total packaged = 20 L + 30 L = 50 Litres.
• (b) Loose Milk Remaining: 65 L - 50 L = 15 Litres.
• (a) Packaged Milk:
- 40 packets × 500 mL = 20,000 mL = 20 Litres.
- 30 bottles × 1 L = 30 Litres.
- Total packaged = 20 L + 30 L = 50 Litres.
• (b) Loose Milk Remaining: 65 L - 50 L = 15 Litres.
CBSE Step-by-Step Marking Scheme
Packets = 20 L; Bottles = 30 L
1.0 M
Total packaged = 50 L
1.0 M
Remaining = 15 L
1.0 M
Q3
Unit Conversions • Litres and Millilitres
Calculate:
(a) Add: 15 L 450 mL + 8 L 850 mL
(b) Subtract: 20 L - 14 L 350 mL
(c) Convert 3,750 mL into Litres and Millilitres.
[3 Marks]
(a) Add: 15 L 450 mL + 8 L 850 mL
(b) Subtract: 20 L - 14 L 350 mL
(c) Convert 3,750 mL into Litres and Millilitres.
✅ Step-by-Step Solution & Working
Calculations:
• (a) Addition: 450 mL + 850 mL = 1,300 mL = 1 L 300 mL. 15 L + 8 L + 1 L = 24 L 300 mL.
• (b) Subtraction: 20 L 000 mL - 14 L 350 mL = 19 L 1,000 mL - 14 L 350 mL = 5 L 650 mL.
• (c) Conversion: 3,750 mL = 3 Litres 750 mL (or 3.75 L).
• (a) Addition: 450 mL + 850 mL = 1,300 mL = 1 L 300 mL. 15 L + 8 L + 1 L = 24 L 300 mL.
• (b) Subtraction: 20 L 000 mL - 14 L 350 mL = 19 L 1,000 mL - 14 L 350 mL = 5 L 650 mL.
• (c) Conversion: 3,750 mL = 3 Litres 750 mL (or 3.75 L).
CBSE Step-by-Step Marking Scheme
Part (a): 24 L 300 mL
1.0 M
Part (b): 5 L 650 mL
1.0 M
Part (c): 3 L 750 mL
1.0 M
Q4
Common Multiples • Number Line Intersections
Find the first three common multiples of 4 and 6. Explain how you find them using lists of multiples and locate their Least Common Multiple (LCM).
[3 Marks]
✅ Step-by-Step Solution & Working
Finding Common Multiples:
1. Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40...
2. Multiples of 6: 6, 12, 18, 24, 30, 36, 42...
3. Common Multiples: Numbers appearing in both lists are 12, 24, 36.
4. LCM: The smallest common multiple is 12. Thus, LCM(4, 6) = 12.
1. Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40...
2. Multiples of 6: 6, 12, 18, 24, 30, 36, 42...
3. Common Multiples: Numbers appearing in both lists are 12, 24, 36.
4. LCM: The smallest common multiple is 12. Thus, LCM(4, 6) = 12.
CBSE Step-by-Step Marking Scheme
Lists of multiples for 4 and 6
1.0 M
Common multiples: 12, 24, 36
1.0 M
LCM = 12
1.0 M
Q5
Factor Trees • Prime Factorization
Draw two different factor trees for the number 60 and write down its prime factorization.
[3 Marks]
✅ Step-by-Step Solution & Working
Factor Trees for 60:
1. Tree A (Starting with 6 × 10):
• 60 = 6 × 10
• 6 splits into 2 × 3
• 10 splits into 2 × 5
• Prime factors: 2, 3, 2, 5.
2. Tree B (Starting with 4 × 15):
• 60 = 4 × 15
• 4 splits into 2 × 2
• 15 splits into 3 × 5
• Prime factors: 2, 2, 3, 5.
3. Prime Factorization: 60 = 2 × 2 × 3 × 5. Both trees yield identical prime factors.
1. Tree A (Starting with 6 × 10):
• 60 = 6 × 10
• 6 splits into 2 × 3
• 10 splits into 2 × 5
• Prime factors: 2, 3, 2, 5.
2. Tree B (Starting with 4 × 15):
• 60 = 4 × 15
• 4 splits into 2 × 2
• 15 splits into 3 × 5
• Prime factors: 2, 2, 3, 5.
3. Prime Factorization: 60 = 2 × 2 × 3 × 5. Both trees yield identical prime factors.
CBSE Step-by-Step Marking Scheme
Two distinct factor trees shown
2.0 M
Prime factorization: 2 × 2 × 3 × 5
1.0 M
Q6
Multiples Problem • Packaging Butter
A dairy packs butter into boxes of 6 packets or boxes of 8 packets.
(a) What is the smallest number of butter packets that can be packed completely in either boxes of 6 or boxes of 8 without leaving any packets out?
(b) How many boxes of 6 and how many boxes of 8 will be needed for that quantity?
[3 Marks]
(a) What is the smallest number of butter packets that can be packed completely in either boxes of 6 or boxes of 8 without leaving any packets out?
(b) How many boxes of 6 and how many boxes of 8 will be needed for that quantity?
✅ Step-by-Step Solution & Working
LCM Application:
• (a) Smallest Number (LCM):
- Multiples of 6: 6, 12, 18, 24, 30...
- Multiples of 8: 8, 16, 24, 32...
- LCM(6, 8) = 24 packets.
• (b) Number of Boxes:
- Boxes of 6 = 24 ÷ 6 = 4 boxes.
- Boxes of 8 = 24 ÷ 8 = 3 boxes.
• (a) Smallest Number (LCM):
- Multiples of 6: 6, 12, 18, 24, 30...
- Multiples of 8: 8, 16, 24, 32...
- LCM(6, 8) = 24 packets.
• (b) Number of Boxes:
- Boxes of 6 = 24 ÷ 6 = 4 boxes.
- Boxes of 8 = 24 ÷ 8 = 3 boxes.
CBSE Step-by-Step Marking Scheme
LCM(6, 8) = 24 packets
1.5 M
4 boxes of 6, 3 boxes of 8
1.5 M
Q7
Factors • Listing & Pairs
(a) Write all the factors of 36.
(b) Write all the factors of 48.
(c) List all the common factors of 36 and 48. What is their Highest Common Factor (HCF)?
[3 Marks]
(b) Write all the factors of 48.
(c) List all the common factors of 36 and 48. What is their Highest Common Factor (HCF)?
✅ Step-by-Step Solution & Working
Factors Analysis:
• (a) Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36.
• (b) Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
• (c) Common Factors: 1, 2, 3, 4, 6, 12.
- Highest Common Factor (HCF): 12.
• (a) Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36.
• (b) Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
• (c) Common Factors: 1, 2, 3, 4, 6, 12.
- Highest Common Factor (HCF): 12.
CBSE Step-by-Step Marking Scheme
Factors of 36 and 48
1.5 M
Common factors & HCF = 12
1.5 M
Q8
Milk Tanker Capacity • Multi-Station Supply
A milk tanker carries 8,000 litres of chilled milk. It delivers 2,450 litres to Society Alpha, 3,180 litres to Society Beta, and the rest to a milk processing plant.
(a) How many litres are delivered to Societies Alpha and Beta combined?
(b) How many litres does the processing plant receive?
[3 Marks]
(a) How many litres are delivered to Societies Alpha and Beta combined?
(b) How many litres does the processing plant receive?
✅ Step-by-Step Solution & Working
Working:
• (a) Societies Combined: 2,450 L + 3,180 L = 5,630 litres.
• (b) Processing Plant: 8,000 L - 5,630 L = 2,370 litres.
• (a) Societies Combined: 2,450 L + 3,180 L = 5,630 litres.
• (b) Processing Plant: 8,000 L - 5,630 L = 2,370 litres.
CBSE Step-by-Step Marking Scheme
Combined societies = 5,630 L
1.5 M
Plant delivery = 2,370 L
1.5 M
Q9
Paneer Yield • Unitary Proportions
At the dairy, 5 litres of buffalo milk yield 1 kg of fresh paneer (cottage cheese).
(a) How many kilograms of paneer can be made from 45 litres of milk?
(b) How many litres of milk are required to prepare 8.5 kg of paneer?
[3 Marks]
(a) How many kilograms of paneer can be made from 45 litres of milk?
(b) How many litres of milk are required to prepare 8.5 kg of paneer?
✅ Step-by-Step Solution & Working
Proportion Calculations:
• (a) Paneer from 45 L: 45 ÷ 5 = 9 kg of paneer.
• (b) Milk for 8.5 kg Paneer: 8.5 × 5 = 42.5 litres of milk.
• (a) Paneer from 45 L: 45 ÷ 5 = 9 kg of paneer.
• (b) Milk for 8.5 kg Paneer: 8.5 × 5 = 42.5 litres of milk.
CBSE Step-by-Step Marking Scheme
45 L gives 9 kg paneer
1.5 M
8.5 kg paneer requires 42.5 L milk
1.5 M
Q10
Measuring Jugs • Filling Volumes
A milkman has only two measuring cans: 200 mL and 500 mL.
(a) How can he measure exactly 900 mL of milk using minimum pours?
(b) How can he measure exactly 1 litre (1,000 mL) using both cans?
[3 Marks]
(a) How can he measure exactly 900 mL of milk using minimum pours?
(b) How can he measure exactly 1 litre (1,000 mL) using both cans?
✅ Step-by-Step Solution & Working
Measuring Jug Solutions:
• (a) For 900 mL:
- Take two 200 mL cans and one 500 mL can: 200 + 200 + 500 = 900 mL (3 pours).
• (b) For 1,000 mL (1 Litre):
- Use the 500 mL can twice: 500 + 500 = 1,000 mL, OR
- Using both cans: one 200 mL can and two 500 mL cans minus one 200 mL can, or simpler: five 200 mL cans.
- Best combination using both cans: two 200 mL + two 300 mL (or two 500 mL cans).
• (a) For 900 mL:
- Take two 200 mL cans and one 500 mL can: 200 + 200 + 500 = 900 mL (3 pours).
• (b) For 1,000 mL (1 Litre):
- Use the 500 mL can twice: 500 + 500 = 1,000 mL, OR
- Using both cans: one 200 mL can and two 500 mL cans minus one 200 mL can, or simpler: five 200 mL cans.
- Best combination using both cans: two 200 mL + two 300 mL (or two 500 mL cans).
CBSE Step-by-Step Marking Scheme
900 mL: 500 mL + 200 mL + 200 mL
1.5 M
1,000 mL: Two 500 mL pours (or equivalent combination)
1.5 M
Q11
Ghee Production • Cost & Profit
A dairy sells pure cow ghee in jars of 500 g at ₹340 per jar.
(a) What is the cost of 1 kg of ghee?
(b) A customer buys 6 jars of ghee. How many kg of ghee does he get, and what is his total bill?
[3 Marks]
(a) What is the cost of 1 kg of ghee?
(b) A customer buys 6 jars of ghee. How many kg of ghee does he get, and what is his total bill?
✅ Step-by-Step Solution & Working
Calculations:
• (a) Cost of 1 kg: 1 kg = 1,000 g = two 500 g jars. Cost = 2 × ₹340 = ₹680 per kg.
• (b) 6 Jars Purchase:
- Total weight = 6 × 500 g = 3,000 g = 3 kg.
- Total bill = 6 × ₹340 = ₹2,040.
• (a) Cost of 1 kg: 1 kg = 1,000 g = two 500 g jars. Cost = 2 × ₹340 = ₹680 per kg.
• (b) 6 Jars Purchase:
- Total weight = 6 × 500 g = 3,000 g = 3 kg.
- Total bill = 6 × ₹340 = ₹2,040.
CBSE Step-by-Step Marking Scheme
1 kg cost = ₹680
1.0 M
Total weight = 3 kg
1.0 M
Total bill = ₹2,040
1.0 M
Q12
Milk Delivery Route • Time Schedule
A milk delivery truck starts at 4:30 AM. It spends 15 minutes at each of its 8 delivery booths and takes 10 minutes to drive between each consecutive booth (7 transit trips).
(a) Find total time spent unloading milk at the booths.
(b) Find total driving time between booths.
(c) At what time does the truck complete all deliveries?
[3 Marks]
(a) Find total time spent unloading milk at the booths.
(b) Find total driving time between booths.
(c) At what time does the truck complete all deliveries?
✅ Step-by-Step Solution & Working
Schedule Breakdown:
• (a) Unloading Time: 8 booths × 15 min = 120 minutes = 2 hours.
• (b) Driving Time: 7 trips × 10 min = 70 minutes (1 hour 10 minutes).
• (c) Completion Time:
- Total elapsed time = 2 h + 1 h 10 min = 3 hours 10 minutes.
- 4:30 AM + 3 h 10 min = 7:40 AM.
• (a) Unloading Time: 8 booths × 15 min = 120 minutes = 2 hours.
• (b) Driving Time: 7 trips × 10 min = 70 minutes (1 hour 10 minutes).
• (c) Completion Time:
- Total elapsed time = 2 h + 1 h 10 min = 3 hours 10 minutes.
- 4:30 AM + 3 h 10 min = 7:40 AM.
CBSE Step-by-Step Marking Scheme
Unloading = 120 min (2 h)
1.0 M
Driving = 70 min
1.0 M
Finish time = 7:40 AM
1.0 M
Q13
Common Factors • Venn Diagram Model
Draw or describe two overlapping circles representing factors of 20 and factors of 30:
(a) List factors in circle for 20.
(b) List factors in circle for 30.
(c) Which numbers are placed in the overlapping shared intersection?
[3 Marks]
(a) List factors in circle for 20.
(b) List factors in circle for 30.
(c) Which numbers are placed in the overlapping shared intersection?
✅ Step-by-Step Solution & Working
Venn Diagram Factors:
• Factors of 20: 1, 2, 4, 5, 10, 20.
• Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30.
• (c) Overlapping Intersection (Common Factors): 1, 2, 5, 10.
- Non-shared in 20 only: 4, 20.
- Non-shared in 30 only: 3, 6, 15, 30.
• Factors of 20: 1, 2, 4, 5, 10, 20.
• Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30.
• (c) Overlapping Intersection (Common Factors): 1, 2, 5, 10.
- Non-shared in 20 only: 4, 20.
- Non-shared in 30 only: 3, 6, 15, 30.
CBSE Step-by-Step Marking Scheme
Factors of 20 identified
1.0 M
Factors of 30 identified
1.0 M
Shared intersection: 1, 2, 5, 10
1.0 M
Q14
Dairy Cattle Feed • Ration Calculation
Each cow on the farm is fed 4 kg of green fodder and 2.5 kg of cattle grain mix twice a day.
(a) How many kilograms of feed does one cow consume in one day?
(b) How many kilograms of feed are required for all 12 cows for one day?
[3 Marks]
(a) How many kilograms of feed does one cow consume in one day?
(b) How many kilograms of feed are required for all 12 cows for one day?
✅ Step-by-Step Solution & Working
Feed Calculations:
• (a) One Cow Daily:
- Per feeding = 4 kg + 2.5 kg = 6.5 kg.
- Twice a day = 6.5 kg × 2 = 13 kg per day.
• (b) 12 Cows Daily: 12 × 13 kg = 156 kg of feed.
• (a) One Cow Daily:
- Per feeding = 4 kg + 2.5 kg = 6.5 kg.
- Twice a day = 6.5 kg × 2 = 13 kg per day.
• (b) 12 Cows Daily: 12 × 13 kg = 156 kg of feed.
CBSE Step-by-Step Marking Scheme
One cow = 13 kg/day
1.5 M
12 cows = 156 kg/day
1.5 M
Q15
Milk Adulteration Test • Density Measurement
A lactometer reading shows pure cow milk density. A sample of 1 litre of pure milk weighs 1,032 grams.
(a) Does 1 litre of milk weigh more or less than 1 litre of water (1,000 g)? By how much?
(b) What is the weight of 5 litres of pure milk in kilograms and grams?
[3 Marks]
(a) Does 1 litre of milk weigh more or less than 1 litre of water (1,000 g)? By how much?
(b) What is the weight of 5 litres of pure milk in kilograms and grams?
✅ Step-by-Step Solution & Working
Density Analysis:
• (a) Comparison: Milk weighs 1,032 g while water weighs 1,000 g. Milk weighs more than water by 32 grams per litre.
• (b) Weight of 5 Litres: 5 × 1,032 g = 5,160 g = 5 kg 160 g.
• (a) Comparison: Milk weighs 1,032 g while water weighs 1,000 g. Milk weighs more than water by 32 grams per litre.
• (b) Weight of 5 Litres: 5 × 1,032 g = 5,160 g = 5 kg 160 g.
CBSE Step-by-Step Marking Scheme
More than water by 32 g
1.5 M
5 litres = 5 kg 160 g
1.5 M
Q16
Prime and Composite Numbers • Classification
Classify the following numbers as Prime or Composite:
11, 15, 19, 21, 29, 33, 37.
Give the definition of a prime number.
[3 Marks]
11, 15, 19, 21, 29, 33, 37.
Give the definition of a prime number.
✅ Step-by-Step Solution & Working
Prime vs Composite Classification:
• Definition: A prime number is a natural number greater than 1 that has exactly two distinct factors: 1 and itself.
• Classification:
- Prime Numbers: 11, 19, 29, 37 (only divisible by 1 and itself).
- Composite Numbers: 15 (3 × 5), 21 (3 × 7), 33 (3 × 11).
• Definition: A prime number is a natural number greater than 1 that has exactly two distinct factors: 1 and itself.
• Classification:
- Prime Numbers: 11, 19, 29, 37 (only divisible by 1 and itself).
- Composite Numbers: 15 (3 × 5), 21 (3 × 7), 33 (3 × 11).
CBSE Step-by-Step Marking Scheme
Definition of prime number
1.0 M
Correct primes: 11, 19, 29, 37
1.0 M
Correct composites: 15, 21, 33
1.0 M
Q17
Dairy Churning • Curd to Buttermilk
To make buttermilk (chaas), a dairy mixes 12 litres of curd with 18 litres of cold water.
(a) What is the total volume of buttermilk prepared?
(b) If buttermilk is served in 250 mL glasses at ₹15 each, how many glasses can be poured?
(c) What is the total collection from selling all the glasses?
[3 Marks]
(a) What is the total volume of buttermilk prepared?
(b) If buttermilk is served in 250 mL glasses at ₹15 each, how many glasses can be poured?
(c) What is the total collection from selling all the glasses?
✅ Step-by-Step Solution & Working
Calculations:
• (a) Total Volume: 12 L + 18 L = 30 Litres.
• (b) Number of Glasses:
- 30 L = 30,000 mL.
- Glasses = 30,000 ÷ 250 = 120 glasses.
• (c) Total Collection: 120 × ₹15 = ₹1,800.
• (a) Total Volume: 12 L + 18 L = 30 Litres.
• (b) Number of Glasses:
- 30 L = 30,000 mL.
- Glasses = 30,000 ÷ 250 = 120 glasses.
• (c) Total Collection: 120 × ₹15 = ₹1,800.
CBSE Step-by-Step Marking Scheme
Total volume = 30 L
1.0 M
Glasses = 120
1.0 M
Collection = ₹1,800
1.0 M
Q18
HOTS • Milk Capacity Riddle
A large milk container is 3/5 full. When 14 litres of milk are added to it, it becomes 4/5 full. What is the total capacity of the container when completely full?
[3 Marks]
✅ Step-by-Step Solution & Working
Fraction & Volume Deduction:
1. Fraction added = 4/5 - 3/5 = 1/5 of the container.
2. Given: 1/5 of the total capacity = 14 litres.
3. Total capacity = 14 × 5 = 70 litres.
4. Verification: 3/5 of 70 = 42 L. Adding 14 L gives 42 + 14 = 56 L. 4/5 of 70 = 56 L. Solution verified!
1. Fraction added = 4/5 - 3/5 = 1/5 of the container.
2. Given: 1/5 of the total capacity = 14 litres.
3. Total capacity = 14 × 5 = 70 litres.
4. Verification: 3/5 of 70 = 42 L. Adding 14 L gives 42 + 14 = 56 L. 4/5 of 70 = 56 L. Solution verified!
CBSE Step-by-Step Marking Scheme
Difference = 1/5 of capacity
1.5 M
Capacity = 70 litres with verification
1.5 M
Q19
Dairy Cooperative Savings
A women's dairy cooperative saves ₹4,200 every week.
(a) How much money do they save in one year (52 weeks)?
(b) If they invest ₹1,50,000 from their savings to purchase a chilling unit, how much savings remain?
[3 Marks]
(a) How much money do they save in one year (52 weeks)?
(b) If they invest ₹1,50,000 from their savings to purchase a chilling unit, how much savings remain?
✅ Step-by-Step Solution & Working
Working:
• (a) Annual Savings: 52 × ₹4,200 = ₹2,18,400.
• (b) Remaining Savings: ₹2,18,400 - ₹1,50,000 = ₹68,400.
• (a) Annual Savings: 52 × ₹4,200 = ₹2,18,400.
• (b) Remaining Savings: ₹2,18,400 - ₹1,50,000 = ₹68,400.
CBSE Step-by-Step Marking Scheme
Annual savings = ₹2,18,400
1.5 M
Remaining savings = ₹68,400
1.5 M
Q20
Number Logic • Co-prime Numbers
(a) What are co-prime numbers?
(b) Are 8 and 15 co-prime? Justify your answer by listing their factors.
[3 Marks]
(b) Are 8 and 15 co-prime? Justify your answer by listing their factors.
✅ Step-by-Step Solution & Working
Co-prime Explanation:
• (a) Definition: Two numbers are co-prime if they have no common factor other than 1 (i.e. HCF = 1).
• (b) Analysis of 8 and 15:
- Factors of 8: 1, 2, 4, 8.
- Factors of 15: 1, 3, 5, 15.
- Common factor: Only 1.
- Conclusion: Yes, 8 and 15 are co-prime numbers.
• (a) Definition: Two numbers are co-prime if they have no common factor other than 1 (i.e. HCF = 1).
• (b) Analysis of 8 and 15:
- Factors of 8: 1, 2, 4, 8.
- Factors of 15: 1, 3, 5, 15.
- Common factor: Only 1.
- Conclusion: Yes, 8 and 15 are co-prime numbers.
CBSE Step-by-Step Marking Scheme
Definition of co-prime (HCF = 1)
1.5 M
Factors listed & affirmative conclusion
1.5 M
Q21
Ice-Cream Factory Milk Usage
An ice-cream unit uses 150 litres of condensed milk daily to produce 600 tubs of ice-cream.
(a) How many millilitres of condensed milk are used per tub?
(b) If production is increased by 50%, how many litres of milk will be required daily?
[3 Marks]
(a) How many millilitres of condensed milk are used per tub?
(b) If production is increased by 50%, how many litres of milk will be required daily?
✅ Step-by-Step Solution & Working
Calculations:
• (a) Milk per Tub: 150 L = 150,000 mL. Milk per tub = 150,000 ÷ 600 = 250 mL per tub.
• (b) 50% Increase: 150 L + (50% of 150 L) = 150 + 75 = 225 litres daily.
• (a) Milk per Tub: 150 L = 150,000 mL. Milk per tub = 150,000 ÷ 600 = 250 mL per tub.
• (b) 50% Increase: 150 L + (50% of 150 L) = 150 + 75 = 225 litres daily.
CBSE Step-by-Step Marking Scheme
250 mL per tub
1.5 M
225 litres with 50% increase
1.5 M
Q22
Mental Arithmetic • Division by 10, 100
Convert millilitres into litres using decimal point shifts:
(a) 4,500 mL
(b) 850 mL
(c) 25 mL
(d) 10,250 mL
[3 Marks]
(a) 4,500 mL
(b) 850 mL
(c) 25 mL
(d) 10,250 mL
✅ Step-by-Step Solution & Working
Shift Decimal 3 Places Left (÷ 1,000):
• (a) 4,500 mL: 4,500 ÷ 1,000 = 4.5 L.
• (b) 850 mL: 850 ÷ 1,000 = 0.85 L.
• (c) 25 mL: 25 ÷ 1,000 = 0.025 L.
• (d) 10,250 mL: 10,250 ÷ 1,000 = 10.25 L.
• (a) 4,500 mL: 4,500 ÷ 1,000 = 4.5 L.
• (b) 850 mL: 850 ÷ 1,000 = 0.85 L.
• (c) 25 mL: 25 ÷ 1,000 = 0.025 L.
• (d) 10,250 mL: 10,250 ÷ 1,000 = 10.25 L.
CBSE Step-by-Step Marking Scheme
Parts (a) & (b): 4.5 L, 0.85 L
1.5 M
Parts (c) & (d): 0.025 L, 10.25 L
1.5 M
Q23
True or False with Arithmetic Reasoning
State True or False with reasoning:
(a) Every multiple of 6 is also a multiple of 2 and 3.
(b) 1 is the smallest prime number.
(c) A capacity of 2 L 50 mL equals 2.5 Litres.
[3 Marks]
(a) Every multiple of 6 is also a multiple of 2 and 3.
(b) 1 is the smallest prime number.
(c) A capacity of 2 L 50 mL equals 2.5 Litres.
✅ Step-by-Step Solution & Working
Justifications:
• (a) True: Since 6 = 2 × 3, any multiple of 6 contains both 2 and 3 as factors, so it is divisible by both 2 and 3.
• (b) False: 1 has only one factor (itself), whereas prime numbers require exactly two distinct factors. The smallest prime number is 2.
• (c) False: 50 mL = 50 ÷ 1,000 = 0.05 L. Therefore 2 L 50 mL = 2.05 Litres (2.5 L is 2 L 500 mL).
• (a) True: Since 6 = 2 × 3, any multiple of 6 contains both 2 and 3 as factors, so it is divisible by both 2 and 3.
• (b) False: 1 has only one factor (itself), whereas prime numbers require exactly two distinct factors. The smallest prime number is 2.
• (c) False: 50 mL = 50 ÷ 1,000 = 0.05 L. Therefore 2 L 50 mL = 2.05 Litres (2.5 L is 2 L 500 mL).
CBSE Step-by-Step Marking Scheme
Part (a): True with prime factor reasoning
1.0 M
Part (b): False (2 is smallest prime)
1.0 M
Part (c): False (2.05 L vs 2.5 L)
1.0 M
Q24
Comprehensive Dairy Farm Review
A dairy cooperative has a cold storage capacity of 15,000 litres. On Monday, it starts with 4,200 L in stock. During the day, it collects 8,400 L from farmers and delivers 7,850 L to bottling plants.
(a) What is the closing stock on Monday evening?
(b) How much unused capacity remains in the storage tank?
[3 Marks]
(a) What is the closing stock on Monday evening?
(b) How much unused capacity remains in the storage tank?
✅ Step-by-Step Solution & Working
Stock Balance:
• (a) Closing Stock:
- Opening stock = 4,200 L.
- Total available = 4,200 + 8,400 = 12,600 L.
- Dispatched = 7,850 L.
- Closing stock = 12,600 - 7,850 = 4,750 litres.
• (b) Unused Capacity: 15,000 L - 4,750 L = 10,250 litres unused.
• (a) Closing Stock:
- Opening stock = 4,200 L.
- Total available = 4,200 + 8,400 = 12,600 L.
- Dispatched = 7,850 L.
- Closing stock = 12,600 - 7,850 = 4,750 litres.
• (b) Unused Capacity: 15,000 L - 4,750 L = 10,250 litres unused.
CBSE Step-by-Step Marking Scheme
Closing stock = 4,750 L
1.5 M
Unused capacity = 10,250 L
1.5 M
🌟 NEP 2020 Competency-Based Case Study
Competency Question: Community Milk Tea Stall Procurement
A tea vendor makes 1 cup of special masala chai using 75 mL of milk. On Sunday, he prepares 400 cups of tea. Milk is bought in 1-litre bottles costing ₹62 each.
Analytical Task: (a) How much milk in litres is needed for 400 cups?
(b) How many whole 1-litre bottles must he buy, and how much will he spend?
(b) How many whole 1-litre bottles must he buy, and how much will he spend?
Evaluation & Model Answer:
Suggested Model Solution:
(a) Total Milk: 400 × 75 mL = 30,000 mL = 30 Litres.
(b) Bottles and Cost: 30 bottles needed ⇒ 30 × ₹62 = ₹1,860.
Rubric: 1M concept identification + 1M step execution + 1M conclusion