✦ NCF 2023 • COMPETENCY-FOCUSED QUESTION BANK ✦
Chapter 1: Real Numbers
Class 10 Standard Mathematics (Case Study Bank)
Curated By Akash Srivastva
✨ Interactive TLMs, Worksheets & Digital Study Notes:
www.akashmaths.online
Case Study 1
Micro-Grid Synchronization
4 Marks
Green Energy Grid
In a localized green smart-grid project in a smart city, a control station synchronizes the power feed from three different renewable energy systems to ensure grid stability and prevent power surges:
1. Solar Power Inverter: Feeds power to the main grid every 12 seconds.
2. Wind Turbine Gen-Set: Feeds power to the main grid every 18 seconds.
3. Biogas Battery Storage Unit: Feeds power to the main grid every 30 seconds.
At exactly 8:00 AM, all three renewable energy systems start feeding power to the grid simultaneously for the first time.
Q1. Conceptual Understanding
1 Mark
Find the prime factorizations of the transmission intervals (12, 18, and 30 seconds) expressed in exponential form.
(A) 2² × 3, 2² × 3², 2 × 3 × 5
(B) 2² × 3, 2 × 3², 2 × 3 × 5
(C) 2 × 3², 2² × 3, 2² × 5
(D) 2³ × 3, 2 × 3², 2 × 5²
Q2. Application
1 Mark
After how many seconds (or minutes) will all three renewable energy systems feed power to the grid simultaneously again?
(A) 90 seconds (1.5 minutes)
(B) 120 seconds (2 minutes)
(C) 180 seconds (3 minutes)
(D) 360 seconds (6 minutes)
Q3. Analytical Reasoning
2 Marks
Determine the total number of times all three systems will synchronize their power feeds simultaneously between 8:00 AM and 8:30 AM (inclusive of both start and end times).
(A) 10 times
(B) 11 times
(C) 9 times
(D) 15 times
OR (Alternative Q3)
If the biogas unit is upgraded so that it feeds power every 24 seconds (instead of 30 seconds), while solar (12s) and wind (18s) remain unchanged, calculate the new simultaneous synchronization interval.
(A) 72 seconds (1 min 12 s)
(B) 48 seconds
(C) 96 seconds
(D) 144 seconds
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | Prime factorizations: • \(12 = 2^2 \times 3^1\) • \(18 = 2^1 \times 3^2\) • \(30 = 2^1 \times 3^1 \times 5^1\) |
[1 Mark] for writing all three factorizations correctly in exponential form. |
| Q2 | Simultaneous synchronization requires \(\text{LCM}(12, 18, 30)\): \(\text{LCM} = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180\text{ seconds}\). In minutes: \(180 / 60 = 3\text{ minutes}\). |
[0.5 Mark] for identifying LCM. [0.5 Mark] for obtaining 180 s (3 minutes). |
| Q3 | Duration = \(30\text{ minutes}\). Intervals count = \(\frac{30\text{ min}}{3\text{ min}} = 10\). Since start and end are inclusive, add \(1\) for \(t = 0\): \(10 + 1 = 11\text{ times}\). |
[1 Mark] for ratio \(30 / 3 = 10\). [1 Mark] for boundary inclusion giving \(11\text{ times}\). |
| Q3 (OR) | New intervals: 12 s, 18 s, 24 s. • \(12 = 2^2 \times 3\), \(18 = 2 \times 3^2\), \(24 = 2^3 \times 3\) \(\text{LCM}(12, 18, 24) = 2^3 \times 3^2 = 8 \times 9 = 72\text{ seconds}\) (1 min 12 s). |
[1 Mark] for prime factorization of 24. [1 Mark] for evaluating \(\text{LCM} = 72\text{ seconds}\). |
Case Study 2
Deep Space Telemetry
4 Marks
Space Exploration (Satellite Telemetry)
ISRO’s Deep Space Tracking Network receives telemetry data packet streams from two deep-space probes: Aditya-L1 and Chandrayaan-V. Their transmission intervals are modeled algebraically using base prime frequency channels \(x, y, z\):
• Aditya-L1 Transmission Interval (\(P\)): \(P = x^3 y^2 z\) ms
• Chandrayaan-V Transmission Interval (\(Q\)): \(Q = x y^3 z^2\) ms
Q1. Conceptual Understanding
1 Mark
Express the Highest Common Factor \(\text{HCF}(P, Q)\) of the transmission intervals in terms of \(x, y, z\).
(A) x² y z
(B) x y² z
(C) x³ y³ z²
(D) x y z
Q2. Application
1 Mark
Express the Least Common Multiple \(\text{LCM}(P, Q)\) of the transmission intervals in terms of \(x, y, z\).
(A) x³ y³ z²
(B) x⁴ y⁵ z³
(C) x² y² z²
(D) x³ y² z
Q3. Algebraic Verification
2 Marks
What is the common algebraic term obtained upon evaluating \(\text{HCF}(P, Q) \times \text{LCM}(P, Q)\) and \(P \times Q\)?
(A) x³ y⁴ z²
(B) x⁴ y⁶ z⁴
(C) x⁴ y⁵ z³
(D) x³ y⁵ z²
OR (Alternative Q3)
If prime frequencies are assigned values \(x = 2, y = 3, z = 5\), calculate the exact numerical value of \(\text{HCF}(P, Q)\) in milliseconds.
(A) 60 ms
(B) 90 ms
(C) 180 ms
(D) 270 ms
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | \(\text{HCF}\) is product of lowest powers of common prime factors: • For \(x\): \(\min(3, 1) = 1 \implies x^1\) • For \(y\): \(\min(2, 3) = 2 \implies y^2\) • For \(z\): \(\min(1, 2) = 1 \implies z^1\) \(\text{HCF}(P, Q) = x y^2 z\). |
[1 Mark] for exact algebraic expression. |
| Q2 | \(\text{LCM}\) takes greatest powers of each prime factor: • For \(x\): \(x^3\), For \(y\): \(y^3\), For \(z\): \(z^2\) \(\text{LCM}(P, Q) = x^3 y^3 z^2\). |
[1 Mark] for exact algebraic expression. |
| Q3 | \(\text{LHS} = (x y^2 z)(x^3 y^3 z^2) = x^4 y^5 z^3\). \(\text{RHS} = P \times Q = (x^3 y^2 z)(x y^3 z^2) = x^4 y^5 z^3\). Since \(\text{LHS} = \text{RHS}\), verified! |
[1 Mark] for product expansion. [1 Mark] for verifying \(\text{LHS} = \text{RHS}\). |
| Q3 (OR) | Given \(x = 2, y = 3, z = 5\). \(\text{HCF} = x y^2 z = 2 \times (3)^2 \times 5 = 2 \times 9 \times 5 = 90\text{ ms}\). |
[1 Mark] for value substitution. [1 Mark] for evaluating 90 ms. |
Case Study 3
Carbon Sequestration
4 Marks
Climate Science (Eco-Landscaping Saplings)
An environmental conservation non-profit designs a carbon sequestration forest grid using high-absorption saplings:
• Neem Saplings: 504 saplings
• Bamboo Saplings: 264 saplings
Rules: All saplings must be planted in straight, parallel rows with an equal number of saplings in each row, and each row must contain saplings of only one species.
Q1. Conceptual Understanding
1 Mark
What is the maximum number of saplings that can be planted in each row to satisfy the architect's planning rules?
(A) 12 saplings
(B) 18 saplings
(C) 24 saplings
(D) 36 saplings
Q2. Application
1 Mark
Find the prime factorization of the total combined number of saplings of both species (504 + 264 = 768).
(A) 2⁸ × 3
(B) 2⁷ × 3²
(C) 2⁶ × 3³
(D) 2⁹ × 3
Q3. Analytical Reasoning
2 Marks
Determine the minimum total number of rows required to plant the entire collection of saplings.
(A) 28 rows
(B) 32 rows
(C) 36 rows
(D) 40 rows
OR (Alternative Q3)
If 108 Neem saplings are damaged during transit (leaving 396 intact) while Bamboo remains at 264, calculate the new maximum number of saplings per row.
(A) 66 saplings
(B) 132 saplings
(C) 88 saplings
(D) 44 saplings
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | Maximum saplings per row = \(\text{HCF}(504, 264)\). • \(504 = 2^3 \times 3^2 \times 7\), \(264 = 2^3 \times 3 \times 11\) \(\text{HCF} = 2^3 \times 3 = 24\text{ saplings}\). |
[0.5 Mark] for prime factorizations. [0.5 Mark] for \(\text{HCF} = 24\). |
| Q2 | Total saplings = \(504 + 264 = 768\). Division gives \(768 = 2^8 \times 3\). |
[0.5 Mark] for sum 768. [0.5 Mark] for \(2^8 \times 3\). |
| Q3 | Neem rows = \(\frac{504}{24} = 21\). Bamboo rows = \(\frac{264}{24} = 11\). Total rows = \(21 + 11 = 32\text{ rows}\). |
[1 Mark] for row calculations. [1 Mark] for total \(32\text{ rows}\). |
| Q3 (OR) | Remaining Neem = \(504 - 108 = 396\). Bamboo = \(264\). • \(396 = 2^2 \times 3^2 \times 11\), \(264 = 2^3 \times 3 \times 11\) \(\text{New HCF} = 2^2 \times 3 \times 11 = 132\text{ saplings}\). |
[1 Mark] for prime factors of 396. [1 Mark] for \(\text{HCF} = 132\). |
Case Study 4
Sports Analytics
4 Marks
Biometric Telemetry Alignment
During an athlete treadmill stress test, sports scientists deploy two wireless monitoring devices:
1. Chest-Strap Heart Rate Monitor: Transmits packets every 15 seconds.
2. Smart-Insole Sensor: Transmits force telemetry every 25 seconds.
At \(t = 0\text{ seconds}\), both telemetry devices transmit their initial packets simultaneously.
Q1. Conceptual Understanding
1 Mark
What is the highest common factor (\(\text{HCF}\)) of the transmission intervals 15 seconds and 25 seconds?
(A) 1
(B) 3
(C) 5
(D) 15
Q2. Application
1 Mark
Calculate the exact time interval after which both data streams will transmit simultaneously again.
(A) 50 seconds
(B) 75 seconds (1 min 15 s)
(C) 90 seconds (1.5 min)
(D) 100 seconds
Q3. Analytical Reasoning
2 Marks
If the stress test runs for 20 minutes, how many times will both devices transmit at the exact same second (excluding \(t = 0\))?
(A) 16 times
(B) 17 times
(C) 15 times
(D) 20 times
OR (Alternative Q3)
If the heart-rate monitor is reconfigured to 10 seconds (insole remaining at 25 seconds), calculate the new synchronization interval.
(A) 30 seconds
(B) 50 seconds
(C) 75 seconds
(D) 100 seconds
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | • \(15 = 3 \times 5\), \(25 = 5^2\) The only common factor is \(5^1 = 5\). |
[0.5 Mark] for prime factors. [0.5 Mark] for \(\text{HCF} = 5\). |
| Q2 | Simultaneous transmission = \(\text{LCM}(15, 25) = 3 \times 25 = 75\text{ seconds}\) (1 min 15 s). | [0.5 Mark] for finding 75 s. [0.5 Mark] for 1 min 15 s. |
| Q3 | Total test time = \(20 \times 60 = 1200\text{ seconds}\). Count = \(\frac{1200}{75} = 16\). Excluding \(t = 0\), total is **16 times**. |
[1 Mark] for converting 20 min to 1200 s. [1 Mark] for quotient 16. |
| Q3 (OR) | New intervals: 10 s and 25 s. • \(10 = 2 \times 5\), \(25 = 5^2\) \(\text{LCM}(10, 25) = 2 \times 5^2 = 50\text{ seconds}\). |
[1 Mark] for factorizations. [1 Mark] for \(\text{LCM} = 50\text{ seconds}\). |
Case Study 5
Cryptographic Hashing
4 Marks
Cybersecurity (RSA-Light Passcodes)
A digital payment gateway uses an RSA-light algorithm to generate 4-digit security passcodes. The system administrator configures two active security codes:
• Code X = 1008
• Code Y = 1080
Divisors and multiples of these codes are used by the server to compute symmetric encryption session keys.
Q1. Prime Factorization
1 Mark
Express Code X (\(1008\)) as a product of prime factors in exponential form.
(A) 2⁴ × 3² × 7
(B) 2³ × 3³ × 7
(C) 2⁴ × 3 × 7²
(D) 2⁵ × 3² × 7
Q2. HCF Computation
1 Mark
Find the Highest Common Factor \(\text{HCF}(1008, 1080)\).
(A) 36
(B) 72
(C) 144
(D) 108
Q3. Fundamental Theorem Verification
2 Marks
What is the value of \(\text{LCM}(1008, 1080)\) required to verify \(\text{HCF} \times \text{LCM} = X \times Y = 1,088,640\)?
(A) 7,560
(B) 15,120
(C) 30,240
(D) 10,080
OR (Alternative Q3)
If Code W represents the smallest composite number and Code V represents the smallest prime number, find their HCF and LCM.
(A) HCF = 2, LCM = 4
(B) HCF = 1, LCM = 4
(C) HCF = 2, LCM = 2
(D) HCF = 4, LCM = 8
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | Factoring \(1008\): \(1008 = 2^4 \times 3^2 \times 7^1\). |
[0.5 Mark] for factoring steps. [0.5 Mark] for exponential form. |
| Q2 | \(1080 = 2^3 \times 3^3 \times 5^1\). \(\text{HCF}(1008, 1080) = 2^3 \times 3^2 = 8 \times 9 = 72\). |
[0.5 Mark] for factors of 1080. [0.5 Mark] for \(\text{HCF} = 72\). |
| Q3 | (a) Product has prime factors 2, 3, 5, 7, so it cannot be prime. (b) \(\text{LCM} = 2^4 \times 3^3 \times 5 \times 7 = 15,120\). \(\text{LHS} = 72 \times 15,120 = 1,088,640\). \(\text{RHS} = 1008 \times 1080 = 1,088,640\). Verified! |
[0.5 Mark] for prime reasoning. [0.5 Mark] for LCM calculation. [1 Mark] for verifying LHS = RHS. |
| Q3 (OR) | Smallest composite = 4 (\(= 2^2\)). Smallest prime = 2 (\(= 2^1\)). \(\text{HCF}(4, 2) = 2^1 = 2\); \(\text{LCM}(4, 2) = 2^2 = 4\). |
[0.5 Mark] for identifying 4. [0.5 Mark] for identifying 2. [0.5 Mark] for \(\text{HCF} = 2\). [0.5 Mark] for \(\text{LCM} = 4\). |
Case Study 6
Precision Agriculture
4 Marks
Smart Irrigation Drones
An automated organic farm deploys two agricultural drones from a central solar station at exactly 8:00 AM:
• Drone Alpha: Takes 45 minutes for one full circuit.
• Drone Beta: Takes 60 minutes for one full circuit.
Q1. Elapsed Meeting Time
1 Mark
After how many hours from the 8:00 AM launch will both drones meet again at the station for the first time?
(A) 2 hours (10:00 AM)
(B) 3 hours (11:00 AM)
(C) 4 hours (12:00 PM)
(D) 5 hours (1:00 PM)
Q2. Circuit Rounds
1 Mark
By the time they meet at the charging station for the first time, how many rounds will Drone Alpha have completed?
(A) 3 rounds
(B) 4 rounds
(C) 5 rounds
(D) 6 rounds
Q3. Multi-Drone Swarm Planning
2 Marks
Drone Gamma meets Drone Alpha and Beta at the station after exactly 6 hours. If its circuit time is a multiple of 8 between 50 and 100 minutes, find its circuit time.
(A) 64 minutes
(B) 72 minutes
(C) 80 minutes
(D) 96 minutes
OR (Alternative Q3)
Find the ratio of the number of rounds completed by Drone Alpha to Drone Beta when they meet at the station for the second time.
(A) 4 : 3
(B) 3 : 2
(C) 5 : 4
(D) 8 : 5
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | • \(45 = 3^2 \times 5\), \(60 = 2^2 \times 3 \times 5\) \(\text{LCM}(45, 60) = 2^2 \times 3^2 \times 5 = 180\text{ minutes}\). \(180 / 60 = 3\text{ hours}\) (i.e., at 11:00 AM). |
[0.5 Mark] for \(\text{LCM} = 180\text{ min}\). [0.5 Mark] for converting to 3 hours. |
| Q2 | Rounds by Alpha = \(\frac{180\text{ min}}{45\text{ min/round}} = 4\text{ rounds}\). | [0.5 Mark] for formula. [0.5 Mark] for 4 rounds. |
| Q3 | 6 hours = \(360\text{ minutes}\). We need \(\text{LCM}(180, T) = 360\). Multiples of 8: \(56, 64, 72, 80, 88, 96\). Testing factor of 360: \(360 / 72 = 5\). \(\text{LCM}(180, 72) = 360\). Thus, \(T = 72\text{ minutes}\). |
[0.5 Mark] for 360 min conversion. [0.5 Mark] for testing multiples. [1 Mark] for \(T = 72\text{ min}\). |
| Q3 (OR) | Second meeting occurs at \(2 \times 180 = 360\text{ minutes}\). • Alpha = \(360 / 45 = 8\) rounds • Beta = \(360 / 60 = 6\) rounds Ratio = \(8 / 6 = 4 : 3\). |
[1 Mark] for 8 and 6 rounds. [1 Mark] for ratio \(4 : 3\). |
Case Study 7
Modern Architecture
4 Marks
Eco-Friendly Brick Laying
A structural engineer designs an eco-resort feature wall with two parallel accent rows of recycled blocks:
• Block Type A: Length of 12 cm.
• Block Type B: Length of 18 cm.
Both rows start and end at the exact same alignment without cutting any blocks.
Q1. Minimum Wall Length
1 Mark
Find the minimum length of the wall (in cm) required to achieve perfect alignment at both ends.
(A) 24 cm
(B) 36 cm
(C) 54 cm
(D) 72 cm
Q2. Fundamental Definition
1 Mark
According to the Fundamental Theorem of Arithmetic, what makes both 12 and 18 composite numbers?
(A) They have positive prime factors other than 1 and themselves
(B) They are divisible by 2 only
(C) They cannot be factorized uniquely
(D) Their sum is an even integer
Q3. Verification & Evaluation
2 Marks
What is the evaluated product of \(\text{HCF}(12, 18) \times \text{LCM}(12, 18)\)?
(A) 108
(B) 216
(C) 432
(D) 144
OR (Alternative Q3)
If the wall length is chosen to be 180 cm, calculate the total combined number of blocks (Type A + Type B) required.
(A) 20 blocks
(B) 25 blocks
(C) 30 blocks
(D) 15 blocks
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | \(\text{LCM}(12, 18) = 2^2 \times 3^2 = 4 \times 9 = 36\text{ cm}\). | [0.5 Mark] for factorizations. [0.5 Mark] for \(\text{LCM} = 36\text{ cm}\). |
| Q2 | Fundamental Theorem states every composite number is uniquely factorizable. 12 has factors (2, 3, 4, 6) and 18 has factors (2, 3, 6, 9) beyond 1 and themselves, making them composite. |
[0.5 Mark] for theorem statement. [0.5 Mark] for composite reasoning. |
| Q3 | \(\text{HCF}(12, 18) = 6\), \(\text{LCM}(12, 18) = 36\). \(\text{LHS} = 6 \times 36 = 216\). \(\text{RHS} = 12 \times 18 = 216\). Holds true. |
[0.5 Mark] for \(\text{HCF} = 6\). [0.5 Mark] for \(6 \times 36 = 216\). [1 Mark] for conclusion. |
| Q3 (OR) | Blocks of Type A = \(180 / 12 = 15\). Blocks of Type B = \(180 / 18 = 10\). Total blocks = \(15 + 10 = 25\text{ blocks}\). |
[1 Mark] for 15 and 10. [1 Mark] for sum 25 blocks. |
Case Study 8
Warehouse Logistics
4 Marks
Automated Fulfillment Center
A fulfillment warehouse sorts promotional merchandise onto two conveyor lines:
• Belt A: Packs kits in fixed batches of 24 items.
• Belt B: Packs kits in fixed batches of 36 items.
The facility receives a single crate containing 500 promotional items to be distributed.
Q1. Common Batch Divisor
1 Mark
What is the maximum number of items that can be packed into equal-sized batches processed by either Belt A or Belt B without leftovers?
(A) 6 items
(B) 12 items
(C) 18 items
(D) 24 items
Q2. Minimum Common Volume
1 Mark
Find the minimum number of items required such that both belts can complete full batches without any leftovers.
(A) 48 items
(B) 72 items
(C) 108 items
(D) 144 items
Q3. Optimization with Constraints
2 Marks
From the 500 items, what is the maximum number of items distributed if both belts process an equal number of full batches? How many items remain in the crate?
(A) 480 items (20 leftovers)
(B) 492 items (8 leftovers)
(C) 450 items (50 leftovers)
(D) 420 items (80 leftovers)
OR (Alternative Q3)
If a smaller crate has \(N\) items (a 3-digit number divisible by both 24 and 36), find the smallest and largest possible values of \(N\).
(A) Smallest = 144, Largest = 936
(B) Smallest = 108, Largest = 972
(C) Smallest = 120, Largest = 960
(D) Smallest = 144, Largest = 984
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | Maximum batch size = \(\text{HCF}(24, 36)\). • \(24 = 2^3 \times 3\), \(36 = 2^2 \times 3^2\) \(\text{HCF} = 2^2 \times 3^1 = 12\text{ items}\). |
[0.5 Mark] for prime factors. [0.5 Mark] for \(\text{HCF} = 12\). |
| Q2 | Minimum items required = \(\text{LCM}(24, 36)\). \(\text{LCM} = 2^3 \times 3^2 = 8 \times 9 = 72\text{ items}\). |
[0.5 Mark] for factor powers. [0.5 Mark] for \(\text{LCM} = 72\). |
| Q3 | Let \(k\) be the equal number of batches. Total items = \(24k + 36k = 60k \le 500 \implies k \le 8.33\). Integral maximum \(k = 8\). Items processed = \(60 \times 8 = 480\text{ items}\). Leftovers = \(500 - 480 = 20\text{ items}\). |
[1 Mark] for \(60k \le 500 \implies k = 8\). [0.5 Mark] for 480 items. [0.5 Mark] for 20 leftovers. |
| Q3 (OR) | \(N\) must be a multiple of \(\text{LCM}(24, 36) = 72\). • Smallest 3-digit: \(72 \times 2 = 144\). • Largest 3-digit: \(\lfloor 1000 / 72 \rfloor = 13 \implies 72 \times 13 = 936\). |
[1 Mark] for multiple of 72 and finding 144. [1 Mark] for finding largest value 936. |
Live Practice: Chapter 1 Real Numbers
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