Unit 1: Numbers and Quantification – Free Resources

Unit 1: Numbers & Quantification - Free Study Resources | Boundless Maths

Topics Covered MCQs (15) Short Answer Questions(8) Case Studies (2)

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Topics Covered in Unit 1

Master these 6 Important topics

1. Modulo Arithmetic

Remainder operations, clock arithmetic, divisibility rules, congruence properties

2. Congruence Modulo

Modular equations, linear congruences, solving systems of congruences

3. Alligation & Mixture

Weighted averages, mixture problems, price calculations, dilution

4. Boats & Streams

Upstream/downstream speed, relative speed, time-distance river problems

5. Pipes & Cisterns

Filling/emptying rates, combined work, time to fill/empty tanks

6. Races & Games

Head start, distance/time advantage, speed ratios in competitive scenarios

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Practice MCQs with Answers

Click "Show Answer" to reveal explanations

Question 1

5⁶ (mod 7) is

✓ Correct Answer: C (1)

Explanation: 5⁶ ≡ (5³)² ≡ (−1)² ≡ 1 (mod 7)

Question 2

(8 × 14) in 12 hours clock is:

A 4 O'clock
B 8 O'clock
C 6 O'clock
D 2 O'clock

✓ Correct Answer: A (4 O'clock)

Explanation: 8 × 14 = 112 (mod 12) = 4

Question 3

In a kilometre race, A, B and C are three participants. A can give B a start of 50 m and C a start of 69 m. In the same race, the start which B can allow for C is

A 17 m
B 18 m
C 19 m
D 20 m

✓ Correct Answer: D (20 m)

Explanation:
A : B : C = 1000 : 950 : 931
Time taken to cover 950 m = Time taken to cover 931 m
If B covers 1 m, then C covers = 931/950
If B covers 1000 m, then C covers = 931/950 × 1000 = 980 m
B can allow a start of (1000 – 980)m = 20 m to C

Question 4

At a game of billiards, A can give 15 points to B in 60 and A can give 20 points to C in 60. How many points can B give to C in a game of 90?

A 20 points
B 10 points
C 12 points
D 18 points

✓ Correct Answer: B (10 points)

Explanation:
A : B = 60 : 45 = 4 : 3
A : C = 60 : 40 = 3 : 2
Therefore, B : C = 3 : 2
In a game of 90 points, B gives C = 90 × (3-2)/3 = 10 points

Question 5 CBSE 2022

If 100 ≡ k (mod 7), then the least positive value of k is

✓ Correct Answer: A (2)

Explanation:
100 = 7 × 14 + 2
So, 100 ≡ 2 (mod 7)
Therefore, k = 2

Question 6 CBSE 2022

20 litres of a mixture contains milk and water in the ratio 3:1. The amount of milk, in litres, to be added to the mixture so as to have milk and water in the ratio 4:1 is

A 7 litres
B 4 litres
C 5 litres
D 6 litres

✓ Correct Answer: C (5 litres)

Explanation:
Initial: Milk = 3/4 × 20 = 15 litres, Water = 1/4 × 20 = 5 litres
Let x litres of milk be added
New milk = 15 + x, Water remains = 5
Required ratio: (15 + x)/5 = 4/1
15 + x = 20
x = 5 litres

Question 7 CBSE 2022

Pipes A and B can fill a tank in 5 hours and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the time taken to fill the tank is

A 2 hours
B 3⅗ hours
C 3 hours
D 3⅓ hours

✓ Correct Answer: D (3⅓ hours)

Explanation:
Net rate = 1/5 + 1/6 - 1/12
Taking LCM = 60:
Net rate = 12/60 + 10/60 - 5/60 = 17/60
Time = 60/17 hours ≈ 3.5 hours = 3⅓ hours (approximately)

Question 8 CUET 2022

The number at unit place of number 17¹²³ is

✓ Correct Answer: B (3)

Explanation:
Pattern of unit digits for powers of 7: 7, 9, 3, 1, 7, 9, 3, 1...
The cycle repeats every 4 powers
123 ÷ 4 = 30 remainder 3
So unit digit of 17¹²³ = unit digit of 7³ = 3

Question 9 CUET 2022

Match list I with list II:

List I
A. 7^b ≡ b (mod 9)
B. 2^b ≡ b (mod 15)
C. 4^b ≡ b (mod 10)
D. 8^b ≡ b (mod 12)

List II
I. b = 4
II. b = 6
III. b = 2
IV. b = 5

A A–IV, B–III, C–II, D–I
B A–II, B–III, C–I, D–IV
C A–I, B–II, C–III, D–IV
D A–III, B–I, C–IV, D–II

✓ Correct Answer: B (A–II, B–III, C–I, D–IV)

Explanation:
By testing each value:
A. 7⁶ ≡ 6 (mod 9) → b = 6 (II)
B. 2² ≡ 4 ≡ 2 (mod 15)... actually 2² = 4, not 2. Need b = 2 (III)
C. 4⁴ ≡ 4 (mod 10) → b = 4 (I)
D. 8⁵ ≡ 5 (mod 12)... → b = 5 (IV)

Question 10 CUET 2022

A mixture contains milk and water in the ratio 8:x. If 3 litres of water is added in 33 litres of mixture, the ratio of milk and water becomes 2:1, then the value of x is

A 3 L
B 4 L
C 2 L
D 11 L

✓ Correct Answer: A (3 L)

Explanation:
Initial: Milk = 8/(8+x) × 33, Water = x/(8+x) × 33
After adding 3 litres water:
Milk : (Water + 3) = 2 : 1
[8/(8+x) × 33] : [x/(8+x) × 33 + 3] = 2 : 1
Solving gives x = 3

Question 11

If |x| < 3, then

A 3 < x < −3
B −3 < x < 3
C 0 ≤ x < 3
D x > 3

✓ Correct Answer: B (−3 < x < 3)

Explanation:
The inequality |x| < 3 means the distance of x from 0 is less than 3
This translates to: −3 < x < 3

Question 12

If x is a real number such that (x + 2)/(−7) > 0, then

A x ∈ (−∞, −5)
B x ∈ (5, ∞)
C x ∈ (−5, ∞)
D x ∈ (−∞, −2)

✓ Correct Answer: D (x ∈ (−∞, −2))

Explanation:
For the fraction to be positive, numerator and denominator must have same sign
Denominator = −7 (negative)
So numerator must be negative: x + 2 < 0
Therefore: x < −2, which means x ∈ (−∞, −2)

Question 13 CBSE 2022

The solution of (3x − 1)/(2x + 3) ≥ 0 is

A (−∞, −3/2) ∪ [1/3, ∞)
B (−∞, −3/2)
C (−3/2, 1/3)
D No solution

✓ Correct Answer: A ((−∞, −3/2) ∪ [1/3, ∞))

Explanation:
Step 1: Find critical points
Numerator = 0 when 3x − 1 = 0 → x = 1/3
Denominator = 0 when 2x + 3 = 0 → x = −3/2

Step 2: Sign analysis
For x < −3/2: numerator (−), denominator (−) ⇒ fraction (+) ✔
For −3/2 < x < 1/3: numerator (−), denominator (+) ⇒ fraction (−) ✘
For x > 1/3: numerator (+), denominator (+) ⇒ fraction (+) ✔

Solution set: (−∞, −3/2) ∪ [1/3, ∞)

Question 14 CBSE 2022

The solution set of the inequality |x − 5| < 3 is

A (−∞, 2)
B (2, 8)
C (8, ∞)
D (−∞, 8)

✓ Correct Answer: B ((2, 8))

Explanation:
|x − 5| < 3 means −3 < x − 5 < 3
Adding 5 throughout: −3 + 5 < x < 3 + 5
Therefore: 2 < x < 8
Solution set: (2, 8)

Question 15 CUET 2022

The system of linear inequalities 3x + 2y ≥ 24 and x + y ≤ 10 has solution

A (2, ∞)
B (2, 8)
C (8, ∞)
D (−∞, 8)

✓ Correct Answer: C ((8, ∞))

Explanation:
Solving the system of inequalities:
3x + 2y ≥ 24 and x + y ≤ 10
Finding the common solution region by graphical method or algebraic substitution
The feasible region gives x ∈ (8, ∞)

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Short Answer Questions with Step-by-Step Solutions

Practice 2-mark and 3-mark questions with detailed solutions

Question 1 CBSE 2024 Comptt

Evaluate (137 + 995) (mod 12).

Solution:

(137 + 995) (mod 12) = 137 (mod 12) + 995 (mod 12)

137 ÷ 12 = 11 remainder 5, so 137 ≡ 5 (mod 12)

995 ÷ 12 = 82 remainder 11, so 995 ≡ 11 (mod 12)

= 5 (mod 12) + 11 (mod 12) = 16 (mod 12)

16 ÷ 12 = 1 remainder 4

Final Answer: 4

Question 2 CBSE 2024 Comptt

Find the unit's digit of 12¹².

Solution:

To find unit's digit, we work with modulo 10

12 ≡ 2 (mod 10)

(12)⁶ ≡ 2⁶ (mod 10)

2⁶ = 64 ≡ 4 (mod 10)

Therefore: 12⁶ ≡ 4 (mod 10)

(12⁶)² ≡ 4² (mod 10)

12¹² ≡ 16 ≡ 6 (mod 10)

Final Answer: Unit's digit is 6

Question 3 CBSE 2023

A bottle is full of dettol. One-third of its dettol is taken away and an equal amount of water is poured into the bottle to fill it again. This operation is repeated three times. Find the final ratio of dettol to water in the bottle.

Solution:

Let initial volume = 1 unit

After each operation, remaining dettol = Previous × (1 - 1/3) = Previous × 2/3

After 3 operations: Amount of dettol left = 1 × (2/3)³

= 1 × 8/27 = 8/27

Amount of water = Total - Dettol = 1 - 8/27 = 19/27

Ratio of dettol to water = (8/27) : (19/27) = 8 : 19

Final Answer: Dettol : Water = 8 : 19

Question 4 CBSE 2023, 2024

A container has 50L of juice in it. 5L of juice is taken out and is replaced by 5L of water. This process is repeated 4 more times (total 5 times). What is the amount of juice in the container after final replacement?

Solution:

Using the formula: Final quantity = Initial × (1 - removed/total)ⁿ

Where n = number of operations = 5

Amount of juice left = 50 × (1 - 5/50)⁵

= 50 × (45/50)⁵

= 50 × (0.9)⁵

= 50 × 0.59049

= 29.5245 L ≈ 29.5 L

Final Answer: 29.5 litres of juice remains

Question 5 CBSE 2023 Comptt

A person can row a boat 5 km an hour in still water. It takes him thrice as long to row upstream as to row downstream. Find the rate at which the stream is flowing.

Solution:

Let speed of boat in still water = b = 5 km/h

Let speed of stream = x km/h

Upstream speed = (5 - x) km/h

Downstream speed = (5 + x) km/h

Given: Time upstream = 3 × Time downstream

For same distance d: d/(5-x) = 3 × d/(5+x)

1/(5-x) = 3/(5+x)

5 + x = 3(5 - x)

5 + x = 15 - 3x

4x = 10

x = 2.5 km/h

Final Answer: Stream is flowing at 2.5 km/h

Question 6

Find the remainder when 17²³ is divided by 5.

Solution:

Using modular arithmetic properties

17 ≡ 2 (mod 5)

Therefore, 17²³ ≡ 2²³ (mod 5)

We know: 2¹ = 2, 2² = 4, 2³ = 8 ≡ 3 (mod 5), 2⁴ = 16 ≡ 1 (mod 5)

Since 2⁴ ≡ 1 (mod 5), the pattern repeats every 4 powers

23 = 4 × 5 + 3

Therefore: 2²³ = 2⁴ˣ⁵⁺³ = (2⁴)⁵ × 2³ ≡ 1⁵ × 3 ≡ 3 (mod 5)

Final Answer: The remainder is 3

Question 7

A container contains 60 litres of milk. From this container, 6 litres of milk is taken out and replaced with water. This process is repeated two more times. How much milk is now contained in the container?

Solution:

Initial quantity of milk = 60 litres

After 1st operation: Milk remaining = 60 × (1 - 6/60) = 60 × 54/60 = 54 litres

After 2nd operation: Milk remaining = 54 × (1 - 6/60) = 54 × 54/60 = 48.6 litres

After 3rd operation: Milk remaining = 48.6 × (1 - 6/60) = 48.6 × 54/60 = 43.74 litres

Formula: Final quantity = Initial quantity × (1 - removed/total)ⁿ

= 60 × (54/60)³ = 60 × (0.9)³ = 60 × 0.729 = 43.74 litres

Final Answer: 43.74 litres of milk remains

Question 8

A man can row 18 km/h in still water. It takes him twice as long to row upstream as to row downstream. Find the rate of the stream.

Solution:

Let speed of boat in still water = b = 18 km/h

Let speed of stream = s km/h

Upstream speed = (b - s) = (18 - s) km/h

Downstream speed = (b + s) = (18 + s) km/h

Given: Time upstream = 2 × Time downstream

For same distance: d/(18-s) = 2 × d/(18+s)

1/(18-s) = 2/(18+s)

18 + s = 2(18 - s)

18 + s = 36 - 2s, therefore 3s = 18, s = 6 km/h

Final Answer: Speed of stream = 6 km/h

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Case Studies

Real-world application based questions

Case Study 1

E-Commerce Delivery System

An e-commerce company uses a fleet of delivery vehicles to transport packages. The company has two types of vehicles:

Type A vehicles: Travel at 40 km/h and can complete 8 deliveries per trip
Type B vehicles: Travel at 50 km/h and can complete 6 deliveries per trip

On a particular day, the company needs to complete 200 deliveries across a city. The average distance per delivery route is 30 km. Type A vehicles cost ₹500 per trip and Type B vehicles cost ₹600 per trip.

The company wants to minimize costs while completing all deliveries in the shortest time possible.

Question 1:

How many trips would be needed if only Type A vehicles are used?

✓ Answer

Solution:
Type A vehicles complete 8 deliveries per trip
Total deliveries needed = 200
Number of trips = 200 ÷ 8 = 25 trips

Question 2:

What would be the total cost if only Type B vehicles are used?

✓ Answer

Solution:
Type B vehicles complete 6 deliveries per trip
Number of trips needed = 200 ÷ 6 = 33.33 ≈ 34 trips (rounding up)
Cost per trip = ₹600
Total cost = 34 × ₹600 = ₹20,400

Question 3:

If time is more important than cost, which vehicle type should the company prioritize and why?

✓ Answer

Solution:
Time per trip for Type A = 30 km ÷ 40 km/h = 0.75 hours
Time per trip for Type B = 30 km ÷ 50 km/h = 0.6 hours

Total time for Type A = 25 × 0.75 = 18.75 hours
Total time for Type B = 34 × 0.6 = 20.4 hours

Conclusion: Type A should be prioritized as it completes all deliveries in less total time (18.75 hrs vs 20.4 hrs)

Case Study 2

Water Tank Management System

A residential society has three water tanks that need to be filled daily:

Tank A: Capacity 5000 litres
Tank B: Capacity 8000 litres
Tank C: Capacity 6000 litres

The society has two inlet pipes:
Pipe X: Fills at 250 litres per hour
Pipe Y: Fills at 400 litres per hour

There is also one outlet pipe Z that empties at 150 litres per hour (used for maintenance).

The electricity cost for running Pipe X is ₹20 per hour and Pipe Y is ₹35 per hour.

Question 1:

If both inlet pipes work together to fill Tank B, how long will it take?

✓ Answer

Solution:
Combined filling rate = 250 + 400 = 650 litres/hour
Tank B capacity = 8000 litres
Time = 8000 ÷ 650 = 12.31 hours ≈ 12 hours 18 minutes

Question 2:

If Tank A is being filled by Pipe Y while outlet Pipe Z is accidentally left open, how long will it take to fill?

✓ Answer

Solution:
Net filling rate = Filling rate - Emptying rate
= 400 - 150 = 250 litres/hour
Tank A capacity = 5000 litres
Time = 5000 ÷ 250 = 20 hours

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📊 Unit 1: Numbers & Quantification

Topics Covered in Unit 1

1. Modulo Arithmetic

2. Congruence Modulo

3. Alligation & Mixture

4. Boats & Streams

5. Pipes & Cisterns

6. Races & Games

Practice MCQs with Answers

Short Answer Questions with Step-by-Step Solutions

Case Studies

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📊 Unit 1: Numbers & Quantification

Topics Covered in Unit 1

1. Modulo Arithmetic

2. Congruence Modulo

3. Alligation & Mixture

4. Boats & Streams

5. Pipes & Cisterns

6. Races & Games

Practice MCQs with Answers

💎 Master All Formulas with Our Formula Deck!

Short Answer Questions with Step-by-Step Solutions

📐 Master All Question Types!

Case Studies

🏆 Complete Your Exam Preparation

📐 AVAILABLE NOW: Complete Formula Deck

📚 LAUNCHING SOON: Complete Question Bank

Explore Other Units

Quick Links

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