(a) (i) Apply addition modulo to positive integers 17 and 13 for modulo 30. (ii) Find subtraction modulo 8 for numbers 11 and 3.

OR

(b) Three pipes A, B and C can together fill a tank in 8 hours. After working at it together for 2 hours, B is closed and A and C fill the remaining part in 9 hours. Determine the time in which pipe B alone can fill the tank.

Numbers, Quantification and Numerical Applications

Modulo Arithmetic / Pipes and Cisterns

Official Answer

This question has two alternatives; either one earns full marks.

Option (a) - Modulo arithmetic:

(i) Addition modulo 30: $17 + 13 = 30$ , and $30 \bmod 30 = 0$ . So $(17 +_{30} 13) = 0$ .
(ii) Subtraction modulo 8: $11 - 3 = 8$ , and $8 \bmod 8 = 0$ . So $(11 -_{8} 3) = 0$ .

Option (b) - Pipes and cisterns:

Combined rate of A, B, C = $1/8$ tank per hour.
Work done in 2 hours together = $2 \times 1/8 = 1/4$ , so remaining = $3/4$ .
A and C fill $3/4$ in 9 hours, so rate of $(A + C) = (3/4) / 9 = 1/12$ per hour.
Rate of B = $\text{rate}(A+B+C) - \text{rate}(A+C) = 1/8 - 1/12 = 1/24$ per hour.

Final answer (b):

Pipe B alone fills the tank in 24 hours.

modulo arithmeticaddition modulosubtraction moduloremainderpipes and cisternsrate of work24 hourscombined rate

Marking Scheme

1Option (a): 1.5 marks for (i) $17 +_{30} 13 = 0$ ; 1.5 marks for (ii) $11 -_{8} 3 = 0$ (with working shown).
2Option (b): 1 mark for combined rate $1/8$ and remaining work $3/4$ ; 1 mark for $\text{rate}(A+C) = 1/12$ ; 1 mark for $\text{rate}(B) = 1/24$ giving 24 hours.
3Accept equivalent correct methods; final numeric answers must match.

Hint

For (a) take the remainder after dividing by the modulus. For (b) subtract the $(A+C)$ rate from the combined $(A+B+C)$ rate to get B's rate.

Quick Oral Answer

For part (a), both $17+13 \bmod 30$ and $11-3 \bmod 8$ give 0 since 30 and 8 are exact multiples of their moduli; for part (b), B's rate is $1/8$ minus $1/12$ , which is $1/24$ , so B alone takes 24 hours.

Analysis & Explanation

The two options test very different Unit 1 skills, so a student picks whichever is stronger for them.

Option (a) - Concept:

In modulo n arithmetic, the result of any operation is the remainder when the ordinary result is divided by n. A remainder of 0 means the result is an exact multiple of the modulus, which is exactly what happens for both parts here (30 is a multiple of 30, 8 is a multiple of 8).

Option (b) - Concept:

Pipes-and-cisterns problems are rate problems: express each pipe's contribution as a fraction of the tank filled per hour, add rates for simultaneous work, and subtract to isolate one pipe. The key insight is $\text{rate}(B) = \text{rate}(A+B+C) - \text{rate}(A+C)$ .

Exam trap:

In (a), students sometimes report 30 or 8 as the answer instead of reducing modulo n to 0. In (b), a frequent error is forgetting that only $3/4$ of the tank remains after the first 2 hours.

Real-world link:

Modulo arithmetic underlies clock time, ISBN/checksum digits, and cryptography; pipes-and-cisterns logic models real filling/draining of reservoirs and pipelines.

Common Mistakes

1In modulo problems, writing the raw sum/difference (30 or 8) instead of reducing it modulo n to get 0.
2In the pipes problem, forgetting that after 2 hours only $3/4$ of the tank is left, not the whole tank.
3Adding times directly (e.g., $8 + 9$ ) instead of adding and subtracting rates of work.

Interesting Facts

Modulo arithmetic is the mathematical backbone of the 12-hour clock and of the check digit in Aadhaar and credit card numbers (Luhn algorithm uses mod 10).

The 'unitary rate' technique used in pipes-and-cisterns dates back to ancient Egyptian and Indian mathematical texts on work-sharing problems.

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Frequently Asked Questions

Why is $17 + 13$ modulo 30 equal to 0 and not 30?

In modulo 30 arithmetic every result is reduced to its remainder on division by 30. Since $17 + 13 = 30$ and 30 is exactly divisible by 30, the remainder is 0. Any multiple of the modulus reduces to 0.

How do you find one pipe's time when only the combined times are given?

Convert every filling scenario into a rate (fraction of tank per hour). Add rates when pipes work together and subtract to isolate one pipe. Here $\text{rate}(B) = \text{rate}(A+B+C) - \text{rate}(A+C) = 1/8 - 1/12 = 1/24$ , so B alone needs 24 hours.