Challenging Tiffins School 11+ Maths Question

I’m diving into a Tiffins School 11+ Maths puzzle that’s been stirring up curiosity and a bit of confusion amongst parents, tutors, and students preparing for the 11+ Tiffins school examination. A special shoutout to Chipo for bringing this intriguing question to our attention. Let’s break it down together, explore the different methods to approach it, and unveil the correct way to solve it. Ready? Let’s get started!

Understanding the Tiffins School 11+ Maths Question

Chipo discovered a mathematics question that initially seemed straightforward but held a bit of a twist. Here’s how the question is presented:

• AB + C = 35
• CB + A = 80

Now, the challenge here is to find out the sum of A, B, and C. Seems simple, right? But here’s where it gets interesting because the interpretation of AB and CB can change the whole approach to solving this problem.

Initial Interpretations and Clarifications

You can think of AB in two different ways:

1. AB could be perceived as a single two-digit number with A being the tens place and B being the units.
2. Alternatively, AB and CB could be interpreted as algebraic expressions where A and B are individual numbers being added and multiplied.

Initially, I thought this was a simple number puzzle with two-digit and one-digit combinations based on the clues given. Indeed, this isn’t an algebra-focused question but a numerical puzzle centered around digit manipulation.

Breaking Down the Problem

Let’s dissect this using the numerical approach:

• For AB + C = 35, let’s consider the largest single-digit number, 9. Subtracting gives us:35 - 9 = 26Which breaks down as:AB = 26andC = 9
• Applying similar subtraction for all numbers from 1 to 9 for C, we can discover possible combinations that satisfy the equation, all ranging from AB = 26 (with C = 9) to AB = 34 (with C = 1).

Further Simplifications & Observations

Since A could be either a 2 or a 3 leading the two-digit numbers (26 to 34), let’s explore CB + A = 80 using the possibilities of A being 2 or 3:

• With A as 2 or 3, you get:CB + 2 = 80 or CB + 3 = 80This assumption simplifies to CB being either 78 or 77. However, since A, B, and C are distinct digits, recurring digits in options (particularly 77) become invalid.

“This problem beautifully demonstrates the fine differences between conventional algebra and digit-based puzzles, which can significantly alter how one perceives and tackles the problem.”

Conclusion: The Solution

Based on the constraints and logical deductions:

• A turns out to be 2.
• B is calculated as 8.
• C is determined to be 7.

These digits uniquely satisfy both equations. Hence, the sum of A, B, and C is:

A + B + C = 2 + 8 + 7 = 17

Visualizing the Problem

This simple illustration should help visual learners understand how the digits interact within the equations provided.

Key Takeaways

• Assess the Problem: Before diving into solving, ensure you understand how the terms are defined within the problem context (e.g., is AB a product of A and B, or a two-digit number?).
• Simplicity is Key: Sometimes, the most straightforward approach can unmask the solution quicker than over-complicated methods.
• Cross-check: Always verify each derived value back into the original equations to ensure consistency across.

This relatively simple question emphasizes the importance of clarifying and comprehending the problem statement fully before attempting a solution. Although understanding the context may take practice, it becomes invaluable, especially in competitive exams like the 11+ where every point counts!

I hope this walkthrough helps you or your child tackle similar questions with confidence. Remember, maths isn’t just about numbers; it’s about understanding the story those numbers tell. See you next time, and all the best on your educational journey!

Watch this video solution to the problem: