Free Online 11+ Maths Tutoring – Open to All
Free Online 11+ Maths Tutoring – Open to All Hi! I’m Rithvik Muthuvelu , a GCSE student at King Edward’s School, Birmingham , and I’m offering free weekly online maths sessions to help students prepare for the 11+ entrance exams . These sessions are open to anyone who wants to improve their maths skills—no school restrictions. What You’ll Get Free weekly online maths classes Focused 11+ preparation : problem-solving, arithmetic, word problems, exam strategies Small-group format for better interaction Ideal for Year 4 and Year 5 students How to Join Weekly Session: Saturdays at 2:00 PM Google Meet Link: https://meet.google.com/nrk-iwmh-gij Contact Email: rithvikmu1@gmail.com If your child is preparing for the 11+ and would like extra support, feel free to join the class or get in touch. Looking forward to helping more students learn and grow! — Rithvik Muthuvelu
Let's call Peter's age "p".
ReplyDeleteGiven that John is twice as old as Peter, John's age is 2p.
In 6 years, Peter will be p + 6 and John will be 2p + 6.
According to the problem, in 6 years, the sum of their ages will be 54. So, we can write the equation:
(p + 6) (Peter's age in 6 years) + (2p + 6) (John's age in 6 years) = 54
Combining like terms:
3p + 12 = 54
Subtracting 12 from both sides:
3p = 42
Dividing both sides by 3:
p = 14
So, Peter's current age is 14 and John's current age is 2 times Peter's age, which is 28.
In summary:
Peter: 14 years old
John: 28 years old