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
Each sheet contains two consecutive page numbers, for example:
ReplyDeleteSheet 1: pages 1 and 2
Sheet 2: pages 3 and 4
Sheet 3: pages 5 and 6
…and so on.
Let’s suppose n sheets are missing. Then there are 2n page numbers missing.
We also know the sum of these missing page numbers is 98.
Let’s look at the sum of page numbers on each missing sheet:
Each sheet has two consecutive numbers: one odd, one even.
For example, Sheet 1: 1 + 2 = 3
Sheet 2: 3 + 4 = 7
Sheet 3: 5 + 6 = 11
…and so on.
So each sheet’s page sum increases by 4:
3, 7, 11, 15, …
That’s an arithmetic sequence:
Sum of an arithmetic sequence = (n/2) × (first term + last term)
We’re told the total sum is 98.
Try adding consecutive sheet sums until we reach 98:
3 (1st sheet)
3 + 7 = 10 (2 sheets)
10 + 11 = 21 (3 sheets)
21 + 15 = 36 (4 sheets)
36 + 19 = 55 (5 sheets)
55 + 23 = 78 (6 sheets)
78 + 27 = 105 (too much)
So try 6 sheets: 3 + 7 + 11 + 15 + 19 + 23 = 78 → not enough
Try 7 sheets: total is 105 → too much
Now check 98 – what group of 2-page numbers add to 98?
Try listing full sheet page pairs:
(1,2) → 3
(3,4) → 7
(5,6) → 11
(7,8) → 15
(9,10) → 19
(11,12) → 23
(13,14) → 27
(15,16) → 31
Total so far: 3 + 7 + 11 + 15 + 19 + 23 + 27 + 31 = 136 (too high)
Now try a different approach:
Let’s try pairs that add up to 98:
Try grouping sheets and check sum:
Sheet 4: (7,8) → 15
Sheet 5: (9,10) → 19
Sheet 6: (11,12) → 23
Sheet 7: (13,14) → 27
Total = 15 + 19 + 23 + 27 = 84
Now add Sheet 8: (15,16) → 31 → total becomes 115 (too much)
Try removing sheet 4 (7+8=15), total becomes:
19 + 23 + 27 = 69
Try adding sheet 3 (5,6) → 11 → total: 80
Add sheet 9: (17,18) → 35 → 80 + 35 = 115 (too much)
Let's go back and solve algebraically.
Let the first page number on the first missing sheet be p.
Then the page numbers on the missing sheets are:
p, p+1
p+2, p+3
...
There are n sheets, so 2n pages.
So the page numbers are:
p, p+1, p+2, p+3, ..., p+2n–2, p+2n–1
These form an arithmetic sequence of 2n terms, starting from p.
Sum of arithmetic sequence:
Sum = (number of terms / 2) × (first + last)
= (2n / 2) × (p + (p + 2n – 1))
= n × (2p + 2n – 1) = 98
Now solve:
n × (2p + 2n – 1) = 98
Try small values of n:
Try n = 3:
3 × (2p + 6 – 1) = 98
3 × (2p + 5) = 98
2p + 5 = 98 ÷ 3 = not whole → skip
Try n = 4:
4 × (2p + 8 – 1) = 98
4 × (2p + 7) = 98
2p + 7 = 24.5 → not whole
Try n = 5:
5 × (2p + 10 – 1) = 98
5 × (2p + 9) = 98
2p + 9 = 19.6 → not whole
Try n = 6:
6 × (2p + 11) = 98
2p + 11 = 98 ÷ 6 = 16.333 → nope
Try n = 7:
7 × (2p + 13) = 98
2p + 13 = 14
2p = 1 → p = 0.5 → invalid
Try n = 2:
2 × (2p + 3) = 98
2p + 3 = 49
2p = 46 → p = 23
Page numbers: 23,24,25,26 → sum = 98 ✅
Final Answer:
2 sheets were missing.