My Pals Are Here Maths Pdf 5a | 2027 |

Sum of Stack B = (\frac{10}{2} \times (18 + 180) = 5 \times 198 = 990). Numbers in both A and B are multiples of both 6 and 9 → multiples of LCM(6,9)=18. From Stack A: multiples of 18 with odd multiplier (18×1=18, 18×3=54, 18×5=90, 18×7=126, 18×9=162) → 5 numbers. From Stack B: multiples of 18 with even multiplier (18×2=36, 18×4=72, 18×6=108, 18×8=144, 18×10=180) → different set! Wait — this means no number is in both A and B , because A requires odd ×6, B requires even ×9. Let’s check 18: A: 6×3 (3 odd, yes), B: 9×2 (2 even, yes) — oh! 18 is in both! So my earlier assumption wrong — 18 satisfies both. But 36? A: 6×6 (6 even → not in A). So intersection is numbers divisible by 18 with multiplier odd for A (×3,×9,×15… no, that's wrong — let's methodically solve.)

She called two students, Lin and Ravi, from the My Pals Are Here Maths 5A class for help. My Pals Are Here Maths Pdf 5a

Miss Lee, the head of the mathematics department, had a problem. The printer in the office had jammed while printing the end-of-year exam papers for Primary 5. When the technician fixed it, the papers printed in a scattered, messy pile—completely out of order. The problem? The pages were numbered from 1 to 180 , but they were stacked in reverse and in chunks. Sum of Stack B = (\frac{10}{2} \times (18

Sum of Stack A = (\frac{15}{2} \times (6 + 180) = 7.5 \times 186 = 1,395). Stack B = 18, 36, 54, …, 180. First term 18, last term 180, common difference 18. From Stack B: multiples of 18 with even

Ravi added, "And now we can reassemble the exam papers correctly."

Sum of intersection: 18+54+90+126+162 = (18+162)=180, (54+126)=180, plus 90 → 180+180+90=450. Stack C = Total − (Sum A + Sum B − Intersection) = 16,290 − (1,395 + 990 − 450) = 16,290 − (2,385 − 450) = 16,290 − 1,935 = 14,355 . Step 7: The twist Lin announced, "Miss Lee, Stack C's total is 14,355."