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primary 6 | Maths
| Data Analysis
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Date Posted: 3 years ago
Views: 483
Let the number of row at 1st = u
Total chairs at 1st = 15u
After removed 6 chairs
Total chairs = 15u - 6
There are u + 9 row of 12 chairs.
Total chairs now
= 12u + 108
Therefore,
15u - 6 = 12u +108
3u = 114
u = 38
So, Total chairs in the end,
= 15u - 6
= 15×38 - 6
= 570 - 6
= 564.
Chairs for briefing = 564
Total chairs at 1st = 15u
After removed 6 chairs
Total chairs = 15u - 6
There are u + 9 row of 12 chairs.
Total chairs now
= 12u + 108
Therefore,
15u - 6 = 12u +108
3u = 114
u = 38
So, Total chairs in the end,
= 15u - 6
= 15×38 - 6
= 570 - 6
= 564.
Chairs for briefing = 564
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similar Q
Date Posted:
3 years ago
Let the number of row at 1st = u
Total chairs at 1st = 15u
After removed 6 chairs
Total chairs = 15u - 6
There are u + 9 row of 12 chairs.
Total chairs now
= 12u + 108
Therefore,
15u - 6 = 12u +108
3u = 114
u = 38
So, Total chairs in the end,
= 15u - 6
= 15×38 - 6
= 570 - 6
= 564.
Chairs for briefing = 564
Total chairs at 1st = 15u
After removed 6 chairs
Total chairs = 15u - 6
There are u + 9 row of 12 chairs.
Total chairs now
= 12u + 108
Therefore,
15u - 6 = 12u +108
3u = 114
u = 38
So, Total chairs in the end,
= 15u - 6
= 15×38 - 6
= 570 - 6
= 564.
Chairs for briefing = 564
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