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Question
primary 6 | Maths
| Percentage
2 Answers Below
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Date Posted: 4 years ago
Views: 243
Mr Yeo = 3000 (120%)
Mr Soon = 100%
120%→ 3000
100% = 100÷120 × 3000 = 2500
After increased same to u%
Mr Yeo = Mr Soon + 590
3000×u/100 = 2500×u + 590
30u = 25u +590
5u = 590
1u = 590÷5
1u = 118%
Therefore, % increased
= 118-100
=18%
Hope this helps
Mr Soon = 100%
120%→ 3000
100% = 100÷120 × 3000 = 2500
After increased same to u%
Mr Yeo = Mr Soon + 590
3000×u/100 = 2500×u + 590
30u = 25u +590
5u = 590
1u = 590÷5
1u = 118%
Therefore, % increased
= 118-100
=18%
Hope this helps
Alternatively, you can work this way:
Mr Yeo = 3000 (120%)
Mr Soon = 100%
120%→ 3000
100% = 100÷120 × 3000 = 2500
The price difference = 3000-2500=500
After increased, the price difference = 590
Total difference increased = 590-500=90
Therefore,% increased
= 90/500x100%
= 18%
Hope this helps
Mr Yeo = 3000 (120%)
Mr Soon = 100%
120%→ 3000
100% = 100÷120 × 3000 = 2500
The price difference = 3000-2500=500
After increased, the price difference = 590
Total difference increased = 590-500=90
Therefore,% increased
= 90/500x100%
= 18%
Hope this helps
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Date Posted:
4 years ago
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Date Posted:
4 years ago
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