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primary 6 | Maths
| Algebra
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Date Posted: 3 years ago
Views: 351
Alternatively,
From here :
5R + 4B = $1120
Rewrite as :
3R + 2B + 2R + 2B = $1120 ①
And,
3R + 2B + 5G = $1120
Rewrite as :
3R + 2B + $500 = $1120 ②
Comparing ① and ②, we see that :
2R + 2B = $500
Multiply by 2,
4R + 4B = $1000
Now compare this to 5R + 4B = $1120
The number of B is now the same.
We can easily see that :
R = $1120 - $1000 = $120
One ring cost $120.
From here :
5R + 4B = $1120
Rewrite as :
3R + 2B + 2R + 2B = $1120 ①
And,
3R + 2B + 5G = $1120
Rewrite as :
3R + 2B + $500 = $1120 ②
Comparing ① and ②, we see that :
2R + 2B = $500
Multiply by 2,
4R + 4B = $1000
Now compare this to 5R + 4B = $1120
The number of B is now the same.
We can easily see that :
R = $1120 - $1000 = $120
One ring cost $120.
See 1 Answer
Key : 1 ring → R , 1 bracelet →B , 1 gold chain → G
From question,
5R + 4B = $1120 ①
3R + 2B + 5G = $1120
Since G = $100, 5G = 5 × $100 = $500
Then,
3R + 2B = $1120 - $500 = $620 ②
What you want to make the same are the number of B. Then compare the R
2 sets of ②: (multiply both sides of equation by 2)
6R + 4B = $1240
Compare this to ①. The number of B is the same. So compare the difference in number of R and the difference in cost.
6R - 5R = $1240 - $1120
R = $120
So one ring cost $120
From question,
5R + 4B = $1120 ①
3R + 2B + 5G = $1120
Since G = $100, 5G = 5 × $100 = $500
Then,
3R + 2B = $1120 - $500 = $620 ②
What you want to make the same are the number of B. Then compare the R
2 sets of ②: (multiply both sides of equation by 2)
6R + 4B = $1240
Compare this to ①. The number of B is the same. So compare the difference in number of R and the difference in cost.
6R - 5R = $1240 - $1120
R = $120
So one ring cost $120
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