J's answer to Ella Weiss's Primary 6 Maths Algebra Singapore question.
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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
Date Posted:
2 years ago
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