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Question
primary 5 | Maths
| Fractions
2 Answers Below
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Date Posted: 2 years ago
Views: 435
total unchanged - internal transfer
【forwards】
at first
M:A : T = 5u : 5u-56 : 10u-56
in the end
M:A : T = 3u : 7u-56 : 10u-56
A-M = 7u-3u-56 = 4u-56 = 72
u = (56+72)÷4 = 32
at first
M → 5u = 5×32 = 160 (Ans)
A → 5u-56 = 160-56 = 104 (Ans)
OR
【backwards】
in the end
M:A : T = 3u : 3u+72 : 6u+72
at first
M:A : T = 5u : 1u+72 : 6u+72
5u = 1u+72 + 56
u = (72 + 56)÷4 = 32
at first
M → 5u = 5×32 = 160 (Ans)
A → u+72 = 32+72
= 160-56 = 104 (Ans)
【forwards】
at first
M:A : T = 5u : 5u-56 : 10u-56
in the end
M:A : T = 3u : 7u-56 : 10u-56
A-M = 7u-3u-56 = 4u-56 = 72
u = (56+72)÷4 = 32
at first
M → 5u = 5×32 = 160 (Ans)
A → 5u-56 = 160-56 = 104 (Ans)
OR
【backwards】
in the end
M:A : T = 3u : 3u+72 : 6u+72
at first
M:A : T = 5u : 1u+72 : 6u+72
5u = 1u+72 + 56
u = (72 + 56)÷4 = 32
at first
M → 5u = 5×32 = 160 (Ans)
A → u+72 = 32+72
= 160-56 = 104 (Ans)
See 2 Answers
done
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Hope this helps
Date Posted:
2 years ago
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