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Primary 4 | Maths
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Date Posted: 4 years ago
Views: 418
method 1
at first
diff. = 2064
in the end
M used 144
so new diff. = 2064 - 144 = 1920
(a)
in the end, M has 1920 more
(b)
4u = 1920
1u = 480
in the end M has 480x5 = 2400
at first, M has 2400+144 = 2544
.
method 2
J unchanged
after
M: 5u
J: 1u
at first
M: 5u + 144
J: 1u
5u + 144 = 1u + 2064
1u = 1920/4 = 480
after
M: 2400
J: 480
diff. = 1920
(a)
in the end, M has 1920 more
(b)
at first, M has 480x5 + 144 = 2544
.
.
.
at first
diff. = 2064
in the end
M used 144
so new diff. = 2064 - 144 = 1920
(a)
in the end, M has 1920 more
(b)
4u = 1920
1u = 480
in the end M has 480x5 = 2400
at first, M has 2400+144 = 2544
.
method 2
J unchanged
after
M: 5u
J: 1u
at first
M: 5u + 144
J: 1u
5u + 144 = 1u + 2064
1u = 1920/4 = 480
after
M: 2400
J: 480
diff. = 1920
(a)
in the end, M has 1920 more
(b)
at first, M has 480x5 + 144 = 2544
.
.
.
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