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secondary 4 | E Maths
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elly
Elly

secondary 4 chevron_right E Maths chevron_right Singapore

thx

Date Posted: 1 year ago
Views: 145
J
J
1 year ago
If you want to solve in only one variable instead of solving simultaneous equations,

Let Min's money at first be $5x
Let Brian's money at first be $3x

Min's money now = $(5x - 22)
Brian's money now = $(3x + 22)

Since the ratio of Min's money to Brian's money now is 3:4, we can rewrite this in fraction form.

(Min's money now) / (Brian's money now) = 3 / 4

(5x - 22) / (3x + 22) = 3 / 4

Cross multiply,

4(5x - 22) = 3(3x + 22)

20x - 88 = 9x + 66

20x - 9x = 66 + 88

11x = 154

x = 154/11
x = 14

Amount of money Min has now

= $(5 × 14 - 22)
= $48
elly
Elly
1 year ago
thx! i understand btr now
J
J
1 year ago
Alternative method (may not be ideal for O Levels since this is a Specimen Paper question).


Total amount is unchanged, internal transfer (This is like a PSLE method)

Make the total units the same for both ratios.

At first

Min $ : Brian $ : Total
5 : 3 : 8
35 : 21 : 56

Now

Min $ : Brian $ : Total
3 : 4 : 7
24 : 32 : 56


Let Min's money at first be $35x

Min's money now = $24x

Change in amount = $35x - $24x = $11x

$11x = $22
x = $22/11
x = 2

Min's money now = $(24 x 2) = $48

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AC Lim
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Hope this helps ♦