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Primary 4 | Maths

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Julie

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Need help for the solution of this question.

Date Posted: 3 months ago
Views: 54
Eric Nicholas K
3 months ago
The extra 3 rulers would have cost \$6, since that is the difference in outcome between “\$2 extra” and “short of \$4”.

3 rulers —> \$6
1 ruler —> \$6 divided by 3 = \$2
38 rulers —> \$2 x 38 = \$76

Since he still has \$2 spare, amount of money he has is \$76 + \$2 = \$78.

Alternatively,

41 rulers —> 2 x \$41 = \$82

He does not have enough money as he is short of \$4.

Amount of money he has is \$82 - \$4 = \$78.
Snell
3 months ago
method 1

cost of 1 ruler: u

amt of money at first: 41u - 4

amt of money at first: 38u + 2

41u - 4= 38u + 2
u = \$2

amt of money at first = 41x2 - 4 = \$78

method 2

cost of 1 ruler
= (4+2)/(41-38) [ sum ÷ difference ]
= 6/3
= \$2

amt of money at first = 38x2 + 2 = \$78

...

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Tip: use simultaneous equations to solve for unknowns

One equation for each statement
Date Posted: 3 months ago
Eric Nicholas K
3 months ago
Primary 4 students have not learnt simultaneous equations yet. These are for secondary 2 students and above.
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done using algebra
Date Posted: 3 months ago 