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Primary 4 | Maths
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Date Posted: 4 years ago
Views: 360
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.
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.
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
...
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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done
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Tip: use simultaneous equations to solve for unknowns
One equation for each statement
One equation for each statement
Date Posted:
4 years ago
Primary 4 students have not learnt simultaneous equations yet. These are for secondary 2 students and above.
done
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done using algebra
Date Posted:
4 years ago
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