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primary 6 | Maths
| Measurement
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Cut ribbon also can be problem...
Date Posted: 1 year ago
Views: 499
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This is similar to a past year PSLE question which involves algebra.
If you notice, there are only 2 sets of unknown length. So we let one of them be X and the other be Y.
So 36 + X = 26 + X + Y = 40 + 2Y
If we look at 36 + X = 26 + X + Y
We can isolate Y and turn the equation into
Y = 36 + X -26 - X or Y = 10
Here, we can also solve X by letting Y = 11
36 + X = 40 + 2(11)
X = 26
Therefore the length of 1 ribbon is 36 + 26 = 62cm.
B) there are 10 cut ins in total. So my cut ins are (5cm x 2) x 10 = 100cm
Next is to find the length of the ribbons that are exposed (for the perimeter)
Which is (62x10)- 4(26) - 4(11) = 472
Perimeter of figure : 472+ 100 = 572cm
My calculation may be wrong for part b as I’m on the phone, but the overall concept is correct.
If you notice, there are only 2 sets of unknown length. So we let one of them be X and the other be Y.
So 36 + X = 26 + X + Y = 40 + 2Y
If we look at 36 + X = 26 + X + Y
We can isolate Y and turn the equation into
Y = 36 + X -26 - X or Y = 10
Here, we can also solve X by letting Y = 11
36 + X = 40 + 2(11)
X = 26
Therefore the length of 1 ribbon is 36 + 26 = 62cm.
B) there are 10 cut ins in total. So my cut ins are (5cm x 2) x 10 = 100cm
Next is to find the length of the ribbons that are exposed (for the perimeter)
Which is (62x10)- 4(26) - 4(11) = 472
Perimeter of figure : 472+ 100 = 572cm
My calculation may be wrong for part b as I’m on the phone, but the overall concept is correct.
Oh my..thanks so much for explaining the steps..i stopped using algebra more than 20 yrs ago..hahaaha..however im still very grateful sir..thanks alot!
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