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primary 5 | Maths
| Area & Volume
3 Answers Below
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Pls assist Q3b
Date Posted: 2 years ago
Views: 462
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assumption method
Date Posted:
2 years ago
done
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idk if it helps lol
Date Posted:
2 years ago
That would be a total of 600 plates but the total number of plates to be delivered is only 500.
$0.80 for each intact plate.
$0.80 × 5 = $4.00
For every 5 plates delivered intact, the company charges him $4.00. But this will be offset by 1 broken plate's penalty of $4.00. The charge is offset/negated by the penalty.
So we group every 5 intact plates and 1 broken plate into 1 set.
5 + 1 = 6 so there are 6 plates in each set.
There are a total of 500 plates to be delivered.
500 ÷ 6 = 83 R2
We can have 83 such sets with 2 leftover. So for 83 sets, there are 83 plates that can be broken. (83 sets × 1 broken plate per set)
The 2 leftover plates must be intact because :
①If 1 was intact and the other broken, the company charges $0.80 for the intact one but pays $4.00 for the broken one. Overall the company has to pay $4.00 - $0.80 = $3.20
②If both are broken the company pays $4.00 × 2 = $8.00
So maximum number of plates that can be broken = 83
$0.80 × 5 = $4.00
For every 5 plates delivered intact, the company charges him $4.00. But this will be offset by 1 broken plate's penalty of $4.00. The charge is offset/negated by the penalty.
So we group every 5 intact plates and 1 broken plate into 1 set.
5 + 1 = 6 so there are 6 plates in each set.
There are a total of 500 plates to be delivered.
500 ÷ 6 = 83 R2
We can have 83 such sets with 2 leftover. So for 83 sets, there are 83 plates that can be broken. (83 sets × 1 broken plate per set)
The 2 leftover plates must be intact because :
①If 1 was intact and the other broken, the company charges $0.80 for the intact one but pays $4.00 for the broken one. Overall the company has to pay $4.00 - $0.80 = $3.20
②If both are broken the company pays $4.00 × 2 = $8.00
So maximum number of plates that can be broken = 83
Assumption method :
① Assume 80 broken plates, 420 intact ones
Penalty = 80 × $4 = $320
Charged amount = 420 × $0.80 = $336
Overall amount earned = $336 - $320 = $16
② Replace one intact plate with a broken plate.
Increase in penalty = $4
Decrease in charged amount = $0.80
Overall decease in amount earned = $4 + $0.80 = $4.80
Replace one plate : $16 - $4.80 = $11.20
Replace a second plate : $11.20 - $4.80 = $6.40
Replace a third plate : 6.40 - $4.80 = $1.60
We can replace 3 intact plates at most.
If we replace another, that would be decreasing by another $4.80.
Instead of earning $1.60 ,
$4.80 - $1.60 = $2.40
The company has to pay a penalty of $2.40 now.
So max number of plates = 80 + 3 = 83
① Assume 80 broken plates, 420 intact ones
Penalty = 80 × $4 = $320
Charged amount = 420 × $0.80 = $336
Overall amount earned = $336 - $320 = $16
② Replace one intact plate with a broken plate.
Increase in penalty = $4
Decrease in charged amount = $0.80
Overall decease in amount earned = $4 + $0.80 = $4.80
Replace one plate : $16 - $4.80 = $11.20
Replace a second plate : $11.20 - $4.80 = $6.40
Replace a third plate : 6.40 - $4.80 = $1.60
We can replace 3 intact plates at most.
If we replace another, that would be decreasing by another $4.80.
Instead of earning $1.60 ,
$4.80 - $1.60 = $2.40
The company has to pay a penalty of $2.40 now.
So max number of plates = 80 + 3 = 83
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