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Elan Coil [88]
3 years ago
10

Calculate the trade discount for 3030 boxes of computer paper if the unit price is ​$14.9314.93 and a single trade discount rate

of 4545​% is allowed.
Mathematics
1 answer:
zhannawk [14.2K]3 years ago
5 0

Answer:

Total trade discount on 30 boxes = $201.555

Step-by-step explanation:

1 box = $14.93

And a trade discount of 45% is allowed on a single box,

Therefore, the discount on one $14.93 box will be

45% × $14.93 = $6.7185

So, if there are now 30 boxes, the trade discount on all 30 boxes will be

30 × (trade discount on one box) = 30 × ($6.7185) = $201.555

Hope this helps!

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On the first day a total of 40 items were sold for $356. Define the variables and write a system of equations to find the number
Alina [70]
<h3><u><em>Question:</em></u></h3>

On the first day, a total of 40 items were sold for $356. Pies cost $10 and cakes cost $8. Define the variables, write a system of equations to find the number of cakes and pies sold, and state how many pies were sold.

<h3><em><u>Answer:</u></em></h3>

The variables are defined as:

"c" represent the number of cakes sold and "p" represent the number of pies sold

The system of equations used are:

c + p = 40 and 8c + 10p = 356

18 pies and 22 cakes were sold

<h3><em><u>Solution:</u></em></h3>

Let "c" represent the number of cakes sold

Let "p" represent the number of pies sold

Cost of 1 pie = $ 10

Cost of 1 cake = $ 8

Given that total of 40 items were sold

number of cakes + number of pies = 40

c + p = 40 ------ eqn 1

<u><em>Given items were sold for $356</em></u>

number of cakes sold x Cost of 1 cake + number of pies sold x Cost of 1 cake = 356

c \times 8 + p \times 10 = 356

8c + 10p = 356  ----- eqn 2

<u><em>Let us solve eqn 1 and eqn 2</em></u>

From eqn 1,

p = 40 - c    ---- eqn 3

Substitute eqn 3 in eqn 2

8c + 10(40 - c) = 356

8c + 400 - 10c = 356

-2c = - 44

c = 22

<em>Substitute c = 22 in eqn 3</em>

p = 40 - c

p = 40 - 22

p = 18

Thus 18 pies and 22 cakes were sold

3 0
3 years ago
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