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zalisa [80]
2 years ago
15

A school bought pens, each costing $1, and pencils, each costing $0.5. The cost of the whole purchase was $220. How many pens an

d pencils were purchased if there were 80 more pencils than pens?
Mathematics
1 answer:
Mkey [24]2 years ago
6 0

Answer:

  • <u>120 pens and 200 pencils.</u>

<u></u>

Explanation:

You can set a system of two equations.

<u>1. Variables</u>

<u />

  • x: number of pens
  • y: number of pencils

<u>2. Cost</u>

  • <em>each pen costs</em> $1, then x pens costs: x
  • <em>each pencil costs</em> $0.5, then y pencil costs: 0.5y

  • Then, the total cost is: x + 0.5y

  • The cost of the whole purchase was $ 220, then the first equation is:

          x + 0.5y = 220 ↔ equation (1)

<u>3. </u><em><u>There were 80 more pencils than pens</u></em>

Then:

  pencils  =      80   +   pens

       ↓                               ↓

       y        =      80   +      x        ↔ equation (2)

<u>4. Solve the system</u>

i) Substitute the equation (2) into the equation (1):

  • x + 0.5(80 + x) = 220

ii) Solve

  • x + 40 + 0.5x = 220
  • 1.5x = 180
  • x = 180/1.5
  • x = 120 pens

iii) Substitute x = 120 into the equation (2)

  • y = 80 + 120
  • y = 200 pencils

Solution: 120 pens and 200 pencils ← answer

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Lesson 1-1: The periodic comet named Johnson has been seen at every return since its discovery in 1949, as shown in the table be
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Answer:

Step-by-step explanation:

Occurrence                 Year seen          Difference

1                                      1949                       -  

2                                     1956               1956 - 1949 = 7

3                                     1963               1963 - 1956 = 7

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2020 - 1970 = 50 (It's not the multiple of 7)

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2 years ago
Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C cont
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Answer:

k= 80%

Step-by-step explanation:

Jar A contains 4*0.45 L acid, and 4 L of a solution  of acid.

Jar B contains 5*0.48 L acid., and 5 L of a solution of acid.

Jar C contains 1*k/100 = k/100 acid, and 1 L of a solution.

50% = 0.5

For jar A.

(2/3)*k/100 L acid  is added to jar A.

Now jar A contains   4*0.45 L + (2/3)*k/100 L acid, and it has (4+2/3)L of a solution.

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We also can find k using jar B.

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[5*0.48 + (1/3)k/100 ]/(5+1/3)= 0.5

This equation also gives k=80%

Check.

We can check at least for jar A.

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L solute/L solution =  [1.8 +(2/3)*0.8] L /(4 2/3) L = 0.5 or 50%  as it is given that jar A has 50% at the end.

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