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GaryK [48]
3 years ago
10

An office manager booked 61 airline tickets. He booked 8 more tickets on Airlines A than Airlines B. On Airlines​ C, he booked 5

more than twice as many tickets as on Airlines B. How many tickets did he book on each​ airline?
Mathematics
1 answer:
Setler79 [48]3 years ago
5 0

Answer:

B = 12

A =  20

C =  29

Step-by-step explanation:

A = tickets on airline A

B = A-8

C = 2B +5

Total = 61

A + B+ C = 61

Taking B = A -8 and writing in terms of A  

A = B+8

Replacing A and C in the equation A + B+ C = 61

( B+8) + B + 2B +5  = 61

Combining like terms

4B +13 = 61

Subtract 13 from each side

4B =13-13 = 61-13

4B =48

Divide by 4

4B /4 = 48/4

B = 12

A = B +8 = 12+8 = 20

C = 2B +5 = 2*12+5 = 24+5 = 29

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Read 2 more answers
A hog producer is feeding soybean meal and corn to his pigs. He needs to know how many pounds he needs to feed of each in order
Mazyrski [523]

Answer:

Feed 1, Soybean meal required =  \dfrac{2000}{3} \text{ lbs.}

Feed 2, Corn meal required = \dfrac{4000}{3} \text{ lbs.}

Step-by-step explanation:

Total feed is <em>1 ton</em> i.e. <em>2000 lbs</em>.

Let <em>x</em> be the amount of Feed 1 required.

Feed 1 has 45\% of protein.

\text{Protein in Feed 1 = }x \times \dfrac{45}{100} ..... (1)

Then, amount of Feed 2 required = (2000 - x) \text{ lbs.}

Feed 2 has 15\% of protein.

\text{Protein in Feed 2 = }(2000 - x) \times \dfrac{15}{100} ..... (2)

As per question, total protein required is 25\% of 2000 lbs .

Adding (1) and (2) and putting it equal to total protein required.

\Rightarrow x \times \dfrac{45}{100} + (2000 - x) \times \dfrac{15}{100} = 2000 \times \dfrac{25}{100}\\\Rightarrow 45x + 30000 - 15x = 50000\\\Rightarrow 30x = 20000\\\Rightarrow x = \dfrac{2000}{3}

Feed 1 required = \dfrac{2000}{3}\text{ lbs .}

Feed 2 required = 2000 - \dfrac{2000}{3}\\

\Rightarrow \dfrac{4000}{3}\text{ lbs.}

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Answer:

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Step-by-step explanation:

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