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Can someone please share the official solution for this question?

Posted from my mobile device
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Official Explanation

Anderson’s order increased from $5 × 100 + $10 × 50 = $1,000 to 1.035 × $1,000 = $1,035. Connolly’s order increased from $5 × 200 + $10 × 200 = $3,000 to 1.03 × $3,000 = $3,090. If x is the percent increase in the cost of belts and y is the percent increase in the cost of hoses, then Anderson’s new invoice is 500x + 500y = 1,035 or, multiplying by 2, 1,000x + 1,000y = 2,070. Connolly’s new order is 1,000x + 2,000y = 3,090. Subtracting the equations tells you 1,000y = 1,020, or y = 1.02, so hoses increased 2%. Substitute to solve for x: 1,000x + 2,000(1.02) = 3,090, and 1,000x = 3,090 – 2,040 = 1,050, so x = 1.05, meaning belts increased 5%.

Answer:
Percent increase in cost of belt: 5%
Percent increase in cost of hose: 2%
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Sajjad1994 Its still not clear how you got this
If x is the percent increase in the cost of belts and y is the percent increase in the cost of hoses, then Anderson’s new invoice is 500x + 500y = 1,035
Please help
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­Affordable: 
- before: \(100B_1 + 50H_1 = 5*100 + 10*50 = 1000\)
- after: \(100B_2 + 50H_2 = 1000 * 1.035 = 1035\)

 Careful
- before: \(200B_1 + 200H_1 = 5*200 + 10*200 = 3000\)
- after: \(200B_2 + 200H_2 = 3000 * 1.03 = 3090\)

\(100B_2 + 50H_2 = 1035\)
\(200B_2 + 100H_2 = 2070\) (2)

\(200B_2 + 200H_2 = 3090\) (1)

(1) - (2)
=> \(100H_2 = 3090 - 2070 = 1020\)

=> \(H_2 = 10.2­\)

=> \(\frac{H_2}{H_1} = \frac{10.2}{10} = 102­\)%­



\(200B_2 = 2070 - 1020 = 1050\)

=> \(B_2 = 5.25\)

=> \(\frac{B_2}{B_1} = \frac{5.25}{5} = 105\)%­

H: 2%
B: 5%­
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Affordable Auto Parts cost ==> 5*100 + 10*50 = 500 + 500 = 1000
Now , Affordable Auto Parts cost = 1035

Lets say , belts increase by x% and hoses increase by y%,
35 = 500 * x/ 100 + 500*y/100

SImilarly ,
1000 + 2000 = 3000
90 = 1000*x/100 + 2000*y/100
Solving ,
5x +5y = 35
x + y= 7
and x + 2y = 9
y= 2
x= 5

Belts = 5%
hoses = 2%­
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