Affordable Auto Parts and Careful Car Repair both have standing

Question

Affordable Auto Parts and Careful Car Repair both have standing monthly orders with the same supplier for belts at $5 each and hoses at $ 10 each. Affordable’s standard shipment is 100 belts and 50 hoses, while Careful’s larger enterprise receives 200 belts and 200 hoses. Recently, the supplier raised prices. Affordable’s total monthly invoice increased 3.5%, but
Careful’s total invoice increased 3%. Find the percent increases in the cost of the belt and the cost of the hose.

Affordable cost is $1000 and new price is 1035
Careful cost is 3000 and new price is 3090
use plug in
belt % increase is 5% & hose is 2%

Archit3110 · Answer

Affordable Auto Parts and Careful Car Repair both have standing monthly orders with the same supplier for belts at $5 each and hoses at $ 10 each. Affordable’s standard shipment is 100 belts and 50 hoses, while Careful’s larger enterprise receives 200 belts and 200 hoses. Recently, the supplier raised prices. Affordable’s total monthly invoice increased 3.5%, but
Careful’s total invoice increased 3%. Find the percent increases in the cost of the belt and the cost of the hose.

Affordable cost is $1000 and new price is 1035
Careful cost is 3000 and new price is 3090
use plug in 
belt % increase is 5% & hose is 2%

Sajjad1994 · Answer

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%

Gemmie · Answer

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%

sayan640 · Answer

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%

Affordable Auto Parts and Careful Car Repair both have standing

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Percent increase in cost of belt	Percent increase in cost of hose
		2%
		2.5%
		3%
		3.25%
		4%
		5%

GMAT Two-part Analysis (TPA) Questions

Affordable Auto Parts and Careful Car Repair both have standing

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