Madison brought $42.50 to the art supply store. She bought a brush, a sketchbook, and a paint set. The brush was 1/4 as much as the sketchbook, and the sketchbook cost 2/3 the cost of the paint set. Madison had $4.00 left over after buying these items.

Amount spent on each item is;

brush = $3.50

sketchbook = $14.00

paint set = $21.00

Completed question:

Madison brought $42.50 to the art supply store. She bought a brush, a sketchbook, and a paint set. The brush was 1/4 as much as the sketchbook, and the sketchbook cost 2/3 the cost of the paint set. Madison had $4.00 left over after buying these items. How much is each item.

Step-by-step explanation:

Let x,y and z represent the price of brush, sketchbook and paint set respectively.

The brush was 1/4 as much as the sketchbook,

x = 1/4 × y

y = 4x 1

the sketchbook cost 2/3 the cost of the paint set

y = 2/3 × z

3y = 2z

z = 3y/2

Substituting equation 1

z = 3(4x)/2 =6x

z = 6x 2

Total amount spent = amount he brought - amount he left with(remainder)

= $42.50 - $4.00

= $38.50

Total amount spent = x+y+z = $38.50

x+y+z = 38.50

Substituting equation 1 and 2

x + 4x + 6x = 38.50

11x = 38.50

x = 38.50/11

x = 3.5

y = 4x = 4(3.5) = 14

z = 6x = 6(3.5) = 21

Amount spent on each item is;

brush = $3.50

sketchbook = $14.00

paint set = $21.00

step-by-step explanation:

30% of $420 is 0.30($420) = $126 and $420 + $126 = $546.

or, 130% of $420 is 1.3($420) = $546.

notice that this is not the $600 original list price with which we started.

why doesn't this work?

the method does not work because 30% of $420 is not the same amount as the 30% of $600. it may seem obvious when stated this way, but it is a very common error made on problems of this type. we cannot just use the same percent and get back to the original amount because we are taking that percent of a different value.

to work problems of this type correctly, it is generally best to write an equation for the situation and then solve.

example: since (100% – 30%)(original price) = (sale price)

70% of the   (original price) is $420.

0.7x   = 420

equation image indicator

x = 600

the original cost was $600.

we may also solve this problem using the two-step method. working the problem in two steps requires us to use more algebra skills. we have to know that x = 1x (identity property of multiplicationtext annotation indicator) and that 1x – 0.3x = 0.7x (subtracting like terms). then we solve the problem as follows:

(original price) – 30%(original price) = sale price

x – 0.3x = 420

0.7x = $420

equation image indicator

x = 600

the original cost was $600.  

note that with both of the above methods we do obtain the correct original amount.

example: after a 5% pay raise, hermione is earning $22,680 per year. what was she earning before the pay raise?

105% of (original salary) = current salary

1.05x = 22,680

equation image indicator

x = 21,600

hermione earned $21,600 per year before the pay raise.

this is an example so maybe you will get how to at least do it!

each side has a different length.  

step-by-step explanation: