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Sec 3.5 Increase and Decrease Problems - PowerPoint PPT Presentation

Sec 3.5 Increase and Decrease Problems . Objectives Learn to identify an increase or decrease problem. Apply the basic diagram for increase or decrease problems. Use the basic percent formula to solve increase or decrease problems. Increase Problems.

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• Objectives

• Learn to identify an increase or decrease problem.

• Apply the basic diagram for increase or decrease problems.

• Use the basic percent formula to solve increase or decrease problems.

The part equals 100% of the base plus some portion of the base.

The part equals 100% of the base plus some portion of the base.

Phrases such as after an increase of,

The part equals 100% of the base plus some portion of the base.

Phrases such as after an increase of, more than,

The part equals 100% of the base plus some portion of the base.

Phrases such as after an increase of, more than, or greater than

The part equals 100% of the base plus some portion of the base.

Phrases such as after an increase of, more than, or greater than often indicate an increase problem.

The part equals 100% of the base plus some portion of the base.

Phrases such as after an increase of, more than, or greater than often indicate an increase problem.

The basic formula for an increase problem is:

The part equals 100% of the base plus some portion of the base.

Phrases such as after an increase of, more than, or greater than often indicate an increase problem.

The basic formula for an increase problem is:

Original value + Increase = New Value

Base Rate of Part

Inc. (after Inc.)

???? 20% \$660

Base plus some portion of the base equals \$660.

????

????

Amt. Of Increase

????

Amt. Of Increase

20% of Base

????

Amt. Of Increase

20% of Base

Sum of Base

and increase is

\$660

(after Inc.) Inc.

\$660 20% ???

100% of Base + 20% of Base = \$660

120% of Base = \$660

120% of Base = \$660

R x B = P

120% of Base = \$660

R x B = P

Hence, R = 120%

P = \$660

B = ???

Hence, R = 120%

P = \$660

B = ???

Thus,

P \$660 \$660

B = ----- = ---------- = ----------- = \$550

R 120% 1.2

So if we take 100% of the base (\$550) + 20% of the base (\$110) we get \$660 (part).

The part equals 100% of the base minus some portion of the base.

The part equals 100% of the base minus some portion of the base.

Phrases such as after a decrease of,

The part equals 100% of the base minus some portion of the base.

Phrases such as after a decrease of, less than,

The part equals 100% of the base minus some portion of the base.

Phrases such as after a decrease of, less than, or after a reduction of

The part equals 100% of the base minus some portion of the base.

Phrases such as after a decrease of, less than, or after a reduction of often indicate a decrease problem.

The part equals 100% of the base minus some portion of the base.

Phrases such as after a decrease of, less than, or after a reduction of often indicate a decrease problem.

The basic formula for a decrease problem is:

The part equals 100% of the base minus some portion of the base, yielding a new value.

Phrases such as after a decrease of, less than, or after a reduction of often indicate a decrease problem.

The basic formula for a decrease problem is:

Original Value - Decrease = New Value

The sale price of a new Palm Pilot, after a 15% decrease, was \$98.38. Find the price of the Palm Pilot before the decrease.

Base Rate of Part

Dec. (after Dec.)

??? 15% \$98.38

Base minus some portion of the base equals \$98.38.

(Part)

(Part)

Amt. of Decrease

(Part)

Amt. of Decrease

15% of Base

(Part)

Amt. of Decrease

15% of Base

Orig. Price minus decrease = price paid

Dec. (after Dec.)

??? 15% \$98.38

100% of Base - 15% of Base = \$98.38

85% of Base = \$98.38

R x B = P

R x B = P

Hence, R = 85%

P = \$98.38

B = ???

R x B = P

Hence, R = 85%

P = \$98.38

B = ???

Thus,

P \$98.38 \$98.38

B = ----- = ---------- = ----------- = \$115.74

R 85% 0.85

So, if we take 100% of the base (\$115.74) minus 15% of the base (\$17.36) we get \$98.38.