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Numbers - Ratio Problem

SOURCE:COMPETITION
Number of Problems: 4.
FOR PRINT ::: (Book)

Problem Num : 1

Type: None

Topic:Numbers

Theme:None
Adjustment# :

Difficulty: 1

Category Ratio Problem

Analysis

Solution/Hint

Problem Num : 2

Type: Application

Topic:Numbers

Theme:Num of Equations
Adjustment# : 0

Difficulty: 1

Category Ratio Problem

Analysis

Solution/Hint

Problem Num : 3

Type:

Topic:Numbers

Adjustment# : 0

Difficulty: 1

'
The ratio $frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ is closest to which of the following numbers?
$ext{(A)} 0.1 qquad ext{(B)} 0.2 qquad ext{(C)} 1 qquad ext{(D)} 5 qquad ext{(E)} 10$
'

Category Ratio Problem

Analysis

Solution/Hint
We factor $frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ as $frac{10^{2000}(1+100)}{10^{2001}(1+1)}=frac{101}{20}$ . As $frac{101}{20}=5.05$ , our answer is $oxed{ ext{(D)} 5 }$ .

Problem Num : 4

Type:

Topic:Numbers

Adjustment# : 0

Difficulty: 1

'
Paula the painter had just enough paint for 30 identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for 25 rooms. How many cans of paint did she use for the 25 rooms?
$mathrm{(A)} 10qquadmathrm{(B)} 12qquadmathrm{(C)} 15qquadmathrm{(D)} 18qquadmathrm{(E)} 25$
'

Category Ratio Problem

Analysis

Solution/Hint
Losing three cans of paint corresponds to being able to paint five fewer rooms. So $frac 35 cdot 25 = oxed{15}$ . The answer is $mathrm{(C)}$ .