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Numbers - Ratio Problem

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

Problem Num : 1

From : NCTM

Type: None

Section:Numbers

Theme:None
Adjustment# :

Difficulty: 1

Category Ratio Problem

Analysis

Solution/Answer

Problem Num : 2

From : NCTM

Type: Application

Section:Numbers

Theme:Num of Equations
Adjustment# : 0

Difficulty: 1

Category Ratio Problem

Analysis

Solution/Answer

Problem Num : 3

From : AMC10

Type:

Section:Numbers

Theme:
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/Answer
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 $oxed{ ext{(D)} 5 }$ .

Answer:

Problem Num : 4

From : AMC10B

Type:

Section:Numbers

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

Answer:

Array ( [0] => 3777 [1] => 3077 [2] => 7753 [3] => 8105 ) 4