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Equation - Word problems

For COMPETITION
Number of Total Problems: 9.
FOR PRINT ::: (Book)

Problem Num : 1

From : AMC10B

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty: 1

'
Chandra pays an on-line service provider a fixed monthly fee plus an hourly charge for connect time. Her December bill was $\text{[math]}$ $12.48$ , but in January her bill was $\text{[math]}$ $17.54$ because she used twice as much connect time as in December. What is the fixed monthly fee?
$mathrm{(A)}$ $2.53 qquadmathrm{(B)}$ $5.06 qquadmathrm{(C)}$ $6.24 qquadmathrm{(D)}$ $7.42 qquadmathrm{(E)}$ $8.77$
'

Category Word problems

Analysis

Solution/Answer
Let $x$ be the fixed fee, and $y$ be the amount she pays for the minutes she used in the first month.
$x+y=12.48$
$x+2y=17.54$
$y=5.06$
$x=7.42$
We want the fixed fee, which is $oxed{ ext{D}}$

Answer:

Problem Num : 2

From : AMC10B

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty: 1

'
The product of three consecutive positive integers is $8$ times their sum. What is the sum of their squares?
$mathrm{(A)} 50qquadmathrm{(B)} 77qquadmathrm{(C)} 110qquadmathrm{(D)} 149qquadmathrm{(E)} 194$
'

Category Word problems

Analysis

Solution/Answer
Let the three consecutive positive integers be $a-1$ , $a$ , and $a+1$ . So, $a(a-1)(a+1)=a^3-a=24a$ . Rearranging and factoring, $a(a+5)(a-5)=0$ , so $a=5$ . Hence, the sum of the squares is $4^2+5^2+6^2=oxed{mathrm{ (B)} 77}$ .

Answer:

Problem Num : 3

From : AMC10

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty: 1

'
Cindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, giving an answer of 43. What would her answer have been had she worked the problem correctly?
$mathrm{(A) } 15qquad mathrm{(B) } 34qquad mathrm{(C) } 43qquad mathrm{(D) } 51qquad mathrm{(E) } 138$

'

Category Word problems

Analysis

Solution/Answer
We work backwards; the number that Cindy started with is $3(43)+9=138$ . Now, the correct result is $frac{138-3}{9}=frac{135}{9}=15$ . Our answer is $oxed{ ext{(A)} 15}$ .

Answer:

Problem Num : 4

From : NCTM

Type: None

Section:Equation

Theme:None
Adjustment# : 0

Difficulty: 1

Category Word problems

Analysis

Solution/Answer

Problem Num : 5

From : AMC12

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty: 1

''
Cindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, giving an answer of 43. What would her answer have been had she worked the problem correctly?
$mathrm{(A) } 15qquad mathrm{(B) } 34qquad mathrm{(C) } 43qquad mathrm{(D) } 51qquad mathrm{(E) } 138$

Category Word problems

Analysis

Solution/Answer
We work backwards; the number that Cindy started with is $3(43)+9=138$ . Now, the correct result is $frac{138-3}{9}=frac{135}{9}=15$ . Our answer is $oxed{ ext{(A)} 15}$ .

Answer:

Problem Num : 6

From : AMC10B

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty: 1

'
A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?
$mathrm{(A) } 10qquad mathrm{(B) } 14qquad mathrm{(C) } 17qquad mathrm{(D) } 20qquad mathrm{(E) } 24$
'

Category Word problems

Analysis

Solution/Answer
Let $x$ be the number of points scored by the Cougars, and $y$ be the number of points scored by the Panthers. The problem is asking for the value of $y$ .
$x+y=34$
$x-y=14$
$2x=48$
$x=24$
$y=10 Rightarrow A$

Answer:

Problem Num : 7

From : AMC10B

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty: 1

'
Twenty percent off 60 is one-third more than what number?
$mathrm{(A)} 16qquadmathrm{(B)} 30qquadmathrm{(C)} 32qquadmathrm{(D)} 36qquadmathrm{(E)} 48$
'

Category Word problems

Analysis

Solution/Answer
Twenty percent less than 60 is $frac 45 cdot 60 = 48$ . One-third more than a number n is $frac 43n$ . Therefore $frac 43n = 48$ and the number is $oxed {36}$ . The answer is $mathrm{(D)}$ .

Answer:

Problem Num : 8

From : AMC10

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty: 1

'
Heather compares the price of a new computer at two different stores. Store $A$ offers $15\%$ off the sticker price followed by a $\text{[math]}$ $90$ rebate, and store $B$ offers $25\%$ off the same sticker price with no rebate. Heather saves $\text{[math]}$ $15$ by buying the computer at store $A$ instead of store $B$ . What is the sticker price of the computer, in dollars?
$mathrm{(A)} 750qquadmathrm{(B)} 900qquadmathrm{(C)} 1000qquadmathrm{(D)} 1050qquadmathrm{(E)} 1500$
'

Category Word problems

Analysis

Solution/Answer
Solution 1

Let the sticker price be $x$ .
The price of the computer is $0.85x-90$ at store $A$ , and $0.75x$ at store $B$ .
Heather saves $\text{[math]}$ $15$ at store $A$ , so $0.85x-90+15=0.75x$ .
Solving, we find $x=750$ , and the thus answer is $mathrm{(A)}$ .

Solution 2

The $\text{[math]}$ $90$ in store $A$ is $\text{[math]}$ $15$ better than the additional $10\%$ off at store $B$ .
Thus the $10\%$ off is equal to $\text{[math]}$ $90$ $-$ $\text{[math]}$ $15$ $=$ $\text{[math]}$ $75$ , and therefore the sticker price is $\text{[math]}$ $750$ .

Answer:

Problem Num : 9

From : NCTM

Type: Understanding

Section:Equation

Theme:None
Adjustment# : 0

Difficulty: 1

Category Word problems

Analysis

Solution/Answer

Array ( [0] => 8211 [1] => 7972 [2] => 7757 [3] => 5966 [4] => 8791 [5] => 8051 [6] => 8107 [7] => 7868 [8] => 3071 ) 9