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1.The width of a rectangular garden is x feet. If 300 feet of fencing is needed

ID: 3121074 • Letter: 1

Question

1.The width of a rectangular garden is x feet. If 300 feet of fencing is needed to enclose the garden, which of the following represents the length of the garden, in feet?

2. Michael wishes to give his son a savings bond that will mature in 8 years. He would likethe value of the savings bond to be $5,000 at maturity. If he can invest in a bond that has an annual interest rate of 4% compounded monthly, which of the following is the best approximation of the amount he should invest?

(A) $3,200 (B) $3,350 (C) $3,500 (D) $3,650

3.An artist drew a rectangle. The artist then modified the rectangle so that the length of the rectangle increased by 30 percent and the width of the rectangle decreased by 40 percent. After the modification, what was the percent decrease in the area of the rectangle?

-_____%

4. A house was purchased for $140,000. Three years later, the value of the house was $155,000. If the value V of the house increased linearly from the date it was purchased, which of the following represents the value, in dollars, of the house t years after the date it was purchased?

(A) V = 140,000 + 15,000t

(B) V = 140,000 + 5,000t

(C) V = 140,000 + 15,000(t – 3)

(D) V = 140,000 + 5,000(t – 3)

5.A painting was purchased for $22 million and sold one year later for $26.4 million. The profit on the most recent sale was what percent of the purchase price?

Explanation / Answer

Solution:

Given that

300 feet of fencing is needed to enclose the garden

therefore perimeter of garden = 300 ft

and width of garden = x ft

let length of garden is l ft

We know that perimeter

2 ( l + x ) = perimeter ( where l = length of rectangle and x = width )

2 (l+x ) = 300

l +x = 150

l = 150- x = Length of rectangular garden

Answer

2.

the value of the savings bond to be $5,000 at maturity

P is principle ie the amount he should invest

rate = 4% = .04

t = 8 year

n= 12 ( compounded monthly)

A = P (1+r/n)^nt

P = 5000 /( 1+0.04/12)96

P = 3632.678

Approximately = $3,650

Option D is correct

(C)

Initial Lenght = l and width = b

after increment of 30% in lenght ( New length )= l +0.3l = 1.3l

and decrese ine 40% (New width) = b-0.4b = 0.6b

Percentage change in area = (area of new rectangle - area of original rectangle) / lb

= ( 1.3l x 0.6b - lb )/ lb x 100

= -22 %

Answer = -22 %

(4)

Option (b) is correct

because as we plug t = 3

we get

V= 140000+15000x3 = 140000+15000= 1550000

Answer

  

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