Expert: Scott A Wilson - 11/1/2008

Question
i've worked out these problems already, i would just like to know what you would get! :)


21. The ratio of the present ages of two museums is 3 to 5.  Twelve years ago, the sum of their ages was less than 56 years.  Find the possibilities for the present age of the younger museum.

a.) Between 18 yr and 32 yr    b.) Between 10 yr and 24 yr

c.)Between 8 yr and 18 yr  or  d.) Between 12 yr and 30 yr


22. Cities A and B are located on an east-west highway at a distance of 244 mi from each other.  A truck left City A at 8:00 A.M. driving 47 mi/h toward City B.  A car left City B at 10:00 A.M. traveling 53 mi/h toward City A.  At what time did the car pass the truck?

a.) 9:30 A.M.   b.) 10:30 A.M.

c.) 11:30 A.M.  d.) 12:30 P.M.


23. Find the solution set of the inequality:
   | 2 y - 3 | < 11

a.) {y: -4 < y < 7}    b.) {y: -2 < y < 8}

c.) {y: 4 < y < 7}     d.) {y: -10 < y < 2}



24. Find the solution set of the inequality:  
   | 4 t + 6 | ≤ 14

a.) {t: 2< t < 5}     b.) {t: -5 ≤ t ≤ 2}

c.) {t: -5 < t < 2}   d.) {t: 2 ≤ t ≤ 5}


25. Find the solution set for the inequality:
   14 > | 4 z - 2 |

a.) {z: -3 < z < 4}    b.) {z: 3< z < -4}
(
c.) z: -3 ≤ z ≤ 4}     d.) {z: 1 ≤ z ≤ 6}


26. Find the solution set for the inequality:
   | n - 2 | ≤ 5

a.) {n: - 3 ≤ n ≤ 7}   b.) {n: -2 ≤ n < 5}

c.) {n: -3 ≤ n < 7}    d.) {n: -1 ≤ n ≤ 4}


27. Solve the equation:  | 7.5 - 5 n | = 2.5

a.)1, -2    b.)2, 3

c.)3, 1     d.)2, 1


28. A rectangle is 3 times as wide as a square.  The rectangle's length is 15 ft shorter than 5 times the square's length.  Given that the sum of the two perimeters must be less than 150 ft, find the set of all possible lengths for the rectangle.

a.){l: l3 ft < l < 50 ft}  b.) {l: 0 ft < l < 30 ft}

c.){l: 5 ft < l < 45 ft}   d.) {l: 10 ft < l < 50 ft}



29. Mr. McTier invests $600 at 9% per year, and $800 at 12% per year.  During what time period will the $1,400 earn between $525 and $975, inclusive, in simple interest?

a.) 2.5 yr ≤ t ≤ 6.5 yr  b.) 3.5 yr ≤ t ≤ 6.5 yr

c.)1.0 yr ≤ t ≤ 4.0 yr   d.) 4.0 yr ≤ t ≤ 6.5 yr



30. A rectangle's length is 12 in. less than 3 times its width.  Find the set of all the possible widths given that the perimeter must be less than 48 in.

a.) {w: 2 in. < w < 4 in.}    b.) {w: 1 in. < w < 5 in.}

c.) {w: 4 in. < w < 9 in.}    d.) {w: 6 in. < w < 14 in.}  

Answer
The ratio of the present ages of two museums is 3 to 5. Twelve years ago, the sum of their ages was less than 56 years. Find the possibilities for the present age of the younger museum.

12 years ago is spoke of, so we can drop b) and c).  If d) were true, 12 years ago museum A would barely have been made (age 0).  There is only one choice left.

a.) Between 18 years and 32 years
b.) Between 10 years and 24 years  
c.) Between 8 years and 18 years
d.) Between 12 years and 30 years


22
Cities A and B are located on an east-west highway at a distance of 244 miles from each other.  A truck left City A at 8:00 A.M. driving 47 miles per hour toward City B. A car left City B at 10:00 A.M. traveling 53 miles per hour hour toward City A. At what time did the car pass the truck?

Distance between cities: 244
Left A: left 8am, 47 mph -> B
Left B: left 10am, 53 mph -> A
A left two hours sooner; at 10,
trucks are 244-2*47 miles apart, which is 150.

Speed of convergence is 47+53=100 mph.
150 miles at 100 mph is 1.5 hours + 10am, which is 11:30am.

a.) 9:30 A.M.
b.) 10:30 A.M.  
c.) 11:30 A.M.
d.) 12:30 P.M.


23
Find the solution set of the inequality:  | 2 y - 3 | < 11

-11 < 2y-3 < 11, so –8 <2y < 14,
so divide –8, 2y, and 14 by 2 and you’ve go the answer.

a.) {y: -4 < y < 7}
b.) {y: -2 < y < 8}
c.) {y: 4 < y < 7}
d.) {y: -10 < y < 2}


24
Find the solution set of the inequality:  | 4 t + 6 | ≤ 14

Take –14 ≤ 4t+6 ≤ 14, subtract 6 from all three term,
and divide all three terms by 4.

a.) {t: 2< t < 5}
b.) {t: -5 ≤ t ≤ 2}  
c.) {t: -5 < t < 2}
d.) {t: 2 ≤ t ≤ 5}


25
Find the solution set for the inequality:  14 > | 4 z - 2 |

-14 < 4z-2 < 14; add 2 to –14, 4z-2, and 14;
Divide all three terms by 4.

a.) {z: -3 < z < 4}
b.) {z: 3< z < -4}
c.) z: -3 ≤ z ≤ 4}
d.) {z: 1 ≤ z ≤ 6}


26. Find the solution set for the inequality: | n - 2 | ≤ 5

Convert to –5 ≤n-2≤5, add 2 to –5, n-2, and 5

a.) {n: - 3 ≤ n ≤ 7}
b.) {n: -2 ≤ n < 5}
c.) {n: -3 ≤ n < 7}
d.) {n: -1 ≤ n ≤ 4}


27. Solve the equation: | 7.5 - 5 n | = 2.5

Solve -7.5 + 5n = 2.5 and 7.5 – 5n = 2.5

a.)1, -2
b.)2, 3
c.)3, 1
d.)2, 1


28. A rectangle is 3 times as wide as a square. The rectangle's length is 15 ft shorter than 5 times the square's length. Given that the sum of the two perimeters must be less than 150 ft, find the set of all possible lengths for the rectangle.

Let L = length of a rectangle
Let W = width of a rectangle
Let S = length of a side of a square

(1) Rectangle 3 times wider: W = 3S
(2) Rectangle 15 short than 5 times: L = 5S – 15
(3) Perimeters: 2W + 2L + 4S  ≤ 150.

From (2), S = (L + 15)/5; from (1), W = 3(L+15)/5
Putting both of these into the perimeter equation gives 2(3(L+15)/5)) + 2L + 4(L+15)/5 ≤ 150.
Multiply entire equation by 5, giving 6L + 90 + 30L + 4L + 60 ≤ 750 =>  40L ≤ 600 => L ≤ 50

I must be missing something.  At least I got it started, but write back again on this problem if assistance is still needed and I might have time to look at it in more detail.

a.){13 ft < L < 50 ft}
b.) {0 ft < L < 30 ft}
c.){5 ft < L < 45 ft}
d.) {10 ft < L < 50 ft}


29. Mr. McTier invests $600 at 9% per year, and $800 at 12% per year. During what time period will the $1,400 earn between $525 and $975, inclusive, in simple interest?

525 ≤ N[(600)(0.09) + (800)(.12)] ≤ 975.; compute (600)(0.09) + (800)(.12), divide all the sides of the equation by the number you get, and you will get what t should lie between.

a.) 2.5 years ≤ t ≤ 6.5 years
b.) 3.5 years ≤ t ≤ 6.5 years
c.)1.0 years ≤ t ≤ 4.0 years
d.) 4.0 years ≤ t ≤ 6.5 years


30. A rectangle's length is 12 in. less than 3 times its width. Find the set of all the possible widths given that the perimeter must be less than 48 in.

L + 12 ≤ 3W and 2L + 2W = 48 => L + W ≤24 => W ≤ 24 – L.
Combining both equations say L + 12 ≤ 72 – 3L, so 4L  ≤ 60, or L ≤15.
From the first and the last, we get 15 + 12 ≤ 3W, so 9 ≤ W.  Also from the first, if L=0, then W=4 (the minimum)  The choice is then W is between 4 and 9.

a.) {w: 2 in. < w < 4 in.}
b.) {w: 1 in. < w < 5 in.}
c.) {w: 4 in. < w < 9 in.}
d.) {w: 6 in. < w < 14 in.}