AB Political Science

Math 1 TF 10:00- 11:30 am

I. AGE PROBLEM

Brian is presently four times older than Sam. Six years later, he will be twice as old as Sam. How old are Brian and Sam? SOLUTION:

Using subscripts variables of BP and SP to denote the age of Brian and Sam at the present time, we may construct the following table: | Present Age| Age Six Years Later|

Brian| BP| BP + 6|

Sam| SP| SP + 6|

Since Brian is presently four times the age of Sam:

BP = 4SP

Knowing that Brian will be twice as old as Sam in six years:

BP + 6 = 2(SP + 6)

Substituting BP = 4SP:

4SP + 6 = 2(SP + 6)

4SP + 6 = 2SP + 12

4SP − 2SP = 12 − 6

2SP = 6

SP = 3 years old

Since BP = 4SP:

BP = 4(3) Years Old

BP = 12 years old

II. NUMBER PROBLEM

"55 bowls of the same size capacity of food.

1. everyone gets their own bowl of soup

2. every two gets one bowl of spaghetti to share

3. every three will get one bowl of salad

4. all r required to have their own helping of salad, spaghetti, and soup."

total number of people is x. Since '" every two gets one bowl of spaghetti to share" then x/2 of the people will have those bowls and since "every three will get one bowl of salad" then x/3 of the people will have those bowls in which they will total 55 bowls. So we have:

x + x/2 + x/3 = 55 ** multiply by 6

6x + 3x + 2x = 330

11x = 330

x = 30

So there are 30 bowls for soup, 30/2 = 15 bowls for spaghetti , and 30/3 = 10 bowls for salad.

Check 30 + 15 + 10 = 55

III. GEOMETRY PROBLEM

A rectangular table top has a perimiter of 18 inches and an area of 20 square inches. Find its dimensions

Let W = width

and L = length

from perimeter:

2(W+L)=18

W+L = 9 (equation 1)

from area:

WL = 20 (equation 2)

solve equation 1 for L:

W+L = 9

L = 9-W

substitute into equation 2 and solve for W:

WL = 20

W(9-W) = 20

9W-W^2 = 20

9W = W^2+20

0 = W^2-9W+20

0 = (W-4)(W-5)

W = {4,5}

Dimensions are: 4 inches by 5 inches

IV. PHYSICS PROBLEM

An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 19.6t + 58.8, where s is in meters. When does the object strike the ground? What is the height (above ground level) when the object smacks into the ground? Well, zero, obviously. So look for the time when the height is s = 0. I'll set s equal to zero, and solve: 0 = –4.9t2 + 19.6t + 58.8

0 = t2 – 4t – 12

0 = (t – 6)(t + 2)

Then t = 6 or t = –2. The second solution is from two seconds before launch, which doesn't make sense in this context. (It makes sense on the graph, because the line crosses the x-axis at –2, but negative time won't work in this word problem.) So "t = –2" is an extraneous solution, and I'll ignore it. The object strikes the ground six seconds after launch.

MIXTURE PROBLEM

Removing From The Solution

John has 20 ounces of a 20% of salt solution. How much water should he evaporate to make it a 30% solution? Solution:

Step 1: Set up a table for water. The water is removed from the original. | original| removed| result|

concentration| | | |

amount| | | |

Step 2: Fill in the table with information given in the question. John has 20 ounces of a 20% of salt solution. How much water should he evaporate to make it a 30% solution? The original concentration of water is 100% – 20% = 80% The resulted concentration of water is 100% – 30% = 70% The water evaporated is 100% water, which is 1 in decimal. Change all the percent to decimals.

Let x = amount of water evaporated. The result would be 20 – x. | original| removed| result|

concentration| 0.8| 1| 0.7|

amount| 20| x| 20 – x|

Step 3: Multiply down each column.

| original| removed| result|

concentration| 0.8| 1| 0.7|

amount| 20| x| 20 – x|

multiply| 0.8 × 20| 1 × x| 0.70(20 – x)|...