[last updated April 11, 1999]

solutions / soluciones

 week XIII/semana X1II: Mar 1 - 5 week XIV/semana X1V: Mar 15 - 19 Egyptian Obelisks: Ap 1 week X /semana X: Jan 25 - 28 week XI /semana XI: Feb 1 - 5 week X11 /semana X1I: Feb 16 - 19 week VII/ semana VII: Jan 4 - 8 week VIII/ semana VIII: Jan. 11-15 week IX/ semana IX: Jan. 18-22 week IV/semana IV: Nov 16-20 week V/semana V: Nov.30 - Dec 4 week VI/semana VI: Dec 7 - 11 week I/semana I: Oct 26-30 week II/semana II: Nov 2-6 week III/semana III: Nov 9-13

Solutions to Week XIV:

Monday:

3 gallons = 1500 square feet

3g = 1500

3/3g = 1500/3

g = 500 square feet

Tuesday:

1 gallon = 400 square feet

g = 400

3g = ?

3 (400) = 1200 square feet

Wednesday:

GIVEN:

g = 400 square feet

FIND:

Must cover 1500 square feet; how many gallons of paint must I buy?

Xg = 1500 /400

X = 3.75 gallons; therefore, I must buy 4 gallons of paint

Solutions to Week XIII:

Monday:

 Area = (Pi) (r2) diameter = 7; r = 7/2 Pi = 22/7 or 3.14

Area = ( 22/7) (7/2)2 = ( 22/7) (7/2) (7/2) = (22) (7) ÷ (2) (2) = 38.5 square inches

Tuesday:

 Area = (Pi) (r2) diameter = 9; r = 9/2 Pi = 22/7 or 3.14

Area = ( 22/7) (9/2)2 = (22/7) (9/2) (9/2) = (22) (81) ÷ (7) (2) (2)= 63.6 square inches

Wednesday:

 Area = (Pi) (r2) diameter = 11; r = 11/2 Pi = 22/7 or 3.14

Area = ( 22/7) (11/2)2 = (22/7) (11/2) (11/2) = (22) (121) ÷ (7) (2) (2)= 95 square inches

Solutions to Week XII:

Tuesday:

Remember: OF usually translates to times in mathematics. Therefore,
1/2 of 1/6 is ( 1/2 ) ( 1/6) = 1/12 of the land is covered with pine trees

You might also find the solution by drawing a picture:
 1/2 1/6
 1/12 1/2 of land 1/2 of land

Wednesday

 16 2/3 - 12 1/4 = make into an improper fraction 50/3- 49/4 = find the common denominator ( 4/4 ) ( 50/3) - ( 3/3 ) ( 49/4) = make the equivalent fractions 200/12- 147/12 = subtract 53/12 = make into a mixed number 4 1/12 tons of coal

Thursday

Remember: OF usually translates to times in mathematics. Therefore,
3/4 of 316 balloons is

( 3/4 ) (316) = (3) (316)/4 = 948/4 = 237 balloons are orange.

Solutions to Week XI:

Monday:

 given: 30 seconds ad costs \$1.6 million find: How many minutes must be sold to make \$1 billion dollars?

1 minute = (2) (\$1,600,000) = \$3,200,000
 \$1,000,000,000 = 312.5 minutes \$3,200,000

312.5 minutes ÷ 60 minutes = 5 hours 12.5 minutes

THEREFORE, the number of minutes required to sell and show \$1 billion of Superbowl ads would take more time than the entire superbowl broadcast lasted, even if they skipped showing the game all together and just showed the ads.

Mr. Sovel was right, but Mrs. Sovel won the bet [dessert at her favorite restaurant on Valentines Day].

Tuesday

 state the problem: (2 1/2) pounds - (3/8) pounds = change all mixed numbers to improper fractions: (5/2) - (3/8) = find the least common denominator: (4/4) (5/2) - (3/8) = subtract the numerators: (20/8) - (3/8) = (20 - 3/8) = (17/8) make a mixed number again: (2 1/8) pounds

Wednesday

 state the problem: (1/2) cup + (1 1/3) cup + (1/4) cup = change all mixed numbers to improper fractions: (1/2) cup + (4/3) cup + (1/4) cup = find the least common denominator: (6/6) (1/2) + (4/4) (4/3) + (3/3) (1/4) = add the numerators: (6/12) +(16/12) + (3/12) = (6 + 16 + 3/12) = (25/12) make a mixed number again: (2 1/12) cups

Thursday

 state the problem: (2 2/3) kiloliters (12) months = change all mixed numbers to improper fractions: (8/3) (12/1) = multiply ((8) (12)/(3) (1)) = (96/3) make a mixed number again: 32 kiloliters per year

Solutions to Week X:

Monday:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610

explanation: a number is the sum of the prior two numbers, that is, the two numbers that come before it.

Tuesday

200 miles divided by 13 miles per gallon

200 miles ÷ 13 miles/gallon = (200) (miles) (1/13) ( gallons/miles ) =

15.38 gallons = 15 gallons for the jeep

Wednesday

200 miles divided by 3/5 miles per gallon

200 miles ÷ 3/5 miles/gallon = (200) (miles) (5/3) ( gallons/miles ) =

1000/3 gallons = 333.33 gallons =

333 gallons for the tank

Solutions to Week IX:

Tuesday:

Two approaches:

First approach:

find the prime factors of each of the numbers:

4 = (2) (2)

15 = (3) (5)

The number that will be divisible [that is, a factor of] into all numbers divisible by both 4 and 15 will either have the factors (2)(3) = 6 OR (2)(5) = 10. Therefore, the numbers 6 and 10 will be such factors. As the list only includes 6, that is the answer.

Second approach:

Set up a table to test several numbers that are divisible by both 4 and 15, then test the various choices;
 60 120 180 6 yes yes yes 8 no yes no 18 no no yes 24 no no no 45 no no yes

6 is the only solution offered that is a yes each time, therefore 6 is the answer.

Wednesday:

(1/2)(1/2) ÷ (1/2) = (1/4) ÷ (1/2) = (1/4) (2/1) = (2/4) = (1/2)

Thursday

The best choice is the one that starts at 1 cent the first day: The total sum earned
would be \$10,737,418.23 earned over a 30 day period.

 Days Amount per day 1 \$0.01 2 \$0.02 3 \$0.04 4 \$0.08 5 \$0.16 6 \$0.32 7 \$0.64 8 \$1.28 9 \$2.56 10 \$5.12 11 \$10.24 12 \$20.48 13 \$40.96 14 \$81.92 15 \$163.84 16 \$327.68 17 \$655.36 18 \$1,310.72 19 \$2,621.44 20 \$5,242.88 21 \$10,485.76 22 \$20,971.52 23 \$41,943.04 24 \$83,886.08 25 \$167,772.16 26 \$335,544.32 27 \$671,088.64 28 \$1,342,177.28 29 \$2,684,354.56 30 \$5,368,709.12 total \$10,737,418.23

Solutions to Week VIII:

Monday:

This is an example of using dimensional analysis to move from one set of units of measurement to another:
 (60 sec/min) (60 min/ hour) (24 hours/ day) = 86,400 sec/ day
 1,000,000,000 seconds = 11,574 days 86,400 seconds/ day

Tuesday

It may be helpful to number the ribbons 1, 2, 3, 4, 5, and 6. Ribbon 1 can be worn with ribbons 2, 3, 4, 5, and 6, giving 5 combinations. Ribbon 2 can be worn with ribbons 3, 4, 5, and 6, giving 4 more combinations. (You would not recount ribbon 2 with ribbon 1, since that was counted in the first step.) Similarly, ribbon 3 can be worn with ribbons 4, 5, and 6, ribbon 4 can be worn with ribbons 5 and 6, and ribbon 5 can be worn with ribbon 6. The total number of combinations is 5 + 4 + 3 + 2 + 1 = 15 combinations.

Thursday

Each man will play 1 game of chess with each of the 5 women. Therefore, each man will play 5 games. Since there are 5 men, there is a total of 25 games played between the men and the women. Each woman will play one game of chess with each of 4 other women. In counting the number of games played, it is important to avoid counting any game more than once. The first woman will play 4 games. The second woman has already played the first woman, so she will play 3 additional games with the other women. Similarly, there are 2 more games played by the third woman, and one more game played by the fourth woman. The games played by the fifth woman have already been counted. So, 5 women play a total of 4 + 3 + 2 + 1, or 10 games. Thus, a total of 35 games of chess have been played

Solutions to Week VII:

Monday:

Two simple approaches might be used.

The first is a ratio:
 6 lemons = X 2 quarts 8 quarts
 (6 lemons) (8 quarts) = X = 48 lemons = 24 lemons 2 quarts 2

The second approach simply finds out how many lemons are needed to make 1 quart [by dividing 2 into both 6 lemons and 2 quarts], then multiplying the result by 8 quarts.

(3) (8) = 24 lemons

Tuesday

This solution/strategy has the following assumptions:
1. the student was born in 1986, therefore turn 18 during 2004
2. the student will turn 18 years old in 2004
3. the date of the assignment [or start date] is January 5, 1999

 Year Days of 1999 (Jan. 5 - Dec. 31) 361 2000 ( (Jan. 1 - Dec. 31) (leap year) 366 2001 (Jan. 1 - Dec. 31) 365 2002 (Jan. 1 - Dec. 31) 365 2003 (Jan. 1 - Dec. 31) 365 subtotal: 1822 Now, count the days in 2004, starting January 1, until you reach your birth date; ADD this to the subtotal. [remember that 2004 is also a leap year] + days from 2004 total days

Wednesday

step 1: subtract 2/3 from 1, in order to find the volume still to be filled. [ 1/3 ]

step 2: to find 1/3 of 15 gallons, multiply the two terms = 5 gallons

Thursday

It takes 10 minutes for one pizza maker to make one pizza.

Solutions to Week VI:

Monday:

A number of solutions approaches may be used
1. create an XY graph, with the X-axis being the distance from shore and the Y-axis being the depth.
2. draw a picture
3. divide the depth by 1.5

9 meters divided by 1.5 meters = 6 meters

Tuesday:

First find out how many 15 minute increments there are in two hours:
1. 2 hours times 60 minutes per hour = 120 minutes
2. 120 minutes divided by 15 minutes = 8

Then multiply 1.5 kilowatts [per 15 minutes] times 8 [the number of 15 minute groupings] = 12 kilowatts

Wednesday

What do we know?
1. 1 dozen = 12
2. regular price is 39¢ each or (12) (39¢) = \$4.68 per dozen
3. special price is 3 for 89¢ or (4) (89¢) = \$3.56 per dozen

Find the difference [subtract] = \$1.12 savings per dozen, at the special price

Solutions to Week V:

Monday

(10 feet)(12 inches per foot) = 120 inches divided by 37 inches per shelf = 3 shelves, with 9 inches of waste, per 10 foot board.

If 15 shelves are needed, divide 15 by 3 [per 10 foot board] = 5 boards needed to make the shelves.

Tuesday

Method 1:
Add up all the weights in frames #1, #2, and #3 = 25.2 kilos. This accounts for all three sizes of chickens, twice each. Then divide this total by 2 =frame #4: 12.6 kilos.

Method 2:
Add frames #1 and #2 =19.1 kilos. Subtract frame #3 [19.1 - 6.1] = 13 kilo or the weight of the two large chickens. Therefore one large chicken weighs 6.5 kilo. Add this to frame #3 for a total of 12.6 kilos.

Wednesday

First find the unknown height [6cm + 9cm =15 cm], then find the unknown length [18 cm + 11 cm = 29 cm]. Now add all the lengths and heights, as if you decided to walk all around the figure = 88cm.

Thursday

Using the information from finding the dimensions of the unknown sides from Wednesday's problem, you may form several different combinations of rectangles. For example, one pair may be [18 cm x 15 cm] + [11 cm x 9 cm] = 369 square centimeters.

Solutions to Week IV:

Monday:

 up \$2.50 + 2.50 up \$3.00 + 3.00 down \$1.25 - 1.25 gain of \$4.25 original cost: \$30.50 current price: \$34.75

Tuesday

 total burn [in seconds] 1145.0 1st stage burn - 86.8 second stage burn [in seconds]: 1058.2

Wednesday

2 people for 4 months [\$645/2 = \$322.50]

3 people for 8 months [\$645/3 = \$215.00]

4(322.5) + 8(215) = 3010
 January \$322.50 February \$322.50 March \$322.50 April \$322.50 May \$215.00 June \$215.00 July \$215.00 August \$215.00 September \$215.00 October \$215.00 November \$215.00 December \$215.00 Merry's cost for one year \$3,010.00

Thursday

 Angler's price \$69.95 less discount - \$10.49 Go Fish Discount price: \$59.46

Solutions to Week III:

Tuesday:

Divide the day pass price by the single ticket price:

\$14.00 ÷ \$3.75 = 3.73, therefore four or more single rides would be less expensive with a day pass.

Thursday:

Note: The question asks about lunches over a 4 week period, therefore:

(2) (\$20) = \$40.00 in allowance

\$7.50 + \$8.25 + \$ 5.25 + \$ 8.75 = \$29.75 spent

Find the difference: \$40.00 - \$29.75 = \$10.25 left over.

Solutions to Week II:

Monday:

(July ÷2) + (August) = total spent

(\$75.80/2) + (\$75.80) = \$37.90 + \$75.80 = \$113.70

Tuesday:

(Monday) + (Tuesday) + 2(Monday + Tuesday) = total boxes moved

(453) + (485) (453 + 485) = 453 + 485 + 938 = 1876 boxes moved

Wednesday:

If the dimensions increased 4 times, to find the original dimensions, you must dived by 4:

19.2 cm /4 = 4.8 cm

25.6 cm /4 = 6.4 cm

Thursday:

Divide the total number of passengers by the total number of buses:

392 / 8 = 49 passengers per bus

Solutions to Week I:

Monday:

note: make all the amounts the same units, therefore:

16 oz.(1 pound) + 12 oz. + 6 =

16 + 12 + 6 = 34 ounces or 2 pounds 2 ounces

Tuesday:

Add the perimeters of each window

note: when adding, be sure to line up the decimals

4.85 m
4.25 m
2.55 m

11.65 m

Wednesday:

Find the total spent, then divide by the number of people:

\$41.50 + \$6.15 = \$47.65

\$47.65 ÷ 3 = \$18.883

Therefore, 2 people paid \$18.88 each and one person paid \$18.89

Thursday:

Find the cost of the computers in one classroom, then multiply for all the classrooms:

(4) (\$865) = \$3,460

(12) (\$3,460) = \$41,520 total