# Answers to Mathematical Problems for the Examination Essay

Answers to the Math Problems for the Examination

1.      Six billion, seven hundred three million, nine hundred thousand ninety

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2.      100,006,000

3.      27 miles per gallon (27 mi/gal)

4.      \$68,977 per year

5.      10,130

6.      21/40

7.      60

8.      One-fourth ( ¼)

9.      6 and 3/7 hours

10.  4 and 1/3

11.  0.033, 0.036, 0.063, 0.066

12.  \$1.00

13.  O.18 (please put a bar at the top of 18)

14.  \$218.56

15.  24.29375

16.  Using proportion, both got 39.6 upon cross multiplication, therefore, nothing is better to be bought.

17.  n=4.5

18.  13.68 cm x 9.12 cm

19.  42

20.  \$333.7

21.  TIP = dinner price x 0.20

22.  600 jump attempts

23.  60%

24.   13.6363

25.  \$18947.50

26.  157.31 mi/hr

27.  97.6 kg

28.  15.64  km / L

29.  5.4 degrees Fahrenheit

30.  264.4mm or 26.44 cm

Detailed Solutions:

1.      Six billion, seven hundred three million, nine hundred thousand ninety

2.      100,006,000

3.      Solution:

Volume of fuel used: 11 gallons

4.      Solution:

\$

5.      Simplified form: 10126 + (1)(4) = 10130

6.      Solution:

;

;

7.      LCD: Factoring all the denominators:

20:       2          *          2          *          5

12:       2          *          2          *                      3

15:                                                      5     *   3

2          *          2          *          5     *   3 = 60

8.      Solution:

(¾ pound of sausage)(1/3) = 3/12 or ¼ pound

9.      Solution:

Time used for the community service: 20 (3/7) = 60/7

Time used for picking up litters on highways: 20(1/4) = 5

Time for tutoring: 20 – 60/7 – 5 =

10.  Solution:

57    inches =56 3/3 inches

–          52 2/3 inches

4 1/3 inches needed to be cut from the longer leg to achieve same length

11.  Answer: 0.033, 0.036, 0.063, 0.066

13.  Solution:

2/11 = 0.181818181818… = 0.18 (please put the bar above 18, thanks!)

14.  Solution:

\$13.66/hour * 16 hours = \$218.56

15.  Solution:

From turkey: (2 3/16)(\$4.95)                         =          \$10.828125

From roast beef: (1 15/16)( \$6.95)    =          \$13.465625

\$24.29375

16.  Solution:

\$39.60 = \$39.60

The buyer will get the same amount based on the proportion/cross multiplication. No item is better to be bought than the other.

17.  Solution:

2 x n = 9;

n = 9/2 = 4.5

18.  35 mm camera: image size: 3.6 cm x 2.4 cm

35 mm negative: image size: 13.8 cm x —?—

Solution:

;

Image size: 13.8 cm x 9.12 cm

19.  Area required for a newly planted shrub: 10.2 sq ft

Total area of the garden: 23.8’ x 18’

à23.8 ft x 18 ft = 428.4 sq ft

Solution:

# of shrubs that can be planted: 428.4 sq ft/10.2 sq ft =  42

20.  Given:

US \$ 36 : 255.6 francs;          rate = 7.1 francs/ US \$

US \$ 47 :    x

Solution: Using ratio and proportion:

;

; .7

21.  Dinner cost: \$ 34.80

Percentage of the dinner cost to be given as TIP: 20 % or 0.20

TIP = (dinner cost)(percentage) = 0.20(dinner cost)

22.  Given: frequency of the horse touching the water upon jumping: 18

% of the total where 18 times was obtained from: 3%

Total number of jumps performed: ?

Solution:  total number of jumps = (frequency) /(% of the total) = 18/0.03 = 600

23.  Given:             # of problems solved: 48;

Total number of problems: 80;

Solution: % correct answer = 48/80 X 100% = 60%

24.  Given:             original price: \$110,000

New price: \$95,000

Difference in price: \$15,000

Solution:         % decrease      = difference/original price

= \$15,000/\$110,000

=13.63 (please put a bar on top of 63)

25.  Base value/Original Price       : \$143,000;

Simple interest rate for 15 years:       13.25% or 0.1325

Interest = (base value) (interest rate)

Solution:         Interest = (\$143,000)(0.1325) = \$18947.50

26.  Conversion of 89.6 ft/s to  x mi/hr

x = (89.6 ) ( ) ( ) ( ) ( ) ( ) ( ) = 61.4 mi/hr

27.  David’s new weight

Original weight:         100 kg

New weight:              2400 g = 2.4 kg

Weight loss:               original weight – new weight = 100 kg – 2.4 kg = 97.6 kg

28.  Requirement: Conversion from mi/gal to km/L

Gas consumption:      36 mi/gal

Solution: (36  = 15.64 km/L

29.  Temperature requirement:      200 °C

Temperature set by Julius:     397.4 °F

Discrepancy in °F: ____

Solution: Convert 200 °C to °F

x °F     = 200 °C (9/5) + 32 = 392 °F

Discrepancy: 397.4 °F – 392 °F = 5.4 °F

30.  (a)Thickness of total pages per book:            3cm = 30mm   (using conversion 1cm:10mm);

(b)Thickness of cover combined per book:   1.4mm + 1.4mm = 2.8mm;

(c) Thickness of a book: (a) + (b) = 30mm + 2.8mm = 32.8mm;

(d) Thickness of eight books: 8 (c) = 8 (32.8mm) = 262.4mm or 26.24cm

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