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Perform Rational Operations - Praxis Core Math

  1. A carpenter has a board that is 813 8\frac{1}{3} feet long. He needs to cut as many pieces as possible that are each 23 \frac{2}{3} feet long. After he cuts the maximum number of full pieces, what is the length, in feet, of the remaining board?
    Answer: 13 \frac{1}{3} First, find how many 23 \frac{2}{3} -foot pieces can be cut from the 813 8\frac{1}{3} -foot board. This is a division problem: 813÷23=253÷23=253×32=252=12.5 8\frac{1}{3} \div \frac{2}{3} = \frac{25}{3} \div \frac{2}{3} = \frac{25}{3} \times \frac{3}{2} = \frac{25}{2} = 12.5 . The carpenter can cut 12 full pieces. The total length of these 12 pieces is 12×23=243=8 12 \times \frac{2}{3} = \frac{24}{3} = 8 feet. The length of the remaining board is the original length minus the length of the cut pieces: 813−8=13 8\frac{1}{3} - 8 = \frac{1}{3} foot.
  2. A full bucket of water weighs 2612 26\frac{1}{2} kilograms. The empty bucket weighs 212 2\frac{1}{2} kilograms. If 13 \frac{1}{3} of the water is removed from the full bucket, what is the total weight of the bucket and the remaining water?
    Answer: 1812 18\frac{1}{2} kilograms First, find the weight of the water alone by subtracting the bucket's weight from the total weight: 2612−212=24 26\frac{1}{2} - 2\frac{1}{2} = 24 kilograms. Next, calculate the amount of water remaining. If 13 \frac{1}{3} is removed, then 1−13=23 1 - \frac{1}{3} = \frac{2}{3} of the water remains. The weight of the remaining water is 23×24=16 \frac{2}{3} \times 24 = 16 kilograms. Finally, find the total weight by adding the weight of the bucket to the weight of the remaining water: 16+212=1812 16 + 2\frac{1}{2} = 18\frac{1}{2} kilograms.
  3. What is the value of the expression (12+13)÷(12−13) (\frac{1}{2} + \frac{1}{3}) \div (\frac{1}{2} - \frac{1}{3}) ?
    Answer: 5 First, evaluate the expression in the first set of parentheses: 12+13=36+26=56 \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} . Next, evaluate the expression in the second set of parentheses: 12−13=36−26=16 \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} . Finally, perform the division: 56÷16=56×61=5 \frac{5}{6} \div \frac{1}{6} = \frac{5}{6} \times \frac{6}{1} = 5 .
  4. A worker is filling a pool. The water level starts at 114 1\frac{1}{4} feet. For 3 hours, the water level rises at a rate of 12 \frac{1}{2} foot per hour. Then, a small leak causes the water level to drop by a total of 13 \frac{1}{3} foot. What is the final water level in feet?
    Answer: 2512 2\frac{5}{12} First, calculate the total rise in water level over 3 hours: 3×12=32 3 \times \frac{1}{2} = \frac{3}{2} feet. Add this to the starting level: 114+32=54+64=114 1\frac{1}{4} + \frac{3}{2} = \frac{5}{4} + \frac{6}{4} = \frac{11}{4} feet. Now, subtract the amount the water level dropped due to the leak: 114−13 \frac{11}{4} - \frac{1}{3} . Find a common denominator, which is 12: 3312−412=2912 \frac{33}{12} - \frac{4}{12} = \frac{29}{12} . Converting this improper fraction to a mixed number gives 2512 2\frac{5}{12} feet.
  5. What is the positive difference between the product of −34 -\frac{3}{4} and 89 \frac{8}{9} and the sum of −34 -\frac{3}{4} and 89 \frac{8}{9} ?
    Answer: 2936 \frac{29}{36} First, calculate the product: −34×89=−2436=−23 -\frac{3}{4} \times \frac{8}{9} = -\frac{24}{36} = -\frac{2}{3} . Next, calculate the sum: −34+89=−2736+3236=536 -\frac{3}{4} + \frac{8}{9} = -\frac{27}{36} + \frac{32}{36} = \frac{5}{36} . The difference between the sum and the product is 536−(−23)=536+23=536+2436=2936 \frac{5}{36} - (-\frac{2}{3}) = \frac{5}{36} + \frac{2}{3} = \frac{5}{36} + \frac{24}{36} = \frac{29}{36} . Since this value is positive, it is the positive difference.
  6. Which of the following is the value of the expression (−1.5)2−(0.5÷−0.2) (-1.5)^2 - (0.5 \div -0.2) ?
    Answer: 4.75 Following the order of operations (PEMDAS/BODMAS): First, evaluate the exponent: (−1.5)2=2.25 (-1.5)^2 = 2.25 . Next, evaluate the division inside the parentheses: 0.5÷−0.2=−2.5 0.5 \div -0.2 = -2.5 . Finally, perform the subtraction: 2.25−(−2.5)=2.25+2.5=4.75 2.25 - (-2.5) = 2.25 + 2.5 = 4.75 .
  7. From a piece of cloth measuring 20 yards, a tailor cuts a piece that is 513 5\frac{1}{3} yards long and another piece that is 712 7\frac{1}{2} yards long. The remaining piece is then divided into two equal parts. What is the length of each part?
    Answer: 3712 3\frac{7}{12} yards First, find the total length of the two pieces cut from the cloth: 513+712=526+736=1256 5\frac{1}{3} + 7\frac{1}{2} = 5\frac{2}{6} + 7\frac{3}{6} = 12\frac{5}{6} yards. Next, find the length of the remaining piece: 20−1256=1966−1256=716 20 - 12\frac{5}{6} = 19\frac{6}{6} - 12\frac{5}{6} = 7\frac{1}{6} yards. Finally, divide the remaining piece into two equal parts: 716÷2=436÷2=436×12=4312 7\frac{1}{6} \div 2 = \frac{43}{6} \div 2 = \frac{43}{6} \times \frac{1}{2} = \frac{43}{12} . Convert this to a mixed number: 3712 3\frac{7}{12} yards.
  8. A machine can produce a batch of widgets in 412 4\frac{1}{2} hours. A new machine can do the same job in 23 \frac{2}{3} of the time. If the new machine starts a batch at 8:00 a.m., at what time will it finish?
    Answer: 11:00 a.m. First, calculate the time it takes the new machine. This is 23 \frac{2}{3} of 412 4\frac{1}{2} hours. Convert the mixed number to an improper fraction: 412=92 4\frac{1}{2} = \frac{9}{2} . Now multiply: 23×92=186=3 \frac{2}{3} \times \frac{9}{2} = \frac{18}{6} = 3 hours. If the machine starts at 8:00 a.m. and runs for 3 hours, it will finish at 11:00 a.m.
  9. A rectangular garden has a length of ( 5125\frac{1}{2} ) feet and a width of ( 3123\frac{1}{2} ) feet. A path that is ( 14\frac{1}{4} ) foot wide is built around the outside of the garden. What is the perimeter, in feet, of the outer edge of the path?
    Answer: 20 The path adds (14)( \frac{1}{4} ) foot to both sides of the length and both sides of the width. The new length is (512+2×14=512+12=6)( 5\frac{1}{2} + 2 \times \frac{1}{4} = 5\frac{1}{2} + \frac{1}{2} = 6 ) feet. The new width is (312+2×14=312+12=4)( 3\frac{1}{2} + 2 \times \frac{1}{4} = 3\frac{1}{2} + \frac{1}{2} = 4 ) feet. The perimeter of the outer edge of the path is calculated as (2×(new length+new width)=2×(6+4)=2×10=20)( 2 \times (\text{new length} + \text{new width}) = 2 \times (6 + 4) = 2 \times 10 = 20 ) feet.
  10. Evaluate the expression 35×(12−2)+710 \frac{3}{5} \times (\frac{1}{2} - 2) + \frac{7}{10} .
    Answer: -0.2 First, evaluate the expression in the parentheses: 12−2=0.5−2=−1.5 \frac{1}{2} - 2 = 0.5 - 2 = -1.5 . Or as a fraction, 12−42=−32 \frac{1}{2} - \frac{4}{2} = -\frac{3}{2} . Next, perform the multiplication: 35×(−32)=−910 \frac{3}{5} \times (-\frac{3}{2}) = -\frac{9}{10} . Finally, perform the addition: −910+710=−210=−15 -\frac{9}{10} + \frac{7}{10} = -\frac{2}{10} = -\frac{1}{5} . As a decimal, this is -0.2.
  11. Over a three-day period, the daily change in a pond's water level was recorded as an increase of 2.5 inches, a decrease of 4.1 inches, and an increase of 0.7 inches. What was the average daily change in the water level for this period?
    Answer: -0.3 inches To find the average daily change, first sum the changes: 2.5+(−4.1)+0.7=−1.6+0.7=−0.9 2.5 + (-4.1) + 0.7 = -1.6 + 0.7 = -0.9 inches. Then, divide the total change by the number of days, which is 3. The average change is −0.9÷3=−0.3 -0.9 \div 3 = -0.3 inches.
  12. Three friends are sharing the cost of a gift that costs $72. The first friend pays 13 \frac{1}{3} of the cost. The second friend pays 58 \frac{5}{8} of the remaining cost. The third friend pays the rest. How much more did the second friend pay than the third friend?
    Answer: $12 First, calculate the first friend's payment: 13×$72=$24 \frac{1}{3} \times \$72 = \$24 . Next, calculate the remaining cost: $72−$24=$48 \$72 - \$24 = \$48 . Then, calculate the second friend's payment: 58×$48=5×$6=$30 \frac{5}{8} \times \$48 = 5 \times \$6 = \$30 . Finally, calculate the third friend's payment, which is the rest of the remaining cost: $48−$30=$18 \$48 - \$30 = \$18 . The question asks how much more the second friend paid than the third friend: $30−$18=$12 \$30 - \$18 = \$12 .
  13. The temperature at 6:00 a.m. was −8.4∘F -8.4^{\circ}F . By noon, the temperature had risen by 15.7∘F 15.7^{\circ}F . By 6:00 p.m., the temperature had dropped by 12 \frac{1}{2} of the temperature at noon. What was the temperature at 6:00 p.m.?
    Answer: 3.65∘F 3.65^{\circ}F First, find the temperature at noon: −8.4+15.7=7.3∘F -8.4 + 15.7 = 7.3^{\circ}F . Next, find the temperature drop by 6:00 p.m. The drop is half of the noon temperature: 12×7.3=3.65∘F \frac{1}{2} \times 7.3 = 3.65^{\circ}F . The temperature at 6:00 p.m. is the noon temperature minus the drop: 7.3−3.65=3.65∘F 7.3 - 3.65 = 3.65^{\circ}F .
  14. An account has a balance of $150.40. A deposit of $49.60 is made. Then, two checks are written: one for $30 and another for 15 \frac{1}{5} of the balance that was in the account after the deposit. What is the final balance in the account?
    Answer: $130.00 First, calculate the balance after the deposit: $150.40+$49.60=$200.00 \$150.40 + \$49.60 = \$200.00 . Next, calculate the amount of the second check, which is 15 \frac{1}{5} of this new balance: 15×$200.00=$40.00 \frac{1}{5} \times \$200.00 = \$40.00 . Now, subtract the amounts of both checks from the balance: $200.00−$30.00−$40.00=$130.00 \$200.00 - \$30.00 - \$40.00 = \$130.00 .
  15. A container holds 20 gallons of water. One-fourth of the water is used for gardening. Then, two-fifths of the remaining water is used for cleaning. How many gallons of water are left in the container?
    Answer: 9 gallons First, calculate the amount of water used for gardening: 14×20=5 \frac{1}{4} \times 20 = 5 gallons. The remaining water is 20−5=15 20 - 5 = 15 gallons. Next, calculate the amount used for cleaning from the remainder: 25×15=6 \frac{2}{5} \times 15 = 6 gallons. The final amount of water left is 15−6=9 15 - 6 = 9 gallons.
  16. A car's gas tank was 78 \frac{7}{8} full. The driver used 12 \frac{1}{2} of the gas in the tank to drive to a meeting. Then, they added 5 gallons of gas to the tank, which was found to be 34 \frac{3}{4} full. What is the total capacity, in gallons, of the gas tank?
    Answer: 16 Let C be the total capacity of the tank. Initially, the tank had 78C \frac{7}{8}C gallons. The driver used half of this amount, so the gas used was 12×78C=716C \frac{1}{2} \times \frac{7}{8}C = \frac{7}{16}C . The amount of gas remaining after the drive was 78C−716C=1416C−716C=716C \frac{7}{8}C - \frac{7}{16}C = \frac{14}{16}C - \frac{7}{16}C = \frac{7}{16}C . After adding 5 gallons, the tank was 34 \frac{3}{4} full. So, 716C+5=34C \frac{7}{16}C + 5 = \frac{3}{4}C . To solve for C, subtract 716C \frac{7}{16}C from both sides: 5=34C−716C=1216C−716C=516C 5 = \frac{3}{4}C - \frac{7}{16}C = \frac{12}{16}C - \frac{7}{16}C = \frac{5}{16}C . Now, solve for C: C=5÷516=5×165=16 C = 5 \div \frac{5}{16} = 5 \times \frac{16}{5} = 16 . The total capacity is 16 gallons.
  17. The product of three numbers is -1. Two of the numbers are 35 \frac{3}{5} and −56 -\frac{5}{6} . What is the third number?
    Answer: 2 Let the third number be x x . The problem states that (35)×(−56)×x=−1 (\frac{3}{5}) \times (-\frac{5}{6}) \times x = -1 . First, multiply the two known numbers: 35×−56=−1530=−12 \frac{3}{5} \times -\frac{5}{6} = -\frac{15}{30} = -\frac{1}{2} . So, the equation is −12×x=−1 -\frac{1}{2} \times x = -1 . To solve for x x , divide both sides by −12 -\frac{1}{2} : x=−1÷(−12)=−1×(−2)=2 x = -1 \div (-\frac{1}{2}) = -1 \times (-2) = 2 .
  18. On a map, the distance between two cities is 614 6\frac{1}{4} inches. If the map scale is 12 \frac{1}{2} inch = 20 miles, what is the actual distance between the two cities in miles?
    Answer: 250 First, find the number of miles per inch on the map. If 12 \frac{1}{2} inch represents 20 miles, then 1 inch represents 20÷12=20×2=40 20 \div \frac{1}{2} = 20 \times 2 = 40 miles. Now, multiply the map distance by this scale factor: 614×40 6\frac{1}{4} \times 40 . Convert the mixed number to an improper fraction: 254 \frac{25}{4} . Then multiply: 254×40=25×404=25×10=250 \frac{25}{4} \times 40 = 25 \times \frac{40}{4} = 25 \times 10 = 250 miles.
  19. If x=−12 x = -\frac{1}{2} and y=23 y = \frac{2}{3} , what is the value of x2−y x^2 - y ?
    Answer: -5/12 Substitute the given values into the expression: (−12)2−(23) (-\frac{1}{2})^2 - (\frac{2}{3}) . First, calculate the square of x: (−12)2=(−12)×(−12)=14 (-\frac{1}{2})^2 = (-\frac{1}{2}) \times (-\frac{1}{2}) = \frac{1}{4} . Now, subtract y from this result: 14−23 \frac{1}{4} - \frac{2}{3} . Find a common denominator, which is 12: 312−812=−512 \frac{3}{12} - \frac{8}{12} = -\frac{5}{12} .
  20. Which of the following expressions has the greatest value?
    Answer: −45÷−23 -\frac{4}{5} \div -\frac{2}{3} To find the expression with the greatest value, we must evaluate each one. A) −45÷−23=−45×−32=1210=1.2 -\frac{4}{5} \div -\frac{2}{3} = -\frac{4}{5} \times -\frac{3}{2} = \frac{12}{10} = 1.2 . B) −12+43=−36+86=56≈0.833 -\frac{1}{2} + \frac{4}{3} = -\frac{3}{6} + \frac{8}{6} = \frac{5}{6} \approx 0.833 . C) 34−(−13)=34+13=912+412=1312≈1.083 \frac{3}{4} - (-\frac{1}{3}) = \frac{3}{4} + \frac{1}{3} = \frac{9}{12} + \frac{4}{12} = \frac{13}{12} \approx 1.083 . D) (−32)×(−23)=1 ( -\frac{3}{2} ) \times ( -\frac{2}{3} ) = 1 . Comparing the results: 1.2, 0.833..., 1.083..., and 1. The greatest value is 1.2, which corresponds to option A.

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