ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Scientific Notation
20 questions · exam conditions
0:00
Scientific NotationQuestion 1 of 20

A microscopic organism doubles its population every hour. If the initial population is 2.5×1032.5 \times 10^{3} organisms, what will the population be after 8 hours?

2.56×1052.56 \times 10^{5} organisms
2.0×1042.0 \times 10^{4} organisms
6.4×1056.4 \times 10^{5} organisms
5.12×1065.12 \times 10^{6} organisms
← Back to quizzes

ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz

ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Scientific Notation

Practice Scientific Notation in ACCUPLACER Quantitative Reasoning, Algebra & Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Scientific Notation, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Quantitative Reasoning, Algebra & Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A microscopic organism doubles its population every hour. If the initial population is 2.5×1032.5 \times 10^{3} organisms, what will the population be after 8 hours?

  1. 2.56×1052.56 \times 10^{5} organisms
  2. 2.0×1042.0 \times 10^{4} organisms
  3. 6.4×1056.4 \times 10^{5} organisms (correct answer)
  4. 5.12×1065.12 \times 10^{6} organisms
Explanation: When you encounter exponential growth problems, you're dealing with quantities that multiply by a constant factor over regular time intervals. Here, the population doubles every hour, so you multiply by 2 each hour. Start with the initial population of 2.5×1032.5 \times 10^{3} organisms. After each hour, multiply by 2:
  • After 1 hour: 2.5×103×22.5 \times 10^{3} \times 2
  • After 2 hours: 2.5×103×222.5 \times 10^{3} \times 2^{2}
  • After 8 hours: 2.5×103×282.5 \times 10^{3} \times 2^{8}
Calculate 28=2562^{8} = 256, so: 2.5×103×256=2.5×256×103=640×103=6.4×1052.5 \times 10^{3} \times 256 = 2.5 \times 256 \times 10^{3} = 640 \times 10^{3} = 6.4 \times 10^{5} This matches answer choice C. Answer A (2.56×1052.56 \times 10^{5}) likely comes from miscalculating 282^{8} as 102.4 instead of 256, or making an error in scientific notation conversion. Answer B (2.0×1042.0 \times 10^{4}) represents linear growth rather than exponential—adding the same amount each hour instead of doubling. This would be 2500×8=20,0002500 \times 8 = 20,000. Answer D (5.12×1065.12 \times 10^{6}) appears to result from using 29=5122^{9} = 512 instead of 28=2562^{8} = 256, suggesting the student counted 9 hours instead of 8. For exponential growth problems, remember the formula: Final Amount = Initial Amount × (Growth Factor)^(Number of Periods). Double-check your exponent calculation—it should equal the number of time periods, not one more.

Question 2

A laboratory needs to dilute a solution from 6.8×10−46.8 \times 10^{-4} molar to 1.7×10−61.7 \times 10^{-6} molar. What is the dilution factor (original concentration ÷ final concentration)?

  1. 1.156×10−91.156 \times 10^{-9}
  2. 2.5×10−32.5 \times 10^{-3}
  3. 4.0×10−24.0 \times 10^{-2}
  4. 4.0×1024.0 \times 10^{2} (correct answer)
Explanation: When you encounter dilution problems, you're working with the relationship between original and final concentrations. The dilution factor tells you how many times more concentrated the original solution was compared to the final solution. To find the dilution factor, you divide the original concentration by the final concentration: 6.8×10−41.7×10−6\frac{6.8 \times 10^{-4}}{1.7 \times 10^{-6}}. When dividing numbers in scientific notation, divide the coefficients and subtract the exponents: 6.81.7×10−4−(−6)=4.0×102\frac{6.8}{1.7} \times 10^{-4-(-6)} = 4.0 \times 10^{2}. This means the original solution was 400 times more concentrated than the final solution. Choice A (1.156×10−91.156 \times 10^{-9}) results from multiplying the concentrations instead of dividing them. This gives you a meaningless product rather than a ratio. Choice B (2.5×10−32.5 \times 10^{-3}) comes from incorrectly dividing final by original concentration, which would give you the inverse of the dilution factor. Choice C (4.0×10−24.0 \times 10^{-2}) happens when you make an error with the exponents during division—likely adding them instead of subtracting, or miscounting the subtraction. Remember that dilution factors should always be greater than 1 when you're going from a more concentrated to a less concentrated solution. If your answer is less than 1, you've likely flipped the fraction or made a calculation error. The dilution factor tells you practically how much you need to dilute: here, you'd take 1 part original solution and add enough solvent to make 400 parts total.

Question 3

If P=4.2×10kP = 4.2 \times 10^k and Q=6×10jQ = 6 \times 10^j, and PQ=7×103\frac{P}{Q} = 7 \times 10^3, what is the value of k−jk-j?

  1. 2
  2. 3
  3. 4 (correct answer)
  4. 5
Explanation: We are given the equation 4.2×10k6×10j=7×103\frac{4.2 \times 10^k}{6 \times 10^j} = 7 \times 10^3. Let's evaluate the left side. Divide the coefficients: 4.2÷6=0.74.2 \div 6 = 0.7. Subtract the exponents: 10k−j10^{k-j}. So, the left side is 0.7×10k−j0.7 \times 10^{k-j}. Now we have 0.7×10k−j=7×1030.7 \times 10^{k-j} = 7 \times 10^3. To compare these, let's put the left side in proper scientific notation: 0.7=7×10−10.7 = 7 \times 10^{-1}. So, (7×10−1)×10k−j=7×10k−j−1(7 \times 10^{-1}) \times 10^{k-j} = 7 \times 10^{k-j-1}. Now we can set the expressions equal: 7×10k−j−1=7×1037 \times 10^{k-j-1} = 7 \times 10^3. The coefficients are equal, so the exponents must be equal: k−j−1=3k-j-1 = 3. Solving for k−jk-j, we get k−j=4k-j = 4.

Question 4

If a number yy is 5×10−75 \times 10^{-7}, what is the reciprocal of yy, expressed in scientific notation?

  1. 5×1075 \times 10^7
  2. 0.2×1070.2 \times 10^7
  3. 2×1072 \times 10^7
  4. 2×1062 \times 10^6 (correct answer)
Explanation: The reciprocal of yy is 1y\frac{1}{y}. So we need to calculate 15×10−7\frac{1}{5 \times 10^{-7}}. This can be separated into 15×110−7\frac{1}{5} \times \frac{1}{10^{-7}}. 15=0.2\frac{1}{5} = 0.2. The reciprocal of 10−710^{-7} is 10710^7. The result is 0.2×1070.2 \times 10^7. To express this in proper scientific notation, we rewrite 0.2 as 2×10−12 \times 10^{-1}. So, (2×10−1)×107=2×10−1+7=2×106(2 \times 10^{-1}) \times 10^7 = 2 \times 10^{-1+7} = 2 \times 10^6.

Question 5

A company's market capitalization was \2.5 \times 10^{11}atthestartoftheyear.Attheendoftheyear,itwasat the start of the year. At the end of the year, it was$1.8 \times 10^{11}$. By how much did the market capitalization decrease?

  1. \7 \times 10^{10}$ (correct answer)
  2. \7 \times 10^{11}$
  3. \0.7 \times 10^{11}$
  4. \7 \times 10^0$
Explanation: To find the decrease, subtract the end-of-year value from the start-of-year value: (2.5×1011)−(1.8×1011)(2.5 \times 10^{11}) - (1.8 \times 10^{11}). Since the exponents are the same, we can subtract the coefficients: 2.5−1.8=0.72.5 - 1.8 = 0.7. The result is 0.7×10110.7 \times 10^{11}. To write this in proper scientific notation, we convert 0.7 to 7×10−17 \times 10^{-1}. So, (7×10−1)×1011=7×10−1+11=7×1010(7 \times 10^{-1}) \times 10^{11} = 7 \times 10^{-1+11} = 7 \times 10^{10}.

Question 6

The product of (5.5×103)(5.5 \times 10^3) and (6×104)(6 \times 10^4) is calculated. What is this product expressed in proper scientific notation?

  1. 3.3×1073.3 \times 10^7
  2. 3.3×1083.3 \times 10^8 (correct answer)
  3. 33×10733 \times 10^7
  4. 3.3×10123.3 \times 10^{12}
Explanation: To find the product, multiply the coefficients and add the exponents. Coefficients: 5.5×6=335.5 \times 6 = 33. Exponents: 103×104=103+4=10710^3 \times 10^4 = 10^{3+4} = 10^7. The result is 33×10733 \times 10^7. To express this in proper scientific notation, the coefficient must be between 1 and 10. We rewrite 33 as 3.3×1013.3 \times 10^1. So, (3.3×101)×107=3.3×101+7=3.3×108(3.3 \times 10^1) \times 10^7 = 3.3 \times 10^{1+7} = 3.3 \times 10^8.

Question 7

The total surface area of Earth is approximately 5.1×1085.1 \times 10^8 square kilometers. The surface area of the Pacific Ocean is approximately 1.65×1081.65 \times 10^8 square kilometers. What is the approximate surface area of Earth that is NOT covered by the Pacific Ocean?

  1. 3.45×1003.45 \times 10^0
  2. 3.45×1073.45 \times 10^7
  3. 3.45×1083.45 \times 10^8 (correct answer)
  4. 4.935×1084.935 \times 10^8
Explanation: We need to subtract the Pacific Ocean's area from Earth's total surface area: (5.1×108)−(1.65×108)(5.1 \times 10^8) - (1.65 \times 10^8). Since the exponents are the same, we subtract the coefficients: 5.1−1.65=3.455.1 - 1.65 = 3.45. The result is 3.45×1083.45 \times 10^8 square kilometers.

Question 8

The distance from Earth to the Moon is approximately 3.84×1053.84 \times 10^5 kilometers. A spacecraft travels from Earth to the Moon and then returns to Earth. What is the total distance traveled by the spacecraft?

  1. 7.68×1007.68 \times 10^0
  2. 3.84×10103.84 \times 10^{10}
  3. 7.68×10107.68 \times 10^{10}
  4. 7.68×1057.68 \times 10^5 (correct answer)
Explanation: The total distance is twice the distance from Earth to the Moon. This can be calculated as 2×(3.84×105)2 \times (3.84 \times 10^5) or by adding the distance to itself. Multiplying, we get (2×3.84)×105=7.68×105(2 \times 3.84) \times 10^5 = 7.68 \times 10^5. The total distance traveled is 7.68×1057.68 \times 10^5 kilometers.

Question 9

The number of stars in the Andromeda Galaxy is estimated to be 1×10121 \times 10^{12}, while the number of stars in our Milky Way galaxy is estimated to be 3×10113 \times 10^{11}. What is the combined total number of stars in both galaxies?

  1. 4×10114 \times 10^{11}
  2. 1.3×10121.3 \times 10^{12} (correct answer)
  3. 4×10124 \times 10^{12}
  4. 3×10233 \times 10^{23}
Explanation: To find the combined total, we add the two numbers: (1×1012)+(3×1011)(1 \times 10^{12}) + (3 \times 10^{11}). We must have a common exponent. Let's use 101210^{12}. We rewrite 3×10113 \times 10^{11} as 0.3×10120.3 \times 10^{12}. Now add the coefficients: 1+0.3=1.31 + 0.3 = 1.3. The result is 1.3×10121.3 \times 10^{12}.

Question 10

Which of the following values is the smallest?

  1. 6.1×10−56.1 \times 10^{-5}
  2. 2.5×10−42.5 \times 10^{-4}
  3. 9.8×10−69.8 \times 10^{-6} (correct answer)
  4. 1.2×10−51.2 \times 10^{-5}
Explanation: To compare numbers in scientific notation, first look at the exponents. The smallest number will have the most negative (smallest) exponent. The exponents are -5, -4, -6, and -5. The smallest exponent is -6. Therefore, 9.8×10−69.8 \times 10^{-6} is the smallest value, regardless of its coefficient being the largest.

Question 11

The approximate mass of Jupiter is 2×10272 \times 10^{27} kilograms, and the approximate mass of the Sun is 2×10302 \times 10^{30} kilograms. The mass of the Sun is approximately how many times the mass of Jupiter?

  1. 0.001
  2. 1
  3. 100
  4. 1,000 (correct answer)
Explanation: To find how many times larger the Sun's mass is, divide the mass of the Sun by the mass of Jupiter: 2×10302×1027\frac{2 \times 10^{30}}{2 \times 10^{27}}. Divide the coefficients: 2÷2=12 \div 2 = 1. Subtract the exponents: 1030−27=10310^{30-27} = 10^3. The result is 1×1031 \times 10^3, which is 1,000.

Question 12

The average human hair has a diameter of about 8×10−58 \times 10^{-5} meters. The diameter of a certain carbon nanotube is 2×10−92 \times 10^{-9} meters. A student calculates 8×10−52×10−9=4×104\frac{8 \times 10^{-5}}{2 \times 10^{-9}} = 4 \times 10^4. What does the result of this calculation represent?

  1. The combined diameter of a human hair and a carbon nanotube.
  2. The difference in diameter between a human hair and a carbon nanotube.
  3. The diameter of a human hair is 40,000 times the diameter of the nanotube. (correct answer)
  4. The diameter of a carbon nanotube is 40,000 times the diameter of a human hair.
Explanation: The calculation divides the diameter of a human hair (the larger value) by the diameter of the carbon nanotube (the smaller value). This division calculates the ratio of the two quantities. The result, 4×1044 \times 10^4 or 40,000, means that the numerator (hair diameter) is 40,000 times larger than the denominator (nanotube diameter).

Question 13

The mass of an electron is 9.1×10−319.1 \times 10^{-31} kg and the mass of a proton is 1.67×10−271.67 \times 10^{-27} kg. What is the combined mass of 3 electrons and 2 protons?

  1. 6.07×10−276.07 \times 10^{-27} kg
  2. 3.61×10−273.61 \times 10^{-27} kg (correct answer)
  3. 2.73×10−302.73 \times 10^{-30} kg
  4. 3.34×10−273.34 \times 10^{-27} kg
Explanation: Mass of 3 electrons: 3×9.1×10−31=2.73×10−303 \times 9.1 \times 10^{-31} = 2.73 \times 10^{-30} kg. Mass of 2 protons: 2×1.67×10−27=3.34×10−272 \times 1.67 \times 10^{-27} = 3.34 \times 10^{-27} kg. Total: 2.73×10−30+3.34×10−27=0.273×10−27+3.34×10−27=3.61×10−272.73 \times 10^{-30} + 3.34 \times 10^{-27} = 0.273 \times 10^{-27} + 3.34 \times 10^{-27} = 3.61 \times 10^{-27} kg. Choice A adds incorrectly. Choice C gives only the electron mass. Choice D gives only the proton mass.

Question 14

A scientist measures the mass of a virus as 4.2×10−184.2 \times 10^{-18} grams and the mass of a bacterium as 9.5×10−139.5 \times 10^{-13} grams. How many times greater is the mass of the bacterium than the mass of the virus?

  1. 2.26×1052.26 \times 10^{5} (correct answer)
  2. 2.26×10−52.26 \times 10^{-5}
  3. 5.3×10305.3 \times 10^{30}
  4. 3.99×10313.99 \times 10^{31}
Explanation: To find how many times greater the bacterium's mass is, divide: 9.5×10−134.2×10−18=9.54.2×10(−13)−(−18)=2.26×105\frac{9.5 \times 10^{-13}}{4.2 \times 10^{-18}} = \frac{9.5}{4.2} \times 10^{(-13)-(-18)} = 2.26 \times 10^{5}. Choice B incorrectly subtracts exponents as (−13)−18=−31(-13) - 18 = -31. Choice C results from multiplying the numbers instead of dividing. Choice D comes from adding exponents instead of subtracting.

Question 15

The wavelength of red light is approximately 7.0×10−77.0 \times 10^{-7} meters, and the wavelength of blue light is approximately 4.5×10−74.5 \times 10^{-7} meters. How many blue light waves would fit in the same distance as 1000 red light waves?

  1. 1400 waves
  2. 643 waves
  3. 1556 waves (correct answer)
  4. 1000 waves
Explanation: When you encounter wavelength problems involving "how many waves fit in a distance," you're dealing with a ratio and proportion situation. The key insight is that shorter wavelengths fit more times into the same distance than longer wavelengths. First, calculate the total distance covered by 1000 red light waves: Distance = 1000 waves × 7.0×10−77.0 \times 10^{-7} meters/wave = 7.0×10−47.0 \times 10^{-4} meters Now find how many blue light waves fit in this same distance by dividing the total distance by the blue wavelength: Number of blue waves = 7.0×10−44.5×10−7=7.04.5×103=1.556×103=1556\frac{7.0 \times 10^{-4}}{4.5 \times 10^{-7}} = \frac{7.0}{4.5} \times 10^{3} = 1.556 \times 10^{3} = 1556 waves This confirms answer C) 1556 waves is correct. Answer A) 1400 waves likely comes from incorrectly multiplying 1000 by the ratio 7.05.0\frac{7.0}{5.0} instead of using the actual blue wavelength. Answer B) 643 waves results from flipping the ratio—dividing by 7.04.5\frac{7.0}{4.5} instead of multiplying, which would tell you how many red waves fit in the blue wave distance. Answer D) 1000 waves ignores the wavelength difference entirely, assuming equal wavelengths. Remember: shorter wavelengths always fit more times into a given distance. Set up your ratio so the shorter wavelength (blue) gives you a larger number of waves than the longer wavelength (red). This directional check helps catch calculation errors.

Question 16

The distance from Earth to the nearest star (other than the Sun) is approximately 4.1×10134.1 \times 10^{13} kilometers. Light travels at 3.0×1083.0 \times 10^{8} meters per second. How many seconds would it take light to travel this distance?

  1. 1.37×1051.37 \times 10^{5}
  2. 1.37×1081.37 \times 10^{8} (correct answer)
  3. 1.23×10211.23 \times 10^{21}
  4. 1.23×1061.23 \times 10^{6}
Explanation: When you encounter problems involving distance, speed, and time with scientific notation, you're working with the fundamental relationship: time = distance ÷ speed. The key is carefully managing units and scientific notation operations. First, convert units so they match. The distance is given in kilometers (4.1×10134.1 \times 10^{13} km), but speed is in meters per second. Convert the distance to meters: 4.1×10134.1 \times 10^{13} km = 4.1×10164.1 \times 10^{16} meters (multiply by 1,000 or 10310^3). Now calculate: time = 4.1×10163.0×108\frac{4.1 \times 10^{16}}{3.0 \times 10^{8}} Divide the coefficients: 4.1÷3.0=1.374.1 ÷ 3.0 = 1.37 Subtract the exponents: 1016÷108=1016−8=10810^{16} ÷ 10^{8} = 10^{16-8} = 10^{8} This gives us 1.37×1081.37 \times 10^{8} seconds, which is answer choice B. Let's examine the wrong answers: Choice A (1.37×1051.37 \times 10^{5}) results from forgetting to convert kilometers to meters—you'd get this if you divided 4.1×10134.1 \times 10^{13} by 3.0×1083.0 \times 10^{8} directly. Choice C (1.23×10211.23 \times 10^{21}) comes from multiplying instead of dividing the numbers. Choice D (1.23×1061.23 \times 10^{6}) combines both the multiplication error and the unit conversion mistake. Study tip: Always check that your units match before calculating, and remember that when dividing numbers in scientific notation, you divide coefficients and subtract exponents. Double-check whether the problem asks for multiplication or division—distance problems often require division to find time.

Question 17

A computer processor operates at 3.2×1093.2 \times 10^{9} cycles per second. If a particular calculation requires 1.6×10121.6 \times 10^{12} cycles, approximately how many seconds will the calculation take?

  1. 5.0×1025.0 \times 10^{2} (correct answer)
  2. 5.0×1035.0 \times 10^{3}
  3. 2.0×10212.0 \times 10^{21}
  4. 5.12×10215.12 \times 10^{21}
Explanation: Time = cycles needed ÷ cycles per second: 1.6×10123.2×109=1.63.2×1012−9=0.5×103=5.0×102\frac{1.6 \times 10^{12}}{3.2 \times 10^{9}} = \frac{1.6}{3.2} \times 10^{12-9} = 0.5 \times 10^{3} = 5.0 \times 10^{2}. Choice B incorrectly keeps the result as 0.5×1030.5 \times 10^{3} without converting to proper scientific notation. Choice C results from multiplying instead of dividing. Choice D comes from multiplying the original numbers incorrectly.

Question 18

The population of a city is 8.4×1058.4 \times 10^{5} people. If the average person consumes 2.3×10−32.3 \times 10^{-3} tons of food per year, what is the total annual food consumption for the city in tons?

  1. 1.932×1031.932 \times 10^{3} (correct answer)
  2. 1.932×1021.932 \times 10^{2}
  3. 19.32×10219.32 \times 10^{2}
  4. 1.932×1081.932 \times 10^{8}
Explanation: Multiply the population by consumption per person: (8.4×105)×(2.3×10−3)=(8.4×2.3)×105+(−3)=19.32×102=1.932×103(8.4 \times 10^{5}) \times (2.3 \times 10^{-3}) = (8.4 \times 2.3) \times 10^{5+(-3)} = 19.32 \times 10^{2} = 1.932 \times 10^{3}. Choice B incorrectly uses 10210^{2} as the final exponent. Choice C fails to convert 19.3219.32 to proper scientific notation. Choice D adds exponents incorrectly as 5+3=85 + 3 = 8.

Question 19

A rectangular field has dimensions 2.4×1032.4 \times 10^{3} meters by 1.8×1021.8 \times 10^{2} meters. If fertilizer must be applied at a rate of 3.5×10−23.5 \times 10^{-2} kg per square meter, how many kilograms of fertilizer are needed?

  1. 1.23×1071.23 \times 10^{7} kg
  2. 1.512×1031.512 \times 10^{3} kg
  3. 4.32×1054.32 \times 10^{5} kg
  4. 1.512×1041.512 \times 10^{4} kg (correct answer)
Explanation: This problem tests your ability to work with scientific notation while solving a practical application involving area and rates. When you see dimensions and rates given in scientific notation, you need to systematically multiply the values while carefully tracking the exponents. First, find the area of the rectangular field by multiplying length times width: (2.4×103)×(1.8×102)(2.4 \times 10^{3}) \times (1.8 \times 10^{2}). Multiply the coefficients: 2.4×1.8=4.322.4 \times 1.8 = 4.32. Add the exponents: 103×102=10510^{3} \times 10^{2} = 10^{5}. So the area is 4.32×1054.32 \times 10^{5} square meters. Next, multiply the area by the fertilizer rate to find total fertilizer needed: (4.32×105)×(3.5×10−2)(4.32 \times 10^{5}) \times (3.5 \times 10^{-2}). Multiply coefficients: 4.32×3.5=15.124.32 \times 3.5 = 15.12. Add exponents: 105×10−2=10310^{5} \times 10^{-2} = 10^{3}. This gives 15.12×10315.12 \times 10^{3}, which equals 1.512×1041.512 \times 10^{4} in proper scientific notation. Looking at the wrong answers: Choice A (1.23×1071.23 \times 10^{7}) results from calculation errors in the multiplication steps. Choice B (1.512×1031.512 \times 10^{3}) occurs if you forget to convert 15.12×10315.12 \times 10^{3} to proper scientific notation. Choice C (4.32×1054.32 \times 10^{5}) is just the area calculation—you stopped before multiplying by the fertilizer rate. The correct answer is D: 1.512×1041.512 \times 10^{4} kg. Strategy tip: In scientific notation problems, work systematically—multiply coefficients separately from exponents, then convert your final answer to proper scientific notation form.

Question 20

The number 9×10159 \times 10^{15} is how many times as large as the number 3×10113 \times 10^{11}?

  1. 3,000
  2. 30,000
  3. 3×1043 \times 10^4 (correct answer)
  4. 3×1053 \times 10^5
Explanation: To find out how many times larger one number is than another, we divide the larger by the smaller: 9×10153×1011\frac{9 \times 10^{15}}{3 \times 10^{11}}. Divide the coefficients: 9÷3=39 \div 3 = 3. Subtract the exponents: 1015−11=10410^{15-11} = 10^4. The result is 3×1043 \times 10^4.