GED Math Quiz: Number Sense

What this quiz covers

This quiz focuses on Number Sense, giving you a quick way to practice the rules, question types, and explanations that matter most for GED Math.

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.

Question 1

1. $\frac{16}{32}$
2. $\frac{1}{5}$
3. $\frac{32}{63}$
4. $\frac{16}{31}$ (correct answer)

Explanation: When you see questions involving powers and series, focus on calculating each term systematically and understanding what the question is asking for as a fraction of the whole. Let's find the cost for each hour: Hour 1 costs $2^1 = 2$ dollars, Hour 2 costs $2^2 = 4$ dollars, Hour 3 costs $2^3 = 8$ dollars, Hour 4 costs $2^4 = 16$ dollars, and Hour 5 costs $2^5 = 32$ dollars. The most expensive hour is the 5th hour at $32. The total bill is $$2 + 4 + 8 + 16 + 32 = 62$ $dollars. Therefore, the fraction spent on the most expensive hour is$ $\frac{32}{62} = \frac{16}{31}$ $. Answer choice A ($$\frac{16}{32}$$) represents a common error where students use only the cost of the most expensive hour as the denominator instead of the total bill. Answer choice B ($$\frac{1}{5}$$) assumes each hour costs the same amount, which ignores the exponential pricing structure entirely. Answer choice C ($$\frac{32}{63}$$) results from miscalculating the total as 63 instead of 62, likely from adding incorrectly or including an extra term. For exponential series problems on the GED, always write out each term explicitly rather than trying to use shortcuts. Double-check your addition of the series terms, and make sure you're answering what's actually being asked—in this case, a fraction where the numerator is the most expensive hour and the denominator is the total cost.

Question 2

1. 12 days
2. 15 days
3. 18 days (correct answer)
4. 27 days

Explanation: When you see a problem asking when two repeating events will coincide again, you're looking for the least common multiple (LCM) of their cycles. This tells you the smallest number of days until both events happen simultaneously again. The marching band practices every 6 days, and the choir practices every 9 days. To find when they'll practice together again, find the LCM of 6 and 9. First, find the prime factorization of each number:

• $6 = 2 \times 3$
• $9 = 3^2$

Question 3

1. 6
2. 14
3. 21
4. 42 (correct answer)

Explanation: Finding the greatest common factor (GCF) means identifying the largest number that divides evenly into both given numbers. When you encounter GCF problems, think about breaking down each number into its prime factors or using the division method. Let's find the GCF of 84 and 126 using prime factorization. First, break down each number:

• $84 = 2^2 × 3 × 7 = 4 × 21$
• $126 = 2 × 3^2 × 7 = 2 × 9 × 7$

Question 4

1. $-0.7,\ -\dfrac{5}{6},\ -\dfrac{2}{3},\ -0.65$
2. $-\dfrac{5}{6},\ -0.7,\ -\dfrac{2}{3},\ -0.65$ (correct answer)
3. $-0.65,\ -\dfrac{2}{3},\ -0.7,\ -\dfrac{5}{6}$
4. $-\dfrac{2}{3},\ -0.65,\ -\dfrac{5}{6},\ -0.7$

Explanation: When comparing negative numbers, remember that they work opposite to positive numbers—the number closest to zero is actually the greatest, while the number farthest from zero is the least. To compare these numbers accurately, convert everything to the same format. Let's change the fractions to decimals: $-\frac{5}{6} = -0.833...$ and $-\frac{2}{3} = -0.666...$ Now you can easily compare: $-0.833$, $-0.65$, $-0.7$, and $-0.666$. On a number line, $-0.833$ is farthest left (smallest), followed by $-0.7$, then $-0.666$, and finally $-0.65$ closest to zero (largest). So the order from least to greatest is: $-\frac{5}{6}, -0.7, -\frac{2}{3}, -0.65$. Choice A incorrectly puts $-0.7$ as the smallest when $-\frac{5}{6}$ is actually smaller. Choice C reverses the entire order, arranging from greatest to least instead of least to greatest. Choice D makes multiple ordering errors, placing $-\frac{2}{3}$ first when it should be third, and putting $-0.7$ last when it should be second. Choice B correctly identifies $-\frac{5}{6}$ as the smallest (farthest from zero) and $-0.65$ as the largest (closest to zero), with the middle values in proper order. Study tip: When ordering negative decimals and fractions, convert everything to decimals first, then remember that for negative numbers, the one with the largest absolute value is actually the smallest number.

Question 5

1. 7
2. 8 (correct answer)
3. 9
4. 6

Explanation: First, find the prime factorization of 72: 72 = 8 × 9 = 2³ × 3². So 2^x × 3^y = 2³ × 3², which means x = 3 and y = 2. Therefore, x + 2y = 3 + 2(2) = 3 + 4 = 8. Choice A (7) might result from calculating x + y + 2 = 3 + 2 + 2. Choice C (9) might come from 3² = 9. Choice D (6) might result from 2x + y = 2(3) + 2 = 8, wait that's not 6. Choice D could come from x + y + 1 = 6.

Question 6

1. $12k$ (correct answer)
2. $12 + k$
3. $12$
4. $\frac{12}{k}$

Explanation: When we multiply 24, 36, and 48 each by k, we get 24k, 36k, and 48k. The GCF of these new numbers is k times the GCF of the original numbers. Since GCF(24, 36, 48) = 12, we have GCF(24k, 36k, 48k) = 12k. This follows from the property that GCF(ka, kb, kc) = k × GCF(a, b, c) for any positive integer k. Choice B (12 + k) incorrectly adds k instead of multiplying. Choice C (12) ignores the effect of multiplying by k. Choice D (12/k) incorrectly divides by k instead of multiplying.

Question 7

1. 36
2. 54
3. 72 (correct answer)
4. 108

Explanation: When you see a problem asking for a number divisible by multiple values, you're looking for the least common multiple (LCM). The batch number must work for all three requirements: divisible by 4, 6, and 9. To find the LCM of 4, 6, and 9, break each number into prime factors: $4 = 2^2$, $6 = 2 \times 3$, and $9 = 3^2$. The LCM uses the highest power of each prime that appears: $2^2 \times 3^2 = 4 \times 9 = 36$. This means any valid batch number must be a multiple of 36. Since we need the smallest batch number greater than 50, we need the first multiple of 36 that exceeds 50. The multiples of 36 are: 36, 72, 108, 144... Since 36 < 50, but 72 > 50, the answer is 72. Let's verify: $72 ÷ 4 = 18$, $72 ÷ 6 = 12$, and $72 ÷ 9 = 8$. All divide evenly. Choice A (36) is the LCM itself, but it's not greater than 50 as required. Choice B (54) might seem reasonable since $54 > 50$, but $54 ÷ 4 = 13.5$, which isn't a whole number, so 54 isn't divisible by 4. Choice D (108) is divisible by all three numbers since it's $36 \times 3$, but it's not the smallest option greater than 50. When solving LCM problems, always check that your answer satisfies the constraint (here, being greater than 50) and verify divisibility by testing each required factor.

Question 8

1. 12
2. 8
3. 6
4. 4 (correct answer)

Explanation: When you're looking for numbers that are multiples of both 6 and 8, you need to find their least common multiple (LCM). Numbers that are multiples of both values will be multiples of their LCM. To find the LCM of 6 and 8, start by finding their prime factorizations: $6 = 2 \times 3$ and $8 = 2^3$. The LCM uses the highest power of each prime factor, so $\text{LCM}(6,8) = 2^3 \times 3 = 24$. This means you're looking for positive multiples of 24 that are less than 100. These are: 24, 48, 72, and 96. That's exactly 4 numbers, confirming answer D. Let's examine why the other choices are wrong. Choice A (12) might come from confusing the LCM with a simpler calculation, or possibly counting multiples of 8 less than 100 (which would be 12 total). Choice B (8) could result from counting multiples of 12 instead of 24, since there are 8 multiples of 12 less than 100. Choice C (6) doesn't correspond to any logical calculation for this problem. The key insight is recognizing that "multiples of both" means you need the LCM, not just the product of the two numbers. Remember: when a question asks for numbers that are multiples of two different values, always find their LCM first. This approach works for any similar problem and helps you avoid the trap of using incorrect intermediate calculations.

Question 9

1. 7
2. 5
3. 6 (correct answer)
4. 4

Explanation: When you encounter expressions with exponents that have the same base, you're working with the laws of exponents. The key insight here is that when dividing powers with the same base, you subtract the exponents: $\frac{a^m}{a^n} = a^{m-n}$. Let's simplify this step by step. You have $\frac{2^{10} \cdot 3^5}{2^7 \cdot 3^2}$, which you can rewrite as $\frac{2^{10}}{2^7} \cdot \frac{3^5}{3^2}$. For the powers of 2: $\frac{2^{10}}{2^7} = 2^{10-7} = 2^3$, so $m = 3$. For the powers of 3: $\frac{3^5}{3^2} = 3^{5-2} = 3^3$, so $n = 3$. Therefore, the simplified expression is $2^3 \cdot 3^3$, and $m + n = 3 + 3 = 6$. Looking at the wrong answers: Choice A (7) might come from incorrectly adding exponents instead of subtracting, getting $2^{17} \cdot 3^7$. Choice B (5) could result from subtracting incorrectly or mixing up the operations. Choice D (4) might come from only considering one base or making arithmetic errors in the subtraction. Study tip: Remember "same base division = subtract exponents" and "same base multiplication = add exponents." When you see fraction form, separate each base and handle the exponents independently. Always double-check your arithmetic, especially when subtracting exponents.