Arithmetic

306 questions

Question 1Question →

Which of the following rational numbers are strictly greater than 35\frac{3}{5}? Select all that apply.

Select all that apply

Show answer & explanation

Answer: 710\frac{7}{10}; 23\frac{2}{3}; 1115\frac{11}{15}

Answer

The fractions greater than 3/5 are 7/10, 2/3, and 11/15.
Converting 3/5 to decimal form yields 0.60. Evaluating the options: 7/10 equals 0.70, 2/3 is approximately 0.667, and 11/15 is approximately 0.733. Because each of these three values is strictly greater than 0.60, all three are correct choices.

Step-by-Step Solution

1
Convert the benchmark fraction 3/5 to decimal form.
3/5 = 0.60.
Decimal conversion provides a clear standard for comparing rational numbers.
2
Convert each given choice to decimal form and compare it to 0.60.
7/10 = 0.70; 2/3 ≈ 0.667; 5/9 ≈ 0.556; 4/7 ≈ 0.571; 11/15 ≈ 0.733.
Direct comparison reveals which decimals exceed 0.60.
3
Select all fractions whose values exceed 0.60.
The values 0.70, 0.667, and 0.733 are all strictly greater than 0.60.
The corresponding fractions 7/10, 2/3, and 11/15 satisfy the given condition.

Key Concept

Comparing Rational Numbers and Fractions
Question 2Question →

A baker has a flour mixture consisting only of wheat flour and rye flour. Currently, wheat flour accounts for 25\frac{2}{5} of the total weight of the mixture. If the baker adds 99 pounds of wheat flour to the mixture, wheat flour will account for 12\frac{1}{2} of the new total weight of the mixture. What was the total weight, in pounds, of the original flour mixture?

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Answer: 45

Answer

The total weight of the original flour mixture was 45 pounds.
The initial total weight of the mixture is 45 pounds. Initially, wheat flour makes up 25×45=18\frac{2}{5} \times 45 = 18 pounds. Adding 9 pounds of wheat flour increases the wheat flour to 18+9=2718 + 9 = 27 pounds and the total weight to 45+9=5445 + 9 = 54 pounds. The new fraction of wheat flour is 2754=12\frac{27}{54} = \frac{1}{2}, which satisfies the given conditions.

Step-by-Step Solution

1
Express the initial weight of wheat flour in terms of the initial total weight WW.
Initial weight of wheat flour = 25W\frac{2}{5}W.
Wheat flour represents 25\frac{2}{5} of the total mixture.
2
Formulate an equation reflecting the addition of 9 pounds of wheat flour.
\frac{\frac{2}{5}W + 9}{W + 9} = \frac{1}{2}
Adding 9 pounds of wheat flour increases both the amount of wheat flour and the total weight of the mixture by 9 pounds.
3
Solve the algebraic equation for WW.
Cross-multiplying gives 2(25W+9)=W+92\left(\frac{2}{5}W + 9\right) = W + 9, which simplifies to 45W+18=W+9\frac{4}{5}W + 18 = W + 9. Subtracting 45W\frac{4}{5}W and 99 from both sides gives 15W=9\frac{1}{5}W = 9, so W=45W = 45.
Isolating WW gives the value of the original total weight.

Key Concept

Setting up and solving equations involving fractional parts when a quantity is added to both the part and the whole.
Question 3Question →

How many positive integers nn less than 3030 satisfy the condition that n2−1n^2 - 1 is divisible by 2424?

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Answer: 10

Answer

10
For n2−1n^2 - 1 to be divisible by 24=8×324 = 8 \times 3, nn must be an odd integer that is not divisible by 3, which means gcd⁡(n,6)=1\gcd(n, 6) = 1. The positive integers less than 30 coprime to 6 are 1, 5, 7, 11, 13, 17, 19, 23, 25, and 29, totaling 10 integers.

Step-by-Step Solution

1
Factor the expression and analyze divisibility requirements for 24.
n2−1=(n−1)(n+1)n^2 - 1 = (n - 1)(n + 1). Since 24=8×324 = 8 \times 3 with gcd⁡(8,3)=1\gcd(8, 3) = 1, n2−1n^2 - 1 must be divisible by both 88 and 33.
Decompose 24 into coprime prime-power factors to evaluate modular conditions independently.
2
Determine the condition for n2−1n^2 - 1 to be divisible by 8.
If nn is even, n2−1n^2 - 1 is odd and cannot be divisible by 8. If nn is odd, let n=2k+1n = 2k + 1; then n2−1=4k(k+1)n^2 - 1 = 4k(k + 1). Since one of kk or k+1k + 1 is always even, 4k(k+1)4k(k + 1) is divisible by 8. Thus, nn must be odd.
Analyze parity requirements for the power of 2.
3
Determine the condition for n2−1n^2 - 1 to be divisible by 3.
If nn is a multiple of 3, n2−1≡−1(mod3)n^2 - 1 \equiv -1 \pmod 3, which is not divisible by 3. If nn is not a multiple of 3, n≡1n \equiv 1 or 2(mod3)2 \pmod 3, so n2≡1(mod3)n^2 \equiv 1 \pmod 3 and n2−1n^2 - 1 is divisible by 3. Thus, nn cannot be a multiple of 3.
Analyze remainder properties modulo 3.
4
Combine the conditions and count valid positive integers n<30n < 30.
The combined requirement is gcd⁡(n,6)=1\gcd(n, 6) = 1 (nn is odd and not a multiple of 3). The positive integers less than 30 satisfying this are 1, 5, 7, 11, 13, 17, 19, 23, 25, and 29. Counting these yields 10 integers.
Enumerate all integers meeting the coprime condition within the specified domain.

Key Concept

Modular arithmetic properties of quadratic expressions and coprimality constraints
Question 4Question →

Let nn be a positive integer whose prime factorization consists only of the prime factors 22 and 33. If nn has exactly 1212 positive divisors and gcd⁡(n,36)=12\gcd(n, 36) = 12, what is the value of nn?

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Answer: 96

Answer

The value of nn is 9696.
Representing n=2a×3bn = 2^a \times 3^b, the number of positive divisors is (a+1)(b+1)=12(a+1)(b+1) = 12. The greatest common divisor gcd⁡(n,36)=gcd⁡(2a×3b,22×32)=2min⁡(a,2)×3min⁡(b,2)=12=22×31\gcd(n, 36) = \gcd(2^a \times 3^b, 2^2 \times 3^2) = 2^{\min(a,2)} \times 3^{\min(b,2)} = 12 = 2^2 \times 3^1. This requires min⁡(a,2)=2  ⟹  a≥2\min(a,2) = 2 \implies a \ge 2 and min⁡(b,2)=1  ⟹  b=1\min(b,2) = 1 \implies b = 1. Substituting b=1b = 1 into (a+1)(1+1)=12(a+1)(1+1) = 12 gives 2(a+1)=122(a+1) = 12, so a=5a = 5. Therefore, n=25×31=32×3=96n = 2^5 \times 3^1 = 32 \times 3 = 96.

Step-by-Step Solution

1
Set up the prime factorization of nn and the divisor count equation.
n=2a×3bn = 2^a \times 3^b and (a+1)(b+1)=12(a + 1)(b + 1) = 12.
Since the prime factors of nn are only 22 and 33, nn must take the form 2a×3b2^a \times 3^b, where the number of positive divisors is (a+1)(b+1)(a+1)(b+1).
2
Analyze the exponent requirements using the greatest common divisor.
a≥2a \ge 2 and b=1b = 1.
gcd⁡(2a×3b,22×32)=2min⁡(a,2)×3min⁡(b,2)=22×31\gcd(2^a \times 3^b, 2^2 \times 3^2) = 2^{\min(a,2)} \times 3^{\min(b,2)} = 2^2 \times 3^1. Matching powers gives min⁡(a,2)=2  ⟹  a≥2\min(a,2) = 2 \implies a \ge 2, and min⁡(b,2)=1  ⟹  b=1\min(b,2) = 1 \implies b = 1.
3
Solve for exponent aa and calculate nn.
a=5a = 5, giving n=25×31=96n = 2^5 \times 3^1 = 96.
Substituting b=1b = 1 into (a+1)(1+1)=12(a+1)(1+1) = 12 gives 2(a+1)=12  ⟹  a=52(a+1) = 12 \implies a = 5, which satisfies a≥2a \ge 2.

Key Concept

Prime exponent rules for GCD and divisor counting
Estimated Time:2m 0s
Question 5Question →

Let K=25×34×53×112K = 2^5 \times 3^4 \times 5^3 \times 11^2. How many positive integer factors of KK are divisible by 300300 but are not divisible by 900900?

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Answer: 24

Answer

24
The correct answer 24 is obtained by analyzing the prime factorizations 300=22×31×52300 = 2^2 \times 3^1 \times 5^2 and 900=22×32×52900 = 2^2 \times 3^2 \times 5^2. Any factor f=2a×3b×5c×11df = 2^a \times 3^b \times 5^c \times 11^d must satisfy a∈{2,3,4,5}a \in \{2, 3, 4, 5\} (4 choices), b=1b = 1 (1 choice, since b≥1b \ge 1 for 300 but b<2b < 2 for 900), c∈{2,3}c \in \{2, 3\} (2 choices), and d∈{0,1,2}d \in \{0, 1, 2\} (3 choices). Multiplying these choices gives 4×1×2×3=244 \times 1 \times 2 \times 3 = 24.

Step-by-Step Solution

1
Express any factor of KK in terms of prime factor exponent constraints.
Any factor ff of KK has the form f=2a×3b×5c×11df = 2^a \times 3^b \times 5^c \times 11^d, where 0≤a≤50 \le a \le 5, 0≤b≤40 \le b \le 4, 0≤c≤30 \le c \le 3, and 0≤d≤20 \le d \le 2.
The prime factors of ff must be subsets of the prime factors of KK with exponents not exceeding those in KK.
2
Find the prime factorizations of 300300 and 900900.
300=22×31×52300 = 2^2 \times 3^1 \times 5^2 and 900=22×32×52900 = 2^2 \times 3^2 \times 5^2.
Divisibility criteria correspond to minimum exponent requirements for each prime factor.
3
Apply the divisibility conditions to determine constraints on each exponent.
For ff to be divisible by 300300, we need a≥2a \ge 2, b≥1b \ge 1, and c≥2c \ge 2. For ff to NOT be divisible by 900900, we must have b<2b < 2. Thus, b=1b = 1.
Combining b≥1b \ge 1 and b<2b < 2 uniquely restricts bb to 11.
4
Count the number of valid choices for each exponent.
a∈{2,3,4,5}a \in \{2, 3, 4, 5\} (44 choices), b∈{1}b \in \{1\} (11 choice), c∈{2,3}c \in \{2, 3\} (22 choices), and d∈{0,1,2}d \in \{0, 1, 2\} (33 choices).
The exponent dd is unrestricted by 300300 or 900900, so it can take any valid power present in KK.
5
Multiply the number of independent choices using the fundamental counting principle.
4×1×2×3=244 \times 1 \times 2 \times 3 = 24.
Each exponent choice can be paired independently to form a unique factor.

Key Concept

Counting Divisors Using Prime Exponent Constraints
Question 6Question →

A beverage mixture is prepared by combining apple juice, orange juice, and grape juice in a ratio of 3:4:53 : 4 : 5 by volume. If a pitcher contains a total of 4848 fluid ounces of this mixture, how many fluid ounces of orange juice are in the pitcher?

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Answer: 1616 fluid ounces

Answer

1616 fluid ounces
The ratio 3:4:53 : 4 : 5 indicates that the total mixture is divided into 3+4+5=123 + 4 + 5 = 12 equal parts. Orange juice accounts for 44 of these 1212 parts, representing 412=13\frac{4}{12} = \frac{1}{3} of the total volume. Taking 13\frac{1}{3} of 4848 fluid ounces gives 1616 fluid ounces.

Step-by-Step Solution

1
Calculate the total number of ratio parts.
Total parts = 3+4+5=123 + 4 + 5 = 12.
To find the fraction of each ingredient in the mixture, we must sum all parts of the ratio.
2
Determine the fraction of the total volume that represents orange juice.
Orange juice fraction = 412=13\frac{4}{12} = \frac{1}{3}.
Orange juice corresponds to 44 parts out of the 1212 total parts.
3
Multiply the orange juice fraction by the total volume.
13×48=16\frac{1}{3} \times 48 = 16 fluid ounces.
Multiplying the part-to-whole fraction by the total volume yields the exact volume of orange juice.

Key Concept

Part-to-Whole Ratio Relationships
Estimated Time:45s
Question 7Question →

If xx and yy are real numbers such that ∣2x−5∣≤9|2x - 5| \le 9 and ∣y+3∣≤4|y + 3| \le 4, what is the maximum possible value of ∣x−2y+1∣|x - 2y + 1|?

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Answer: 2222

Answer

The maximum possible value of ∣x−2y+1∣|x - 2y + 1| is 22.
The correct answer is 22 because solving the absolute value inequalities gives −2≤x≤7-2 \le x \le 7 and −7≤y≤1-7 \le y \le 1. To maximize x−2y+1x - 2y + 1, we take the largest possible value of xx (77), the smallest possible value of yy (−7-7, making −2y=14-2y = 14), and add 11, obtaining 7+14+1=227 + 14 + 1 = 22. Since the lower bound of x−2y+1x - 2y + 1 is −3-3, the maximum absolute value is ∣22∣=22|22| = 22.

Step-by-Step Solution

1
Find the range of possible values for xx from the inequality ∣2x−5∣≤9|2x - 5| \le 9.
−9≤2x−5≤9  ⟹  −4≤2x≤14  ⟹  −2≤x≤7-9 \le 2x - 5 \le 9 \implies -4 \le 2x \le 14 \implies -2 \le x \le 7.
An absolute value inequality ∣A∣≤B|A| \le B unwraps to −B≤A≤B-B \le A \le B.
2
Find the range of possible values for yy from the inequality ∣y+3∣≤4|y + 3| \le 4.
−4≤y+3≤4  ⟹  −7≤y≤1-4 \le y + 3 \le 4 \implies -7 \le y \le 1.
Unwrapping the absolute value inequality gives the upper and lower bounds for yy on the real number line.
3
Determine the range of possible values for −2y-2y.
Multiplying −7≤y≤1-7 \le y \le 1 by −2-2 and reversing the inequality signs yields −2(1)≤−2y≤−2(−7)  ⟹  −2≤−2y≤14-2(1) \le -2y \le -2(-7) \implies -2 \le -2y \le 14.
Multiplying an inequality by a negative number reverses the direction of the inequality signs.
4
Combine the ranges to find the minimum and maximum bounds for z=x−2y+1z = x - 2y + 1.
Minimum z=(−2)+(−2)+1=−3z = (-2) + (-2) + 1 = -3; Maximum z=7+14+1=22z = 7 + 14 + 1 = 22. Thus, −3≤x−2y+1≤22-3 \le x - 2y + 1 \le 22.
Adding the individual minimums gives the absolute minimum, and adding the individual maximums gives the absolute maximum.
5
Calculate the maximum value of the absolute value ∣x−2y+1∣|x - 2y + 1|.
max(∣x−2y+1∣)=max(∣−3∣,∣22∣)=22\\max(|x - 2y + 1|) = \\max(|-3|, |22|) = 22.
The absolute value of a quantity ranging from −3-3 to 2222 reaches its maximum distance from zero at 2222.

Key Concept

Properties of Real Numbers, Number Line Inequalities, and Absolute Value Operations
Estimated Time:2m 30s
Question 8Question →

A laboratory mixture is created by combining two solutions containing a specific tracer compound. Solution X has a volume of 2.5×1032.5 \times 10^3 milliliters and a tracer concentration of 4.8×10−74.8 \times 10^{-7} grams per milliliter. Solution Y has a volume of 7.5×1037.5 \times 10^3 milliliters and a tracer concentration of 1.6×10−61.6 \times 10^{-6} grams per milliliter. What is the total mass, in grams, of the tracer compound present in the combined mixture?

Show answer & explanation

Answer: 1.32×10−21.32 \times 10^{-2}

Answer

1.32×10−21.32 \times 10^{-2} grams
The mass in Solution X is (2.5×103)×(4.8×10−7)=1.2×10−3(2.5 \times 10^3) \times (4.8 \times 10^{-7}) = 1.2 \times 10^{-3} grams, and the mass in Solution Y is (7.5×103)×(1.6×10−6)=1.2×10−2(7.5 \times 10^3) \times (1.6 \times 10^{-6}) = 1.2 \times 10^{-2} grams. Rewriting 1.2×10−31.2 \times 10^{-3} as 0.12×10−20.12 \times 10^{-2} allows direct addition: (0.12+1.2)×10−2=1.32×10−2(0.12 + 1.2) \times 10^{-2} = 1.32 \times 10^{-2} grams.

Step-by-Step Solution

1
Calculate the mass of the tracer in Solution X
MX=(2.5×103)×(4.8×10−7)=12.0×10−4=1.2×10−3M_X = (2.5 \times 10^3) \times (4.8 \times 10^{-7}) = 12.0 \times 10^{-4} = 1.2 \times 10^{-3} grams
Mass equals volume multiplied by concentration.
2
Calculate the mass of the tracer in Solution Y
MY=(7.5×103)×(1.6×10−6)=12.0×10−3=1.2×10−2M_Y = (7.5 \times 10^3) \times (1.6 \times 10^{-6}) = 12.0 \times 10^{-3} = 1.2 \times 10^{-2} grams
Mass equals volume multiplied by concentration.
3
Sum the tracer masses and adjust to scientific notation
Mtotal=1.2×10−3+1.2×10−2=0.12×10−2+1.2×10−2=1.32×10−2M_{\text{total}} = 1.2 \times 10^{-3} + 1.2 \times 10^{-2} = 0.12 \times 10^{-2} + 1.2 \times 10^{-2} = 1.32 \times 10^{-2} grams
To add terms in scientific notation, convert them to have matching exponents before adding coefficients.

Key Concept

Arithmetic operations with Scientific Notation and Decimals
Estimated Time:1m 30s
Question 9Question →

A sequence a1,a2,a3,…a_1, a_2, a_3, \dots is defined by an=1n+n+2a_n = \frac{1}{\sqrt{n} + \sqrt{n+2}} for all positive integers nn. If S=∑n=198anS = \sum_{n=1}^{98} a_n, which of the following is closest to the value of SS when rounded to the nearest tenth?

Show answer & explanation

Answer: 8.8

Answer

The value of SS rounded to the nearest tenth is 8.8.
The term ana_n simplifies to n+2−n2\frac{\sqrt{n+2} - \sqrt{n}}{2} upon rationalizing the denominator. Summing from n=1n=1 to 9898 causes all middle terms to cancel out, leaving 99+100−1−22\frac{\sqrt{99} + \sqrt{100} - \sqrt{1} - \sqrt{2}}{2}. Substituting 100=10\sqrt{100} = 10, 1=1\sqrt{1} = 1, 99≈9.95\sqrt{99} \approx 9.95, and 2≈1.41\sqrt{2} \approx 1.41 gives approximately 8.778.77, which rounds to 8.8.

Step-by-Step Solution

1
Rationalize the general term ana_n
an=1n+n+2⋅n+2−nn+2−n=n+2−n(n+2)−n=n+2−n2a_n = \frac{1}{\sqrt{n} + \sqrt{n+2}} \cdot \frac{\sqrt{n+2} - \sqrt{n}}{\sqrt{n+2} - \sqrt{n}} = \frac{\sqrt{n+2} - \sqrt{n}}{(n+2) - n} = \frac{\sqrt{n+2} - \sqrt{n}}{2}
Eliminating radicals from the denominator reveals the underlying telescoping structure of the sequence.
2
Expand the summation S=∑n=198anS = \sum_{n=1}^{98} a_n
S=12[(3−1)+(4−2)+(5−3)+⋯+(99−97)+(100−98)]S = \frac{1}{2} \left[ (\sqrt{3} - \sqrt{1}) + (\sqrt{4} - \sqrt{2}) + (\sqrt{5} - \sqrt{3}) + \dots + (\sqrt{99} - \sqrt{97}) + (\sqrt{100} - \sqrt{98}) \right]
Writing out initial and final terms demonstrates which terms cancel.
3
Simplify the telescoping sum
S=99+100−1−22=99+10−1−22=9+99−22S = \frac{\sqrt{99} + \sqrt{100} - \sqrt{1} - \sqrt{2}}{2} = \frac{\sqrt{99} + 10 - 1 - \sqrt{2}}{2} = \frac{9 + \sqrt{99} - \sqrt{2}}{2}
All intermediate terms cancel out, leaving two positive boundary terms and two negative boundary terms.
4
Estimate square root values and perform rounding
Since 99≈9.94987\sqrt{99} \approx 9.94987 and 2≈1.41421\sqrt{2} \approx 1.41421, S≈9+9.94987−1.414212=17.535662=8.76783≈8.8S \approx \frac{9 + 9.94987 - 1.41421}{2} = \frac{17.53566}{2} = 8.76783 \approx 8.8
Evaluating the radicals to two decimal places allows accurate rounding to the nearest tenth.

Key Concept

Telescoping Series Summation and Square Root Estimation
Question 10Question →

If x>1x > 1 and xx=x2\sqrt{x^{\sqrt{x}}} = x^2, what is the value of xx?

Show answer & explanation

Answer: 1616

Answer

16
By rewriting the square root as an exponent of 1/21/2, the left side becomes xx2x^{\frac{\sqrt{x}}{2}}. Since the base xx is greater than 1, set the exponents equal to each other: x2=2\frac{\sqrt{x}}{2} = 2, which gives x=4\sqrt{x} = 4. Squaring both sides yields x=16x = 16.

Step-by-Step Solution

1
Express the radical on the left side of the equation as a fractional exponent.
xx=(xx)12=xx2\sqrt{x^{\sqrt{x}}} = (x^{\sqrt{x}})^{\frac{1}{2}} = x^{\frac{\sqrt{x}}{2}}
The square root rule states that ak=a1k\sqrt[k]{a} = a^{\frac{1}{k}}.
2
Set the exponent of the left side equal to the exponent of the right side.
x2=2\frac{\sqrt{x}}{2} = 2
Since the bases are equal (x>1x > 1), their corresponding exponents must be equal.
3
Solve for x\sqrt{x} by multiplying both sides by 2.
x=4\sqrt{x} = 4
Isolating the radical term allows determination of the root value.
4
Square both sides to solve for xx.
x=42=16x = 4^2 = 16
Squaring a principal square root yields the underlying radicand.

Key Concept

Combining Fractional Exponents and Radical Expressions
Question 11Question →

A renewable energy facility distributes stored electricity among three battery banks: Bank 1, Bank 2, and Bank 3. Initially, the ratio of the energy stored in Bank 1 to Bank 2 to Bank 3 is 3:4:53 : 4 : 5.

To balance the system, two sequential transfers of energy are performed without any energy loss:
1. Energy is transferred from Bank 3 to Bank 1 such that the ratio of the energy in Bank 1 to Bank 2 becomes 5:45 : 4.
2. Energy is then transferred from Bank 2 to Bank 3 such that the ratio of the energy in Bank 2 to Bank 3 becomes 1:41 : 4.

If Bank 3 contains 60 MWh60\text{ MWh} more energy after the second transfer than it did initially, what was the total amount of energy, in MWh, stored across all three battery banks?

Show answer & explanation

Answer: 1,200 MWh1,200\text{ MWh}

Answer

The total energy stored across all three battery banks is 1,200 MWh1,200\text{ MWh}.
The correct answer is 1,200 MWh1,200\text{ MWh}. By setting up algebraic expressions for the energy in each bank relative to total energy TT, we track how each transfer modifies individual bank totals while conserving total energy. The intermediate transfer reduces Bank 3 from 2560T\frac{25}{60}T to 1560T\frac{15}{60}T, and the subsequent transfer increases it to 2860T\frac{28}{60}T. The resulting difference of 360T=120T\frac{3}{60}T = \frac{1}{20}T equals 60 MWh60\text{ MWh}, solving directly to T=1,200 MWhT = 1,200\text{ MWh}.

Step-by-Step Solution

1
Express initial energy quantities in terms of total energy TT
Bank 1 = 312T\frac{3}{12}T, Bank 2 = 412T\frac{4}{12}T, Bank 3 = 512T=2560T\frac{5}{12}T = \frac{25}{60}T
The ratio 3:4:53 : 4 : 5 sums to 1212 total parts.
2
Calculate energy amounts after the first transfer (Bank 3 to Bank 1)
Bank 2 remains 412T\frac{4}{12}T; Bank 1 becomes 512T\frac{5}{12}T; Bank 3 becomes T−512T−412T=312TT - \frac{5}{12}T - \frac{4}{12}T = \frac{3}{12}T
The new ratio of Bank 1 to Bank 2 is 5:45 : 4, and Bank 2's energy did not change.
3
Calculate energy amounts after the second transfer (Bank 2 to Bank 3)
Combined energy in Banks 2 and 3 = 412T+312T=712T\frac{4}{12}T + \frac{3}{12}T = \frac{7}{12}T. Final Bank 3 = 45×712T=2860T\frac{4}{5} \times \frac{7}{12}T = \frac{28}{60}T
Bank 1 remains unchanged at 512T\frac{5}{12}T. The remaining energy is divided between Bank 2 and Bank 3 in a 1:41 : 4 ratio.
4
Determine the net change in Bank 3 and solve for total energy TT
Net change = 2860T−2560T=360T=120T\frac{28}{60}T - \frac{25}{60}T = \frac{3}{60}T = \frac{1}{20}T. Since 120T=60 MWh\frac{1}{20}T = 60\text{ MWh}, T=1,200 MWhT = 1,200\text{ MWh}
Bank 3 ended with 60 MWh60\text{ MWh} more than its initial amount.

Key Concept

Multi-stage ratio rebalancing and conservation of total quantity
Question 12Question →

Let pp and qq be integers such that p<0<qp < 0 < q, (−1)p+(−1)q=0(-1)^p + (-1)^q = 0, and p2q+pp^2 q + p is an even integer. Which of the following expressions must be negative?

Show answer & explanation

Answer: pqp^q

Answer

pqp^q
From (−1)p+(−1)q=0(-1)^p + (-1)^q = 0, pp and qq must have opposite parities. Factoring p2q+pp^2 q + p into p(pq+1)p(pq + 1) shows that if pp were odd, qq would be even, leading to an odd product odd×odd=odd\text{odd} \times \text{odd} = \text{odd}. Because p2q+pp^2 q + p is given as even, pp must be an even integer and qq must be an odd integer. Given p<0p < 0, pp is a negative even integer, and qq is a positive odd integer. The expression pqp^q represents a negative base raised to an odd exponent, which is guaranteed to be negative.

Step-by-Step Solution

1
Determine the parities of pp and qq using the equation (−1)p+(−1)q=0(-1)^p + (-1)^q = 0.
One of pp or qq is even and the other is odd.
For the sum (−1)p+(−1)q(-1)^p + (-1)^q to equal 00, one term must equal 11 and the other must equal −1-1, meaning pp and qq have opposite parities.
2
Analyze the parity of p2q+p=p(pq+1)p^2 q + p = p(pq + 1) to determine which variable is even.
pp must be even, and qq must be odd.
If pp were odd, qq would be even, making pqpq even, pq+1pq+1 odd, and p(pq+1)p(pq+1) odd. Since p2q+pp^2 q + p is even, pp must be even and qq must be odd.
3
Evaluate the sign of pqp^q using sign rules for exponents.
pq<0p^q < 0
pp is negative (p<0p < 0) and qq is a positive odd integer. A negative number raised to an odd power is always negative.

Key Concept

Parity and sign rules for negative bases and integer exponents
Question 13Question →

If xx is a real number such that 16x+34=8x−1\sqrt[4]{16^{x+3}} = 8^{x-1}, what is the value of xx?

Show answer & explanation

Answer: 3

Answer

The value of xx is 33.
To solve 16x+34=8x−1\sqrt[4]{16^{x+3}} = 8^{x-1}, express both sides with the base 2. The left side simplifies to (24)x+34=24(x+3)4=2x+3\sqrt[4]{(2^4)^{x+3}} = 2^{\frac{4(x+3)}{4}} = 2^{x+3}. The right side simplifies to (23)x−1=23(x−1)=23x−3(2^3)^{x-1} = 2^{3(x-1)} = 2^{3x-3}. Equating the exponents yields x+3=3x−3x + 3 = 3x - 3, which solves to 2x=62x = 6, giving x=3x = 3.

Step-by-Step Solution

1
Rewrite 16 and 8 using prime base 2
16=2416 = 2^4 and 8=238 = 2^3
Converting terms to a common base allows direct comparison of exponents.
2
Simplify the left-hand side radical expression
16x+34=(24)x+34=24(x+3)4=2x+3\sqrt[4]{16^{x+3}} = \sqrt[4]{(2^4)^{x+3}} = 2^{\frac{4(x+3)}{4}} = 2^{x+3}
The nn-th root amn\sqrt[n]{a^m} is equivalent to am/na^{m/n}.
3
Simplify the right-hand side exponential expression
8x−1=(23)x−1=23(x−1)=23x−38^{x-1} = (2^3)^{x-1} = 2^{3(x-1)} = 2^{3x-3}
Applying the exponent power rule (am)n=am⋅n(a^m)^n = a^{m \cdot n} requires multiplying 33 by (x−1)(x - 1).
4
Equate the exponents and solve for xx
x+3=3x−3  ⟹  2x=6  ⟹  x=3x + 3 = 3x - 3 \implies 2x = 6 \implies x = 3
When au=ava^u = a^v for a>0a > 0 and a≠1a \neq 1, it follows that u=vu = v.

Key Concept

Solving exponential equations using prime base factorization and radical conversion rules
Question 14Question →

How many positive integers less than 100100 are divisible by both 44 and 66, but are NOT divisible by 88?

Show answer & explanation

Answer: 4

Answer

The correct numerical answer is 4.
To be divisible by both 4 and 6, an integer must be a multiple of LCM(4,6)=12\text{LCM}(4, 6) = 12. The positive integers less than 100 that are multiples of 12 are 12, 24, 36, 48, 60, 72, 84, and 96 (8 integers). Among these, those divisible by 8 are multiples of LCM(12,8)=24\text{LCM}(12, 8) = 24, which are 24, 48, 72, and 96 (4 integers). Subtracting the excluded integers yields 8−4=48 - 4 = 4.

Step-by-Step Solution

1
Find the least common multiple of 4 and 6.
LCM(4, 6) = 12
An integer divisible by both 4 and 6 must be a multiple of their least common multiple.
2
Count positive integers less than 100 that are multiples of 12.
The multiples are 12, 24, 36, 48, 60, 72, 84, and 96, giving 8 integers.
The largest multiple of 12 strictly less than 100 is 96 (12 × 8).
3
Identify multiples of 12 that are also divisible by 8.
Since LCM(12, 8) = 24, these are the multiples of 24: 24, 48, 72, and 96, giving 4 integers.
Any integer divisible by both 12 and 8 must be a multiple of 24.
4
Subtract the excluded integers from the total count.
8 - 4 = 4
We exclude the multiples of 8 from the set of multiples of 12.

Key Concept

Divisibility, Least Common Multiple (LCM), and Set Exclusion
Question 15Question →

What is the least positive integer nn that leaves a remainder of 33 when divided by 77, a remainder of 44 when divided by 55, and is divisible by 99?

Show answer & explanation

Answer: 234

Answer

The least positive integer satisfying all three conditions is 234.
To find the least positive integer nn that satisfies n≡3(mod7)n \equiv 3 \pmod{7}, n≡4(mod5)n \equiv 4 \pmod{5}, and n≡0(mod9)n \equiv 0 \pmod{9}, we first find a general expression for integers meeting the first two conditions. Checking values of 5m+45m + 4 modulo 7 gives 2424 as the smallest positive integer matching both. The combined condition is n≡24(mod35)n \equiv 24 \pmod{35}, or n=35k+24n = 35k + 24. Requiring 35k+2435k + 24 to be divisible by 9 gives 8k+6≡0(mod9)8k + 6 \equiv 0 \pmod{9}, which simplifies to k≡6(mod9)k \equiv 6 \pmod{9}. The smallest non-negative integer value for kk is 66, leading to n=35(6)+24=234n = 35(6) + 24 = 234.

Step-by-Step Solution

1
Set up system of modular congruences for the remainders
n≡3(mod7)n \equiv 3 \pmod{7}, n≡4(mod5)n \equiv 4 \pmod{5}, and n≡0(mod9)n \equiv 0 \pmod{9}
Translates the remainder and divisibility conditions into mathematical equations.
2
Combine the first two congruences using the Chinese Remainder Theorem approach
n≡24(mod35)n \equiv 24 \pmod{35}, so n=35k+24n = 35k + 24 for an integer k≥0k \ge 0
Since lcm(5,7)=35\text{lcm}(5, 7) = 35, the solutions to the combined system repeat every 35 integer values.
3
Enforce the divisibility condition by 9 on n=35k+24n = 35k + 24
35k+24≡0(mod9)  ⟹  8k+6≡0(mod9)  ⟹  −k+6≡0(mod9)  ⟹  k≡6(mod9)35k + 24 \equiv 0 \pmod{9} \implies 8k + 6 \equiv 0 \pmod{9} \implies -k + 6 \equiv 0 \pmod{9} \implies k \equiv 6 \pmod{9}
Reduces coefficients modulo 9 to find the values of kk that make nn a multiple of 9.
4
Calculate the smallest positive integer nn corresponding to k=6k = 6
n=35(6)+24=234n = 35(6) + 24 = 234
Choosing k=6k = 6 yields the smallest non-negative integer for kk that satisfies all conditions.

Key Concept

Simultaneous congruences and divisibility constraints
Question 16Question →

If nn is an integer that is divisible by 1515, which of the following statements MUST also be true? Select all such statements.

Select all that apply

Show answer & explanation

Answer: nn is divisible by 33; nn is divisible by 55; n+15n + 15 is divisible by 1515

Answer

The statements asserting that nn is divisible by 33, nn is divisible by 55, and n+15n + 15 is divisible by 1515 must all be true.
Any integer divisible by 1515 must be divisible by all factors of 1515, which includes 33 and 55. Furthermore, adding 1515 to a multiple of 1515 yields another multiple of 1515, so n+15n + 15 is also divisible by 1515.

Step-by-Step Solution

1
Analyze the prime factor decomposition of the divisor
The prime factorization of 1515 is 3×53 \times 5. Therefore, any integer nn that is a multiple of 1515 can be expressed as n=15k=3×5×kn = 15k = 3 \times 5 \times k for some integer kk.
Understanding factor relationships allows evaluation of divisibility rules.
2
Evaluate divisibility of nn by 33 and 55
Since n=3(5k)n = 3(5k), nn is divisible by 33. Since n=5(3k)n = 5(3k), nn is divisible by 55. Both statements are always true.
Any multiple of a composite number is also a multiple of that number's factors.
3
Evaluate the expression n+15n + 15
n+15=15k+15=15(k+1)n + 15 = 15k + 15 = 15(k + 1). Since k+1k + 1 is an integer, n+15n + 15 is a multiple of 1515. This statement is always true.
Adding a multiple of 1515 to another multiple of 1515 yields a multiple of 1515.
4
Test counterexamples for remaining statements
For n=15n = 15: 1515 is not divisible by 3030, so divisibility by 3030 is not guaranteed. Additionally, 1515 is odd, so being an even integer is not guaranteed.
A statement must hold for all possible values of nn to be necessarily true.

Key Concept

Divisibility by a composite number implies divisibility by all of its factors, and multiples of a number remain multiples when another multiple of that number is added.
Question 17Question →

On the real number line, point PP represents the real number xx, point QQ represents 77, and point RR represents −5-5. If the distance between PP and QQ is equal to 33 times the distance between PP and RR, what is the sum of all possible values of xx?

Show answer & explanation

Answer: −13-13

Answer

The sum of all possible values of xx is −13-13.
The distance between xx and 77 is ∣x−7∣|x - 7| and the distance between xx and −5-5 is ∣x+5∣|x + 5|. Equating ∣x−7∣=3∣x+5∣|x - 7| = 3|x + 5| leads to two equations: x−7=3(x+5)x - 7 = 3(x + 5) giving x=−11x = -11, and x−7=−3(x+5)x - 7 = -3(x + 5) giving x=−2x = -2. Adding both solutions yields (−11)+(−2)=−13(-11) + (-2) = -13.

Step-by-Step Solution

1
Set up the distance equation using absolute value notation
The distance between P(x)P(x) and Q(7)Q(7) is ∣x−7∣|x - 7|, and the distance between P(x)P(x) and R(−5)R(-5) is ∣x−(−5)∣=∣x+5∣|x - (-5)| = |x + 5|. The problem specifies that ∣x−7∣=3∣x+5∣|x - 7| = 3|x + 5|.
Distance between two points aa and bb on a real number line is expressed as ∣a−b∣|a - b|.
2
Solve Case 1 where x−7x - 7 and x+5x + 5 have the same sign
x−7=3(x+5)  ⟹  x−7=3x+15  ⟹  −22=2x  ⟹  x=−11x - 7 = 3(x + 5) \implies x - 7 = 3x + 15 \implies -22 = 2x \implies x = -11.
When both absolute value expressions have identical signs, ∣x−7∣=3∣x+5∣|x - 7| = 3|x + 5| simplifies directly to x−7=3(x+5)x - 7 = 3(x + 5).
3
Solve Case 2 where x−7x - 7 and x+5x + 5 have opposite signs
x−7=−3(x+5)  ⟹  x−7=−3x−15  ⟹  4x=−8  ⟹  x=−2x - 7 = -3(x + 5) \implies x - 7 = -3x - 15 \implies 4x = -8 \implies x = -2.
When the absolute value expressions have opposite signs, ∣x−7∣=3∣x+5∣|x - 7| = 3|x + 5| simplifies to x−7=−3(x+5)x - 7 = -3(x + 5).
4
Calculate the sum of all possible values of xx
(−11)+(−2)=−13(-11) + (-2) = -13.
Summing the two solutions gives the final required value.

Key Concept

Distance on a number line and absolute value equations
Estimated Time:1m 30s
Question 18Question →
If xx is a positive integer such that
4x+1⋅52x+4x⋅52x+1=30,000\sqrt{4^{x+1} \cdot 5^{2x} + 4^x \cdot 5^{2x+1}} = 30,000
what is the value of xx?
Show answer & explanation

Answer: 4

Answer

The value of xx is 4.
Factoring the common exponential term 4x⋅52x4^x \cdot 5^{2x} inside the radical yields (4x⋅52x)(4+5)=9⋅4x⋅(52)x=9⋅(4⋅25)x=9⋅100x=9⋅102x=3⋅10x\sqrt{(4^x \cdot 5^{2x})(4 + 5)} = \sqrt{9 \cdot 4^x \cdot (5^2)^x} = \sqrt{9 \cdot (4 \cdot 25)^x} = \sqrt{9 \cdot 100^x} = \sqrt{9 \cdot 10^{2x}} = 3 \cdot 10^x. Setting 3⋅10x=30,0003 \cdot 10^x = 30,000 gives 10x=10,000=10410^x = 10,000 = 10^4, which means x=4x = 4.

Step-by-Step Solution

1
Separate the addition in exponents using exponent rules.
4x+1⋅52x=4x⋅41⋅52x4^{x+1} \cdot 5^{2x} = 4^x \cdot 4^1 \cdot 5^{2x} and 4x⋅52x+1=4x⋅52x⋅514^x \cdot 5^{2x+1} = 4^x \cdot 5^{2x} \cdot 5^1.
Applying the product rule of exponents am+n=am⋅ana^{m+n} = a^m \cdot a^n prepares terms for factoring.
2
Factor out the common expression 4x⋅52x4^x \cdot 5^{2x} from the sum inside the radical.
4x+1⋅52x+4x⋅52x+1=(4x⋅52x)(4+5)=9⋅4x⋅52x4^{x+1} \cdot 5^{2x} + 4^x \cdot 5^{2x+1} = (4^x \cdot 5^{2x})(4 + 5) = 9 \cdot 4^x \cdot 5^{2x}.
Factoring converts the sum under the square root into a single product.
3
Combine terms with powers into base 10.
4x⋅52x=4x⋅(52)x=4x⋅25x=(4⋅25)x=100x=102x4^x \cdot 5^{2x} = 4^x \cdot (5^2)^x = 4^x \cdot 25^x = (4 \cdot 25)^x = 100^x = 10^{2x}.
Using power of a power (am)n=amn(a^m)^n = a^{mn} and power of a product anbn=(ab)na^n b^n = (ab)^n simplifies the expression into powers of 10.
4
Take the square root of the simplified product.
9⋅102x=9⋅102x=3⋅10x\sqrt{9 \cdot 10^{2x}} = \sqrt{9} \cdot \sqrt{10^{2x}} = 3 \cdot 10^x.
Applying the product rule for radicals ab=ab\sqrt{ab} = \sqrt{a}\sqrt{b} and halving the exponent (102x)1/2=10x(10^{2x})^{1/2} = 10^x.
5
Equate the simplified expression to 30,000 and solve for xx.
3⋅10x=30,000  ⟹  10x=10,000  ⟹  10x=104  ⟹  x=43 \cdot 10^x = 30,000 \implies 10^x = 10,000 \implies 10^x = 10^4 \implies x = 4.
Dividing both sides by 3 isolates 10x10^x, and matching exponential bases gives x=4x = 4.

Key Concept

Exponent Rules and Radical Simplification
Estimated Time:2m 0s
Question 19Question →

If xx and yy are real numbers such that ∣x+2∣≤3|x + 2| \le 3 and ∣y−5∣≤2|y - 5| \le 2, which of the following could be the value of ∣x−y∣|x - y|? Select all such values.

Select all that apply

Show answer & explanation

Answer: 4; 8; 12

Answer

The possible values for ∣x−y∣|x - y| are 4, 8, and 12.
Solving the inequality ∣x+2∣≤3|x + 2| \le 3 gives −5≤x≤1-5 \le x \le 1, and solving ∣y−5∣≤2|y - 5| \le 2 gives 3≤y≤73 \le y \le 7. The minimum possible value of x−yx - y occurs at −5−7=−12-5 - 7 = -12, and the maximum value occurs at 1−3=−21 - 3 = -2. Thus, x−yx - y lies entirely in the interval [−12,−2][-12, -2]. Taking absolute values shows that ∣x−y∣|x - y| must lie in the interval [2,12][2, 12]. The numbers 4, 8, and 12 all fall within this interval and are valid solutions.

Step-by-Step Solution

1
Solve the absolute value inequality for xx.
−3≤x+2≤3  ⟹  −5≤x≤1-3 \le x + 2 \le 3 \implies -5 \le x \le 1
Unpack ∣x+2∣≤3|x + 2| \le 3 into a compound inequality and isolate xx.
2
Solve the absolute value inequality for yy.
−2≤y−5≤2  ⟹  3≤y≤7-2 \le y - 5 \le 2 \implies 3 \le y \le 7
Unpack ∣y−5∣≤2|y - 5| \le 2 into a compound inequality and isolate yy.
3
Determine the minimum and maximum possible values of x−yx - y.
Minimum x−y=−5−7=−12x - y = -5 - 7 = -12; Maximum x−y=1−3=−2x - y = 1 - 3 = -2.
To minimize x−yx - y, take the smallest xx and largest yy. To maximize x−yx - y, take the largest xx and smallest yy.
4
Find the range of ∣x−y∣|x - y|.
2≤∣x−y∣≤122 \le |x - y| \le 12
Since −12≤x−y≤−2-12 \le x - y \le -2, taking the absolute value yields values in the closed interval [2,12][2, 12].
5
Evaluate the choices against the interval [2,12][2, 12].
4, 8, and 12 lie inside [2,12][2, 12], whereas 1 and 15 do not.
Any real number in [2,12][2, 12] is a achievable value for ∣x−y∣|x - y|.

Key Concept

Real Numbers, Number Line, and Absolute Value Inequalities
Question 20Question →

Suppose aa, bb, and cc are integers such that a<0<b<ca < 0 < b < c. If a(b−c)a(b - c) is an odd integer and a2b+bca^2 b + b c is an even integer, which of the following expressions MUST be a positive even integer?

Show answer & explanation

Answer: b(c−a)b(c - a)

Answer

b(c−a)b(c - a) is guaranteed to be a positive even integer.
The expression b(c−a)b(c - a) consists of bb (which is positive and even) multiplied by (c−a)(c - a) (which is positive and even, as subtracting a negative odd number from a positive odd number yields a positive even number). The product of two positive even integers is always a positive even integer.

Step-by-Step Solution

1
Determine the signs of variables aa, bb, and cc.
a<0a < 0 (negative), b>0b > 0 (positive), and c>0c > 0 (positive).
Directly given by the inequality a<0<b<ca < 0 < b < c.
2
Analyze parity from the given condition that a(b−c)a(b - c) is odd.
aa is odd and (b−c)(b - c) is odd.
A product of two integers is odd if and only if both factors are odd.
3
Analyze parity from the second condition a2b+bc=b(a2+c)a^2 b + b c = b(a^2 + c) being even.
bb must be even, and cc must be odd.
Since aa is odd, a2a^2 is odd. If bb were odd, then a2+ca^2 + c would need to be even (making cc odd), but if both bb and cc were odd, b−cb - c would be even, contradicting step 2. Therefore, bb must be even. Since bb is even and b−cb - c is odd, cc must be odd.
4
Evaluate the sign and parity of b(c−a)b(c - a).
c−a=odd−(negative odd)=positive evenc - a = \text{odd} - (\text{negative odd}) = \text{positive even}. Since bb is positive even, b(c−a)=positive even×positive even=positive evenb(c - a) = \text{positive even} \times \text{positive even} = \text{positive even}.
Subtracting a negative number yields addition (c−a>0c - a > 0), and subtracting an odd integer from an odd integer produces an even integer.

Key Concept

Even-Odd Parity and Integer Sign Properties under Multiplication and Subtraction
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