Tüm alıştırma soruları

2195 soru

Soru 981Soru

A web hosting company charges each enterprise client a one-time fixed setup fee plus a constant monthly maintenance fee per server. A client operating 44 servers pays a total of $1,100\$1,100 for the setup fee and the first 66 months of server maintenance. A client operating 99 servers pays a total of $1,850\$1,850 for the setup fee and the first 66 months of server maintenance. What is the one-time fixed setup fee, in dollars?

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Cevap: 500

Cevap

500
The fixed setup fee is $500\$500. Letting SS represent the fixed setup fee and MM represent the 6-month maintenance fee per server, the two given scenarios yield S+4M=1100S + 4M = 1100 and S+9M=1850S + 9M = 1850. Subtracting the first equation from the second gives 5M=7505M = 750, which simplifies to M=150M = 150. Substituting M=150M = 150 into S+4(150)=1100S + 4(150) = 1100 gives S+600=1100S + 600 = 1100, so S=500S = 500.

Adım Adım Çözüm

1
Define variables for the unknown fixed cost and per-server cost, and construct the system of linear equations.
Let SS be the fixed setup fee in dollars and MM be the 6-month maintenance fee per server in dollars. The equations are S+4M=1100S + 4M = 1100 and S+9M=1850S + 9M = 1850.
Modeling the situational relationships as a linear system allows isolated solution of each unknown.
2
Subtract the two linear equations to eliminate the fixed fee SS and solve for MM.
5M=750    M=1505M = 750 \implies M = 150.
Since the coefficient of SS is 1 in both equations, elimination by subtraction directly isolates MM.
3
Substitute the value of MM back into the first equation to solve for SS.
S+4(150)=1100    S+600=1100    S=500S + 4(150) = 1100 \implies S + 600 = 1100 \implies S = 500.
Replacing MM with 150 yields a linear equation in one variable for the setup fee.

Anahtar Kavram

Linear Equations in One and Two Variables
Tahmini Süre:1m 30s
Soru 982Soru

A pharmacist has 8080 milliliters of a topical liquid formulation that is 5%5\% active ingredient by volume. How many milliliters of pure active ingredient must the pharmacist add to this formulation so that the resulting mixture is 24%24\% active ingredient by volume?

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Cevap: 20

Cevap

The pharmacist must add 2020 milliliters of pure active ingredient.
Adding 2020 mL of pure active ingredient increases the amount of solute from 44 mL to 2424 mL and the total mixture volume from 8080 mL to 100100 mL, yielding a final concentration of 24100=24%\frac{24}{100} = 24\%.

Adım Adım Çözüm

1
Determine initial amount of solute
Initial active ingredient volume = 80×0.05=480 \times 0.05 = 4 mL
Before adding pure active ingredient, the solution contains 5%5\% active ingredient of the total 8080 mL volume.
2
Formulate algebraic expressions for total active ingredient and total volume
Total active ingredient = 4+x4 + x mL; Total mixture volume = 80+x80 + x mL
Adding xx mL of pure active ingredient increases both the solute volume and the total mixture volume by xx.
3
Set up and solve the mixture concentration equation
4+x80+x=625    100+25x=480+6x    19x=380    x=20\frac{4 + x}{80 + x} = \frac{6}{25} \implies 100 + 25x = 480 + 6x \implies 19x = 380 \implies x = 20
Setting the solute ratio equal to the target concentration of 24%24\% yields a linear equation in xx.

Anahtar Kavram

Mixture Concentration and Algebraic Dilution/Fortification
Soru 983Soru

A corporation has three regional branches: Branch X, Branch Y, and Branch Z. The ratio of the number of employees in Branch X to Branch Y is 1:31 : 3. The average monthly salary of employees in Branch Y is 40%40\% higher than the average monthly salary of employees in Branch X. Branch Z has twice as many employees as Branch Y, and the average monthly salary of employees in Branch Z is x%x\% lower than the combined average monthly salary of employees in Branches X and Y. If the overall average monthly salary across all three branches combined is $910\$910, and the average monthly salary in Branch X is $1000\$1{}000, what is the value of xx?

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Cevap: 50

Cevap

50
The combined average salary of Branches X and Y is calculated by weighting Branch X's salary ($1000\$1{}000) with 1 unit of weight and Branch Y's salary ($1400\$1{}400) with 3 units of weight, giving 1000+42004=$1300\frac{1{}000 + 4{}200}{4} = \$1{}300. For the entire company of 1010 units of employees with an average salary of $910\$910, the total payroll is 10×910=$910010 \times 910 = \$9{}100. Subtracting the combined payroll of Branches X and Y ($5200\$5{}200) leaves $3900\$3{}900 for Branch Z's 66 units of employees. Thus, Branch Z's average salary is 39006=$650\frac{3{}900}{6} = \$650. Comparing $650\$650 to $1300\$1{}300 gives a percentage decrease of 13006501300×100%=50%\frac{1{}300 - 650}{1{}300} \times 100\% = 50\%.

Adım Adım Çözüm

1
Determine the average monthly salary of Branch Y.
Branch Y average salary = 1000×1.40=$14001{}000 \times 1.40 = \$1{}400.
Branch Y's average salary is given as 40%40\% higher than Branch X's average salary of $1000\$1{}000.
2
Express employee counts in terms of a single variable nn.
Branch X has nn employees, Branch Y has 3n3n employees, and Branch Z has 2×3n=6n2 \times 3n = 6n employees. Total employees = n+3n+6n=10nn + 3n + 6n = 10n.
The ratio of Branch X to Branch Y employees is 1:31:3, and Branch Z has twice as many employees as Branch Y.
3
Calculate the combined average salary of Branches X and Y.
Combined average salary SXY=n(1000)+3n(1400)n+3n=5200n4n=$1300S_{XY} = \frac{n(1{}000) + 3n(1{}400)}{n + 3n} = \frac{5{}200n}{4n} = \$1{}300.
The combined average of two sets is total combined earnings divided by total combined size.
4
Calculate total company payroll and determine Branch Z's average salary.
Total payroll = 10n×910=9100n10n \times 910 = 9{}100n. Branch Z payroll = 9100n5200n=3900n9{}100n - 5{}200n = 3{}900n. Average salary for Branch Z SZ=3900n6n=$650S_Z = \frac{3{}900n}{6n} = \$650.
Total payroll is total employees times overall mean. Subtracting the combined payroll of X and Y yields Branch Z's payroll.
5
Compute the percentage by which Branch Z's average salary is lower than the combined average salary of Branches X and Y.
x = \frac{1{}300 - 650}{1{}300} \times 100\% = 50\%.
Percentage decrease is calculated as (Base Value - New Value) / Base Value.

Anahtar Kavram

Weighted Averages of Combined Sets
Tahmini Süre:2m 0s
Soru 984Soru

How many integer values of xx satisfy the inequality x25x6|x^2 - 5x| \le 6?

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Cevap: 8

Cevap

The correct answer is 8.
To solve x25x6|x^2 - 5x| \le 6, express it as 6x25x6-6 \le x^2 - 5x \le 6. Solving x25x60x^2 - 5x - 6 \le 0 gives [1,6][-1, 6], while solving x25x+60x^2 - 5x + 6 \ge 0 gives (,2][3,)(-\infty, 2] \cup [3, \infty). Taking their intersection yields the solution set [1,2][3,6][-1, 2] \cup [3, 6]. The integers contained in this set are 1,0,1,2,3,4,5,6-1, 0, 1, 2, 3, 4, 5, 6, which gives 8 distinct integer values.

Adım Adım Çözüm

1
Rewrite the absolute value inequality
6x25x6-6 \le x^2 - 5x \le 6
For any real expression AA and constant k0k \ge 0, Ak|A| \le k is equivalent to kAk-k \le A \le k.
2
Solve the upper bound condition x25x6x^2 - 5x \le 6
1x6-1 \le x \le 6
Subtract 6 from both sides to obtain x25x60x^2 - 5x - 6 \le 0. Factoring gives (x6)(x+1)0(x - 6)(x + 1) \le 0.
3
Solve the lower bound condition x25x6x^2 - 5x \ge -6
x2x \le 2 or x3x \ge 3
Add 6 to both sides to obtain x25x+60x^2 - 5x + 6 \ge 0. Factoring gives (x2)(x3)0(x - 2)(x - 3) \ge 0.
4
Combine the solution sets
[1,2][3,6][-1, 2] \cup [3, 6]
The intersection of [1,6][-1, 6] with (,2][3,)(-\infty, 2] \cup [3, \infty) is the set of intervals [1,2][-1, 2] and [3,6][3, 6].
5
Count all integer solutions in the combined set
8 integer values
The integer values in [1,2][-1, 2] are 1,0,1,2-1, 0, 1, 2 (4 integers), and in [3,6][3, 6] are 3,4,5,63, 4, 5, 6 (4 integers), totaling 4+4=84 + 4 = 8 integers.

Anahtar Kavram

Solving quadratic absolute value inequalities using compound inequality decomposition.
Soru 985Soru

A specialized coffee roastery produces three custom blends—Roast Alpha, Roast Beta, and Roast Gamma—using three varieties of single-origin beans: Grade A, Grade B, and Grade C.

- One batch of Roast Alpha requires 3 kg of Grade A, 1 kg of Grade B, and 2 kg of Grade C beans, and has a total raw material cost of 64.OnebatchofRoastBetarequires1kgofGradeA,4kgofGradeB,and2kgofGradeCbeans,andhasatotalrawmaterialcostof64. - One batch of Roast Beta requires 1 kg of Grade A, 4 kg of Grade B, and 2 kg of Grade C beans, and has a total raw material cost of 64.
- One batch of Roast Gamma requires 2 kg of Grade A, 2 kg of Grade B, and 5 kg of Grade C beans, and has a total raw material cost of $90.

What is the cost, in dollars, of 1 kg of Grade A beans?

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Cevap: 12

Cevap

The cost of 1 kg of Grade A beans is 12 dollars.
Setting up the 3-variable linear system 3x+y+2z=643x + y + 2z = 64, x+4y+2z=64x + 4y + 2z = 64, and 2x+2y+5z=902x + 2y + 5z = 90 allows us to eliminate zz by subtracting the second equation from the first, yielding 2x3y=02x - 3y = 0, or x=1.5yx = 1.5y. Substituting this relationship back into the system leads to y=8y = 8 and x=12x = 12. Thus, 1 kg of Grade A beans costs 12 dollars.

Adım Adım Çözüm

1
Formulate linear equations representing the total cost of each coffee blend batch.
Let xx be the cost per kg of Grade A beans, yy be the cost per kg of Grade B beans, and zz be the cost per kg of Grade C beans:
(1)3x+y+2z=64(2)x+4y+2z=64(3)2x+2y+5z=90\begin{aligned} (1)\quad 3x + y + 2z &= 64 \\ (2)\quad x + 4y + 2z &= 64 \\ (3)\quad 2x + 2y + 5z &= 90 \end{aligned}
Translating the word problem into a system of 3 linear equations with 3 variables.
2
Eliminate variable zz by subtracting Equation (2) from Equation (1).
(3x+y+2z)(x+4y+2z)=6464    2x3y=0    x=1.5y(3x + y + 2z) - (x + 4y + 2z) = 64 - 64 \implies 2x - 3y = 0 \implies x = 1.5y
Since both equations (1) and (2) contain the term +2z+2z, subtracting them removes zz directly and provides a simple relation between xx and yy.
3
Substitute x=1.5yx = 1.5y into Equation (1) and Equation (3) to obtain a system in terms of yy and zz.
From Equation (1):
3(1.5y)+y+2z=64    5.5y+2z=64    11y+4z=128(4)3(1.5y) + y + 2z = 64 \implies 5.5y + 2z = 64 \implies 11y + 4z = 128 \quad (4)
From Equation (3):
2(1.5y)+2y+5z=90    5y+5z=90    y+z=18    z=18y2(1.5y) + 2y + 5z = 90 \implies 5y + 5z = 90 \implies y + z = 18 \implies z = 18 - y
Reducing the system from 3 variables down to 2 variables.
4
Substitute z=18yz = 18 - y into Equation (4) to solve for yy, and subsequently calculate xx.
11y+4(18y)=128    7y+72=128    7y=56    y=811y + 4(18 - y) = 128 \implies 7y + 72 = 128 \implies 7y = 56 \implies y = 8
Using x=1.5yx = 1.5y:
x=1.5×8=12x = 1.5 \times 8 = 12
Solving the single-variable linear equation for yy, then substituting back to find the required cost xx for Grade A beans.

Anahtar Kavram

Solving a 3-Variable System of Linear Equations via Variable Elimination
Soru 986Soru

A specialty coffee roaster creates a signature espresso blend by combining Arabica coffee beans costing 18.00perkilogramwithRobustacoffeebeanscosting18.00 per kilogram with Robusta coffee beans costing 12.00 per kilogram. The final blend weighs 50 kilograms and has an overall average cost of $15.60 per kilogram. How many kilograms of Arabica beans are included in the blend?

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Cevap: 30

Cevap

The blend contains 30 kilograms of Arabica coffee beans.
To determine the required quantity of Arabica beans, establish the total cost equation: 18A+12(50A)=50(15.60)18A + 12(50 - A) = 50(15.60). Expanding the terms gives 18A+60012A=78018A + 600 - 12A = 780, which simplifies to 6A=1806A = 180, yielding A=30A = 30 kilograms.

Adım Adım Çözüm

1
Define variables and relate component weights
Let AA represent the mass of Arabica beans in kilograms. The mass of Robusta beans is (50A)(50 - A) kilograms.
The sum of the individual component weights must equal the total blend weight of 50 kilograms.
2
Formulate the weighted average total cost equation
18A+12(50A)=15.60×50=78018A + 12(50 - A) = 15.60 \times 50 = 780
The combined monetary cost of both bean types equals the total value of the 50 kg blend.
3
Solve for the unknown variable AA
6A+600=780    6A=180    A=306A + 600 = 780 \implies 6A = 180 \implies A = 30
Simplifying the algebraic linear equation determines the exact weight of Arabica beans.

Anahtar Kavram

Weighted Averages in Applied Contexts
Soru 987Soru

In a plane, there are nn points such that no three points are collinear. If the number of distinct triangles that can be formed using these points as vertices is exactly 55 times the number of distinct line segments that can be formed by joining pairs of these points, what is the value of nn?

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Cevap: 1717

Cevap

The correct value of nn is 1717.
The number of distinct triangles formed by nn points (no three collinear) is (n3)=n(n1)(n2)6\binom{n}{3} = \frac{n(n-1)(n-2)}{6}, and the number of line segments is (n2)=n(n1)2\binom{n}{2} = \frac{n(n-1)}{2}. Setting (n3)=5(n2)\binom{n}{3} = 5 \binom{n}{2} gives n(n1)(n2)6=5n(n1)2\frac{n(n-1)(n-2)}{6} = \frac{5n(n-1)}{2}. Dividing both sides by n(n1)2\frac{n(n-1)}{2} simplifies the equation to n23=5\frac{n-2}{3} = 5, yielding n2=15n - 2 = 15, so n=17n = 17.

Adım Adım Çözüm

1
Express the number of triangles and line segments using combinations.
The number of triangles formed by selecting 33 non-collinear points from nn points is (n3)=n(n1)(n2)6\binom{n}{3} = \frac{n(n-1)(n-2)}{6}. The number of line segments formed by selecting 22 points from nn points is (n2)=n(n1)2\binom{n}{2} = \frac{n(n-1)}{2}.
Order does not matter when selecting vertices for triangles or endpoints for line segments.
2
Set up the given equation relating the two quantities.
n(n1)(n2)6=5×n(n1)2\frac{n(n-1)(n-2)}{6} = 5 \times \frac{n(n-1)}{2}
The problem states that the number of triangles is 55 times the number of line segments.
3
Simplify the equation for n3n \ge 3.
Divide both sides by n(n1)2\frac{n(n-1)}{2} to get n23=5\frac{n-2}{3} = 5, which simplifies to n2=15n - 2 = 15.
Since n3n \ge 3, n(n1)0n(n-1) \ne 0, so dividing by common terms is valid.
4
Solve for nn.
n = 15 + 2 = 17
Adding 22 to both sides isolates nn.

Anahtar Kavram

Combinations in Geometry
Soru 988Soru

A data set consists of 99 positive integers. The set has a unique mode of 1414, a median of 1212, and an arithmetic mean of 1111. If the range of the data set is 1010, what is the maximum possible value of the largest integer in the set?

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Cevap: 1616

Cevap

The maximum possible value of the largest integer in the set is 1616.
The correct answer states that the maximum possible value is 1616. By ordering the 99 terms x1x2x9x_1 \le x_2 \le \dots \le x_9, the sum of all terms must equal 9999, with median x5=12x_5 = 12 and x9=x1+10x_9 = x_1 + 10. If x9=16x_9 = 16, then x1=6x_1 = 6, and we can construct a valid set {6,6,8,9,12,14,14,14,16}\{6, 6, 8, 9, 12, 14, 14, 14, 16\} where 1414 is the unique mode appearing 33 times. Trying a larger value such as 1717 forces x1=7x_1 = 7, which makes it impossible to maintain 1414 as the unique mode without violating the total sum of 9999.

Adım Adım Çözüm

1
Express the total sum and set structure using ordered variables.
Let the 99 positive integers in non-decreasing order be x1x2x3x4x5x6x7x8x9x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6 \le x_7 \le x_8 \le x_9. The total sum is 9×11=999 \times 11 = 99. The median is x5=12x_5 = 12. The range is x9x1=10    x9=x1+10x_9 - x_1 = 10 \implies x_9 = x_1 + 10.
Establishing the position of the median and the exact sum provides structural bounds for maximizing x9x_9.
2
Analyze the frequency requirements for the unique mode.
Since 14>1214 > 12 (the median), the number 1414 must lie in the upper half of the set (x6,x7,x8,x9x_6, x_7, x_8, x_9). To be a unique mode, 1414 must appear at least twice. If x9=16x_9 = 16, then x1=6x_1 = 6.
Maximizing x9x_9 is equivalent to maximizing x1x_1, so we test the largest possible values for x1x_1 and verify the uniqueness of the mode 1414.
3
Evaluate x1=7x_1 = 7 (which would give x9=17x_9 = 17).
If x1=7x_1 = 7, then x9=17x_9 = 17. The sum equation becomes 7+x2+x3+x4+12+x6+x7+x8+17=99    x2+x3+x4+x6+x7+x8=637 + x_2 + x_3 + x_4 + 12 + x_6 + x_7 + x_8 + 17 = 99 \implies x_2 + x_3 + x_4 + x_6 + x_7 + x_8 = 63. Since x2,x3,x47x_2, x_3, x_4 \ge 7 and x6,x7,x812x_6, x_7, x_8 \ge 12, achieving this sum while keeping 1414 as the unique mode is impossible (it would require 77 to appear 44 times, making 77 the mode instead of 1414). Thus, x9x_9 cannot be 1717 or greater.
Testing x9=17x_9 = 17 shows a violation of the unique mode condition.
4
Construct a valid set for x1=6x_1 = 6 (giving x9=16x_9 = 16).
If x1=6x_1 = 6 and x9=16x_9 = 16, consider the set {6,6,8,9,12,14,14,14,16}\{6, 6, 8, 9, 12, 14, 14, 14, 16\}. Sum = 6+6+8+9+12+14+14+14+16=996+6+8+9+12+14+14+14+16 = 99. Median = 1212. Unique mode = 1414 (frequency 33). Range = 166=1016 - 6 = 10. All conditions are satisfied.
Constructing an explicit valid data set proves that 1616 is attainable and is the maximum.

Anahtar Kavram

Maximizing elements in a constrained discrete data set using mean, median, mode, and range properties
Soru 989Soru

A commercial real estate developer acquired a property for $P\$P. Over the first year, the market value of the property increased by 25%25\%. During the second year, the property's value decreased by x%x\%. In the third year, the value increased again by 20%20\% relative to its value at the end of the second year. If the final value of the property at the end of the third year was 14%14\% greater than the original acquisition price $P\$P, what is the value of xx?

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Cevap: 2424

Cevap

The value of xx is 24.
To find the net effect of successive percentage changes, express each period's change as a multiplier of the value from the preceding period. A 25%25\% increase corresponds to a multiplier of 1.251.25, an x%x\% decrease corresponds to a multiplier of (1x100)\left(1 - \frac{x}{100}\right), and a 20%20\% increase corresponds to a multiplier of 1.201.20. Combined, these produce a final multiplier of 1.25×1.20×(1x100)=1.50(1x100)1.25 \times 1.20 \times \left(1 - \frac{x}{100}\right) = 1.50\left(1 - \frac{x}{100}\right). Setting this equal to the net 14%14\% total increase (1.141.14) yields 1.50(1x100)=1.141.50\left(1 - \frac{x}{100}\right) = 1.14, which simplifies to 1x100=0.761 - \frac{x}{100} = 0.76, giving x=24x = 24.

Adım Adım Çözüm

1
Express each sequential period's value using multiplier notation.
At Year 1 end: V1=1.25PV_1 = 1.25P. At Year 2 end: V2=1.25P(1x100)V_2 = 1.25P \left(1 - \frac{x}{100}\right). At Year 3 end: V3=1.25P(1x100)×1.20V_3 = 1.25P \left(1 - \frac{x}{100}\right) \times 1.20.
Successive percent changes compound on the intermediate values of each period, not on the original base price.
2
Equate the overall combined growth multiplier to the given total net change.
1.25×1.20×(1x100)=1.14    1.50×(1x100)=1.141.25 \times 1.20 \times \left(1 - \frac{x}{100}\right) = 1.14 \implies 1.50 \times \left(1 - \frac{x}{100}\right) = 1.14
The final value is given as 14%14\% greater than PP, which corresponds to a net multiplier of 1.141.14.
3
Solve the algebraic equation for xx.
1x100=1.141.50=0.76    x100=0.24    x=241 - \frac{x}{100} = \frac{1.14}{1.50} = 0.76 \implies \frac{x}{100} = 0.24 \implies x = 24
Subtracting 0.760.76 from 11 yields the fractional decrease of 0.240.24, which equals 24%24\%.

Anahtar Kavram

Successive Percent Change and Multipliers
Tahmini Süre:2m 0s
Soru 990Soru

If aa, bb, and cc are non-zero real numbers such that a2b3c<0a^2 b^3 c < 0, ab5c2>0\frac{a}{b^5 c^2} > 0, and ac>bca c > b c, which of the following expressions MUST be true?

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Cevap: bac<0\frac{b - a}{c} < 0

Cevap

The expression \(\frac{b - a}{c} < 0\) MUST be true.
The condition a2b3c<0a^2 b^3 c < 0 requires bc<0b c < 0 because a2>0a^2 > 0. The condition ab5c2>0\frac{a}{b^5 c^2} > 0 requires ab>0a b > 0 because c2>0c^2 > 0. Combining these shows that aa and bb share the same sign, whereas cc has the opposite sign. Consequently, ac<0a c < 0. From ac>bca c > b c, subtracting bcb c yields (ab)c>0(a - b) c > 0. Multiplying by 1-1 gives (ba)c<0(b - a) c < 0, and dividing by c2>0c^2 > 0 produces bac<0\frac{b - a}{c} < 0, which MUST be true in all cases.

Adım Adım Çözüm

1
Determine the relative signs of bb and cc using a2b3c<0a^2 b^3 c < 0.
bc<0b c < 0, meaning bb and cc have opposite signs.
Since a0a \neq 0, a2>0a^2 > 0 always. Dividing a2b3c<0a^2 b^3 c < 0 by a2a^2 yields b3c<0b^3 c < 0. Because b3b^3 has the same sign as bb, bc<0b c < 0.
2
Determine the relative signs of aa and bb using ab5c2>0\frac{a}{b^5 c^2} > 0.
ab>0a b > 0, meaning aa and bb have the same sign.
Since c0c \neq 0, c2>0c^2 > 0 always. Multiplying by c2c^2 gives ab5>0\frac{a}{b^5} > 0, which implies aa and b5b^5 have the same sign. Thus ab>0a b > 0.
3
Determine the relationship between aa and cc.
ac<0a c < 0, meaning aa and cc have opposite signs.
Since aa and bb have the same sign (ab>0a b > 0) while bb and cc have opposite signs (bc<0b c < 0), aa and cc must have opposite signs.
4
Analyze the inequality ac>bca c > b c.
\(\frac{b - a}{c} < 0\)
Rearranging ac>bca c > b c gives acbc>0    (ab)c>0a c - b c > 0 \implies (a - b) c > 0. Multiplying both sides by 1-1 flips the inequality: (ba)c<0(b - a) c < 0. Dividing by the strictly positive quantity c2c^2 yields (ba)cc2<0    bac<0\frac{(b - a) c}{c^2} < 0 \implies \frac{b - a}{c} < 0.

Anahtar Kavram

Deducing sign relationships of variables in inequalities and algebraic transformations without assuming positive signs.
Tahmini Süre:2m 0s
Soru 991Soru

In an arithmetic sequence, the sum of the first 44 terms is 2828 and the sum of the first 88 terms is 8888. What is the 10th10\text{th} term of this sequence?

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Cevap: 2222

Cevap

The 10th term of the sequence is 22.
Using the arithmetic series sum formula Sn=n2(2a1+(n1)d)S_n = \frac{n}{2}(2a_1 + (n-1)d), the conditions yield two linear equations: 2a1+3d=142a_1 + 3d = 14 and 2a1+7d=222a_1 + 7d = 22. Subtracting these equations gives 4d=84d = 8, so d=2d = 2. Substituting d=2d = 2 back into 2a1+3(2)=142a_1 + 3(2) = 14 yields a1=4a_1 = 4. The 10th term is then calculated as a1+9d=4+9(2)=22a_1 + 9d = 4 + 9(2) = 22.

Adım Adım Çözüm

1
Express the given sums using the arithmetic series sum formula Sn=n2(2a1+(n1)d)S_n = \frac{n}{2}(2a_1 + (n-1)d).
For S4=28S_4 = 28: 42(2a1+3d)=28    2a1+3d=14\frac{4}{2}(2a_1 + 3d) = 28 \implies 2a_1 + 3d = 14.
For S8=88S_8 = 88: 82(2a1+7d)=88    2a1+7d=22\frac{8}{2}(2a_1 + 7d) = 88 \implies 2a_1 + 7d = 22.
Setting up linear equations in terms of the first term a1a_1 and common difference dd allows us to solve for both sequence parameters.
2
Subtract the first equation from the second equation to solve for dd.
(2a1+7d)(2a1+3d)=2214    4d=8    d=2(2a_1 + 7d) - (2a_1 + 3d) = 22 - 14 \implies 4d = 8 \implies d = 2.
Eliminating a1a_1 isolates the common difference dd.
3
Substitute d=2d = 2 back into the first equation to solve for a1a_1.
2a1+3(2)=14    2a1+6=14    2a1=8    a1=42a_1 + 3(2) = 14 \implies 2a_1 + 6 = 14 \implies 2a_1 = 8 \implies a_1 = 4.
Finding a1a_1 completes the essential parameters of the sequence.
4
Calculate the 10th term using the formula an=a1+(n1)da_n = a_1 + (n-1)d.
a10=4+(101)(2)=4+18=22a_{10} = 4 + (10 - 1)(2) = 4 + 18 = 22.
Evaluating the formula at n=10n = 10 provides the target term.

Anahtar Kavram

Arithmetic sequence term and series sum formulas.
Tahmini Süre:2m 0s
Soru 992Soru

A software company audited 120120 applications for compliance with two protocols: Accessibility (Protocol A) and Security (Protocol B). The audit revealed that 7575 applications complied with Protocol A, 6060 applications complied with Protocol B, and 3535 applications complied with neither protocol. If an application selected at random from the audited group is known to comply with Protocol A, what is the probability that it also complies with Protocol B?

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Cevap: 23\frac{2}{3}

Cevap

23\frac{2}{3}
Out of the 120 total applications, 35 comply with neither protocol, leaving 85 that comply with at least one. Using the inclusion-exclusion principle (75+6085=5075 + 60 - 85 = 50), exactly 50 applications comply with both protocols. Since the application is already known to comply with Protocol A, the denominator is restricted to the 75 applications in Protocol A. The conditional probability is therefore 50/75=2/350 / 75 = 2/3.

Adım Adım Çözüm

1
Determine the number of applications that comply with at least one protocol.
Total applications minus those complying with neither: 12035=85120 - 35 = 85.
The total sample space is partitioned into applications complying with at least one protocol and those complying with neither.
2
Find the number of applications complying with both Protocol A and Protocol B using the Principle of Inclusion-Exclusion.
AB=A+BAB=75+6085=50|A \cap B| = |A| + |B| - |A \cup B| = 75 + 60 - 85 = 50.
Summing the counts of Protocol A and Protocol B counts applications in both protocols twice.
3
Calculate the conditional probability P(BA)P(B|A).
P(BA)=ABA=5075=23P(B|A) = \frac{|A \cap B|}{|A|} = \frac{50}{75} = \frac{2}{3}.
Given that the selected application complies with Protocol A, the sample space is restricted to A=75|A| = 75.

Anahtar Kavram

Conditional Probability with Overlapping Sets
Tahmini Süre:1m 30s
Soru 993Soru

A sequence consists of kk consecutive positive odd integers. The arithmetic mean of the 33 largest integers in the sequence is 2929. If the sum of all kk integers in the sequence is 207207, what is the value of kk?

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Cevap: 9

Cevap

The total number of terms in the sequence, kk, is 9.
The arithmetic mean of 3 consecutive odd integers is their middle term, so the three largest terms are 27, 29, and 31. The largest term is 31. Using the sum formula for an arithmetic progression, S=ka+312=207S = k \cdot \frac{a + 31}{2} = 207. Expressing the first term as a=312(k1)=332ka = 31 - 2(k - 1) = 33 - 2k yields k(32k)=207k(32 - k) = 207, which simplifies to k232k+207=0k^2 - 32k + 207 = 0. The roots are k=9k = 9 and k=23k = 23. Because all integers in the sequence must be positive, a=332k>0a = 33 - 2k > 0, requiring k16k \le 16. Therefore, k=9k = 9.

Adım Adım Çözüm

1
Determine the largest integer in the sequence.
The largest integer is 3131.
For any 3 consecutive odd integers, the arithmetic mean is equal to the middle integer. Since the mean is 2929, the three largest integers are 27,29,3127, 29, 31, so the maximum term is 3131.
2
Express the smallest term aa in terms of kk.
a=332ka = 33 - 2k.
The kk-th term of a consecutive odd integer sequence starting at aa is given by 31=a+2(k1)    a=332k31 = a + 2(k - 1) \implies a = 33 - 2k.
3
Apply the positivity constraint.
k16k \le 16.
Since all terms are positive integers, the smallest term must satisfy a1    332k1    k16a \ge 1 \implies 33 - 2k \ge 1 \implies k \le 16.
4
Set up and solve the sum equation for kk.
k=9k = 9.
The sum of an arithmetic sequence is S=k×a+L2S = k \times \frac{a + L}{2}. Substituting S=207S = 207, L=31L = 31, and a=332ka = 33 - 2k gives 207=k×(332k)+312=k(32k)    k232k+207=0207 = k \times \frac{(33 - 2k) + 31}{2} = k(32 - k) \implies k^2 - 32k + 207 = 0. Factoring yields (k9)(k23)=0(k - 9)(k - 23) = 0. Since k16k \le 16, k=9k = 9.

Anahtar Kavram

The sum of a sequence of consecutive evenly-spaced numbers equals the number of terms multiplied by the average of the first and last terms.
Tahmini Süre:1m 45s
Soru 994Soru

A regional commercial airline compiled operational data for all scheduled flights across its Eastern Division during 2025. Analysis of the logs revealed that every flight experiencing a departure delay exceeding 45 minutes had undergone unscheduled engine maintenance earlier that same day. Additionally, none of the flights that underwent unscheduled engine maintenance earlier that day arrived at their destination on schedule.

Which of the following can be properly inferred from the passage above?

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Cevap: Every flight in the Eastern Division during 2025 that experienced a departure delay exceeding 45 minutes failed to arrive at its destination on schedule.

Cevap

Every flight in the Eastern Division during 2025 that experienced a departure delay exceeding 45 minutes failed to arrive at its destination on schedule.
The correct answer directly synthesizes the two stated factual premises using transitive deduction: every flight with a departure delay exceeding 45 minutes underwent unscheduled engine maintenance, and every flight that underwent unscheduled engine maintenance failed to arrive on schedule. Therefore, any flight with a departure delay exceeding 45 minutes necessarily failed to arrive on schedule.

Adım Adım Çözüm

1
Identify the explicit premises given in the stimulus.
Premise 1: Departure delay > 45 min → Unscheduled engine maintenance. Premise 2: Unscheduled engine maintenance → NOT arrive on schedule.
Inference questions require strictly combining stated premises without introducing unstated assumptions.
2
Apply transitive conditional logic to connect the premises.
Departure delay > 45 min → Unscheduled engine maintenance → NOT arrive on schedule.
If A implies B, and B implies C, then A must imply C.
3
Evaluate the choices to find the one that MUST BE TRUE based solely on this deduction.
The statement that any flight delayed by over 45 minutes failed to arrive on schedule is a direct, inescapable logical deduction.
An inference must be guaranteed by the premises, whereas assumptions and speculations add unproven claims.

Anahtar Kavram

Distinguishing strict logical inferences (Must Be True) from unstated causal assumptions and out-of-scope speculations.
Soru 995Soru

An autonomous electric shuttle completed a 240-mile test run from Point A to Point B. For the first 120 miles, the shuttle traveled at a constant speed of vv miles per hour. For the remaining 120 miles, due to battery efficiency controls, the shuttle traveled at a reduced speed equal to 23v\frac{2}{3}v miles per hour. On the return trip along the exact same 240-mile route from Point B to Point A, the shuttle maintained a uniform speed of v+10v + 10 miles per hour. If the total time for the outbound trip from Point A to Point B was 2 hours longer than the total time for the return trip from Point B to Point A, what was the shuttle's average speed, in miles per hour, for the outbound trip from Point A to Point B?

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Cevap: 40

Cevap

40 miles per hour
To find the average speed for the outbound trip, calculate the total time taken for both 120-mile legs. The first half takes 120v\frac{120}{v} hours, and the second half at 23v\frac{2}{3}v speed takes 12023v=180v\frac{120}{\frac{2}{3}v} = \frac{180}{v} hours, giving a total outbound time of 300v\frac{300}{v} hours. Setting up the time difference equation against the return trip gives 300v240v+10=2\frac{300}{v} - \frac{240}{v + 10} = 2. Solving this quadratic equation yields v=50v = 50 mph. Substituting v=50v = 50 into the total outbound time formula gives 30050=6\frac{300}{50} = 6 hours. Finally, dividing the total distance of 240 miles by 6 hours gives an average speed of 40 miles per hour.

Adım Adım Çözüm

1
Express the total outbound travel time in terms of vv.
Toutbound=120v+12023v=120v+180v=300vT_{\text{outbound}} = \frac{120}{v} + \frac{120}{\frac{2}{3}v} = \frac{120}{v} + \frac{180}{v} = \frac{300}{v} hours.
Time equals distance divided by speed for each 120-mile segment.
2
Express the return travel time and set up the time-difference equation.
300v240v+10=2\frac{300}{v} - \frac{240}{v + 10} = 2.
The return trip covers 240 miles at speed v+10v + 10 mph and takes 2 hours less than the outbound trip.
3
Solve the algebraic equation for vv.
Simplifying 150v120v+10=1\frac{150}{v} - \frac{120}{v + 10} = 1 leads to v220v1500=0v^2 - 20v - 1500 = 0, giving positive root v=50v = 50.
Factoring (v50)(v+30)=0(v - 50)(v + 30) = 0 yields v=50v = 50 mph since speed must be positive.
4
Calculate total outbound time and outbound average speed.
Total outbound time = 30050=6\frac{300}{50} = 6 hours; Average speed = 2406=40\frac{240}{6} = 40 mph.
Average speed is defined as total distance divided by total elapsed time.

Anahtar Kavram

Average Speed across Multi-Leg Trips (Total Distance / Total Time)
Tahmini Süre:2m 30s
Soru 996Soru

The table below shows the distribution of scores achieved by 5050 candidates on a professional certification assessment:

ScoreNumber of Candidates
606044
70701111
80802020
90901010
10010055

What is the interquartile range (IQR) of the assessment scores?

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Cevap: 2020

Cevap

The interquartile range of the assessment scores is 2020.
To find the interquartile range (IQR), first find Q1Q_1 (25th25\text{th} percentile) and Q3Q_3 (75th75\text{th} percentile). With 5050 total candidates, Q1Q_1 is situated around position 12.512.5, which falls into the score of 7070. Q3Q_3 is situated around position 37.537.5, which falls into the score of 9090. Subtracting Q1Q_1 from Q3Q_3 yields 9070=2090 - 70 = 20.

Adım Adım Çözüm

1
Calculate the cumulative frequency distribution to find score positions.
Score 6060: candidates 11 to 44; Score 7070: candidates 55 to 1515; Score 8080: candidates 1616 to 3535; Score 9090: candidates 3636 to 4545; Score 100100: candidates 4646 to 5050. Total N=50N = 50.
Cumulative frequencies identify the precise position of ranked scores.
2
Determine the first quartile (Q1Q_1), which represents the 25th25\text{th} percentile.
The 25th25\text{th} percentile corresponds to the 0.25×50=12.5th0.25 \times 50 = 12.5\text{th} position. Looking at the cumulative frequencies, candidate positions 55 through 1515 all scored 7070, so Q1=70Q_1 = 70.
The first quartile marks the score boundary below which 25%25\% of the dataset falls.
3
Determine the third quartile (Q3Q_3), which represents the 75th75\text{th} percentile.
The 75th75\text{th} percentile corresponds to the 0.75×50=37.5th0.75 \times 50 = 37.5\text{th} position. Candidate positions 3636 through 4545 all scored 9090, so Q3=90Q_3 = 90.
The third quartile marks the score boundary below which 75%75\% of the dataset falls.
4
Compute the interquartile range (IQR=Q3Q1)(\text{IQR} = Q_3 - Q_1).
\text{IQR} = 90 - 70 = 20.
The interquartile range measures the spread of the middle 50%50\% of the distribution.

Anahtar Kavram

Interquartile Range (IQR) and Quartile Positions in Frequency Distributions
Soru 997Soru

At the beginning of 2021, a biotechnology firm allocated a fixed annual budget to its principal research laboratory. In 2022, the laboratory's budget was increased by 20%20\% relative to its 2021 budget. In 2023, the budget was reduced by 15%15\% from its 2022 level. In 2024, the budget was increased by 25%25\% over its 2023 level. If the budget in 2024 exceeded the budget in 2021 by $55,000\$55,000, what was the laboratory's budget in 2021, in dollars?

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Cevap: 200000

Cevap

200,000 dollars
The initial 2021 budget BB undergoes successive percentage changes over three years. A 20%20\% increase in 2022 results in 1.20B1.20B. A 15%15\% decrease in 2023 yields 1.20B×0.85=1.02B1.20B \times 0.85 = 1.02B. A 25%25\% increase in 2024 yields 1.02B×1.25=1.275B1.02B \times 1.25 = 1.275B. The net increase over the initial budget is 1.275BB=0.275B1.275B - B = 0.275B. Setting 0.275B=55,0000.275B = 55,000 gives B=55,0000.275=200,000B = \frac{55,000}{0.275} = 200,000 dollars.

Adım Adım Çözüm

1
Represent the annual budgets sequentially in terms of the initial 2021 budget BB
2022 budget = 1.20B1.20B, 2023 budget = 1.20B×0.85=1.02B1.20B \times 0.85 = 1.02B, 2024 budget = 1.02B×1.25=1.275B1.02B \times 1.25 = 1.275B
Calculate successive percentage changes sequentially by multiplying the respective multipliers for each period
2
Calculate the net change in budget from 2021 to 2024
1.275BB=0.275B1.275B - B = 0.275B
Determine how much the final year's budget exceeds the initial base budget
3
Set up and solve the linear equation for BB
0.275B=55,000    B=55,0000.275=200,0000.275B = 55,000 \implies B = \frac{55,000}{0.275} = 200,000
Equate the net algebraic difference to the given dollar amount to solve for the initial 2021 budget

Anahtar Kavram

Successive Percent Change and Base Value Tracking
Soru 998Soru

Let N=504N = 504. If MM is the smallest positive integer such that N×MN \times M is a perfect cube, what is the value of MM?

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Cevap: 147

Cevap

The smallest positive integer MM such that 504×M504 \times M is a perfect cube is 147.
For an integer to be a perfect cube, the exponent of each prime factor in its prime factorization must be a multiple of 3. Prime factorizing 504 yields 23×32×712^3 \times 3^2 \times 7^1. The exponent of 2 is 3 (already a multiple of 3). The exponent of 3 is 2, which requires 1 additional factor of 3 to reach 3. The exponent of 7 is 1, which requires 2 additional factors of 7 (727^2) to reach 3. Therefore, the minimum value for MM is 31×72=3×49=1473^1 \times 7^2 = 3 \times 49 = 147.

Adım Adım Çözüm

1
Express 504 as a product of its prime factors.
504=23×32×71504 = 2^3 \times 3^2 \times 7^1
Decomposing NN into prime factors allows analysis of the exponents required for perfect power conditions.
2
Apply the prime exponent rule for perfect cubes.
Every prime factor in N×MN \times M must have an exponent that is a multiple of 3.
A number KK is a perfect cube if and only if K=p13a×p23b×K = p_1^{3a} \times p_2^{3b} \times \dots
3
Calculate the missing prime factors needed to complete the cube.
M=332×731=31×72M = 3^{3-2} \times 7^{3-1} = 3^1 \times 7^2
To minimize MM, we raise each prime to the smallest non-negative power that rounds the existing exponent up to the nearest multiple of 3.
4
Evaluate the value of MM.
M=3×49=147M = 3 \times 49 = 147
Direct arithmetic computation.

Anahtar Kavram

Prime Factorization and Exponent Requirements for Perfect Powers
Tahmini Süre:1m 30s
Soru 999Soru

The positive integer nn has a prime factorization of the form 2a×3b×5c2^a \times 3^b \times 5^c, where aa, bb, and cc are positive integers. If nn is divisible by both 12 and 15, and nn has exactly 24 positive integer divisors, what is the least possible value of nn?

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Cevap: 360

Cevap

The least possible value of nn is 360.
To find the least value of n=2a×3b×5cn = 2^a \times 3^b \times 5^c divisible by 12=22×312 = 2^2 \times 3 and 15=3×515 = 3 \times 5, we require a2a \ge 2, b1b \ge 1, and c1c \ge 1. The number of positive divisors is (a+1)(b+1)(c+1)=24(a+1)(b+1)(c+1) = 24. Testing valid factor triples for 24 with a+13a+1 \ge 3, b+12b+1 \ge 2, and c+12c+1 \ge 2 gives possible values n=480n = 480 (from a=5,b=1,c=1a=5, b=1, c=1), n=360n = 360 (from a=3,b=2,c=1a=3, b=2, c=1), n=540n = 540 (from a=2,b=3,c=1a=2, b=3, c=1), and n=600n = 600 (from a=3,b=1,c=2a=3, b=1, c=2). The minimum among these valid integers is 360.

Adım Adım Çözüm

1
Determine the minimum exponent constraints from divisibility conditions.
Since nn is divisible by 12=22×3112 = 2^2 \times 3^1, we must have a2a \ge 2 and b1b \ge 1. Since nn is divisible by 15=31×5115 = 3^1 \times 5^1, we must have b1b \ge 1 and c1c \ge 1. Combining these, a2a \ge 2, b1b \ge 1, and c1c \ge 1.
Divisibility requires that the prime factorization of nn contains at least the prime powers present in the prime factorizations of 12 and 15.
2
Set up the total divisor count equation.
The total number of positive integer divisors of n=2a×3b×5cn = 2^a \times 3^b \times 5^c is given by (a+1)(b+1)(c+1)=24(a+1)(b+1)(c+1) = 24.
The divisor counting formula states that for a number with prime factorization p1e1p2e2p_1^{e_1} p_2^{e_2} \dots, the total number of positive divisors is (e1+1)(e2+1)(e_1+1)(e_2+1)\dots
3
Find all integer factor triples (a+1,b+1,c+1)(a+1, b+1, c+1) that multiply to 24 subject to a+13a+1 \ge 3, b+12b+1 \ge 2, and c+12c+1 \ge 2.
The valid factorizations of 24 into three factors meeting the bounds are:
- (6,2,2)    (a,b,c)=(5,1,1)(6, 2, 2) \implies (a, b, c) = (5, 1, 1), giving n=25×31×51=480n = 2^5 \times 3^1 \times 5^1 = 480
- (4,3,2)    (a,b,c)=(3,2,1)(4, 3, 2) \implies (a, b, c) = (3, 2, 1), giving n=23×32×51=360n = 2^3 \times 3^2 \times 5^1 = 360
- (3,4,2)    (a,b,c)=(2,3,1)(3, 4, 2) \implies (a, b, c) = (2, 3, 1), giving n=22×33×51=540n = 2^2 \times 3^3 \times 5^1 = 540
- (4,2,3)    (a,b,c)=(3,1,2)(4, 2, 3) \implies (a, b, c) = (3, 1, 2), giving n=23×31×52=600n = 2^3 \times 3^1 \times 5^2 = 600
To minimize n=2a×3b×5cn = 2^a \times 3^b \times 5^c, we must evaluate all valid permutations of exponents consistent with the divisor product constraint.
4
Compare the resulting values of nn to find the minimum.
Comparing 480, 360, 540, and 600, the smallest value is 360.
Assigning the larger exponent 3 to the smallest prime base 2 and exponent 2 to prime base 3 minimizes the overall product.

Anahtar Kavram

Divisor Counting Formula and Prime Factorization Constraints
Tahmini Süre:2m 0s
Soru 1000Soru

An executive board consisting of 8 distinct members needs to form a subcommittee of 4 members. However, board members Alex and Blair refuse to serve on the subcommittee together unless board member Morgan is also selected. How many different 4-member subcommittees can be formed under these conditions?

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Cevap: 60

Cevap

60
The solution uses complementary counting. First, calculate the total possible 4-member subcommittees from 8 members without restrictions, which is (84)=70\binom{8}{4} = 70. Second, identify the restricted scenario that is not allowed: Alex and Blair are both selected, but Morgan is excluded. In this invalid scenario, 2 spots are taken by Alex and Blair, Morgan is excluded from consideration, leaving 2 spots to be filled from the remaining 5 board members, which equals (52)=10\binom{5}{2} = 10 invalid subcommittees. Subtracting the invalid subcommittees from the total gives 7010=6070 - 10 = 60 valid subcommittees.

Adım Adım Çözüm

1
Calculate the total number of ways to choose a 4-member subcommittee from 8 members without restrictions.
The total unrestricted combinations is (84)=8×7×6×54×3×2×1=70\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70.
Since the order of selecting members into a subcommittee does not matter, use the combination formula (nk)\binom{n}{k}.
2
Determine the condition under which a subcommittee selection is invalid.
A subcommittee is invalid if and only if both Alex and Blair are selected AND Morgan is excluded.
Alex and Blair agree to serve together only if Morgan is also present. Thus, having Alex and Blair together without Morgan violates the condition.
3
Calculate the number of invalid subcommittees.
The number of invalid subcommittees is (52)=5×42×1=10\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10.
Alex and Blair take 2 of the 4 spots, and Morgan cannot take any spot. The remaining 2 spots must be filled from the remaining 83=58 - 3 = 5 members.
4
Subtract the invalid subcommittees from the total unrestricted subcommittees using complementary counting.
7010=6070 - 10 = 60 valid subcommittees.
Complementary counting yields the total number of subcommittees that satisfy the restriction.

Anahtar Kavram

Combinations with Restrictions and Complementary Counting
Tahmini Süre:2m 0s
ÖncekiSayfa 50 / 110Sonraki
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