Tüm alıştırma soruları

387 soru

Soru 1Soru

A rectangular tabletop measuring 126 centimeters126\text{ centimeters} by 180 centimeters180\text{ centimeters} is to be completely covered by non-overlapping, identical square tiles of the maximum possible side length, such that no tiles need to be cut. How many such square tiles are required to cover the entire tabletop?

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Cevap: 70

Cevap

70 tiles are required to cover the tabletop.
To cover a 126 cm×180 cm126\text{ cm} \times 180\text{ cm} surface with identical square tiles of maximum side length without cutting, the side length of each square tile must be the greatest common divisor of 126126 and 180180, which is 18 cm18\text{ cm}. Dividing the surface dimensions by 18 cm18\text{ cm} gives 77 tiles along one side and 1010 tiles along the other, yielding a total of 7070 tiles.

Adım Adım Çözüm

1
Determine the prime factorizations of the tabletop dimensions 126 and 180.
126=21×32×71126 = 2^1 \times 3^2 \times 7^1 and 180=22×32×51180 = 2^2 \times 3^2 \times 5^1
Prime factorization allows systematic extraction of the greatest common divisor.
2
Calculate the Greatest Common Divisor (GCD) of 126 and 180.
GCD(126,180)=2min(1,2)×3min(2,2)=21×32=18\text{GCD}(126, 180) = 2^{\min(1,2)} \times 3^{\min(2,2)} = 2^1 \times 3^2 = 18
The maximum side length of a square tile that fits evenly without cutting is the GCD of the two dimensions.
3
Find the total number of tiles required.
\text{Total tiles} = \left(\frac{126}{18}\right) \times \left(\frac{180}{18}\right) = 7 \times 10 = 70
The total area divided by the area of one tile gives the number of tiles needed.

Anahtar Kavram

Greatest Common Divisor (GCD) application in geometric spatial partitioning
Soru 2Soru

Set SS consists of all consecutive integers from m-m to nn, inclusive, where mm and nn are positive integers with n>mn > m. If set SS contains exactly 2525 integers and the sum of all integers in set SS is 7575, what is the value of mm?

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Cevap: 9

Cevap

The value of mm is 99.
To find mm, use the two fundamental properties of consecutive integer sets: term count and set sum. The number of terms from m-m to nn inclusive is n(m)+1=n+m+1=25n - (-m) + 1 = n + m + 1 = 25, which yields n+m=24n + m = 24. The sum of an evenly spaced set is the product of the number of terms and the arithmetic mean of the smallest and largest terms: 25×m+n2=7525 \times \frac{-m + n}{2} = 75, which simplifies to nm=6n - m = 6. Subtracting nm=6n - m = 6 from n+m=24n + m = 24 gives 2m=182m = 18, so m=9m = 9.

Adım Adım Çözüm

1
Set up the equation for the number of terms in the set.
n+m=24n + m = 24
The number of integers from m-m to nn inclusive is n(m)+1=n+m+1=25n - (-m) + 1 = n + m + 1 = 25.
2
Set up the equation for the sum of the integers in the set.
nm=6n - m = 6
The sum of an arithmetic progression is given by number of terms×mean=25×m+n2=75\text{number of terms} \times \text{mean} = 25 \times \frac{-m + n}{2} = 75, leading to nm2=3\frac{n - m}{2} = 3.
3
Solve for mm using the two linear equations.
m=9m = 9
Subtracting nm=6n - m = 6 from n+m=24n + m = 24 yields 2m=182m = 18, giving m=9m = 9.

Anahtar Kavram

Properties of consecutive integer sets: inclusive term counting (nstart+1n - \text{start} + 1) and set sum calculation (count×mean\text{count} \times \text{mean}).
Tahmini Süre:1m 30s
Soru 3Soru

A fuel tank initially contains 6060 gallons of a fuel mixture that is 10%10\% ethanol and 90%90\% gasoline by volume. A mechanic removes xx gallons of this mixture and replaces it with an equal volume of pure ethanol to obtain a new mixture that is 25%25\% ethanol by volume. What is the value of xx?

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Cevap: 10

Cevap

The volume of fuel mixture that must be removed and replaced with pure ethanol is 1010 gallons.
Replacing 1010 gallons of the 10%10\% ethanol fuel with pure ethanol removes 11 gallon of ethanol and adds 1010 gallons of pure ethanol. The total ethanol in the tank becomes 61+10=156 - 1 + 10 = 15 gallons, which represents exactly 25%25\% of the total 6060-gallon volume.

Adım Adım Çözüm

1
Calculate the initial volume of ethanol in the tank
Ethanol volume = 0.10×60=60.10 \times 60 = 6 gallons.
Establishes the starting quantity of the solute.
2
Express the amount of ethanol after removal and replacement in terms of xx
Final ethanol volume = 60.10x+x=6+0.90x6 - 0.10x + x = 6 + 0.90x gallons.
Removing xx gallons of fuel removes 10%10\% ethanol, while adding xx gallons of pure ethanol adds 100%100\% ethanol.
3
Set up an equation using the target ethanol concentration
6+0.90x=0.25×60=156 + 0.90x = 0.25 \times 60 = 15.
The final 6060-gallon mixture must contain 25%25\% ethanol by volume.
4
Solve the equation for xx
0.90x=9    x=100.90x = 9 \implies x = 10.
Isolates xx to find the required replacement volume.

Anahtar Kavram

Dilution and fluid replacement in mixture problems
Soru 4Soru

If xx and yy are integers such that 5x1-5 \le x \le -1 and 2y62 \le y \le 6, what is the minimum possible value of xyx+y\frac{x - y}{x + y}?

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Cevap: -11

Cevap

The minimum possible value of the expression is -11.
The correct answer is -11 because the numerator xyx - y is negative for all allowed values of xx and yy. To minimize a negative fraction, the denominator x+yx + y must be positive and minimized (equal to 1), while the numerator must be as negative as possible. Choosing x=5x = -5 and y=6y = 6 gives a numerator of 11-11 and a denominator of 11, producing the minimum value of -11.

Adım Adım Çözüm

1
Evaluate the sign of the numerator
Since xx is negative (x1x \le -1) and yy is positive (y2y \ge 2), xyx - y is always negative (xy3x - y \le -3).
Subtracting a positive integer from a negative integer yields a negative result.
2
Determine the conditions for minimizing a negative fraction
To obtain the minimum (most negative) value, the denominator x+yx + y must be a positive integer and as small as possible.
A negative numerator divided by a positive denominator yields a negative quotient. Dividing by a smaller positive number yields a quotient with larger absolute value, making it smaller (more negative).
3
Identify the smallest positive denominator and solve for variables
The smallest positive integer for x+yx + y is 11. Setting y=1xy = 1 - x gives xy=2x1x - y = 2x - 1.
Since xx and yy are integers, x+yx + y must be an integer.
4
Maximize the magnitude of the negative numerator
Using the lowest boundary x=5x = -5 gives y=6y = 6. The value of the expression is 565+6=11\frac{-5 - 6}{-5 + 6} = -11.
Evaluating at x=5x = -5 and y=6y = 6 gives numerator 11-11 and denominator 11, resulting in 11-11.

Anahtar Kavram

Minimizing algebraic fractions involving signed numbers
Soru 5Soru

Let m=2a5311bm = 2^a \cdot 5^3 \cdot 11^b and n=245c112n = 2^4 \cdot 5^c \cdot 11^2, where aa, bb, and cc are positive integers. If the greatest common divisor of mm and nn is 23521122^3 \cdot 5^2 \cdot 11^2, and the least common multiple of mm and nn has exactly 180180 positive integer divisors, what is the value of a+b+ca + b + c?

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Cevap: 13

Cevap

The value of a+b+ca + b + c is 1313.
To find a,b,ca, b, c, analyze prime factor exponents. The GCD uses minimum exponents: min(a,4)=3    a=3\min(a,4)=3 \implies a=3, min(3,c)=2    c=2\min(3,c)=2 \implies c=2, and min(b,2)=2    b2\min(b,2)=2 \implies b \ge 2. The LCM uses maximum exponents: 24,53,11b2^4, 5^3, 11^b. The number of divisors of the LCM is (4+1)(3+1)(b+1)=20(b+1)=180(4+1)(3+1)(b+1) = 20(b+1) = 180, yielding b=8b=8. Adding a+b+c=3+8+2=13a+b+c = 3+8+2 = 13.

Adım Adım Çözüm

1
Use the prime exponent rule for greatest common divisor to find aa and cc, and bound bb.
a=3a = 3, c=2c = 2, and b2b \ge 2.
The greatest common divisor takes the minimum exponent for each prime factor: min(a,4)=3    a=3\min(a, 4) = 3 \implies a = 3, min(3,c)=2    c=2\min(3, c) = 2 \implies c = 2, and min(b,2)=2    b2\min(b, 2) = 2 \implies b \ge 2.
2
Determine the prime factorization of lcm(m,n)\text{lcm}(m, n).
lcm(m,n)=245311b\text{lcm}(m, n) = 2^4 \cdot 5^3 \cdot 11^b.
The least common multiple takes the maximum exponent for each prime factor: max(3,4)=4\max(3, 4) = 4, max(3,2)=3\max(3, 2) = 3, and max(b,2)=b\max(b, 2) = b since b2b \ge 2.
3
Apply the total number of divisors formula to solve for bb.
b=8b = 8.
The number of positive integer divisors of 245311b2^4 \cdot 5^3 \cdot 11^b is (4+1)(3+1)(b+1)=20(b+1)=180(4+1)(3+1)(b+1) = 20(b+1) = 180, which yields b+1=9b+1 = 9 so b=8b = 8.
4
Calculate the sum a+b+ca + b + c.
a+b+c=13a + b + c = 13.
Summing the values a=3a=3, b=8b=8, and c=2c=2 gives 3+8+2=133 + 8 + 2 = 13.

Anahtar Kavram

GCD and LCM prime factor exponent rules and total divisor formula
Soru 6Soru

What is the remainder when the expression 745322+8157^{45} \cdot 3^{22} + 8^{15} is divided by 1010?

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Cevap: 5

Cevap

The remainder is 5.
Dividing any number by 10 leaves a remainder equal to the units digit of that number. By analyzing the units digit pattern (cyclicity of period 4) for powers of 7, 3, and 8: 74571=7(mod10)7^{45} \equiv 7^1 = 7 \pmod{10}, 32232=9(mod10)3^{22} \equiv 3^2 = 9 \pmod{10}, and 81583=2(mod10)8^{15} \equiv 8^3 = 2 \pmod{10}. The expression simplifies to (7×9)+2=63+2=65(7 \times 9) + 2 = 63 + 2 = 65, which has a units digit of 5. Therefore, the remainder when divided by 10 is 5.

Adım Adım Çözüm

1
Relate remainder modulo 10 to units digit cyclicity.
Finding the remainder when an expression is divided by 10 is equivalent to finding its units digit.
Any positive integer NN can be expressed as 10k+r10k + r, where rr is the units digit and the remainder when NN is divided by 10.
2
Determine the units digit of 7457^{45}.
7457(mod10)7^{45} \equiv 7 \pmod{10}.
Powers of 7 repeat their units digits in a cycle of length 4 (7, 9, 3, 1). Dividing the exponent 45 by 4 gives a remainder of 1, so 7457^{45} has the same units digit as 71=77^1 = 7.
3
Determine the units digit of 3223^{22}.
3229(mod10)3^{22} \equiv 9 \pmod{10}.
Powers of 3 repeat their units digits in a cycle of length 4 (3, 9, 7, 1). Dividing the exponent 22 by 4 gives a remainder of 2, so 3223^{22} has the same units digit as 32=93^2 = 9.
4
Determine the units digit of the product 7453227^{45} \cdot 3^{22}.
7453223(mod10)7^{45} \cdot 3^{22} \equiv 3 \pmod{10}.
The product of the units digits is 7×9=637 \times 9 = 63, which has a units digit of 3.
5
Determine the units digit of 8158^{15}.
8152(mod10)8^{15} \equiv 2 \pmod{10}.
Powers of 8 repeat their units digits in a cycle of length 4 (8, 4, 2, 6). Dividing the exponent 15 by 4 gives a remainder of 3, so 8158^{15} has the same units digit as 83=5128^3 = 512, which ends in 2.
6
Combine the results to find the final remainder modulo 10.
5
Adding the units digit of the first term (3) and the second term (2) gives 3+2=53 + 2 = 5.

Anahtar Kavram

Units Digit Cyclicity and Modular Arithmetic
Soru 7Soru

If 5x5x2=60055^{x} - 5^{x-2} = 600\sqrt{5}, what is the value of xx?

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Cevap: 4.5

Cevap

The value of xx is 4.5.
Factoring out 5x25^{x-2} converts the left side into 5x2(251)=245x25^{x-2}(25 - 1) = 24 \cdot 5^{x-2}. Dividing 6005600\sqrt{5} by 24 gives 25525\sqrt{5}, which equals 52.55^{2.5}. Setting x2=2.5x - 2 = 2.5 gives x=4.5x = 4.5.

Adım Adım Çözüm

1
Factor out the smallest power of 5, which is 5x25^{x-2}, from the left side of the equation.
5x2(521)=60055^{x-2}(5^2 - 1) = 600\sqrt{5}
Factoring isolates the constant multiplier from the variable power term.
2
Calculate the numerical value inside the parentheses.
245x2=600524 \cdot 5^{x-2} = 600\sqrt{5}
521=251=245^2 - 1 = 25 - 1 = 24.
3
Divide both sides of the equation by 24.
5x2=2555^{x-2} = 25\sqrt{5}
Isolating 5x25^{x-2} gives 600524=255\frac{600\sqrt{5}}{24} = 25\sqrt{5}.
4
Rewrite 25525\sqrt{5} as a single exponential expression with base 5.
5x2=5250.5=52.55^{x-2} = 5^2 \cdot 5^{0.5} = 5^{2.5}
Using the product rule for exponents, 5251/2=52+0.5=52.55^2 \cdot 5^{1/2} = 5^{2 + 0.5} = 5^{2.5}.
5
Equate the exponents since the bases are identical.
x2=2.5    x=4.5x - 2 = 2.5 \implies x = 4.5
For any positive base b1b \neq 1, if bm=bnb^m = b^n, then m=nm = n.

Anahtar Kavram

Exponents, Roots, and Powers of Integers
Soru 8Soru

Let n=2a3b5n = 2^a \cdot 3^b \cdot 5, where aa and bb are positive integers. If nn has a total of 24 positive divisors and has an equal number of even positive divisors and odd positive divisors, what is the value of bb?

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Cevap: 5

Cevap

5
For n=2a3b51n = 2^a \cdot 3^b \cdot 5^1, the total number of divisors is (a+1)(b+1)(2)=24(a + 1)(b + 1)(2) = 24, giving (a+1)(b+1)=12(a + 1)(b + 1) = 12. The odd divisors are formed strictly from 3b513^b \cdot 5^1, giving (b+1)(2)(b + 1)(2) odd divisors. The even divisors require at least one factor of 2, giving a(b+1)(2)a(b + 1)(2) even divisors. Setting even and odd divisor counts equal gives 2a(b+1)=2(b+1)2a(b + 1) = 2(b + 1), which reduces to a=1a = 1. Substituting a=1a = 1 into (a+1)(b+1)=12(a + 1)(b + 1) = 12 gives 2(b+1)=122(b + 1) = 12, leading directly to b=5b = 5.

Adım Adım Çözüm

1
Set up the formula for total positive divisors of nn.
(a+1)(b+1)(1+1)=24    (a+1)(b+1)=12(a + 1)(b + 1)(1 + 1) = 24 \implies (a + 1)(b + 1) = 12
The total number of divisors of p1e1p2e2pkekp_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} is (e1+1)(e2+1)(ek+1)(e_1 + 1)(e_2 + 1) \cdots (e_k + 1).
2
Determine the counts of odd and even positive divisors.
Odd divisors = 2(b+1)2(b + 1), Even divisors = 2a(b+1)2a(b + 1)
Odd divisors cannot contain any factors of 2 (exponent of 2 is 0). Even divisors must contain at least one factor of 2 (exponent of 2 can be 1,2,,a1, 2, \dots, a).
3
Equate the number of odd and even divisors to find aa.
2a(b+1)=2(b+1)    a=12a(b + 1) = 2(b + 1) \implies a = 1
Dividing both sides by 2(b+1)2(b + 1) (which is positive since b1b \ge 1) leaves a=1a = 1.
4
Solve for bb using the total divisor relation.
(1+1)(b+1)=12    2(b+1)=12    b=5(1 + 1)(b + 1) = 12 \implies 2(b + 1) = 12 \implies b = 5
Substituting a=1a = 1 into (a+1)(b+1)=12(a + 1)(b + 1) = 12 yields b=5b = 5.

Anahtar Kavram

Counting total, odd, and even positive divisors using prime factorization exponents
Soru 9Soru

If nn is a positive integer such that 6n+3152n110n+19n+1=8100\frac{6^{n+3} \cdot 15^{2n-1}}{10^{n+1} \cdot 9^{n+1}} = 8{}100, what is the value of nn?

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Cevap: 4

Cevap

4
Rewriting all composite bases (6,15,10,96, 15, 10, 9) into prime bases (2,3,52, 3, 5) simplifies the equation to 43n5n2=8,1004 \cdot 3^n \cdot 5^{n-2} = 8,100. Dividing by 4 yields 3n5n2=2,025=34523^n \cdot 5^{n-2} = 2,025 = 3^4 \cdot 5^2, which gives n=4n = 4.

Adım Adım Çözüm

1
Express each composite base in the fraction using its prime factor decomposition.
Numerator: (23)n+3(35)2n1=2n+33n+332n152n1=2n+333n+252n1(2 \cdot 3)^{n+3} \cdot (3 \cdot 5)^{2n-1} = 2^{n+3} \cdot 3^{n+3} \cdot 3^{2n-1} \cdot 5^{2n-1} = 2^{n+3} \cdot 3^{3n+2} \cdot 5^{2n-1}.
Denominator: (25)n+1(32)n+1=2n+15n+132n+2(2 \cdot 5)^{n+1} \cdot (3^2)^{n+1} = 2^{n+1} \cdot 5^{n+1} \cdot 3^{2n+2}.
Breaking composite numbers down into prime factors allows terms with identical bases to be combined using exponent rules.
2
Simplify the fraction by subtracting the denominator exponents from the numerator exponents for each prime base.
2n+333n+252n12n+132n+25n+1=2(n+3)(n+1)3(3n+2)(2n+2)5(2n1)(n+1)=223n5n2=43n5n2\frac{2^{n+3} \cdot 3^{3n+2} \cdot 5^{2n-1}}{2^{n+1} \cdot 3^{2n+2} \cdot 5^{n+1}} = 2^{(n+3)-(n+1)} \cdot 3^{(3n+2)-(2n+2)} \cdot 5^{(2n-1)-(n+1)} = 2^2 \cdot 3^n \cdot 5^{n-2} = 4 \cdot 3^n \cdot 5^{n-2}.
Applying the quotient rule for exponents: axay=axy\frac{a^x}{a^y} = a^{x-y}.
3
Set the simplified expression equal to 8,1008,100 and solve for nn.
43n5n2=8,100    3n5n2=2,0254 \cdot 3^n \cdot 5^{n-2} = 8,100 \implies 3^n \cdot 5^{n-2} = 2,025. Prime factorization of 2,025=8125=34522,025 = 81 \cdot 25 = 3^4 \cdot 5^2. Equating exponents gives n=4n = 4 (and n2=2n-2 = 2, which confirms n=4n=4).
Because 3 and 5 are distinct prime bases, the unique prime factorization theorem guarantees that 3n5n2=34523^n \cdot 5^{n-2} = 3^4 \cdot 5^2 implies n=4n = 4.

Anahtar Kavram

Simplifying exponential expressions using prime factorization and quotient laws
Tahmini Süre:2m 0s
Soru 10Soru

For any positive integer nn, let f(n)f(n) denote the product of all distinct prime factors of nn. For example, f(12)=2×3=6f(12) = 2 \times 3 = 6. What is the value of f(263452243254)f(2^6 \cdot 3^4 \cdot 5^2 - 2^4 \cdot 3^2 \cdot 5^4)?

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Cevap: 330

Cevap

330
To find f(263452243254)f(2^6 \cdot 3^4 \cdot 5^2 - 2^4 \cdot 3^2 \cdot 5^4), first factor out the greatest common factor 2432522^4 \cdot 3^2 \cdot 5^2. This yields 243252(223252)=243252(3625)=243252112^4 \cdot 3^2 \cdot 5^2 (2^2 \cdot 3^2 - 5^2) = 2^4 \cdot 3^2 \cdot 5^2 (36 - 25) = 2^4 \cdot 3^2 \cdot 5^2 \cdot 11. The distinct prime factors present in this expression are 2, 3, 5, and 11. Multiplying these distinct prime factors gives 2×3×5×11=3302 \times 3 \times 5 \times 11 = 330.

Adım Adım Çözüm

1
Factor out the greatest common term 2432522^4 \cdot 3^2 \cdot 5^2 from 2634522432542^6 \cdot 3^4 \cdot 5^2 - 2^4 \cdot 3^2 \cdot 5^4
243252(223252)2^4 \cdot 3^2 \cdot 5^2 \cdot (2^2 \cdot 3^2 - 5^2)
Factoring out common prime powers simplifies the expression and avoids large calculations.
2
Evaluate the arithmetic expression inside the parentheses
223252=4925=3625=112^2 \cdot 3^2 - 5^2 = 4 \cdot 9 - 25 = 36 - 25 = 11
11 is itself a prime number.
3
Write the full prime factorization of the overall number
2432521112^4 \cdot 3^2 \cdot 5^2 \cdot 11^1
All bases (2, 3, 5, 11) are prime numbers, giving the complete prime factorization.
4
Multiply each distinct prime factor together to evaluate f(n)f(n)
2×3×5×11=3302 \times 3 \times 5 \times 11 = 330
The function f(n)f(n) takes the product of each unique prime factor exactly once.

Anahtar Kavram

Prime Factorization and Product of Distinct Prime Factors
Soru 11Soru

A set SS consists of kk consecutive integers. The sum of the first mm integers in set SS is 3434, and the sum of the last mm integers in set SS is 7474. If the sum of all kk integers in set SS is 189189, what is the value of kk?

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Cevap: 14

Cevap

The total number of integers in set S is 14.
By applying the property that the arithmetic mean of an evenly spaced set is the average of the mean of its first mm elements and the mean of its last mm elements, we establish that the mean of the set is 54m\frac{54}{m}. Since the total sum is 189189, k54m=189k \cdot \frac{54}{m} = 189, which yields k=3.5mk = 3.5m. Substituting this ratio into the difference between the two subset sums m(km)=40m(k-m) = 40 gives 2.5m2=402.5m^2 = 40, so m=4m = 4 and k=14k = 14.

Adım Adım Çözüm

1
Set up algebraic expressions for the sums of the first m terms and last m terms.
Let the set be S={a,a+1,,a+k1}S = \{a, a+1, \dots, a+k-1\}. The first mm terms sum to S1=ma+m(m1)2=34S_1 = m a + \frac{m(m-1)}{2} = 34. The last mm terms sum to S2=m(a+km)+m(m1)2=74S_2 = m(a+k-m) + \frac{m(m-1)}{2} = 74.
Consecutive integer sums can be represented by the starting term and the number of terms.
2
Subtract the sum of the first m terms from the sum of the last m terms.
S2S1=m(a+km)ma=m(km)=7434=40S_2 - S_1 = m(a+k-m) - ma = m(k-m) = 74 - 34 = 40.
Subtracting eliminates the initial term aa and quadratic term m(m1)2\frac{m(m-1)}{2}, giving a clean relationship between mm and kk.
3
Determine the arithmetic mean of set S using subset averages.
The average of the first mm terms is 34m\frac{34}{m} and the average of the last mm terms is 74m\frac{74}{m}. The average of the entire set is the midpoint of these two averages: Mean=12(34m+74m)=54m\text{Mean} = \frac{1}{2}\left(\frac{34}{m} + \frac{74}{m}\right) = \frac{54}{m}.
In any evenly spaced set, the overall median/mean is equal to the average of the lower-bound subset mean and upper-bound subset mean.
4
Relate the total sum to the set size k and overall mean.
Stotal=k×Mean    189=k(54m)    km=18954=3.5    k=3.5mS_{total} = k \times \text{Mean} \implies 189 = k \left(\frac{54}{m}\right) \implies \frac{k}{m} = \frac{189}{54} = 3.5 \implies k = 3.5m.
The sum of a set of consecutive integers is always equal to the number of terms times the mean of the set.
5
Solve for m and k using the system of equations.
Substitute k=3.5mk = 3.5m into m(km)=40    m(2.5m)=40    2.5m2=40    m2=16    m=4m(k-m) = 40 \implies m(2.5m) = 40 \implies 2.5m^2 = 40 \implies m^2 = 16 \implies m = 4. Thus, k=3.5×4=14k = 3.5 \times 4 = 14.
Since m>0m > 0, taking the positive square root gives m=4m = 4, which leads directly to k=14k = 14.

Anahtar Kavram

Average and Sum Equivalences in Evenly Spaced Sets
Soru 12Soru

How many integer values of yy satisfy the inequality 2y7y+2|2y - 7| \le |y + 2|?

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Cevap: 8

Cevap

The total number of integer values of yy satisfying the inequality is 8.
Squaring both sides of 2y7y+2|2y - 7| \le |y + 2| gives 3y232y+4503y^2 - 32y + 45 \le 0, which factors into (3y5)(y9)0(3y - 5)(y - 9) \le 0. The solution range for yy is 53y9\frac{5}{3} \le y \le 9. Since 531.67\frac{5}{3} \approx 1.67, the integer values of yy satisfying this range are 2,3,4,5,6,7,8,92, 3, 4, 5, 6, 7, 8, 9, giving a total of 8 integers.

Adım Adım Çözüm

1
Square both sides of the inequality 2y7y+2|2y - 7| \le |y + 2|
(2y7)2(y+2)2(2y - 7)^2 \le (y + 2)^2
Since both sides of an absolute value expression are non-negative, squaring both sides maintains the inequality direction.
2
Expand terms and move all terms to the left side
3y232y+4503y^2 - 32y + 45 \le 0
Expanding gives 4y228y+49y2+4y+44y^2 - 28y + 49 \le y^2 + 4y + 4. Subtracting (y2+4y+4)(y^2 + 4y + 4) from both sides produces the standard quadratic inequality.
3
Factor the quadratic expression to find critical points
(3y5)(y9)0(3y - 5)(y - 9) \le 0, yielding 53y9\frac{5}{3} \le y \le 9
The roots are y=53y = \frac{5}{3} and y=9y = 9. A quadratic with a positive leading coefficient is non-positive between its roots.
4
Determine all integers within the range [53,9]\left[\frac{5}{3}, 9\right]
2,3,4,5,6,7,8,92, 3, 4, 5, 6, 7, 8, 9 (8 integers total)
Because 531.67\frac{5}{3} \approx 1.67, the smallest integer within the range is 2 and the largest is 9.

Anahtar Kavram

Solving absolute value inequalities of the form AB|A| \le |B| by squaring both sides and determining integer solutions.
Soru 13Soru

An investor allocated a total principal of $25,000\$25,000 between two savings accounts, Account X and Account Y. Account X earns simple interest at an annual rate of 8%8\%, while Account Y earns interest at an annual rate of 10%10\%, compounded annually. If after 22 years the total interest earned from Account Y exceeds the total interest earned from Account X by $1,550\$1,550, what was the amount, in dollars, invested in Account Y?

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Cevap: 15000

Cevap

The amount invested in Account Y was $15,000 dollars.
Let PYP_Y be the principal invested in Account Y. Since the total investment is $25,000\$25,000, the principal invested in Account X is 25,000PY25,000 - P_Y. Account X earns simple interest over 2 years equal to (25,000PY)×0.08×2=4,0000.16PY(25,000 - P_Y) \times 0.08 \times 2 = 4,000 - 0.16 P_Y. Account Y earns compound interest over 2 years equal to PY×((1+0.10)21)=0.21PYP_Y \times ((1 + 0.10)^2 - 1) = 0.21 P_Y. The problem states that the interest from Account Y exceeds that from Account X by $1,550\$1,550, giving the equation 0.21PY(4,0000.16PY)=1,5500.21 P_Y - (4,000 - 0.16 P_Y) = 1,550. Combining like terms yields 0.37PY=5,5500.37 P_Y = 5,550, which simplifies to PY=15,000P_Y = 15,000.

Adım Adım Çözüm

1
Express the principal of Account X in terms of Account Y
Principal of Account X = 25,000PY25,000 - P_Y
The total sum invested across both accounts is $25,000\$25,000.
2
Calculate the interest earned from Account X over 2 years
IX=4,0000.16PYI_X = 4,000 - 0.16 P_Y
Simple interest is calculated as I=P×r×t=(25,000PY)×0.08×2I = P \times r \times t = (25,000 - P_Y) \times 0.08 \times 2.
3
Calculate the compound interest earned from Account Y over 2 years
IY=0.21PYI_Y = 0.21 P_Y
The compound multiplier for 2 years at 10%10\% is (1.10)2=1.21(1.10)^2 = 1.21, giving a interest percentage of 1.211=0.211.21 - 1 = 0.21 or 21%21\%.
4
Set up and simplify the equation for the difference in interest
0.37PY=5,5500.37 P_Y = 5,550
Subtracting IXI_X from IYI_Y yields 0.21PY(4,0000.16PY)=1,5500.21 P_Y - (4,000 - 0.16 P_Y) = 1,550, which simplifies to 0.37PY4,000=1,5500.37 P_Y - 4,000 = 1,550.
5
Solve for PYP_Y
PY=15,000P_Y = 15,000
Dividing 5,5505,550 by 0.370.37 gives 15,00015,000.

Anahtar Kavram

Combining Simple and Compound Interest Linear Equations
Soru 14Soru

An analyst at an investment firm evaluates a portfolio of 160160 corporate bonds for two potential risk factors: credit rating downgrade risk and liquidity risk. The evaluation reveals that 6060 bonds have credit rating downgrade risk, 7272 bonds have liquidity risk, and 6464 bonds have neither risk factor. If a bond is selected at random from those in the portfolio that have at least one of the two risk factors, what is the probability that it has credit rating downgrade risk? Express your answer as a decimal.

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Cevap: 0.625

Cevap

The probability is 0.625 (or 5/8).
To calculate the probability that a bond has credit rating downgrade risk given that it has at least one risk factor, the sample space must be restricted to bonds with at least one risk factor. Out of 160160 bonds, 6464 have neither risk factor, leaving 16064=96160 - 64 = 96 bonds with at least one risk factor. All 6060 bonds with credit rating downgrade risk are part of this group. The required conditional probability is 6096=58=0.625\frac{60}{96} = \frac{5}{8} = 0.625.

Adım Adım Çözüm

1
Calculate the size of the restricted sample space (bonds with at least one risk factor).
Total bonds with at least one risk factor = 160 - 64 = 96 bonds.
The condition specifies that the selection is made only from bonds having at least one risk factor.
2
Identify the number of favorable outcomes within this restricted sample space.
Number of bonds with credit rating downgrade risk = 60.
All 60 bonds with credit rating downgrade risk inherently possess at least one risk factor, so they lie entirely within the restricted sample space.
3
Compute the conditional probability P(Downgrade Risk | At Least One Risk).
60 / 96 = 5 / 8 = 0.625.
Conditional probability requires dividing the count of favorable outcomes by the count of the restricted sample space.

Anahtar Kavram

Conditional Probability and Sample Space Restriction
Soru 15Soru

How many integer values of xx satisfy the inequality 2x574||2x - 5| - 7| \leq 4?

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Cevap: 10

Cevap

The total number of integer values of xx satisfying the inequality is 10.
To solve 2x574||2x - 5| - 7| \leq 4, rewrite the inequality without the outer absolute value as 42x574-4 \leq |2x - 5| - 7 \leq 4. Adding 7 across all parts yields 32x5113 \leq |2x - 5| \leq 11. The inequality 2x511|2x - 5| \leq 11 simplifies to 3x8-3 \leq x \leq 8. The inequality 2x53|2x - 5| \geq 3 simplifies to x1x \leq 1 or x4x \geq 4. Intersecting these two regions gives the set of real numbers x[3,1][4,8]x \in [-3, 1] \cup [4, 8]. The integer solutions within [3,1][-3, 1] are 3,2,1,0,1-3, -2, -1, 0, 1 (5 integers), and within [4,8][4, 8] are 4,5,6,7,84, 5, 6, 7, 8 (5 integers). The total number of valid integer solutions is 5+5=105 + 5 = 10.

Adım Adım Çözüm

1
Unfold the outer absolute value expression.
42x574-4 \leq |2x - 5| - 7 \leq 4
An inequality of the form UC|U| \leq C with C>0C > 0 is equivalent to CUC-C \leq U \leq C.
2
Isolate the inner absolute value expression by adding 7 throughout.
32x5113 \leq |2x - 5| \leq 11
Adding a constant to all parts preserves the direction of the inequality.
3
Solve the upper bound 2x511|2x - 5| \leq 11.
3x8-3 \leq x \leq 8
112x511-11 \leq 2x - 5 \leq 11 adds 5 to give 62x16-6 \leq 2x \leq 16, which divides by 2 to yield 3x8-3 \leq x \leq 8.
4
Solve the lower bound 2x53|2x - 5| \geq 3.
x1x \leq 1 or x4x \geq 4
An inequality UC|U| \geq C splits into UCU \geq C (2x53    x42x - 5 \geq 3 \implies x \geq 4) or UCU \leq -C (2x53    x12x - 5 \leq -3 \implies x \leq 1).
5
Find the intersection of the upper and lower bound conditions and count the integers.
10 integer solutions: {3,2,1,0,1,4,5,6,7,8}\{-3, -2, -1, 0, 1, 4, 5, 6, 7, 8\}.
Combining 3x8-3 \leq x \leq 8 with (x1x \leq 1 or x4x \geq 4) produces two disjoint intervals [3,1][-3, 1] and [4,8][4, 8], containing 5 integers each.

Anahtar Kavram

Solving nested absolute value inequalities using double inequalities and boundary region intersections.
Soru 16Soru

What is the sum of all integer values of xx that satisfy the inequality x26x+5<2x2|x^2 - 6x + 5| < 2x - 2?

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Cevap: 15

Cevap

The sum of all integer values of xx satisfying the inequality is 1515.
Factoring both sides gives (x1)(x5)<2(x1)|(x - 1)(x - 5)| < 2(x - 1). Since the absolute value is non-negative, the right-hand side requires 2x2>0    x>12x - 2 > 0 \implies x > 1. Under x>1x > 1, the term x1x - 1 is strictly positive, allowing us to simplify to x5<2|x - 5| < 2, which yields 3<x<73 < x < 7. The integer solutions are 4,5,4, 5, and 66, and their sum is 1515.

Adım Adım Çözüm

1
Determine the necessary condition for the right-hand side of the inequality.
Because the left-hand side x26x+5|x^2 - 6x + 5| is non-negative for all real xx, the right-hand side must be strictly positive. Thus, 2x2>0    x>12x - 2 > 0 \implies x > 1.
An absolute value expression cannot be strictly less than a zero or negative quantity.
2
Factor the quadratic expression inside the absolute value and the linear expression on the right.
(x1)(x5)<2(x1)|(x - 1)(x - 5)| < 2(x - 1).
Factoring reveals a common linear factor (x1)(x - 1) on both sides.
3
Simplify the inequality using the condition x>1x > 1.
Since x>1x > 1, we know x1>0x - 1 > 0, so x1=x1|x - 1| = x - 1. Splitting the product inside the absolute value gives (x1)x5<2(x1)(x - 1)|x - 5| < 2(x - 1). Dividing both sides by (x1)(x - 1) yields x5<2|x - 5| < 2.
Dividing an inequality by a strictly positive number preserves the direction of the inequality sign.
4
Solve the simplified absolute value inequality and sum the integer solutions.
x5<2    2<x5<2    3<x<7|x - 5| < 2 \implies -2 < x - 5 < 2 \implies 3 < x < 7. The integer solutions strictly within this range are x=4,5,x = 4, 5, and 66. Their sum is 4+5+6=154 + 5 + 6 = 15.
The integers strictly between 33 and 77 are 44, 55, and 66.

Anahtar Kavram

Solving Quadratic Absolute Value Inequalities via Domain Constraints and Factoring
Tahmini Süre:2m 0s
Soru 17Soru

A manufacturing shop produces two models of custom bicycle frames: Standard and Deluxe. Producing each Standard frame requires 22 hours of welding and 11 hour of painting. Producing each Deluxe frame requires 33 hours of welding and 22 hours of painting. During a single week, the shop logged a total of 130130 hours of welding and 7575 hours of painting for these two models. How many Deluxe bicycle frames were produced during that week?

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Cevap: 20

Cevap

The number of Deluxe bicycle frames produced during that week is 20.
Translating the resource limitations into a system of two linear equations yields 2x+3y=1302x + 3y = 130 for welding hours and x+2y=75x + 2y = 75 for painting hours, where xx and yy represent the number of Standard and Deluxe frames respectively. Expressing xx in terms of yy from the painting equation gives x=752yx = 75 - 2y. Substituting this expression into the welding equation gives 2(752y)+3y=1302(75 - 2y) + 3y = 130, which simplifies to 150y=130150 - y = 130, giving y=20y = 20 Deluxe frames.

Adım Adım Çözüm

1
Define variables for the unknowns.
Let xx be the number of Standard bicycle frames produced and yy be the number of Deluxe bicycle frames produced.
Assigning variables allows us to translate the problem into algebraic expressions.
2
Set up a system of linear equations.
Welding constraint: 2x+3y=1302x + 3y = 130
Painting constraint: x+2y=75x + 2y = 75
Each constraint represents the sum of hours spent on Standard and Deluxe frames for that process.
3
Solve the system using elimination or substitution.
Multiply the painting equation by 22: 2x+4y=1502x + 4y = 150.
Subtract the welding equation (2x+3y=1302x + 3y = 130) from this result: (2x+4y)(2x+3y)=150130y=20(2x + 4y) - (2x + 3y) = 150 - 130 \Rightarrow y = 20.
Eliminating xx directly solves for yy, which is the requested quantity (Deluxe frames).

Anahtar Kavram

Solving systems of two linear equations with two variables by substitution or elimination.
Tahmini Süre:1m 30s
Soru 18Soru

A company recorded the number of daily sales over a 7-day week. The numbers of daily sales, when arranged in ascending order, are 12,15,18,x,22,y,3012, 15, 18, x, 22, y, 30. If the median of the 7 daily sales figures is 2020 and the arithmetic mean is 2121, what is the value of yy?

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Cevap: 30

Cevap

The value of yy is 30.
For an ordered set of 7 numbers, the median is the 4th term, which means x=20x = 20. The total sum of the set is found by multiplying the number of terms by the mean: 7×21=1477 \times 21 = 147. Adding all known terms (12+15+18+20+22+3012 + 15 + 18 + 20 + 22 + 30) gives 117. Subtracting 117 from 147 yields y=30y = 30.

Adım Adım Çözüm

1
Determine the value of xx using the median definition.
x=20x = 20
Since the 7 numbers are given in ascending order, the median is the middle (4th) term.
2
Calculate the total sum of all 7 numbers.
Total sum = 147147
The sum of elements in a set equals the number of elements multiplied by the arithmetic mean (7×21=1477 \times 21 = 147).
3
Sum all known numbers and set up an equation for yy.
117+y=147117 + y = 147
12+15+18+20+22+30=11712 + 15 + 18 + 20 + 22 + 30 = 117.
4
Solve for yy.
y=30y = 30
Subtract 117 from 147 to isolate yy.

Anahtar Kavram

Using properties of median and arithmetic mean to find missing terms in an ordered data set.
Soru 19Soru

At an agricultural cooperative, 0.450.45 of the total fruit harvested by weight consisted of apples, and 38\frac{3}{8} of the remaining fruit weight consisted of pears. If the remaining 550550 kilograms of fruit consisted entirely of oranges, what was the total weight, in kilograms, of the fruit harvested?

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Cevap: 1600

Cevap

1600
Subtracting the apple portion leaves 55% (or 11/20) of the total harvest. Since pears make up 3/8 of this remainder, oranges make up the remaining 5/8 of the 11/20 portion, which simplifies to 11/32 of the total harvest. Setting 11/32 of the total harvest equal to 550 kg yields a total weight of 1,600 kg.

Adım Adım Çözüm

1
Find the fraction of the total harvest left after subtracting the apples.
The remaining portion is 1 - 0.45 = 0.55, which equals 11/20 of the total harvest.
Apples account for 0.45 of the harvest.
2
Calculate the portion of the harvest represented by oranges.
Since pears are 3/8 of the remainder, oranges are 5/8 of the remainder. Thus, oranges are (5/8) * (11/20) = 11/32 of the total harvest.
The non-apple harvest consists only of pears and oranges.
3
Solve for the total weight using the given weight of oranges.
Total weight = 550 * (32 / 11) = 1600 kg.
11/32 of the total weight equals 550 kilograms.

Anahtar Kavram

Multi-step arithmetic combining decimals and fractions to solve remaining-quantity word problems
Soru 20Soru

What is the exponent of 33 in the prime factorization of the integer S=25!+26!+27!S = 25! + 26! + 27!?

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Cevap: 16

Cevap

The exponent of 33 in the prime factorization of SS is 1616.
Factoring 25!25! from the sum gives S=25!(1+26+26×27)=25!(729)=25!×36S = 25!(1 + 26 + 26 \times 27) = 25!(729) = 25! \times 3^6. Applying Legendre's formula to 25!25! yields 25/3+25/9=8+2=10\lfloor 25/3 \rfloor + \lfloor 25/9 \rfloor = 8 + 2 = 10 factors of 33. Adding the 66 factors of 33 from 729=36729 = 3^6 gives a total exponent of 10+6=1610 + 6 = 16.

Adım Adım Çözüm

1
Factor out 25!25! from the sum S=25!+26!+27!S = 25! + 26! + 27!
S=25!(1+26+26×27)S = 25! \left(1 + 26 + 26 \times 27\right)
Factoring out the greatest common factorial term 25!25! converts the sum into a product of 25!25! and an integer factor.
2
Simplify the expression inside the parentheses
1+26+702=729=361 + 26 + 702 = 729 = 3^6
Using algebraic simplification, 1+26(1+27)=1+26(28)=729=272=(33)2=361 + 26(1 + 27) = 1 + 26(28) = 729 = 27^2 = (3^3)^2 = 3^6.
3
Compute the exponent of 33 in 25!25! using Legendre's formula
E3(25!)=253+259+2527=8+2+0=10E_3(25!) = \lfloor \frac{25}{3} \rfloor + \lfloor \frac{25}{9} \rfloor + \lfloor \frac{25}{27} \rfloor = 8 + 2 + 0 = 10
Legendre's formula counts the total prime factors of 33 contributed by all multiples of 3,9,27,3, 9, 27, \dots up to 2525.
4
Combine the exponent of 33 from 25!25! and the factor 729729
Total exponent of 3=10+6=163 = 10 + 6 = 16
Since S=25!×36=(310×k)×36=316×kS = 25! \times 3^6 = (3^{10} \times k) \times 3^6 = 3^{16} \times k (where 3k3 \nmid k), the total exponent of 33 is 10+6=1610 + 6 = 16.

Anahtar Kavram

Exponent of a prime in factorial expressions using Legendre's formula and algebraic factoring
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