Arithmetic

306 soru

Soru 161Soru

If pp, p+2p + 2, and p+4p + 4 are all prime numbers, what is the value of p2+5p^2 + 5?

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Cevap: 14

Cevap

The value of p2+5p^2 + 5 is 14.
For any three consecutive odd integers pp, p+2p + 2, and p+4p + 4, exactly one of them must be a multiple of 3. Because all three expressions represent prime numbers, the term divisible by 3 must be equal to 3 (since 3 is the only prime divisible by 3). Setting p=3p = 3 gives p+2=5p + 2 = 5 and p+4=7p + 4 = 7, both of which are prime. Any choice of p>3p > 3 forces either p+2p + 2 or p+4p + 4 to be a multiple of 3 greater than 3, making it composite. Thus p=3p = 3 is uniquely determined, and evaluating p2+5p^2 + 5 gives 32+5=143^2 + 5 = 14.

Adım Adım Çözüm

1
Analyze the possible remainders when the prime pp is divided by 3.
Any positive integer pp can be written in one of three forms: 3k3k, 3k+13k + 1, or 3k+23k + 2 for some integer kk.
Dividing any integer by 3 leaves a remainder of 0, 1, or 2.
2
Test each remainder case for the expressions pp, p+2p + 2, and p+4p + 4.
If p=3k+1p = 3k + 1, then p+2=3k+3=3(k+1)p + 2 = 3k + 3 = 3(k + 1), which is divisible by 3. If p=3k+2p = 3k + 2, then p+4=3k+6=3(k+2)p + 4 = 3k + 6 = 3(k + 2), which is divisible by 3.
Among any three consecutive odd numbers of the form p,p+2,p+4p, p+2, p+4, exactly one of them must be a multiple of 3.
3
Deduce the unique value of pp.
The only way all three numbers pp, p+2p + 2, and p+4p + 4 can be prime is if the one divisible by 3 is equal to 3 itself, which forces p=3p = 3.
The only prime number divisible by 3 is 3 itself; any larger multiple of 3 is composite.
4
Evaluate the target expression p2+5p^2 + 5 using p=3p = 3.
32+5=9+5=143^2 + 5 = 9 + 5 = 14.
Substitute the uniquely determined value p=3p = 3 into the given algebraic expression.

Anahtar Kavram

Divisibility properties of consecutive odd integers and prime number definitions
Tahmini Süre:1m 30s
Soru 162Soru

On the real number line, pp and qq are real numbers such that 32p5|3 - 2p| \le 5 and q+4=2|q + 4| = 2. What is the maximum possible value of p2q|p^2 - q|?

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Cevap: 22

Cevap

22
Solving 32p5|3 - 2p| \le 5 gives 532p5    1p4-5 \le 3 - 2p \le 5 \implies -1 \le p \le 4, so p2p^2 can range from 00 up to 1616. Solving q+4=2|q + 4| = 2 yields q=2q = -2 or q=6q = -6. To maximize p2q|p^2 - q|, we combine the maximum possible value of p2p^2 (1616) with q=6q = -6, obtaining 16(6)=22|16 - (-6)| = 22.

Adım Adım Çözüm

1
Solve the absolute value inequality 32p5|3 - 2p| \le 5 for pp.
1p4-1 \le p \le 4
532p5    82p2-5 \le 3 - 2p \le 5 \implies -8 \le -2p \le 2. Dividing by 2-2 and reversing the inequality direction gives 1p4-1 \le p \le 4.
2
Determine the range of possible values for p2p^2.
0p2160 \le p^2 \le 16
Since pp spans from 1-1 to 44 (which includes 00), the minimum square is 02=00^2 = 0 and the maximum square is 42=164^2 = 16.
3
Solve the absolute value equation q+4=2|q + 4| = 2 for qq.
q=2q = -2 or q=6q = -6
q+4=2    q=2q + 4 = 2 \implies q = -2, and q+4=2    q=6q + 4 = -2 \implies q = -6.
4
Find the combination of p2p^2 and qq that maximizes p2q|p^2 - q|.
Maximum value is 22
Pairing p2=16p^2 = 16 with q=6q = -6 yields 16(6)=22|16 - (-6)| = 22, which is greater than 16(2)=18|16 - (-2)| = 18.

Anahtar Kavram

Absolute value inequalities, real number bounds, and distance optimization.
Soru 163Soru

A chemical laboratory starts with a liquid solution having a total volume of 3.2×1043.2 \times 10^{-4} cubic meters. The solution is partitioned equally into 400400 identical micro-vials. Subsequently, heat treatment causes the volume of liquid in each micro-vial to evaporate, reducing its volume by 75%75\%. What is the final volume of liquid remaining in a single micro-vial, expressed in scientific notation?

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Cevap: 2.0×1072.0 \times 10^{-7} cubic meters

Cevap

2.0×1072.0 \times 10^{-7} cubic meters
Dividing the initial total volume of 3.2×1043.2 \times 10^{-4} cubic meters by 400400 (4×1024 \times 10^2) gives an initial volume of 8.0×1078.0 \times 10^{-7} cubic meters per micro-vial. Reducing this volume by 75%75\% means 25%25\% of the liquid remains. Multiplying 8.0×1078.0 \times 10^{-7} by 0.250.25 yields 2.0×1072.0 \times 10^{-7} cubic meters.

Adım Adım Çözüm

1
Express the number of micro-vials in scientific notation.
400=4×102400 = 4 \times 10^2
Converting into scientific notation simplifies division involving powers of 10.
2
Calculate the initial volume per micro-vial before evaporation.
3.2×1044×102=(3.24)×1042=0.8×106=8.0×107\frac{3.2 \times 10^{-4}}{4 \times 10^2} = \left(\frac{3.2}{4}\right) \times 10^{-4 - 2} = 0.8 \times 10^{-6} = 8.0 \times 10^{-7} cubic meters
Equal partitioning requires dividing the total volume by the total number of containers.
3
Determine the remaining fraction of liquid after a 75%75\% volume reduction.
Remaining fraction =100%75%=25%=0.25= 100\% - 75\% = 25\% = 0.25
A reduction by 75%75\% leaves 25%25\% of the liquid volume in the vial.
4
Multiply the volume per micro-vial by the remaining fraction.
8.0×107×0.25=2.0×1078.0 \times 10^{-7} \times 0.25 = 2.0 \times 10^{-7} cubic meters
Applying the remaining fraction yields the final volume in standard scientific notation format a×10na \times 10^n where 1a<101 \le a < 10.

Anahtar Kavram

Operations with Decimals, Place Value, and Scientific Notation
Tahmini Süre:2m 0s
Soru 164Soru

What is the value of 27+2724\frac{2^7 + 2^7}{2^4}?

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Cevap: 16

Cevap

16
Combining the numerator yields 27+27=2(27)=282^7 + 2^7 = 2(2^7) = 2^8. Dividing 282^8 by 242^4 using the quotient rule gives 284=24=162^{8-4} = 2^4 = 16.

Adım Adım Çözüm

1
Simplify the numerator by factoring out common terms
27+27=2×27=21+7=282^7 + 2^7 = 2 \times 2^7 = 2^{1+7} = 2^8
Adding two identical exponential terms is equivalent to multiplying one term by 2.
2
Apply the quotient rule of exponents
2824=284=24\frac{2^8}{2^4} = 2^{8-4} = 2^4
When dividing exponential expressions with the same base, subtract the exponent of the denominator from the exponent of the numerator.
3
Evaluate the power
2^4 = 16
Compute the numerical value of 2 raised to the 4th power.

Anahtar Kavram

Combining like exponential terms and applying the quotient rule of exponents.
Soru 165Soru

A municipal library system allocated a total annual acquisition budget of BB dollars between fiction and non-fiction titles. In the first half of the year, 60%60\% of the total budget was spent, with 70%70\% of that spent amount allocated to fiction books and the remainder to non-fiction books. In the second half of the year, the remaining budget was fully spent such that equal dollar amounts were allocated to fiction and non-fiction books. What percentage of the total annual budget BB was spent on fiction books?

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Cevap: 62%62\%

Cevap

The total percentage of budget BB spent on fiction books is 62%62\%.
To find the total percentage of budget BB spent on fiction books, we analyze the two periods separately relative to the total budget BB. In the first half, 60%60\% of BB is spent, and 70%70\% of this spent amount goes to fiction, giving 0.70×0.60B=0.42B0.70 \times 0.60 B = 0.42 B (42%42\% of BB). In the second half, the remaining 40%40\% of BB is spent equally between fiction and non-fiction, so fiction receives 50%50\% of 40%40\%, which is 0.50×0.40B=0.20B0.50 \times 0.40 B = 0.20 B (20%20\% of BB). Adding both periods yields 42%+20%=62%42\% + 20\% = 62\% of the total annual budget BB.

Adım Adım Çözüm

1
Calculate the percentage of total budget BB spent on fiction in the first half of the year.
0.70×0.60B=0.42B0.70 \times 0.60 B = 0.42 B, which is 42%42\% of BB.
The library spent 60%60\% of budget BB, and 70%70\% of that spent portion went to fiction books.
2
Determine the remaining budget percentage for the second half of the year.
100%60%=40%100\% - 60\% = 40\% of BB.
The remaining budget is the total budget minus the portion spent during the first half.
3
Calculate the percentage of total budget BB spent on fiction in the second half of the year.
0.50×0.40B=0.20B0.50 \times 0.40 B = 0.20 B, which is 20%20\% of BB.
The remaining 40%40\% budget was split equally between fiction and non-fiction.
4
Sum the fiction expenditures from both halves of the year.
42%+20%=62%42\% + 20\% = 62\% of BB.
Adding the contributions from both periods gives the total fraction of budget BB allocated to fiction books.

Anahtar Kavram

Multi-stage percentage change and part-of-a-part base shift calculations.
Tahmini Süre:1m 30s
Soru 166Soru

If nn is a positive integer that is divisible by 66 but not divisible by 44, and n2n^2 has exactly 1515 positive divisors, what is the value of nn?

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Cevap: 1818

Cevap

The value of nn is 1818.
Because nn is divisible by 66, it must have prime factors 22 and 33. Since nn is not divisible by 44, the exponent of 22 in nn is exactly 11, meaning n2n^2 has 222^2 as a factor. The divisor count formula for n2n^2 requires (2+1)(2b+1)=15(2+1)(2b+1) = 15, which gives 2b+1=52b+1 = 5, so b=2b=2. Thus, n=2132=18n = 2^1 \cdot 3^2 = 18.

Adım Adım Çözüm

1
Analyze the prime factorization structure of nn based on given divisibility conditions.
Since nn is divisible by 6=236 = 2 \cdot 3, its prime factorization must contain at least one factor of 22 and one factor of 33. Since nn is not divisible by 4=224 = 2^2, the exponent of 22 in the prime factorization of nn must be exactly 11. Thus, n=213bkn = 2^1 \cdot 3^b \cdot k, where kk contains no factors of 22 or 33.
Establishing the exponent of 22 narrows down the search space for prime factor exponents.
2
Express n2n^2 in terms of its prime factors and write the formula for its number of positive divisors.
n2=2232bk2n^2 = 2^2 \cdot 3^{2b} \cdot k^2. The number of positive divisors of n2n^2 is given by d(n2)=(2+1)(2b+1)d(k2)=3(2b+1)d(k2)=15d(n^2) = (2 + 1)(2b + 1) \cdot d(k^2) = 3(2b + 1) \cdot d(k^2) = 15.
The total number of positive divisors of a number p1a1p2a2p_1^{a_1} p_2^{a_2} \dots is (a1+1)(a2+1)(a_1 + 1)(a_2 + 1) \dots.
3
Solve for bb and determine if kk has any additional prime factors.
Dividing 1515 by 33 gives (2b+1)d(k2)=5(2b + 1) \cdot d(k^2) = 5. Since 55 is prime, we must have d(k2)=1d(k^2) = 1 (meaning k=1k = 1) and 2b+1=52b + 1 = 5, which yields 2b=4    b=22b = 4 \implies b = 2.
Determining the exponent of 33 fixes the exact prime factorization of nn.
4
Calculate nn.
n=2132=18n = 2^1 \cdot 3^2 = 18.
Multiplying the prime factors together yields the target integer.

Anahtar Kavram

Prime Factorization and Divisor Count Formula
Tahmini Süre:1m 30s
Soru 167Soru

At a research institute, an annual equipment budget is distributed among three departments. First, 25\frac{2}{5} of the total budget is allocated to the Biology department. Next, 38\frac{3}{8} of the remaining budget is allocated to the Chemistry department. The Physics department receives the remaining balance of $45,000\$45,000. What was the institute's total annual equipment budget?

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Cevap: $120,000

Cevap

$120,000
Subtracting the Biology department's share of 25\frac{2}{5} leaves 35\frac{3}{5} of the total budget. The Chemistry department receives 38\frac{3}{8} of this remainder, leaving 58\frac{5}{8} of the remainder for the Physics department. Multiplying these fractions yields 58×35=38\frac{5}{8} \times \frac{3}{5} = \frac{3}{8} of the total budget for Physics. Equating 38\frac{3}{8} of the total budget to $45,000\$45,000 gives a total budget of $120,000\$120,000.

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1
Determine the fraction of the budget remaining after the Biology allocation.
The remaining fraction is 125=351 - \frac{2}{5} = \frac{3}{5} of the total budget.
The Biology department receives 25\frac{2}{5} of the overall budget.
2
Calculate the fraction of the total budget allocated to Physics.
The Physics department receives 138=581 - \frac{3}{8} = \frac{5}{8} of the remaining budget, which equals 58×35=38\frac{5}{8} \times \frac{3}{5} = \frac{3}{8} of the total budget.
The Chemistry department receives 38\frac{3}{8} of the remaining budget, leaving 58\frac{5}{8} of that remainder for Physics.
3
Solve for the total budget BB.
\frac{3}{8}B = 45,000 \implies B = 45,000 \times \frac{8}{3} = 120,000.
The Physics allocation of $45,000\$45,000 represents 38\frac{3}{8} of the total budget BB.

Anahtar Kavram

Sequential Fraction of Remainder Calculations
Tahmini Süre:1m 30s
Soru 168Soru

A commercial trucking fleet monitored its vehicle fuel efficiency (in miles per gallon) and average fuel price (in dollars per gallon) over a two-year period. In Year 1, vehicle fuel efficiency increased by 25%25\%, while the price of fuel increased by 15%15\%. In Year 2, vehicle fuel efficiency increased by an additional 20%20\% relative to Year 1, while the price of fuel decreased by 10%10\% relative to Year 1. Assuming the total annual miles driven per vehicle remained constant throughout the two-year period, by what percent did the total annual fuel cost per vehicle change from the beginning of Year 1 to the end of Year 2?

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Cevap: 31%31\% decrease

Cevap

A 31%31\% decrease
The total fuel cost is given by Cost=Distance×PriceEfficiency\text{Cost} = \text{Distance} \times \frac{\text{Price}}{\text{Efficiency}}. After Year 1, efficiency becomes 1.25E01.25 E_0 and price becomes 1.15P01.15 P_0. After Year 2, efficiency increases to 1.25E0×1.20=1.50E01.25 E_0 \times 1.20 = 1.50 E_0, while price changes to 1.15P0×0.90=1.035P01.15 P_0 \times 0.90 = 1.035 P_0. The new cost ratio is 1.0351.50=0.69\frac{1.035}{1.50} = 0.69, meaning the new cost is 69%69\% of the original cost. This represents a net decrease of 100%69%=31%100\% - 69\% = 31\%.

Adım Adım Çözüm

1
Formulate the fuel cost relationship in terms of distance, efficiency, and price.
Initial cost C0=D×P0E0C_0 = D \times \frac{P_0}{E_0}, where DD is distance driven, P0P_0 is initial price per gallon, and E0E_0 is initial fuel efficiency in miles per gallon.
Total fuel used is DE0\frac{D}{E_0}, so multiplying by price per gallon gives total cost.
2
Calculate the updated fuel efficiency E2E_2 and price P2P_2 at the end of Year 2 using successive percentage changes.
E2=E0×1.25×1.20=1.50E0E_2 = E_0 \times 1.25 \times 1.20 = 1.50 E_0 and P2=P0×1.15×0.90=1.035P0P_2 = P_0 \times 1.15 \times 0.90 = 1.035 P_0.
Compounding changes sequentially: Year 1 efficiency factor is 1.251.25, Year 2 factor is 1.201.20; Year 1 price factor is 1.151.15, Year 2 factor is 0.900.90.
3
Compute the final fuel cost C2C_2 as a fraction of the initial cost C0C_0.
C2=D×1.035P01.50E0=1.0351.50×C0=0.69C0C_2 = D \times \frac{1.035 P_0}{1.50 E_0} = \frac{1.035}{1.50} \times C_0 = 0.69 C_0.
Substituting the expressions for P2P_2 and E2E_2 into the fuel cost formula.
4
Determine the net percentage change from C0C_0 to C2C_2.
\text{Percent Change} = \frac{0.69 C_0 - C_0}{C_0} \times 100\% = -31\%,representinga, representing a 31\%$ decrease.
Subtracting 1.001.00 from 0.690.69 yields 0.31-0.31, which corresponds to a 31%31\% reduction.

Anahtar Kavram

Successive Percentage Changes and Rate Relationships
Soru 169Soru

A positive integer nn has a prime factorization of the form 2x×3y×7z2^x \times 3^y \times 7^z, where xx, yy, and zz are positive integers. The greatest common divisor of nn and 420420 is 8484, and the least common multiple of nn and 420420 is 8,8208,820. What is the total number of positive integer factors of nn?

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Cevap: 2727

Cevap

The total number of positive integer factors of nn is 2727.
By prime factorizing 420420, 8484, and 8,8208,820, we find that 420=22×31×51×71420 = 2^2 \times 3^1 \times 5^1 \times 7^1, gcd(n,420)=22×31×71\gcd(n, 420) = 2^2 \times 3^1 \times 7^1, and lcm(n,420)=22×32×51×72\text{lcm}(n, 420) = 2^2 \times 3^2 \times 5^1 \times 7^2. Comparing the minimum and maximum powers for each prime factor shows that n=22×32×72n = 2^2 \times 3^2 \times 7^2. Applying the divisor counting formula gives (2+1)(2+1)(2+1)=27(2+1)(2+1)(2+1) = 27.

Adım Adım Çözüm

1
Express all given values in their prime factorized forms.
420=22×31×51×71420 = 2^2 \times 3^1 \times 5^1 \times 7^1, gcd(n,420)=84=22×31×71\gcd(n, 420) = 84 = 2^2 \times 3^1 \times 7^1, and lcm(n,420)=8,820=22×32×51×72\text{lcm}(n, 420) = 8,820 = 2^2 \times 3^2 \times 5^1 \times 7^2.
Finding the prime factorizations allows comparison of prime exponent bounds for GCD and LCM.
2
Determine the exponents xx, yy, and zz for n=2x×3y×7zn = 2^x \times 3^y \times 7^z.
For prime 22: max(x,2)=2\max(x, 2) = 2 and min(x,2)=2    x=2\min(x, 2) = 2 \implies x = 2.
For prime 33: max(y,1)=2\max(y, 1) = 2 and min(y,1)=1    y=2\min(y, 1) = 1 \implies y = 2.
For prime 77: max(z,1)=2\max(z, 1) = 2 and min(z,1)=1    z=2\min(z, 1) = 1 \implies z = 2.
GCD takes the minimum exponent of each prime factor, while LCM takes the maximum exponent.
3
Calculate the total number of positive factors of n=22×32×72n = 2^2 \times 3^2 \times 7^2.
Total factors =(x+1)(y+1)(z+1)=(2+1)(2+1)(2+1)=3×3×3=27= (x + 1)(y + 1)(z + 1) = (2 + 1)(2 + 1)(2 + 1) = 3 \times 3 \times 3 = 27.
The number of positive divisors of p1ap2bpkkp_1^{a} p_2^{b} \dots p_k^{k} is (a+1)(b+1)(k+1)(a+1)(b+1)\dots(k+1).

Anahtar Kavram

Relationship between prime factorization, greatest common divisor (GCD), least common multiple (LCM), and the number of positive divisors.
Soru 170Soru

Consider the expression N=0.00048×(2.5×107)1.2×102N = \frac{0.00048 \times (2.5 \times 10^7)}{1.2 \times 10^{-2}}. Which of the following expressions are equal to NN? Select all such values.

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Cevap: 1.0×1061.0 \times 10^6; (4.0×104)×25(4.0 \times 10^4) \times 25; 5.0×1035.0×109\frac{5.0 \times 10^{-3}}{5.0 \times 10^{-9}}

Cevap

The expressions equal to NN are 1.0×1061.0 \times 10^6, (4.0×104)×25(4.0 \times 10^4) \times 25, and 5.0×1035.0×109\frac{5.0 \times 10^{-3}}{5.0 \times 10^{-9}}.
Evaluating NN simplifies to 1.0×1061.0 \times 10^6. The expression stating 1.0×1061.0 \times 10^6 is an exact scientific notation match. The expression (4.0×104)×25(4.0 \times 10^4) \times 25 computes to 40,000×25=1,000,00040,000 \times 25 = 1,000,000. The fraction 5.0×1035.0×109\frac{5.0 \times 10^{-3}}{5.0 \times 10^{-9}} evaluates to 1.0×103(9)=1.0×1061.0 \times 10^{-3 - (-9)} = 1.0 \times 10^6.

Adım Adım Çözüm

1
Express all numbers in the stem's numerator in standard scientific notation
0.00048=4.8×1040.00048 = 4.8 \times 10^{-4} and the numerator becomes (4.8×104)×(2.5×107)(4.8 \times 10^{-4}) \times (2.5 \times 10^7)
Standardizing numbers into scientific notation simplifies operations involving powers of 10
2
Multiply the coefficients and combine the powers of 10 in the numerator
(4.8×2.5)×104+7=12×103=1.2×104(4.8 \times 2.5) \times 10^{-4 + 7} = 12 \times 10^3 = 1.2 \times 10^4
Coefficients are multiplied directly while exponents are added according to index laws
3
Divide the simplified numerator by the denominator
1.2×1041.2×102=(1.21.2)×104(2)=1.0×106=1,000,000\frac{1.2 \times 10^4}{1.2 \times 10^{-2}} = \left(\frac{1.2}{1.2}\right) \times 10^{4 - (-2)} = 1.0 \times 10^6 = 1,000,000
Subtracting a negative exponent in the denominator is equivalent to adding its positive value
4
Evaluate each given option to determine equivalence to 1.0×1061.0 \times 10^6
The expressions representing 1.0×1061.0 \times 10^6 are 1.0×1061.0 \times 10^6, (4.0×104)×25(4.0 \times 10^4) \times 25, and 5.0×1035.0×109\frac{5.0 \times 10^{-3}}{5.0 \times 10^{-9}}
Direct comparison verifies which options match the calculated value of NN

Anahtar Kavram

Operations with Decimals and Scientific Notation
Soru 171Soru

If xx is a real number such that x4=16x^4 = 16, which of the following values could be equal to x3x^3? Select all that apply.

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Cevap: 8-8; 88

Cevap

The values 8-8 and 88 are the possible values of x3x^3.
Solving x4=16x^4 = 16 gives two real solutions, x=2x = 2 and x=2x = -2, because raising any real number to an even power yields a non-negative result. Cubing each solution gives 23=82^3 = 8 and (2)3=8(-2)^3 = -8. Thus, both 8-8 and 88 are correct values for x3x^3.

Adım Adım Çözüm

1
Find all real solutions for xx in the equation x4=16x^4 = 16.
x=2x = 2 or x=2x = -2.
Taking the fourth root of both sides gives x=164=2|x| = \sqrt[4]{16} = 2, so xx can be positive or negative.
2
Calculate x3x^3 for the positive root x=2x = 2.
23=82^3 = 8.
Cubing a positive number yields a positive result.
3
Calculate x3x^3 for the negative root x=2x = -2.
(2)3=8(-2)^3 = -8.
Cubing a negative number yields a negative result because an odd exponent preserves the sign.

Anahtar Kavram

Even and odd power rules for positive and negative real bases
Soru 172Soru

If N=24×33×52N = 2^4 \times 3^3 \times 5^2, how many positive integer factors of NN are divisible by 66 but not divisible by 1212?

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Cevap: 9

Cevap

The number of positive integer factors of NN divisible by 6 but not by 12 is 9.
Any factor of N=24×33×52N = 2^4 \times 3^3 \times 5^2 is of the form 2a×3b×5c2^a \times 3^b \times 5^c. For the factor to be divisible by 66, we must have a1a \ge 1 and b1b \ge 1. For it to not be divisible by 12=22×3112 = 2^2 \times 3^1, we must have a<2a < 2. Therefore, aa must equal 11. The possible choices for aa are 11 value (11), for bb are 33 values (1,2,31, 2, 3), and for cc are 33 values (0,1,20, 1, 2). Multiplying these options gives 1×3×3=91 \times 3 \times 3 = 9.

Adım Adım Çözüm

1
Express the prime factorization structure of a factor of NN.
Any factor of NN takes the form 2a×3b×5c2^a \times 3^b \times 5^c with bounds 0a40 \le a \le 4, 0b30 \le b \le 3, and 0c20 \le c \le 2.
Divisors of a number are formed by taking prime factors with exponents between zero and their maximum powers in the original number.
2
Determine exponent constraints for divisibility by 6 and non-divisibility by 12.
Divisibility by 66 requires a1a \ge 1 and b1b \ge 1. Non-divisibility by 1212 requires a<2a < 2. Thus, a=1a = 1 exactly.
A factor must contain at least one factor of 2 and one factor of 3 to be a multiple of 6, but containing two or more factors of 2 makes it a multiple of 12.
3
Count combinations of choices for the exponents.
Exponent aa has 11 option (a=1a = 1), bb has 33 options (b{1,2,3}b \in \{1, 2, 3\}), and cc has 33 options (c{0,1,2}c \in \{0, 1, 2\}). Total =1×3×3=9= 1 \times 3 \times 3 = 9.
Applying the fundamental counting principle by multiplying the number of choices for independent prime factor exponents.

Anahtar Kavram

Counting Divisors with Prime Factorization and Divisibility Constraints
Soru 173Soru

At the beginning of Year 1, an investor divided a sum of money between two portfolio accounts, Account X and Account Y, such that the initial balance of Account X was 25%25\% greater than the initial balance of Account Y.

Over a two-year period:
- Account X earned compound interest at a constant annual rate of 20%20\% per year.
- Account Y decreased in value by 10%10\% during Year 1, and then increased in value by r%r\% during Year 2.

If the total combined value of both accounts at the end of Year 2 was 32%32\% greater than the total combined initial balance at the beginning of Year 1, what is the value of rr?

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Cevap: 3030

Cevap

The value of rr is 3030.
The correct value is 3030. Account X starts at 125%125\% of Account Y's initial value (125125 vs 100100, totaling 225225). Compounding Account X at 20%20\% per year for 2 years yields 125×1.202=180125 \times 1.20^2 = 180. Account Y drops 10%10\% to 9090 in Year 1. For the overall total to reach 225×1.32=297225 \times 1.32 = 297, Account Y must reach 297180=117297 - 180 = 117 at the end of Year 2. The percentage increase from 9090 to 117117 is 1179090×100%=2790×100%=30%\frac{117 - 90}{90} \times 100\% = \frac{27}{90} \times 100\% = 30\%.

Adım Adım Çözüm

1
Define initial account balances using a suitable variable base
Let initial balance of Account Y be Y0=100Y_0 = 100 units. Initial balance of Account X is X0=1.25×100=125X_0 = 1.25 \times 100 = 125 units. Combined initial balance is T0=125+100=225T_0 = 125 + 100 = 225 units.
Establishing a standard numerical base simplifies tracking multi-account percentage changes.
2
Calculate the value of Account X at the end of Year 2
X2=125×(1+0.20)2=125×1.44=180X_2 = 125 \times (1 + 0.20)^2 = 125 \times 1.44 = 180 units.
Account X compounds annually at 20%20\% for 2 full years.
3
Express the value of Account Y at the end of Year 2 in terms of rr
After Year 1: Y1=100×(10.10)=90Y_1 = 100 \times (1 - 0.10) = 90 units. After Year 2: Y2=90×(1+r100)Y_2 = 90 \times \left(1 + \frac{r}{100}\right) units.
Account Y first loses 10%10\% of its initial value, creating a new base of 9090 units for the Year 2 growth rate.
4
Determine the required total combined final value
T2=225×(1+0.32)=225×1.32=297T_2 = 225 \times (1 + 0.32) = 225 \times 1.32 = 297 units.
The overall portfolio value grew by 32%32\% relative to the initial combined balance of 225225 units.
5
Set up and solve the combined equation for rr
180+90(1+r100)=297    90(1+r100)=117    1+r100=1.30    r=30180 + 90 \left(1 + \frac{r}{100}\right) = 297 \implies 90 \left(1 + \frac{r}{100}\right) = 117 \implies 1 + \frac{r}{100} = 1.30 \implies r = 30.
Equating the sum of individual final account balances to the target portfolio balance yields the exact value of rr.

Anahtar Kavram

Successive percent changes on shifting bases combined with compound interest calculations across multiple accounts.
Soru 174Soru

An electronics retailer evaluated its annual revenue over a four-year period. In Year 2, the revenue increased by 25%25\% compared to Year 1. In Year 3, the revenue decreased by 20%20\% compared to Year 2. In Year 4, the revenue increased by 15%15\% compared to Year 3.

Which of the following statements regarding the retailer's revenue must be true? Select all such statements.

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Cevap: The revenue in Year 3 is equal to the revenue in Year 1.; The overall percentage change in revenue from Year 1 to Year 4 is a 15%15\% increase.; The revenue in Year 2 is 25%25\% greater than the revenue in Year 3.

Cevap

The statements asserting that Year 3 revenue equals Year 1 revenue, that the overall revenue change from Year 1 to Year 4 is a 15% increase, and that Year 2 revenue is 25% greater than Year 3 revenue are all true.
Let the revenue in Year 1 be RR. Revenue in Year 2 is 1.25R1.25R. Revenue in Year 3 is 1.25R×0.80=1.00R1.25R \times 0.80 = 1.00R, which equals the revenue in Year 1. Revenue in Year 4 is 1.00R×1.15=1.15R1.00R \times 1.15 = 1.15R, representing a net 15%15\% increase from Year 1. Furthermore, Year 2 revenue (1.25R1.25R) compared to Year 3 revenue (1.00R1.00R) is calculated as 1.25R1.00R1.00R=0.25\frac{1.25R - 1.00R}{1.00R} = 0.25, or a 25%25\% increase. Therefore, the first three statements are mathematically true.

Adım Adım Çözüm

1
Define variables and calculate Year 2 revenue relative to Year 1
Let Year 1 revenue be RR. Year 2 revenue = R×(1+0.25)=1.25RR \times (1 + 0.25) = 1.25R.
A 25% increase corresponds to a multiplier of 1.25.
2
Calculate Year 3 revenue relative to Year 2 and Year 1
Year 3 revenue = 1.25R×(10.20)=1.25R×0.80=1.00R1.25R \times (1 - 0.20) = 1.25R \times 0.80 = 1.00R.
A 20% decrease applies to the new base value of 1.25R1.25R.
3
Calculate Year 4 revenue relative to Year 3 and Year 1
Year 4 revenue = 1.00R×(1+0.15)=1.15R1.00R \times (1 + 0.15) = 1.15R.
A 15% increase applies to Year 3 revenue, which equals RR.
4
Evaluate each statement against the calculated values
Year 3 revenue (1.00R1.00R) equals Year 1 revenue (RR). Year 4 revenue (1.15R1.15R) is 15% greater than Year 1 revenue (RR). Year 2 revenue (1.25R1.25R) is 25% greater than Year 3 revenue (1.00R1.00R).
Determines which statements hold true mathematically.

Anahtar Kavram

Successive Percent Changes and Base Value Shifting
Soru 175Soru

Let rr and ss be rational numbers such that 0<r<120 < r < \frac{1}{2} and 1<s<21 < s < 2. Which of the following inequalities MUST be true? Select all that apply.

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Cevap: rs<1r \cdot s < 1; \frac{s}{r} > 2

Cevap

The correct inequalities that must be true are rs<1r \cdot s < 1 and sr>2\frac{s}{r} > 2.
The inequality stating that the product of rr and ss is less than 11 is guaranteed because multiplying the maximum possible bounds (12\frac{1}{2} and 22) gives 11. The inequality stating that the ratio of ss to rr is greater than 22 is guaranteed because the numerator is strictly greater than 11 and the denominator is strictly less than 12\frac{1}{2}.

Adım Adım Çözüm

1
Analyze rs<1r \cdot s < 1
Since r<12r < \frac{1}{2} and s<2s < 2, and both are positive rational numbers, multiplying their upper limits gives rs<(12)(2)=1r \cdot s < \left(\frac{1}{2}\right)(2) = 1. This statement MUST be true.
Properties of inequality multiplication for positive rational numbers.
2
Analyze sr>2\frac{s}{r} > 2
Since s>1s > 1 and r<12r < \frac{1}{2}, taking the ratio gives sr>11/2=2\frac{s}{r} > \frac{1}{1/2} = 2. This statement MUST be true.
Dividing a larger positive number by a fraction less than 1/21/2 amplifies the quotient.
3
Test counterexamples for remaining options
For sr>1s - r > 1, pick s=1.2s = 1.2 and r=0.4sr=0.81r = 0.4 \Rightarrow s - r = 0.8 \le 1 (False). For rs>14\frac{r}{s} > \frac{1}{4}, pick r=0.1r = 0.1 and s=1.6rs=116<14s = 1.6 \Rightarrow \frac{r}{s} = \frac{1}{16} < \frac{1}{4} (False). For r+s>32r + s > \frac{3}{2}, pick r=0.1r = 0.1 and s=1.1r+s=1.2<1.5s = 1.1 \Rightarrow r + s = 1.2 < 1.5 (False).
A single valid counterexample disproves a 'must be true' statement.

Anahtar Kavram

Properties of Rational Numbers and Inequalities
Soru 176Soru

Let nn be an integer such that 15n15-15 \le n \le 15. How many integer values of nn satisfy both of the following conditions?

1. (1)n2+n+1<0(-1)^{n^2 + n + 1} < 0
2. (n3)(n+4)2<0(n - 3)(n + 4)^2 < 0

Cevabı ve açıklamayı göster

Cevap: 17

Cevap

17
Condition 1 is satisfied by every integer because n2+n=n(n+1)n^2 + n = n(n+1) is always even (as the product of two consecutive integers), which makes n2+n+1n^2 + n + 1 always odd and (1)odd=1<0(-1)^{\text{odd}} = -1 < 0. Condition 2 requires (n3)(n+4)2<0(n - 3)(n + 4)^2 < 0. Since (n+4)2>0(n+4)^2 > 0 for all n4n \neq -4, this reduces to n<3n < 3 while excluding n=4n = -4. Within the range 15n15-15 \le n \le 15, there are 18 integers strictly less than 3, and removing n=4n = -4 leaves 17 valid values.

Adım Adım Çözüm

1
Determine the parity of the exponent n2+n+1n^2 + n + 1
n2+n+1n^2 + n + 1 is always odd for any integer nn, making (1)n2+n+1=1<0(-1)^{n^2 + n + 1} = -1 < 0 unconditionally true.
The product of consecutive integers n(n+1)n(n+1) is always even, so adding 1 results in an odd number.
2
Solve the sign inequality (n3)(n+4)2<0(n - 3)(n + 4)^2 < 0
n<3n < 3 with n4n \neq -4.
A squared expression is strictly positive except when its base is zero. At n=4n = -4, the product becomes 0, violating the strict inequality.
3
Count the integer solutions in the range 15n15-15 \le n \le 15
17 integers satisfy both conditions.
There are 18 integers less than 3 in the interval [15,15][-15, 15], and excluding n=4n = -4 gives 181=1718 - 1 = 17.

Anahtar Kavram

Parity of consecutive integer products and sign rules for squared terms in inequalities
Tahmini Süre:2m 0s
Soru 177Soru

In a given fiscal quarter, a retail company increased the unit price of its primary product by p%p\% while the total quantity of the product sold decreased by p%p\%, where p>0p > 0. As a result of these two concurrent changes, the total revenue generated from the product experienced a net decrease of exactly 4%4\%. Which of the following statements must be true? Select all that apply.

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Cevap: The value of pp is 2020.; The final revenue generated from the product is equal to 96%96\% of the initial revenue.; If the unit price had instead decreased by p%p\% and the quantity sold had increased by p%p\%, total revenue would still have decreased by 4%4\%.

Cevap

The correct statements are that the value of p is 20, the final revenue is 96% of the initial revenue, and reversing the direction of the percentage changes would still result in a 4% decrease.
The correct options accurately identify that p=20p = 20, that the final revenue is 96%96\% of the initial value, and that reversing the roles of increase and decrease produces the exact same net multiplier of 0.960.96. Using the difference of squares identity (1+x)(1x)=1x2(1 + x)(1 - x) = 1 - x^2, the combined multiplier for price and quantity shifts is 1(p/100)2=0.961 - (p/100)^2 = 0.96. Solving gives p2=400p^2 = 400, so p=20p = 20. Additionally, (10.20)(1+0.20)=0.96(1 - 0.20)(1 + 0.20) = 0.96, showing symmetry in net outcome.

Adım Adım Çözüm

1
Express the final revenue as a function of initial revenue, price change, and quantity change.
Final Revenue = Initial Revenue ×(1+p100)×(1p100)\times (1 + \frac{p}{100}) \times (1 - \frac{p}{100})
Total revenue is Price ×\times Quantity, so independent percentage changes multiply.
2
Apply the difference of squares formula and set equal to the given net revenue decrease.
(1+p100)(1p100)=1p210000=10.04=0.96(1 + \frac{p}{100})(1 - \frac{p}{100}) = 1 - \frac{p^2}{10000} = 1 - 0.04 = 0.96
A 4%4\% decrease leaves 96%96\% or 0.960.96 of the original value.
3
Solve the algebraic equation for pp.
p210000=0.04    p2=400    p=20\frac{p^2}{10000} = 0.04 \implies p^2 = 400 \implies p = 20
Isolating pp demonstrates that p=20p = 20.
4
Evaluate alternative scenario with price decrease and quantity increase.
(120100)(1+20100)=(0.80)(1.20)=0.96(1 - \frac{20}{100})(1 + \frac{20}{100}) = (0.80)(1.20) = 0.96
Multiplication is commutative, yielding the exact same 4%4\% net decrease.

Anahtar Kavram

Successive Percent Changes and Base Shift Multipliers
Soru 178Soru

A laboratory technician prepares a cleaning solution by mixing a concentrated formula and distilled water in a ratio of 2:72 : 7 by volume. If the technician uses 1414 liters of the concentrated formula, how many liters of distilled water are required?

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Cevap: 4949

Cevap

4949 liters of distilled water
The ratio of concentrated formula to distilled water is given as 2:72 : 7. Since 1414 liters of concentrated formula represents 22 ratio parts (14=2×714 = 2 \times 7), each ratio part equals 77 liters. Multiplying the 77 ratio parts of distilled water by 77 liters gives 7×7=497 \times 7 = 49 liters.

Adım Adım Çözüm

1
Set up the proportion relating concentrated formula to distilled water
Concentrated FormulaDistilled Water=27=14x\frac{\text{Concentrated Formula}}{\text{Distilled Water}} = \frac{2}{7} = \frac{14}{x}
The given ratio is part-to-part (22 parts formula for every 77 parts water).
2
Solve for the unknown volume xx using cross-multiplication
2x=714    2x=982 \cdot x = 7 \cdot 14 \implies 2x = 98
Cross-multiplying equates the products of the diagonal terms in a proportion.
3
Divide by 22 to isolate xx
x=49x = 49
Isolating xx provides the required amount of distilled water in liters.

Anahtar Kavram

Direct Part-to-Part Proportions
Soru 179Soru

An investment fund managed two separate accounts, Account X and Account Y, over a two-year period.

- At the start of Year 1, Account X held $100,000\$100,000 and Account Y held $150,000\$150,000.
- During Year 1, Account X increased in value by 20%20\%, while Account Y decreased in value by x%x\%.
- During Year 2, Account X decreased in value by x%x\%, while Account Y increased in value by 20%20\%.
- At the end of Year 2, the combined total value of both accounts was $243,000\$243,000.

Which of the following statements must be true? Select all such statements.

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Cevabı ve açıklamayı göster

Cevap: The value of xx is equal to 1919.; At the end of Year 2, Account X experienced a net decrease of 2.8%2.8\% relative to its initial value.; At the end of Year 1, Account Y had a greater dollar value than Account X.

Cevap

The correct statements are those asserting that x=19x = 19, that Account X experienced a net decrease of 2.8%2.8\% relative to its initial value, and that at the end of Year 1 Account Y had a greater dollar value than Account X.
The statement specifying that x=19x = 19 is correct because solving the combined account equation $300,000(1x/100)=$243,000\$300,000(1 - x/100) = \$243,000 yields 1x/100=0.811 - x/100 = 0.81, giving x=19x = 19. The statement regarding Account X experiencing a net decrease of 2.8%2.8\% is correct because 1.20×0.81=0.9721.20 \times 0.81 = 0.972, which corresponds to a 2.8%2.8\% decrease from initial value. The statement asserting that Account Y had a greater dollar value at the end of Year 1 than Account X is correct because Account Y was worth $121,500\$121,500 compared to $120,000\$120,000 for Account X.

Adım Adım Çözüm

1
Express the value of each account at the end of Year 1 in terms of xx.
Account X value = $100,000×(1+0.20)=$120,000\$100,000 \times (1 + 0.20) = \$120,000. Account Y value = $150,000×(1x/100)\$150,000 \times (1 - x/100).
Calculate the intermediate values after Year 1 percentage changes.
2
Express the value of each account at the end of Year 2 and sum them to form an equation for the combined total.
Account X end of Year 2 = $120,000(1x/100)\$120,000(1 - x/100). Account Y end of Year 2 = $150,000(1x/100)(1+0.20)=$180,000(1x/100)\$150,000(1 - x/100)(1 + 0.20) = \$180,000(1 - x/100). Combined total = ($120,000+$180,000)(1x/100)=$300,000(1x/100)=$243,000(\$120,000 + \$180,000)(1 - x/100) = \$300,000(1 - x/100) = \$243,000.
Use the given final total value of $243,000\$243,000 to solve for xx.
3
Solve for xx.
1x/100=243,000/300,000=0.81    x/100=0.19    x=191 - x/100 = 243,000 / 300,000 = 0.81 \implies x/100 = 0.19 \implies x = 19.
Find the numerical value of xx.
4
Evaluate each statement against the calculated values.
1. x=19x = 19 is true.
2. Net change multiplier for Account X = 1.20×0.81=0.9721.20 \times 0.81 = 0.972, representing a 2.8%2.8\% decrease (true).
3. End of Year 1 values: Account X = $120,000\$120,000, Account Y = $150,000×0.81=$121,500\$150,000 \times 0.81 = \$121,500. Since $121,500>$120,000\$121,500 > \$120,000, Account Y was greater (true).
4. Account Y net change multiplier = 0.81×1.20=0.9720.81 \times 1.20 = 0.972, identical to Account X, so percentage decreases are equal (false).
5. Overall combined percent change = (243,000250,000)/250,000=7,000/250,000=2.8%(243,000 - 250,000)/250,000 = -7,000/250,000 = -2.8\% (false).
Determine which statements are mathematically true.

Anahtar Kavram

Successive Percentage Changes and Base Shifts in Multi-Asset Portfolios
Soru 180Soru

A positive integer nn yields a remainder of 77 when divided by 1212. What is the remainder when the expression 5n+35n + 3 is divided by 66?

Cevabı ve açıklamayı göster

Cevap: 22

Cevap

The remainder when 5n+35n + 3 is divided by 66 is 22.
Since nn leaves a remainder of 77 when divided by 1212, we can write n=12k+7n = 12k + 7 for some integer k0k \ge 0. Substituting this into 5n+35n + 3 yields 5(12k+7)+3=60k+385(12k + 7) + 3 = 60k + 38. Factoring out 66 gives 6(10k+6)+26(10k + 6) + 2. Because 6(10k+6)6(10k + 6) is divisible by 66, the remainder of the expression when divided by 66 is 22.

Adım Adım Çözüm

1
Express nn in terms of the division algorithm for divisor 1212.
n=12k+7n = 12k + 7 for some non-negative integer kk
An integer that leaves a remainder of 77 when divided by 1212 can be represented as 12k+712k + 7.
2
Substitute the expression for nn into 5n+35n + 3 and simplify.
5(12k+7)+3=60k+35+3=60k+385(12k + 7) + 3 = 60k + 35 + 3 = 60k + 38
Algebraic expansion allows us to analyze the entire expression modulo 66.
3
Determine the remainder of 60k+3860k + 38 when divided by 66.
60k+38=6(10k+6)+260k + 38 = 6(10k + 6) + 2
Since 60k+3660k + 36 is an exact multiple of 66, the leftover term 22 is the remainder.

Anahtar Kavram

Properties of Integers and Remainder Arithmetic
Tahmini Süre:1m 30s
ÖncekiSayfa 9 / 16Sonraki
Arithmetic Alıştırma Soruları — GRE General Test — Sayfa 9 | Examkin