Tüm alıştırma soruları

231 soru

Soru 61Soru

A laboratory vessel contains a liquid solution consisting of water and ethanol, where water makes up 38\frac{3}{8} of the total volume. First, 15\frac{1}{5} of the total volume of the solution is drained and replaced with an equal volume of pure ethanol. Next, 14\frac{1}{4} of the resulting mixture is evaporated, removing water and ethanol in proportion to their presence. Finally, pure water is added to fill the vessel back to its original total volume. What fraction of the final solution is water? Express your answer as a decimal.

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Cevap: 0.475

Cevap

The fraction of the final solution that is water is 0.475 (or 19/40).
By following the multi-step fractional changes to the liquid volume, the remaining water prior to refilling is 940\frac{9}{40} of the original capacity. Refilling the missing 14\frac{1}{4} (or 1040\frac{10}{40}) volume with pure water yields 1940=0.475\frac{19}{40} = 0.475 of the total solution as water.

Adım Adım Çözüm

1
Track water content after the initial replacement.
Water fraction becomes 310\frac{3}{10} of the original volume.
Let the original total volume be VV. Initially, water volume is 38V\frac{3}{8}V. Removing 15\frac{1}{5} of the solution leaves 45\frac{4}{5} of the original solution, so the water volume becomes 45×38V=310V\frac{4}{5} \times \frac{3}{8}V = \frac{3}{10}V. Replacing the removed volume with pure ethanol brings total volume back to VV, with water occupying 310V\frac{3}{10}V.
2
Track water content after evaporation.
Water volume becomes 940V\frac{9}{40}V and total solution volume becomes 34V\frac{3}{4}V.
Evaporating 14\frac{1}{4} of the solution leaves 34\frac{3}{4} of the mixture intact. The remaining water volume is 34×310V=940V\frac{3}{4} \times \frac{3}{10}V = \frac{9}{40}V.
3
Calculate the final water fraction after refilling with pure water.
Final water volume is 1940V=0.475V\frac{19}{40}V = 0.475V.
To restore the total volume from 34V\frac{3}{4}V back to VV, an amount equal to V34V=14VV - \frac{3}{4}V = \frac{1}{4}V of pure water is added. Adding this to the existing water gives 940V+14V=940V+1040V=1940V\frac{9}{40}V + \frac{1}{4}V = \frac{9}{40}V + \frac{10}{40}V = \frac{19}{40}V. Dividing by total volume VV yields 1940=0.475\frac{19}{40} = 0.475.

Anahtar Kavram

Sequential fractional reduction and component tracking
Tahmini Süre:2m 30s
Soru 62Soru

Two automated timers are set to chime at regular intervals. The first timer chimes every 1212 minutes, and the second timer chimes every 1818 minutes. If both timers chime simultaneously at 12:00 PM, how many minutes will pass before they next chime at the exact same time?

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Cevap: 36

Cevap

36 minutes
The time interval between simultaneous chimes is given by the least common multiple (LCM) of the two individual intervals, 1212 minutes and 1818 minutes. Using prime factorization, 12=22×312 = 2^2 \times 3 and 18=2×3218 = 2 \times 3^2. The LCM takes the highest power of each prime present: 22×32=4×9=362^2 \times 3^2 = 4 \times 9 = 36. Thus, 3636 minutes will elapse before the timers chime together again.

Adım Adım Çözüm

1
Find the prime factorizations of 12 and 18
12=22×312 = 2^2 \times 3 and 18=2×3218 = 2 \times 3^2
Expressing numbers as products of prime factors allows us to easily compute the least common multiple.
2
Calculate the Least Common Multiple (LCM)
LCM(12,18)=2max(2,1)×3max(1,2)=22×32=36\text{LCM}(12, 18) = 2^{\max(2,1)} \times 3^{\max(1,2)} = 2^2 \times 3^2 = 36
The next simultaneous occurrence of periodic events is given by the LCM of their periods.

Anahtar Kavram

Least Common Multiple (LCM) of periodic events
Soru 63Soru

A community library received a donation of 120120 new books. On Monday, 38\frac{3}{8} of the donated books were cataloged. On Tuesday, 13\frac{1}{3} of the remaining uncataloged books were cataloged. How many of the donated books remain to be cataloged?

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Cevap: 50

Cevap

50 books remain to be cataloged.
First, find the number of books cataloged on Monday: 3/8 of 120 is 45 books. Subtracting 45 from 120 leaves 75 books uncataloged. On Tuesday, 1/3 of those remaining 75 books were cataloged, which is 25 books. Subtracting 25 from 75 gives 50 books remaining to be cataloged.

Adım Adım Çözüm

1
Calculate the number of books cataloged on Monday.
45 books
Multiply the initial total of 120 books by the fraction cataloged on Monday, which is 3/8.
2
Calculate the uncataloged books remaining after Monday.
75 books
Subtract the 45 cataloged books from the initial 120 books.
3
Calculate the number of books cataloged on Tuesday.
25 books
Multiply the remaining 75 books by the fraction cataloged on Tuesday, which is 1/3.
4
Calculate the final number of uncataloged books.
50 books
Subtract the 25 books cataloged on Tuesday from the 75 uncataloged books remaining after Monday.

Anahtar Kavram

Fraction of a Remaining Quantity
Tahmini Süre:50s
Soru 64Soru

What is the largest two-digit positive integer nn such that when nn is divided by 44, the remainder is 33, and when nn is divided by 55, the remainder is 22?

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Cevap: 87

Cevap

The largest two-digit positive integer satisfying both remainder conditions is 87.
Any integer satisfying both remainder requirements must be of the form n=20m+7n = 20m + 7 for an integer mm. Testing values for mm shows that m=4m = 4 yields n=87n = 87, which is the largest two-digit integer fitting this rule.

Adım Adım Çözüm

1
Set up modular arithmetic equations for the given remainder conditions.
n3(mod4)n \equiv 3 \pmod 4 and n2(mod5)n \equiv 2 \pmod 5
Translate the verbal description of remainders into mathematical congruence relations.
2
Combine the congruence relations to find the general form of nn.
n=20m+7n = 20m + 7 for non-negative integers mm
The least common multiple of 44 and 55 is 2020, meaning solutions repeat every 2020 units.
3
Find the maximum integer mm that produces a two-digit integer.
For m=4m = 4, n=87n = 87. For m=5m = 5, n=107n = 107.
Two-digit integers are strictly less than 100100.

Anahtar Kavram

Simultaneous Remainders and Divisibility Cycles
Tahmini Süre:1m 30s
Soru 65Soru

In a manufacturing facility, three automated machines—Machine A, Machine B, and Machine C—produce components at constant individual rates. Machine A working alone completes 310\frac{3}{10} of a standard daily order in 33 hours. Machine B working alone completes 25\frac{2}{5} of the order in 66 hours. Machine C working alone completes 14\frac{1}{4} of the order in 55 hours.

Initially, all three machines work together for 33 hours. At the end of 33 hours, Machine A malfunctions and stops, while Machines B and C continue working together without interruption until the daily order is finished. How many total hours does it take from the start to complete the entire daily order?

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Cevap: 6

Cevap

The total time required from the start to complete the entire daily order is 6 hours.
Each machine's rate per hour is found by dividing the given fraction by hours: Machine A is 110\frac{1}{10}, B is 115\frac{1}{15}, and C is 120\frac{1}{20}. Summing these gives an initial combined rate of 1360\frac{13}{60} per hour. In the first 3 hours, the three machines complete 3×1360=13203 \times \frac{13}{60} = \frac{13}{20} of the order, leaving 11320=7201 - \frac{13}{20} = \frac{7}{20} of the order remaining. When Machine A stops, the remaining combined rate of B and C is 115+120=760\frac{1}{15} + \frac{1}{20} = \frac{7}{60} per hour. Dividing the remaining 720\frac{7}{20} by 760\frac{7}{60} yields 3 additional hours. Adding the initial 3 hours gives a total duration of 6 hours.

Adım Adım Çözüm

1
Determine individual hourly work rates for each machine.
Machine A rate = 110\frac{1}{10} order/hr, Machine B rate = 115\frac{1}{15} order/hr, Machine C rate = 120\frac{1}{20} order/hr.
Divide the fraction of the job completed by the duration in hours to find the unit rate for each machine.
2
Calculate the combined rate of all three machines and the fraction completed in the first 3 hours.
Combined rate = 1360\frac{13}{60} order/hr; Work completed in 3 hours = 1320\frac{13}{20}, leaving 720\frac{7}{20} of the order unfinished.
Sum the three rates using a common denominator of 60, then multiply by 3 hours to find the total work done during the first stage.
3
Calculate the combined rate of Machines B and C, and find the additional time required to complete the remaining fraction.
Combined rate of B and C = 760\frac{7}{60} order/hr; Additional time needed = 3 hours.
Divide the remaining fraction 720\frac{7}{20} by the combined rate 760\frac{7}{60} to determine the remaining hours needed.
4
Add the initial duration and additional duration to obtain the total time.
Total time = 3+3=63 + 3 = 6 hours.
The question asks for the total elapsed time from the start of the job.

Anahtar Kavram

Adding rational fractions with different denominators to solve multi-stage work and rate problems.
Soru 66Soru

An employee's hourly wage increased from $20.00\$20.00 to $25.00\$25.00. What was the percent increase in the employee's hourly wage?

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Cevap: 25

Cevap

The percent increase in the employee's hourly wage is 25%25\%.
The percent increase is calculated using the formula Percent Increase=New ValueOriginal ValueOriginal Value×100%\text{Percent Increase} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\%. Substituting the given values gives 252020×100%=520×100%=25%\frac{25 - 20}{20} \times 100\% = \frac{5}{20} \times 100\% = 25\%.

Adım Adım Çözüm

1
Calculate the absolute increase in hourly wage
Increase = $25.00$20.00=$5.00\$25.00 - \$20.00 = \$5.00
Percent change is calculated relative to the absolute change from the initial value.
2
Calculate the percent change using the original wage as the base
Percent Increase=$5.00$20.00×100%=25%\text{Percent Increase} = \frac{\$5.00}{\$20.00} \times 100\% = 25\%
The formula for percent change is Amount of ChangeOriginal Amount×100%\frac{\text{Amount of Change}}{\text{Original Amount}} \times 100\%.

Anahtar Kavram

Percent Change Formula
Soru 67Soru

If mm is an even integer and nn is an odd integer, what is the value of (1)m+(1)n(-1)^m + (-1)^n?

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Cevap: 0

Cevap

0
Since mm is an even integer, (1)m=1(-1)^m = 1. Since nn is an odd integer, (1)n=1(-1)^n = -1. Adding 11 and 1-1 yields 00.

Adım Adım Çözüm

1
Evaluate (1)m(-1)^m for an even integer mm
(1)m=1(-1)^m = 1
Raising 1-1 to an even integer power always results in 11.
2
Evaluate (1)n(-1)^n for an odd integer nn
(1)n=1(-1)^n = -1
Raising 1-1 to an odd integer power always results in 1-1.
3
Sum the two evaluated expressions
1+(1)=01 + (-1) = 0
Combining 11 and 1-1 equals 00.

Anahtar Kavram

Exponent sign rules for even and odd powers
Soru 68Soru

In a community library, 415\frac{4}{15} of the total book collection consists of fiction books, 25\frac{2}{5} consists of non-fiction books, and the remaining 240 books are children's books. If 38\frac{3}{8} of the fiction books are hardcovers and the rest are softcovers, how many softcover fiction books are in the library?

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Cevap: 120

Cevap

The total number of softcover fiction books in the library is 120.
To find the number of softcover fiction books, first calculate the total number of books in the library. The combined fraction of fiction and non-fiction books is 415+25=415+615=1015=23\frac{4}{15} + \frac{2}{5} = \frac{4}{15} + \frac{6}{15} = \frac{10}{15} = \frac{2}{3}. Therefore, children's books represent the remaining 123=131 - \frac{2}{3} = \frac{1}{3} of the collection. Since 13\frac{1}{3} of the total equals 240 books, the total collection is 240×3=720240 \times 3 = 720 books. Next, compute the total number of fiction books: 415×720=192\frac{4}{15} \times 720 = 192. Finally, since 38\frac{3}{8} of fiction books are hardcovers, the remaining fraction of softcovers is 138=581 - \frac{3}{8} = \frac{5}{8}. Multiplying 58×192\frac{5}{8} \times 192 gives 120 softcover fiction books.

Adım Adım Çözüm

1
Find the combined fraction of fiction and non-fiction books, then determine the fraction of children's books.
Combined fraction = 415+615=1015=23\frac{4}{15} + \frac{6}{15} = \frac{10}{15} = \frac{2}{3}. Children's books fraction = 123=131 - \frac{2}{3} = \frac{1}{3}.
The sum of all non-overlapping fractions comprising the entire collection must equal 1.
2
Calculate the total number of books in the library.
Total collection = 240÷13=720240 \div \frac{1}{3} = 720 books.
The 240 children's books represent exactly 13\frac{1}{3} of the whole collection.
3
Determine the total count of fiction books.
Fiction books = 415×720=192\frac{4}{15} \times 720 = 192 books.
Fiction books make up 415\frac{4}{15} of the total 720 books.
4
Calculate the number of softcover fiction books.
Softcover fiction books = (138)×192=58×192=120\left(1 - \frac{3}{8}\right) \times 192 = \frac{5}{8} \times 192 = 120 books.
If 38\frac{3}{8} of the fiction subset are hardcovers, the remaining 58\frac{5}{8} of that subset are softcovers.

Anahtar Kavram

Solving multi-step rational number word problems by combining fractions and determining fractional parts of a subset.
Tahmini Süre:1m 30s
Soru 69Soru

Two positive integers aa and bb have prime factorizations of the form a=2x×3y×5za = 2^x \times 3^y \times 5^z and b=2u×3v×5wb = 2^u \times 3^v \times 5^w, where x,y,z,u,v,wx, y, z, u, v, w are non-negative integers. The greatest common divisor (GCD) of aa and bb is 180180, and the least common multiple (LCM) of aa and bb is 54,00054,000. If aa has exactly 2424 positive divisors, what is the value of bb?

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Cevap: 18000

Cevap

18000
Prime factorizing GCD(a,b)=180=22×32×51\text{GCD}(a,b) = 180 = 2^2 \times 3^2 \times 5^1 and LCM(a,b)=54,000=24×33×53\text{LCM}(a,b) = 54,000 = 2^4 \times 3^3 \times 5^3 reveals that the pair of exponents for bases 2,3,52, 3, 5 are {2,4}\{2, 4\}, {2,3}\{2, 3\}, and {1,3}\{1, 3\} respectively. Applying the total divisor formula (x+1)(y+1)(z+1)=24(x+1)(y+1)(z+1) = 24 for aa, x+1x+1 must divide 2424. Since x{2,4}x \in \{2, 4\}, xx must be 22 (since 55 does not divide 2424). This leaves (y+1)(z+1)=8(y+1)(z+1) = 8. With y{2,3}y \in \{2, 3\}, y+1y+1 must be 44 (since 33 does not divide 88), giving y=3y = 3 and z=1z = 1. Thus, a=22×33×51a = 2^2 \times 3^3 \times 5^1. The remaining exponent choices for bb are u=4u = 4, v=2v = 2, and w=3w = 3, giving b=24×32×53=18,000b = 2^4 \times 3^2 \times 5^3 = 18,000.

Adım Adım Çözüm

1
Factorize GCD and LCM into prime bases.
GCD = 2^2 * 3^2 * 5^1; LCM = 2^4 * 3^3 * 5^3.
GCD represents the minimum prime exponents and LCM represents the maximum prime exponents of a and b.
2
Set up candidate exponent pairs for primes 2, 3, and 5.
{x, u} = {2, 4}, {y, v} = {2, 3}, and {z, w} = {1, 3}.
For any prime p, min(exp_p(a), exp_p(b)) = exp_p(GCD) and max(exp_p(a), exp_p(b)) = exp_p(LCM).
3
Solve for x, y, and z using the divisor count formula d(a) = (x+1)(y+1)(z+1) = 24.
x = 2, y = 3, z = 1.
If x = 4, then x+1 = 5, which is not a factor of 24. Thus x = 2. Then (y+1)(z+1) = 8. Since y in {2, 3}, y+1 = 4 implies y = 3 and z = 1.
4
Determine the exponents of b and calculate b.
u = 4, v = 2, w = 3, so b = 2^4 * 3^2 * 5^3 = 18,000.
Complementary values from the sets {x,u}, {y,v}, and {z,w} give the exponents for b.

Anahtar Kavram

Relationship between prime factor exponents in GCD, LCM, and divisor counting functions.
Soru 70Soru

Let NN be a positive integer with exactly 1212 positive divisors. If the sum of the distinct prime factors of NN is 1212 and NN is not divisible by 44, what is the least possible value of NN?

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Cevap: 126

Cevap

126
The least possible value of NN is 126. The only sets of distinct prime factors summing to 12 are {5,7}\{5, 7\} and {2,3,7}\{2, 3, 7\}. Using {2,3,7}\{2, 3, 7\} produces smaller candidates. Because NN is not divisible by 4, the exponent of 2 must be 1. The total divisor condition (1+1)(b+1)(c+1)=12(1+1)(b+1)(c+1) = 12 requires (b+1)(c+1)=6(b+1)(c+1) = 6, giving exponent pairs (1,2)(1, 2) or (2,1)(2, 1) for bases 3 and 7. Assigning the exponent 2 to 3 and 1 to 7 minimizes NN, giving 213271=1262^1 \cdot 3^2 \cdot 7^1 = 126.

Adım Adım Çözüm

1
Find all sets of distinct prime factors that sum to 12.
The possible sets of distinct prime factors are {5,7}\{5, 7\} and {2,3,7}\{2, 3, 7\}.
Testing combinations of prime numbers (2,3,5,7,11,2, 3, 5, 7, 11, \dots): 5+7=125 + 7 = 12 and 2+3+7=122 + 3 + 7 = 12 are the only valid sets of distinct primes summing to 12.
2
Analyze the set of prime factors {2,3,7}\{2, 3, 7\} under the condition that NN is not divisible by 4.
The prime factorization is N=2a3b7cN = 2^a \cdot 3^b \cdot 7^c, where a=1a = 1.
Since 2 is a prime factor of NN, a1a \ge 1. Because NN is not divisible by 4 (222^2), aa must be strictly less than 2. Thus, a=1a = 1.
3
Determine the exponents bb and cc using the total number of positive divisors.
The number of divisors is (1+1)(b+1)(c+1)=12(1+1)(b+1)(c+1) = 12, which simplifies to (b+1)(c+1)=6(b+1)(c+1) = 6. The possible integer pairs (b,c)(b, c) for b,c1b, c \ge 1 are (1,2)(1, 2) and (2,1)(2, 1).
The divisor count formula for N=p1e1p2e2pkekN = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k} is (e1+1)(e2+1)(ek+1)=12(e_1+1)(e_2+1)\dots(e_k+1) = 12.
4
Calculate values of NN for these exponent pairs.
For (b,c)=(1,2)(b, c) = (1, 2), N=213172=294N = 2^1 \cdot 3^1 \cdot 7^2 = 294. For (b,c)=(2,1)(b, c) = (2, 1), N=213271=126N = 2^1 \cdot 3^2 \cdot 7^1 = 126.
To make NN as small as possible, assign the larger exponent to the smaller prime base (3271<31723^2 \cdot 7^1 < 3^1 \cdot 7^2).
5
Compare with candidate values from the alternative set of prime factors {5,7}\{5, 7\}.
For {5,7}\{5, 7\}, N=5372=6125N = 5^3 \cdot 7^2 = 6125 or 5571=218755^5 \cdot 7^1 = 21875, both of which are much larger than 126.
The combination (a+1)(b+1)=12(a+1)(b+1) = 12 yields exponents of 3 and 2 (or 5 and 1), resulting in much higher prime powers.

Anahtar Kavram

Divisor count formula combined with prime factorization constraints
Soru 71Soru

If x4=9|x - 4| = 9 and x<0x < 0, what is the value of xx?

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Cevap: -5

Cevap

The value of xx is 5-5.
The absolute value equation x4=9|x - 4| = 9 specifies that the distance between xx and 44 on the real number line is equal to 99. Moving 99 units to the left of 44 gives 49=54 - 9 = -5, and moving 99 units to the right gives 4+9=134 + 9 = 13. Because xx is specified to be negative (x<0x < 0), the correct value of xx is 5-5.

Adım Adım Çözüm

1
Set up the two linear equations representing the absolute value equation x4=9|x - 4| = 9.
x4=9x - 4 = 9 or x4=9x - 4 = -9
By definition, A=B|A| = B (where B0B \ge 0) implies A=BA = B or A=BA = -B.
2
Solve for xx in both equations.
x=13x = 13 or x=5x = -5
Adding 44 to both sides of x4=9x - 4 = 9 gives x=13x = 13, and adding 44 to both sides of x4=9x - 4 = -9 gives x=5x = -5.
3
Select the value of xx that satisfies the given condition x<0x < 0.
x=5x = -5
The value 1313 is positive, while 5-5 is negative and fulfills x<0x < 0.

Anahtar Kavram

Absolute Value as Distance and Solving Absolute Value Equations
Tahmini Süre:45s
Soru 72Soru

At the beginning of the year, a cloud storage service had a certain number of active paid subscriptions. During the first six months of the year, the number of active subscriptions increased by 20%20\%. During the second six months of the year, the number of active subscriptions decreased by 15%15\% relative to the total at the middle of the year. If the service had 12,24012,240 active subscriptions at the end of the year, what was the number of active subscriptions at the beginning of the year?

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Cevap: 12000

Cevap

12000
To find the initial subscription count, model the successive percentage changes sequentially. Increasing the initial amount NN by 20%20\% yields 1.20N1.20N. Decreasing that intermediate total by 15%15\% multiplies 1.20N1.20N by 0.850.85, giving 1.02N1.02N. Setting 1.02N=12,2401.02N = 12,240 gives N=12,000N = 12,000.

Adım Adım Çözüm

1
Express the mid-year subscription total in terms of the initial subscription count NN.
Mid-year subscriptions =N×(1+0.20)=1.20N= N \times (1 + 0.20) = 1.20N
A 20%20\% increase scales the original quantity by a factor of 1.201.20.
2
Express the end-of-year subscription total in terms of NN after a 15%15\% decrease from the mid-year total.
End-of-year subscriptions =1.20N×(10.15)=1.20N×0.85=1.02N= 1.20N \times (1 - 0.15) = 1.20N \times 0.85 = 1.02N
A 15%15\% decrease reduces the new base quantity to 85%85\% of its mid-year value.
3
Set the algebraic expression for the final total equal to the given numerical total and solve for NN.
1.02N=12,240    N=12,0001.02N = 12,240 \implies N = 12,000
Dividing the final amount by the net multiplier 1.021.02 yields the original starting value.

Anahtar Kavram

Successive Percentage Changes and Base Shift
Tahmini Süre:1m 30s
Soru 73Soru

If 2x×43=292^x \times 4^3 = 2^9, what is the value of xx?

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Cevap: 3

Cevap

The value of xx is 3.
By rewriting 434^3 as (22)3=26(2^2)^3 = 2^6, the equation becomes 2x26=292^x \cdot 2^6 = 2^9. Applying the product rule gives 2x+6=292^{x+6} = 2^9, which simplifies to x+6=9x + 6 = 9, yielding x=3x = 3.

Adım Adım Çözüm

1
Convert all terms to base 2
43=(22)3=22×3=264^3 = (2^2)^3 = 2^{2 \times 3} = 2^6
To apply exponent rules, expressions should share a common base.
2
Apply product rule of exponents to the left side of the equation
2x×26=2x+62^x \times 2^6 = 2^{x+6}
When multiplying exponential terms with the same base, add their exponents: aman=am+na^m \cdot a^n = a^{m+n}.
3
Equate the exponents from both sides
x+6=9x + 6 = 9
Since the bases are identical and non-zero, the exponents must be equal.
4
Solve the linear equation for xx
x=3x = 3
Subtract 6 from both sides of the equation.

Anahtar Kavram

Combining exponential terms with equal bases using power rules (aman=am+na^m \cdot a^n = a^{m+n} and (am)n=amn(a^m)^n = a^{mn}).
Soru 74Soru

A high-precision instrument measures the mass of a single micro-particle PP as 4.8×1084.8 \times 10^{-8} grams and a single micro-particle QQ as 8.0×1098.0 \times 10^{-9} grams. A sample consists of a combination of PP and QQ particles in a ratio of 33 particles of PP for every 55 particles of QQ. If the total mass of the sample is 9.2×1059.2 \times 10^{-5} grams, what is the total number of particles in the sample?

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Cevap: 4000

Cevap

The total number of particles in the sample is 4,000.
Aligning powers of 10 shows that one P particle weighs 4.8×1084.8 \times 10^{-8} g and one Q particle weighs 0.8×1080.8 \times 10^{-8} g. A combined unit of 3 P particles and 5 Q particles has a mass of 3(4.8×108)+5(0.8×108)=18.4×108=1.84×1073(4.8 \times 10^{-8}) + 5(0.8 \times 10^{-8}) = 18.4 \times 10^{-8} = 1.84 \times 10^{-7} g. Dividing the total sample mass of 9.2×1059.2 \times 10^{-5} g by 1.84×1071.84 \times 10^{-7} g yields 500 units. Since each unit contains 8 particles (3 + 5), the total number of particles is 500×8=4,000500 \times 8 = 4,000.

Adım Adım Çözüm

1
Convert the mass of particle Q so that it shares the same exponent (10810^{-8}) as particle P.
Mass of single particle Q = 0.8×1080.8 \times 10^{-8} grams.
Aligning powers of 10 is necessary before performing addition of masses.
2
Find the combined mass of a fundamental ratio group consisting of 3 particles of P and 5 particles of Q.
Mass of one ratio group = 3(4.8×108)+5(0.8×108)=14.4×108+4.0×108=18.4×108=1.84×1073(4.8 \times 10^{-8}) + 5(0.8 \times 10^{-8}) = 14.4 \times 10^{-8} + 4.0 \times 10^{-8} = 18.4 \times 10^{-8} = 1.84 \times 10^{-7} grams.
Determines the mass contributed by each set of 8 particles.
3
Divide the total mass of the sample by the mass of a single ratio group.
Number of groups = 9.2×1051.84×107=9.21.84×102=5×102=500\frac{9.2 \times 10^{-5}}{1.84 \times 10^{-7}} = \frac{9.2}{1.84} \times 10^2 = 5 \times 10^2 = 500 groups.
Determines how many full ratio sets of particles make up the sample.
4
Multiply the number of groups by the total number of particles contained in each group (3+5=83 + 5 = 8).
Total particles = 500×8=4000500 \times 8 = 4000.
Yields the total count of individual particles in the sample.

Anahtar Kavram

Operations with scientific notation, decimal place value alignment, and weighted proportional sums.
Soru 75Soru

If xx is a real number that satisfies the equation x+6x9+x6x9=10\sqrt{x + 6\sqrt{x - 9}} + \sqrt{x - 6\sqrt{x - 9}} = 10, what is the value of xx?

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Cevap: 34

Cevap

34
Using the substitution u=x90u = \sqrt{x - 9} \ge 0, we have x=u2+9x = u^2 + 9. The expressions under the square roots become x+6x9=u2+6u+9=(u+3)2x + 6\sqrt{x - 9} = u^2 + 6u + 9 = (u + 3)^2 and x6x9=u26u+9=(u3)2x - 6\sqrt{x - 9} = u^2 - 6u + 9 = (u - 3)^2. Taking square roots gives (u+3)2+(u3)2=(u+3)+u3=10\sqrt{(u + 3)^2} + \sqrt{(u - 3)^2} = (u + 3) + |u - 3| = 10. For u3u \ge 3, this simplifies to (u+3)+(u3)=10    2u=10    u=5(u + 3) + (u - 3) = 10 \implies 2u = 10 \implies u = 5. Finally, substituting u=5u = 5 back yields x=52+9=34x = 5^2 + 9 = 34.

Adım Adım Çözüm

1
Define a variable substitution to simplify the nested radical structure.
Let u=x9u = \sqrt{x - 9} where u0u \ge 0. Squaring both sides gives u2=x9u^2 = x - 9, so x=u2+9x = u^2 + 9.
This substitution allows the expressions inside the outer square roots to be rewritten as polynomials in terms of uu.
2
Rewrite the expressions under each square root as perfect square trinomials.
x+6x9=(u2+9)+6u=(u+3)2x + 6\sqrt{x - 9} = (u^2 + 9) + 6u = (u + 3)^2 and x6x9=(u2+9)6u=(u3)2x - 6\sqrt{x - 9} = (u^2 + 9) - 6u = (u - 3)^2.
Expressing terms as perfect squares allows the outer radicals to be simplified.
3
Simplify the square root expressions using absolute values.
(u+3)2+(u3)2=(u+3)+u3=10\sqrt{(u + 3)^2} + \sqrt{(u - 3)^2} = (u + 3) + |u - 3| = 10.
For any real number aa, a2=a\sqrt{a^2} = |a|. Since u0u \ge 0, u+3>0u + 3 > 0, so u+3=u+3|u + 3| = u + 3.
4
Solve the absolute value equation across valid domain intervals.
If u3u \ge 3, u3=u3|u - 3| = u - 3, giving (u+3)+(u3)=10    2u=10    u=5(u + 3) + (u - 3) = 10 \implies 2u = 10 \implies u = 5. If 0u<30 \le u < 3, u3=3u|u - 3| = 3 - u, giving (u+3)+(3u)=610(u + 3) + (3 - u) = 6 \neq 10 (no solution). Thus, u=5u = 5.
Splitting into cases based on the definition of absolute value isolates the valid root.
5
Substitute u=5u = 5 back into the expression for xx.
x=52+9=25+9=34x = 5^2 + 9 = 25 + 9 = 34.
Converting from uu back to xx provides the solution to the original equation.

Anahtar Kavram

Simplifying nested radicals by completing the square under the radical sign and applying the identity a2=a\sqrt{a^2} = |a|.
Soru 76Soru

What is the least integer value of xx that satisfies the inequality 4x511|4x - 5| \le 11?

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Cevap: -1

Cevap

The least integer value of xx that satisfies the inequality is 1-1.
Expanding the absolute value inequality 4x511|4x - 5| \le 11 produces 114x511-11 \le 4x - 5 \le 11. Adding 5 across all sections yields 64x16-6 \le 4x \le 16, and dividing by 4 results in 1.5x4-1.5 \le x \le 4. The set of integers within this range is {1,0,1,2,3,4}\{-1, 0, 1, 2, 3, 4\}, making 1-1 the least integer value.

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1
Rewrite the absolute value inequality as a double inequality.
114x511-11 \le 4x - 5 \le 11
An absolute value inequality of the form ua|u| \le a for a0a \ge 0 is equivalent to aua-a \le u \le a.
2
Add 5 to all parts of the compound inequality.
64x16-6 \le 4x \le 16
To isolate xx, first add 5 to eliminate the constant term 5-5.
3
Divide all parts by 4.
1.5x4-1.5 \le x \le 4
Dividing by a positive constant preserves the direction of the inequality signs.
4
Select the minimum integer contained within the solution interval [1.5,4][-1.5, 4].
1-1
The integer values satisfying 1.5x4-1.5 \le x \le 4 are 1,0,1,2,3,4-1, 0, 1, 2, 3, 4. The smallest among these is 1-1.

Anahtar Kavram

Linear inequalities involving absolute value
Soru 77Soru

A coffee shop owner creates a 3030-pound blend of coffee by mixing Bean X, which costs $8\$8 per pound, with Bean Y, which costs $14\$14 per pound. If the final mixture costs $10\$10 per pound, how many pounds of Bean X are in the mixture?

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Cevap: 20

Cevap

The mixture contains 20 pounds of Bean X.
The quantity of Bean X required is 20 pounds. Setting up the cost equation 8x+14(30x)=3008x + 14(30 - x) = 300 simplifies to 6x=120-6x = -120, giving x=20x = 20.

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1
Define the unknown variables representing the quantities of each component.
Let xx be the weight of Bean X in pounds. The weight of Bean Y is (30x)(30 - x) pounds.
Since the total weight of the mixture is 30 pounds, expressing Bean Y in terms of xx creates a single-variable system.
2
Set up a linear equation based on total financial value.
8x+14(30x)=10×308x + 14(30 - x) = 10 \times 30
The total cost of Bean X plus the total cost of Bean Y equals the total cost of the combined 30-pound mixture.
3
Solve the algebraic equation for xx.
8x+42014x=300    6x=120    x=208x + 420 - 14x = 300 \implies -6x = -120 \implies x = 20
Combining like terms isolates the variable xx to yield the required quantity of Bean X.

Anahtar Kavram

Linear mixture modeling and single-variable algebraic modeling
Soru 78Soru

Circle C1C_1 has radius r1r_1 and Circle C2C_2 has radius r2r_2. An arc on Circle C1C_1 subtended by a central angle of θ1\theta_1^\circ has the exact same length as an arc on Circle C2C_2 subtended by a central angle of θ2\theta_2^\circ. The sector formed by this arc in Circle C1C_1 has an area of 54π54\pi, and the sector formed by this arc in Circle C2C_2 has an area of 36π36\pi. If θ1+θ2=150\theta_1 + \theta_2 = 150, what is the value of r1r_1?

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Cevap: 18

Cevap

The radius r1r_1 of Circle C1C_1 is 18.
Using the relationship A=12LrA = \frac{1}{2} L r, the ratio of the two sector areas gives r1r2=54π36π=1.5\frac{r_1}{r_2} = \frac{54\pi}{36\pi} = 1.5, so r2=23r1r_2 = \frac{2}{3} r_1. Expressing arc length as L=108πr1L = \frac{108\pi}{r_1}, the central angles are θ1=19440r12\theta_1 = \frac{19440}{r_1^2} and θ2=29160r12\theta_2 = \frac{29160}{r_1^2}. Setting their sum equal to 150150 leads to 48600r12=150\frac{48600}{r_1^2} = 150, giving r12=324r_1^2 = 324 and r1=18r_1 = 18.

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1
Relate sector area to arc length and radius
A1=12Lr1=54πA_1 = \frac{1}{2} L r_1 = 54\pi and A2=12Lr2=36πA_2 = \frac{1}{2} L r_2 = 36\pi
The area of a sector with arc length LL and radius rr is given by A=θ360πr2=12LrA = \frac{\theta}{360}\pi r^2 = \frac{1}{2} L r.
2
Find the ratio of r1r_1 to r2r_2
r1r2=54π36π=32    r2=23r1\frac{r_1}{r_2} = \frac{54\pi}{36\pi} = \frac{3}{2} \implies r_2 = \frac{2}{3}r_1
Dividing the first area equation by the second cancels out 12L\frac{1}{2}L.
3
Express central angles in terms of r1r_1
θ1=19440r12\theta_1 = \frac{19440}{r_1^2} and θ2=29160r12\theta_2 = \frac{29160}{r_1^2}
Since L=108πr1L = \frac{108\pi}{r_1}, substituting into θ1=L2πr1×360\theta_1 = \frac{L}{2\pi r_1} \times 360 yields θ1=19440r12\theta_1 = \frac{19440}{r_1^2}, and substituting into θ2=L2π(23r1)×360\theta_2 = \frac{L}{2\pi (\frac{2}{3}r_1)} \times 360 yields θ2=29160r12\theta_2 = \frac{29160}{r_1^2}.
4
Solve for r1r_1 using the angle sum equation
r1=18r_1 = 18
Summing the angles gives 19440+29160r12=48600r12=150    r12=324    r1=18\frac{19440 + 29160}{r_1^2} = \frac{48600}{r_1^2} = 150 \implies r_1^2 = 324 \implies r_1 = 18.

Anahtar Kavram

Relationship between arc length, radius, central angle, and sector area
Tahmini Süre:2m 30s
Soru 79Soru

If xx and yy are real numbers that satisfy the inequalities 32x9|3 - 2x| \le 9 and 4y+313|4y + 3| \le 13, what is the maximum possible value of the expression 3x4y|3x - 4y|?

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Cevap: 34

Cevap

The maximum possible value of 3x4y|3x - 4y| is 3434.
To find the maximum possible value of 3x4y|3x - 4y|, we first solve for the ranges of xx and yy from their respective absolute value inequalities. From 32x9|3 - 2x| \le 9, we get 932x9    3x6-9 \le 3 - 2x \le 9 \implies -3 \le x \le 6. From 4y+313|4y + 3| \le 13, we get 134y+313    4y2.5-13 \le 4y + 3 \le 13 \implies -4 \le y \le 2.5. To maximize 3x4y3x - 4y, we take the maximum value of 3x3x (3×6=183 \times 6 = 18) and the maximum value of 4y-4y (4×4=16-4 \times -4 = 16), giving 18+16=3418 + 16 = 34. To minimize 3x4y3x - 4y, we take the minimum value of 3x3x (3×3=93 \times -3 = -9) and the minimum value of 4y-4y (4×2.5=10-4 \times 2.5 = -10), giving 910=19-9 - 10 = -19. The expression 3x4y3x - 4y ranges from 19-19 to 3434, so the maximum possible magnitude 3x4y|3x - 4y| is max(19,34)=34\max(|-19|, |34|) = 34.

Adım Adım Çözüm

1
Solve the inequality 32x9|3 - 2x| \le 9 for xx.
3x6-3 \le x \le 6
Unpacking the absolute value yields 932x9-9 \le 3 - 2x \le 9. Subtracting 33 gives 122x6-12 \le -2x \le 6. Dividing by 2-2 and reversing the inequality signs produces 3x6-3 \le x \le 6.
2
Solve the inequality 4y+313|4y + 3| \le 13 for yy.
4y2.5-4 \le y \le 2.5
Unpacking the absolute value yields 134y+313-13 \le 4y + 3 \le 13. Subtracting 33 gives 164y10-16 \le 4y \le 10. Dividing by 44 gives 4y2.5-4 \le y \le 2.5.
3
Find the range of possible values for 3x3x and 4y-4y.
93x18-9 \le 3x \le 18 and 104y16-10 \le -4y \le 16
Multiplying 3x6-3 \le x \le 6 by 33 gives 93x18-9 \le 3x \le 18. Multiplying 4y2.5-4 \le y \le 2.5 by 4-4 and flipping the signs gives 104y16-10 \le -4y \le 16.
4
Combine the bounds for 3x3x and 4y-4y to find the range for 3x4y3x - 4y.
193x4y34-19 \le 3x - 4y \le 34
The minimum value of 3x4y3x - 4y is (9)+(10)=19(-9) + (-10) = -19. The maximum value of 3x4y3x - 4y is 18+16=3418 + 16 = 34.
5
Determine the maximum absolute value 3x4y|3x - 4y| over the interval [19,34][-19, 34].
34
The absolute value of any number in the interval [19,34][-19, 34] ranges from 00 to max(19,34)=34\max(|-19|, |34|) = 34.

Anahtar Kavram

Absolute Value Inequalities and Expression Bounding
Soru 80Soru

Lines L1L_1 and L2L_2 intersect at point PP to form an acute angle of 5454^\circ. Line MM passes through point PP and is perpendicular to line L1L_1. Ray RR originates at point PP and lies in the interior of one of the obtuse angles formed by L1L_1 and L2L_2. If Ray RR bisects the angle formed between line MM and line L2L_2, what is the measure, in degrees, of the acute angle formed by Ray RR and line L1L_1?

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Cevap: 72

Cevap

The measure of the acute angle formed by Ray RR and line L1L_1 is 7272^\circ.
Lines L1L_1 and L2L_2 form an obtuse angle of 18054=126180^\circ - 54^\circ = 126^\circ. Perpendicular line MM consumes 9090^\circ of this angle relative to L1L_1, leaving 3636^\circ between line MM and line L2L_2. Bisecting this remaining angle gives 1818^\circ. Ray RR therefore lies 1818^\circ away from perpendicular line MM, making an acute angle of 9018=7290^\circ - 18^\circ = 72^\circ with line L1L_1.

Adım Adım Çözüm

1
Determine the measure of the obtuse angle between lines L1L_1 and L2L_2.
The obtuse angle measures 18054=126180^\circ - 54^\circ = 126^\circ.
Adjacent angles along intersecting straight lines are supplementary and sum to 180180^\circ.
2
Find the angle between perpendicular line MM and line L2L_2.
The angle between line MM and line L2L_2 within the obtuse region is 12690=36126^\circ - 90^\circ = 36^\circ.
Line MM is perpendicular to line L1L_1, taking up 9090^\circ of the 126126^\circ obtuse angle.
3
Determine the angle formed by Ray RR after bisecting the 3636^\circ angle.
The angle between Ray RR and line MM is 36/2=1836^\circ / 2 = 18^\circ.
An angle bisector divides an angle into two equal parts.
4
Calculate the acute angle between Ray RR and line L1L_1.
The acute angle formed between Ray RR and line L1L_1 is 9018=7290^\circ - 18^\circ = 72^\circ.
Line MM forms a 9090^\circ angle with line L1L_1. Subtracting the 1818^\circ offset created by Ray RR yields the acute angle of 7272^\circ.

Anahtar Kavram

Supplementary angles, perpendicular lines, and angle bisectors
Tahmini Süre:2m 0s
ÖncekiSayfa 4 / 12Sonraki
Tüm alıştırma soruları — GRE General Test | Examkin