Arithmetic

306 soru

Soru 61Soru

An artisan bakery produces a signature flour blend by combining wheat, rye, and oat flour. Initially, the ratio of wheat flour to rye flour by weight is 5:35:3, and the ratio of rye flour to oat flour by weight is 4:14:1. The baker prepares an initial batch of this mixture weighing exactly 140140 pounds. To adjust the recipe for a special order, pure oat flour is added to the batch until oat flour accounts for exactly 20%20\% of the total weight of the new mixture. How many pounds of pure oat flour must the baker add?

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Cevap: 20

Cevap

20 pounds
To solve this problem, first express the ratios of wheat, rye, and oat flour with a common term for rye. Since Wheat : Rye = 5 : 3 (or 20 : 12) and Rye : Oat = 4 : 1 (or 12 : 3), the combined ratio Wheat : Rye : Oat is 20 : 12 : 3, giving 35 total parts. In a 140-pound batch, each part equals 4 pounds, meaning the batch contains 12 pounds of oat flour and 128 pounds of non-oat flour. When additional oat flour is added, the non-oat weight stays fixed at 128 pounds. For oat flour to be 20% of the new mixture, non-oat flour must be 80%. Dividing 128 by 0.80 gives a new total weight of 160 pounds. The new oat weight is 20% of 160, which is 32 pounds. Subtracting the initial 12 pounds yields 20 pounds of added oat flour.

Adım Adım Çözüm

1
Combine the two given ratios into a single three-part ratio.
Wheat : Rye = 5:3=20:125 : 3 = 20 : 12, and Rye : Oat = 4:1=12:34 : 1 = 12 : 3. Thus, Wheat : Rye : Oat = 20:12:320 : 12 : 3.
Scaling the ratios so that the common element (Rye) has equal parts allows unification into a single ratio.
2
Calculate the initial weight of each component in the 140-pound batch.
Total ratio parts = 20+12+3=3520 + 12 + 3 = 35 parts. Each part is 140/35=4140 / 35 = 4 pounds. Initial Oat flour = 3×4=123 \times 4 = 12 pounds. Non-oat flour (Wheat + Rye) = (20+12)×4=128(20 + 12) \times 4 = 128 pounds.
Determining the weight of the constant non-oat portion is key to solving mixture adjustment problems.
3
Set up an equation for the new mixture where oat flour makes up 20% of the total weight.
Since non-oat flour remains unchanged at 128128 pounds, it represents 80%80\% of the new total weight. New Total Weight = 128/0.80=160128 / 0.80 = 160 pounds.
If oat flour is 20% of the new total, the non-oat components must constitute the remaining 80%.
4
Calculate the weight of oat flour added.
New Oat Weight = 20%20\% of 160=32160 = 32 pounds. Added Oat Flour = 3212=2032 - 12 = 20 pounds.
Subtracting the initial oat weight from the final required oat weight yields the amount added.

Anahtar Kavram

Combining multi-part ratios and solving ratio adjustment problems by holding unchanged components constant.
Soru 62Soru

A financial forecasting model predicts monthly sales revenue using a sequence where the initial revenue in month 1 is R1=$80000R_1 = \$80{}000, and the revenue for each subsequent month nn (for n2n \ge 2) is calculated using the recursive formula Rn=0.75Rn1+5000R_n = 0.75 R_{n-1} + 5{}000. If each monthly revenue figure RnR_n is rounded to the nearest hundred dollars before summing, what is the estimated total revenue, in dollars, for the first 4 months combined?

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Cevap: 244100

Cevap

The estimated total revenue for the first 4 months combined is 244,100 dollars.
By evaluating the recursive formula Rn=0.75Rn1+5000R_n = 0.75 R_{n-1} + 5{}000 for months 1 through 4, we obtain R1=80000R_1 = 80{}000, R2=65000R_2 = 65{}000, R3=53750R_3 = 53{}750, and R4=45312.50R_4 = 45{}312.50. Rounding each to the nearest hundred yields 8000080{}000, 6500065{}000, 5380053{}800, and 4530045{}300, respectively. Summing these four rounded figures gives 244100244{}100.

Adım Adım Çözüm

1
Find the first term R1R_1 and its rounded value.
R1=80000R_1 = 80{}000, which rounds to 8000080{}000.
Given initial term.
2
Calculate the second term R2R_2 using the recursive formula and round it.
R2=0.75(80000)+5000=65000R_2 = 0.75(80{}000) + 5{}000 = 65{}000, which rounds to 6500065{}000.
Apply R2=0.75R1+5000R_2 = 0.75 R_1 + 5{}000.
3
Calculate the third term R3R_3 using unrounded R2R_2 and round to the nearest hundred.
R3=0.75(65000)+5000=53750R_3 = 0.75(65{}000) + 5{}000 = 53{}750, which rounds to 5380053{}800.
Apply R3=0.75R2+5000R_3 = 0.75 R_2 + 5{}000 and round to the nearest 100.
4
Calculate the fourth term R4R_4 using unrounded R3R_3 and round to the nearest hundred.
R4=0.75(53750)+5000=45312.50R_4 = 0.75(53{}750) + 5{}000 = 45{}312.50, which rounds to 4530045{}300.
Apply R4=0.75R3+5000R_4 = 0.75 R_3 + 5{}000 and round to the nearest 100.
5
Add the four rounded terms together.
80000+65000+53800+45300=24410080{}000 + 65{}000 + 53{}800 + 45{}300 = 244{}100.
Compute total estimated revenue as specified.

Anahtar Kavram

Estimation, Rounding, and Sequences
Soru 63Soru

In an industrial mechanical system, Gear AA has 1616 teeth, Gear BB has 2424 teeth, and Gear CC has 4040 teeth. Gear AA is meshed directly with Gear BB, and Gear BB is meshed directly with Gear CC. If Gear AA rotates at a constant speed of 150150 revolutions per minute (rpm\text{rpm}), what is the rotational speed, in revolutions per minute (rpm\text{rpm}), of Gear CC?

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Cevap: 60

Cevap

The rotational speed of Gear C is 60 rpm.
For meshed gears, the linear speed of teeth at the point of contact must be identical. Thus, the product of the number of teeth and rotational speed remains constant (NASA=NCSCN_A S_A = N_C S_C). Substituting the known values gives 16×150=40×SC16 \times 150 = 40 \times S_C, yielding 2,400=40SC2,400 = 40 S_C, so SC=60S_C = 60 rpm.

Adım Adım Çözüm

1
Determine the inverse proportional relationship between number of gear teeth and rotational speed.
The product of teeth count and rotational speed is constant across directly meshed gears: NA×SA=NB×SB=NC×SCN_A \times S_A = N_B \times S_B = N_C \times S_C.
Directly meshed gears engage tooth for tooth, meaning they pass the same total number of teeth per unit time.
2
Calculate the total tooth displacement rate per minute from Gear A.
16×150=2,40016 \times 150 = 2,400 teeth per minute.
Gear A has 16 teeth and completes 150 revolutions per minute.
3
Calculate the rotational speed of Gear C.
SpeedC=2,40040=60\text{Speed}_C = \frac{2,400}{40} = 60 rpm.
Gear C has 40 teeth, so dividing the total tooth displacement rate by 40 yields its revolutions per minute.

Anahtar Kavram

Inverse Proportionality in Gear Rates
Tahmini Süre:1m 15s
Soru 64Soru

How many positive integer factors of 3636 are prime numbers?

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Cevap: 22

Cevap

There are 22 positive integer factors of 3636 that are prime numbers: 22 and 33.
Listing all positive integer factors of 3636 yields 1,2,3,4,6,9,12,18,1, 2, 3, 4, 6, 9, 12, 18, and 3636. Evaluating each factor against the definition of a prime number (an integer greater than 11 divisible only by 11 and itself) shows that only 22 and 33 are prime. Therefore, there are exactly 22 prime factors.

Adım Adım Çözüm

1
Find all positive integer factors of 3636.
The positive integer factors of 3636 are 1,2,3,4,6,9,12,18,1, 2, 3, 4, 6, 9, 12, 18, and 3636.
Dividing 3636 by integers from 11 to 3636 yields integer quotients for these nine numbers.
2
Identify which of these factors meet the definition of a prime number.
The prime factors are 22 and 33.
A prime number is an integer greater than 11 with exactly two distinct positive divisors: 11 and itself. The integer 11 is not prime, while 4,6,9,12,18,4, 6, 9, 12, 18, and 3636 are composite.
3
Count the identified prime factors.
The total count is 22.
There are exactly two distinct prime numbers (22 and 33) in the list of factors.

Anahtar Kavram

Prime Factors and Definition of Prime Numbers
Tahmini Süre:45s
Soru 65Soru

When the positive integer nn is divided by 77, the remainder is 44. What is the remainder when n+15n + 15 is divided by 77?

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Cevap: 5

Cevap

5
Adding 15 to nn increases the remainder by 15. The original remainder is 4, so the new sum of remainders is 4+15=194 + 15 = 19. Dividing 19 by 7 gives a quotient of 2 and a remainder of 5.

Adım Adım Çözüm

1
Express nn algebraically based on the given remainder rule.
n=7k+4n = 7k + 4 for some non-negative integer kk
By the division algorithm, any integer divided by 7 with remainder 4 can be written as a multiple of 7 plus 4.
2
Substitute nn into the expression n+15n + 15.
n+15=7k+4+15=7k+19n + 15 = 7k + 4 + 15 = 7k + 19
We need to find the remainder of this new quantity when divided by 7.
3
Extract the largest multiple of 7 from 19.
7k+19=7k+14+5=7(k+2)+57k + 19 = 7k + 14 + 5 = 7(k + 2) + 5
Grouping multiples of 7 isolates the remaining constant term.
4
Identify the final remainder.
The remainder is 5.
7(k+2)7(k + 2) is completely divisible by 7, leaving 5 as the remainder.

Anahtar Kavram

Remainder arithmetic and divisibility properties
Soru 66Soru

Which of the following integers is divisible by both 33 and 44?

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Cevap: 7272

Cevap

7272
For an integer to be divisible by both 33 and 44, it must satisfy the rules for both divisors. For 7272, the sum of the digits is 7+2=97 + 2 = 9, which is divisible by 33. Additionally, 72÷4=1872 \div 4 = 18, which is an integer. Thus, 7272 is divisible by both 33 and 44.

Adım Adım Çözüm

1
Recall the divisibility rule for 3
An integer is divisible by 3 if the sum of its digits is divisible by 3.
This allows quick evaluation of divisibility by 3.
2
Recall the divisibility rule for 4
An integer is divisible by 4 if its last two digits form a number divisible by 4.
This allows quick evaluation of divisibility by 4.
3
Test 7272 against both divisibility rules
Sum of digits of 7272 is 7+2=97 + 2 = 9, which is divisible by 33. Also, 72÷4=1872 \div 4 = 18, which is an integer.
7272 meets both required criteria.

Anahtar Kavram

Divisibility Rules for 3 and 4
Tahmini Süre:45s
Soru 67Soru

If xx and yy are positive integers such that 3x+5y=1013x + 5y = 101 and xx is a prime number, what is the maximum possible value of the product xyxy?

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Cevap: 170

Cevap

170
The equation 3x+5y=1013x + 5y = 101 requires 1013x101 - 3x to be a multiple of 5. For positive integers xx and yy, xx must be less than 34 and leave a remainder of 2 when divided by 5 (meaning xx ends in 2 or 7). Testing the primes meeting this condition yields three valid pairs: (2,19)(2, 19) with product 38, (7,16)(7, 16) with product 112, and (17,10)(17, 10) with product 170. Thus, 170 is the maximum product.

Adım Adım Çözüm

1
Isolate 5y5y and analyze modular divisibility by 5.
5y=1013x5y = 101 - 3x. For yy to be an integer, 1013x101 - 3x must be divisible by 5. Since 1011(mod5)101 \equiv 1 \pmod 5, we require 3x1(mod5)3x \equiv 1 \pmod 5, which simplifies to x2(mod5)x \equiv 2 \pmod 5.
Properties of modular arithmetic determine which values of xx yield integer values for yy.
2
Determine the boundary constraints and prime condition for xx.
Since yy is a positive integer (y1y \ge 1), 3x<101    x333x < 101 \implies x \le 33. The prime numbers x33x \le 33 that satisfy x2(mod5)x \equiv 2 \pmod 5 (or end in 2 or 7) are x=2x = 2, x=7x = 7, and x=17x = 17.
Only primes satisfying both the inequality bound and divisibility criteria need to be tested.
3
Calculate the corresponding values of yy and products xyxy for each prime candidate.
For x=2x = 2: 5y=1016=95    y=195y = 101 - 6 = 95 \implies y = 19, product xy=2×19=38xy = 2 \times 19 = 38.
For x=7x = 7: 5y=10121=80    y=165y = 101 - 21 = 80 \implies y = 16, product xy=7×16=112xy = 7 \times 16 = 112.
For x=17x = 17: 5y=10151=50    y=105y = 101 - 51 = 50 \implies y = 10, product xy=17×10=170xy = 17 \times 10 = 170.
Evaluating all valid candidate pairs allows identifying the absolute maximum product.
4
Identify the maximum product.
The maximum possible value of xyxy is 170.
Comparing 38, 112, and 170 confirms that 170 is the greatest value.

Anahtar Kavram

Linear Diophantine equations with prime integer constraints
Soru 68Soru

If kk is a positive integer less than 100100 such that kk is divisible by 66 but not divisible by 88, what is the greatest possible value of kk?

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Cevap: 90

Cevap

90
The positive multiples of 6 less than 100 in descending order are 96, 90, 84, etc. The largest candidate, 96, is divisible by 8 (96=8×1296 = 8 \times 12), so it is excluded by the conditions. The next largest candidate, 90, is divisible by 6 (90=6×1590 = 6 \times 15) but leaves a remainder of 2 when divided by 8 (90=8×11+290 = 8 \times 11 + 2). Thus, 90 is the greatest integer satisfying all conditions.

Adım Adım Çözüm

1
Identify the largest multiples of 6 that are less than 100.
The multiples of 6 less than 100 in descending order are 96, 90, 84, 78, ...
Finding the greatest possible value requires testing candidates starting from the largest possible multiple of 6 below 100.
2
Test 96 against the constraint of not being divisible by 8.
96 / 8 = 12, so 96 is divisible by 8.
The question specifies that k must NOT be divisible by 8, ruling out 96.
3
Test the next candidate, 90, against the constraint.
90 / 8 = 11 R 2, so 90 is not divisible by 8.
90 meets all specified criteria: it is a positive integer less than 100, divisible by 6, and not divisible by 8.

Anahtar Kavram

Properties of Multiples and Divisibility Constraints
Soru 69Soru

Let mm and nn be positive integers such that m2nm^2 n is divisible by 7272 and mn2m n^2 is divisible by 108108. What is the minimum possible value of the product mnmn?

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Cevap: 36

Cevap

36
The product (m2n)(mn2)=(mn)3(m^2 n)(m n^2) = (mn)^3 must be divisible by 72×108=25×3572 \times 108 = 2^5 \times 3^5. For (mn)3(mn)^3 to be a valid cube of an integer, the exponents of 2 and 3 in its prime factorization must be multiples of 3 greater than or equal to 5. The smallest such multiples of 3 are 6, meaning (mn)326×36=(36)3(mn)^3 \ge 2^6 \times 3^6 = (36)^3, so mn36mn \ge 36. Setting m=6m = 6 and n=6n = 6 confirms that 3636 satisfies all requirements.

Adım Adım Çözüm

1
Express the given divisibility conditions in terms of prime factorizations.
72=23×3272 = 2^3 \times 3^2 and 108=22×33108 = 2^2 \times 3^3. Therefore, m2nm^2 n is a multiple of 23×322^3 \times 3^2, and mn2m n^2 is a multiple of 22×332^2 \times 3^3.
Prime factorization allows us to analyze the minimum exponent requirements for 2 and 3.
2
Multiply the two expressions to find a lower bound on (mn)3(mn)^3.
(m2n)(mn2)=m3n3=(mn)3(m^2 n)(m n^2) = m^3 n^3 = (mn)^3 must be divisible by (23×32)(22×33)=25×35(2^3 \times 3^2)(2^2 \times 3^3) = 2^5 \times 3^5.
Combining the expressions yields a perfect cube (mn)3(mn)^3.
3
Determine the minimum prime powers needed for mnmn.
Since (mn)3(mn)^3 is a perfect cube divisible by 25×352^5 \times 3^5, the exponents of 2 and 3 in (mn)3(mn)^3 must be multiples of 3 that are at least 5. The smallest such exponents are 6 for both 2 and 3. Thus, (mn)326×36=(22×32)3=363(mn)^3 \ge 2^6 \times 3^6 = (2^2 \times 3^2)^3 = 36^3, so mn36mn \ge 36.
Exponents in a perfect cube factorization must be multiples of 3.
4
Verify that mn=36mn = 36 is achievable with integer values of mm and nn.
Setting m=6m = 6 and n=6n = 6 gives mn=36mn = 36. Then m2n=63=216=3×72m^2 n = 6^3 = 216 = 3 \times 72 (divisible by 72) and mn2=63=216=2×108m n^2 = 6^3 = 216 = 2 \times 108 (divisible by 108).
Constructing valid integers mm and nn confirms that 36 is achievable.

Anahtar Kavram

Properties of Integers and Prime Factorization
Soru 70Soru

If kk is an integer greater than 11 such that k3kk^3 - k is a multiple of 240240, what is the least possible value of kk?

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Cevap: 9

Cevap

The least possible value of kk is 9.
Factoring k3kk^3 - k gives (k1)k(k+1)(k-1)k(k+1), representing the product of three consecutive integers. The prime factorization of 240240 is 24×3×52^4 \times 3 \times 5. To be divisible by 240240, the product (k1)k(k+1)(k-1)k(k+1) must contain at least four factors of 2, one factor of 3, and one factor of 5. Testing integers k>1k > 1: for k=9k=9, the product is (8)(9)(10)=720(8)(9)(10) = 720, which is 3×2403 \times 240. Checking all integers 1<k<91 < k < 9 confirms that no smaller integer satisfies the condition.

Adım Adım Çözüm

1
Factor the algebraic expression k3kk^3 - k.
k3k=k(k21)=(k1)k(k+1)k^3 - k = k(k^2 - 1) = (k-1)k(k+1).
This expresses the polynomial as a product of three consecutive integers.
2
Determine the prime factorization of 240.
240=24×3×5=16×3×5240 = 2^4 \times 3 \times 5 = 16 \times 3 \times 5.
For (k1)k(k+1)(k-1)k(k+1) to be a multiple of 240, the product must be divisible by 16, 3, and 5 simultaneously.
3
Test candidate values of k>1k > 1 to find the smallest valid integer.
For k=9k=9, (k1)k(k+1)=(8)(9)(10)=720(k-1)k(k+1) = (8)(9)(10) = 720, which equals 3×2403 \times 240.
Testing smaller values: k=5k=5 yields 120120, k=6k=6 yields 210210, k=7k=7 yields 336336, and k=8k=8 yields 504504. None of these are divisible by 240.

Anahtar Kavram

Divisibility rules and prime factor distribution across consecutive integers
Tahmini Süre:2m 0s
Soru 71Soru

When the positive integer mm is divided by 88, the remainder is 55. What is the remainder when 3m+73m + 7 is divided by 88?

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Cevap: 66

Cevap

The remainder is 66.
Since mm leaves a remainder of 55 when divided by 88, we can evaluate 3m+73m + 7 modulo 88 by replacing mm with 55. Calculating 3(5)+73(5) + 7 gives 2222. Dividing 2222 by 88 gives a quotient of 22 with a remainder of 66 because 22=8×2+622 = 8 \times 2 + 6.

Adım Adım Çözüm

1
Express integer mm using the division algorithm.
m=8k+5m = 8k + 5 for some non-negative integer kk.
Dividing mm by 88 yields a remainder of 55.
2
Substitute the expression for mm into 3m+73m + 7.
3m+7=3(8k+5)+7=24k+15+7=24k+223m + 7 = 3(8k + 5) + 7 = 24k + 15 + 7 = 24k + 22.
This expresses the new quantity algebraically in terms of kk.
3
Determine the remainder when 24k+2224k + 22 is divided by 88.
24k+22=8(3k+2)+624k + 22 = 8(3k + 2) + 6, so the remainder is 66.
The term 8(3k+2)8(3k + 2) is a multiple of 88, leaving 66 as the non-negative remainder less than 88.

Anahtar Kavram

Remainder properties of linear combinations of integers
Tahmini Süre:45s
Soru 72Soru

For all positive integers nn, what is the greatest integer that MUST divide n5nn^5 - n?

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Cevap: 30

Cevap

The greatest integer that must divide n5nn^5 - n for all positive integers nn is 30.
Factoring the expression yields n5n=(n1)n(n+1)(n2+1)n^5 - n = (n-1)n(n+1)(n^2+1). The factor (n1)n(n+1)(n-1)n(n+1) consists of three consecutive integers, ensuring divisibility by both 2 and 3 (and thus by 6). Furthermore, the units digit of n5n^5 is always identical to the units digit of nn for any positive integer nn, which guarantees that n5nn^5 - n is divisible by 5. Since 6 and 5 share no common factors, n5nn^5 - n must be divisible by 6×5=306 \times 5 = 30. Testing n=2n = 2 gives 252=302^5 - 2 = 30, proving that 30 is the largest integer that divides the expression for all positive integers nn.

Adım Adım Çözüm

1
Factor the algebraic expression n5nn^5 - n.
n5n=n(n41)=n(n21)(n2+1)=(n1)n(n+1)(n2+1)n^5 - n = n(n^4 - 1) = n(n^2 - 1)(n^2 + 1) = (n - 1)n(n + 1)(n^2 + 1).
Factoring helps isolate product terms with known divisibility rules.
2
Analyze divisibility by 2 and 3.
The sub-expression (n1)n(n+1)(n - 1)n(n + 1) is the product of three consecutive integers.
Among any three consecutive integers, at least one is divisible by 2 and exactly one is divisible by 3. Therefore, (n1)n(n+1)(n - 1)n(n + 1) is always divisible by 2×3=62 \times 3 = 6.
3
Analyze divisibility by 5.
By Fermat's Little Theorem (or analyzing last-digit repeating cycles of powers), for any integer nn, n5n(mod5)n^5 \equiv n \pmod 5, which implies 55 divides n5nn^5 - n.
The units digit of n5n^5 is always equal to the units digit of nn, so n5nn^5 - n always ends in 0 or 5, making it a multiple of 5.
4
Combine the common prime factors and evaluate the maximum lower bound.
Since 2, 3, and 5 are pairwise coprime, any number divisible by 2, 3, and 5 must be divisible by 2×3×5=302 \times 3 \times 5 = 30. Evaluating at n=2n = 2 yields 252=302^5 - 2 = 30, showing no integer greater than 30 can divide n5nn^5 - n for all nn.
The greatest common divisor across all generated values of n5nn^5 - n is 30.

Anahtar Kavram

Divisibility properties of consecutive integers and power mod rules
Tahmini Süre:1m 30s
Soru 73Soru

A container originally holds a liquid mixture in which 25\frac{2}{5} of the total volume is pure juice. After 66 liters of pure juice are added to the container, pure juice makes up 12\frac{1}{2} of the new total volume. How many liters of mixture were originally in the container?

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Cevap: 30

Cevap

30 liters
The original mixture volume VV contains 25V\frac{2}{5}V liters of pure juice. When 66 liters of pure juice are added, the new volume of juice becomes 25V+6\frac{2}{5}V + 6 and the new total volume becomes V+6V + 6. Setting 25V+6=12(V+6)\frac{2}{5}V + 6 = \frac{1}{2}(V + 6) leads to 110V=3\frac{1}{10}V = 3, giving V=30V = 30 liters.

Adım Adım Çözüm

1
Define variables for the original volume and pure juice volume.
Let VV represent the original volume of the mixture in liters. The original volume of pure juice is 25V\frac{2}{5}V.
Expressing the initial quantity of juice in terms of the unknown total volume establishes the baseline algebraic relationship.
2
Set up an equation incorporating the added quantity of pure juice.
New juice volume =25V+6= \frac{2}{5}V + 6, and new total volume =V+6= V + 6. The equation is 25V+6=12(V+6)\frac{2}{5}V + 6 = \frac{1}{2}(V + 6).
Adding pure juice increases both the pure juice component and the total volume of the mixture by 66 liters.
3
Solve the linear equation for VV.
Expanding the right side gives 25V+6=12V+3\frac{2}{5}V + 6 = \frac{1}{2}V + 3. Subtracting 25V\frac{2}{5}V from both sides gives 63=(1225)V3=110VV=306 - 3 = \left(\frac{1}{2} - \frac{2}{5}\right)V \Rightarrow 3 = \frac{1}{10}V \Rightarrow V = 30.
Finding a common denominator for the fractions 12=510\frac{1}{2} = \frac{5}{10} and 25=410\frac{2}{5} = \frac{4}{10} allows solving for VV directly.

Anahtar Kavram

Solving linear equations involving fractions of whole quantities and mixture relationships.
Tahmini Süre:1m 0s
Soru 74Soru

If aa and bb are positive integers such that 15a15a is divisible by 2424 and 10b10b is divisible by 1414, which of the following statements MUST be true? Select all such statements.

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Cevap: aa is divisible by 88; The product abab is divisible by 5656

Cevap

The statements asserting that aa is divisible by 88 and that the product abab is divisible by 5656 must be true.
Because 24=23×324 = 2^3 \times 3 and 15=3×515 = 3 \times 5, aa must provide three factors of 22, meaning aa is divisible by 88. Similarly, 14=2×714 = 2 \times 7 and 10=2×510 = 2 \times 5, so bb must provide a factor of 77, meaning bb is divisible by 77. Multiplying the minimum properties (aa is a multiple of 88 and bb is a multiple of 77) guarantees that abab is a multiple of 5656.

Adım Adım Çözüm

1
Analyze the condition 15a15a is divisible by 2424
Prime factorizations: 15=3×515 = 3 \times 5 and 24=23×324 = 2^3 \times 3. For 2424 to divide 15a15a, aa must supply 23=82^3 = 8, so aa is a multiple of 88.
Determines the prime factors required for aa.
2
Analyze the condition 10b10b is divisible by 1414
Prime factorizations: 10=2×510 = 2 \times 5 and 14=2×714 = 2 \times 7. For 1414 to divide 10b10b, bb must supply 77, so bb is a multiple of 77.
Determines the prime factors required for bb.
3
Evaluate each statement against the deduced minimum properties of aa and bb
Since aa is a multiple of 88, the statement regarding aa being divisible by 88 is always true. Since aa is a multiple of 88 and bb is a multiple of 77, abab is a multiple of 8×7=568 \times 7 = 56. Counterexamples disprove the remaining choices (b=7b=7 disproves divisibility of bb by 1414 and evenness of a+ba+b; a=8a=8 disproves divisibility of aa by 2424).
Identifies which statements MUST be true for all valid values of aa and bb.

Anahtar Kavram

Divisibility and Prime Factorization Constraints
Soru 75Soru

When the positive integer nn is divided by 77, the remainder is 44. What is the remainder when n+18n + 18 is divided by 77?

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Cevap: 1

Cevap

The remainder when n+18n + 18 is divided by 77 is 11.
Because nn leaves a remainder of 44 when divided by 77, nn can be represented as 7k+47k + 4. Adding 1818 gives n+18=7k+22n + 18 = 7k + 22. Since 7k7k is a multiple of 77, the remainder of (7k+22)÷7(7k + 22) \div 7 depends entirely on 22÷722 \div 7. Dividing 2222 by 77 gives a quotient of 33 with a remainder of 11.

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1
Represent the positive integer nn algebraically based on the given remainder.
n=7k+4n = 7k + 4 for some non-negative integer kk.
By the division algorithm, any integer nn divided by 77 with remainder 44 can be written as 7k+47k + 4.
2
Add 1818 to nn.
n+18=(7k+4)+18=7k+22n + 18 = (7k + 4) + 18 = 7k + 22.
Substitute 7k+47k + 4 for nn in the expression n+18n + 18.
3
Determine the remainder when 7k+227k + 22 is divided by 77.
Since 7k7k is divisible by 77, the remainder is 22(mod7)=122 \pmod 7 = 1.
Dividing 2222 by 77 yields a quotient of 33 and a remainder of 11 (22=7×3+122 = 7 \times 3 + 1).

Anahtar Kavram

Remainder Properties under Addition
Soru 76Soru

If nn is a positive integer such that nn is divisible by 1212 and n2n^2 is divisible by 180180, what is the least possible value of nn?

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Cevap: 60

Cevap

60
The correct answer is 60. The prime factorization of 12 is 22×312^2 \times 3^1, requiring nn to have at least two factors of 2 and one factor of 3. The prime factorization of 180 is 22×32×512^2 \times 3^2 \times 5^1. For n2n^2 to be divisible by 180, nn must contribute a factor of 5 (which becomes 525^2 in n2n^2). Multiplying the minimal prime factors yields 22×31×51=602^2 \times 3^1 \times 5^1 = 60.

Adım Adım Çözüm

1
Analyze the prime factorization required for nn to be divisible by 12
Since 12=22×3112 = 2^2 \times 3^1, any positive integer nn divisible by 12 must contain at least 222^2 and 313^1 in its prime factorization.
An integer must contain all prime factors of its divisor with at least equal exponents.
2
Analyze the prime factorization required for n2n^2 to be divisible by 180
Since 180=22×32×51180 = 2^2 \times 3^2 \times 5^1, n2n^2 must contain at least 222^2, 323^2, and 515^1. Because n2n^2 doubles all prime exponents of nn, nn must contribute at least 515^1, which makes the factor of 5 in n2n^2 equal to 525^2.
The exponent of any prime factor in a perfect square n2n^2 must be even.
3
Combine the minimal prime factor requirements for nn
The minimum prime factorization for nn is 22×31×51=4×3×5=602^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60.
Taking the minimum required power of each prime factor yields the smallest positive integer fulfilling both conditions.

Anahtar Kavram

Prime Factorization and Divisibility of Powers
Soru 77Soru

When the positive integer nn is divided by 7, the remainder is 3, and when nn is divided by 11, the remainder is 5. If nn is a three-digit integer less than 200 that is divisible by 6, what is the value of nn?

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Cevap: 192

Cevap

The value of nn is 192.
To find nn, combine the remainder requirements n3(mod7)n \equiv 3 \pmod 7 and n5(mod11)n \equiv 5 \pmod{11}. The smallest positive integer solution is 3838. Because lcm(7,11)=77\text{lcm}(7, 11) = 77, all valid integers take the form n=77m+38n = 77m + 38. Restricting nn to three-digit numbers less than 200 gives candidates n=115n = 115 (m=1m=1) and n=192n = 192 (m=2m=2). Among these, only 192 is divisible by 6 (since 192=6×32192 = 6 \times 32).

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1
Formulate remainder congruences for the given conditions
n3(mod7)n \equiv 3 \pmod{7} and n5(mod11)n \equiv 5 \pmod{11}
Dividing nn by 7 leaves remainder 3, and dividing by 11 leaves remainder 5.
2
Find the smallest positive integer satisfying both congruences
n=38n = 38
Checking values 11k+511k + 5: 5,16,27,385, 16, 27, 38. 38÷7=538 \div 7 = 5 remainder 3, so 38 satisfies both conditions.
3
Determine the general solution for nn using the Chinese Remainder Theorem logic
n=77m+38n = 77m + 38 for integer m0m \ge 0
Since 7 and 11 are coprime, the solutions repeat every lcm(7,11)=77\text{lcm}(7, 11) = 77.
4
Evaluate candidate values for nn such that 100n<200100 \le n < 200
For m=1m=1, n=115n = 115; for m=2m=2, n=192n = 192
These are the only three-digit integers less than 200 of the form 77m+3877m + 38.
5
Apply the final constraint that nn must be divisible by 6
192192 is divisible by 6 (192=6×32192 = 6 \times 32)
115115 is odd, so it is not divisible by 6. 192192 is even and the sum of its digits (1+9+2=121+9+2=12) is a multiple of 3, so it is divisible by 6.

Anahtar Kavram

Simultaneous congruences and combined divisibility rules (Chinese Remainder Theorem)
Tahmini Süre:2m 0s
Soru 78Soru

Let N=2a3b5cN = 2^a \cdot 3^b \cdot 5^c, where aa, bb, and cc are positive integers such that NN is a multiple of 88. If N2N^2 has 105105 distinct positive divisors and N6\frac{N}{6} has 1212 distinct positive divisors, how many distinct positive divisors does 10N10N have?

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Cevap: 45

Cevap

The number of distinct positive divisors of 10N10N is 45.
The correct answer is 45. Given N=2a3b5cN = 2^a \cdot 3^b \cdot 5^c, the number of positive divisors of N2=22a32b52cN^2 = 2^{2a} \cdot 3^{2b} \cdot 5^{2c} is (2a+1)(2b+1)(2c+1)=105(2a+1)(2b+1)(2c+1) = 105. The unique factor decomposition of 105 into three factors greater than 1 is 3×5×73 \times 5 \times 7. Because NN is a multiple of 8, a3a \ge 3, so 2a+172a+1 \ge 7, forcing 2a+1=72a+1 = 7, which means a=3a = 3. The remaining factor set {2b+1,2c+1}={3,5}\{2b+1, 2c+1\} = \{3, 5\} gives two cases for (b,c)(b, c): (2,1)(2, 1) or (1,2)(1, 2). For N6=2a13b15c\frac{N}{6} = 2^{a-1} \cdot 3^{b-1} \cdot 5^c, the divisor count is ab(c+1)=3b(c+1)=12a \cdot b \cdot (c+1) = 3 \cdot b \cdot (c+1) = 12, so b(c+1)=4b(c+1) = 4. Substituting (b,c)=(2,1)(b, c) = (2, 1) yields 2×2=42 \times 2 = 4, which satisfies this relation. Therefore, (a,b,c)=(3,2,1)(a, b, c) = (3, 2, 1). Now 10N=2a+13b5c+1=24325210N = 2^{a+1} \cdot 3^b \cdot 5^{c+1} = 2^4 \cdot 3^2 \cdot 5^2. Its total number of positive divisors is (4+1)(2+1)(2+1)=5×3×3=45(4+1)(2+1)(2+1) = 5 \times 3 \times 3 = 45.

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1
Express the divisor count formula for N2N^2
d(N2)=(2a+1)(2b+1)(2c+1)=105d(N^2) = (2a+1)(2b+1)(2c+1) = 105
For an integer with prime factorization p1e1p2e2p_1^{e_1} p_2^{e_2} \cdots, the number of divisors is (e1+1)(e2+1)(e_1+1)(e_2+1)\cdots. Here N2=22a32b52cN^2 = 2^{2a} \cdot 3^{2b} \cdot 5^{2c}.
2
Determine the value of aa using the divisibility condition
a=3a = 3
Since NN is a multiple of 8=238 = 2^3, we must have a3a \ge 3, which implies 2a+172a+1 \ge 7. The prime factorization of 105105 into three odd factors greater than 11 is 3×5×73 \times 5 \times 7. Thus, 2a+1=72a+1 = 7, giving a=3a = 3.
3
Solve for exponents bb and cc using the divisor count of N6\frac{N}{6}
b=2b = 2 and c=1c = 1
Since N6=2a13b15c\frac{N}{6} = 2^{a-1} \cdot 3^{b-1} \cdot 5^c, its divisor count is ab(c+1)=3b(c+1)=12a \cdot b \cdot (c+1) = 3b(c+1) = 12, so b(c+1)=4b(c+1) = 4. From step 2, {2b+1,2c+1}={3,5}\{2b+1, 2c+1\} = \{3, 5\}, so either (b,c)=(2,1)(b, c) = (2, 1) or (b,c)=(1,2)(b, c) = (1, 2). Testing (b,c)=(2,1)(b, c) = (2, 1) gives 2(1+1)=42(1+1) = 4, which satisfies the equation.
4
Calculate the prime factorization and divisor count of 10N10N
d(10N)=(4+1)(2+1)(2+1)=45d(10N) = (4+1)(2+1)(2+1) = 45
Since 10N=25(233251)=24325210N = 2 \cdot 5 \cdot (2^3 \cdot 3^2 \cdot 5^1) = 2^4 \cdot 3^2 \cdot 5^2, applying the divisor formula yields 5×3×3=455 \times 3 \times 3 = 45.

Anahtar Kavram

Divisor Count Formula & Prime Factorization Constraints
Tahmini Süre:2m 30s
Soru 79Soru

A storage tank is initially 23\frac{2}{3} full of liquid. After 14\frac{1}{4} of the liquid inside the tank is drained out, 1515 gallons of liquid are added to the tank. If the tank is then 56\frac{5}{6} full, what is the total capacity of the tank, in gallons?

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Cevap: 45

Cevap

45 gallons
The correct answer of 45 gallons is found by calculating the liquid remaining after draining 14\frac{1}{4} of the initial 23C\frac{2}{3} C, which leaves 12C\frac{1}{2} C. Adding 15 gallons brings the total volume to 56C\frac{5}{6} C. Solving 56C12C=15\frac{5}{6} C - \frac{1}{2} C = 15 gives 13C=15\frac{1}{3} C = 15, so C=45C = 45.

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1
Determine the fraction of liquid in the tank after draining.
The initial amount of liquid is 23C\frac{2}{3} C, where CC is the total capacity. Draining 14\frac{1}{4} of this liquid leaves 114=341 - \frac{1}{4} = \frac{3}{4} of the initial liquid. Thus, the remaining liquid is 34×23C=12C\frac{3}{4} \times \frac{2}{3} C = \frac{1}{2} C.
The fraction drained applies to the existing liquid volume, not the total container capacity.
2
Set up an equation incorporating the added liquid and final tank fraction.
12C+15=56C\frac{1}{2} C + 15 = \frac{5}{6} C
Adding 1515 gallons to the remaining liquid yields a volume equal to 56\frac{5}{6} of the total capacity.
3
Solve for the total capacity CC.
Subtracting 12C\frac{1}{2} C (which is 36C\frac{3}{6} C) from both sides gives 15=56C36C=26C=13C15 = \frac{5}{6} C - \frac{3}{6} C = \frac{2}{6} C = \frac{1}{3} C. Multiplying both sides by 33 yields C=45C = 45.
Isolating CC determines the overall volume of the tank.

Anahtar Kavram

Multi-step operations with fractions and fractional parts of a whole
Tahmini Süre:1m 30s
Soru 80Soru

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
ÖncekiSayfa 4 / 16Sonraki
Arithmetic Alıştırma Soruları — GRE General Test — Sayfa 4 | Examkin