Basic Numeracy

295 questions

Question 81Question

Three environmental monitoring sensors transmit data packets at regular intervals of 83\frac{8}{3} seconds, 125\frac{12}{5} seconds, and 247\frac{24}{7} seconds, respectively. If all three sensors transmit a packet simultaneously at exactly 10:00:00 AM, how many seconds will pass before they next transmit a packet together at the exact same instant?

Show answer & explanation

Answer: 2424

Answer

24 seconds
The sensors will sync up again at a time that is a multiple of all three individual intervals. To find this, we calculate the LCM of the fractions 83\frac{8}{3}, 125\frac{12}{5}, and 247\frac{24}{7}. The mathematical rule for the LCM of fractions is the LCM of the numerators divided by the HCF of the denominators. We find LCM(8,12,24)=24\text{LCM}(8, 12, 24) = 24 and HCF(3,5,7)=1\text{HCF}(3, 5, 7) = 1. Thus, the total LCM is 241=24\frac{24}{1} = 24 seconds.

Step-by-Step Solution

1
Identify the mathematical requirement
The time until the next simultaneous transmission is the Least Common Multiple (LCM) of the three interval times.
Simultaneous events that repeat at regular intervals align perfectly at the LCM of those individual intervals.
2
Apply the mathematical formula for the LCM of fractions
LCM of fractions=LCM of numeratorsHCF of denominators\text{LCM of fractions} = \frac{\text{LCM of numerators}}{\text{HCF of denominators}}
Calculating the LCM for fractions requires evaluating the numerators and denominators cross-layer using this specific formula.
3
Calculate the LCM of the numerators
LCM(8,12,24)=24\text{LCM}(8, 12, 24) = 24
The smallest integer that is perfectly divisible by 8, 12, and 24 is 24.
4
Calculate the HCF of the denominators
HCF(3,5,7)=1\text{HCF}(3, 5, 7) = 1
The numbers 3, 5, and 7 are all prime numbers and share no common divisor other than 1.
5
Compute the final result
241=24\frac{24}{1} = 24 seconds
Dividing the LCM of the numerators by the HCF of the denominators gives the synchronized time interval.

Key Concept

Calculating the Lowest Common Multiple (LCM) of fractional values for synchronization problems.
Question 82Question

A local water reservoir was initially filled to exactly 0.30.\overline{3} of its maximum capacity. After several days of heavy rainfall, an additional 14.414.4 million liters of water flowed into the reservoir, bringing the water level to 0.60.6 of its maximum capacity.

Currently, a nearby town consumes water from the reservoir at a steady rate of 1121\frac{1}{2} million liters per day. Additionally, an undiscovered leak at the base of the reservoir drains water at a constant rate of 0.050.05 million liters per hour.

If the reservoir were completely full and no further water was added, in how many days would it become completely empty?

Show answer & explanation

Answer: 20

Answer

20
The total capacity of the reservoir is found by equating the fractional increase (3513=415\frac{3}{5} - \frac{1}{3} = \frac{4}{15}) to the volume added (14.414.4 million liters), yielding 5454 million liters. The total daily water depletion is the sum of the town's daily usage (1.51.5 million liters) and the leak converted to a daily rate (0.05×24=1.20.05 \times 24 = 1.2 million liters), giving 2.72.7 million liters per day. Dividing 5454 by 2.72.7 gives exactly 2020 days.

Step-by-Step Solution

1
Convert the decimal fill levels into fractions and compute their difference.
Initial level = 13\frac{1}{3}. Final level = 35\frac{3}{5}. Difference = 3513=415\frac{3}{5} - \frac{1}{3} = \frac{4}{15}.
Working with exact fractions avoids rounding errors from recurring decimals and simplifies finding the proportion of water added.
2
Calculate the maximum capacity of the reservoir.
415×Capacity=14.4    Capacity=14.4×154=54\frac{4}{15} \times \text{Capacity} = 14.4 \implies \text{Capacity} = 14.4 \times \frac{15}{4} = 54 million liters.
The difference in the fractional water level represents exactly the volume of rain added.
3
Calculate the total volume of water lost per day.
Town = 1.51.5 million L/day. Leak = 0.05×24=1.20.05 \times 24 = 1.2 million L/day. Total = 2.72.7 million L/day.
The rates must be in the same time unit (days) before they can be accurately combined.
4
Determine the time required to empty the completely full reservoir.
54÷2.7=2054 \div 2.7 = 20 days.
Dividing the total capacity by the combined daily rate of depletion gives the time in days.

Key Concept

Fractions and Decimals
Estimated Time:3m 0s
Question 83Question

During a numerical assessment, a student is asked to calculate a final parameter NN by determining three specific mathematical components:

- Component AA: The mathematically correct positive remainder when 27-27 is divided by 44.
- Component BB: The unit digit of the number 21202^{120}.
- Component CC: The value of the arithmetic expression 8+12÷48 + 12 \div 4.

The parameter is then calculated using the formula N=A+B+C+7N = A + B + C + 7. Assuming the student evaluates all components correctly, which of the following best describes the classification of the number NN?

Show answer & explanation

Answer: An odd perfect square

Answer

The mathematically correct evaluation of the formula yields 25, which is classified as an odd perfect square.
When correctly calculated, Component A is 1, Component B is 6, and Component C is 11. Substituting these into the given formula yields N=1+6+11+7=25N = 1 + 6 + 11 + 7 = 25. Because 25 is an odd number and can be expressed as 525^2, it correctly falls under the classification of an odd perfect square.

Step-by-Step Solution

1
Calculate Component A (the positive remainder of 27÷4-27 \div 4)
A=1A = 1
By definition of division, 27=4×(7)+1-27 = 4 \times (-7) + 1. The remainder must be a non-negative integer less than the divisor.
2
Calculate Component B (the unit digit of 21202^{120})
B=6B = 6
The unit digit of powers of 2 follows a cycle of 4 (2, 4, 8, 6). Since 120 is perfectly divisible by 4, the unit digit corresponds to the 4th position in the cycle, which is 6.
3
Calculate Component C (the value of 8+12÷48 + 12 \div 4)
C=11C = 11
According to the BODMAS rule, division must be performed before addition. First, 12÷4=312 \div 4 = 3. Then, 8+3=118 + 3 = 11.
4
Calculate the final parameter NN using the formula N=A+B+C+7N = A + B + C + 7
N=1+6+11+7=25N = 1 + 6 + 11 + 7 = 25
Substituting the correctly evaluated components yields 2525, which is equal to 525^2 (an odd perfect square).

Key Concept

Classification of numbers based on proper mathematical operations, modular arithmetic, and cyclicity rules.
Question 84Question

An interior designer is planning to cover the lengths of three different hallways with square marble tiles. The lengths of the three hallways are 154\frac{15}{4} meters, 256\frac{25}{6} meters, and 358\frac{35}{8} meters. To avoid cutting any marble tiles, the designer wants to use the largest possible square tiles that can perfectly measure these lengths.

What is the maximum possible side length of each square tile?

Show answer & explanation

Answer: 524\frac{5}{24} meters

Answer

524\frac{5}{24} meters
The maximum possible side length that perfectly measures all three hallway lengths is found by calculating the Highest Common Factor (HCF) of the three fractions. According to the mathematical rule for fractions, this is the HCF of the numerators (15,25,3515, 25, 35) divided by the LCM of the denominators (4,6,84, 6, 8), which yields 524\frac{5}{24} meters.

Step-by-Step Solution

1
Identify the mathematical operation required.
Calculate the Highest Common Factor (HCF) of the three fractions to find the maximum possible side length.
The tiles must perfectly measure the lengths without being cut, meaning the tile length must be a common factor. 'Largest possible' indicates the Highest Common Factor.
2
Recall the formula for finding the HCF of fractions.
HCF of fractions = HCF of numeratorsLCM of denominators\frac{\text{HCF of numerators}}{\text{LCM of denominators}}
This is the standard rule for determining the highest common divisor among fractional values.
3
Calculate the HCF of the numerators.
Numerators are 1515, 2525, and 3535. Their HCF is 55.
The largest integer that perfectly divides 1515, 2525, and 3535 is 55.
4
Calculate the LCM of the denominators.
Denominators are 44, 66, and 88. Their LCM is 2424.
The smallest integer that is a multiple of 44, 66, and 88 is 2424.
5
Apply the results to the formula.
524\frac{5}{24} meters.
Substituting the computed values into the fraction formula provides the final answer.

Key Concept

HCF of fractions
Question 85Question

What is the remainder when 4154^{15} is divided by 55?

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Answer: 44

Answer

The remainder when 4154^{15} is divided by 55 is 44.
Since 41(mod5)4 \equiv -1 \pmod 5, raising to the 15th power gives (1)15=1(mod5)(-1)^{15} = -1 \pmod 5. Adding the divisor 55 converts the negative remainder 1-1 into the valid positive remainder 44.

Step-by-Step Solution

1
Express the base number in modular form relative to the divisor.
4=514 = 5 - 1, which means 41(mod5)4 \equiv -1 \pmod 5.
Expressing the base as 1-1 modulo 55 simplifies exponent calculation.
2
Raise the modular congruence to the power of 15.
415(1)15(mod5)=1(mod5)4^{15} \equiv (-1)^{15} \pmod 5 = -1 \pmod 5.
An odd power of 1-1 evaluates to 1-1.
3
Convert the negative remainder into a positive remainder.
1+5=4-1 + 5 = 4.
Standard remainders in integer division must be non-negative integers less than the divisor.

Key Concept

Remainder Theorem and Negative Remainder Conversion
Question 86Question

If the six-digit number 45231x45231x is completely divisible by 99, what is the value of the digit xx?

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Answer: 3

Answer

3
According to the divisibility rule for 9, a number is divisible by 9 if the sum of its digits is a multiple of 9. For the number 45231x45231x, the sum of the digits is 4+5+2+3+1+x=15+x4 + 5 + 2 + 3 + 1 + x = 15 + x. The smallest multiple of 9 that is greater than or equal to 15 is 18. Setting 15+x=1815 + x = 18 gives x=3x = 3.

Step-by-Step Solution

1
Find the sum of the known digits in the given number.
4+5+2+3+1=154 + 5 + 2 + 3 + 1 = 15
The divisibility rule for 9 requires analyzing the sum of all digits.
2
Formulate the condition for divisibility by 9.
15+x15 + x must be a multiple of 9.
Including the unknown unit digit xx, the total digit sum is 15+x15 + x.
3
Solve for the single-digit integer xx where 0x90 \le x \le 9.
x=3x = 3
The smallest multiple of 9 greater than or equal to 15 is 18, giving 15+x=1815 + x = 18, so x=3x = 3.

Key Concept

Divisibility Rule for 9

Alternative Method

Dividing 452,310 by 9 yields 50,256 with a remainder of 6. To make the number divisible by 9, the remaining amount needed is 96=39 - 6 = 3, so the unit digit xx must be 3.
Estimated Time:45s
Question 87Question

A municipal corporation is distributing its annual infrastructure development fund. Exactly 0.50.\overline{5} of the total fund is allocated to urban housing projects. From the remaining fund, 37\frac{3}{7} is assigned to public transportation. The rest of the fund is distributed equally between two sectors: renewable energy and public parks. If the amount allocated to renewable energy is $3.2\$3.2 million, what is the total annual infrastructure development fund?

Show answer & explanation

Answer: $25.2\$25.2 million

Answer

$25.2\$25.2 million
The correct sequence evaluates the recurring decimal 0.50.\overline{5} as 59\frac{5}{9}. This leaves 49\frac{4}{9} of the fund. Transport takes 37\frac{3}{7} of this remainder, which is 1263\frac{12}{63}. The new remainder is 491263=1663\frac{4}{9} - \frac{12}{63} = \frac{16}{63}. This is split equally, meaning renewable energy gets 863\frac{8}{63} of the total. Solving 863x=3.2\frac{8}{63}x = 3.2 gives x=25.2x = 25.2.

Step-by-Step Solution

1
Convert the recurring decimal to a fraction.
0.5=590.\overline{5} = \frac{5}{9}
Fractional form is required to perform precise subsequent operations without rounding errors.
2
Calculate the remaining fraction of the fund after the housing allocation.
159=491 - \frac{5}{9} = \frac{4}{9}
The next allocation is taken from the remaining fund, not the total fund.
3
Determine the fraction of the total fund allocated to public transportation.
37×49=1263=421\frac{3}{7} \times \frac{4}{9} = \frac{12}{63} = \frac{4}{21}
Multiplying the proportion (37\frac{3}{7}) by the available remainder (49\frac{4}{9}) yields its share of the total.
4
Calculate the new remaining fund for the last two sectors.
49421=28631263=1663\frac{4}{9} - \frac{4}{21} = \frac{28}{63} - \frac{12}{63} = \frac{16}{63}
Subtracting the transportation allocation from the previous remainder gives the final undistributed portion.
5
Determine the fraction of the total fund allocated to renewable energy.
12×1663=863\frac{1}{2} \times \frac{16}{63} = \frac{8}{63}
The problem states the final remainder is split equally between two sectors.
6
Equate the fraction to the given monetary value and solve for the total fund (xx).
863x=3.2    x=3.2×638=0.4×63=25.2\frac{8}{63}x = 3.2 \implies x = 3.2 \times \frac{63}{8} = 0.4 \times 63 = 25.2
This establishes the algebraic relationship to find the initial total.

Key Concept

Successive Fractional Operations and Recurring Decimals

Alternative Method

Instead of calculating absolute remaining fractions through subtraction at each step, you can chain the remaining proportions through direct multiplication: Total × (1 - 5/9) × (1 - 3/7) × (1/2) = Total × (4/9) × (4/7) × (1/2) = Total × (8/63). Setting (8/63) equal to 3.2 million quickly yields 25.2 million.
Estimated Time:2m 30s
Question 88Question

A student is reviewing fundamental number properties and evaluating several arithmetic scenarios. Identify which of the following assertions regarding number classification are mathematically valid. (Select all that apply)

Select all that apply

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Answer: The product of any non-zero rational number and an irrational number is always an irrational number.; Every non-terminating, recurring decimal expansion can be expressed as a rational number.

Answer

The mathematically valid assertions are that the product of a non-zero rational and an irrational number is always irrational, and that every non-terminating, recurring decimal is a rational number.
The assertions stating that the product of a non-zero rational and an irrational number is irrational, and that recurring decimals are rational, are fundamental and mathematically sound properties of the real number system.

Step-by-Step Solution

1
Evaluate the first assertion regarding the product of rational and irrational numbers.
The statement is determined to be valid.
Multiplying a ratio of integers (non-zero) by a non-repeating decimal inherently yields a non-repeating decimal.
2
Evaluate the arithmetic expression in the second assertion.
The expression equals 1717, which is a prime number.
Following order of operations, multiplication precedes addition (3×5=153 \times 5 = 15, then 15+2=1715 + 2 = 17).
3
Determine the unit digit for the third assertion.
The unit digit is 66, which is not a perfect square.
Because 4040 is a multiple of 44 (the cyclicity of base 22), the unit digit corresponds to the 44 th power in the cycle (24=162^4 = 16).
4
Evaluate the fourth assertion regarding recurring decimals.
The statement is determined to be valid.
All repeating decimals can be mathematically converted into a fraction format (e.g., 0.333...0.333... becomes 1/31/3), satisfying the definition of a rational number.

Key Concept

Classification of Rational and Irrational Numbers
Question 89Question

An inventory analyst is auditing two specific electronic components in a warehouse. He calculates that the product of their exact unit quantities is 32,17532,175, and the Highest Common Factor (HCF) of these two quantities is 1515. If there are more than 100100 units of each component currently in stock, what is the total combined quantity of both components?

Show answer & explanation

Answer: 360

Answer

The correct total combined quantity is 360 units.
By representing the numbers as 15x15x and 15y15y, we determine that 15x×15y=32,17515x \times 15y = 32,175, which simplifies to xy=143xy = 143. Factoring 143143 into co-prime pairs gives (1,143)(1, 143) and (11,13)(11, 13). Multiplying these by the HCF (1515) gives potential quantities of (15,2145)(15, 2145) and (165,195)(165, 195). Only the pair (165,195)(165, 195) satisfies the condition that both quantities must be greater than 100100. Their sum is 165+195=360165 + 195 = 360.

Step-by-Step Solution

1
Express the two unknown quantities mathematically using their HCF.
Let the quantities be 15x15x and 15y15y, where xx and yy are co-prime integers.
Since the HCF of the two numbers is 1515, both numbers must be multiples of 1515. Their remaining factors (xx and yy) cannot share any common factors other than 11.
2
Formulate an equation using the given product of the quantities.
15x×15y=32,17515x \times 15y = 32,175
The problem states the product of the two component quantities is equal to 32,17532,175.
3
Solve the equation for the product of the co-prime variables xx and yy.
225xy=32,175xy=32,175225=143225xy = 32,175 \Rightarrow xy = \frac{32,175}{225} = 143
Isolating xyxy simplifies the problem to finding two co-prime factors that multiply to 143143.
4
Identify all co-prime factor pairs of 143143.
The co-prime pairs are (1,143)(1, 143) and (11,13)(11, 13).
We must list all integer pairs that multiply to 143143 and verify they share no common divisors.
5
Calculate the possible original quantities and apply the boundary constraints.
Pair 1 gives (15,2145)(15, 2145). Pair 2 gives (165,195)(165, 195). Because the problem states both quantities are >100>100, we must choose (165,195)(165, 195).
The constraint strictly eliminates the first pair, as 1515 is not greater than 100100.
6
Sum the valid quantities.
165+195=360165 + 195 = 360
The question asks for the total combined quantity of both components.

Key Concept

The relationship between the Highest Common Factor (HCF) and the product of two numbers, utilizing co-prime factor pairs.
Question 90Question

A mathematician is formally classifying a set of numbers and analyzing their properties. Evaluate the mathematical validity of the following three propositions:

I. The number 00 is an even integer, but it is considered neither positive nor negative.
II. The fraction 227\frac{22}{7} is an irrational number because it is commonly used to represent the constant π\pi.
III. The absolute difference between any two distinct odd prime numbers is always an even composite number.

Which of the propositions given above is/are mathematically correct?

Show answer & explanation

Answer: Only Proposition I

Answer

Only Proposition I is mathematically correct.
Proposition I is correct because 0 is an integer that is divisible by 2 (making it even) and is strictly the neutral boundary between positive and negative values. Proposition II is incorrect because 22/7 is expressed as a ratio of two integers, making it a rational number, despite being used as an approximation for pi. Proposition III is incorrect because the difference between the odd prime numbers 5 and 3 is exactly 2, which is an even prime number rather than a composite number.

Step-by-Step Solution

1
Evaluate Proposition I regarding the properties of the number 0.
The number 0 is an integer, is divisible by 2 without a remainder (making it even), and serves as the boundary between positive and negative numbers. Therefore, Proposition I is true.
To verify the fundamental classification and parity of zero.
2
Evaluate Proposition II regarding the classification of the fraction 22/7.
The value 22/7 is a ratio of two integers (22 and 7). Any number that can be expressed as a fraction of integers with a non-zero denominator is a rational number. Therefore, Proposition II is false.
To distinguish between a rational fraction and the actual irrational constant (pi) it often approximates.
3
Evaluate Proposition III regarding the absolute difference between distinct odd prime numbers.
Consider the distinct odd prime numbers 5 and 3. Their absolute difference is |5 - 3| = 2. Since 2 is a prime number, the difference is not always an even composite number. Therefore, Proposition III is false.
To test the universal claim with a basic mathematical counterexample.

Key Concept

Classification of Rational vs. Irrational Numbers, Prime Numbers, and Properties of Zero
Question 91Question

A digital signal processing unit transmits three distinct telemetry data packets in continuous periodic cycles. The transmission cycles for the three packets take 1415\frac{14}{15} seconds, 75\frac{7}{5} seconds, and 6320\frac{63}{20} seconds, respectively. If all three packets are transmitted simultaneously at the beginning of a diagnostic test, after how many seconds will all three packets be transmitted together again for the first time?

Show answer & explanation

Answer: 1265\frac{126}{5}

Answer

1265\frac{126}{5}
To determine when all packets will transmit simultaneously again, we calculate the Least Common Multiple (LCM) of their cycle periods. The LCM of fractions is defined as the LCM of their numerators divided by the HCF of their denominators. Given the numerators (14, 7, 63) and denominators (15, 5, 20), we find LCM(14, 7, 63) = 126 and HCF(15, 5, 20) = 5. Dividing these yields the exact time of 1265\frac{126}{5} seconds.

Step-by-Step Solution

1
Identify the mathematical operation required.
The problem asks for the next simultaneous transmission of periodic cycles, which requires finding the Least Common Multiple (LCM) of the given time intervals: 1415\frac{14}{15}, 75\frac{7}{5}, and 6320\frac{63}{20}.
Repeating simultaneous events align at time intervals that are common multiples of their individual periods. The first alignment corresponds to the least common multiple.
2
Apply the rule for finding the LCM of fractions.
The correct mathematical formula is LCM=LCM of numeratorsHCF of denominators\text{LCM} = \frac{\text{LCM of numerators}}{\text{HCF of denominators}}.
This is the required procedure for finding the LCM of rational numbers expressed as simplified fractions.
3
Determine the LCM of the numerators.
The numerators are 1414, 77, and 6363. Their prime factorizations are 14=2×714 = 2 \times 7, 7=77 = 7, and 63=32×763 = 3^2 \times 7. Thus, LCM(14,7,63)=2×32×7=126\text{LCM}(14, 7, 63) = 2 \times 3^2 \times 7 = 126.
The LCM must consist of the highest power of all prime factors present among the numerators.
4
Determine the HCF of the denominators.
The denominators are 1515, 55, and 2020. Their prime factorizations are 15=3×515 = 3 \times 5, 5=55 = 5, and 20=22×520 = 2^2 \times 5. The Highest Common Factor(15,5,20)=5\text{Highest Common Factor}(15, 5, 20) = 5.
The HCF is the largest prime factor shared by all the denominators.
5
Calculate the final LCM of the transmission cycles.
Divide the LCM of the numerators by the HCF of the denominators to obtain exactly 1265\frac{126}{5} seconds.
Applying the values derived in the previous steps to the fraction formula yields the exact coincidence time.

Key Concept

Calculating the LCM of fractions to synchronize periodic events.
Question 92Question

During a system stress test, four data processing modules record their average latency per transaction (in seconds) as given below. Arrange the modules in ascending order based on their latency times (from fastest to slowest).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct ascending sequence from fastest to slowest latency is Module Y, followed by Module X, then Module W, and finally Module Z.
By converting all latency values into extended decimals, we find Module Y (0.25656...0.25656...) is the smallest, followed by Module X (0.25714...0.25714...), then Module W (0.25757...0.25757...), and finally Module Z (0.25777...0.25777...). Therefore, the correct ascending order is Module Y, Module X, Module W, Module Z.

Step-by-Step Solution

1
Calculate the exact value for Module W using the HCF of fractions formula.
HCF = HCF(34,85)LCM(33,22)=1766\frac{\text{HCF}(34, 85)}{\text{LCM}(33, 22)} = \frac{17}{66}. Converted to a decimal, this is 0.25757...0.25757... (0.2570.2\overline{57}).
To standardize the mathematical expression into a comparable decimal format.
2
Convert the value for Module X into a decimal.
9350.2571428...\frac{9}{35} \approx 0.2571428...
Decimals allow for direct digit-by-digit comparison.
3
Expand the recurring decimal for Module Y to several decimal places.
0.256=0.2565656...0.2\overline{56} = 0.2565656...
To provide enough precision to compare digits past the third decimal place.
4
Convert the recurring decimal in Module Z to a fraction, multiply, and convert back to a decimal.
1.28=1281290=11690=58451.2\overline{8} = \frac{128-12}{90} = \frac{116}{90} = \frac{58}{45}. Then, 5845×15=58225\frac{58}{45} \times \frac{1}{5} = \frac{58}{225}. Converting to decimal gives 0.25777...0.25777... (0.2570.25\overline{7}).
To determine its exact decimal value for accurate comparison against the other modules.
5
Compare the decimal values digit by digit from left to right to establish the ascending order.
0.25656...0.25656... (Y) < 0.25714...0.25714... (X) < 0.25757...0.25757... (W) < 0.25777...0.25777... (Z).
The question asks for the fastest to slowest modules, which corresponds to the shortest to longest latency times (ascending numerical order).

Key Concept

Advanced comparison of fractions, recurring decimals, and the application of HCF/LCM rules for fractions.

Alternative Method

Find a common denominator for all fractions: 66 for W (17/66), 35 for X (9/35), 990 for Y (254/990), and 225 for Z (58/225). The LCM of 66, 35, 990, and 225 is 34650. Convert all fractions to this denominator to compare the exact numerators, though decimal conversion to the fourth digit is faster.
Estimated Time:2m 30s
Question 93Question

Consider all positive two-digit integers where the sum of their digits is exactly 1010. How many of these integers are prime numbers?

Show answer & explanation

Answer: 3

Answer

3
By listing all two-digit numbers with a digit sum of 10 (19, 28, 37, 46, 55, 64, 73, 82, 91) and testing them for primality, we find that only 19, 37, and 73 are prime. The number 91 is composite (7×137 \times 13), resulting in exactly 3 prime numbers meeting the condition.

Step-by-Step Solution

1
Identify all two-digit positive integers whose digits sum to 10.
The numbers are 19, 28, 37, 46, 55, 64, 73, 82, and 91.
Systematically listing the pairs of digits (from 1 to 9) that add up to 10 ensures no valid numbers are missed.
2
Evaluate each number to determine if it is prime or composite.
Even numbers (28, 46, 64, 82) and multiples of 5 (55) are composite. The remaining odd numbers to check are 19, 37, 73, and 91.
Applying basic divisibility rules for 2 and 5 quickly eliminates obvious composite numbers.
3
Test the remaining odd numbers for primality.
19, 37, and 73 are prime numbers. 91 is a composite number because 91=7×1391 = 7 \times 13.
A prime number has exactly two distinct positive divisors: 1 and itself. 91 is a common trap as its divisibility by 7 is not always immediately obvious.
4
Count the total number of prime numbers identified.
There are exactly 3 prime numbers in the set: 19, 37, and 73.
This directly answers the specific question asked in the stem.

Key Concept

Identification of Prime and Composite Numbers
Question 94Question

An environmental agency uses three autonomous drones to patrol a protected reserve. The drones fly on continuous looping routes. Drone X completes one full route in 4215\frac{42}{15} hours, Drone Y completes a route in 3520\frac{35}{20} hours, and Drone Z completes a route in 6330\frac{63}{30} hours.

If all three drones depart simultaneously from the base station, after how many hours will they all meet at the base station again for the first time?

Show answer & explanation

Answer: 42

Answer

42
The drones will meet again at a time that is a common multiple of all their individual cycle times. The very first time this happens is represented by the Least Common Multiple (LCM). When calculating the LCM of fractions, it is mathematically required to reduce them to their simplest terms first: 145\frac{14}{5}, 74\frac{7}{4}, and 2110\frac{21}{10}. Using the correct formula, LCM(14,7,21)HCF(5,4,10)\frac{\text{LCM}(14, 7, 21)}{\text{HCF}(5, 4, 10)}, we get 421=42\frac{42}{1} = 42 hours.

Step-by-Step Solution

1
Identify the mathematical operation required.
The convergence time is the Least Common Multiple (LCM) of the three cycle times.
The drones will meet again at a time that is a common multiple of all their individual route completion times.
2
Simplify the given fractions to their lowest terms.
4215145\frac{42}{15} \rightarrow \frac{14}{5}; 352074\frac{35}{20} \rightarrow \frac{7}{4}; 63302110\frac{63}{30} \rightarrow \frac{21}{10}
The formula for the LCM of fractions requires all fractions to be in their simplest form to yield the correct result.
3
Calculate the LCM of the simplified numerators.
LCM(14,7,21)=42\text{LCM}(14, 7, 21) = 42
The numerator of the resulting fraction must be divisible by all original numerators.
4
Calculate the HCF of the simplified denominators.
HCF(5,4,10)=1\text{HCF}(5, 4, 10) = 1
The denominator of the resulting fraction must evenly divide all original denominators.
5
Compute the final fraction.
421=42\frac{42}{1} = 42 hours
Applying the formula LCM of NumeratorsHCF of Denominators\frac{\text{LCM of Numerators}}{\text{HCF of Denominators}} gives the exact time of the next simultaneous meeting.

Key Concept

Calculating the Least Common Multiple (LCM) of fractions, emphasizing the critical prerequisite of simplifying the fractions first.
Estimated Time:2m 30s
Question 95Question

Consider the following statements regarding the properties of numbers:

Statement 1: The integer 00 is neither positive nor negative, but it is classified as an even rational number.
Statement 2: The addition of any rational number and any irrational number always results in an irrational number.
Statement 3: The fraction 227\frac{22}{7} is an irrational number because it represents the exact value of π\pi.

Which of the statements given above is/are mathematically correct?

Show answer & explanation

Answer: 1 and 2 only

Answer

The correct answer includes Statement 1 and Statement 2 only.
The correct answer correctly identifies that Statement 1 and Statement 2 are mathematically sound truths, while Statement 3 is fundamentally flawed. Statement 1 holds because 00 satisfies the definition of an even number (2n2n) and a rational number (pq\frac{p}{q}). Statement 2 holds due to mathematical contradiction proofs regarding sums. Statement 3 is false because 227\frac{22}{7} is a rational number by definition, despite being an approximation for the irrational π\pi.

Step-by-Step Solution

1
Evaluate the mathematical properties of 00 in Statement 1.
Statement 1 is valid.
Zero has no sign (neither positive nor negative). It is an even integer because it can be expressed as 2×02 \times 0. It is a rational number because it can be written as 01\frac{0}{1}.
2
Analyze the closure property of addition for Statement 2.
Statement 2 is valid.
If a rational number rr is added to an irrational number xx, the sum must be irrational. (If r+x=qr + x = q where qq is rational, then x=qrx = q - r, meaning xx would be rational, which contradicts the premise).
3
Examine the classification of 227\frac{22}{7} in Statement 3.
Statement 3 is invalid.
The fraction 227\frac{22}{7} is the ratio of two integers, which strictly defines it as a rational number. It is merely a common numerical approximation for π\pi, not the exact irrational value of π\pi itself.

Key Concept

Core properties of rational/irrational numbers and fundamental definitions of integers like zero.
Question 96Question

An industrial chemical plant stores a specialized solvent in three large cylindrical vats. The first vat contains 53.353.\overline{3} liters, the second contains 71.171.\overline{1} liters, and the third contains 26.626.\overline{6} liters of the solvent. The plant manager wants to completely transfer the solvent from all three vats into identical, smaller drums such that every drum is completely filled, no solvent is left over in any vat, and solvents from different vats are not mixed. What should be the maximum possible capacity of each drum?

Show answer & explanation

Answer: 8.88.\overline{8} liters

Answer

The maximum possible capacity of each drum is 8.88.\overline{8} liters.
To find the maximum identical capacity that leaves no remainder, we must calculate the Highest Common Factor (HCF) of the three volumes. By converting the recurring decimals to fractions (1603\frac{160}{3}, 6409\frac{640}{9}, 803\frac{80}{3}) and applying the fraction HCF formula (HCF of numeratorsLCM of denominators\frac{\text{HCF of numerators}}{\text{LCM of denominators}}), we obtain 809\frac{80}{9}, which perfectly translates to 8.88.\overline{8} liters.

Step-by-Step Solution

1
Convert the given recurring decimals representing the solvent volumes into their simplest fractional forms.
53.3=53+39=160353.\overline{3} = 53 + \frac{3}{9} = \frac{160}{3}, 71.1=71+19=640971.\overline{1} = 71 + \frac{1}{9} = \frac{640}{9}, and 26.6=26+69=80326.\overline{6} = 26 + \frac{6}{9} = \frac{80}{3}.
Fractional forms are required to accurately and properly compute the highest common factor (HCF) of non-integer values.
2
Identify the mathematical operation required based on the physical constraints described in the problem.
We must calculate the Highest Common Factor (HCF) of the three volumes: 1603\frac{160}{3}, 6409\frac{640}{9}, and 803\frac{80}{3}.
The solvent must be divided equally without any remainders, meaning the drum size must be a common factor of all three initial volumes, and the problem asks for the 'maximum possible capacity'.
3
Apply the standard formula for finding the HCF of multiple fractions.
The formula is: HCF=HCF of numeratorsLCM of denominators\text{HCF} = \frac{\text{HCF of numerators}}{\text{LCM of denominators}}.
This formula ensures the resulting fraction will evenly divide each of the original fractions without leaving a remainder.
4
Calculate the HCF of the numerators (160160, 640640, 8080) and the LCM of the denominators (33, 99, 33).
HCF(160,640,80)=80\text{HCF}(160, 640, 80) = 80, and LCM(3,9,3)=9\text{LCM}(3, 9, 3) = 9. Thus, the HCF of the fractions is 809\frac{80}{9}.
8080 is the largest integer dividing 160160, 640640, and 8080. 99 is the smallest integer divisible by 33, 99, and 33.
5
Convert the resulting fraction back into a recurring decimal.
809=8+89=8.8\frac{80}{9} = 8 + \frac{8}{9} = 8.\overline{8} liters.
The final calculated capacity should match the formatting style of the given options.

Key Concept

Calculating the Highest Common Factor (HCF) of recurring decimals by converting them to fractions and using the fraction HCF rule.
Estimated Time:2m 30s
Question 97Question

A boutique chocolatier is preparing special gift assortments. They have three large blocks of premium cocoa weighing 212\frac{21}{2} kg, 354\frac{35}{4} kg, and 498\frac{49}{8} kg. They must mold these blocks into the largest possible identical solid chocolate bars such that no cocoa is left over from any of the three blocks. What should be the exact weight of each chocolate bar?

Show answer & explanation

Answer: 78\frac{7}{8} kg

Answer

The exact weight of each chocolate bar should be 78\frac{7}{8} kg.
To find the maximum identical weight that can perfectly divide all three cocoa blocks, we must calculate the Highest Common Factor (HCF) of the three fractions. The formula dictates finding the HCF of the numerators (21, 35, 49), which is 7, and dividing it by the LCM of the denominators (2, 4, 8), which is 8. This results in the correct weight of 7/8 kg.

Step-by-Step Solution

1
Identify the mathematical operation required.
The problem asks for the 'largest possible identical' divisions of the blocks with no remainders, meaning we must find the Highest Common Factor (HCF) of the given fractional weights.
HCF provides the maximum uniform size that can perfectly divide a given set of quantities.
2
Recall the formula for finding the HCF of fractions.
HCF of fractions = (HCF of Numerators) / (LCM of Denominators).
This is the standard mathematical rule for calculating the greatest common divisor for fractional values.
3
Calculate the HCF of the numerators: 21, 35, and 49.
The highest common factor for 21, 35, and 49 is 7.
21 = 3 × 7; 35 = 5 × 7; 49 = 7 × 7. The greatest shared prime factor is 7.
4
Calculate the LCM of the denominators: 2, 4, and 8.
The lowest common multiple for 2, 4, and 8 is 8.
8 is a multiple of both 2 and 4, making it the smallest common denominator.
5
Apply the calculated values to the fraction formula.
78\frac{7}{8} kg.
Combining the calculated numerator HCF and denominator LCM yields the final maximum weight.

Key Concept

HCF of Fractions
Question 98Question

When evaluating the fundamental categories and properties of the real number system, which of the following statements are mathematically correct?

Select all that apply

Show answer & explanation

Answer: The square root of any prime number inherently belongs to the set of irrational numbers.; Every integer can be expressed as a rational number in the form pq\frac{p}{q}, where qq is a non-zero integer.

Answer

The mathematically correct statements are that the square root of any prime number is irrational, and every integer can be expressed as a rational number.
The correct statements accurately identify that prime numbers lack perfect square roots, making their roots inherently irrational, and that all integers are essentially a subset of rational numbers since any integer can be expressed as a fraction with a denominator of one.

Step-by-Step Solution

1
Analyze the statement regarding prime numbers and irrationality.
Since prime numbers only have two distinct positive divisors, they are never perfect squares. Therefore, their square roots cannot be simplified into rational fractions, making the statement correct.
To verify the relationship between primes and algebraic irrationality.
2
Evaluate the properties of zero given in the second statement.
Zero is determined to be neutral (neither positive nor negative) and even (since 0=2×00 = 2 \times 0). Therefore, the statement claiming it is positive and neither odd nor even is false.
To check for misconceptions regarding the classification and parity of zero.
3
Assess the definition of rational numbers applied to integers.
An integer zz can always be written as z1\frac{z}{1}. Since 11 is a non-zero integer, this satisfies the pq\frac{p}{q} definition of rational numbers, making the statement correct.
To confirm the subset relationship between integers and rational numbers.
4
Examine the classification of π\pi.
While 227\frac{22}{7} is a common approximation, π\pi cannot be written as a simple fraction of integers. It is irrational, making the statement false.
To differentiate between an exact mathematical constant and its heuristic approximation.

Key Concept

Classification of Numbers and Fundamental Properties
Question 99Question

A large agricultural cooperative is dividing a massive tract of land for different crops. They allocate 0.4285710.\overline{428571} of the total land to cultivate sunflowers, 0.160.1\overline{6} of the total land to cultivate maize, and 0.050.0\overline{5} of the total land to cultivate organic vegetables. The remaining land, which measures exactly 110110 hectares, is preserved as a wildlife reserve. What is the total area of the tract of land, in hectares?

Show answer & explanation

Answer: 315

Answer

315
The total area is found by properly converting all recurring decimals into exact fractions (3/73/7, 1/61/6, and 1/181/18), summing them to find the total allocated land (41/6341/63), determining the remaining land fraction (22/6322/63), and setting it equal to the given 110110 hectares. Solving for the whole yields exactly 315315 hectares.

Step-by-Step Solution

1
Convert the pure recurring decimal 0.4285710.\overline{428571} into a simplified fraction.
The fraction is 37\frac{3}{7}.
Recognizing that 17=0.142857\frac{1}{7} = 0.\overline{142857}, we can multiply by 33 to get 0.4285710.\overline{428571}. Alternatively, using the algebraic method: 428571999999=37\frac{428571}{999999} = \frac{3}{7}.
2
Convert the mixed recurring decimals 0.160.1\overline{6} and 0.050.0\overline{5} into fractions.
0.16=16190=1590=160.1\overline{6} = \frac{16-1}{90} = \frac{15}{90} = \frac{1}{6} and 0.05=5090=590=1180.0\overline{5} = \frac{5-0}{90} = \frac{5}{90} = \frac{1}{18}.
To operate with mixed recurring decimals, subtract the non-repeating part from the entire number, and place it over a denominator consisting of 9s (for repeating digits) followed by 0s (for non-repeating digits after the decimal point).
3
Calculate the total fraction of land allocated to the three crops.
37+16+118=37+3+118=37+418=37+29=27+1463=4163\frac{3}{7} + \frac{1}{6} + \frac{1}{18} = \frac{3}{7} + \frac{3+1}{18} = \frac{3}{7} + \frac{4}{18} = \frac{3}{7} + \frac{2}{9} = \frac{27+14}{63} = \frac{41}{63}.
Finding a common denominator (6363) allows us to sum the individual crop fractions to determine the total proportion of cultivated land.
4
Determine the fraction representing the wildlife reserve and calculate the total land area.
Reserve fraction = 14163=22631 - \frac{41}{63} = \frac{22}{63}. Total Area = 110×6322=315110 \times \frac{63}{22} = 315 hectares.
The unallocated fraction represents the reserve area. Setting this fraction of the total area (TT) equal to 110110 hectares (2263×T=110\frac{22}{63} \times T = 110) gives the final answer.

Key Concept

Fractions and Decimals
Estimated Time:2m 30s
Question 100Question

A pharmaceutical laboratory uses three automated dispensers to add chemical reagents into a continuous reaction vessel. Dispenser X adds a drop every 149\frac{14}{9} seconds, Dispenser Y every 356\frac{35}{6} seconds, and Dispenser Z every 4912\frac{49}{12} seconds. If all three dispensers release a drop simultaneously, what is the minimum time interval (in seconds) until they all release a drop together again?

Show answer & explanation

Answer: 4903\frac{490}{3}

Answer

The minimum time interval is 4903\frac{490}{3} seconds.
The correct answer is found by applying the LCM formula for fractions. Since the events repeat over time, their next simultaneous occurrence happens at a time that is a common multiple of all three periods. By calculating the LCM of the numerators (14,35,4949014, 35, 49 \rightarrow 490) and dividing it by the HCF of the denominators (9,6,1239, 6, 12 \rightarrow 3), we get 4903\frac{490}{3} seconds.

Step-by-Step Solution

1
Identify the mathematical operation required.
We need to find the Lowest Common Multiple (LCM) of the three fractional time intervals.
Simultaneous periodic events repeat exactly at the LCM of their individual periods.
2
State the formula for the LCM of fractions.
LCM of fractions = LCM of numeratorsHCF of denominators\frac{\text{LCM of numerators}}{\text{HCF of denominators}}.
This standard formula ensures the resulting value is a multiple of each original fraction.
3
Calculate the LCM of the numerators.
The numerators are 1414, 3535, and 4949. Their prime factorizations are 14=2×714 = 2 \times 7, 35=5×735 = 5 \times 7, and 49=7249 = 7^2. The LCM is 2×5×72=4902 \times 5 \times 7^2 = 490.
The LCM must include the highest power of all prime factors present in the numbers.
4
Calculate the HCF of the denominators.
The denominators are 99, 66, and 1212. Their prime factorizations are 9=329 = 3^2, 6=2×36 = 2 \times 3, and 12=22×312 = 2^2 \times 3. The highest common factor is 33.
The HCF is the largest positive integer that divides each of the numbers without leaving a remainder.
5
Apply the formula to find the final LCM.
LCM = 4903\frac{490}{3}.
Substituting the calculated numerator LCM and denominator HCF into the fraction formula.

Key Concept

Lowest Common Multiple (LCM) of fractions for periodic events
Estimated Time:1m 30s
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