Basic Numeracy Practice Questions — State PSC Exam — Page 7

Question 121Question →

An artisan is preparing metal rods for a custom fence. Two existing rods, which measure $\frac{15}{4}$ meters and $\frac{25}{8}$ meters in length, must be cut into identical smaller pieces of the maximum possible length without wasting any material. What should be the length of each piece?

\frac{75}{4}

meters

\frac{5}{4}

meters

\frac{5}{8}

meters

\frac{75}{8}

meters

Show answer & explanation

Answer: $\frac{5}{8}$ meters

Answer

\frac{5}{8}

meters

The maximum possible length of each piece is found by calculating the HCF of the two fractional lengths. The correct formula is to divide the HCF of the numerators (15 and 25) by the LCM of the denominators (4 and 8). The HCF of 15 and 25 is 5, and the LCM of 4 and 8 is 8, resulting in exactly

\frac{5}{8}

meters.

Step-by-Step Solution

1

Identify the mathematical operation required for the scenario.

Finding the maximum possible identical length from two given lengths requires calculating their Highest Common Factor (HCF).

The pieces must be of equal length and as large as possible without leaving any remainder.

2

State the formula for finding the HCF of fractions.

\text{HCF of fractions} = \frac{\text{HCF of numerators}}{\text{LCM of denominators}}

This is the standard rule for determining the greatest common divisor of rational numbers.

3

Calculate the HCF of the numerators (15 and 25).

The factors of 15 are 1, 3, 5, 15. The factors of 25 are 1, 5, 25. The highest common factor is 5.

The numerator of our final answer must be the HCF of the original numerators.

4

Calculate the LCM of the denominators (4 and 8).

The multiples of 8 (8, 16, 24...) are already divisible by 4. Thus, the least common multiple is 8.

The denominator of our final answer must be the LCM of the original denominators.

5

Construct the final fraction.

\frac{5}{8}

meters.

Combining the results from the previous steps yields the correct maximum length.

Key Concept

Highest Common Factor (HCF) of fractions

Question 122Question →

Consider the algebraic fraction below:

\frac{5^{n+2} - 5^n}{5^{n+1} - 5^n}

Which of the following represents the simplified exact value of this expression?

4

5

6

25

Show answer & explanation

Answer: 6

Answer

6

The expression can be systematically simplified by factoring out the lowest common power of the base, which is

5^n

, from both the numerator and the denominator. The numerator becomes

5^n(5^2 - 1) = 5^n(24)

. The denominator becomes

5^n(5^1 - 1) = 5^n(4)

. Canceling the common factor

5^n

from the top and bottom leaves

24 / 4

, which equals 6.

Step-by-Step Solution

1

Identify the greatest common factor in both the numerator and the denominator.

The common factor is

5^n

.

Factoring out the lowest power of the base simplifies algebraic expressions involving exponential terms.

2

Factor out

5^n

from the numerator.

5^{n+2} - 5^n = 5^n(5^2 - 1)

Applying the exponent rule

a^{x+y} = a^x \times a^y

in reverse allows us to extract the common term.

3

Factor out

5^n

from the denominator.

5^{n+1} - 5^n = 5^n(5^1 - 1)

Applying the same extraction process to the bottom expression.

4

Substitute the factored forms back into the fraction and cancel the common

5^n

term.

\frac{5^n(25 - 1)}{5^n(5 - 1)} = \frac{24}{4}

The

5^n

multiplier in the numerator and denominator cancel each other out, leaving basic arithmetic values.

5

Perform the final arithmetic division.

24 \div 4 = 6

Simplifying the remaining numerical fraction yields the final exact value.

Key Concept

Simplifying exponential expressions by factoring out common base powers.

Estimated Time:1m 15s

Question 123Question →

Consider the set of the first 100 positive integers (from 1 to 100 inclusive).

An integer $N$ from this set satisfies all of the following three conditions simultaneously:
1. $N$ is a composite number.
2. $N$ is neither divisible by 2 nor divisible by 3.
3. The square root of $N$ is an irrational number.

What is the total number of possible values for $N$ ?

Show answer & explanation

Answer: 7

Answer

There are exactly 7 values for N that satisfy all three conditions.

The complete set of numbers not divisible by 2 or 3 from 1 to 100 contains 33 integers. Excluding the number 1 (neither prime nor composite) and the 23 prime numbers leaves exactly 9 composite numbers. From these 9 composites, 25 and 49 must be excluded because their square roots are 5 and 7 (rational numbers). This leaves 7 valid integers.

Step-by-Step Solution

1

Determine the total number of integers from 1 to 100 that are neither divisible by 2 nor divisible by 3.

There are 33 such numbers.

Using the inclusion-exclusion principle: there are 50 multiples of 2, 33 multiples of 3, and 16 multiples of 6. Multiples of 2 or 3 = 50 + 33 - 16 = 67. The remaining numbers are 100 - 67 = 33.

2

Filter the 33 remaining integers to find those that are composite.

There are 9 composite numbers: 25, 35, 49, 55, 65, 77, 85, 91, and 95.

Of the 33 numbers, 1 is neither prime nor composite. There are 25 primes up to 100, and excluding 2 and 3 leaves 23 primes. Thus, the composites are 33 - 1 - 23 = 9. These are the products of primes 5 and greater.

3

Eliminate numbers from the composite list whose square roots are rational.

Remove 25 and 49.

Condition 3 requires the square root of N to be irrational. An integer has a rational square root if and only if it is a perfect square. In our list, 25 and 49 are perfect squares.

4

Count the final remaining valid numbers.

7 numbers remain.

Subtracting the 2 perfect squares from the 9 composite numbers leaves 7 numbers that satisfy all three conditions.

Key Concept

Classification of numbers combining prime/composite definitions, divisibility principles, and properties of rational and irrational roots.

Question 124Question →

A logistics manager is packing identical relief kits into crates. When she attempts to pack them in equal batches of $16$ , $24$ , $30$ , or $36$ kits per crate, there are always exactly $8$ kits left over. However, if she packs them in batches of exactly $19$ kits per crate, there are zero kits left over. What is the least possible total number of relief kits she could be packing?

Show answer & explanation

Answer: 2888

Answer

The least possible total number of relief kits is 2888.

The correct answer is derived by finding the general form of a number that leaves a remainder of 8 when divided by 16, 24, 30, and 36. This form is

720k + 8

. By applying the final condition that the total number must be exactly divisible by 19, we find the smallest valid multiplier is

k=4

. Substituting this back gives

720(4) + 8 = 2888

.

Step-by-Step Solution

1

Establish the relationship for the total number of kits based on the remainders.

The number of kits,

N

, leaves a remainder of

8

when divided by

16

,

24

,

30

, and

36

. Thus,

N = \text{LCM}(16, 24, 30, 36) \times k + 8

.

Any number that leaves the same remainder when divided by multiple divisors can be expressed as a multiple of their least common multiple plus that remainder.

2

Calculate the least common multiple (LCM) of

16

,

24

,

30

, and

36

.

The prime factorizations are

16 = 2^4

,

24 = 2^3 \times 3

,

30 = 2 \times 3 \times 5

, and

36 = 2^2 \times 3^2

. The LCM is the product of the highest powers:

2^4 \times 3^2 \times 5 = 720

.

The LCM is required to find the base repeating cycle for the division condition.

3

Formulate the exact divisibility condition.

Substitute the LCM into the equation to get

N = 720k + 8

. The problem states

N

is exactly divisible by

19

, so

(720k + 8) \equiv 0 \pmod{19}

.

This applies the second constraint of the problem to find the specific multiplier

k

.

4

Simplify the modular arithmetic equation to solve for

k

.

Divide

720

by

19

to find the remainder:

720 = 19 \times 37 + 17

. Substitute

17

for

720

to get

(17k + 8) \equiv 0 \pmod{19}

. This can be written as

(-2k + 8) \equiv 0 \pmod{19}

.

Simplifying large numbers using modulo properties makes finding the integer

k

manageable.

5

Find the smallest positive integer

k

that satisfies the equation.

Solving

-2k + 8 = 0

yields

2k = 8

, which means

k = 4

. Checking:

17(4) + 8 = 68 + 8 = 76

, and

76 \div 19 = 4

, which is exactly divisible.

Finding the smallest valid

k

ensures we calculate the least possible total number of kits.

6

Calculate the final total number of kits.

N = 720(4) + 8 = 2880 + 8 = 2888

.

Substituting

k = 4

back into the original formula for

N

provides the final numerical answer.

Key Concept

Finding a specific numerical value based on multiple simultaneous divisibility and remainder conditions using Least Common Multiple (LCM) and modular arithmetic.

Question 125Question →

Identify all two-digit prime numbers where both the tens digit and the units digit are strictly prime numbers. What is the sum of the largest and the smallest numbers that meet this criterion?

Show answer & explanation

Answer: 96

Answer

96

The correct answer requires finding the intersection of two distinct sets: two-digit numbers formed entirely by prime digits, and two-digit numbers that are mathematically prime. This resulting set is {23, 37, 53, 73}. Adding the minimum value (23) and maximum value (73) yields 96.

Step-by-Step Solution

1

Identify valid single digits.

The single-digit primes available for use are 2, 3, 5, and 7.

The problem states that both individual digits of the target number must be prime.

2

Determine valid units digits.

The units digit can only be 3 or 7.

If the units digit is 2, the number is even. If the units digit is 5, the number is a multiple of 5. Both cases result in a composite two-digit number.

3

List all potential combinations.

The possible combinations are 23, 33, 53, 73, 27, 37, 57, and 77.

These are generated by pairing any prime tens digit {2, 3, 5, 7} with the valid prime units digits {3, 7}.

4

Eliminate composite numbers from the list.

The valid primes are 23, 37, 53, and 73. The numbers 27, 33, 57, and 77 are removed.

27, 33, and 57 are divisible by 3 (sum of digits is a multiple of 3). 77 is divisible by 7.

5

Calculate the final sum.

23 + 73 = 96.

The question asks for the sum of the smallest valid number (23) and the largest valid number (73).

Key Concept

Classification of prime digits and prime numbers.

Question 126Question →

A botanist is preparing nutrient solutions. She has three different liquid nutrient extracts measuring $\frac{24}{5}$ liters, $\frac{32}{3}$ liters, and $\frac{40}{7}$ liters. She wants to distribute them entirely into identical small sample vials of maximum possible capacity, such that no extract is left over and the extracts are not mixed. What should be the maximum capacity of each sample vial?

\frac{8}{105}

liters

\frac{1}{480}

liters

480

liters

8

liters

Show answer & explanation

Answer: $\frac{8}{105}$ liters

Answer

The correct maximum capacity is

\frac{8}{105}

liters.

To find the maximum identical capacity that can exactly measure out each of the given quantities without mixing, we must calculate the Highest Common Factor (HCF) of the given fractions. By applying the formula for the HCF of fractions, we divide the HCF of the numerators (8) by the Least Common Multiple (LCM) of the denominators (105), resulting in

\frac{8}{105}

liters.

Step-by-Step Solution

1

Identify the mathematical operation required.

We need to find the Highest Common Factor (HCF) of the three fractions to determine the maximum equal capacity.

The vials must have the maximum possible capacity and perfectly measure out all three volumes without mixing them or leaving any remainder.

2

Recall the formula for the HCF of fractions.

HCF of fractions = (HCF of numerators) / (LCM of denominators).

This is the standard algebraic rule for determining the greatest common divisor of fractional values.

3

Calculate the HCF of the numerators: 24, 32, and 40.

The prime factorizations are

24 = 2^3 \times 3

,

32 = 2^5

, and

40 = 2^3 \times 5

. The highest common factor is

2^3 = 8

.

The HCF of the numerators forms the numerator of our final answer.

4

Calculate the LCM of the denominators: 5, 3, and 7.

Since 5, 3, and 7 are all prime numbers, their LCM is their product:

5 \times 3 \times 7 = 105

.

The LCM of the denominators forms the denominator of our final answer.

5

Construct the final fraction.

The capacity is

\frac{8}{105}

liters.

Dividing the HCF of the numerators by the LCM of the denominators gives the final HCF of the original fractions.

Key Concept

HCF and LCM of fractions

Question 127Question →

Consider the following statements regarding the classification of numbers:

I. The number $0$ is an even integer, but it is classified as neither positive nor negative.
II. The fraction $\frac{22}{7}$ is an irrational number because it is commonly used as the value of the constant $\pi$ .
III. The product of any two distinct irrational numbers is always an irrational number.

Which of the statements given above is/are correct?

I and II only

I and III only

I only

II and III only

Show answer & explanation

Answer: I only

Answer

Only the first statement is correct.

Only the first statement is true. Zero is an even integer because it is an integer multiple of

2

(

0 = 2 \times 0

), and it sits at the origin of the number line, meaning it is neither positive nor negative. The second statement is false because any number that can be expressed as the quotient of two integers, such as

\frac{22}{7}

, is rational by definition. The third statement is false because the product of distinct irrational numbers can be rational (for instance,

\sqrt{2} \times 8\sqrt{2} = 16

).

Step-by-Step Solution

1

Evaluate the parity and sign classification of zero in Statement I.

Zero is divisible by

2

without a remainder (

0 = 2 \times 0

), making it an even integer. It sits exactly between the negative and positive numbers on the number line, so it is strictly neither positive nor negative. Statement I is true.

To verify fundamental integer properties regarding the origin point.

2

Analyze the definition of the fraction in Statement II.

A rational number is any number that can be expressed as a ratio of two integers (

\frac{p}{q}

, where

q \neq 0

). Since

22

and

7

are integers,

\frac{22}{7}

is a rational number. It is merely an approximation of

\pi

, not its exact value. Statement II is false.

To test the distinction between a rational approximation and an irrational constant.

3

Test the closure property of irrational numbers under multiplication for Statement III.

Multiplying two distinct irrational numbers, such as

(2 - \sqrt{3})

and

(2 + \sqrt{3})

, yields

4 - 3 = 1

. Alternatively,

\sqrt{2} \times \sqrt{8} = \sqrt{16} = 4

. Since

1

and

4

are rational numbers, the product of distinct irrationals is not always irrational. Statement III is false.

To determine if irrational numbers are closed under multiplication.

Key Concept

Classification of Real Numbers: Parity of Zero, Rational vs Irrational Definitions, and Closure Properties

Estimated Time:1m 15s

Question 128Question →

Evaluate the following statements regarding the properties and classification of numbers. Which of the statements are mathematically correct? (Select all that apply)

Select all that apply

The product of the irrational numbers

\sqrt{8}

and

\sqrt{18}

is a composite integer.

The unit digit of the exponential number

2^{24}

is an odd integer.

The sum of any two distinct odd prime numbers is always an even composite integer.

The remainder obtained when

-17

is divided by

5

is an even integer.

Show answer & explanation

Answer: The product of the irrational numbers $\sqrt{8}$ and $\sqrt{18}$ is a composite integer.; The sum of any two distinct odd prime numbers is always an even composite integer.

Answer

The statement regarding the product of

\sqrt{8}

and

\sqrt{18}

being a composite integer, and the statement that the sum of any two distinct odd prime numbers is an even composite integer, are both mathematically correct.

The statement about the irrational product is correct because

\sqrt{8} \times \sqrt{18} = 12

, and 12 is a composite integer. The statement regarding prime numbers is correct because adding any two odd numbers always yields an even number. Since the smallest distinct odd primes are 3 and 5 (summing to 8), their sum will always be an even integer strictly greater than 2, which guarantees it is composite.

Step-by-Step Solution

1

Evaluate the product of the given irrational numbers.

\sqrt{8} \times \sqrt{18} = \sqrt{144} = 12

. Since 12 has factors other than 1 and itself, it is a composite integer.

To verify if the statement about the irrational product correctly classifies the resulting number.

2

Calculate the unit digit of

2^{24}

.

The exponent 24 is perfectly divisible by 4 (

24 \pmod 4 = 0

). We use the 4th power in the cycle:

2^4 = 16

. The unit digit is 6, which is even.

To determine the parity of the unit digit for the relevant statement.

3

Analyze the sum of two distinct odd prime numbers.

An odd integer added to another odd integer yields an even integer. The smallest possible sum is

3 + 5 = 8

. Any even integer greater than 2 is composite.

To mathematically prove the validity of the statement concerning prime sums.

4

Determine the remainder when

-17

is divided by 5.

Using modulo arithmetic,

-17 = 5 \times (-4) + 3

. The true positive remainder is 3, which is an odd integer.

To correctly evaluate the division operation and classify the remainder for the modulo statement.

Key Concept

Classification of Numbers and Number Properties

Estimated Time:1m 30s

Question 129Question →

Calculate the lowest common multiple (LCM) of the fractions $\frac{3}{4}$ , $\frac{9}{10}$ , and $\frac{15}{16}$ . Express your final answer as a decimal.

Show answer & explanation

Answer: 22.5

Answer

22.5

The lowest common multiple (LCM) of a set of fractions is found by dividing the LCM of their numerators by the highest common factor (HCF) of their denominators. For the numerators

3

,

9

, and

15

, the LCM is

45

. For the denominators

4

,

10

, and

16

, the HCF is

2

. Thus, the LCM of the fractions is

\frac{45}{2}

, which equals

22.5

in decimal form.

Step-by-Step Solution

1

Recall the mathematical formula for finding the LCM of multiple fractions.

\text{LCM of fractions} = \frac{\text{LCM of numerators}}{\text{HCF of denominators}}

This formula is the standard method to determine the lowest common multiple when dealing with fractional values.

2

Identify the numerators and calculate their lowest common multiple (LCM).

The numerators are

3

,

9

, and

15

. Their LCM is

45

(since

45

is the smallest number perfectly divisible by

3

,

9

, and

15

).

The numerator of the final fraction requires the LCM of all the given numerators.

3

Identify the denominators and calculate their highest common factor (HCF).

The denominators are

4

,

10

, and

16

. Their HCF is

2

(since

2

is the largest number that divides

4

,

10

, and

16

without a remainder).

The denominator of the final fraction requires the HCF of all the given denominators.

4

Apply the calculated values to the fraction LCM formula.

\text{LCM} = \frac{45}{2}

Combining the results from the previous steps yields the LCM in fractional form.

5

Convert the resulting fraction into a decimal format.

\frac{45}{2} = 22.5

The question explicitly requires the final answer to be expressed as a decimal.

Key Concept

Calculating the LCM of fractions using the specific formula: LCM of numerators divided by the HCF of denominators.

Question 130Question →

A digital communications system transmits data packets of three specific sizes: $\frac{35}{18}$ MB, $\frac{7}{4}$ MB, and $\frac{49}{24}$ MB. To optimize the network buffer, the engineers must define a standardized base unit size. Every transmitted packet must be an exact multiple of this base unit. To maximize efficiency, what is the largest possible size for this base unit?

\frac{7}{72}

MB

\frac{245}{2}

MB

\frac{7}{2}

MB

\frac{245}{72}

MB

Show answer & explanation

Answer: $\frac{7}{72}$ MB

Answer

The largest possible size for the base unit is

\frac{7}{72}

MB.

To find the largest possible base unit that perfectly divides all three packet sizes, the Highest Common Factor (HCF) of the fractions must be calculated. The formula for the HCF of fractions is HCF(Numerators) / LCM(Denominators). The numerators are 35, 7, and 49, which have an HCF of 7. The denominators are 18, 4, and 24, which have an LCM of 72. Thus, the correct largest base unit is 7/72 MB.

Step-by-Step Solution

1

Identify the required mathematical operation.

Find the Highest Common Factor (HCF) of the three fractions:

\frac{35}{18}

,

\frac{7}{4}

, and

\frac{49}{24}

.

The problem asks for the 'largest possible size' that perfectly divides the sizes of all given packets, which is the definition of HCF.

2

Calculate the HCF of the numerators.

The numerators are 35, 7, and 49. Their HCF is 7.

This value forms the numerator of the final fraction according to the formula.

3

Calculate the Least Common Multiple (LCM) of the denominators.

The denominators are 18, 4, and 24. Their LCM is 72.

This value forms the denominator of the final fraction according to the formula.

4

Combine the results to find the final HCF of the fractions.

\frac{7}{72}

MB.

Applying the formula: HCF of fractions =

\frac{\text{HCF of Numerators}}{\text{LCM of Denominators}}

.

Key Concept

Highest Common Factor (HCF) of Fractions

Question 131Question →

Consider the largest three-digit natural number $N$ which, when successively divided by $6$ , $7$ , and $8$ , leaves remainders of $4$ , $3$ , and $5$ respectively. What is the true positive remainder when the mathematical expression $E = 42^{63} - N$ is divided by $13$ ?

3

7

8

6

Show answer & explanation

Answer: 7

Answer

The correct remainder is 7.

The correct answer is derived by first reconstructing the number through successive division logic to find

N = 904

. Next, applying Fermat's Little Theorem simplifies

42^{63} \pmod{13}

to

1

. The expression becomes

1 - 7 = -6 \pmod{13}

, which correctly maps to a true positive remainder of

7

.

Step-by-Step Solution

1

Determine the algebraic form of the number

N

based on the successive division conditions.

N = 336 \cdot Q_3 + 232

By working backward from the final quotient

Q_3

, we construct the relations:

Q_2 = 8 \cdot Q_3 + 5

,

Q_1 = 7 \cdot Q_2 + 3

, and

N = 6 \cdot Q_1 + 4

. Substituting these yields the general form for

N

.

2

Find the largest three-digit natural number

N

.

N = 904

Substitute increasing integer values for

Q_3

. For

Q_3 = 0

,

N = 232

. For

Q_3 = 1

,

N = 568

. For

Q_3 = 2

,

N = 904

. For

Q_3 = 3

,

N = 1240

(which is four digits). Thus,

904

is the largest valid three-digit number.

3

Calculate the remainder of

42^{63}

divided by

13

.

42^{63} \equiv 1 \pmod{13}

Since

42 = 13 \times 3 + 3

,

42 \equiv 3 \pmod{13}

. By Fermat's Little Theorem,

3^{12} \equiv 1 \pmod{13}

. Breaking down the exponent:

63 = 12 \times 5 + 3

. Thus,

3^{63} \equiv (3^{12})^5 \cdot 3^3 \equiv 1^5 \cdot 27 \equiv 27 \pmod{13}

. Finally,

27 \pmod{13} = 1

.

4

Calculate the remainder of

N

divided by

13

and evaluate the full expression

E

.

E \equiv 7 \pmod{13}

904 = 13 \times 69 + 7

, so

N \equiv 7 \pmod{13}

. Substituting the remainders into the expression:

E \equiv 1 - 7 = -6 \pmod{13}

. To find the true positive remainder, add the divisor:

-6 + 13 = 7

.

Key Concept

Combining Successive Division modeling with Modular Exponentiation and Negative Remainder conversion.

Question 132Question →

If the expression $\frac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}}$ can be expressed in the form $a - b\sqrt{6}$ where $a$ and $b$ are rational numbers, what is the exact value of $a + b$ ?

Show answer & explanation

Answer: 7

Answer

The correct value is 7.

The correct answer is derived by multiplying the numerator and denominator by the conjugate

3\sqrt{2} - 2\sqrt{3}

. This rationalizes the denominator to

6

. Expanding the numerator gives

30 - 12\sqrt{6}

. Dividing the numerator by

6

yields

5 - 2\sqrt{6}

. Setting this equal to

a - b\sqrt{6}

identifies

a = 5

and

b = 2

, giving a final sum of

7

.

Step-by-Step Solution

1

Multiply the numerator and denominator by the conjugate of the denominator,

3\sqrt{2} - 2\sqrt{3}

.

\frac{(3\sqrt{2} - 2\sqrt{3})^2}{(3\sqrt{2} + 2\sqrt{3})(3\sqrt{2} - 2\sqrt{3})}

This process, known as rationalizing the denominator, removes the surds from the bottom of the fraction.

2

Expand the numerator using the binomial square formula

(x-y)^2 = x^2 - 2xy + y^2

.

(3\sqrt{2})^2 - 2(3\sqrt{2})(2\sqrt{3}) + (2\sqrt{3})^2 = 18 - 12\sqrt{6} + 12 = 30 - 12\sqrt{6}

Expanding the squared binomial simplifies the top part of the fraction.

3

Expand the denominator using the difference of squares formula

(x+y)(x-y) = x^2 - y^2

.

(3\sqrt{2})^2 - (2\sqrt{3})^2 = 18 - 12 = 6

This guarantees that the denominator becomes a rational number.

4

Divide the terms in the numerator by the denominator.

\frac{30 - 12\sqrt{6}}{6} = 5 - 2\sqrt{6}

Simplifying the fraction allows us to match it to the given form

a - b\sqrt{6}

.

5

Equate the simplified expression to

a - b\sqrt{6}

and solve for

a + b

.

a = 5

,

b = 2

, and

a + b = 7

By direct comparison of rational and irrational parts, we determine the values of

a

and

b

to find their sum.

Key Concept

Rationalizing the denominator using conjugates and expanding binomial expressions involving surds.

Question 133Question →

An inventory tracking system logs the total number of specialized microchips stored in a warehouse as exactly $5 \times 14^{33} - 3 \times 28^{50}$ units. These microchips are being shipped out in standard crates that hold exactly $13$ units each. The remaining microchips that do not make a full crate will be set aside for quality assurance testing. How many microchips will be set aside?

Show answer & explanation

Answer: 6

Answer

6 microchips will be set aside for quality assurance testing.

By evaluating the modulo

13

for each term independently using exponent cyclicity, the expression

5 \times 14^{33}

reduces to

5

, and

3 \times 28^{50}

reduces to

12

. Subtracting these yields an intermediate remainder of

-7

. Since remainders representing physical quantities must be non-negative, we add the divisor (

13

) to

-7

, resulting in a final positive remainder of

6

.

Step-by-Step Solution

1

Formulate the modulo expression.

(5 \times 14^{33} - 3 \times 28^{50}) \pmod{13}

Finding the remaining chips after filling crates of 13 is equivalent to calculating the remainder when the total is divided by 13.

2

Reduce the base of the first term.

5 \times 1^{33} = 5 \pmod{13}

14

leaves a remainder of

1

when divided by

13

. Any power of

1

remains

1

.

3

Reduce the base of the second term.

3 \times 2^{50} \pmod{13}

28

divided by

13

is

2

with a remainder of

2

.

4

Apply Fermat's Little Theorem to evaluate

2^{50} \pmod{13}

.

2^{12} \equiv 1 \pmod{13}

, so

2^{50} \equiv 2^2 \equiv 4 \pmod{13}

Powers of

2

repeat every

12

modulo

13

. Since

50 = 12 \times 4 + 2

, the remainder of the exponent is

2

.

5

Evaluate the full second term.

3 \times 4 = 12 \pmod{13}

Multiply the evaluated exponent component by the original coefficient.

6

Subtract to find the initial remainder.

5 - 12 = -7

Substitute the evaluated remainders back into the full expression.

7

Convert the negative remainder into a positive value.

-7 + 13 = 6

A physical count must be a positive integer. Adding the divisor to a negative remainder gives the true positive remainder.

Key Concept

Modular arithmetic properties, Fermat's Little Theorem, and conversion of negative remainders.

Question 134Question →

An agronomist is comparing the soil moisture retention rates of four different agricultural plots. The rates, measured in liters per cubic meter per day, are given below. Arrange the plots in ascending order of their moisture retention rates.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct ascending order is Plot Beta, Plot Delta, Plot Gamma, and then Plot Alpha.

By converting all rates to extended decimals, we get: Plot Beta (

1.444...

), Plot Delta (

1.450

), Plot Gamma (

1.4545...

), and Plot Alpha (

1.4555...

). Comparing them digit-by-digit from left to right establishes the strictly ascending order: Beta, Delta, Gamma, Alpha.

Step-by-Step Solution

1

Convert the fraction into a decimal to create a uniform format for comparison.

Plot Beta:

\frac{13}{9} = 1.4444...

Decimal forms allow for direct digit-by-digit comparison of magnitudes.

2

Expand all values to at least four decimal places to clearly see the repeating patterns.

Plot Alpha:

1.4555...

Plot Beta:

1.4444...

Plot Gamma:

1.4545...

Plot Delta:

1.4500...

Truncating or estimating too early can hide the subtle differences between terminating and repeating decimals.

3

Compare the tenths and hundredths places.

All values have a

4

in the tenths place. In the hundredths place, Plot Beta has a

4

, while the others have a

5

. Thus, Plot Beta is the smallest.

Comparing digits from left to right establishes the overall magnitude.

4

Compare the thousandths place for the remaining three plots.

Plot Delta has

0

, Plot Gamma has

4

, and Plot Alpha has

5

.

This establishes the final order for the remaining values: Delta (

1.450...

) < Gamma (

1.4545...

) < Alpha (

1.4555...

).

Key Concept

Comparing terminating decimals, recurring decimals, and fractions by converting them to expanded decimal forms.

Question 135Question →

In a highly automated manufacturing plant, three different robotic assembly lines are synchronized by a central control system. Line A completes a specific micro-calibration cycle every $\frac{4}{15}$ of a minute. Line B completes its calibration cycle every $\frac{6}{25}$ of a minute, and Line C completes its cycle every $\frac{8}{35}$ of a minute. If all three lines complete a calibration cycle simultaneously at a given moment, what is the shortest time interval that must elapse before all three lines complete their calibration cycles together again?

\frac{2}{525}

minutes

\frac{8}{175}

minutes

\frac{24}{5}

minutes

\frac{2}{5}

minutes

Show answer & explanation

Answer: $\frac{24}{5}$ minutes

Answer

The shortest time interval is

\frac{24}{5}

minutes.

To find when independent periodic events next occur simultaneously, we must calculate their Lowest Common Multiple (LCM). For fractions, the LCM is found by dividing the LCM of the numerators by the HCF of the denominators. The LCM of the numerators (

4, 6, 8

) is

24

, and the HCF of the denominators (

15, 25, 35

) is

5

. Therefore, the overall LCM is

\frac{24}{5}

.

Step-by-Step Solution

1

Identify the mathematical operation required.

We need to find the Lowest Common Multiple (LCM) of the three given time intervals (

\frac{4}{15}

,

\frac{6}{25}

,

\frac{8}{35}

) to determine when the periodic events will next coincide.

The LCM represents the smallest common interval of time into which each individual cycle can fit perfectly.

2

Recall the formula for finding the LCM of fractions.

\text{LCM of fractions} = \frac{\text{LCM of numerators}}{\text{HCF of denominators}}

.

This standard formula is required when working with periodic fractional amounts.

3

Find the LCM of the numerators.

The numerators are

4

,

6

, and

8

. The lowest common multiple of

4, 6, \text{ and } 8

is

24

.

Multiples of

8

are

8, 16, 24

.

24

is the smallest multiple evenly divisible by both

4

and

6

.

4

Find the Highest Common Factor (HCF) of the denominators.

The denominators are

15

,

25

, and

35

. The highest common factor of

15, 25, \text{ and } 35

is

5

.

The common prime factor shared by

3 \times 5

,

5 \times 5

, and

7 \times 5

is

5

.

5

Apply the values to the fraction LCM formula.

\frac{\text{LCM}(4, 6, 8)}{\text{HCF}(15, 25, 35)} = \frac{24}{5}

.

Combining the calculated numerator and denominator provides the final fractional interval.

Key Concept

Lowest Common Multiple (LCM) of Fractions

Question 136Question →

A civil engineering team is installing three different types of sensor cables along a newly constructed bridge. The standard supplier rolls for these cables come in lengths of $\frac{55}{2}$ meters, $\frac{77}{4}$ meters, and $\frac{121}{5}$ meters. To ensure modular replacement without any waste, the team must cut all the rolls into smaller, equal-length segments. What is the maximum possible length, in meters, of each segment?

Show answer & explanation

Answer: 0.55

Answer

0.55 meters

The problem asks for the maximum equal length that can divide all three roll lengths without any remainder, which mathematically corresponds to the Highest Common Factor (HCF). Using the standard formula for fractions: HCF = HCF(55, 77, 121) / LCM(2, 4, 5) = 11 / 20 = 0.55 meters.

Step-by-Step Solution

1

Determine the correct mathematical operation based on the problem context.

The requirement to cut multiple rolls into equal segments of maximum possible length with zero waste indicates the need to find the Highest Common Factor (HCF) of the three lengths.

HCF is used to find the largest common quantity that can perfectly divide multiple numbers.

2

State the standard formula for finding the HCF of fractions.

HCF = \frac{\text{HCF of numerators}}{\text{LCM of denominators}}

This is the mathematical rule for finding the HCF of rational numbers.

3

Calculate the HCF of the numerators.

The numerators are

55

,

77

, and

121

. Their prime factorizations are

5 \times 11

,

7 \times 11

, and

11 \times 11

. The HCF is

11

.

The largest common prime factor across all three numerators is 11.

4

Calculate the LCM of the denominators.

The denominators are

2

,

4

, and

5

. The lowest common multiple is

20

.

20 is the smallest positive integer divisible by 2, 4, and 5.

5

Compute the final decimal value.

\frac{11}{20} = 0.55

Converting the fraction to a decimal yields the exact required length.

Key Concept

Highest Common Factor (HCF) of Fractions

Question 137Question →

At a metropolitan central station, three express commuter trains depart simultaneously at 6:00 AM on Monday. The Green Line completes its round trip every $2 \frac{1}{4}$ hours, the Blue Line every $3 \frac{3}{8}$ hours, and the Red Line every $4 \frac{1}{2}$ hours. Assuming continuous operation without delays, how many hours will it take for all three trains to depart simultaneously from the central station again?

13 \frac{1}{2}

1 \frac{1}{8}

3 \frac{3}{8}

12

Show answer & explanation

Answer: $13 \frac{1}{2}$

Answer

The correct time is

13 \frac{1}{2}

hours, found by taking the Least Common Multiple (LCM) of the three fractional time intervals.

To find the next simultaneous departure, the Least Common Multiple (LCM) of the round-trip times is needed. Converting the times to improper fractions gives

\frac{9}{4}

,

\frac{27}{8}

, and

\frac{9}{2}

. The LCM of fractions is found by dividing the LCM of the numerators (which is 27) by the Highest Common Factor (HCF) of the denominators (which is 2). This gives

\frac{27}{2}

, or

13 \frac{1}{2}

hours.

Step-by-Step Solution

1

Convert the mixed numbers to improper fractions.

Green Line:

\frac{9}{4}

hours, Blue Line:

\frac{27}{8}

hours, Red Line:

\frac{9}{2}

hours.

Improper fractions are required to accurately calculate the Least Common Multiple (LCM) of fractions.

2

Identify the formula for finding the LCM of fractions.

\text{LCM of fractions} = \frac{\text{LCM of numerators}}{\text{HCF of denominators}}

This formula allows us to find the lowest common multiple for non-integer values.

3

Calculate the LCM of the numerators.

The numerators are 9, 27, and 9. Their LCM is 27.

27 is the smallest number perfectly divisible by 9, 27, and 9.

4

Calculate the HCF of the denominators.

The denominators are 4, 8, and 2. Their HCF is 2.

2 is the largest number that perfectly divides 4, 8, and 2.

5

Apply the values to the fraction LCM formula.

\frac{27}{2} = 13 \frac{1}{2}

hours.

Dividing the LCM of numerators by the HCF of denominators yields the final LCM of the fractions.

Key Concept

Calculating the Least Common Multiple (LCM) of fractions to solve periodic synchronization problems.

Question 138Question →

What is the value of the mathematical expression $24 + 4 \times \left[ 15 - \left\{ 12 \div 4 + (8 - \overline{5 - 2}) \right\} \right]$ ?

52

196

68

44

Show answer & explanation

Answer: 52

Answer

The simplified value of the expression is 52.

Following the VBODMAS hierarchy: first evaluate the bar line

\overline{5-2} = 3

, then the round brackets

(8-3) = 5

, then division inside the curly braces

12 \div 4 = 3

, followed by addition inside curly braces

3+5 = 8

. Next, subtract within square brackets

15-8 = 7

. Finally, multiply before adding:

4 \times 7 = 28

, and

24 + 28 = 52

.

Step-by-Step Solution

1

Evaluate the vinculum bar

\overline{5 - 2}

5 - 2 = 3

According to VBODMAS rules, operations under a vinculum (bar) have the highest priority among bracket operations.

2

Evaluate the expression inside the round brackets

(8 - 3)

8 - 3 = 5

Round brackets (parentheses) are simplified next.

3

Evaluate the expression inside the curly braces

\{12 \div 4 + 5\}

12 \div 4 = 3

, then

3 + 5 = 8

Inside the curly braces, division takes precedence over addition.

4

Evaluate the expression inside the square brackets

[15 - 8]

15 - 8 = 7

Square brackets are solved after round and curly brackets.

5

Perform multiplication before final addition

4 \times 7 = 28

Multiplication takes precedence over addition outside the brackets.

6

Perform the final addition

24 + 28 = 52

Complete the final addition step to obtain the result.

Key Concept

VBODMAS Rule (Vinculum, Brackets, Orders, Division, Multiplication, Addition, Subtraction)

Question 139Question →

What is the simplified value of the following expression?

25\% \text{ of } 160 - \left[ 12 \div \left\{ 3\frac{1}{5} + \left( 1.8 - \overline{0.6 + 0.4} \right) \right\} \right] \times 5

25

185

27.5

39.4

Show answer & explanation

Answer: 25

Answer

25

Evaluating the expression following the standard BODMAS priority (Vinculum

\rightarrow

Brackets

\rightarrow

Orders/Percentage

\rightarrow

Division

\rightarrow

Multiplication

\rightarrow

Subtraction):
1. Vinculum:

\overline{0.6 + 0.4} = 1

2. Round brackets:

1.8 - 1 = 0.8

3. Curly brackets:

3\frac{1}{5} + 0.8 = 3.2 + 0.8 = 4

4. Square brackets:

12 \div 4 = 3

5. Percentage:

25\% \text{ of } 160 = 40

6. Final calculation:

40 - (3 \times 5) = 40 - 15 = 25

.
Hence, the correct value is 25.

Step-by-Step Solution

1

Evaluate the expression under the vinculum bar

\overline{0.6 + 0.4}

0.6 + 0.4 = 1.0

The vinculum (bar bracket) takes the highest priority in the hierarchy of brackets.

2

Evaluate the innermost round brackets

(1.8 - 1.0)

1.8 - 1.0 = 0.8

Round brackets must be resolved next after the vinculum.

3

Convert mixed fraction

3\frac{1}{5}

to decimal and evaluate the curly brackets

\left\{ 3.2 + 0.8 \right\}

3\frac{1}{5} = 3.2

, so

3.2 + 0.8 = 4.0

Curly brackets are evaluated after resolving all inner terms.

4

Evaluate the square brackets

\left[ 12 \div 4.0 \right]

12 \div 4 = 3

Square brackets enclose the remaining multi-bracket expression.

5

Evaluate the 'of' percentage operation

25\% \text{ of } 160

\frac{25}{100} \times 160 = 40

'Of' / percentage operations take precedence over standard multiplication and subtraction.

6

Perform multiplication before subtraction in

40 - 3 \times 5

3 \times 5 = 15

, then

40 - 15 = 25

According to BODMAS, multiplication precedes subtraction.

Key Concept

Hierarchy of Brackets and BODMAS Rule

Estimated Time:2m 0s

Question 140Question →

What is the unit digit of the expression $432^{216} - 517^{142} - 734^{121} \times 283^{105}$ ?

5

0

1

4

Show answer & explanation

Answer: 5

Answer

The unit digit of the given expression is 5.

Evaluating each component:

432^{216}

ends in 6 (since 216 is divisible by 4, giving

2^4 = 16

),

517^{142}

ends in 9 (

142 \pmod 4 = 2

, giving

7^2 = 49

), and

734^{121} \times 283^{105}

ends in

4 \times 3 = 12 \rightarrow 2

. Combining these gives

(6 - 9 - 2) = -5 \equiv 5 \pmod{10}

. Thus, the option with value 5 is correct.

Step-by-Step Solution

1

Determine the unit digit of the first term

432^{216}

Unit digit is 6

The base unit digit is 2, which has a cyclicity of 4 (pattern: 2, 4, 8, 6). Dividing exponent 216 by 4 gives a remainder of 0. When the remainder is 0, we use the 4th power:

2^4 = 16

, so the unit digit is 6.

2

Determine the unit digit of the second term

517^{142}

Unit digit is 9

The base unit digit is 7, which has a cyclicity of 4 (pattern: 7, 9, 3, 1). Dividing exponent 142 by 4 gives a remainder of 2. Taking

7^2 = 49

, the unit digit is 9.

3

Determine the unit digit of the product term

734^{121} \times 283^{105}

Unit digit is 2

For

734^{121}

, base 4 with an odd exponent gives unit digit 4. For

283^{105}

, base 3 with exponent

105 \pmod 4 = 1

gives unit digit 3. Multiplying their unit digits:

4 \times 3 = 12

, so the unit digit is 2.

4

Combine the unit digits following order of operations (BODMAS)

Unit digit is 5

We compute

(6 - 9 - 2) \pmod{10}

. Evaluating left-to-right:

6 - 9 = -3 \equiv 7 \pmod{10}

(by adding 10 to handle borrowing). Then

7 - 2 = 5

.

Key Concept

Unit digit calculation using cyclicity of numbers and modular arithmetic rules under BODMAS

Estimated Time:2m 0s

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