Question 1

There are just two ways in which $5$ may be expressed as the sum of two different positive (non-zero) integers, namely $5 = 4 + 1 = 3 + 2$. In how many ways, $9$ can be expressed as the sum of two different positive (non-zero) integers?

Accepted Answer

### Concept & Logic The problem asks for the number of ways to partition a number into two distinct positive integers. To find all unique pairs, we systematically list pairs starting from the highest possible first integer and lowest second integer, stopping when pairs repeat or become identical. ### Step-by-Step Solution **Given:** * Target sum = $9$ * Conditions: Two numbers must be positive (non-zero) and different. **Calculation / Deduction:** Let the two integers be $x$ and $y$ such that $x + y = 9$ and $x > y > 0$. Let's list the possible values: 1. If $y = 1$, then $x = 8$ ($8 + 1 = 9$) 2. If $y = 2$, then $x = 7$ ($7 + 2 = 9$) 3. If $y = 3$, then $x = 6$ ($6 + 3 = 9$) 4. If $y = 4$, then $x = 5$ ($5 + 4 = 9$) If we continue to $y = 5$, $x = 4$, this is the same pair as $(5, 4)$ just reversed. Since the order of addition does not create a new set of numbers, we stop here. Total valid pairs = $4$. ### Exam Strategy & Shortcut For a given integer $N$, the number of ways to express it as the sum of two distinct positive integers is given by $\frac{N-1}{2}$ if $N$ is odd, and $\frac{N-2}{2}$ if $N$ is even. Here $N = 9$ (odd), so the number of ways is $\frac{9-1}{2} = 4$. This formula gives the answer instantly without listing. ### Common Pitfall A common mistake is treating $(8+1)$ and $(1+8)$ as two different ways. The question asks for the combination of two numbers, not permutations, so order doesn't matter. Another error is including zero ($9+0$), but the question specifically restricts to non-zero integers. ### Final Answer Therefore, the correct answer is 4.

Question 2

$P$ and $Q$ are two positive integers such that $PQ = 64$. Which of the following cannot be the value of $P + Q$?

Accepted Answer

### Concept & Formula When the product of two variables is given as a constant, their sum can take various values depending on their specific factor pair. We need to find all factor pairs of the given product to determine all possible sums. ### Step-by-Step Solution **Given:** * $P$ and $Q$ are positive integers. * $PQ = 64$ **Calculation / Deduction:** First, identify all pairs of positive integers $(P, Q)$ whose product is $64$. The factor pairs of $64$ are: 1. $1 	imes 64 = 64 \implies P+Q = 1+64 = 65$ 2. $2 	imes 32 = 64 \implies P+Q = 2+32 = 34$ 3. $4 	imes 16 = 64 \implies P+Q = 4+16 = 20$ 4. $8 	imes 8 = 64 \implies P+Q = 8+8 = 16$ Comparing these possible sums ($65$, $34$, $20$, $16$) with the given options ($16$, $20$, $35$, $65$), we can see that $35$ is not in our list of possible sums. ### Exam Strategy & Shortcut To quickly check options, mentally deduce the factors of $64$. Notice that $35$ is very close to $34$ (which comes from $32+2$). Since the distance between factors grows rapidly ($32$ to $64$, $16$ to $32$), the resulting sums jump directly from $34$ to $65$. Thus, $35$ is clearly impossible. ### Common Pitfall Students might forget that $P$ and $Q$ can be equal (like $8$ and $8$) unless the question specifically restricts them to "distinct" integers. If they improperly excluded $8+8=16$, they might get confused, but here they can be the same, and $35$ is definitively the impossible value. ### Final Answer Therefore, the correct answer is 35.

Question 3

If $x + y + z = 9$ and both $y$ and $z$ are positive integers greater than zero, then the maximum value $x$ can take is

Accepted Answer

### Concept & Logic In an equation with multiple variables summing to a constant, to maximize one variable, you must minimize all other variables subject to the given boundary constraints. ### Step-by-Step Solution **Given:** * $x + y + z = 9$ * $y$ and $z$ are strictly positive integers ($y > 0$, $z > 0$). **Calculation / Deduction:** To find the maximum possible value for $x$, we must assign the absolute minimum possible values to $y$ and $z$. Since $y$ and $z$ are positive integers, the smallest integer value they can take is $1$. Let $y = 1$ and $z = 1$. Substitute these minimum values into the equation: $$x + 1 + 1 = 9$$ $$x + 2 = 9$$ $$x = 7$$ ### Exam Strategy & Shortcut When dealing with "maximize one term in a sum" problems, immediately peg the other terms to their absolute minimum boundary conditions. Positive integer means minimum is $1$. So, $9 - 1 - 1 = 7$. This can be solved mentally in under 5 seconds. ### Common Pitfall The most frequent mistake is assuming "positive integer" includes $0$, leading to $y=0$, $z=0$, and $x=9$, which is incorrect. Zero is a neutral integer—neither positive nor negative. Always remember positive integers strictly start from $1$. ### Final Answer Therefore, the correct answer is 7.

Question 4

What is the sum of the squares of the digits from $1$ to $9$?

Accepted Answer

### Concept & Formula The sum of the squares of the first $n$ natural numbers follows a specific algebraic series formula. Instead of manually squaring and adding each digit, applying this formula yields the result directly and safely. The standard formula is: $$\Sigma n^2 = \frac{n(n+1)(2n+1)}{6}$$ ### Step-by-Step Solution **Given:** * We need to find the sum of squares of digits from $1$ to $9$. * $n = 9$ **Calculation / Deduction:** Substitute $n = 9$ into the sum of squares formula: $$\Sigma 9^2 = \frac{9(9+1)(2(9)+1)}{6}$$ $$\Sigma 9^2 = \frac{9 	imes 10 	imes 19}{6}$$ Simplify the fraction by dividing $9$ and $6$ by $3$: $$\Sigma 9^2 = \frac{3 	imes 10 	imes 19}{2}$$ Divide $10$ by $2$: $$\Sigma 9^2 = 3 	imes 5 	imes 19$$ $$\Sigma 9^2 = 15 	imes 19$$ Calculate the final product: $$15 	imes 19 = 15 	imes (20 - 1) = 300 - 15 = 285$$ ### Exam Strategy & Shortcut For competitive exams, memorizing the sums of squares up to $10$ is a massive time-saver. The sum of squares up to $10$ is famously $385$. To find the sum up to $9$, simply subtract the square of $10$: $385 - 100 = 285$. This guarantees an instantaneous answer. ### Common Pitfall A common error is confusing the sum of squares formula with the square of the sum formula $(\frac{n(n+1)}{2})^2$. Using the wrong formula for $n=9$ would give $45^2 = 2025$, leading to confusion and wasted time when it doesn't match the options. ### Final Answer Therefore, the correct answer is 285.

Question 5

If $n$ is an integer between $20$ and $80$, then any of the following could be $n + 7$ except

Accepted Answer

### Concept & Logic When establishing the possible values of an expression involving an inequality, we must substitute the boundary values of the given constraint to find the new bounds. Alternatively, we can work backwards from the options to see which one violates the initial condition. ### Step-by-Step Solution **Given:** * $n$ is an integer between $20$ and $80$. This strictly means $20 < n < 80$. * We need to identify which option falls outside the range of $n + 7$. **Calculation / Deduction:** First, establish the strict integer range for $n$: $21 \le n \le 79$ Next, find the corresponding range for $n + 7$ by adding $7$ to all parts of the inequality: $$21 + 7 \le n + 7 \le 79 + 7$$ $$28 \le n + 7 \le 86$$ This means the valid range for $n + 7$ must be anywhere between $28$ and $86$ inclusive. Let's check the options against this mathematical boundary: (a) $47$: Inside range (b) $58$: Inside range (c) $84$: Inside range (d) $88$: Outside range ($88 > 86$) ### Exam Strategy & Shortcut Instead of calculating and shifting the full inequality range, test the extreme options backwards. The largest option is $88$. If $n+7 = 88$, then $n = 81$. The question explicitly states $n$ is between $20$ and $80$. Since $81$ is greater than $80$, option (d) is immediately invalid. Reverse-checking options is highly efficient here. ### Common Pitfall Students often misinterpret "between 20 and 80" as inclusive ($20 \le n \le 80$). However, even if incorrectly interpreted inclusively, $n+7$ would max out at $80+7 = 87$. $88$ is still definitively impossible. The primary pitfall is wasting time doing forward calculation instead of using the reverse-check shortcut. ### Final Answer Therefore, the correct answer is 88.

Question 6

Which one of the following is the correct sequence in respect of the Roman numerals: $C$, $D$, $L$ and $M$?

Accepted Answer

### Concept & Logic The problem requires knowledge of the integer values corresponding to standard Roman numerals to arrange them in descending order. ### Step-by-Step Solution **Given:** * The Roman numerals to evaluate are $C$, $D$, $L$, and $M$. **Calculation / Deduction:** First, recall the integer values for each Roman numeral: * $L = 50$ * $C = 100$ * $D = 500$ * $M = 1000$ Now, arrange these values in strictly descending order (from highest to lowest): $1000 > 500 > 100 > 50$ Substituting the corresponding numerals back into the inequality gives: $M > D > C > L$ ### Exam Strategy & Shortcut Use the common mnemonic "LCD Monitor" to remember the ascending order of the larger Roman numerals ($L=50, C=100, D=500, M=1000$). Since the question asks for the correct sequence and options are formatted with "greater than" signs, simply reverse the mnemonic order to get $M > D > C > L$. ### Common Pitfall A frequent mistake is confusing the value of $C$ and $L$. Students sometimes associate $L$ with "large" or forget that $C$ stands for "Century" (100). Reversing these two leads to an incorrect sequence. ### Final Answer Therefore, the correct answer is **$M > D > C > L$**.

Question 7

If the numbers from $1$ to $24$, which are divisible by $2$ are arranged in descending order, which number will be at the 8th place from the bottom?

Accepted Answer

### Concept & Logic When an arithmetic progression is arranged in descending order, counting from the bottom is mathematically identical to counting the same sequence in its natural ascending order. ### Step-by-Step Solution **Given:** * The set of numbers is integers from $1$ to $24$ that are divisible by $2$ (even numbers). * The arrangement is in descending order (highest to lowest). * We need to find the 8th number from the bottom. **Calculation / Deduction:** Identify the even numbers between $1$ and $24$: $2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24$ Arranging them in descending order forms this sequence: $24, 22, 20, 18, 16, 14, 12, 10, 8, 6, 4, 2$ The "bottom" of a descending list is simply the smallest number, which is $2$. Counting 8 positions up from the bottom (i.e., treating $2$ as the 1st position): 1st: 2 2nd: 4 3rd: 6 ... The $n$-th number from the bottom is simply $2 	imes n$. For the 8th place: $2 	imes 8 = 16$. ### Exam Strategy & Shortcut Skip generating the descending list altogether. Recognize that "8th from the bottom in a descending list of even numbers" is exactly the same as asking for the "8th even number starting from 2". Thus, simply calculate $8 	imes 2 = 16$. ### Common Pitfall The most common mistake is counting from the top instead of the bottom. If a student calculates the 8th place from the top, they would perform $24 - (8-1) 	imes 2 = 10$, and incorrectly choose option (a). Always double-check the direction of the positional request. ### Final Answer Therefore, the correct answer is **16**.

Question 8

$2 - 2 + 2 - 2 + \dots 101$ terms = $x$

Accepted Answer

### Concept & Strategy For an alternating series where identical values are repeatedly added and subtracted, the sum depends entirely on whether the total number of terms is even or odd. Pairs of terms will cancel each other out to equal zero. ### Step-by-Step Solution **Given:** * The series is $2 - 2 + 2 - 2 + \dots$ * The total number of terms is $101$. **Calculation / Deduction:** Let's group the terms in the sequence into adjacent pairs: $(2 - 2) + (2 - 2) + (2 - 2) + \dots$ Each pair evaluates to $0$. Since each pair requires 2 terms, we divide the total number of terms by 2 to find how many complete pairs exist: $101 \div 2 = 50$ pairs with a remainder of $1$. This means there are exactly $50$ pairs of $(2 - 2)$ which all sum to $0$. The remainder of $1$ signifies that there is exactly one unpaired term left at the very end of the sequence. Because the series starts with a positive $2$, the 101st term will also be a positive $2$. Sum $= 0 + 0 + \dots + 0 + 2 = 2$. ### Exam Strategy & Shortcut For any alternating series in the format $a - a + a - a \dots$, instantly look at the term count $N$. If $N$ is even, the sum is $0$. If $N$ is odd, the sum is the first term $a$. Since $101$ is odd and the first term is $2$, the answer is instantly $2$. ### Common Pitfall Students often glance at the alternating $2$s, assume they all cancel out into infinity, and mistakenly select $0$. This happens when they fail to note that $101$ is an odd number, leaving one uncancelled term. ### Final Answer Therefore, the correct answer is **2**.

Question 9

98th term of the infinite series $1, 2, 3, 4, 1, 2, 3, 4, 1, 2, \dots$ is

Accepted Answer

### Concept & Formula To find a specific term in a periodic, repeating sequence, use modular arithmetic. Divide the target term number by the length of the repeating cycle to find the remainder, which points to the exact position within the base cycle. ### Step-by-Step Solution **Given:** * The infinite sequence is $1, 2, 3, 4, 1, 2, 3, 4 \dots$ * Target position $= 98$th term. **Calculation / Deduction:** Identify the repeating block (the cycle) of the series. The sequence repeats the block "$1, 2, 3, 4$". The length of this cycle is $4$ terms. To find the 98th term, divide 98 by the cycle length (4): $98 \div 4 = 24$ with a remainder of $2$. This mathematical result means the full block of "$1, 2, 3, 4$" repeats exactly $24$ times (accounting for $96$ terms). The 97th term begins a new cycle (starting at $1$). The 98th term is the 2nd number in the cycle. Looking at the base cycle "$1, 2, 3, 4$", the 2nd number is $2$. ### Exam Strategy & Shortcut Use the divisibility rule for 4 (look at the last two digits, which is just 98 here). The closest multiple of 4 below 98 is 96. Simply subtract: $98 - 96 = 2$. A remainder of 2 always means the answer is the 2nd term in the repeating sequence. ### Common Pitfall A common mistake is confusing the quotient with the remainder. A student might calculate $98 \div 4 = 24.5$ and blindly guess "4" because of the 4 in the quotient, or they might calculate a remainder of 0 if their division is flawed, leading them to pick the last term of the cycle. ### Final Answer Therefore, the correct answer is **2**.

Question 10

If $x, y, z$ be the digits of a number beginning from the left, the number is

Accepted Answer

### Concept & Formula The algebraic representation of a number relies on the decimal place value system. Each digit in a number is multiplied by a power of 10 corresponding to its position (units, tens, hundreds). ### Step-by-Step Solution **Given:** * The digits of a three-digit number, reading from left to right, are $x$, $y$, and $z$. **Calculation / Deduction:** In a standard base-10 number system, the positions from right to left are units ($10^0$), tens ($10^1$), hundreds ($10^2$), and so on. Because the sequence $x, y, z$ begins from the left, the positions are assigned as follows: * $x$ is in the hundreds place. * $y$ is in the tens place. * $z$ is in the units (ones) place. To find the mathematical value of the number, multiply each digit by its respective place value and sum them up: $	ext{Value} = (x 	imes 100) + (y 	imes 10) + (z 	imes 1)$ $	ext{Value} = 100x + 10y + z$ ### Exam Strategy & Shortcut Whenever translating digits into algebraic values, strictly map them to standard place values ($1, 10, 100$). A left-to-right reading of a 3-digit number universally translates to $100(	ext{first}) + 10(	ext{second}) + 1(	ext{third})$. Match this structure immediately to option (d). ### Common Pitfall The most common trap is selecting $xyz$. In arithmetic, placing digits next to each other forms a number (like $456$), but in algebra, placing variables next to each other indicates multiplication ($x 	imes y 	imes z$). $xyz$ is the product of the digits, not the value of the number itself. ### Final Answer Therefore, the correct answer is **$100x + 10y + z$**.

Question 11

If $x, y, z$ and $w$ be the digits of a number beginning from the left, the number is

Accepted Answer

### Concept & Formula The algebraic representation of a number relies on the standard base-10 place value system. Each digit is multiplied by a power of 10 corresponding to its position. ### Step-by-Step Solution **Given:** * The digits of a four-digit number, reading from left to right, are $x$, $y$, $z$, and $w$. **Calculation / Deduction:** In a standard base-10 number system, the positions from right to left are units ($10^0$), tens ($10^1$), hundreds ($10^2$), and thousands ($10^3$). Because the sequence $x, y, z, w$ begins from the left, the positions are assigned as follows: * $x$ is in the thousands place. * $y$ is in the hundreds place. * $z$ is in the tens place. * $w$ is in the units (ones) place. To find the mathematical value of the number, multiply each digit by its respective place value and sum them up: $$	ext{Value} = (x 	imes 10^3) + (y 	imes 10^2) + (z 	imes 10^1) + (w 	imes 10^0)$$ $$	ext{Value} = 10^3x + 10^2y + 10z + w$$ ### Exam Strategy & Shortcut When translating digits into algebraic values from left to right, instantly assign the highest power of 10 to the first variable. Since it's a 4-digit number, the first power is $10^3$ assigned to $x$. Option (d) perfectly matches this descending power structure. ### Common Pitfall The most common mistake is choosing option (a) $xyzw$. In arithmetic, placing digits next to each other forms a number (like $4567$), but in algebra, placing variables side-by-side indicates multiplication ($x 	imes y 	imes z 	imes w$). Option (c) is also a trap where the powers of 10 are reversed. ### Final Answer Therefore, the correct answer is **$10^3x + 10^2y + 10z + w$**.

Question 12

If $n$ and $p$ are both odd numbers, which of the following is an even number?

Accepted Answer

### Concept & Logic Parity rules dictate how even and odd numbers interact during basic arithmetic operations. The sum of two odd numbers is always an even number. ### Step-by-Step Solution **Given:** * $n$ is an odd number. * $p$ is an odd number. **Calculation / Deduction:** Let's evaluate the options using the fundamental rules of parity: * **Option (a) $n + p$**: The sum of two odd numbers is always even (Odd + Odd = Even). * **Option (b) $n + p + 1$**: Since $n + p$ is even, adding 1 to an even number results in an odd number (Even + 1 = Odd). * **Option (c) $np + 2$**: The product of two odd numbers ($np$) is odd. Adding an even number (2) to an odd number results in an odd number (Odd + Even = Odd). * **Option (d) $np$**: The product of two odd numbers is always odd (Odd $	imes$ Odd = Odd). ### Exam Strategy & Shortcut Instead of thinking abstractly, instantly substitute the smallest positive odd numbers to test the options. Let $n = 1$ and $p = 3$. (a) $1 + 3 = 4$ (Even) (b) $1 + 3 + 1 = 5$ (Odd) (c) $(1 	imes 3) + 2 = 5$ (Odd) (d) $1 	imes 3 = 3$ (Odd) Option (a) is the only even result. ### Common Pitfall A common pitfall is wasting time trying to prove the algebra (e.g., setting $n = 2k+1$ and $p = 2m+1$) when a simple substitution method provides an immediate, foolproof verification of parity rules under time constraints. ### Final Answer Therefore, the correct answer is **$n + p$**.

Question 13

For the integer $n$, if $n^3$ is odd, then which of the following statements are true? I. $n$ is odd. II. $n^2$ is odd. III. $n^2$ is even.

Accepted Answer

### Concept & Logic The parity of an integer is preserved when raised to a positive integer power. This means if a number is odd, any positive power of that number will be odd, and conversely, if a positive power of an integer is odd, the base integer itself must be odd. ### Step-by-Step Solution **Given:** * $n$ is an integer. * $n^3$ is an odd number. **Calculation / Deduction:** First, determine the parity of $n$ based on $n^3$. The cube of an even number is even (e.g., $2^3 = 8$). The cube of an odd number is odd (e.g., $3^3 = 27$). Since $n^3$ is odd, the base $n$ MUST be an odd integer. * This makes **Statement I ($n$ is odd)** True. Next, evaluate $n^2$. Since we established that $n$ is odd, $n^2$ is the product of two odd numbers (Odd $	imes$ Odd). The product of two odd numbers is always odd (e.g., $3^2 = 9$). * This makes **Statement II ($n^2$ is odd)** True. * Consequently, **Statement III ($n^2$ is even)** must be False. Statements I and II are the only true statements. ### Exam Strategy & Shortcut Pick a small odd perfect cube, like $27$. Then $n = 3$. Check the statements rapidly: $n = 3$ (odd - Statement I is true), $n^2 = 9$ (odd - Statement II is true, Statement III is false). Both I and II are verified in under 5 seconds. ### Common Pitfall A common mistake is confusing the rules of parity for addition with those for multiplication, leading students to mistakenly believe that squaring an odd number makes it even. Always fall back to testing with the number 3 if unsure. ### Final Answer Therefore, the correct answer is **I and II only**.

Question 14

If $(n - 1)$ is an odd number, what are the two other odd numbers nearest to it?

Accepted Answer

### Concept & Logic Consecutive odd numbers are always separated by a constant difference of exactly 2. To find the nearest odd numbers to any given odd expression, you simply subtract 2 for the preceding odd number and add 2 for the succeeding odd number. ### Step-by-Step Solution **Given:** * The core expression is $(n - 1)$. * $(n - 1)$ is an odd number. **Calculation / Deduction:** To find the immediately preceding odd number, subtract $2$ from the given odd number: $$	ext{Previous Odd} = (n - 1) - 2$$ $$	ext{Previous Odd} = n - 3$$ To find the immediately succeeding odd number, add $2$ to the given odd number: $$	ext{Next Odd} = (n - 1) + 2$$ $$	ext{Next Odd} = n + 1$$ Therefore, the two nearest odd numbers are $n - 3$ and $n + 1$. ### Exam Strategy & Shortcut Use real numbers to bypass the algebra. Let $(n - 1) = 5$, which means $n = 6$. The nearest odd numbers to $5$ are $3$ and $7$. Now substitute $n = 6$ into the options. Option (c) gives $6 - 3 = 3$ and $6 + 1 = 7$. This matches our numerical logic perfectly and guarantees accuracy. ### Common Pitfall Students often misread the question and add or subtract 1 instead of 2. Doing so yields $(n - 2)$ and $n$, which represent consecutive integers (an even/odd mix), rather than consecutive *odd* numbers. ### Final Answer Therefore, the correct answer is **$n - 3, n + 1$**.

Question 15

Which of the following is always odd?

Accepted Answer

### Concept & Logic The behavior of odd and even numbers under basic arithmetic operations follows strict parity rules. Understanding these foundational rules allows you to eliminate incorrect operations immediately. ### Step-by-Step Solution **Given:** * We need to find the operation between two odd numbers that always results in an odd number. **Calculation / Deduction:** Let's evaluate each arithmetic operation: * **Option (a) Sum:** When you add two odd numbers, the "extra" units pair up to form an even number. (e.g., $3 + 5 = 8$). Always even. * **Option (b) Difference:** Subtracting an odd number from another odd number removes the "oddness", leaving an even result. (e.g., $7 - 3 = 4$). Always even. * **Option (c) Product:** An odd number has no factors of 2. Multiplying two numbers with no factors of 2 results in a product that also has no factors of 2. (e.g., $3 	imes 5 = 15$). Always odd. ### Exam Strategy & Shortcut Whenever a question asks "which of the following is always [even/odd]", bypass mental abstractions and quickly plug in small prime numbers like $3$ and $5$. Sum: $3+5=8$ (Even). Difference: $5-3=2$ (Even). Product: $3 	imes 5=15$ (Odd). Option (c) is the only one that yields an odd result. This technique takes 3 seconds. ### Common Pitfall A standard trap is overthinking the generalized algebra, such as testing $(2k+1)(2m+1) = 4km + 2k + 2m + 1$, when a simple numerical substitution is far faster and less prone to algebraic errors under exam pressure. ### Final Answer Therefore, the correct answer is **Product of two odd numbers**.

Question 16

If $x$ is an odd integer, then which of the following is true?

Accepted Answer

### Concept & Logic The properties of odd and even numbers dictate the parity of algebraic expressions. When determining parity, substituting a base odd integer simplifies the algebraic rules. ### Step-by-Step Solution * **Given:** $x$ is an odd integer. * **Deduction:** Let's assume a simple odd integer for $x$, such as $x = 1$. * **Calculation:** * Test Option (a): $5(1) - 2 = 3$ (This is odd, so the statement is false). * Test Option (b): $5(1)^2 + 2 = 5(1) + 2 = 7$ (This is odd, so the statement is true). * Test Option (c): $5(1)^2 + 3 = 5(1) + 3 = 8$ (This is even, so the statement is false). ### Exam Strategy & Shortcut **Value Substitution:** Never try to memorize complex parity rules (like $Odd 	imes Odd + Even = Odd$) under time pressure. Immediately substitute $x = 1$ (for odd) or $x = 2$ (for even) to instantly verify the correct option. ### Common Pitfall Students sometimes pick an overly large or negative odd number (like $-3$ or $7$), which introduces unnecessary arithmetic complexity and increases the chance of minor calculation errors. Stick to $1$. ### Final Answer Therefore, the correct answer is **$5x^2 + 2$ is odd**.

Question 17

If $a$ and $b$ are two numbers such that $ab = 0$, then

Accepted Answer

### Concept & Rule This problem tests the **Zero Product Property** of real numbers. $$ab = 0 \implies a = 0 	ext{ or } b = 0$$ ### Step-by-Step Solution * **Given:** The product of two numbers $a$ and $b$ is zero ($ab = 0$). * **Deduction:** For any multiplication to result in zero, at least one of the factors must be zero. * It is possible that $a = 0$ while $b$ is any non-zero number (e.g., $0 	imes 5 = 0$). * It is possible that $b = 0$ while $a$ is any non-zero number (e.g., $7 	imes 0 = 0$). * It is also possible that both are zero ($0 	imes 0 = 0$). * Therefore, the condition covers $a = 0$, or $b = 0$, or both. ### Exam Strategy & Shortcut This is a direct theoretical question. Recognize the exact definition of the zero product property to immediately eliminate the options that use "and" exclusively or require one variable to be strictly non-zero. ### Common Pitfall A common mistake is selecting "$a = 0$ and $b = 0$" assuming both must be zero, or missing the "or both" condition. The mathematical "or" is inclusive, meaning it accounts for either or both. ### Final Answer Therefore, the correct answer is **$a = 0$ or $b = 0$ or both**.

Question 18

If $A, B, C, D$ are numbers in increasing order and $D, B, E$ are numbers in decreasing order, then which one of the following sequences need neither be in a decreasing nor in an increasing order?

Accepted Answer

### Concept & Logic This is an inequalities and relative ordering problem. We must establish a master chain of inequalities by combining the given sequences. ### Step-by-Step Solution * **Given:** 1) $A, B, C, D$ are in increasing order: $A < B < C < D$ 2) $D, B, E$ are in decreasing order: $D > B > E$ (which rewrites to $E < B < D$) * **Deduction:** Combine the knowns relative to $B$: * $A < B$ * $E < B$ * $B < C < D$ * Let's evaluate the options to find the unordered sequence: * (a) $E, C, D$: Since $E < B$ and $B < C < D$, we know $E < C < D$. This is strictly increasing. * (b) $E, B, C$: Since $E < B$ and $B < C$, we know $E < B < C$. This is strictly increasing. * (c) $D, B, A$: Since $A < B < D$, reversing it gives $D > B > A$. This is strictly decreasing. * (d) $A, E, C$: We know $A < B$ and $E < B$. However, no relationship is given between $A$ and $E$. One could be larger than the other. Thus, the order between $A$ and $E$ is undefined. ### Exam Strategy & Shortcut **Variable Assignment:** Assign simple numbers to fit the rules. Let $B = 10, C = 20, D = 30$. For $A < B$, let $A = 5$. For $E < B$, let $E = 8$. Now check (d): $A, E, C \implies 5, 8, 20$ (increasing). Now change $E$ to $2$ (since $2 < 10$). Check (d): $A, E, C \implies 5, 2, 20$ (neither strictly increasing nor decreasing). Since it breaks, this is the answer. ### Common Pitfall Assuming that because $A$ is the "first" variable mentioned, it must be the absolute smallest. The problem only states $A < B$; $E$ could easily be smaller than $A$. ### Final Answer Therefore, the correct answer is **$A, E, C$**.

Question 19

If $m, n, o, p$ and $q$ are integers, then $m (n + o) (p - q)$ must be even when which of the following is even?

Accepted Answer

### Concept & Rule The product of multiple integers is even if at least **one** of the integers (or factors) in the multiplication is even. ### Step-by-Step Solution * **Given:** The expression is $m 	imes (n + o) 	imes (p - q)$. This represents three distinct factors multiplied together. * **Deduction:** For the entire product to *always* be even regardless of the other variables, one of the direct factors must definitively be even. * Let's evaluate the options: * If $p$ is even (Option b), $q$ could be odd, making $(p - q)$ odd. If the other factors are odd, the product is odd. * If $m + n$ is even (Option c) or $n + p$ is even (Option d), it doesn't guarantee that any of the explicit factors $m$, $(n + o)$, or $(p - q)$ are even. * If $m$ is even (Option a), then an even number is being multiplied by the rest of the expression. $Even 	imes Integer = Even$. ### Exam Strategy & Shortcut Identify the standalone multipliers in the expression. Here, $m$ is a standalone factor. If a standalone factor is even, the entire product is mathematically forced to be even, making it the most direct and undeniable condition. ### Common Pitfall Overthinking the properties of addition/subtraction inside the parentheses (like trying to figure out if $n+o$ can be forced to be even). Focus on the overarching operation, which is multiplication. ### Final Answer Therefore, the correct answer is **$m$**.

Question 20

If $n$ is a negative number, then which of the following is the least?

Accepted Answer

### Concept & Logic When working with negative numbers, multiplying by a positive constant scales the number further into the negatives, making its value "lesser" (further left on the number line). Squaring a negative number yields a positive result. ### Step-by-Step Solution * **Given:** $n$ is a negative number ($n < 0$). * **Deduction:** Let's substitute a simple negative integer, like $n = -2$, to evaluate the magnitude of each option. * **Calculation:** * Option (a): $0$ * Option (b): $-(-2) = 2$ (Positive) * Option (c): $2(-2) = -4$ (Negative) * Option (d): $(-2)^2 = 4$ (Positive) * Comparing the results: $-4 < 0 < 2 < 4$. The smallest value is $-4$, which corresponds to the expression $2n$. ### Exam Strategy & Shortcut **Conceptual Elimination:** A negative number ($n$) multiplied by a positive integer ($2$) yields a negative number ($2n$). Option (b) negates a negative, yielding a positive. Option (d) squares a negative, yielding a positive. Since $2n$ is the only expression that remains strictly negative, it must be less than $0$ and the positive options. ### Common Pitfall Confusing "least" with "closest to zero". Students sometimes pick $0$ or $-n$ thinking about absolute value. Remember that on a number line, $-4$ is smaller (less) than $-2$. ### Final Answer Therefore, the correct answer is **$2n$**.

Number System Questions

Questions

Practice smarter