Question 1

The smallest prime number, that is the fifth term of an increasing arithmetic sequence in which all the four preceding terms are also prime, is

Accepted Answer

### Concept & Logic For five consecutive terms in an increasing arithmetic progression (A.P.) to all be prime, the common difference ($d$) must be chosen carefully to avoid multiples of 2, 3, and 5. If a prime sequence has 5 terms and doesn't start with 5, one of the terms will inevitably be a multiple of 5 (and therefore composite). Thus, the sequence must start exactly with the prime number 5. Furthermore, to avoid multiples of 2 and 3, the common difference $d$ must be a multiple of $2 	imes 3 = 6$. ### Step-by-Step Solution * **Determine the starting term:** As established, the sequence must begin with $a = 5$. * **Determine the common difference:** To keep the subsequent numbers prime, $d$ must be a multiple of 6. Let's test the smallest possible valid difference, $d = 6$. * **Generate the sequence:** Term 1: $5$ (Prime) Term 2: $5 + 6 = 11$ (Prime) Term 3: $11 + 6 = 17$ (Prime) Term 4: $17 + 6 = 23$ (Prime) Term 5: $23 + 6 = 29$ (Prime) * **Verify the condition:** All five terms are prime numbers. The fifth term is 29. ### Exam Strategy & Shortcut Instead of deriving the sequence from scratch, work backwards from the given options. Since the options represent the *fifth* term, subtract standard common differences ($d=2, d=4, d=6$) to see if you hit a composite number. For option (a) 17: Sequence ending in 17 with $d=2$ is 9, 11, 13, 15, 17 (9 and 15 are composite). For option (b) 29: Subtracting $d=6$ repeatedly gives 23, 17, 11, 5. All are primes. You have found the answer instantly. ### Common Pitfall Testing sequences with a common difference of 2 or 4. In any set of 3 or more consecutive odd numbers separated by 2 or 4, one will always be a multiple of 3 (e.g., 3, 5, 7, 9). You must use a common difference that is a multiple of 6 to step over the multiples of 3. ### Final Answer **Therefore, the correct answer is 29.**

Question 2

The number of prime numbers between 301 and 320 are

Accepted Answer

### Concept & Strategy To count the prime numbers in a specific range, list the odd numbers and systematically eliminate multiples of 3, 5, and then test the remaining numbers against larger primes (7, 11, 13, 17). ### Step-by-Step Solution * **List potential primes:** The odd numbers between 301 and 320 are: 301, 303, 305, 307, 309, 311, 313, 315, 317, 319. * **Eliminate obvious composites:** - Multiples of 5 (ends in 5): **305, 315**. - Multiples of 3 (sum of digits is divisible by 3): $303$ ($3+0+3=6$), $309$ ($3+0+9=12$). * **Test remaining numbers (301, 307, 311, 313, 317, 319):** - **301:** Divisible by 7 ($301 = 7 	imes 43$). - **307:** Not divisible by 7, 11, 13, or 17. **(Prime)** - **311:** Not divisible by 7, 11, 13, or 17. **(Prime)** - **313:** Not divisible by 7, 11, 13, or 17. **(Prime)** - **317:** Not divisible by 7, 11, 13, or 17. **(Prime)** - **319:** Divisible by 11 ($319 = 11 	imes 29$). * **Count the primes:** There are exactly four prime numbers: 307, 311, 313, and 317. ### Exam Strategy & Shortcut Quick elimination is key. You can instantly cross out 303, 305, 309, and 315. For the rest, knowing the divisibility rule for 11 (alternating sum) quickly eliminates 319 ($3 - 1 + 9 = 11$). Recognizing that $301 = 300 + 1$ isn't helpful, but seeing it as $280 + 21$ immediately shows it is a multiple of 7. This leaves only the true primes to verify. ### Common Pitfall Mistakenly counting 301 or 319 as prime numbers because they do not fall to the basic divisibility rules of 2, 3, or 5. Always remember to test division by 7 and 11 for numbers up to 400. ### Final Answer **Therefore, the correct answer is 4.**

Question 3

Consider the following statements:

1. If $p > 2$ is a prime, then it can be written as $4n + 1$ or $4n + 3$ for a suitable natural number $n$.

2. If $p > 2$ is a prime, then $(p - 1)(p + 1)$ is always divisible by 4.

Of these statements,

Accepted Answer

### Concept & Logic This question tests fundamental properties of odd prime numbers, specifically their representation modulo 4 and the algebraic properties of consecutive even numbers. ### Step-by-Step Solution * **Analyze Statement 1:** "If $p > 2$ is a prime, then it can be written as $4n + 1$ or $4n + 3$..." Since $p > 2$ is prime, $p$ must be an odd number. When any odd integer is divided by 4, the remainder is always either 1 or 3. Therefore, every odd number can be represented as $4n + 1$ or $4n + 3$ (where $n$ is an integer $\ge 0$). Thus, Statement 1 is generally accepted as true in this context. * **Analyze Statement 2:** "If $p > 2$ is a prime, then $(p - 1)(p + 1)$ is always divisible by 4." Since $p$ is an odd prime, both $(p - 1)$ and $(p + 1)$ must be even numbers. They are consecutive even numbers (e.g., if $p=5$, they are 4 and 6). In any pair of consecutive even numbers, one is a multiple of 2, and the other is a multiple of 4. Their product will always be a multiple of $2 	imes 4 = 8$. Since the product is always divisible by 8, it is inherently always divisible by 4. Statement 2 is true. ### Exam Strategy & Shortcut Instead of proving the algebra, simply test the statements with the smallest odd prime, $p = 3$. Statement 1: $3 = 4(0) + 3$. (True, fitting the $4n+3$ format). Statement 2: $(3 - 1)(3 + 1) = 2 	imes 4 = 8$. 8 is divisible by 4. (True). Test $p = 5$: Statement 1: $5 = 4(1) + 1$. (True). Statement 2: $(5 - 1)(5 + 1) = 4 	imes 6 = 24$. 24 is divisible by 4. (True). Both hold up immediately. ### Common Pitfall Getting tripped up by the technical definition of a "natural number" in Statement 1. If natural numbers are strictly $\{1, 2, 3...\}$ (excluding 0), then $p=3$ cannot be formed. However, in aptitude exams, this statement is testing the $modulo\ 4$ property of odd primes, and $n$ is routinely treated as a non-negative integer. Overthinking the $n=0$ edge case will lead you to pick the wrong option. ### Final Answer **Therefore, the correct answer is (1) and (2) are true.**

Question 4

What is the first value of $n$ for which $n^2 + n + 41$ is not a prime?

Accepted Answer

### Concept & Formula The expression given is Euler's famous prime-generating polynomial: $$P(n) = n^2 + n + 41$$ This formula is renowned for producing prime numbers for all integer values from $n = 0$ up to $n = 39$. However, it fails at the value equal to the constant term minus one. ### Step-by-Step Solution * **Analyze the Polynomial:** We need to find an $n$ where $n^2 + n + 41$ can be factored, meaning it shares a common divisor and is no longer prime. * **Test the critical value $n = 40$:** Substitute $n = 40$ into the expression: $$P(40) = 40^2 + 40 + 41$$ Factor out the common term (40) from the first two terms: $$P(40) = 40(40 + 1) + 41$$ $$P(40) = 40(41) + 41$$ Now factor out the common 41: $$P(40) = 41(40 + 1)$$ $$P(40) = 41 	imes 41 = 41^2$$ * **Conclusion:** Since the result is $41^2$, the number is a perfect square and thus composite, not prime. Therefore, 40 is the first value where the formula fails. ### Exam Strategy & Shortcut You don't need to calculate large squares to solve this. Look at the constant term in the polynomial, which is 41. If you plug in $n = 41$, every term is a multiple of 41 ($41^2 + 41 + 41$), making it composite. By the same factorization logic, plugging in $n = 40$ creates $40(41) + 41$, which also factors cleanly by 41. The answer will always relate directly to the constant term. ### Common Pitfall Attempting to brute force the calculation for options (a), (b), and (c). Manually calculating and then verifying the primality of $P(10) = 151$ or $P(20) = 461$ wastes a massive amount of time. Recognize the algebraic pattern instead. ### Final Answer **Therefore, the correct answer is 40.**

Question 5

Let $X_k = (p_1 p_2 \dots p_k) + 1$, where $p_1, p_2, \dots, p_k$ are the first $k$ primes.

Consider the following:

1. $X_k$ is a prime number.

2. $X_k$ is a composite number.

3. $X_k + 1$ is always an even number.

Which of the above …

Accepted Answer

### Concept & Logic This problem relies on Euclid's proof of the infinitude of primes and basic even/odd number properties. The product of the first $k$ primes, plus one, generates a number that is not divisible by any of those initial primes. ### Step-by-Step Solution * **Analyze Statements 1 & 2:** The formula is $X_k = (p_1 	imes p_2 	imes \dots 	imes p_k) + 1$. While $X_k$ is prime for the first few values of $k$ (e.g., $X_1 = 2+1=3$, $X_2 = 2 	imes 3 + 1 = 7$), it is not *always* prime. For example, when $k=6$: $$X_6 = (2 	imes 3 	imes 5 	imes 7 	imes 11 	imes 13) + 1 = 30031$$ $30031$ is composite because $30031 = 59 	imes 509$. Because $X_k$ can be prime or composite depending on $k$, both sweeping statements "1. $X_k$ is a prime number" and "2. $X_k$ is a composite number" are false. * **Analyze Statement 3:** The expression is $X_k + 1$. Substitute $X_k$: $$X_k + 1 = (p_1 	imes p_2 	imes \dots 	imes p_k) + 1 + 1$$ $$X_k + 1 = (p_1 	imes p_2 	imes \dots 	imes p_k) + 2$$ Since the first prime $p_1$ is always 2, the product $(p_1 \dots p_k)$ is always an even number. $$	ext{Even Number} + 2 = 	ext{Even Number}$$ Therefore, $X_k + 1$ is unconditionally an even number. Statement 3 is correct. ### Exam Strategy & Shortcut You don't need to memorize the $X_6$ exception to solve this. Because options 1 and 2 make absolute claims that contradict each other, suspect they are both conditionally true and therefore false as general rules. Focus on Statement 3. You know $p_1 = 2$, so the product block is even. $X_k$ is (Even + 1) = Odd. Thus, $X_k + 1$ is (Odd + 1) = Even. This guarantees Statement 3 is correct. ### Common Pitfall Confusing the equation in Euclid's proof with a "prime generating formula". Euclid's proof merely shows $X_k$ has prime factors *larger* than $p_k$, it does not guarantee $X_k$ itself is prime. Assuming Statement 1 is true is the most common error. ### Final Answer **Therefore, the correct answer is 3 only.**

Question 6

$6 	imes 3 (3 - 1)$ is equal to

Accepted Answer

### Concept & Strategy The concept required to solve this problem is the fundamental order of operations, commonly known by the acronym BODMAS (Brackets, Orders, Division, Multiplication, Addition, Subtraction). $$ 	ext{BODMAS Rule applies strictly from left to right for multiplication/division} $$ ### Step-by-Step Solution * **Given Expression:** $6 	imes 3 (3 - 1)$ * **Step 1:** Solve the expression inside the brackets first. $(3 - 1) = 2$ * **Step 2:** Substitute the resolved bracket back into the expression. $6 	imes 3 	imes 2$ * **Step 3:** Perform multiplication from left to right. $18 	imes 2 = 36$ ### Exam Strategy & Shortcut For straightforward arithmetic expressions like this, process it purely mentally. Recognize that the bracket simplifies to $2$, and multiply the remaining small integers in your head: $6 	imes 3 = 18$, and $18 	imes 2 = 36$. Writing down steps costs precious seconds. ### Common Pitfall The most common mistake is misinterpreting the implied multiplication outside the bracket. Some students might multiply $6 	imes 3$ to get $18$, and then erroneously subtract $1$ instead of treating the bracket as a grouped entity to be resolved first. Always prioritize the bracketed terms. ### Final Answer Therefore, the correct answer is **36**.

Question 7

$1234 + 2345 - 3456 + 4567 = x$

Accepted Answer

### Concept & Strategy This question tests basic addition and subtraction of large numbers. The strategy is to either group numbers with the same mathematical sign or use the unit digit methodology to quickly eliminate incorrect options. ### Step-by-Step Solution * **Given:** Find the value of $x$ in $1234 + 2345 - 3456 + 4567$. * **Step 1:** Group all the positive numbers together to make addition easier. $(1234 + 2345 + 4567) - 3456$ * **Step 2:** Calculate the sum of the positive terms. $1234 + 2345 = 3579$ $3579 + 4567 = 8146$ * **Step 3:** Subtract the negative term from the obtained sum. $8146 - 3456 = 4690$ ### Exam Strategy & Shortcut Use the **Unit Digit and Tens Digit Method** to save time. * Unit digit calculation: $4 + 5 - 6 + 7 = 10$. The unit digit is $0$, with a carry of $1$. * Tens digit calculation: $3 + 4 - 5 + 6 + 1 (	ext{carry}) = 9$. The last two digits must be $90$. Looking at the options, only (a) and (d) end in $90$. A quick scan of the thousands places ($1+2-3+4 = 4$) confirms the answer should be in the 4000s, verifying $4690$. ### Common Pitfall A frequent mistake is performing the operations strictly sequentially without proper carrying/borrowing, especially when subtracting a larger component mid-sequence. Grouping positive terms first prevents negative intermediate steps. ### Final Answer Therefore, the correct answer is **4690**.

Question 8

$5566 - 7788 + 9988 = x + 4444$

Accepted Answer

### Concept & Formula This problem requires transposing terms across an algebraic equation to isolate the unknown variable $x$. $$ 	ext{If } A + B = x + C, 	ext{ then } x = A + B - C $$ ### Step-by-Step Solution * **Given:** The equation $5566 - 7788 + 9988 = x + 4444$. * **Step 1:** Transpose $4444$ to the left side of the equation to isolate $x$. $x = 5566 - 7788 + 9988 - 4444$ * **Step 2:** Group terms logically to simplify mental calculations. Notice that $(9988 - 7788)$ is an easy subtraction. $x = 5566 + (9988 - 7788) - 4444$ * **Step 3:** Perform the grouped subtraction. $x = 5566 + 2200 - 4444$ * **Step 4:** Add the positive terms. $x = 7766 - 4444$ * **Step 5:** Perform the final subtraction. $x = 3322$ ### Exam Strategy & Shortcut Observe the digits. Almost all numbers are repeating multiples of $1111$. You can temporarily factor out $1111$ to simplify: $1111 	imes (5 - 7 + 9 - 4) = x$ Calculate the simple bracket: $5 - 7 + 9 - 4 = 14 - 11 = 3$. Multiply back: $1111 	imes 3 = 3333$. Wait, let's re-verify: $5566$ is $1111 	imes 5$ ? No, $5566$ is $1111 	imes 5.01...$ Ah, the numbers are $5566, 7788, 9988, 4444$. They are multiples of $11$, not $1111$. Let's stick to the fast unit digit tracking: $6 - 8 + 8 - 4 = 2$. Tens: $6 - 8 + 8 - 4 = 2$. Hundreds: $5 - 7 + 9 - 4 = 3$. Thousands: $5 - 7 + 9 - 4 = 3$. Result is precisely $3322$. ### Common Pitfall Students often waste time doing sequential borrowing and carrying on large numbers instead of recognizing grouping opportunities (like $9988 - 7788 = 2200$) or checking last digits. ### Final Answer Therefore, the correct answer is **3322**.

Question 9

$38649 - 1624 - 4483 = x$

Accepted Answer

### Concept & Strategy When multiple numbers are being subtracted from a primary number, the most efficient approach is to sum all the negative components first, and then perform a single subtraction operation. $$ A - B - C = A - (B + C) $$ ### Step-by-Step Solution * **Given:** The expression $38649 - 1624 - 4483$. * **Step 1:** Group the numbers that need to be subtracted. $-(1624 + 4483)$ * **Step 2:** Calculate the sum of these subtrahends. $1624 + 4483 = 6107$ * **Step 3:** Subtract this combined total from the primary number. $38649 - 6107$ * **Step 4:** Perform the final subtraction. $38649 - 6107 = 32542$ * **Step 5:** Compare $32542$ with the given options: (a) 32425, (b) 32452, (c) 34522, (d) 35422. Since it matches none of them, the correct choice is "None of these". ### Exam Strategy & Shortcut Track the last two digits to quickly eliminate options. Sum of last two digits of subtrahends: $24 + 83 = 107$ (so the effective subtraction ends in $07$). Subtract from main number: $49 - 07 = 42$. The correct answer must end in $42$. Options (b) $32452$ and (d) $35422$ end in $42$. Now check the hundreds digit: $6 - (6+4) = 6 - 10 = -4$. Borrowing $1$ makes it $16 - 10 = 5$ in the hundreds place. The last three digits are $542$. None of the options match $542$, immediately select "None of these". ### Common Pitfall Choosing an option that "looks close" out of panic. Students might calculate $32542$, see option (b) $32452$, assume they made a small error, and mark it. Trust your calculation. ### Final Answer Therefore, the correct answer is **None of these**.

Question 10

$884697 - 773697 - 102479 = x$

Accepted Answer

### Concept & Strategy This is a multi-step subtraction problem. The fastest approach relies on recognizing specific number patterns to simplify mental math, grouping terms to reduce the number of borrowing steps. ### Step-by-Step Solution * **Given:** The continuous subtraction $884697 - 773697 - 102479$. * **Step 1:** Notice that the first two numbers share the same last three digits ($697$). Subtracting them sequentially is much faster than summing the negative components first. $884697 - 773697$ * **Step 2:** Calculate the difference. Since $697 - 697 = 000$, just subtract the thousands: $884 - 773 = 111$. The result is $111000$. * **Step 3:** Subtract the final number from this intermediate result. $111000 - 102479$ * **Step 4:** Perform the subtraction. You can break it down logically: $111000 - 102000 = 9000$ $9000 - 479 = 8521$ ### Exam Strategy & Shortcut Once you see $884697 - 773697 = 111000$, look at the last number: $102479$. You just need to calculate $111000 - 102479$. Use the **complement method** for the last three digits: $1000 - 479 = 521$. The answer must end in $521$. Looking at the options, only (c) ends in $521$. You can mark it instantly without completing the full calculation. ### Common Pitfall A common error is summing $773697$ and $102479$ first. While mathematically sound, it involves heavier carrying and borrowing, consuming far more time than simply recognizing the $697$ cancellation in the first step. ### Final Answer Therefore, the correct answer is **8521**.

Question 11

$10531 + 4813 - 728 = x 	imes 87$

Accepted Answer

### Concept & Strategy This question tests standard simplification rules combined with algebraic isolation. You must first resolve the complex arithmetic on the left-hand side (LHS) of the equation before dividing by the coefficient on the right-hand side (RHS). $$ 	ext{If } A + B - C = x 	imes D, 	ext{ then } x = \frac{A + B - C}{D} $$ ### Step-by-Step Solution * **Given:** The equation $10531 + 4813 - 728 = x 	imes 87$. * **Step 1:** Add the positive terms on the left side. $10531 + 4813 = 15344$ * **Step 2:** Subtract the negative term from the sum. $15344 - 728 = 14616$ * **Step 3:** Substitute the simplified LHS back into the equation. $14616 = x 	imes 87$ * **Step 4:** Isolate $x$ by dividing the LHS by $87$. $x = \frac{14616}{87}$ * **Step 5:** Perform the division (or use the shortcut below to avoid long division). $x = 168$ ### Exam Strategy & Shortcut **Unit Digit Analysis** is the fastest way to crack this. 1. Look at the unit digits on the left side: $1 + 3 - 8 ightarrow 4 - 8$. Since we can't subtract $8$ from $4$, borrow a $1$ (making it $14$). $14 - 8 = 6$. So, the entire left side resolves to a number ending in $6$. 2. Look at the right side: $x 	imes 87$. The unit digit of $x$ multiplied by $7$ must yield a number ending in $6$. What times $7$ ends in $6$? Only $8$ ($8 	imes 7 = 56$). 3. Check the options. Only option (a) $168$ ends in $8$. You can confidently mark it without performing the $14616 \div 87$ long division. ### Common Pitfall The most common mistake is attempting the long division ($14616 \div 87$) under time pressure, which often leads to arithmetic errors and wastes precious minutes. Always look for unit digit shortcuts before doing heavy division. ### Final Answer Therefore, the correct answer is **168**.

Question 12

What is $394$ times $113$?

Accepted Answer

### Concept & Strategy This is a direct multiplication problem. However, instead of traditional vertical multiplication, applying the **distributive property** allows you to break the numbers into easily calculable chunks. $$ A 	imes (B + C) = (A 	imes B) + (A 	imes C) $$ ### Step-by-Step Solution * **Given:** The expression $394 	imes 113$. * **Step 1:** Break down $113$ into its expanded base-10 form to make multiplication simpler. $113 = 100 + 10 + 3$ * **Step 2:** Apply the distributive property. $394 	imes (100 + 10 + 3) = (394 	imes 100) + (394 	imes 10) + (394 	imes 3)$ * **Step 3:** Calculate the individual parts. $394 	imes 100 = 39400$ $394 	imes 10 = 3940$ $394 	imes 3 = 1182$ * **Step 4:** Sum the calculated parts. $39400 + 3940 + 1182 = 44522$ ### Exam Strategy & Shortcut Use the **Digital Sum (Digital Root)** method to verify the answer instantly. 1. Find the digital sum of $394$: $3 + 9 + 4 = 16 ightarrow 1 + 6 = 7$. 2. Find the digital sum of $113$: $1 + 1 + 3 = 5$. 3. Multiply their digital sums: $7 	imes 5 = 35 ightarrow 3 + 5 = 8$. The correct option must have a digital sum of $8$. Let's check the options: (a) $44402 ightarrow 4+4+4+0+2 = 14 ightarrow 5$ (b) $44522 ightarrow 4+4+5+2+2 = 17 ightarrow 8$ (Match!) (c) $44632 ightarrow 4+4+6+3+2 = 19 ightarrow 1$ (d) $44802 ightarrow 4+4+8+0+2 = 18 ightarrow 9$ Option (b) is the only valid answer. ### Common Pitfall Writing out the traditional three-tier multiplication stack increases the likelihood of addition misalignment or carrying errors. Mental math using distribution or digital sums is far safer and faster. ### Final Answer Therefore, the correct answer is **44522**.

Question 13

$1260 \div 14 \div 9 = x$

Accepted Answer

### Concept & Rule This question tests the **Continuous Division Rule**. When an expression involves multiple consecutive division operations, they must be executed strictly from left to right. Alternatively, you can convert consecutive divisions into a single multiplication in the denominator. $$ A \div B \div C = \frac{A}{B 	imes C} $$ ### Step-by-Step Solution * **Given:** The expression $1260 \div 14 \div 9$. * **Step 1:** Convert the consecutive divisions into a single fraction to avoid decimal complications. $x = \frac{1260}{14 	imes 9}$ * **Step 2:** Multiply the terms in the denominator. $14 	imes 9 = 126$ * **Step 3:** Substitute the denominator back into the fraction. $x = \frac{1260}{126}$ * **Step 4:** Perform the final simplification. $x = 10$ ### Exam Strategy & Shortcut Memorizing multiplication tables up to $20$ is crucial for banking exams. If you know that $14 	imes 9 = 126$, you can look at $1260 \div 14 \div 9$ and immediately see the relationship. You are essentially evaluating $1260 \div 126$. The digits perfectly align, leaving just the zero, making the answer $10$ in less than two seconds. ### Common Pitfall The most dangerous mistake here is dividing right-to-left. A student might incorrectly calculate $14 \div 9$ first, or arbitrarily rearrange the order. Division is **not commutative or associative**. Always calculate left-to-right or use the denominator rule. ### Final Answer Therefore, the correct answer is **10**.

Question 14

$136 	imes 12 	imes 8 = x$

Accepted Answer

### Concept & Strategy This is a sequential multiplication problem. Because multiplication is associative, you can group the numbers in whatever order makes the calculation easiest. The goal is to reach a number close to a base of $100$ to utilize mental math. $$ A 	imes B 	imes C = A 	imes (B 	imes C) $$ ### Step-by-Step Solution * **Given:** The expression $136 	imes 12 	imes 8$. * **Step 1:** Group the smaller numbers first to create a single multiplier. $12 	imes 8 = 96$ * **Step 2:** The expression simplifies to $136 	imes 96$. * **Step 3:** Use the distributive property, recognizing that $96$ is close to $100$. $136 	imes (100 - 4)$ * **Step 4:** Multiply across the bracket. $(136 	imes 100) - (136 	imes 4)$ $13600 - 544$ * **Step 5:** Perform the subtraction. $13600 - 544 = 13056$ * **Step 6:** Compare the correct result ($13056$) with the given options: (a) 12066, (b) 13046, (c) 13064, (d) 13066. Since it does not appear, the correct choice is "None of these". ### Exam Strategy & Shortcut Use **Unit Digit Analysis** to eliminate options rapidly. Multiply the unit digits of the given numbers: $6 	imes 2 	imes 8$. $6 	imes 2 = 12$ (ends in $2$). $2 	imes 8 = 16$ (ends in $6$). The final answer MUST end in $6$. This instantly eliminates option (c) $13064$. Next, estimate the magnitude: $136 	imes 100 = 13600$. Since we are multiplying by $96$ (which is $4$ less than $100$), the answer must be slightly less than $13600$. $136 	imes 4$ is roughly $500$. So, $13600 - 500 = 13100$. None of the remaining options (12066, 13046, 13066) match the exact calculation of $13056$. ### Common Pitfall When students calculate $13056$ and see options like $13046$ or $13066$, they often assume they made a small arithmetic error and guess one of the close options. Trust your precise math and boldly select "None of these" if your calculation is solid. ### Final Answer Therefore, the correct answer is **None of these**.

Question 15

$8888 + 848 + 88 - x = 7337 + 737$

Accepted Answer

### Concept & Formula This problem tests the ability to manipulate large algebraic equations by transposing terms. The most efficient method is to simplify both sides of the equation independently before isolating the variable. $$ 	ext{If } \Sigma L - x = \Sigma R, 	ext{ then } x = \Sigma L - \Sigma R $$ ### Step-by-Step Solution * **Given:** $8888 + 848 + 88 - x = 7337 + 737$. * **Step 1:** Calculate the sum on the Left-Hand Side (LHS). $8888 + 848 + 88$ First add the smaller numbers: $848 + 88 = 936$. Then add to the main number: $8888 + 936 = 9824$. * **Step 2:** Calculate the sum on the Right-Hand Side (RHS). $7337 + 737 = 8074$ * **Step 3:** Substitute the sums back into the original equation. $9824 - x = 8074$ * **Step 4:** Transpose $x$ and the RHS sum to isolate $x$. $x = 9824 - 8074$ * **Step 5:** Perform the final subtraction. $x = 1750$ ### Exam Strategy & Shortcut Instead of full addition, check the **last two digits**. LHS last two digits: $88 + 48 + 88 = 224$ (so it ends in $24$). RHS last two digits: $37 + 37 = 74$ (ends in $74$). Equation for last two digits: $24 - x = 74$. To subtract, borrow $100$: $124 - x = 74 ightarrow x = 124 - 74 = 50$. The answer must end in $50$. Unfortunately, all options (a, b, c, d) end in $50$. Let's step it up and check the **hundreds digit** without summing everything. LHS hundreds sum: $800 + 800 + 0 = 1600$. Plus the carry of $2$ from the tens ($224$), so $1800$. RHS hundreds sum: $300 + 700 = 1000$. Equation for hundreds: $18 - x = 10 ightarrow x = 8$ (or $7$ depending on borrowing in the exact calculation). Let's just do the last three digits for safety: LHS: $888 + 848 + 088 = 1824$. RHS: $337 + 737 = 1074$. $1824 - 1074 = 750$. The answer must end in $750$. Only option (d) $1750$ matches. ### Common Pitfall A common error is improper alignment of place values when performing manual addition, specifically adding $88$ to the hundreds column instead of the tens column. Always align numbers right-to-left before adding. ### Final Answer Therefore, the correct answer is **1750**.

Question 16

$414 	imes x 	imes 7 = 127512$

Accepted Answer

### Concept & Strategy This algebraic equation requires isolating the unknown variable $x$. However, completely resolving the multiplication and subsequent division is highly time-consuming. The fastest approach is utilizing the unit digit methodology. $$ A 	imes x 	imes B = C $$ ### Step-by-Step Solution * **Given:** The equation $414 	imes x 	imes 7 = 127512$. * **Step 1:** Multiply the known terms on the left side. $414 	imes 7 = 2898$ * **Step 2:** Substitute back into the equation. $2898 	imes x = 127512$ * **Step 3:** Isolate $x$ by dividing both sides by $2898$. $x = \frac{127512}{2898}$ * **Step 4:** Perform the division. $x = 44$ ### Exam Strategy & Shortcut Use **Unit Digit Analysis** to bypass the long division entirely. Look at the given equation: $414 	imes x 	imes 7 = 127512$. 1. Multiply the unit digits of the known numbers on the left: $4 	imes 7 = 28$. The left side multiplier has a unit digit of $8$. 2. The right side of the equation ends in $2$ ($127512$). 3. Therefore, $8 	imes (	ext{unit digit of } x)$ must result in a number ending in $2$. 4. Check multiplication tables for $8$: $8 	imes 4 = 32$ and $8 	imes 9 = 72$. So $x$ must end in $4$ or $9$. 5. Look at the options: (a) 36, (b) 40, (c) 44, (d) 48. Only option (c) ends in $4$. You can mark it immediately. ### Common Pitfall The most common mistake is performing the full long division ($127512 \div 2898$) under time pressure. This not only wastes valuable minutes but also heavily increases the probability of a calculation error. ### Final Answer Therefore, the correct answer is **44**.

Question 17

Product of $82540027$ and $43253$ is

Accepted Answer

### Concept & Logic This problem asks for the product of two extremely large numbers. In banking and aptitude exams, such questions are intentionally designed to test your observational skills rather than your ability to perform tedious multiplication. ### Step-by-Step Solution * **Given:** Multiply $82540027$ by $43253$. * **Step 1:** Extract the unit digits of both numbers. Number 1 unit digit: $7$ Number 2 unit digit: $3$ * **Step 2:** Multiply the unit digits together. $7 	imes 3 = 21$ * **Step 3:** Identify the final unit digit of the product. The product must end in $1$. * **Step 4:** Evaluate the given options based on their last digit. (a) ends in 1 (b) ends in 2 (c) ends in 3 (d) ends in 4 * **Step 5:** Since only one option matches the required unit digit, it must be the correct answer. ### Exam Strategy & Shortcut Do not even write the numbers down. The moment you see options that are identical except for their final digit, simply multiply the last digits of the given numbers ($7 	imes 3 = 21$). The answer ends in $1$. Option (a) is the only one. Total time taken: 2 seconds. ### Common Pitfall A major pitfall is initiating the long multiplication out of habit or panic, not realizing the options differ strictly by a single digit at the very end. Always scan the options before starting a heavy calculation. ### Final Answer Therefore, the correct answer is **3570103787831**.

Question 18

$(46351 - 36418 - 4505) \div x = 1357$

Accepted Answer

### Concept & Strategy This question evaluates your ability to resolve multi-term brackets following the BODMAS rule, followed by transposing variables in a division equation. Grouping numbers with the same sign simplifies the bracket resolution. $$ 	ext{If } \frac{A}{x} = B, 	ext{ then } x = \frac{A}{B} $$ ### Step-by-Step Solution * **Given:** The equation $(46351 - 36418 - 4505) \div x = 1357$. * **Step 1:** Simplify the expression inside the parentheses. Sum the negative numbers first to avoid sequential subtraction. $-(36418 + 4505) = -40923$ * **Step 2:** Subtract this sum from the primary positive number. $46351 - 40923 = 5428$ * **Step 3:** Substitute the resolved bracket back into the main equation. $5428 \div x = 1357$ * **Step 4:** Rearrange the equation to solve for $x$. $x = \frac{5428}{1357}$ * **Step 5:** Perform the division to find the final value. $x = 4$ ### Exam Strategy & Shortcut Use **Unit Digit Tracking** combined with mental math to solve this in seconds. 1. Find the unit digit of the bracket: $1 - 8 - 5$. Borrow to make it positive: $11 - 8 = 3$, and $3 - 5 ightarrow 13 - 5 = 8$. The bracket evaluates to a number ending in $8$. 2. Set up the unit digit equation: $8 \div x = 7 \implies 8 = 7 	imes x$. 3. What unit digit multiplied by $7$ yields a number ending in $8$? $7 	imes 4 = 28$. So, $x$ must end in $4$. 4. Look at the options: only option (c) is $4$. Verify mentally: $1350 	imes 4 \approx 5400$, which matches our magnitude. ### Common Pitfall Students often subtract sequentially ($46351 - 36418$, then subtract $4505$), which increases the chance of borrowing errors. Grouping the negatives ($36418 + 4505$) is structurally safer. ### Final Answer Therefore, the correct answer is **4**.

Question 19

$6 	imes 66 	imes 666 = x$

Accepted Answer

### Concept & Logic This is a sequential multiplication problem involving repeating digits. The most efficient way to handle it is to multiply the smaller factors first, and then apply distribution techniques to the final large multiplication to reduce cognitive load. ### Step-by-Step Solution * **Given:** The expression $6 	imes 66 	imes 666$. * **Step 1:** Multiply the two smaller numbers. $6 	imes 66 = 396$ * **Step 2:** Substitute back into the expression. $396 	imes 666$ * **Step 3:** Use the distributive property to make mental math easier. Notice that $396$ is very close to $400$. $(400 - 4) 	imes 666$ * **Step 4:** Distribute the multiplication across the terms. $(400 	imes 666) - (4 	imes 666)$ * **Step 5:** Calculate the individual parts. $400 	imes 666 = 266400$ $4 	imes 666 = 2664$ * **Step 6:** Perform the final subtraction. $266400 - 2664 = 263736$ ### Exam Strategy & Shortcut You can eliminate options rapidly using the **Unit Digit** and **Last Two Digits** methods. 1. **Unit digit:** $6 	imes 6 	imes 6 = 216$. The answer must end in $6$. This instantly eliminates option (b) $263763$. 2. **Last two digits:** From $396 	imes 666$, we can use $(400 - 4) 	imes 666$. The last two digits will be determined entirely by $-4 	imes 66 = -264$. Subtracting this from a hundreds base (like $400 - 264$), the last two digits will be $100 - 64 = 36$. Options (a), (c), and (d) all end in $36$. 3. **Digital Sum:** The digital root of $6$ is $6$. The digital root of $66$ is $3$. The digital root of $666$ is $9$. Any number multiplied by a multiple of $9$ results in a digital sum of $9$. (a) $2+6+3+3+7+6 = 27 = 9$. (c) $2+6+3+7+3+6 = 27 = 9$. At this point, the $400 - 4$ distribution shortcut shown in the main steps is the definitive way to confirm $263736$ without full column multiplication. ### Common Pitfall Writing this out vertically as $666 	imes 66$ and then multiplying by $6$ again is highly prone to alignment and carry-over mistakes due to the repeating '6's confusing the eye. Leveraging the $(400 - 4)$ base is much safer. ### Final Answer Therefore, the correct answer is **263736**.

Question 20

If you subtract $-1$ from $+1$, what will be the result?

Accepted Answer

### Concept & Rule This question tests a fundamental principle of signed numbers (integers). Subtracting a negative integer is mathematically equivalent to adding its positive counterpart. $$ A - (-B) = A + B $$ ### Step-by-Step Solution * **Given:** Subtract $-1$ from $+1$. * **Step 1:** Translate the English phrasing into a mathematical equation. "Subtract $A$ from $B$" translates to $B - A$. $(+1) - (-1)$ * **Step 2:** Apply the signed number rule. The two negative signs (the subtraction operation and the negative value of the number) cancel each other out to become a positive. $1 + 1$ * **Step 3:** Perform the simple addition. $1 + 1 = 2$ ### Exam Strategy & Shortcut Simply visualize the number line. If you are at $+1$, and you subtract a positive number, you move left. But subtracting a negative number acts as a "double reverse," meaning you move to the right. Starting at $+1$ and moving right by $1$ unit places you at $+2$. ### Common Pitfall The most frequent error is misreading the English phrasing "subtract from." Many students write down $(-1) - (+1) = -2$, selecting option (a). Always remember that the value coming *after* the word "from" is your starting number (the minuend). ### Final Answer Therefore, the correct answer is **2**.

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