Question 1

A 3-digit number $4a3$ is added to another 3-digit number $984$ to give the four-digit number $13b7$, which is divisible by $11$. Then, $(a + b)$ is

Accepted Answer

### Concept & Logic The divisibility rule of $11$ states that the difference between the sum of digits at odd places and the sum of digits at even places must be either $0$ or a multiple of $11$. ### Step-by-Step Solution * **Given:** * First number: $4a3$ * Second number: $984$ * Sum: $13b7$, which is divisible by $11$. * **Calculation / Deduction:** 1. Add the numbers to find the relation between $a$ and $b$: $$4a3 + 984 = 13b7$$ Looking at the columns: Units: $3 + 4 = 7$ Tens: $a + 8 = b \implies b - a = 8$ Hundreds: $4 + 9 = 13$ 2. Apply the divisibility rule of $11$ to $13b7$: Sum of digits at odd places (1st and 3rd from right): $7 + 3 = 10$ Sum of digits at even places (2nd and 4th from right): $b + 1$ Difference: $10 - (b + 1) = 9 - b$ 3. For $13b7$ to be divisible by $11$, the result $(9 - b)$ must be $0$ or a multiple of $11$. Since $b$ is a single digit ($0 \le b \le 9$), the only possible value is when $9 - b = 0$, giving $b = 9$. 4. Substitute $b = 9$ into our tens column equation: $a + 8 = 9 \implies a = 1$ 5. Calculate the required value $(a + b)$: $a + b = 1 + 9 = 10$ ### Exam Strategy & Shortcut You can quickly solve this by directly testing the divisibility rule on $13b7$. Sum of alternating digits must be equal: $1 + b = 3 + 7 \implies b = 9$. Once $b=9$, you know from the addition that $a+8=9 \implies a=1$. Their sum is $10$. This skips writing out the full addition columns and solves the problem in 10 seconds. ### Common Pitfall A common mistake is incorrectly evaluating the alternating sum from left-to-right versus right-to-left, or forgetting that $a$ and $b$ must be single digits between $0$ and $9$, which strictly limits the possible multiples of $11$ to just $0$. ### Final Answer Therefore, the correct answer is 10.

Question 2

If $$ \begin{array}{l} ab \overline{ ) 252 ( } ba \ \ \ \ \ \ \underline{ 24 } \ \ \ \ \ \ \ \ 12 \ \ \ \ \ \ \ \underline{ 12 } \ \ \ \ \ \ \ \ 	imes \end{array} $$ the values of $a$ and $b$ are

Accepted Answer

### Concept & Logic This problem relies on reverse-engineering the steps of long division, recognizing that each subtraction step represents the divisor multiplied by a single digit of the quotient. ### Step-by-Step Solution * **Given:** * Divisor: $ab$ (a 2-digit number) * Quotient: $ba$ (a 2-digit number) * Dividend: $252$ * **Calculation / Deduction:** 1. In the first step of the long division, the divisor ($ab$) is multiplied by the first digit of the quotient ($b$) to get $24$. $$ab 	imes b = 24$$ 2. After subtracting $24$ from $25$, the remainder is $1$, and the next digit ($2$) is brought down to form $12$. 3. In the second step, the divisor ($ab$) is multiplied by the second digit of the quotient ($a$) to yield $12$. $$ab 	imes a = 12$$ 4. Since $a$ and $b$ are individual digits, $ab$ is a two-digit number. The only two-digit number that can multiply by an integer to yield $12$ is $12$ itself. Therefore, $ab = 12$, which means $a = 1$ and $b = 2$. 5. Let's verify this against the first equation: If $a = 1$ and $b = 2$, does $12 	imes 2 = 24$? Yes. ### Exam Strategy & Shortcut Option Elimination is the fastest method here. Test Option (a) where $a=1, b=2$: Divisor is $12$, quotient is $21$. First step of division: $12 	imes 2 = 24$. Second step: $12 	imes 1 = 12$. This perfectly matches the visual representation in the question. You can mark (a) and move on in 15 seconds. ### Common Pitfall Students often misinterpret the quotient variables. The quotient is $ba$, meaning the *first* division step uses the tens digit $b$, and the *second* step uses the units digit $a$. Reversing these would lead to confusion and an inability to satisfy $ab 	imes a = 12$. ### Final Answer Therefore, the correct answer is 1, 2.

Question 3

$$ \begin{array}{r} * \ * \ * \ 	imes \ \ \ * \ \hline 8 \ * \ * \ 1 \ \hline \end{array} $$ In the above multiplication problem, $*$ is equal to

Accepted Answer

### Concept & Logic In missing-digit multiplication problems where the same symbol represents the unknown, all instances of that symbol must be the *exact same digit*. We solve this by analyzing unit digit constraints. ### Step-by-Step Solution * **Given:** * A 3-digit number consisting of identical digits ($d d d$) is multiplied by the same digit ($d$). * The product is a 4-digit number starting with $8$, having the same middle digits, and ending in $1$ ($8 d d 1$). * **Calculation / Deduction:** 1. Let the asterisk $*$ be the digit $d$. The equation becomes: $$d d d 	imes d = 8 d d 1$$ 2. Focus on the unit digit of the multiplication. The unit digit of $d 	imes d$ must end in $1$. The only single digits that yield a unit digit of $1$ when squared are $1$ ($1 	imes 1 = 1$) and $9$ ($9 	imes 9 = 81$). 3. Test $d = 1$: $$111 	imes 1 = 111$$ This yields a 3-digit number, which contradicts the 4-digit result starting with 8. 4. Test $d = 9$: $$999 	imes 9 = 8991$$ This perfectly matches the structure $8 * * 1$, where the middle digits are also $9$. ### Exam Strategy & Shortcut Use the Unit Digit Strategy and Option Elimination. The product ends in $1$. Looking at the options, $1 	imes 1 = 1$, $3 	imes 3 = 9$, $7 	imes 7 = 49$, and $9 	imes 9 = 81$. So, the answer must be $1$ or $9$. Since the product starts with $8$, the multiplier must be large. $999 	imes 9 = 8991$. Done. ### Common Pitfall Assuming that the asterisks can represent *different* digits is the most common pitfall. If a question asks "$*$ is equal to", it definitively implies a single, uniform value for all asterisks. ### Final Answer Therefore, the correct answer is 9.

Question 4

If $*$ means adding $6$ times the second number to the first number, then $(1 * 2) * 3$ equals

Accepted Answer

### Concept & Formula This question tests logical operators, where an arbitrary symbol is defined as a specific mathematical sequence. You must strictly follow the order of operations (BODMAS), computing inner brackets first. ### Step-by-Step Solution * **Given:** * The operator rule: $x * y = x + 6y$ * The expression to evaluate: $(1 * 2) * 3$ * **Calculation / Deduction:** 1. Evaluate the expression inside the parentheses first: $(1 * 2)$. Here, first number $= 1$, second number $= 2$. $$1 * 2 = 1 + 6(2) = 1 + 12 = 13$$ 2. Substitute the result back into the original expression: $(1 * 2) * 3$ becomes $13 * 3$. 3. Apply the operator rule again to $13 * 3$. Here, first number $= 13$, second number $= 3$. $$13 * 3 = 13 + 6(3) = 13 + 18 = 31$$ ### Exam Strategy & Shortcut Simply write the operations explicitly as algebraic substitutions: $(1 + 6 	imes 2) + 6 	imes 3 = 13 + 18 = 31$. It is a straightforward calculation that requires no special tricks, just careful attention to not mix up the "first" and "second" numbers. ### Common Pitfall A frequent mistake is ignoring the parentheses and attempting to evaluate $2 * 3$ first, or misapplying the rule by adding $6$ times the *first* number to the second, which would completely change the outcome. ### Final Answer Therefore, the correct answer is 31.

Question 5

If $1 	imes 2 	imes 3 	imes ........ 	imes n$ is denoted by $\lfloor n$, then $\lfloor 8 - \lfloor 7 - \lfloor 6$ is equal to

Accepted Answer

### Concept & Formula This question tests the fundamental property of factorials. The symbol $\lfloor n$ is an older notation representing $n!$ (n factorial). The core formula used to simplify factorials is: $$n! = n 	imes (n-1)!$$ ### Step-by-Step Solution * **Given:** * The expression: $\lfloor 8 - \lfloor 7 - \lfloor 6$ (which translates to $8! - 7! - 6!$) * **Calculation / Deduction:** 1. To subtract these terms easily, express the larger factorials in terms of the smallest factorial in the expression ($6!$). $$8! = 8 	imes 7 	imes 6! = 56 	imes 6!$$ $$7! = 7 	imes 6!$$ 2. Substitute these expanded forms back into the original expression: $$(56 	imes 6!) - (7 	imes 6!) - (1 	imes 6!)$$ 3. Factor out the common term, $6!$: $$6! 	imes (56 - 7 - 1)$$ 4. Simplify the numbers inside the bracket: $$56 - 7 - 1 = 48$$ So, the expression becomes $48 	imes 6!$. 5. Look at the options. We need to factor $48$ into $6 	imes 8$ to match the format of the given answers: $$48 	imes 6! = 6 	imes 8 	imes \lfloor 6$$ ### Exam Strategy & Shortcut Whenever you see addition or subtraction of consecutive factorials, immediately factor out the smallest one. $8! - 7! - 6! ightarrow 6!(8 	imes 7 - 7 - 1) = 6!(56 - 8) = 6! 	imes 48$. Scanning the options, only option (c) has $6 	imes 8 = 48$ multiplied by $\lfloor 6$. This takes under 20 seconds. ### Common Pitfall Students unfamiliar with the old factorial notation $\lfloor n$ might mistake it for an absolute value, floor function, or right-angle bracket, leading to nonsensical arithmetic like $8 - 7 - 6$. Always recognize this symbol in Indian competitive exams as a factorial. ### Final Answer Therefore, the correct answer is $6 	imes 8 	imes \lfloor 6$.

Question 6

The highest power of $9$ dividing $99!$ completely is

Accepted Answer

### Concept & Formula To find the highest power of a composite number (like $9$) in a factorial ($n!$), we must first break it down into its prime factors. Since $9 = 3^2$, we need to find the highest power of the prime number $3$ in $99!$, and then halve it. We use Legendre's Formula to find the highest power of a prime $p$ in $n!$: $$E_p(n!) = \lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor + \lfloor \frac{n}{p^3} \rfloor + ...$$ ### Step-by-Step Solution * **Given:** * Factorial: $99!$ * Divisor: $9$ (which is $3^2$) * **Calculation / Deduction:** 1. Find the highest power of $3$ in $99!$ using Legendre's Formula by successively dividing $99$ by $3$ and adding the quotients (ignoring remainders). 2. $\lfloor 99 / 3 \rfloor = 33$ 3. $\lfloor 33 / 3 \rfloor = 11$ 4. $\lfloor 11 / 3 \rfloor = 3$ 5. $\lfloor 3 / 3 \rfloor = 1$ 6. Total power of $3 = 33 + 11 + 3 + 1 = 48$. So, $99!$ contains $3^{48}$. 7. Since we need the highest power of $9$, and $9 = 3^2$, we group the powers of $3$ in pairs. 8. $3^{48} = (3^2)^{24} = 9^{24}$ ### Exam Strategy & Shortcut Use the successive division method on the prime factor $3$. Divide $99$ by $3$, write the integer quotient below it, and repeat until the quotient is $0$. Sum the quotients: $33 + 11 + 3 + 1 = 48$. Since $9$ is $3$ squared, immediately divide $48$ by $2$ to get $24$. This takes less than $20$ seconds. ### Common Pitfall The most common mistake is applying Legendre's formula directly with $9$ (i.e., $99/9 = 11$, $11/9 = 1 \implies 12$). Legendre's formula *only* works for prime numbers. You must always reduce the divisor to its prime factors first. ### Final Answer Therefore, the correct answer is 24.

Question 7

For an integer $n$, $n! = n(n - 1) (n - 2)............. 3.2.1.$  Then, $1! + 2! + 3! + ........ + 100!$ when divided by $5$ leaves remainder

Accepted Answer

### Concept & Logic In modular arithmetic, the remainder of a sum is equal to the sum of the remainders. For any factorial $n!$ where $n \ge 5$, the expansion includes the factor $5$ (e.g., $5! = 5 	imes 4 	imes 3 	imes 2 	imes 1$). Therefore, all factorials from $5!$ onwards are perfectly divisible by $5$ and leave a remainder of $0$. ### Step-by-Step Solution * **Given:** * Expression: $1! + 2! + 3! + ... + 100!$ * Divisor: $5$ * **Calculation / Deduction:** 1. Break the series into two parts: terms before $5!$ and terms from $5!$ onwards. $$ (1! + 2! + 3! + 4!) + (5! + 6! + ... + 100!) $$ 2. Analyze the second part. Since $5!, 6!, 7!$, etc., all contain $5$ as a multiplier, their remainders when divided by $5$ are strictly $0$. 3. Calculate the exact value of the first part: $1! = 1$ $2! = 2$ $3! = 6$ $4! = 24$ 4. Sum these initial terms: $1 + 2 + 6 + 24 = 33$ 5. Find the remainder when $33$ is divided by $5$. $33 / 5 = 6$ with a remainder of $3$. ### Exam Strategy & Shortcut Recognize immediately that $5!$ and beyond contribute nothing to the remainder when dividing by $5$. Simply compute $1 + 2 + 6 + 24 = 33$, and visually see that $33$ ends in $3$. Since numbers ending in $0$ or $5$ are divisible by $5$, the remainder is just the units digit $3$. ### Common Pitfall Students often get intimidated by the $100!$ and attempt to find a complex pattern or formula for the entire series, wasting time. Remembering that factorials quickly absorb all small prime factors is key. ### Final Answer Therefore, the correct answer is 3.

Question 8

The number of prime factors in the expression $6^{10} \cdot 7^{17} \cdot 11^{27}$ is equal to

Accepted Answer

### Concept & Strategy To find the total number of prime factors in a product of exponents, you must first prime factorize every base. Once all bases are prime, the total number of prime factors is simply the sum of all the exponents. ### Step-by-Step Solution * **Given:** * The mathematical expression: $6^{10} \cdot 7^{17} \cdot 11^{27}$ * **Calculation / Deduction:** 1. Inspect the bases: $6$, $7$, and $11$. 2. Base $7$ and base $11$ are already prime numbers. 3. Base $6$ is a composite number. Prime factorize it: $$6 = 2 	imes 3$$ 4. Substitute this back into the expression using exponent rules $(a \cdot b)^n = a^n \cdot b^n$: $$6^{10} = (2 	imes 3)^{10} = 2^{10} \cdot 3^{10}$$ 5. Rewrite the full expression with only prime bases: $$2^{10} \cdot 3^{10} \cdot 7^{17} \cdot 11^{27}$$ 6. The exponents represent how many times each prime factor appears. Add the exponents to find the total count: $$Total Prime Factors = 10 + 10 + 17 + 27 = 64$$ ### Exam Strategy & Shortcut Scan the bases. $7$ and $11$ are prime, contributing $17$ and $27$ directly. $6$ breaks into two primes ($2$ and $3$). Thus, the power of $10$ applies twice ($2 	imes 10 = 20$). Add them up: $20 + 17 + 27 = 64$. You can do this entirely mentally. ### Common Pitfall The most frequent error is ignoring the base $6$ and simply adding the given exponents: $10 + 17 + 27 = 54$. Always ensure every single base is reduced to prime numbers before summing powers. ### Final Answer Therefore, the correct answer is 64.

Question 9

What is the number of prime factors contained in the product $30^7 	imes 22^5 	imes 34^{11}$?

Accepted Answer

### Concept & Strategy To count the total prime factors in an exponential product, you must break down every composite base into its prime building blocks. Apply the exponent to each individual prime, and then sum all the resulting exponents. ### Step-by-Step Solution * **Given:** * The product expression: $30^7 	imes 22^5 	imes 34^{11}$ * **Calculation / Deduction:** 1. Prime factorize each composite base independently: $30 = 2 	imes 3 	imes 5$ (3 prime factors) $22 = 2 	imes 11$ (2 prime factors) $34 = 2 	imes 17$ (2 prime factors) 2. Apply the original exponents to these prime factorizations using the rule $(abc)^n = a^n \cdot b^n \cdot c^n$: $30^7 = (2 	imes 3 	imes 5)^7 = 2^7 	imes 3^7 	imes 5^7$ $22^5 = (2 	imes 11)^5 = 2^5 	imes 11^5$ $34^{11} = (2 	imes 17)^{11} = 2^{11} 	imes 17^{11}$ 3. Count the total prime factors by summing all the new exponents: From $30^7$: $7 + 7 + 7 = 21$ factors From $22^5$: $5 + 5 = 10$ factors From $34^{11}$: $11 + 11 = 22$ factors 4. Calculate the final sum: $$Total = 21 + 10 + 22 = 53$$ ### Exam Strategy & Shortcut Count the number of prime components in each base and multiply by the exponent. $30$ has $3$ components ($2,3,5$): $3 	imes 7 = 21$. $22$ has $2$ components ($2,11$): $2 	imes 5 = 10$. $34$ has $2$ components ($2,17$): $2 	imes 11 = 22$. Sum $= 21 + 10 + 22 = 53$. This method skips rewriting the entire algebraic string. ### Common Pitfall A common error is to just prime factorize the base and forget to distribute the exponent to *every* prime factor. For instance, treating $30^7$ as just $7$ prime factors instead of $21$. ### Final Answer Therefore, the correct answer is 53.

Question 10

What number multiplied by $48$ will give the same product as $173$ multiplied by $240$?

Accepted Answer

### Concept & Formula This is a direct application of algebraic equivalence. We are asked to set up a simple linear equation where two products are equal to each other. $$x \cdot A = B \cdot C$$ ### Step-by-Step Solution * **Given:** * Let the unknown number be $x$. * First product expression: $x 	imes 48$ * Second product expression: $173 	imes 240$ * **Calculation / Deduction:** 1. Equate the two expressions according to the problem statement: $$x 	imes 48 = 173 	imes 240$$ 2. Isolate $x$ by dividing both sides by $48$: $$x = \frac{173 	imes 240}{48}$$ 3. Simplify the fraction before multiplying to save time. Notice that $240$ is a multiple of $48$. $240 / 48 = 5$ 4. Substitute this back into the equation: $$x = 173 	imes 5$$ 5. Calculate the final multiplication: $$173 	imes 5 = 865$$ ### Exam Strategy & Shortcut Instead of blindly multiplying $173$ by $240$ (which wastes time and introduces arithmetic error risk), always look for cancellation first. $240$ is exactly $5$ times $48$. So, the unknown number is just $5$ times $173$. Even faster: $173 	imes 10 = 1730$. Half of $1730$ is $865$. ### Common Pitfall The worst approach here is calculating $173 	imes 240 = 41520$ and then trying to do long division by $48$. While it yields the correct answer, it consumes excessive time on a timed competitive exam. Always simplify fractions first. ### Final Answer Therefore, the correct answer is 865.

Question 11

A positive number, which when added to $1000$, gives a sum which is greater than when it is multiplied by $1000$. This positive integer is

Accepted Answer

### Concept & Logic This problem sets up a basic linear inequality comparing the addition and multiplication of an unknown variable with a constant. ### Step-by-Step Solution * **Given:** * Let the positive integer be $x$. * Condition: The sum of the number and $1000$ is strictly greater than their product. * **Calculation / Deduction:** 1. Translate the given condition into an algebraic inequality: $$x + 1000 > 1000x$$ 2. Isolate the terms with $x$ on one side by subtracting $x$ from both sides: $$1000 > 999x$$ 3. Solve for $x$ by dividing both sides by $999$: $$x < \frac{1000}{999}$$ 4. Evaluate the resulting fraction: $\frac{1000}{999}$ is slightly greater than $1$ (approximately $1.001$). 5. The problem explicitly states $x$ is a "positive integer". The only positive integer strictly less than $1.001$ is $1$. ### Exam Strategy & Shortcut Option verification is incredibly fast here and avoids algebra entirely. Test the smallest option first. Let's try $x = 1$. The sum is $1 + 1000 = 1001$. The product is $1 	imes 1000 = 1000$. Since $1001 > 1000$, option (a) satisfies the condition immediately. You can quickly see that for any integer $2$ or greater, the multiplication side will grow much faster ($2000 > 1002$). ### Common Pitfall Students might overcomplicate the phrasing and set up an equality ($=$) instead of an inequality ($>$). Additionally, ignoring the constraint that the number must be a "positive integer" could cause confusion if one tries to solve the inequality abstractly without restricting the domain of $x$. ### Final Answer Therefore, the correct answer is 1.

Question 12

$7$ is added to a certain number ; the sum is multiplied by $5$ ; the product is divided by $9$ and $3$ is subtracted from the quotient. Thus, if the remainder left is $12$, what was the original number?

Accepted Answer

### Concept & Logic This problem involves a straightforward sequence of arithmetic operations. The most efficient way to find the unknown starting number is to apply "reverse engineering," working backwards from the final result using inverse operations. ### Step-by-Step Solution * **Given:** * Let the original number be $x$. * Sequence of operations: Add $7$, multiply by $5$, divide by $9$, subtract $3$. * Final result (referred to as "remainder left" in the text) is $12$. * **Calculation / Deduction:** * **Method 1: Algebraic Setup** 1. Write the sequence as a single equation: $$\frac{(x + 7) 	imes 5}{9} - 3 = 12$$ 2. Solve for $x$. First, add $3$ to both sides: $$\frac{(x + 7) 	imes 5}{9} = 15$$ 3. Multiply both sides by $9$: $$(x + 7) 	imes 5 = 135$$ 4. Divide both sides by $5$: $$x + 7 = 27$$ 5. Subtract $7$: $$x = 20$$ * **Method 2: Reverse Operations (Mental Math)** 1. Start with final result: $12$. 2. Inverse of subtracting $3$ is adding $3$: $12 + 3 = 15$. 3. Inverse of dividing by $9$ is multiplying by $9$: $15 	imes 9 = 135$. 4. Inverse of multiplying by $5$ is dividing by $5$: $135 / 5 = 27$. 5. Inverse of adding $7$ is subtracting $7$: $27 - 7 = 20$. ### Exam Strategy & Shortcut Using the inverse operation method (Method 2) is the fastest strategy. You can execute steps 1 through 5 entirely in your head without writing a single algebraic equation. Read the prompt backward and invert the verbs: $12 ightarrow +3 ightarrow 	imes 9 ightarrow \div 5 ightarrow -7 ightarrow 20$. ### Common Pitfall The word "remainder" in the final sentence is poorly phrased in the original text; it means the *final mathematical result*, not the remainder of a division operation. If a student interprets "remainder" as a modulo operation, they will be unable to construct the proper equation. ### Final Answer Therefore, the correct answer is 20.

Question 13

Symbiosis runs a Corporate Training Programme. At the end of running the first programme, its total takings were ₹ $38950$. There were more than $45$ but less than $100$ participants. What was the participant fee for the programme?

Accepted Answer

### Concept & Strategy The total revenue (takings) is the product of the number of participants and the fee per participant. Because the number of participants must be a whole number, the correct participant fee must be a perfect divisor of the total takings. ### Step-by-Step Solution * **Given:** * Total takings = ₹ $38950$ * Number of participants ($n$) is an integer where $45 < n < 100$. * Options for fee ($f$): $410$, $450$, $500$, $510$. * Formula: $n 	imes f = 38950 \implies n = \frac{38950}{f}$ * **Calculation / Deduction:** 1. We need to find which option perfectly divides $38950$ to yield an integer between $46$ and $99$. 2. **Test Option (c) 500:** $38950 / 500 = 389.5 / 5$. This will not result in a whole number. Reject. 3. **Test Option (b) 450:** $38950 / 450 = 3895 / 45$. Since $3895$ does not end in $0$, it is not perfectly divisible by $45$ (which requires divisibility by both $5$ and $9$). Sum of digits of $3895 = 25$, not divisible by $9$. Reject. 4. **Test Option (d) 510:** $38950 / 510 = 3895 / 51$. $51$ is $17 	imes 3$. Since the sum of digits of $3895$ is $25$, it's not divisible by $3$, so it can't be divisible by $51$. Reject. 5. **Test Option (a) 410:** $38950 / 410 = 3895 / 41$. Let's divide: $41 	imes 100 = 4100$. Since $3895$ is close to $4100$, let's try $41 	imes 95$. $41 	imes (100 - 5) = 4100 - 205 = 3895$. Thus, the number of participants is exactly $95$. 6. Check the constraint: Is $95$ between $45$ and $100$? Yes. ### Exam Strategy & Shortcut Use divisibility rules to eliminate options rapidly. You know $n = 38950 / f$. Ignore the trailing zeros: you are essentially dividing $3895$ by $41$, $45$, $50$, or $51$. Since $3895$ doesn't end in $0$, it cannot be divided evenly by $50$. Since the sum of digits ($3+8+9+5 = 25$) is not a multiple of $3$, it cannot be divided by $45$ (needs $9$) or $51$ (needs $3$). By process of elimination, only $410$ remains. You don't even need to calculate $3895 / 41 = 95$ to know it's the right answer. ### Common Pitfall Attempting to formulate complex algebraic equations or guess the number of participants first is a massive time sink. The fastest route is recognizing the integer constraint and testing the given options as divisors. ### Final Answer Therefore, the correct answer is ₹ 410.

Question 14

The sum of four consecutive even numbers $A, B, C$ and $D$ is $180$. What is the sum of the set of next four consecutive even numbers?

Accepted Answer

### Concept & Logic Consecutive even numbers are spaced exactly $2$ units apart (e.g., $x, x+2, x+4, x+6$). When shifting to the "next set" of four consecutive even numbers, each individual number in the new set is a specific, constant distance away from its corresponding counterpart in the original set. ### Step-by-Step Solution * **Given:** * Set 1: $A, B, C, D$ (four consecutive even numbers). * $A + B + C + D = 180$. * **Calculation / Deduction:** * **Method 1: Standard Algebra** 1. Let the numbers be $x, x+2, x+4, x+6$. 2. Set up the equation: $x + (x+2) + (x+4) + (x+6) = 180$. 3. Combine terms: $4x + 12 = 180$. 4. Solve for $x$: $4x = 168 \implies x = 42$. 5. The first set is $42, 44, 46, 48$. 6. The "next four" consecutive even numbers continue the sequence: $50, 52, 54, 56$. 7. Sum the new set: $50 + 52 + 54 + 56 = 212$. ### Exam Strategy & Shortcut Use the "Shift" strategy. You do not need to find the actual numbers to solve this! Think about how the sets align: Set 1: $A$, $B$, $C$, $D$ Set 2: $A+8$, $B+8$, $C+8$, $D+8$ Why $+8$? The next even number after $D$ is $D+2$. Since there are four numbers in the set, the first number of the *new* set is $4 	imes 2 = 8$ units larger than the first number of the *old* set. Therefore, every single number in the new set is exactly $8$ greater than its counterpart. Total increase $= 4 	ext{ numbers} 	imes 8 = 32$. New Sum $= 	ext{Old Sum} + 32 = 180 + 32 = 212$. This takes 5 seconds. ### Common Pitfall The most common trap is wasting time executing the full algebraic calculation to find $x = 42$. While it yields the correct answer, it consumes precious time that could be saved by applying the logical shift shortcut. Another minor mistake is finding the new set of numbers but making an arithmetic error when adding $50+52+54+56$. ### Final Answer Therefore, the correct answer is 212.

Question 15

A young girl counted in the following way on the fingers of her left hand. She started calling the thumb $1$, the index finger $2$, middle finger $3$, ring finger $4$, little finger $5$, then reversed direction, calling the ring finger $6$, middle f…

Accepted Answer

### Concept & Logic This is a classic periodic sequence (or cyclic pattern) problem. To find the position of a large number in a repeating sequence, you must identify the length of one complete cycle and then use modular arithmetic (finding the remainder) to locate the target number's position within that cycle. ### Step-by-Step Solution * **Given:** * Target number to reach: $1994$. * The counting pattern across $5$ fingers, reversing at the ends. * **Calculation / Deduction:** 1. Map the numbers to the fingers to find the full repeating cycle: $1 ightarrow$ Thumb $2 ightarrow$ Index $3 ightarrow$ Middle $4 ightarrow$ Ring $5 ightarrow$ Little (Reverse happens here) $6 ightarrow$ Ring $7 ightarrow$ Middle $8 ightarrow$ Index $9 ightarrow$ Thumb (Reverse happens here, cycle repeats) 2. Notice that the pattern begins anew on the Thumb at $9$, $17$, $25$, etc. The numbers assigned to the Thumb are $1, 9, 17...$ which follow the format $8k + 1$. 3. This means the entire counting sequence operates on a cycle of exactly $8$ steps. 4. To find where $1994$ lands, divide it by the cycle length ($8$) and find the remainder. $$1994 \div 8$$ 5. Execute the division: $8 	imes 200 = 1600$. $1994 - 1600 = 394$. $8 	imes 40 = 320$. $394 - 320 = 74$. $8 	imes 9 = 72$. $74 - 72 = 2$. So, $1994 = (249 	imes 8) + 2$. 6. The remainder is $2$. 7. Match the remainder to the mapped cycle in step 1. A remainder of $2$ corresponds to position $2$ in the cycle, which is the Index finger. ### Exam Strategy & Shortcut Instead of long division, use the divisibility rule for $8$ to find the remainder instantly. A number is divisible by $8$ if its last three digits are divisible by $8$. Look at the last three digits of $1994$, which are $994$. $994 \div 8 = 124$ with a remainder of $2$ ($124 	imes 8 = 992$). Since the remainder is $2$, look at the 2nd step of the cycle: $1$ is thumb, $2$ is index finger. You arrive at the answer in seconds. ### Common Pitfall A very common mistake is assuming the cycle length is $10$ (because we count "back and forth" across $5$ fingers) or $5$ (the number of unique fingers). Mapping out the numbers explicitly until you hit the thumb a second time (at $9$) is crucial to proving the cycle length is actually $8$. ### Final Answer Therefore, the correct answer is index finger.

Question 16

Given $n = 1 + x$ and $x$ is the product of four consecutive integers. Then which of the following is true?  I. $n$ is an odd integer. II. $n$ is prime. III. $n$ is a perfect square.

Accepted Answer

### Concept & Logic The key insight is that the product of any four consecutive integers plus $1$ always results in the square of an odd integer. We can represent four consecutive integers as $k$, $(k+1)$, $(k+2)$, and $(k+3)$. ### Step-by-Step Solution * **Given:** $x = k(k+1)(k+2)(k+3)$ $n = x + 1$ * **Calculation / Deduction:** Group the outermost and innermost terms together to expand the expression efficiently: $x = [k(k+3)] 	imes [(k+1)(k+2)]$ $x = (k^2 + 3k) 	imes (k^2 + 3k + 2)$ Let $a = k^2 + 3k$. The equation simplifies to: $x = a(a + 2) = a^2 + 2a$ Now, calculate $n$: $n = x + 1$ $n = a^2 + 2a + 1$ $n = (a + 1)^2$ Since $n = (a + 1)^2$, $n$ is clearly a **perfect square** (Statement III is true). To check if it is odd: $a = k^2 + 3k = k(k+3)$. For any integer $k$, either $k$ is even or $(k+3)$ is even. Thus, their product $a$ is always **even**. If $a$ is even, then $(a + 1)$ is **odd**. The square of an odd number is always an **odd integer** (Statement I is true). Since perfect squares (greater than 1) have at least three factors (1, the square root, and themselves), $n$ cannot be prime (Statement II is false). ### Exam Strategy & Shortcut **Option Elimination via Substitution:** Under time pressure, pick the easiest set of consecutive integers: $1, 2, 3, 4$. * $x = 1 	imes 2 	imes 3 	imes 4 = 24$ * $n = 1 + 24 = 25$ * Check conditions for $25$: It is odd (I is true), it is NOT prime (II is false), and it is a perfect square ($5^2$) (III is true). * Therefore, Both I and III are correct. ### Common Pitfall Students often waste time trying to algebraically prove the statements without substituting simple numbers, or they test with $0$ ($0, 1, 2, 3 ightarrow x = 0, n = 1$), which can create ambiguity about whether $1$ acts as a prime or standard odd perfect square in this specific phrasing. Using positive consecutive integers like $1, 2, 3, 4$ is safer. ### Final Answer **Therefore, the correct answer is Both I and III are correct.**

Question 17

If $x + y = 15$ and $xy = 56$, then what is the value of $x^2 + y^2$?

Accepted Answer

### Concept & Formula This problem is a direct application of standard algebraic identities. The sum of squares can be derived algebraically using the expansion of the square of a sum. $$ (x + y)^2 = x^2 + y^2 + 2xy $$ ### Step-by-Step Solution * **Given:** $x + y = 15$ $xy = 56$ * **Calculation / Deduction:** Rearrange the algebraic identity to isolate the target expression, $x^2 + y^2$: $$ x^2 + y^2 = (x + y)^2 - 2xy $$ Substitute the given values into the formula: $$ x^2 + y^2 = (15)^2 - 2(56) $$ $$ x^2 + y^2 = 225 - 112 $$ $$ x^2 + y^2 = 113 $$ ### Exam Strategy & Shortcut **Number Guessing (Factor Pairing):** When the numbers are small, try to guess the integer values of $x$ and $y$ that satisfy both equations simultaneously. * Factors of $56$: $1 	imes 56$, $2 	imes 28$, $4 	imes 14$, $7 	imes 8$. * Check which pair adds up to $15$: $7 + 8 = 15$. * So, $x = 7$ and $y = 8$ (or vice versa). * Calculate $x^2 + y^2 = 7^2 + 8^2 = 49 + 64 = 113$. This completely bypasses algebraic expansion. ### Common Pitfall A frequent mistake is calculating $(x + y)^2$ and forgetting to subtract the $2xy$ term, leading a student to mistakenly assume the answer is $225$, or subtracting just $xy$ instead of $2xy$. ### Final Answer **Therefore, the correct answer is 113.**

Question 18

Given that $(1^2 + 2^2 + 3^2 + .... + 20^2) = 2870$, the value of $(2^2 + 4^2 + 6^2 + .... + 40^2)$ is

Accepted Answer

### Concept & Logic When every term in a series is multiplied by a constant, the sum of the series is also multiplied by that constant. In a series of squares, factoring out a common base requires squaring that common factor. $$ (ab)^2 = a^2 	imes b^2 $$ ### Step-by-Step Solution * **Given:** $1^2 + 2^2 + 3^2 + .... + 20^2 = 2870$ Target Series: $2^2 + 4^2 + 6^2 + .... + 40^2$ * **Calculation / Deduction:** Notice that every term in the target series is an even number squared. We can rewrite each term as a multiple of $2$: $$ (2 	imes 1)^2 + (2 	imes 2)^2 + (2 	imes 3)^2 + .... + (2 	imes 20)^2 $$ Apply the exponent rule $(ab)^2 = a^2 	imes b^2$: $$ (2^2 	imes 1^2) + (2^2 	imes 2^2) + (2^2 	imes 3^2) + .... + (2^2 	imes 20^2) $$ Factor out the common term $2^2$ (which is $4$): $$ 4 	imes (1^2 + 2^2 + 3^2 + .... + 20^2) $$ Substitute the known value of the original series ($2870$): $$ 4 	imes 2870 $$ $$ = 11480 $$ ### Exam Strategy & Shortcut **Ratio Method:** Recognize immediately that the second series represents the square of $2$ times the first series. The multiplier isn't $2$, it is $2^2 = 4$. Simply multiply the given total by $4$. $$ 2870 	imes 4 = 11480 $$ ### Common Pitfall The most common mistake is assuming that doubling the base numbers simply doubles the final sum. Students often hastily calculate $2870 	imes 2 = 5740$, failing to realize that squaring a doubled number quadruples its value. ### Final Answer **Therefore, the correct answer is 11480.**

Question 19

The value of $5^2 + 6^2 + .... + 10^2 + 20^2$ is

Accepted Answer

### Concept & Formula To find the sum of squares of a continuous segment of natural numbers, we use the standard formula for the sum of the squares of the first $n$ natural numbers, and subtract the missing initial terms. Note that the final term $20^2$ acts as a disconnected standalone addition to the primary series. $$ \sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6} $$ ### Step-by-Step Solution * **Given Series:** $(5^2 + 6^2 + .... + 10^2) + 20^2$ * **Calculation / Deduction:** First, find the sum of the continuous series from $5^2$ to $10^2$. This is equivalent to finding the sum from $1^2$ to $10^2$, and subtracting the sum from $1^2$ to $4^2$. Sum from $1$ to $10$ ($n=10$): $$ \frac{10(11)(21)}{6} = \frac{2310}{6} = 385 $$ Sum from $1$ to $4$ ($n=4$): $$ \frac{4(5)(9)}{6} = \frac{180}{6} = 30 $$ Sum from $5^2$ to $10^2$: $$ 385 - 30 = 355 $$ Now, add the standalone $20^2$ term at the end of the expression: $$ 20^2 = 400 $$ $$ Total = 355 + 400 = 755 $$ ### Exam Strategy & Shortcut **Unit Digit Check:** If you are short on time, calculate the unit digits of the squares mentally. $5^2 \rightarrow 5$ $6^2 \rightarrow 6$ $7^2 \rightarrow 9$ $8^2 \rightarrow 4$ $9^2 \rightarrow 1$ $10^2 \rightarrow 0$ $20^2 \rightarrow 0$ Summing these unit digits: $5 + 6 + 9 + 4 + 1 + 0 + 0 = 25$. The final unit digit must be $5$. Looking at the options, both $755$ and $765$ end in $5$, narrowing it down to a 50/50 guess, but calculating the actual arithmetic is safe and fast enough here. ### Common Pitfall Students often skim the question and mistakenly assume it is a continuous series from $5^2$ all the way to $20^2$. Paying close attention to the punctuation and exact numbers given is critical; the series explicitly stops at $10^2$ and then appends a singular $20^2$. ### Final Answer **Therefore, the correct answer is 755.**

Question 20

Given that $1 + 2 + 3 + 4 + .... + 10 = 55$, then the sum $6 + 12 + 18 + 24 + .... + 60$ is equal to

Accepted Answer

### Concept & Logic When evaluating an arithmetic progression, check if the series is simply a scalar multiple of a simpler, known series. Factoring out the greatest common divisor simplifies the calculation entirely. ### Step-by-Step Solution * **Given:** $1 + 2 + 3 + 4 + .... + 10 = 55$ Target Series: $6 + 12 + 18 + 24 + .... + 60$ * **Calculation / Deduction:** Observe the terms in the target series. Each term is an exact multiple of $6$. Factor out the constant multiplier ($6$) from the entire series: $$ 6 	imes (1) + 6 	imes (2) + 6 	imes (3) + .... + 6 	imes (10) $$ $$ = 6 	imes (1 + 2 + 3 + 4 + .... + 10) $$ The bracketed expression is exactly the series provided in the given statement. Substitute the known value ($55$) into the equation: $$ = 6 	imes 55 $$ Calculate the final product: $$ 6 	imes 50 = 300 $$ $$ 6 	imes 5 = 30 $$ $$ 300 + 30 = 330 $$ ### Exam Strategy & Shortcut **Direct Multiplication:** The core of aptitude tests is pattern recognition. Immediately notice that $6 = 1 	imes 6$, and $60 = 10 	imes 6$. The whole series is scaled by $6$. You just need to multiply the given total sum by the scalar: $55 	imes 6 = 330$. ### Common Pitfall A common time-wasting pitfall is ignoring the given hint ($1$ to $10 = 55$) and manually applying the arithmetic progression sum formula $S = \frac{n}{2}(a + l)$ to the target series. While it yields the right answer, it eats up valuable seconds in an exam environment. ### Final Answer **Therefore, the correct answer is 330.**

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