Question 1

Two consecutive even positive integers, sum of the squares of which is 1060, are

Accepted Answer

Concept & Logic: This problem maps to forming a quadratic equation based on consecutive even numbers. However, in an exam setting, relying on standard quadratic factoring is slow; understanding unit digits and perfect squares is the primary logical key. Consecutive even integers can be algebraically represented as x and x + 2. Step-by-Step Solution: • Given: The sum of the squares of two consecutive even integers is 1060. • Let the numbers be x and x + 2. • Set up the mathematical equation: x^2 + (x + 2)^2 = 1060 • Expand the binomial: x^2 + x^2 + 4x + 4 = 1060 2x^2 + 4x - 1056 = 0 • Simplify by dividing by 2: x^2 + 2x - 528 = 0 • Calculation: Factor the quadratic equation. We need two numbers that multiply to -528 and add to 2. (x + 24)(x - 22) = 0 • Since the question specifies *positive* integers, x = -24 is invalid. Therefore, x = 22. • The consecutive even number is x + 2 = 24. Exam Strategy & Shortcut: Never solve the quadratic equation in the exam. Use Option Elimination via Unit Digits. We need the sum of the squares to end in a 0 (1060). Let's check the unit digits of the squares of the options: • (a) 12 and 14: 2^2 + 4^2 = 4 + 16 = 20. Unit digit is 0. Sum is 144 + 196 = 340. Too small. • (b) 20 and 22: 0^2 + 2^2 = 0 + 4 = 4. Unit digit is 4, not 0. Reject immediately. • (c) 22 and 24: 2^2 + 4^2 = 4 + 16 = 20. Unit digit is 0. Let's estimate: 20^2 + 20^2 = 800. 25^2 + 25^2 = 1250. The answer must be closely bounded between these. Let's calculate: 484 + 576 = 1060. This matches perfectly. • (d) 15 and 18 are not consecutive even integers. Reject immediately. Common Pitfall: The main pitfall is wasting 2-3 minutes trying to manually factorize large numbers like 528 during the quadratic step. Another mistake is forgetting the definition of "consecutive even" and representing the numbers as x and x+1. Final Answer: Therefore, the correct answer is 22 and 24.

Question 2

Consider the following statements for the sequence of numbers given below:  11, 111, 1111, 11111, ....  1. Each number can be expressed in the form $(4m + 3)$, where $m$ is a natural number. 2. Some numbers are squares.  Which of the above statement…

Accepted Answer

Concept & Logic: This problem tests two distinct number system concepts: the divisibility rule for 4 and the unit digit/remainder properties of perfect squares. • Divisibility by 4: A number is divisible by 4 if its last two digits form a number divisible by 4. • Properties of Squares: Any perfect square, when divided by 4, will *always* leave a remainder of exactly 0 (if it is an even square) or 1 (if it is an odd square). A perfect square can never leave a remainder of 2 or 3 when divided by 4. Step-by-Step Solution: • Evaluate Statement 1: Can they be expressed as (4m + 3)? • The numbers in the sequence all end in "11". • When you divide 11 by 4, the quotient is 2 and the remainder is 3. (11 = 4 × 2 + 3). • For any larger number ending in 11 (like 111, 1111), the hundreds, thousands, etc., places are all multiples of 100. Since 100 is perfectly divisible by 4, the remainder of the entire number depends *only* on the last two digits (11). • Therefore, every number in this sequence leaves a remainder of 3 when divided by 4. • This means they can all be written as 4m + 3. Statement 1 is correct. • Evaluate Statement 2: Are some numbers squares? • As proven in Statement 1, every number in this sequence is of the form 4m + 3. • By the fundamental properties of perfect squares, no perfect square can be of the form 4m + 3 (they must be 4m or 4m+1). • Therefore, none of these numbers can possibly be perfect squares. Statement 2 is incorrect. Exam Strategy & Shortcut: For Statement 1, don't try to prove it algebraically for the whole sequence. Just test the first term: 11 = 4(2) + 3. Test the second: 111 = 4(27) + 3. It holds true. For Statement 2, remember the universal rule: Perfect squares only end in 0, 1, 4, 5, 6, or 9. Since these numbers end in 1, they *could* theoretically be squares based on the unit digit, but the modulo 4 rule (4m+3) instantly disqualifies them without any calculation. Common Pitfall: Students often get confused by the unit digit "1" in Statement 2. Since 121 (11^2) ends in 1, and 81 (9^2) ends in 1, students guess that somewhere in the infinite sequence, one of these numbers might be a square. Relying on unit digits alone is a trap; always use modulo/remainder rules to verify. Final Answer: Therefore, the correct answer is 1 only.

Question 3

Directions: Read the table carefully and answer the following question.  The table shows the data about distance (km) covered from one point to another point by the vehicle used between each point and speed of each vehicle (km/h). Note: (i) If one p…

Accepted Answer

Concept & Factorization: This problem requires reverse-engineering a multiplication equation by testing single-digit factors of a given product. PQR × X = 2634 Step-by-Step Solution: 1. Find single-digit factors of 2634: We need X to be a single digit (1 \le X \le 9) such that dividing 2634 by X results in a three-digit integer (PQR). - If X = 2: 2634 / 2 = 1317 (This is a 4-digit number, so X 
eq 2) - If X = 3: 2634 / 3 = 878. Check condition: Is the tens digit (7) equal to half of X (1.5)? No. - If X = 6: 2634 / 6 = 439. Check condition: Is the tens digit (3) equal to half of X (6 / 2 = 3)? Yes! 2. We have identified the numbers: PQR = 439 and X = 6. 3. Evaluate the given statements about 439: - (i) Unit digit is 9: The unit digit of 439 is indeed 9. (True) - (ii) It is a prime number: The square root of 439 is approximately 20.9. Testing prime factors up to 19 (2, 3, 5, 7, 11, 13, 17, 19): 439 is not divisible by any of them. Thus, it is prime. (True) - (iii) Sum of the digits is 18: 4 + 3 + 9 = 16 
eq 18. (False) 4. Therefore, only statements (i) and (ii) are true. Exam Strategy & Shortcut: To quickly find single-digit factors of 2634, use divisibility rules. The number ends in 4 (even, so divisible by 2). Sum of digits 2+6+3+4 = 15 (divisible by 3). Since it is divisible by both 2 and 3, it is definitely divisible by 6. Testing 6 first directly gives you the correct framework. Common Pitfall: Failing to verify prime numbers correctly. Remember you only need to test divisibility by prime numbers up to the square root of the target number to definitively prove it is prime. Final Answer: Therefore, the correct answer is Only (i) and (ii).

Question 4

A four-digit odd number 'M' whose unit digit is not 5 when divided by a prime number gives resultant as a three-digit number 'N' whose tens place digit is 0. The sum of digits of M is 8 while that of N is 5. If M < 3400 then which of the following s…

Accepted Answer

Concept & Number Properties: This problem requires synthesizing multiple number properties: divisibility, prime numbers, digit sums, and parity (odd/even). Let M = N × p, where p is a prime number. Since M is odd, both factors N and p must strictly be odd numbers. Step-by-Step Solution: 1. Determine the possibilities for N: N is a 3-digit number. Tens place of N = 0. Sum of digits of N = 5. Possible configurations for N: 104, 203, 302, 401. Since M is odd, N must be odd. This eliminates 104 and 302. N is either 203 or 401. 2. Test prime multipliers for N = 401: M = 401 × p (where p is an odd prime, so p \in \{3, 5, 7, \dots\}) - If p=3 \implies M = 1203. Sum of digits = 1+2+0+3 = 6 (Does not equal 8). - If p=5 \implies M = 2005. Unit digit is 5 (Not allowed by the rules). - If p=7 \implies M = 2807. Sum of digits = 2+8+0+7 = 17 (Does not equal 8). 3. Test prime multipliers for N = 203: M = 203 × p - If p=3 \implies M = 609 (Not a 4-digit number). - If p=5 \implies M = 1015. Unit digit is 5 (Not allowed). - If p=7 \implies M = 1421. Sum of digits = 1+4+2+1 = 8. Unit digit is 1 (not 5). M < 3400. This perfectly satisfies all conditions! So, M = 1421 and N = 203. 4. Evaluate the statements: - Statement I: Product of digits of M = 1 × 4 × 2 × 1 = 8. Sum of digits of N = 2 + 0 + 3 = 5. Is 8 three more than 5? Yes (8 = 5 + 3). (TRUE) - Statement II: Hundred place digit of M is 4. Hundred place digit of N is 2. Difference = 4 - 2 = 2. (TRUE) - Statement III: Sum of M and N = 1421 + 203 = 1624. This is not 1729. (FALSE) Statements I and II are correct. Exam Strategy & Shortcut: Limit your search space immediately. The condition "Tens digit is 0" and "Sum is 5" restricted N to only four numbers instantly. Identifying that N MUST be odd halves the work. Always process the most restrictive constraints first to rapidly reach the correct numbers. Common Pitfall: A frequent error is forgetting that if a product (M) is odd, ALL of its integer factors must be odd. Students might waste time testing even numbers for N or p. Final Answer: Therefore, the correct answer is Only (I) and (II).

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