Question 1

Find the greatest number of $4$ digits which when divided by $4$, $5$, $6$, $7$ and $8$ leaves $1$, $2$, $3$, $4$ and $5$ as remainders.

Accepted Answer

Concept & Strategy: Common difference K = Divisor - Remainder = 3. Formula: N = Multiple of LCM - 3. Step-by-Step Solution: • Find LCM of (4, 5, 6, 7, 8): LCM = 840. • Greatest 4-digit number: 9999 \div 840 = 11 remainder 759. 9999 - 759 = 9240. • Apply constant difference: 9240 - 3 = 9237. Exam Strategy & Shortcut: Check options: (Option + 3) must be divisible by 840. 9237 + 3 = 9240. 9240 \div 840 = 11. Correct. Common Pitfall: Forgetting the final subtraction of 3. Always subtract the common difference K at the end. Final Answer: Therefore, the correct answer is 9237.

Question 2

Find the least number which when divided by 16, 18, 20 and 25 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.

Accepted Answer

Concept & Formula: To find a number that leaves a remainder r when divided by multiple divisors a, b, c, we use the Least Common Multiple (LCM): \Number = \LCM(a, b, c) \× k + r where k is a positive integer chosen such that the condition for the secondary divisor is fully satisfied. Step-by-Step Solution: Given: * Divisors: 16, 18, 20, 25 • Remainder in each case: 4 • Complete divisibility by: 7 Calculation: 1. Find the LCM of 16, 18, 20, and 25: • 16 = 2^4 • 18 = 2 \× 3^2 • 20 = 2^2 \× 5 • 25 = 5^2 • \LCM = 2^4 \× 3^2 \× 5^2 = 16 \× 9 \× 25 = 3600 2. Express the required number in the standard form: \Number = 3600k + 4 3. Test values of k such that (3600k + 4) is perfectly divisible by 7: • Simplify 3600 \pmod 7: 3600 = 7 \× 514 + 2 • So, the expression becomes: 2k + 4 • For k = 1 \implies 2(1) + 4 = 6 (not divisible by 7) • For k = 2 \implies 2(2) + 4 = 8 (not divisible by 7) • For k = 3 \implies 2(3) + 4 = 10 (not divisible by 7) • For k = 4 \implies 2(4) + 4 = 12 (not divisible by 7) • For k = 5 \implies 2(5) + 4 = 14 (divisible by 7) 4. Substitute k = 5 into the original expression: \Number = 3600 \× 5 + 4 = 18000 + 4 = 18004 Exam Strategy & Shortcut: Option Elimination: Check the options for divisibility by 7 directly. • 17004: 17004 \div 7 = 2429.14 (No) • 18000: Not possible since it leaves a remainder of 0 when divided by 20 and 25, not 4. • 18002: 18002 \div 7 = 2571.71 (No) • 18004: 18004 \div 7 = 2572 (Yes) Only 18004 satisfies the divisibility condition for 7. Common Pitfall: Students often forget to find the specific value of k and mark a value like 3600 + 4 = 3604 if it is in the options, ignoring the divisibility test for the number 7. Always verify the secondary divisibility condition. Final Answer: Therefore, the correct answer is 18004.

Question 3

An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 a.m. The next time, when they would beep together at the earliest is

Accepted Answer

Concept & Formula: Simultaneous events occurring at different regular intervals repeat together at time durations that are common multiples of their individual intervals. The earliest next occurrence is found using the LCM: \Time Interval = \LCM(\Interval_1, \Interval_2) Step-by-Step Solution: Given: * Interval of first device = 60 seconds • Interval of second device = 62 seconds • Initial simultaneous beep time = 10:00 a.m. Calculation: 1. Find the LCM of 60 and 62: • 60 = 2^2 \× 3 \× 5 • 62 = 2 \× 31 • \LCM = 2^2 \× 3 \× 5 \× 31 = 60 \× 31 = 1860 \ seconds 2. Convert the total seconds into minutes: \Minutes = \(1860/60) = 31 \ minutes 3. Add this interval to the initial starting time: \Next Time = 10:00 \ a.m. + 31 \ minutes = 10:31 \ a.m. Exam Strategy & Shortcut: Instead of computing the large multiplication fully, note that: \LCM(60, 62) = 60 \× \(62/\HCF)(60, 62) = 60 \× \(62/2) = 60 \× 31 \ seconds Dividing by 60 to get minutes immediately cancels out the 60, leaving exactly 31 minutes. Add 31 minutes directly to 10 a.m. to get 10:31 a.m. Common Pitfall: A frequent error is forgetting to convert the calculated LCM from seconds to minutes, or making a manual error during the multiplication step (60 \× 31). Final Answer: Therefore, the correct answer is 10.31 a.m.

Question 4

Four different electronic devices make a beep after every 30 minutes, 1 hour, $1\frac{1}{2}$ hour and 1 hour 45 minutes respectively. All the devices beeped together at 12 noon. They will again beep together at

Accepted Answer

Concept & Formula: To find when multiple periodic events will synchronize again, we calculate the Least Common Multiple (LCM) of their individual periods. Next Simultaneous Event = Initial Time + LCM(t_1, t_2, t_3, \dots) Crucially, all time units must be converted to the same metric (e.g., minutes) before calculating the LCM. Step-by-Step Solution: Given: • Device 1: 30 minutes • Device 2: 1 hour = 60 minutes • Device 3: 1(1/2) hour = 90 minutes • Device 4: 1 hour 45 minutes = 105 minutes • Start Time: 12 noon Calculation: 1. Find the LCM of 30, 60, 90, and 105: • 30 = 2 × 3 × 5 • 60 = 2^2 × 3 × 5 • 90 = 2 × 3^2 × 5 • 105 = 3 × 5 × 7 2. Take the highest power of each prime factor: • LCM = 2^2 × 3^2 × 5 × 7 = 4 × 9 × 5 × 7 • LCM = 1260 minutes 3. Convert the LCM back into hours: Hours = (1260/60) = 21 hours 4. Add 21 hours to the starting time (12 noon): • 12 noon + 12 hours = 12 midnight • 12 midnight + 9 hours = 9 a.m. (the next day) Exam Strategy & Shortcut: Instead of full prime factorization, use the step-by-step LCM method: • LCM(30, 60) = 60 • LCM(60, 90) = 180 • Now find LCM(180, 105) • Both are divisible by 15: 180 = 15 × 12 and 105 = 15 × 7. • Since 12 and 7 are co-prime, LCM = 15 × 12 × 7 = 180 × 7 = 1260 minutes. Dividing 1260 by 60 mentally (126 / 6) gives 21 hours instantly. Common Pitfall: Failing to correctly convert mixed time formats (1(1/2) hours or 1 hour 45 minutes) strictly into minutes. Trying to find the LCM using decimal hours (like 1.5 or 1.75) without applying fraction LCM rules will lead to incorrect or messy calculations. Final Answer: Therefore, the correct answer is 9 a.m.

Question 5

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they meet again at the starting…

Accepted Answer

Concept & Strategy: For objects moving in a circle, the time required to meet again precisely at the starting point is determined by calculating the Least Common Multiple (LCM) of the individual lap times. Meeting Time = LCM(T_1, T_2, T_3) Step-by-Step Solution: Given: • Lap time for A = 252 seconds • Lap time for B = 308 seconds • Lap time for C = 198 seconds Calculation: 1. Perform prime factorization for each number to find the LCM: • 252 = 2 × 126 = 2^2 × 63 = 2^2 × 3^2 × 7 • 308 = 2 × 154 = 2^2 × 77 = 2^2 × 7 × 11 • 198 = 2 × 99 = 2 × 3^2 × 11 2. Identify the highest power of each prime factor present (2, 3, 7, 11): • Highest power of 2 is 2^2 • Highest power of 3 is 3^2 • Highest power of 7 is 7^1 • Highest power of 11 is 11^1 3. Multiply these together to find the LCM: LCM = 2^2 × 3^2 × 7 × 11 LCM = 4 × 9 × 7 × 11 LCM = 36 × 77 = 2772 seconds 4. Convert the total seconds into minutes and seconds: (2772/60) = 46.2 minutes • The whole number is 46 minutes. • The remainder is 0.2 minutes. Convert back to seconds: 0.2 × 60 = 12 seconds. • Total time = 46 minutes and 12 seconds. Exam Strategy & Shortcut: Look closely at the numbers: 252, 308, 198. They are all multiples of 2. Even better, they all seem related to 11 and 9 rules. 198 is obviously a multiple of 9 and 11. 308 is 28 × 11. 252 is a multiple of 9. If you factor out common multipliers iteratively: LCM(252, 308, 198) = 2 × LCM(126, 154, 99) You can quickly build the factor tree without doing full primes for each if you recognize 11 and 7 as key factors, ultimately yielding 2772. Another shortcut: 2772 ends in a 2. When divided by 60, the remainder must end in a 2 (2772 \pmod{10} = 2). The only option ending in a "2" in the seconds column is option (d) 12 seconds. Common Pitfall: During long multiplication like 36 × 77, simple arithmetic errors can cascade, leading to the wrong total seconds. Additionally, students sometimes divide by 60 and see 46.2 on their scratchpad and mistakenly think it means "46 minutes and 20 seconds", rather than correctly doing 0.2 × 60 = 12 seconds. Final Answer: Therefore, the correct answer is 46 minutes 12 seconds.

Question 6

Three wheels can complete 40, 24 and 16 revolutions per minute respectively. There is a red spot on each wheel that touches the ground at time zero. After how much time, all these spots will simultaneously touch the ground again?

Accepted Answer

Concept & Strategy: To find when multiple repeating cyclical events synchronize again, we must calculate the Least Common Multiple (LCM) of their individual time periods. First, we must convert the given speeds (revolutions per minute) into the time taken for a *single* revolution. Time per revolution = \frac{Total Time}{Number of Revolutions} For fractions, the LCM is calculated as: LCM of fractions = \frac{LCM of Numerators}{HCF of Denominators} Step-by-Step Solution: Given: • Speed of Wheel 1 = 40 rev/min • Speed of Wheel 2 = 24 rev/min • Speed of Wheel 3 = 16 rev/min Calculation: 1. Find the time taken for one complete revolution for each wheel in seconds (since options are primarily in seconds and minutes, seconds is a better base unit): • Wheel 1: 60 seconds / 40 rev = (60/40) = (3/2) seconds • Wheel 2: 60 seconds / 24 rev = (60/24) = (5/2) seconds • Wheel 3: 60 seconds / 16 rev = (60/16) = (15/4) seconds 2. Calculate the LCM of these fractional time periods ((3/2), (5/2), (15/4)): • Find the LCM of the numerators (3, 5, 15): The lowest common multiple of 3, 5, and 15 is 15. • Find the HCF of the denominators (2, 2, 4): The highest common factor of 2, 2, and 4 is 2. 3. Apply the fraction formula: Simultaneous Time = (15/2) seconds 4. Convert to a mixed fraction to match the options: (15/2) seconds = 7.5 seconds = 7(1/2) sec Exam Strategy & Shortcut: Always quickly convert "rate" to "time" before finding the LCM. Taking the LCM of 40, 24, and 16 gives 240, which would incorrectly imply minutes or revolutions. By converting to seconds first ((3/2), (5/2), (15/4)), you immediately see the numerator must be a multiple of 15, guiding you directly to 15/2. Common Pitfall: The most dangerous pitfall is finding the LCM of the given numbers (40, 24, 16) directly. This gives the number of *revolutions* until they align, not the *time*. Always ensure you are finding the LCM of the actual time intervals between the events. Final Answer: Therefore, the correct answer is 7(1/2) sec.

Question 7

A pendulum strikes 5 times in 3 seconds and another pendulum strikes 7 times in 4 seconds. If both pendulums start striking at the same time, how many clear strikes can be listened in 1 minute?

Accepted Answer

Concept & Logic: To find the total number of "clear strikes" heard from two different pendulums, we must use the principle of inclusion-exclusion. A clear strike is heard whenever *either* pendulum strikes. If they strike at the exact same moment, they blend into one clear strike. |A \cup B| = |A| + |B| - |A \cap B| Where |A| and |B| are the number of strikes of each pendulum, and |A \cap B| is the number of simultaneous strikes. Step-by-Step Solution: Given: • Pendulum A rate: 5 strikes in 3 seconds \implies 1 strike every (3/5) seconds. • Pendulum B rate: 7 strikes in 4 seconds \implies 1 strike every (4/7) seconds. • Total time window: 1 minute = 60 seconds. • (Note: Based on standard exam interpretations matching the options, we evaluate the strikes occurring *within* the 60 seconds, excluding the initial simultaneous strike at t=0). Calculation: 1. Calculate individual strikes in 60 seconds: • Pendulum A strikes: (60/3/5) = 60 × (5/3) = 100 strikes • Pendulum B strikes: (60/4/7) = 60 × (7/4) = 105 strikes 2. Calculate the interval of simultaneous strikes: • Find the LCM of the time intervals ((3/5) and (4/7)). • LCM = \frac{LCM(3, 4)}{HCF(5, 7)} = (12/1) = 12 seconds. • They strike together every 12 seconds. 3. Calculate the number of simultaneous strikes in 60 seconds: • Simultaneous strikes = (60/12) = 5 strikes. 4. Calculate the total clear strikes heard: • Total Clear Strikes = 100 + 105 - 5 = 200. Exam Strategy & Shortcut: Instead of writing out fraction formulas, think of it as rates: Pendulum A does (5/3) strikes/sec. In 60s, it does 60 × (5/3) = 100. Pendulum B does (7/4) strikes/sec. In 60s, it does 60 × (7/4) = 105. Total independent strikes = 205. The overlap happens at the LCM of their fractional periods ((3/5) and (4/7)), which is 12 seconds. 60 / 12 = 5 overlaps. Subtract the overlaps: 205 - 5 = 200. Common Pitfall: A major point of confusion is whether to count the strike at t=0. If you calculate (60 / interval) + 1, you get 101, 106, and 6 overlaps, yielding 201. Since 201 is not an option, the question framework clearly asks for the strikes that occur *after* the initial release during the 1-minute span. Always let the options guide the strictness of the boundary conditions. Final Answer: Therefore, the correct answer is 200.

Question 8

Find the HCF of 132, 204 and 228.

Accepted Answer

Concept & Strategy: The Highest Common Factor (HCF) of a set of numbers is the largest number that divides all of them without leaving a remainder. The fastest way to find the HCF of large numbers is using the Difference Method: The HCF of any two numbers must also be a factor of their difference. HCF(a, b) must divide |a - b| Step-by-Step Solution: Given: • Numbers: 132, 204, and 228. Calculation: 1. Find the differences between the adjacent numbers: • Difference 1: 204 - 132 = 72 • Difference 2: 228 - 204 = 24 2. The HCF of the three numbers must be a factor of the smallest difference, which is 24. 3. Look at the given options to find factors of 24: • (a) 12 - Is a factor of 24. • (b) 18 - Not a factor of 24. • (c) 6 - Is a factor of 24. • (d) 21 - Not a factor of 24. 4. Test the highest possible candidate first, which is 12: • 132 \div 12 = 11 (Divisible) • 204 \div 12 = 17 (Divisible) • 228 \div 12 = 19 (Divisible) 5. Since 11, 17, and 19 are prime to each other (co-prime), 12 is indeed the highest common factor. Exam Strategy & Shortcut: Use the difference method combined with option elimination. The difference between 228 and 204 is exactly 24. The HCF *cannot* be larger than 24, and it *must* divide 24 perfectly. Looking at the options (12, 18, 6, 21), 18 and 21 are instantly eliminated. Between 12 and 6, always test the larger number first since you are looking for the *Highest* Common Factor. Common Pitfall: Resorting to full prime factorization for all three numbers (132 = 2^2 × 3 × 11, etc.) takes too much time under exam conditions and increases the chance of minor calculation errors. Use the difference method to save time. Final Answer: Therefore, the correct answer is 12.

Question 9

If three numbers are $2a$, $5a$ and $7a$, what will be their LCM?

Accepted Answer

Concept & Formula: The Least Common Multiple (LCM) of algebraic terms is found by multiplying the LCM of their numerical coefficients by the highest power of each variable present. LCM(cx, cy) = LCM(c) × Highest Power of Variables Step-by-Step Solution: Given: • The three numbers are 2a, 5a, and 7a. Calculation: 1. Separate the numerical coefficients from the variables: • Coefficients: 2, 5, 7 • Variable part: a, a, a 2. Find the LCM of the numerical coefficients: • Since 2, 5, and 7 are all prime numbers, their LCM is simply their product. • LCM(2, 5, 7) = 2 × 5 × 7 = 70 3. Find the LCM of the variable parts: • The highest power of the variable 'a' across all three terms is just a^1 (or a). 4. Combine both parts: • Total LCM = 70 × a = 70a Exam Strategy & Shortcut: Recognize immediately that 2, 5, and 7 share no common factors (they are mutually prime). Therefore, you just multiply them (2 × 5 × 7 = 70). Because the variable 'a' is common to all terms and raised to the first power in each, it simply tags along once. 70 × a = 70a. Common Pitfall: A very common mistake in algebra is multiplying the variables together as if they were independent distinct terms, resulting in 70a^3. Remember, LCM looks for the *highest power* of the common base present in any single term, it does not multiply the bases together. Final Answer: Therefore, the correct answer is 70a.

Question 10

A number $x$ is divided by 7. When this number is divided by 8, 12 and 16. It leaves a remainder 3 in each case. The least value of $x$ is:

Accepted Answer

Concept & Strategy: To find a number that leaves the same remainder when divided by multiple divisors, we use the LCM (Least Common Multiple) of those divisors. The general form of the number is N = LCM(a, b, c) × k + R. We then test integer values of k to satisfy the secondary condition (divisibility by 7). Step-by-Step Solution: • Given: Divisors are 8, 12, and 16. Remainder R = 3. The number must be a multiple of 7 (implied by "divided by 7" without a remainder, a common phrasing quirk for "divisible by 7" in exams). • Step 1: Find the LCM of 8, 12, and 16. • 8 = 2^3 • 12 = 2^2 × 3 • 16 = 2^4 • LCM(8, 12, 16) = 2^4 × 3 = 48 • Step 2: Express the number in the general form. • Number x = 48k + 3 • Step 3: Find the least value of k for which x is divisible by 7. • For k = 1: x = 48(1) + 3 = 51 (Not divisible by 7) • For k = 2: x = 48(2) + 3 = 99 (Not divisible by 7) • For k = 3: x = 48(3) + 3 = 147. And 147 \div 7 = 21. (Divisible by 7) Exam Strategy & Shortcut: You can completely bypass the LCM calculation by using Option Elimination. The question states the number is divisible by 7. Test the given options directly: • (a) 148 \div 7 = 21.14 (No) • (b) 149 \div 7 = 21.28 (No) • (c) 150 \div 7 = 21.42 (No) • (d) 147 \div 7 = 21 (Yes) Only 147 is a perfect multiple of 7. Common Pitfall: Students often find the LCM (48) and just add 3 to get 51, forgetting to check the divisibility condition for 7. Always verify that your final answer satisfies *all* constraints given in the problem. Final Answer: Therefore, the correct answer is 147.

Question 11

Three plots having areas 110, 130 and 190 square metres are to be subdivided into flower beds of equal size. If the breadth of a bed is 2 metre, the maximum length of a bed can be

Accepted Answer

Concept & Highest Common Factor (HCF): To divide different total areas into equal-sized beds of maximum possible area, find the Highest Common Factor (HCF) of the given total areas. Area of Bed = Length × Breadth Step-by-Step Solution: 1. Find the maximum area of a single flower bed: - The areas of the plots are 110, 130, and 190 sq. m. - The maximum area of an equal-sized bed must perfectly divide all three plot areas. - Maximum Area = HCF(110, 130, 190). - The HCF of 110, 130, and 190 is 10 (since 11 × 10 = 110, 13 × 10 = 130, 19 × 10 = 190, and 11, 13, 19 are prime to each other). - So, the maximum area of a flower bed is 10 sq. m. 2. Find the maximum length of the bed: - The breadth of the bed is given as 2 m. - Area = Length × Breadth \Rightarrow 10 = Length × 2. - Length = (10/2) = 5 m. Exam Strategy & Shortcut: Recognize instantly that dividing distinct groups into maximum equal components requires the HCF. The trailing zero in 110, 130, 190 makes 10 the obvious HCF. Since Area = 10 and Breadth = 2, Length = 5. This can be solved mentally in 5 seconds. Common Pitfall: A common mistake is trying to find the LCM instead of the HCF, or forgetting to divide the resulting HCF (area) by the given breadth to find the length. Final Answer: Therefore, the correct answer is 5 m.

Question 12

The circumferences of the front and rear wheels of a bicycle are 3.5 m and 3 m respectively. If the vehicle is moving at a speed of 15 m/sec, the shortest time in which both the wheels will make a whole number of turns is

Accepted Answer

Concept & Formula: Both wheels will complete a whole number of turns when the total distance covered is a common multiple of their circumferences. The shortest distance is their Least Common Multiple (LCM). Time = \frac{Distance}{Speed} Step-by-Step Solution: • Given: • Circumference of front wheel (C_1) = 3.5 m = (7/2) m • Circumference of rear wheel (C_2) = 3 m • Speed (S) = 15 m/sec • Find the LCM of the circumferences: • To find the LCM of a fraction and an integer: LCM((7/2), (3/1)) = \frac{LCM of numerators (7, 3)}{HCF of denominators (2, 1)} • LCM(7, 3) = 21 • HCF(2, 1) = 1 • Shortest distance = (21/1) = 21 m • Calculate the time taken: Time = \frac{21 m}{15 m/sec} = (7/5) seconds (7/5) = 1.4 seconds Exam Strategy & Shortcut: Multiply by 10 to work with integers: LCM of 35 and 30 is 210. Divide by 10 to revert to meters: 21m. Time = (21/15) = 1.4 sec. This avoids fraction rules entirely. Common Pitfall: Miscalculating the LCM of a decimal number. Always convert decimals to fractions or scale them up to integers before taking the LCM. Final Answer: Therefore, the correct answer is 1.4 seconds.

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