Q: Find the square root of 15625.

Introduction / Context:This problem asks for the exact square root of 15625. Recognizing perfect squares and using factorization shortens the solution. Such questions are typical in number system and speed math sections and help improve mental calculation skills. Given Data / Assumptions:We are given the number 15625.We must find a positive number whose square equals 15625.We are looking for an exact integer square root, not an approximation. Concept / Approach:A good approach is to factor 15625 into prime factors or recognize it as a power of 5. Many powers of 5 end in 25, and 125 squared is a well known value. We can test 125 and confirm by direct multiplication, or factor systematically to be certain. Step-by-Step Solution:Step 1: Notice that 15625 ends with 625, which suggests a relationship with powers of 5.Step 2: Recall that 5^4 = 625 and 5^6 = 15625.Step 3: Therefore, 15625 = 5^6.Step 4: The square root of 5^6 is 5^(6 / 2) = 5^3.Step 5: Compute 5^3 = 5 × 5 × 5 = 125.Step 6: Thus, the square root of 15625 is 125. Verification / Alternative check:Verify by squaring 125 directly. Compute 125 × 125. First, 125 × 100 = 12500. Next, 125 × 25 = 125 × (100 / 4) = 12500 / 4 = 3125. Add these results: 12500 + 3125 = 15625. Since 125 squared equals 15625 exactly, our square root is correct. Why Other Options Are Wrong:If we square 135, we get 18225, which is larger than 15625. Squaring 145 gives 21025, also too large. Squaring 150 gives 22500, much larger than 15625. None of these match 15625, so they cannot be the square root. Common Pitfalls:Some learners guess nearby values and do not verify by squaring, which can lead to mistakes. Others may not recognize 15625 as a power of 5 and attempt lengthy factorization. Remembering common squares such as 125^2 = 15625 can speed up calculations significantly. Final Answer:The square root of 15625 is 125.

Question 1

If a positive integer n leaves a remainder of 2 when divided by 5, which of the following statements must always be true?

Accepted Answer

Introduction / Context:This question tests understanding of modular arithmetic and how remainders behave when we add constants to a number. We are told how n behaves when divided by 5 and asked which statement must hold for all such integers n. The idea is to write n in a standard modular form and examine how expressions like n + 1 or n + 3 behave. Given Data / Assumptions:n is a positive integer.When n is divided by 5, the remainder is 2.Symbolically, this means n ≡ 2 (mod 5).We must identify which statement is always true for every such n. Concept / Approach:If n leaves remainder 2 when divided by 5, we can write n in the form n = 5k + 2 for some integer k. We then substitute this expression into each option. For each statement, we check whether it holds for all integers k or fails for at least one choice of k. A statement is a must-true property only if it holds for every possible k. Step-by-Step Solution:Step 1: Write n as n = 5k + 2, where k is an integer.Step 2: Examine option D, which states that n + 3 is divisible by 5.Step 3: Compute n + 3 = (5k + 2) + 3 = 5k + 5.Step 4: Factor this as 5(k + 1). Since k + 1 is an integer, n + 3 is always a multiple of 5.Step 5: Therefore, n + 3 is divisible by 5 for all integers k. Verification / Alternative check:Test a few specific values. For k = 1, n = 7. Then n + 3 = 10, which is divisible by 5. For k = 3, n = 17, and n + 3 = 20, again divisible by 5. For k = 5, n = 27, and n + 3 = 30, still divisible by 5. These examples support the algebraic reasoning that n + 3 is always divisible by 5. Why Other Options Are Wrong:The statement that n is odd is not always true. For k = 2, n = 12, which is even. The claim that n + 1 cannot be a prime is false because for k = 2, n + 1 = 13, which is a prime number. The statement about (n + 2) divided by 7 having remainder 2 is not guaranteed; the remainder depends on k and can vary, so it is not a must-true condition. Common Pitfalls:A common mistake is to test only one or two values and assume a pattern without algebraic proof. Another error is to confuse what is sometimes true with what must always be true. Writing n as 5k + 2 and analyzing each option symbolically ensures the conclusion is valid for all possible integers n. Final Answer:The statement that must always hold is that n + 3 is divisible by 5.

Question 2

If x and y are real numbers and the quotient x / y is an integer, which of the following statements need NOT always be true?

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Introduction / Context:This logical reasoning question explores what must or must not be true when a quotient x / y is known to be an integer. It is important to distinguish between conditions that are necessary for the quotient to be an integer and additional conditions that may or may not hold, such as the sign of the numbers. Given Data / Assumptions:The quotient x / y is an integer.x and y are real numbers, with y not equal to zero.We must identify which statement is not required to be true in every such situation. Concept / Approach:If x / y is an integer, then x can be written as y times some integer k, that is, x = k y. From this, we can infer that x is divisible by y and that y cannot be zero. However, nothing in this relationship forces x and y to be positive, because the integer k can be positive or negative and y can be positive or negative as long as their product produces x. We examine each statement in this light. Step-by-Step Solution:Step 1: From x / y being an integer, write x = k y, where k is an integer and y is nonzero.Step 2: This directly implies that x is divisible by y, so statement A must always be true.Step 3: Because division by zero is undefined, y must be a nonzero number for x / y to even exist, so statement B must also always be true.Step 4: Statement C simply repeats the given information that x / y is an integer, so it is trivially true whenever the condition is satisfied.Step 5: Statement D claims that both x and y are positive. This is not necessary. For example, if x = −6 and y = −3, then x / y = 2, which is still an integer, even though neither x nor y is positive. Verification / Alternative check:Test another example for statement D. Take x = 10 and y = −5. Then x / y = −2, which is also an integer. Here x is positive and y is negative, again contradicting the claim that both must be positive. These counterexamples show that statement D need not always hold, even though the quotient is an integer. Why Other Options Are Wrong:Statement A must hold because if x / y = k (integer), then x = k y, so x is exactly divisible by y. Statement B is necessary because dividing by zero is impossible. Statement C restates the given condition and is therefore always true under the problem assumption. They cannot be the answers to a need not always true question. Common Pitfalls:Some learners assume that integer division requires positive numbers, which is not correct. Integers include negative values and zero, and the quotient of two negative numbers can be a positive integer. Confusing optional conditions like positivity with necessary conditions like nonzero divisor often leads to incorrect choices. Final Answer:The statement that need not always be true is that both x and y are positive.

Question 3

If you write down all the numbers from 1 to 100, how many times will the digit 4 be written in total (counting all positions in each number)?

Accepted Answer

Introduction / Context:This counting question focuses on digit frequency in a range of numbers. You must count how many times the digit 4 appears when writing all numbers from 1 to 100. This checks attention to detail and understanding of place value, as the digit can appear in the tens place or the units place. Given Data / Assumptions:We consider all integers from 1 to 100 inclusive.We must count every occurrence of the digit 4 in every position.This includes occurrences in both the tens place and the units place. Concept / Approach:The most efficient strategy is to count occurrences separately in the tens place and the units place. For each place, we identify how many numbers in the range have a 4 in that specific position. Because some numbers, such as 44, contain the digit twice, we must ensure these are counted correctly in each place. Step-by-Step Solution:Step 1: Count the numbers between 1 and 100 that have 4 in the tens place.Step 2: The tens digit is 4 for numbers from 40 to 49 inclusive. There are 10 such numbers: 40, 41, 42, 43, 44, 45, 46, 47, 48 and 49.Step 3: Therefore, there are 10 occurrences of the digit 4 in the tens place.Step 4: Next, count numbers with 4 in the units place.Step 5: These are 4, 14, 24, 34, 44, 54, 64, 74, 84 and 94. Again, there are 10 such numbers.Step 6: Each of these contributes one occurrence of 4 in the units place, so there are 10 occurrences in the units place.Step 7: The number 44 has a 4 in both tens and units places, and we have correctly counted it twice, once in each place.Step 8: Total occurrences of the digit 4 = 10 (tens) + 10 (units) = 20. Verification / Alternative check:We can list all numbers containing the digit 4 and check the total occurrences. In the tens place they are 40 to 49 (10 occurrences). In the units place they are all numbers ending in 4, a set of 10 numbers. Counting each 4 explicitly yields the same total of 20, confirming our calculation. Why Other Options Are Wrong:The value 11 would result from incorrectly counting only the numbers that contain a 4 at least once and not counting multiple occurrences. The values 14 and 17 are intermediate and might arise from partial counting of tens or units contributions. None of these reflect the full breakdown into tens and units positions. Common Pitfalls:Many learners forget that 44 has two occurrences of the digit 4 and treat it as a single occurrence. Others may count either only the tens place or only the units place, but not both, leading to undercounting. Breaking the problem into clear place value counts helps avoid these errors. Final Answer:The digit 4 is written a total of 20 times when listing all numbers from 1 to 100.

Question 4

Two bus tickets from city K to city L and three bus tickets from city K to city M cost Rs. 77 in total. Three tickets from city K to L and two tickets from city K to M cost Rs. 73 in total. What are the individual ticket fares from city K to L and f…

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Introduction / Context:This is a classic linear equations word problem involving two unknown fares. We know the total cost of different combinations of bus tickets between city K and two other cities L and M. Using these totals, we can set up a system of equations to find the individual ticket prices. Given Data / Assumptions:Let the fare from K to L be x rupees.Let the fare from K to M be y rupees.Two tickets K to L and three tickets K to M cost 77 rupees.Three tickets K to L and two tickets K to M cost 73 rupees. Concept / Approach:We translate the word statements into algebraic equations in variables x and y. There will be two linear equations, which we can solve using substitution or elimination. Once the values of x and y are found, we match them with the correct option that lists fares for L and M in the correct order. Step-by-Step Solution:Step 1: From the first statement, write 2x + 3y = 77.Step 2: From the second statement, write 3x + 2y = 73.Step 3: Use the elimination method. Multiply the first equation by 3: 6x + 9y = 231.Step 4: Multiply the second equation by 2: 6x + 4y = 146.Step 5: Subtract the second resulting equation from the first: (6x + 9y) − (6x + 4y) = 231 − 146.Step 6: This gives 5y = 85, so y = 17.Step 7: Substitute y = 17 into 3x + 2y = 73 to find x.Step 8: Compute 3x + 2 × 17 = 73, so 3x + 34 = 73 and 3x = 39, hence x = 13. Verification / Alternative check:Check the first condition with x = 13 and y = 17. Two tickets to L and three to M cost 2 × 13 + 3 × 17 = 26 + 51 = 77, matching the problem statement. Check the second condition: three tickets to L and two to M cost 3 × 13 + 2 × 17 = 39 + 34 = 73, also matching the given total. This confirms that the fares are correct. Why Other Options Are Wrong:The pair 17 and 19 would give totals 2 × 17 + 3 × 19 = 89 and 3 × 17 + 2 × 19 = 89, which do not match 77 and 73. The pairs 23 and 17, and 13 and 19 similarly fail when substituted into the given equations. Only 13 and 17 satisfy both cost equations simultaneously. Common Pitfalls:Errors often occur when setting up the equations, such as misplacing coefficients or mixing up which fare corresponds to which city. Another mistake is performing elimination incorrectly, leading to wrong values for x or y. Writing equations carefully and checking each substitution step helps avoid these issues. Final Answer:The fare from K to L is Rs. 13 and the fare from K to M is Rs. 17.

Question 5

Which of the following numbers is greater than 4: 5, −5, −1/2 or −25?

Accepted Answer

Introduction / Context:This question checks very basic understanding of the number line and comparison of positive and negative numbers. You are asked to determine which given number is greater than 4. While simple, such questions ensure that fundamental concepts are solid. Given Data / Assumptions:The numbers to compare with 4 are 5, −5, −1/2 and −25.We must find which of these is strictly greater than 4.All numbers are real and can be compared on a standard number line. Concept / Approach:Positive numbers are greater than any negative numbers. Among positive numbers, those with larger magnitude are greater. Negative numbers lie to the left of zero on the number line and are always less than positive numbers. Fractions between −1 and 0 are still negative and less than any positive integer. Step-by-Step Solution:Step 1: Note that 5 is a positive integer greater than zero.Step 2: The numbers −5 and −25 are negative integers, both less than zero.Step 3: The number −1/2 is a negative fraction, also less than zero.Step 4: Because 4 is a positive integer, any negative number is automatically less than 4.Step 5: Therefore, −5, −1/2 and −25 are all less than 4.Step 6: Compare 5 and 4: 5 is one unit to the right of 4 on the number line, so 5 is greater than 4. Verification / Alternative check:Visualize or sketch a number line with points marked for −25, −5, −1/2, 0, 4 and 5. It is clear that −25, −5 and −1/2 lie to the left of 0, while 4 and 5 lie to the right. Since 5 is to the right of 4, it is greater. None of the negative numbers come close to 4 on the positive side of the line. Why Other Options Are Wrong:The number −5 is negative and therefore less than any positive number including 4. Similarly, −1/2 and −25 are negative and lie well to the left of 4 on the number line. None of these can be greater than 4. Only 5 is both positive and greater than 4. Common Pitfalls:Although rare at this level, some learners misinterpret the relative size of negative numbers, thinking that −25 is greater because its magnitude is larger. It is important to remember that on the number line, being further left means being smaller, so −25 is less than −5. Keeping a mental picture of the number line avoids such confusion. Final Answer:The only number greater than 4 among the options is 5.

Question 6

Which of the following lists shows the correct ascending order (from smallest to largest) of the numbers 1/3, 4/15 and 0.33?

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Introduction / Context:This question asks you to compare fractions and a decimal and then arrange them in ascending order. It tests your ability to convert between forms or to compare numbers using a common baseline, such as decimal approximations. Given Data / Assumptions:The numbers to compare are 1/3, 4/15 and 0.33.We must order them from smallest to largest.Standard decimal value comparisons apply. Concept / Approach:A straightforward method is to convert all numbers to decimals or to convert them to a common denominator. Here, converting fractions to decimal form is quick. We approximate their decimal values and compare them. It is important to recall that 1/3 is approximately 0.3333 repeating, and 0.33 is close but slightly less. Step-by-Step Solution:Step 1: Convert 1/3 to a decimal. 1 divided by 3 is approximately 0.3333 repeating.Step 2: Convert 4/15 to a decimal. Compute 4 divided by 15, which is approximately 0.2666 repeating.Step 3: The decimal 0.33 is already given and equals 0.33 exactly.Step 4: Compare the decimal values: 0.2666..., 0.33 and 0.3333....Step 5: Clearly, 0.2666... (which corresponds to 4/15) is the smallest.Step 6: Next, 0.33 is slightly less than 0.3333..., so 0.33 is in the middle.Step 7: The largest value is 0.3333..., corresponding to 1/3.Step 8: Therefore, the correct ascending order is 4/15, 0.33, 1/3. Verification / Alternative check:To verify using fractions, express each number with a common denominator, for example 300. Then 1/3 equals 100/300, 4/15 equals 80/300 and 0.33 is approximately 99/300. The order 80/300 < 99/300 < 100/300 confirms that 4/15 is smallest, 0.33 is next and 1/3 is largest. Why Other Options Are Wrong:Any option that places 1/3 before 4/15 is incorrect because 1/3 is larger than 4/15. Options that put 0.33 after 1/3 mistakenly treat them as equal or reversed, ignoring that 0.33 is slightly less than 0.3333 repeating. Only the ordering 4/15, 0.33, 1/3 matches the correct comparison. Common Pitfalls:Some learners assume 0.33 equals exactly 1/3, which is not true, and therefore may place them in the wrong order. Others might compare the fractions without converting and mis-handle the denominators. Using decimal approximations or a common denominator carefully avoids such errors. Final Answer:The correct ascending order is 4/15, 0.33, 1/3.

Question 7

What number should come next in the series 7, 4.5, 5.5, 12, 49, ?

Accepted Answer

Introduction / Context:This is a number series question where you must identify the pattern connecting consecutive terms and then extend it to find the next term. The series 7, 4.5, 5.5, 12, 49, ? may not follow a simple arithmetic or geometric progression, so we look for a pattern in how each term is generated from the previous one. Given Data / Assumptions:Given series: 7, 4.5, 5.5, 12, 49, ?We must determine the next term in this sequence.All terms are real numbers and may involve multiplication and addition patterns. Concept / Approach:We examine how each term arises from the previous one. One useful strategy is to look for multiplicative relationships combined with small additions. After testing a few possibilities, we notice that each term is approximately the previous term multiplied by a factor that doubles each time, plus 1. The factor sequence itself forms a pattern: 0.5, 1, 2, 4, then 8 for the next step. Step-by-Step Solution:Step 1: From 7 to 4.5: 7 × 0.5 + 1 = 3.5 + 1 = 4.5.Step 2: From 4.5 to 5.5: 4.5 × 1 + 1 = 4.5 + 1 = 5.5.Step 3: From 5.5 to 12: 5.5 × 2 + 1 = 11 + 1 = 12.Step 4: From 12 to 49: 12 × 4 + 1 = 48 + 1 = 49.Step 5: Observe that the multipliers are 0.5, 1, 2 and 4, which double each time.Step 6: The next multiplier should therefore be 8, continuing this doubling pattern.Step 7: Apply this rule to 49: 49 × 8 + 1 = 392 + 1 = 393.Step 8: So the next term in the series is 393. Verification / Alternative check:Verify the pattern consistency: each step is previous term times a factor plus 1, and the factors follow the sequence 1/2, 1, 2, 4, 8. This internal pattern is smooth and consistent, which strongly indicates the intended rule. No other reasonable simple pattern fits all the given terms as neatly. Why Other Options Are Wrong:The values 396, 354 and 367 do not arise from applying the identified rule of multiplying by 8 and adding 1. For example, 49 × 8 + 1 equals 393 exactly, not any of the other numbers. Alternative ad hoc patterns would require changing the rule mid-series, which is not typical in standard test questions. Common Pitfalls:Students may initially try only arithmetic differences or ratios and not notice the combined multiply then add pattern. Some might also miss the factor sequence doubling from 0.5 to 1 to 2 to 4. It helps to systematically test both additive and multiplicative relationships rather than giving up after the simplest attempts. Final Answer:The next number in the series is 393.

Question 8

Ten is added to a certain number. The resulting sum is multiplied by 7, the product is divided by 5, and then 5 is subtracted from the quotient. The final result is half of 88. What is the original number?

Accepted Answer

Introduction / Context:This problem is a typical algebraic puzzle based on a sequence of operations. Starting from an unknown number, we perform several steps and are given the final result. The task is to reverse these steps using algebra and solve for the original number. Such questions test the ability to translate verbal descriptions into equations. Given Data / Assumptions:Let the original number be N.Ten is added to N.The sum is multiplied by 7.The product is divided by 5.Five is subtracted from the quotient.The final result equals half of 88, which is 44. Concept / Approach:We convert the verbal process into a single algebraic expression in N and set it equal to 44. Then we solve the resulting linear equation step by step. It is important to keep the order of operations exactly as described and to simplify carefully. Step-by-Step Solution:Step 1: Start with N.Step 2: Add 10 to get N + 10.Step 3: Multiply this sum by 7 to get 7(N + 10).Step 4: Divide the product by 5 to get [7(N + 10)] / 5.Step 5: Subtract 5 from this quotient: [7(N + 10)] / 5 − 5.Step 6: According to the problem, this expression equals half of 88, which is 44.Step 7: Set up the equation: [7(N + 10)] / 5 − 5 = 44.Step 8: Add 5 to both sides: [7(N + 10)] / 5 = 49.Step 9: Multiply both sides by 5: 7(N + 10) = 49 × 5 = 245.Step 10: Divide both sides by 7: N + 10 = 245 / 7 = 35.Step 11: Subtract 10 from both sides: N = 35 − 10 = 25. Verification / Alternative check:Check the answer by following the operations with N = 25. Add 10 to get 35. Multiply by 7 to get 245. Divide by 5 to obtain 49. Subtract 5 to get 44, which is exactly half of 88. This confirms that the original number 25 fits the described process. Why Other Options Are Wrong:For N = 20, the process gives ((20 + 10) × 7) / 5 − 5 = (30 × 7) / 5 − 5 = 210 / 5 − 5 = 42 − 5 = 37, not 44. For N = 21, we get ((21 + 10) × 7) / 5 − 5 = 217 / 5 − 5, which is not an integer and not 44. For N = 30, the result is ((30 + 10) × 7) / 5 − 5 = 280 / 5 − 5 = 56 − 5 = 51. None match the required outcome. Common Pitfalls:Some learners reverse the order of operations or incorrectly handle the division and subtraction steps. Others may forget to convert half of 88 to 44 before forming the equation. Carefully writing the sequence as nested algebraic operations prevents such errors. Final Answer:The original number is 25.

Question 9

How many integers from 700 to 950 inclusive are neither divisible by 3 nor divisible by 7?

Accepted Answer

Introduction / Context:This counting problem uses the inclusion exclusion principle to find how many numbers in a given range are not divisible by certain divisors. The range is from 700 to 950 inclusive, and we must exclude numbers divisible by either 3 or 7. Such problems are common in aptitude exams and reinforce understanding of multiples and overlaps between sets. Given Data / Assumptions:Range of integers: 700 to 950 inclusive.We want numbers not divisible by 3 and not divisible by 7.We will count and subtract those divisible by 3 or 7 using inclusion exclusion. Concept / Approach:First, count the total numbers in the range. Then count how many are divisible by 3 and how many are divisible by 7. These counts overlap for numbers divisible by both 3 and 7, that is, by 21. The count of numbers divisible by 3 or 7 is count(3) + count(7) − count(21). Finally, subtract this from the total to find numbers divisible by neither. Step-by-Step Solution:Step 1: Total numbers from 700 to 950 inclusive: 950 − 700 + 1 = 251.Step 2: Count multiples of 3 in this range.Step 3: First multiple of 3 not less than 700 is 702 (3 × 234).Step 4: Last multiple of 3 not greater than 950 is 948 (3 × 316).Step 5: Number of multiples of 3 is 316 − 234 + 1 = 83.Step 6: Count multiples of 7 in the range.Step 7: First multiple of 7 not less than 700 is 700 (7 × 100).Step 8: Last multiple of 7 not greater than 950 is 945 (7 × 135).Step 9: Number of multiples of 7 is 135 − 100 + 1 = 36.Step 10: Count multiples of 21 (LCM of 3 and 7) to correct for double counting.Step 11: First multiple of 21 not less than 700 is 714 (21 × 34).Step 12: Last multiple of 21 not greater than 950 is 945 (21 × 45).Step 13: Number of multiples of 21 is 45 − 34 + 1 = 12.Step 14: Numbers divisible by 3 or 7: 83 + 36 − 12 = 107.Step 15: Numbers divisible by neither 3 nor 7: 251 − 107 = 144. Verification / Alternative check:As a quick sanity check, note that about one third of numbers are multiples of 3 and about one seventh are multiples of 7, with some overlap. From 251 numbers, having roughly 107 excluded is reasonable, leaving about 144, which matches the exact computed value. Spot checking a few numbers near boundaries confirms the first and last multiples for each divisor are chosen correctly. Why Other Options Are Wrong:The option 107 is the count of numbers that are divisible by 3 or 7, not the count of numbers that avoid both. The values 141, 145 and other close numbers would reflect minor miscalculations in counting multiples or applying inclusion exclusion incorrectly, such as forgetting to subtract the multiples of 21. Common Pitfalls:Common mistakes include forgetting to add 1 when counting terms inclusively, misidentifying the first or last multiples in the range, or neglecting to subtract the overlap of multiples of 21. Systematically computing each count and writing out each step reduces these errors. Final Answer:The number of integers from 700 to 950 that are neither divisible by 3 nor by 7 is 144.

Question 10

Find the square root of 15625.

Accepted Answer

Introduction / Context:This problem asks for the exact square root of 15625. Recognizing perfect squares and using factorization shortens the solution. Such questions are typical in number system and speed math sections and help improve mental calculation skills. Given Data / Assumptions:We are given the number 15625.We must find a positive number whose square equals 15625.We are looking for an exact integer square root, not an approximation. Concept / Approach:A good approach is to factor 15625 into prime factors or recognize it as a power of 5. Many powers of 5 end in 25, and 125 squared is a well known value. We can test 125 and confirm by direct multiplication, or factor systematically to be certain. Step-by-Step Solution:Step 1: Notice that 15625 ends with 625, which suggests a relationship with powers of 5.Step 2: Recall that 5^4 = 625 and 5^6 = 15625.Step 3: Therefore, 15625 = 5^6.Step 4: The square root of 5^6 is 5^(6 / 2) = 5^3.Step 5: Compute 5^3 = 5 × 5 × 5 = 125.Step 6: Thus, the square root of 15625 is 125. Verification / Alternative check:Verify by squaring 125 directly. Compute 125 × 125. First, 125 × 100 = 12500. Next, 125 × 25 = 125 × (100 / 4) = 12500 / 4 = 3125. Add these results: 12500 + 3125 = 15625. Since 125 squared equals 15625 exactly, our square root is correct. Why Other Options Are Wrong:If we square 135, we get 18225, which is larger than 15625. Squaring 145 gives 21025, also too large. Squaring 150 gives 22500, much larger than 15625. None of these match 15625, so they cannot be the square root. Common Pitfalls:Some learners guess nearby values and do not verify by squaring, which can lead to mistakes. Others may not recognize 15625 as a power of 5 and attempt lengthy factorization. Remembering common squares such as 125^2 = 15625 can speed up calculations significantly. Final Answer:The square root of 15625 is 125.

Question 11

The average value of the property owned by Agil, Mugilan and Anitha is Rs. 130 crore. Agil's property value is Rs. 20 crore more than Mugilan's property value, and Anitha's property value is Rs. 50 crore more than Agil's property value. What is the …

Accepted Answer

Introduction / Context:This is an averages and linear equations question based on three related quantities. We know the average property value of three people and the differences between their individual property values. Using this information, we can set up algebraic equations to determine each person's property value and specifically find Anitha's share. Given Data / Assumptions:Average property value of Agil, Mugilan and Anitha is Rs. 130 crore.Therefore, the total combined property value is 3 × 130 = Rs. 390 crore.Let Mugilan's property value be M crore.Agil's property value is Rs. 20 crore more than Mugilan's, so Agil's property is M + 20 crore.Anitha's property value is Rs. 50 crore more than Agil's property value. Concept / Approach:We express all three property values in terms of Mugilan's property M. Then we use the total property value based on the average to form an equation in M. Solving this equation gives M, after which we compute Agil's and Anitha's property values using the given differences. Step-by-Step Solution:Step 1: Write Mugilan's property as M.Step 2: Agil's property is M + 20.Step 3: Anitha's property is 50 more than Agil's, so Anitha's property is (M + 20) + 50 = M + 70.Step 4: Total property value of all three is M + (M + 20) + (M + 70) = 3M + 90.Step 5: From the given average, total property is 3 × 130 = 390.Step 6: Set up the equation: 3M + 90 = 390.Step 7: Subtract 90 from both sides: 3M = 300.Step 8: Divide both sides by 3: M = 100.Step 9: Agil's property is M + 20 = 100 + 20 = 120.Step 10: Anitha's property is M + 70 = 100 + 70 = 170. Verification / Alternative check:Check the totals. With Mugilan at 100, Agil at 120 and Anitha at 170, the total property is 100 + 120 + 170 = 390 crore. Dividing by 3 gives an average of 130 crore, which matches the given information. Anitha's property is indeed Rs. 50 crore more than Agil's and Agil's is Rs. 20 crore more than Mugilan's, confirming that the relationships are satisfied. Why Other Options Are Wrong:If Anitha had 120 crore, the differences with the others could not satisfy the given relationships and the average would not be 130 crore. Similarly, 100 crore or 150 crore for Anitha lead to totals that do not equal 390 crore with the stated differences. Only 170 crore fits both the average and the difference conditions. Common Pitfalls:Common mistakes include misinterpreting which person has more property and by how much, or setting up the total incorrectly. Some may forget to multiply the average by the number of people to get the total. Properly labelling each person's property in terms of a single variable avoids confusion. Final Answer:The value of Anitha's property is Rs. 170 crore.

Question 12

What is the largest five digit number that is exactly divisible by 95?

Accepted Answer

Introduction / Context:This problem asks for the largest five digit number that is a multiple of 95. It combines understanding of place value for five digit numbers with divisibility and simple multiplication. The aim is to find the highest multiple of 95 that does not exceed 99999. Given Data / Assumptions:The largest five digit number is 99999.We want the greatest multiple of 95 that is less than or equal to 99999.All numbers involved are integers. Concept / Approach:If we divide 99999 by 95 and ignore any remainder, the integer part of the quotient tells us how many full multiples of 95 fit into 99999. Multiplying that integer by 95 gives the required largest multiple that is still within the five digit range. Step-by-Step Solution:Step 1: Compute the approximate quotient of 99999 divided by 95.Step 2: Start with 95 × 1000 = 95000, which is well below 99999.Step 3: The difference 99999 − 95000 = 4999 shows how much more we can add.Step 4: Compute 95 × 50 = 4750 and 95 × 52 = 4940.Step 5: Add 52 to 1000 to get 1052. Then 95 × 1052 = 95000 + 4940 = 99940.Step 6: If we try 95 × 1053, we get 99940 + 95 = 100035, which exceeds 99999 and is no longer a five digit number.Step 7: Thus, 99940 is the largest five digit multiple of 95. Verification / Alternative check:Check the divisibility of 99940 by 95. Divide 99940 by 5 first: 99940 / 5 = 19988. Next, check divisibility of 19988 by 19. Multiply 19 by 1000 to get 19000, and 19 by 526 to get 9994, so 19 × 1052 = 19988. Therefore 99940 = 95 × 1052 exactly. Since 95 × 1053 overshoots 99999, 99940 is indeed the largest valid multiple. Why Other Options Are Wrong:The number 99935 is not divisible by 95; dividing by 5 gives 19987, which is not a multiple of 19. The numbers 99936 and 99933 are not multiples of 5 at all because they do not end in 0 or 5. Therefore they cannot be multiples of 95, which is 5 × 19. Only 99940 satisfies both the divisibility and the maximum size conditions. Common Pitfalls:Some learners attempt random trial and error close to 99999 without using division, which may miss the exact multiple. Others forget that a multiple of 95 must end in 0 or 5 due to the factor 5. Using the stepwise method of first multiplying 95 by a convenient base like 1000 and then adjusting upward is efficient and reliable. Final Answer:The largest five digit number divisible by 95 is 99940.

Question 13

Evaluate the expression 934 + 7217 - 9115 and find the resulting integer value.

Accepted Answer

Introduction / Context:This question checks basic arithmetic skills involving addition and subtraction of integers. The expression 934 + 7217 - 9115 combines two additions and one subtraction, and you are required to compute the final integer value. Such problems appear frequently in aptitude exams to test accuracy, speed and comfort with simple calculations, especially when the result can be negative. Given Data / Assumptions: - The expression to be evaluated is 934 + 7217 - 9115. - All numbers are integers. - Standard rules of arithmetic for addition and subtraction apply. - The operations are performed from left to right. Concept / Approach:The key idea is to compute the expression step by step while keeping track of signs. First add the two positive numbers, then subtract the last term. Working in stages reduces errors and makes the calculation easier to handle mentally. It is also important to be comfortable with negative results when the number being subtracted is larger than the current total. Step-by-Step Solution: Step 1: Start with the expression 934 + 7217 - 9115. Step 2: Add the first two numbers: 934 + 7217. Step 3: Compute 934 + 7000 = 7934, then 7934 + 217 = 8151. So 934 + 7217 = 8151. Step 4: Now subtract 9115 from this result: 8151 - 9115. Step 5: Since 9115 is larger than 8151, the result will be negative. Step 6: Find the difference between 9115 and 8151: 9115 - 8151 = 964. Step 7: Because we are doing 8151 - 9115, the answer is -964. Verification / Alternative check:You can check the result by rearranging the expression slightly. Write it as (934 - 9115) + 7217. First compute 934 - 9115 = -8181. Then add 7217: -8181 + 7217 = -(8181 - 7217) = -(964) = -964. This alternative path leads to the same answer, confirming the calculation. Why Other Options Are Wrong:The values -974, -864 and -946 come from possible mistakes in subtraction, such as misaligning digits or miscomputing 9115 - 8151. None of these numbers equal the correct difference of 964 in magnitude, so they cannot be correct for the original expression. Common Pitfalls:Common errors include adding all three numbers instead of subtracting the last term, or forgetting that the final answer can be negative. Another frequent mistake is mishandling borrowing in multi digit subtraction, which leads to wrong differences like 954 or 974. Carefully performing the intermediate steps and double checking the difference between 9115 and 8151 helps avoid these mistakes. Final Answer:The value of the expression 934 + 7217 - 9115 is -964, which matches option B.

Question 14

For how many integer values of n does the expression 4n + 7 take a value that is greater than 1 and less than 200?

Accepted Answer

Introduction / Context:This question tests your understanding of linear expressions and inequalities. You are given an expression in terms of an integer variable n, namely 4n + 7, and asked to count how many integer values of n make this expression lie strictly between 1 and 200. Problems like this are common in aptitude tests to check comfort with inequality manipulation and counting integers in a range. Given Data / Assumptions: - The expression is 4n + 7. - n must be an integer (it can be negative, zero or positive). - The value of 4n + 7 must be greater than 1 and less than 200. - We need to count all integer n that satisfy these conditions. Concept / Approach:The main idea is to convert the condition on 4n + 7 into a double inequality involving n. Once we isolate n, we obtain a range of real numbers. We then identify all integers lying within this open interval. Finally, we count how many such integers exist using a simple counting formula. Throughout, we treat the inequality carefully to preserve the strict greater than and less than signs. Step-by-Step Solution: Step 1: Start from the condition 1 < 4n + 7 < 200. Step 2: Subtract 7 from all three parts of the inequality to isolate the term with n: 1 - 7 < 4n < 200 - 7. Step 3: Simplify the bounds: -6 < 4n < 193. Step 4: Divide all parts by 4 (a positive number, so inequality directions do not change): -6 / 4 < n < 193 / 4. Step 5: Simplify the fractions: -1.5 < n < 48.25. Step 6: Since n must be an integer, n can take any integer value strictly greater than -1.5 and strictly less than 48.25. Step 7: The smallest integer greater than -1.5 is -1. Step 8: The largest integer less than 48.25 is 48. Step 9: Count the integers from -1 to 48 inclusive: total = 48 - (-1) + 1 = 50. Verification / Alternative check:To verify, check the boundary values. For n = -1, 4n + 7 = 4(-1) + 7 = 3, which is between 1 and 200, so -1 is included. For n = 48, 4n + 7 = 192 + 7 = 199, which is also between 1 and 200, so 48 is included. For n = -2, 4n + 7 = -8 + 7 = -1, which is not greater than 1, so -2 is excluded. For n = 49, 4n + 7 = 203, which is larger than 200, so 49 is excluded. This confirms that -1 through 48 are exactly the valid integers, giving 50 values. Why Other Options Are Wrong:Options 48, 49 and 51 result from miscounting the integer range or from mistakenly including or excluding one of the boundary values. For example, stopping at 47 instead of 48 gives 49 values, while incorrectly starting at 0 gives 49 as well. Only 50 matches the careful inequality analysis. Common Pitfalls:Typical mistakes include using greater than or equal to instead of strict greater than, which changes whether boundary values are allowed. Another common error is to forget that n can be negative, leading to counting only positive integers. Writing the inequality properly and explicitly listing a few values near the boundaries helps prevent such errors. Final Answer:The number of integer values of n for which 4n + 7 lies between 1 and 200 is 50, corresponding to option C.

Question 15

Let x and y be positive numbers related by x = y - (50 / y). If the value of y is doubled in this equation, how will the value of x change?

Accepted Answer

Introduction / Context:This problem tests your understanding of how changing one variable affects another when the two are related through a non linear expression. The relationship x = y - (50 / y) involves both y and its reciprocal. You are asked to predict what happens to x when y is doubled, without substituting any particular numerical value for y. This type of question is common in algebra and data sufficiency style problems. Given Data / Assumptions: - x and y are both greater than 0. - They satisfy the equation x = y - (50 / y). - We consider what happens when y is replaced by 2y. - We must compare the new value of x with its original value. Concept / Approach:Instead of guessing or plugging in many sample values, we can derive a general expression for the new value of x after doubling y. Then we compare it algebraically to the original x. We will find the difference and see whether it is positive, zero, or negative, and then also compare the new x with 2x to determine whether the increase is more than double. Working symbolically keeps the reasoning valid for all positive y instead of just specific numerical examples. Step-by-Step Solution: Step 1: Original relation is x = y - 50 / y. Step 2: After doubling y, the new value of y is 2y. Step 3: Substitute 2y into the expression to get the new x, call it x'. Step 4: Compute x': x' = 2y - 50 / (2y) = 2y - 25 / y. Step 5: Write the original x alongside: x = y - 50 / y. Step 6: Find the change in x by subtracting: x' - x = (2y - 25 / y) - (y - 50 / y). Step 7: Simplify: x' - x = 2y - 25 / y - y + 50 / y = y + 25 / y. Step 8: Since y is positive, both y and 25 / y are positive, so y + 25 / y is greater than 0. Step 9: Therefore x' - x > 0, which means x increases when y is doubled. Step 10: Now compare x' with 2x. Compute 2x = 2y - 100 / y. Step 11: Compare x' and 2x: x' - 2x = (2y - 25 / y) - (2y - 100 / y) = 75 / y. Step 12: Since y is positive, 75 / y is greater than 0, so x' > 2x; the new x is more than double the old x. Verification / Alternative check:To build intuition, try a specific positive value like y = 5. Then x = 5 - 50 / 5 = 5 - 10 = -5. Doubling y to 10 gives x' = 10 - 50 / 10 = 10 - 5 = 5. Here x changed from -5 to 5, which is an increase of 10, clearly more than double. Trying another value, say y = 10, confirms the pattern: x = 10 - 5 = 5, while x' = 20 - 2.5 = 17.5, which is again more than twice 5. This matches the algebraic result that x' > 2x for all positive y. Why Other Options Are Wrong:The value of x clearly does not decrease, because x' - x = y + 25 / y is positive, so option A is incorrect. It does not stay the same since the difference is not zero. Option C claiming an increase four fold would require x' = 4x, which is not supported by the general formula. Only option D, that x increases to more than double, matches the inequality x' > 2x proved above. Common Pitfalls:A common mistake is to think that doubling y will simply double x, because students expect a linear relationship. Here the presence of the reciprocal term 50 / y makes the relationship non linear. Another error is to test only one example and then generalize without algebraic proof; depending on the chosen value, this may give limited insight. Systematically deriving x' and comparing it with x and 2x is the most reliable method. Final Answer:When y is doubled, the value of x increases to more than double its original value, which corresponds to option D.

Question 16

The average of three consecutive even numbers is A. If the next five consecutive even numbers are also included, what will be the average of these eight even numbers in terms of A?

Accepted Answer

Introduction / Context:This question tests your understanding of averages and how they change when additional terms are added to a set, especially for sequences with a clear pattern like consecutive even numbers. You are given the average of three consecutive even numbers as A, and then asked to find the new average when the next five consecutive even numbers are included. This combines knowledge of arithmetic progressions and the concept of mean. Given Data / Assumptions: - The first three numbers are consecutive even integers. - Their average is A. - The next five consecutive even numbers following these three are added. - We now have a total of eight consecutive even numbers. - We must express the new average in terms of A. Concept / Approach:Consecutive even numbers form an arithmetic progression with common difference 2. The average of equally spaced numbers is the middle term (if the count is odd) or the average of the two middle terms (if the count is even). We use the fact that for three consecutive even numbers, the middle number is the average, and then describe all eight numbers relative to that middle number. Finally, we compute the average of the new set using the first and last terms or by summing relative offsets. Step-by-Step Solution: Step 1: Let the three consecutive even numbers be n - 2, n and n + 2. Step 2: The average of these three numbers is (n - 2 + n + n + 2) / 3 = 3n / 3 = n. Step 3: We are told that this average is A, so n = A. Step 4: The next five consecutive even numbers are obtained by adding 2 successively: A + 4, A + 6, A + 8, A + 10 and A + 12. Step 5: Altogether, the eight numbers are: A - 2, A, A + 2, A + 4, A + 6, A + 8, A + 10 and A + 12. Step 6: These eight numbers form an arithmetic progression with first term A - 2 and last term A + 12, and common difference 2. Step 7: The average of an arithmetic progression is equal to the average of the first and last terms. Step 8: Compute the new average: ((A - 2) + (A + 12)) / 2 = (2A + 10) / 2 = A + 5. Verification / Alternative check:You can verify the formula with an example. Suppose the original three numbers are 4, 6 and 8. Their average is (4 + 6 + 8) / 3 = 6, so A = 6. The next five even numbers are 10, 12, 14, 16 and 18. The full list of eight numbers is 4, 6, 8, 10, 12, 14, 16, 18. Their sum is 88, and the average is 88 / 8 = 11, which equals A + 5 = 6 + 5. This confirms the derived expression. Why Other Options Are Wrong:The options A + 3, A + 4 and A + 7 come from incorrect assumptions about how far the overall center shifts when more numbers are added. For instance, simply adding a fixed constant or only looking at the next number ignores the influence of the entire group of eight. Only A + 5 matches the exact calculation of the new mean using first and last terms. Common Pitfalls:A common mistake is to assume that the new average is just the average of A and the last new number, which is not correct for combined sets. Another pitfall is to miscount the number of terms or forget that the original average A equals the middle number of the first three. Carefully listing the actual numbers in terms of A and using the arithmetic progression average formula avoids these errors. Final Answer:The average of all eight consecutive even numbers is A + 5, which corresponds to option C.

Question 17

The average of 8 consecutive integers is 23/2. What is the average of the first three integers in this sequence?

Accepted Answer

Introduction / Context:This question involves consecutive integers and averages. You are told that the average of 8 consecutive integers is 23/2, and you need to find the average of just the first three numbers in that sequence. This tests your ability to represent consecutive integers algebraically, use the mean formula and reason about symmetry in evenly spaced sequences. Given Data / Assumptions: - There are 8 consecutive integers. - Their average is 23/2. - The integers are equally spaced with a difference of 1. - We want the average of the first three integers in this sequence. Concept / Approach:Consecutive integers can be represented as k, k + 1, k + 2, ..., k + 7. For consecutive terms, the average is simply the mid point, which is equal to the average of the first and last terms. Once we determine the first integer k using the given overall average, we can identify the first three integers and compute their average directly. This approach avoids finding all eight values individually by focusing on the pattern and properties of arithmetic sequences. Step-by-Step Solution: Step 1: Let the 8 consecutive integers be k, k + 1, k + 2, k + 3, k + 4, k + 5, k + 6 and k + 7. Step 2: The average of these 8 integers is (first + last) / 2 = (k + (k + 7)) / 2 = (2k + 7) / 2. Step 3: We are told that this average equals 23/2. Step 4: Set up the equation (2k + 7) / 2 = 23 / 2. Step 5: Since denominators are the same and non zero, equate the numerators: 2k + 7 = 23. Step 6: Solve for k: 2k = 23 - 7 = 16, so k = 8. Step 7: Therefore the 8 consecutive integers are 8, 9, 10, 11, 12, 13, 14 and 15. Step 8: The first three integers are 8, 9 and 10. Step 9: Compute their average: (8 + 9 + 10) / 3 = 27 / 3 = 9. Verification / Alternative check:We can verify by checking the given average. The sum of all 8 integers is 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 = 92. The average is then 92 / 8 = 11.5, which is 23/2, matching the condition. Since the first three numbers are definitely 8, 9 and 10, their average must be 9, confirming our answer. Why Other Options Are Wrong:Option B, 19/2, would correspond to 9.5, which is not the average of 8, 9 and 10. Option C, 8, is the first integer but not the average of the first three. Option D, 10, is the third integer but again not the average. Only 9 is consistent with the arithmetic mean of the first three terms. Common Pitfalls:Students sometimes confuse the given average of all 8 numbers with the average of a subset, assuming they are the same. Others may incorrectly take the average of just the first and second integers or make algebra mistakes when solving for k. Carefully setting up the representation of consecutive integers and using the formula for the mean helps avoid these issues. Final Answer:The average of the first three integers in the sequence is 9, which corresponds to option A.

Question 18

What digit should replace the blank so that the number 275_476 becomes exactly divisible by 11?

Accepted Answer

Introduction / Context:This problem focuses on the divisibility rule for 11 and asks you to find a missing digit in a large number. The number 275_476 contains a blank digit that must be chosen so that the entire number is divisible by 11. Such questions are common in competitive exams to test knowledge of divisibility rules and comfort with handling multi digit numbers without full long division. Given Data / Assumptions: - The number has the form 275x476, where x is a single digit from 0 to 9. - The complete number must be divisible by 11. - We must determine which digit x satisfies this condition. Concept / Approach:The divisibility rule for 11 states that a number is divisible by 11 if and only if the difference between the sum of the digits in odd positions and the sum of the digits in even positions (counted from the left) is a multiple of 11, including 0. We apply this rule to the seven digit number 275x476, derive an expression in terms of the unknown digit x and then solve for x using simple modular reasoning. Step-by-Step Solution: Step 1: Write the digits of 275x476 explicitly: 2, 7, 5, x, 4, 7, 6. Step 2: Label the positions from the left: position 1 is 2, 2 is 7, 3 is 5, 4 is x, 5 is 4, 6 is 7 and 7 is 6. Step 3: Identify odd positions (1, 3, 5, 7) and even positions (2, 4, 6). Step 4: Sum the digits in odd positions: 2 + 5 + 4 + 6 = 17. Step 5: Sum the digits in even positions: 7 + x + 7 = 14 + x. Step 6: Compute the difference between these sums: 17 - (14 + x) = 3 - x. Step 7: For divisibility by 11, this difference must be a multiple of 11. The simplest possibility is 3 - x = 0, which gives x = 3. Step 8: Other multiples like 3 - x = 11 or 3 - x = -11 would lead to x = -8 or x = 14, which are not valid digits. Step 9: Therefore the only valid digit that works is x = 3. Verification / Alternative check:Substitute x = 3 into the number to get 2753476. Recalculate the divisibility rule. Sum of odd position digits remains 17. Sum of even position digits becomes 7 + 3 + 7 = 17. The difference is 17 - 17 = 0, which is a multiple of 11. This confirms that 2753476 is divisible by 11. No other option digit satisfies this condition without breaking the rule. Why Other Options Are Wrong:For x = 6, 4 or 2, the difference 3 - x becomes -3, -1 or 1 respectively, none of which are multiples of 11. Hence the resulting numbers are not divisible by 11. These values represent common incorrect guesses when the divisibility rule is not applied correctly. Common Pitfalls:Typical mistakes include adding all digits together instead of separating odd and even positions or miscounting the positions from the left. Sometimes students take the absolute value of the difference and compare it to 11 without allowing 0, which is also a multiple of 11. Carefully labeling positions and writing separate sums avoids these errors. Final Answer:The missing digit that makes 275_476 divisible by 11 is 3, which corresponds to option D.

Question 19

Which one of the following numbers is irrational (that is, not rational)?

Accepted Answer

Introduction / Context:This question examines your understanding of rational and irrational numbers. A rational number can be written as a ratio of two integers, whereas an irrational number cannot be expressed in such a form and has a non terminating, non repeating decimal expansion. You are given several expressions involving square roots and asked to identify which one is irrational. Given Data / Assumptions: - Option A: 3√8. - Option B: 3√64. - Option C: √64. - Option D: 24. - Standard definitions of rational and irrational numbers apply. Concept / Approach:The key concept is that square roots of perfect squares are integers and therefore rational. Square roots of non perfect squares are irrational. Multiplying an irrational number by a non zero rational number still gives an irrational number. We evaluate each option by simplifying the square roots where possible and then classifying the result as rational or irrational based on these rules. Step-by-Step Solution: Step 1: Consider √64. Since 64 is a perfect square (8 * 8 = 64), √64 = 8, which is an integer and therefore rational. Step 2: Consider 3√64. We already know √64 = 8, so 3√64 = 3 * 8 = 24, which is also an integer and rational. Step 3: Option D is 24 explicitly, which is clearly a rational integer. Step 4: Now consider 3√8. Here the radicand 8 is not a perfect square. We can write √8 as √(4 * 2) = √4 * √2 = 2√2. Step 5: Then 3√8 = 3 * 2√2 = 6√2. Step 6: The number √2 is known to be irrational, and multiplying it by the non zero rational number 6 keeps the product irrational. Step 7: Therefore 3√8 (equal to 6√2) is irrational, while all other options simplify to rational numbers. Verification / Alternative check:You can further verify by approximating the values. √2 is about 1.414..., so 6√2 is about 8.485..., which has a non terminating and non repeating decimal expansion, consistent with being irrational. In contrast, √64 is exactly 8, 3√64 is exactly 24 and 24 is clearly a whole number; all of these have finite decimal expansions and are rational. This confirms the classification derived from simplifying the square roots. Why Other Options Are Wrong:Option B, 3√64, equals 24 and is rational. Option C, √64, equals 8 and is also rational. Option D is simply 24, a rational integer. None of these can be categorized as irrational. Only option A involves the square root of a non perfect square and remains irrational after simplification. Common Pitfalls:Some learners may think that any expression with a square root is automatically irrational, which is not true for perfect squares like 64. Others may forget to simplify radicals like √8 and fail to recognize that 3√8 contains √2 as a factor. Remember that the critical factor is whether the number under the root is a perfect square, not just the presence of the radical sign. Final Answer:The irrational number among the given options is 3√8, which corresponds to option A.

Question 20

30% of a number exceeds 25% of the same number by 27. What is the value of the number?

Accepted Answer

Introduction / Context:This question tests basic percentage and algebra skills. You are told that 30 percent of a number is greater than 25 percent of the same number by 27. From this information, you must determine the original number. Such problems are very common in quantitative aptitude exams and involve translating verbal statements into algebraic equations using percentage notation. Given Data / Assumptions: - Let the unknown number be N. - 30% of N is written as 30/100 * N. - 25% of N is written as 25/100 * N. - The difference between these two quantities is 27. Concept / Approach:The key idea is to express the given percentage statements in algebraic form and then solve a simple linear equation. Since both percentages refer to the same number N, their difference is just a fixed percentage of N. Once we compute that percentage, we can equate it to 27 and solve for N by basic division. Working systematically minimizes errors and makes the computation straightforward. Step-by-Step Solution: Step 1: Let the number be N. Step 2: Write 30% of N as (30 / 100) * N = 0.30N. Step 3: Write 25% of N as (25 / 100) * N = 0.25N. Step 4: According to the question, 30% of N exceeds 25% of N by 27. That is, 0.30N - 0.25N = 27. Step 5: Simplify the left side: 0.30N - 0.25N = 0.05N. Step 6: So we have 0.05N = 27. Step 7: Solve for N by dividing both sides by 0.05: N = 27 / 0.05. Step 8: Note that 0.05 = 5 / 100 = 1 / 20, so dividing by 0.05 is the same as multiplying by 20. Step 9: Compute N = 27 * 20 = 540. Verification / Alternative check:Check the answer by substituting N = 540 back into the original statements. Compute 30% of 540: (30 / 100) * 540 = 0.30 * 540 = 162. Compute 25% of 540: (25 / 100) * 540 = 0.25 * 540 = 135. The difference 162 - 135 is 27, which matches the condition in the question. This confirms that N = 540 is correct. Why Other Options Are Wrong:If N = 270, then 30% is 81 and 25% is 67.5, giving a difference of 13.5, not 27. For N = 108, 30% is 32.4 and 25% is 27, giving difference 5.4. For N = 90, 30% is 27 and 25% is 22.5, difference 4.5. None of these differences equal 27, so these options are incorrect. Common Pitfalls:Some learners mistakenly add the percentages instead of subtracting them or miscalculate the percentage values. Others might assume that the difference 5% directly equals 27, forgetting that it is 5% of N, not 5% of 100. Carefully translating the verbal description into the equation 0.05N = 27 prevents these errors. Final Answer:The value of the number is 540, which corresponds to option A.

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