Q: In place value notation, what is the numerical value of 28 tens?

Introduction / Context: Place value is a core concept in the number system. It tells us how the position of a digit in a number determines its value in terms of ones, tens, hundreds and so on. Understanding statements such as 28 tens or 9 tens is important for interpreting numbers correctly in arithmetic reasoning questions. Given Data / Assumptions: - We are given the expression 28 tens. - One ten is equal to 10 units (or 10 ones). - We need to convert 28 tens into its equivalent numerical value in simple digits. Concept / Approach: The phrase 28 tens means 28 groups of 10. Therefore, the total value is 28 multiplied by 10. In place value terms, a digit in the tens place is worth 10 times its face value. Here we treat 28 as the count of tens and convert this into units by multiplication. Step-by-Step Solution: Step 1: Understand that 1 ten = 10 units. Step 2: Therefore, 28 tens = 28 * 10 units. Step 3: Multiply 28 by 10: 28 * 10 = 280. Step 4: Hence, the numerical value of 28 tens is 280. Verification / Alternative Check: You can think in terms of place value in a number like 280. The digit 8 is in the tens place and contributes 8 * 10 = 80, and the digit 2 in the hundreds place contributes 2 * 100 = 200, for a total of 280. Another way is to imagine writing 28 as 20 + 8 and then calculating 10 * 28 to verify that the answer is 280. Why Other Options Are Wrong: Option A (28) ignores the place value of tens and treats tens as ones. Option B (2) is far too small and does not correspond to 28 groups of anything. Option C (2800) comes from mistakenly multiplying by 100 instead of 10. Common Pitfalls: A common error is to confuse tens and hundreds, leading to multiplication by 100 instead of 10. Some learners also misread 28 tens as simply the number 28 rather than a count of tens. Carefully translating verbal place value statements into multiplication helps avoid these mistakes. Final Answer: The value of 28 tens is 280.

Q: In place value notation, 9 tens and 10 ones together are equal to which number?

Introduction / Context: Understanding expressions like 9 tens and 10 ones helps to build a strong foundation in place value. These expressions describe a number in terms of groups of tens and individual ones. Translating such descriptions into standard numerical form is a basic but important skill in arithmetic reasoning. Given Data / Assumptions: - We have 9 tens. - We also have 10 ones. - We must combine these to find the total number in standard numeric form. Concept / Approach: A ten represents 10 units, and a one represents 1 unit. So 9 tens means 9 * 10, and 10 ones means 10 * 1. Adding these values gives the total number. Sometimes, after addition, regrouping may show that the number of ones can be converted into more tens. Step-by-Step Solution: Step 1: Convert tens to units. 9 tens = 9 * 10 = 90 units. Step 2: Convert ones to units. 10 ones = 10 * 1 = 10 units. Step 3: Add the units: 90 + 10 = 100 units. Step 4: Therefore, 9 tens and 10 ones together equal 100. Verification / Alternative Check: You can also think of 9 tens and 10 ones as 9 tens plus 1 ten and 0 ones, because 10 ones equal 1 ten. Thus, you get 10 tens, which is 100. This confirms the result in another way and shows how regrouping works in the place value system. Why Other Options Are Wrong: Option A (910) comes from concatenating 9 and 10 as digits rather than adding their place values. Option B (90) counts only the tens and ignores the ones. Option C (91) would correspond to 9 tens and 1 one, not 9 tens and 10 ones. Common Pitfalls: Learners sometimes read tens and ones as separate digits and write them side by side, instead of converting them to total units. To avoid this, always translate phrases like 9 tens and 10 ones into multiplication and addition, then simplify. Final Answer: The expression 9 tens and 10 ones equals 100.

Question 1

In a row of boys, there are 6 boys between A and B, and A is the first boy in the row. There are 3 boys between B and C. If there are 12 boys standing after C, then what is the minimum possible number of boys in the row?

Accepted Answer

Introduction / Context:This question is about positions in a single row of boys. It describes the relative positions of three boys A, B and C and asks for the minimum number of boys in the entire row. Solving this type of problem requires interpreting the phrases "boys between" correctly and considering different possible arrangements to find the smallest consistent total. Given Data / Assumptions:A is the first boy in the row, so his position from the left is 1.There are 6 boys between A and B.There are 3 boys between B and C.There are 12 boys after C, meaning to the right of C in the row.We are asked for the minimum possible total number of boys in the row.We assume all positions are filled by exactly one boy and there are no gaps. Concept / Approach:First, convert "boys between" into distances in terms of positions. If there are 6 boys between A and B, the difference in their positions is 6 + 1 = 7. Next, compute the possible positions for C relative to B, taking into account that there are 3 boys between B and C. C could be to the right or to the left of B. We must also account for the 12 boys after C and then select the arrangement that yields the smallest total number of boys while satisfying all conditions. Step-by-Step Solution:Step 1: Since A is the first boy, A is at position 1 from the left.Step 2: There are 6 boys between A and B, so the distance between their positions is 6 + 1 = 7. Because A is at position 1, B must be at position 1 + 7 = 8.Step 3: There are 3 boys between B and C, so the distance between their positions is 3 + 1 = 4. Thus C can be at position 8 + 4 = 12 to the right of B or at position 8 - 4 = 4 to the left of B.Step 4: Next, there are 12 boys after C. If C is at position 12, then there are 12 boys to the right, so the total length of the row would be 12 (position of C) + 12 = 24.Step 5: If C is instead at position 4, then, with 12 boys after C, the last boy is at position 4 + 12 = 16, so the row has 16 boys.Step 6: Both arrangements are consistent with the conditions, but the question asks for the minimum possible number of boys, so we choose the smaller total of 16.Step 7: Therefore the minimum possible number of boys in the row is 16. Verification / Alternative check:We can quickly sketch the arrangement that gives 16 boys. Place A at position 1 and B at position 8, leaving positions 2 to 7 between them, which contains 6 boys, satisfying the first condition. Place C at position 4 so that positions 5, 6 and 7 lie between C and B, giving 3 boys between B and C. Finally, place 12 boys after C, occupying positions 5 through 16. All conditions are satisfied with a total of 16 boys. Why Other Options Are Wrong:A total of 20 or 24 boys is possible in some configurations, but 16 is smaller and still satisfies all conditions, so those larger totals are not the minimum. A total of 18 boys would require a different position for C or B that would break the given spacing constraints. Therefore those alternatives do not correspond to the minimum valid arrangement. Common Pitfalls:Students may misinterpret "boys between" and simply add or subtract incorrectly, or they may forget to consider both possible sides for C relative to B. Another mistake is to ignore the requirement of minimum total and accept the first valid arrangement rather than checking for a smaller consistent arrangement. Final Answer:The minimum possible number of boys in the row is 16.

Question 2

The weights of four boxes are 70, 100, 20 and 40 kilograms. Which of the following cannot be the total weight, in kilograms, of any combination of these boxes if each box is used at most once?

Accepted Answer

Introduction / Context:This is another subset sum problem involving four box weights. We are given several candidate totals and must determine which one cannot be obtained by summing the weights of some subset of the boxes, with the restriction that each box can be used at most once. Problems like this test systematic enumeration or logical reasoning with sums. Given Data / Assumptions:The four box weights are 70 kilograms, 100 kilograms, 20 kilograms and 40 kilograms.We can choose any subset of these four boxes, possibly just one, several or all four.Each box can appear at most once in any chosen subset.The candidate totals are 230, 190, 160 and 200 kilograms.We need to find which total is impossible to obtain from any allowed combination. Concept / Approach:Because there are only four boxes, one direct approach is to list all possible subset sums. Alternatively, we can reason about each option by trying to express it as a sum of some combination of the four weights. If a total cannot be matched by any choice of 70, 100, 20 and 40 without repetition, then that total is the impossible one. We will use a mix of both methods for clarity. Step-by-Step Solution:Step 1: Start with simple combinations. Single box totals are 70, 100, 20 and 40 kilograms.Step 2: Consider combinations of two boxes: 70 + 100 = 170, 70 + 20 = 90, 70 + 40 = 110, 100 + 20 = 120, 100 + 40 = 140 and 20 + 40 = 60. The distinct two box totals are 60, 90, 110, 120, 140 and 170.Step 3: Consider combinations of three boxes: 70 + 100 + 20 = 190, 70 + 100 + 40 = 210, 70 + 20 + 40 = 130 and 100 + 20 + 40 = 160. The three box totals are 190, 210, 130 and 160.Step 4: Consider all four boxes together: 70 + 100 + 20 + 40 = 230.Step 5: Collect the distinct totals that are possible: 20, 40, 60, 70, 90, 100, 110, 120, 130, 140, 160, 170, 190, 210 and 230.Step 6: Compare these with the given options. We see that 230 is possible (all four boxes), 190 is possible (70 + 100 + 20), and 160 is possible (100 + 20 + 40).Step 7: The total 200 kilograms does not appear in the list of achievable sums, so it cannot be the sum of any subset of the given weights. Verification / Alternative check:We can also reason about 200 directly. The largest total less than or equal to 200 using the two largest weights is 100 + 70 + 40 = 210, which is already too large. So we cannot use all three of 70, 100 and 40 together. Trying 100 + 70 + 20 gives 190, which is close but not 200. Using 100 + 40 + 20 gives 160. Using 70 + 40 + 20 gives 130. No combination of three boxes gives 200, and clearly no combination of two or one box gives 200, since the possible one and two box sums are already listed. This confirms that 200 is impossible. Why Other Options Are Wrong:230 kilograms is the sum of all four boxes: 70 + 100 + 20 + 40 = 230. 190 kilograms is the sum of 70, 100 and 20. 160 kilograms is the sum of 100, 40 and 20. Because each of these totals can be formed by a valid combination of the given weights, they are not the required impossible total. Common Pitfalls:It is easy to overlook certain combinations or to assume that a total is impossible without a complete check. Some learners might try only one or two combinations for 200 and conclude too quickly. Systematically listing sums or using logical bounds, such as the maximum and near maximum totals, helps avoid such errors. Final Answer:The total weight that cannot be formed from any combination of the four boxes is 200 kilograms.

Question 3

In basic arithmetic with signed integers, a negative number plus a positive number equals what type of result when only their signs are known?

Accepted Answer

Introduction / Context: When students first encounter arithmetic with positive and negative integers, they often try to apply oversimplified rules such as negative plus positive is always negative or always positive. In reality, the result of adding a negative number and a positive number depends on the relative sizes of their absolute values. This question checks whether you understand that signs alone are not enough to determine the final sign of the sum. Given Data / Assumptions: - One addend is a negative integer. - The other addend is a positive integer. - No information is given about their exact numerical values or magnitudes. - We are asked what the result will be based only on this sign information. Concept / Approach: The general form of such a sum is -a + b, where a and b are positive real numbers or integers. The sign of the result is decided by comparing a and b. If the negative number has greater magnitude, the result is negative. If the positive number has greater magnitude, the result is positive. If the magnitudes are equal, the result is zero. Without knowing specific values, we cannot fix a single outcome. Step-by-Step Solution: Step 1: Represent the negative number as -a and the positive number as b, where a > 0 and b > 0. Step 2: Consider the sum S = -a + b. Step 3: If a > b, then the negative part dominates and S = -(a - b), which is negative. Step 4: If a < b, then the positive part dominates and S = b - a, which is positive. Step 5: If a = b, then S = -a + a = 0, so the sum is exactly zero. Step 6: Since the problem does not specify the sizes of a and b, all three outcomes are possible. Verification / Alternative Check: You can verify this by simple examples: - Example 1: -7 + 2 = -5, which is negative. - Example 2: -3 + 10 = 7, which is positive. - Example 3: -4 + 4 = 0, which is zero. These cases show clearly that different numeric choices give different types of results, even though every sum is negative plus positive. Why Other Options Are Wrong: Option A (A negative number) is wrong because sometimes the positive number is larger in magnitude, giving a positive result instead of a negative one. Option B (A positive number) is wrong for the same reason; sometimes the negative number dominates, making the sum negative. Option C (Zero) is wrong because zero occurs only in the special case when the magnitudes are exactly equal, which is not guaranteed here. Common Pitfalls: A common mistake is to memorize a rule that a negative plus a positive is always positive or always negative. Another error is to focus only on the sign of one number and ignore magnitudes. Correct integer arithmetic always compares absolute values when combining opposite signs. Final Answer: Because the magnitudes of the negative and positive numbers are not specified and all three outcomes are possible, the result cannot be determined from the signs alone.

Question 4

When two parallel lines are cut by a transversal, are the alternate exterior angles always congruent?

Accepted Answer

Introduction / Context: In Euclidean geometry, when two parallel lines are intersected by a third line, called a transversal, several special angle pairs are formed. These include corresponding angles, alternate interior angles and alternate exterior angles. This question tests your understanding of the property of alternate exterior angles when the two lines are parallel. Given Data / Assumptions: - There are two distinct straight lines in a plane. - These two lines are stated to be parallel to each other. - A transversal line cuts both parallel lines, forming several angles at the points of intersection. - We focus on alternate exterior angles formed outside the region between the two parallel lines, on opposite sides of the transversal. Concept / Approach: In Euclidean geometry, when a transversal intersects two parallel lines, alternate exterior angles are always equal in measure, or congruent. This is closely related to the fact that corresponding angles are equal and that the sum of the angles on a straight line is 180 degrees. Parallelism ensures consistent angle relationships everywhere the transversal crosses. Step-by-Step Solution: Step 1: Draw two parallel lines and mark them as line l and line m. Step 2: Draw a transversal that intersects both lines at distinct points. Step 3: Label one exterior angle on the first line above or below the parallel lines. Step 4: Identify the angle on the other line that lies outside the two lines and on the opposite side of the transversal. This is the alternate exterior angle. Step 5: Use the parallel line postulate and properties of corresponding angles to see that these two alternate exterior angles have equal measure. Step 6: Conclude that for parallel lines, alternate exterior angles are congruent in every such configuration. Verification / Alternative Check: You can verify this with a protractor on an accurate drawing or using dynamic geometry software. Fix two lines as parallel and draw a transversal at different slopes. Measure one alternate exterior angle and then its partner. No matter how the transversal is placed, as long as the lines remain parallel, the pair of alternate exterior angles will always have the same measure. Why Other Options Are Wrong: Option B is wrong because alternate exterior angles are not always supplementary; they are equal in measure when lines are parallel. Supplementary angles are a different angle pair type. Option C is wrong because it ignores the fundamental theorem about parallel lines and transversals that guarantees equality of these angles. Option D is wrong because if the lines intersect, they are not parallel and the standard parallel line angle relationships do not hold in the same way. Common Pitfalls: Students often confuse alternate exterior angles with alternate interior angles or corresponding angles. Another frequent mistake is assuming that all angle pairs formed by a transversal are supplementary, which is not correct. Recognizing the position (inside or outside) and side of the transversal is key to identifying angle types correctly. Final Answer: For two parallel lines cut by a transversal, the alternate exterior angles are always congruent.

Question 5

In Euclidean geometry, for similar triangles, which statement correctly describes the relationship between their corresponding angles?

Accepted Answer

Introduction / Context: Similarity of triangles is a central concept in Euclidean geometry and appears frequently in school mathematics, competitive exams and many real world applications such as map reading and architectural scale models. This question checks whether you can recall the precise definition of similar triangles and the relationship between their corresponding angles. Given Data / Assumptions: - We are working in standard Euclidean geometry. - Two triangles are given to be similar, not congruent unless explicitly stated. - The question asks about the relationship between corresponding angles of these similar triangles. - No specific side lengths or angle measures are provided, so the answer must rely on the formal definition of similarity. Concept / Approach: By definition, two triangles are similar if their corresponding angles are equal in measure and the ratios of corresponding sides are constant. Equal angles ensure that the overall shape is the same, while proportional sides allow the triangles to be scaled versions of each other. This is different from congruent triangles, which have both equal angles and equal side lengths, meaning they are identical in size and shape. Step-by-Step Solution: Step 1: Recall the definition of similar triangles: they have equal corresponding angles and proportional corresponding sides. Step 2: Identify the three pairs of corresponding angles in the two similar triangles. Step 3: Because the triangles are similar, each pair of corresponding angles must have exactly the same measure. Step 4: Note that their sides generally differ by a constant scale factor, so side lengths are not automatically equal. Step 5: Recognize that equality of all sides and angles is the definition of congruence, not similarity. Verification / Alternative Check: Consider a small triangle with side lengths 3, 4 and 5 units and a larger triangle with side lengths 6, 8 and 10 units. The angles in both triangles remain the same because each larger side is exactly twice the corresponding smaller side. The triangles are similar with a scale factor of 2, but not congruent because the side lengths differ. However, each corresponding angle pair has the same measure, illustrating the core property being tested. Why Other Options Are Wrong: Option B is wrong because similarity does not require equal side lengths, only proportional ones. Option C describes congruent triangles, where both angles and sides match exactly, which is stronger than similarity. Option D is incorrect because areas of similar triangles are related by the square of the scale factor and are generally not equal unless the scale factor is 1; equality of area does not define similarity. Common Pitfalls: A frequent confusion is mixing up similar and congruent triangles. Some learners also think that equal areas imply similarity, which is not true. Always remember that for similarity, the critical condition is that corresponding angles are equal and corresponding sides are in the same ratio. Final Answer: For similar triangles in Euclidean geometry, corresponding angles are equal in measure.

Question 6

In computer science, what is the time complexity of the linear search algorithm in the worst case for searching an element in an unsorted list of size n?

Accepted Answer

Introduction / Context: Linear search is one of the simplest searching algorithms taught in introductory programming and data structures courses. It is often used on small or unsorted collections of data. Knowing its time complexity is important for analyzing algorithm efficiency and for comparing it with more advanced search methods like binary search. Given Data / Assumptions: - We have an unsorted list or array containing n elements. - We wish to search for a given target value in this list. - The algorithm used is linear search, which checks elements one by one. - The question specifically asks for the time complexity in the worst case. Concept / Approach: In linear search, we start from the first element and compare it with the target. If it does not match, we move to the next element, and so on, until we either find the target or exhaust the list. In the worst case, the target is either at the last position or not present at all, forcing us to make n comparisons. When expressing algorithmic time complexity, we use Big O notation, which focuses on the highest order term as n becomes large. Step-by-Step Solution: Step 1: Let the list contain n elements, indexed from 1 to n. Step 2: In linear search, the algorithm compares the target with element 1, then element 2, and so on. Step 3: In the best case, the target is at the first position, and only one comparison is needed. Step 4: In the worst case, the algorithm must compare the target with all n elements before concluding success or failure. Step 5: Therefore, the total number of comparisons in the worst case is proportional to n. Step 6: In Big O notation, a quantity proportional to n is written as O(n). Verification / Alternative Check: You can verify this by implementing linear search in code and counting comparisons for different positions of the target. As the list grows from 10 to 100 to 1000 elements, you will observe that the worst case number of comparisons grows roughly linearly with n. No nested loops are involved, so the complexity is not quadratic or logarithmic. Why Other Options Are Wrong: Option A (O(n^2)) would imply a nested loop, which linear search does not have. Option B (O(n log n)) is typical of more complex algorithms like efficient sorting, not a simple search through one list. Option D (O(log n)) applies to algorithms like binary search on sorted data, which repeatedly halves the search interval. Linear search does not halve the search space; it checks elements sequentially. Common Pitfalls: A common mistake is to confuse linear search with binary search or to assume any search can be done in logarithmic time. Time complexity always depends on the structure of the algorithm and the properties of the data, such as whether it is sorted. Final Answer: The worst case time complexity of linear search on an unsorted list of size n is O(n).

Question 7

In basic combinatorics, what is the value of the binomial coefficient 6 choose 3, which counts the number of ways to choose 3 objects from 6 distinct objects?

Accepted Answer

Introduction / Context: The expression 6 choose 3, written as 6C3 or C(6, 3), appears frequently in combinatorics, probability and many aptitude tests. It represents the number of distinct subsets of size 3 that can be formed from a set of 6 distinct elements, when order does not matter. Understanding how to evaluate such expressions using factorials is a fundamental skill. Given Data / Assumptions: - We have 6 distinct objects in total. - We want to select 3 of these objects at a time. - The order of selection does not matter, only the chosen group matters. - We are asked to compute the value of 6C3. Concept / Approach: The binomial coefficient nC r is defined by the formula nC r = n! / (r! * (n - r)!). For 6C3, n is 6 and r is 3. Using factorial values 6! = 720 and 3! = 6, and (6 - 3)! = 3! = 6, we can substitute into the formula and simplify. Often, instead of computing full factorials, we cancel common factors to make the calculation quicker and less error prone. Step-by-Step Solution: Step 1: Write the general formula for combinations: nC r = n! / (r! * (n - r)!). Step 2: Substitute n = 6 and r = 3 to get 6C3 = 6! / (3! * 3!). Step 3: Expand 6! as 6 * 5 * 4 * 3 * 2 * 1 and 3! as 3 * 2 * 1. Step 4: Cancel common factors to simplify: 6! / (3! * 3!) = (6 * 5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * (3 * 2 * 1)). Step 5: After cancellation, you can simplify directly: 6C3 = (6 * 5 * 4) / (3 * 2 * 1) = 120 / 6 = 20. Step 6: Conclude that there are 20 distinct ways to choose 3 objects out of 6. Verification / Alternative Check: As a check, you might list combinations in a systematic way, such as fixing the smallest element and choosing pairs for the rest. Although listing becomes impractical for large n, for small n like 6, you can count and confirm that exactly 20 unique triplets are possible. This matches the value found using the formula and confirms the calculation. Why Other Options Are Wrong: Option B (30) could come from confusing combinations with permutations or from miscalculations in factorial simplification. Option C (18) may result from incorrect cancellation or forgetting one combination. Option D (24) is typical when someone mistakenly uses permutations 4P3 or another unrelated expression instead of 6C3. Common Pitfalls: Learners sometimes confuse nC r with nP r, or they fail to cancel factorial terms correctly. Another common issue is thinking that order matters for combinations, which is not correct. Remember that combinations count selections where the arrangement does not matter. Final Answer: The value of the binomial coefficient 6 choose 3 is 20.

Question 8

The price of commodity X increases by Rs 60 every year, while the price of commodity Y increases by Rs 20 every year. If in 2002 the price of commodity X was Rs 320 and the price of commodity Y was Rs 740, in which future year will commodity X cost …

Accepted Answer

Introduction / Context: This is a typical arithmetic reasoning question involving linear growth in prices over time. Each commodity increases in price by a fixed amount every year, so the situation can be modeled using simple linear equations. The goal is to determine the year when the price of commodity X exceeds the price of commodity Y by Rs 60. Given Data / Assumptions: - In the base year 2002, price of commodity X = Rs 320. - In 2002, price of commodity Y = Rs 740. - Every year, the price of X increases by Rs 60. - Every year, the price of Y increases by Rs 20. - We assume the annual increases remain constant over the years. - We must find the calendar year when X is exactly Rs 60 more expensive than Y. Concept / Approach: Because both prices increase linearly with time, we can represent them with linear expressions in terms of t, the number of years after 2002. Then we set up an equation that expresses the required condition: price of X = price of Y + 60. Solving this equation for t gives the number of years after 2002, from which we can find the actual year. Step-by-Step Solution: Step 1: Let t be the number of years after 2002. Step 2: Write the price of commodity X after t years: X(t) = 320 + 60 * t. Step 3: Write the price of commodity Y after t years: Y(t) = 740 + 20 * t. Step 4: We want X(t) to be Rs 60 more than Y(t), so set up the equation: 320 + 60 * t = 740 + 20 * t + 60. Step 5: Simplify the right side: 740 + 20 * t + 60 = 800 + 20 * t. Step 6: The equation is now 320 + 60 * t = 800 + 20 * t. Step 7: Bring all terms involving t to one side: 60 * t - 20 * t = 800 - 320. Step 8: This gives 40 * t = 480. Step 9: Solve for t: t = 480 / 40 = 12. Step 10: The required year is 2002 + 12 = 2014. Verification / Alternative Check: Check the prices in 2014: t = 12. Then X(12) = 320 + 60 * 12 = 320 + 720 = Rs 1040. Y(12) = 740 + 20 * 12 = 740 + 240 = Rs 980. The difference is 1040 - 980 = Rs 60, which matches the requirement. Checking any earlier year gives a smaller difference, so 2014 is indeed the first year when X is Rs 60 more expensive than Y. Why Other Options Are Wrong: Option A (2011), Option C (2012) and Option D (2017) do not satisfy the difference of exactly Rs 60 when calculated with the given annual increments. They either give a smaller or larger price gap between X and Y. Common Pitfalls: Students sometimes mix up the direction of the difference, setting X(t) + 60 = Y(t) instead of X(t) = Y(t) + 60, or they plug in options without forming the equation. Another common mistake is forgetting to check the year relative to the base year, leading to incorrect addition or subtraction of t from 2002. Final Answer: Commodity X will cost Rs 60 more than commodity Y in the year 2014.

Question 9

Two people A and B start from the same place: A starts at 11:00 a.m. and walks at 3.5 km per hour, while B starts at 1:00 p.m. and walks at 1 km per hour for the first hour, 2 km per hour for the next hour, 3 km per hour for the next hour and so on.…

Accepted Answer

Introduction / Context: This problem is a relative speed question with a twist: one person walks at a constant speed, while the other starts later and increases speed in steps every hour. Such questions are common in arithmetic reasoning sections of competitive exams because they test the ability to handle nonuniform motion and to set up equations across time intervals. Given Data / Assumptions: - Person A starts at 11:00 a.m. with a constant speed of 3.5 km per hour. - Person B starts from the same point at 1:00 p.m., which is 2 hours after A. - B walks at 1 km per hour from 1:00 p.m. to 2:00 p.m. - B walks at 2 km per hour from 2:00 p.m. to 3:00 p.m. - B walks at 3 km per hour from 3:00 p.m. to 4:00 p.m., and so on, increasing by 1 km per hour each hour. - Both move along the same straight path in the same direction. Concept / Approach: Let t be the number of hours after 11:00 a.m. When t is large enough, B will have walked several hours, each at different speeds. We compute the distance A has covered as a simple linear function of t, and the distance B has covered as the sum of distances from each hourly segment. We then find the time t when both distances are equal, which gives the catch-up time. Step-by-Step Solution: Step 1: Let t be hours after 11:00 a.m. Step 2: Distance covered by A is A(t) = 3.5 * t. Step 3: B starts at t = 2 (1:00 p.m.). For t less than 2, B has not started and cannot catch up. Step 4: For t hours after 11:00 a.m., with t at least 2, B has walked t - 2 hours, but at different speeds each hour. From 1:00 p.m. to 2:00 p.m. the speed is 1 km per hour, from 2:00 p.m. to 3:00 p.m. it is 2 km per hour, from 3:00 p.m. to 4:00 p.m. it is 3 km per hour, and so on. Step 5: For t between 9 and 10 hours after 11:00 a.m. (that is, between 8:00 p.m. and 9:00 p.m.), B has already completed whole hours at speeds 1, 2, 3, 4, 5, 6 and 7 km per hour after starting. The total distance from those full hours is 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 km. Step 6: In the current hour segment between 8:00 p.m. and 9:00 p.m., B walks at 8 km per hour. So for t between 9 and 10, B(t) = 28 + 8 * (t - 9). Step 7: Set A(t) equal to B(t) for t between 9 and 10: 3.5 * t = 28 + 8 * (t - 9). Step 8: Expand and simplify: 3.5 * t = 28 + 8 * t - 72, so 3.5 * t = 8 * t - 44. Step 9: Rearranging gives 8 * t - 3.5 * t = 44, so 4.5 * t = 44. Step 10: Solve for t: t = 44 / 4.5 ≈ 9.7777 hours after 11:00 a.m. Step 11: Convert 0.7777 hours to minutes by multiplying by 60: 0.7777 * 60 ≈ 46.7 minutes. Step 12: Therefore, the catch-up time is approximately 11:00 a.m. + 9 hours 47 minutes = 8:47 p.m. Verification / Alternative Check: Check A’s distance at t ≈ 9.78 hours: A(t) ≈ 3.5 * 9.7777 ≈ 34.22 km. For B, the distance is 28 + 8 * (t - 9) ≈ 28 + 8 * 0.7777 ≈ 34.22 km. The distances agree, confirming that B has caught A at approximately 8:47 p.m. Among the options, 8 : 47 pm is closest to this computed time. Why Other Options Are Wrong: Options B, C and D correspond to earlier times at which B’s cumulative distance is still less than A’s distance, since A has had a two hour head start and B’s speed increases gradually. Only by about 8:47 p.m. does B accumulate enough distance to match A. Common Pitfalls: Students often assume B’s speed increase can be averaged and use a single average speed, which gives an inaccurate answer. Another common error is to ignore the two hour head start of A, or to assume that B’s speed is constant during the whole journey. Final Answer: B will catch up with A at approximately 8 : 47 pm.

Question 10

At the rate of 9,000 punches per hour, how much total time is required to punch 1,500 cards, each card having 45 columns to be punched?

Accepted Answer

Introduction / Context: This is a work and time problem that involves a constant rate of punching columns on cards. Each card has a fixed number of columns, and there is a known rate of punches per hour. By calculating the total number of punches needed and then dividing by the given rate, we can find how long the task will take. Given Data / Assumptions: - Number of cards to be punched = 1,500. - Each card has 45 columns that must be punched. - The punching machine operates at 9,000 punches per hour. - The machine maintains this rate steadily throughout the task. Concept / Approach: The total work can be measured in number of punches required. The rate of work is given as punches per hour. Time can be found by the basic relation time = total work / rate. After computing the time in hours, we can convert the fractional part of an hour to minutes for a clearer answer. Step-by-Step Solution: Step 1: Compute the total number of punches needed. Step 2: Each card has 45 columns, so total punches = 1,500 * 45. Step 3: Multiply: 1,500 * 45 = 67,500 punches. Step 4: The punching rate is 9,000 punches per hour. Step 5: Use time = total work / rate, so time in hours = 67,500 / 9,000. Step 6: Simplify the fraction: 67,500 / 9,000 = 7.5 hours. Step 7: Convert 0.5 hours to minutes: 0.5 * 60 = 30 minutes. Step 8: Therefore, the required time is 7 hours and 30 minutes. Verification / Alternative Check: As a check, consider how many punches are done in 7.5 hours at 9,000 punches per hour. That is 9,000 * 7.5 = 67,500 punches, which matches exactly the required number of punches. Therefore, the time calculation is consistent. If you tried 8 hours instead, you would get 72,000 punches, which is more than necessary, showing that 8 hours is not the minimal correct answer. Why Other Options Are Wrong: Option A (8 hrs) corresponds to 72,000 punches, overshooting the requirement. Option B (7 hrs 15 min) is 7.25 hours, giving 9,000 * 7.25 = 65,250 punches, which is short of 67,500 punches. Option D (8 hrs 20 min) is even more than 8 hours and produces far more punches than needed. Common Pitfalls: A typical mistake is to miscalculate the total number of punches or to divide incorrectly, sometimes forgetting to convert fractional hours into minutes. Another error is to multiply the rate by the number of cards instead of number of columns, which ignores the structure of the problem. Final Answer: The time required to punch all the cards is 7 hrs 30 min.

Question 11

A child's kiddy bank contains only Rs 1 coins, 50 paise coins and 25 paise coins. What is the total amount in the bank, given that: (1) the total number of coins is 20, and (2) the numbers of 50 paise and 25 paise coins are in the ratio 7 : 3?

Accepted Answer

Introduction / Context: This is a classic data sufficiency question. Instead of directly calculating the amount of money, you must decide whether the given statements provide enough information to determine the total amount uniquely. The coins involved are Rs 1 coins, 50 paise coins and 25 paise coins, and the statements concern the total number of coins and a ratio between two coin types. Given Data / Assumptions: - The bank contains only Rs 1, 50 paise and 25 paise coins. - Statement (1): The total number of coins is 20. - Statement (2): The numbers of 50 paise and 25 paise coins are in the ratio 7 : 3. - We need to know whether we can find a unique total amount from these statements. Concept / Approach: Let x be the number of Rs 1 coins, y the number of 50 paise coins and z the number of 25 paise coins. Statement (1) gives an equation involving x, y and z. Statement (2) gives a ratio between y and z, which we can express in terms of a single parameter. We then see whether combined statements lead to a unique solution or multiple solutions. If more than one possible combination of coin counts satisfies both statements, then the total amount cannot be determined uniquely. Step-by-Step Solution: Step 1: Let x = number of Rs 1 coins, y = number of 50 paise coins, z = number of 25 paise coins. Step 2: From statement (1), x + y + z = 20. Step 3: From statement (2), y : z = 7 : 3, so let y = 7k and z = 3k for some positive integer k. Step 4: Substitute into x + y + z = 20 to get x + 7k + 3k = 20, so x + 10k = 20, which gives x = 20 - 10k. Step 5: For x, y and z to be nonnegative integers, k can be 0, 1 or 2. Step 6: For k = 0: x = 20, y = 0, z = 0, total amount = 20 rupees. Step 7: For k = 1: x = 10, y = 7, z = 3, total amount = 10 * 1 + 7 * 0.5 + 3 * 0.25 = 10 + 3.5 + 0.75 = 14.25 rupees. Step 8: For k = 2: x = 0, y = 14, z = 6, total amount = 0 + 14 * 0.5 + 6 * 0.25 = 7 + 1.5 = 8.5 rupees. Step 9: Thus, multiple different total amounts (20, 14.25 and 8.5) satisfy both statements (1) and (2). Verification / Alternative Check: We have explicitly constructed three different valid combinations of coins satisfying all conditions. Since each gives a different total amount, the information is clearly not sufficient to fix one unique value. Trying to use only statement (1) or only statement (2) obviously produces even more possible combinations. Why Other Options Are Wrong: Option A is wrong because statement (1) alone gives only x + y + z = 20, which has infinitely many integer solutions and therefore cannot determine a unique amount. Option B is wrong because the ratio alone, without the total number of coins or total amount, is insufficient. Option C is wrong because even combining both statements leads to multiple valid solutions and multiple total amounts. Common Pitfalls: A common mistake is to assume that a given ratio and a total count automatically produce a unique solution, without actually checking whether variables can take multiple integer values. Another pitfall is to ignore the possibility that k can take more than one integer value in the ratio representation. Final Answer: Even when both statements are used together, the total amount cannot be determined uniquely, so even both statements (1) and (2) together are not sufficient to answer the question.

Question 12

In real number arithmetic, what is the multiplicative inverse (reciprocal) of 0?

Accepted Answer

Introduction / Context: The concept of a multiplicative inverse, also called a reciprocal, is fundamental in arithmetic and algebra. For most nonzero real numbers, this is a straightforward idea, but zero is an important exception. This question checks whether you understand why zero behaves differently and why its multiplicative inverse does not exist in the real number system. Given Data / Assumptions: - We are working within the real numbers. - The multiplicative inverse of a number a is a number b such that a * b = 1. - The question asks specifically about the multiplicative inverse of 0. - There is no special extension like complex infinity being considered as a real number. Concept / Approach: For any nonzero real number a, its multiplicative inverse is 1 / a, because a * (1 / a) = 1. However, for a = 0, the expression 1 / 0 is undefined in the real number system. There is no real number b such that 0 * b = 1, because 0 multiplied by any real number is always 0. Therefore, zero has no multiplicative inverse. Step-by-Step Solution: Step 1: Recall the definition of multiplicative inverse: for a number a, its inverse b satisfies a * b = 1. Step 2: Substitute a = 0 into this definition. We look for a real number b such that 0 * b = 1. Step 3: Use the property of zero multiplication: 0 multiplied by any real number b gives 0, so 0 * b = 0. Step 4: Compare 0 * b = 0 with the required value 1. There is no real number b that changes this result to 1. Step 5: Therefore, the equation 0 * b = 1 has no solution in the real numbers, and zero has no multiplicative inverse. Verification / Alternative Check: Consider the reciprocal notation 1 / a. For any nonzero a, 1 / a is defined and finite. But for a = 0, 1 / 0 is undefined, because division by zero is not permitted in real arithmetic. Graphically, the function f(x) = 1 / x has a vertical asymptote at x = 0, and it never takes a finite real value at that point, reinforcing that no reciprocal of 0 exists. Why Other Options Are Wrong: Option A (0) is wrong because 0 * 0 = 0, not 1, so 0 cannot be its own multiplicative inverse. Option B (1) is wrong because 0 * 1 = 0, not 1. Option C (Infinity) is not a real number; in the real number system, infinity is not a valid numerical value that can be used to satisfy the equation 0 * b = 1. Multiplying 0 by any finite or infinite quantity does not meaningfully produce 1 in standard real arithmetic. Common Pitfalls: Students sometimes confuse the additive and multiplicative identities, or they think of division by zero as producing infinity. In standard real number arithmetic, division by zero is simply undefined, and infinity is not treated as a real number that can serve as a multiplicative inverse. Final Answer: In the real number system, the multiplicative inverse of 0 does not exist.

Question 13

In basic geometry, how many right angles does a square have?

Accepted Answer

Introduction / Context: Squares are among the most familiar shapes in plane geometry. They are special quadrilaterals that are both rectangles and rhombuses. Understanding the angle properties of a square is essential for many basic geometry questions, including those involving area, perimeter and symmetry. Given Data / Assumptions: - A square is a quadrilateral with four equal sides. - All interior angles of a square are right angles. - A right angle measures exactly 90 degrees. - We are asked how many right angles are present in a square. Concept / Approach: By definition, a square is a special rectangle. A rectangle is a quadrilateral with four right angles. Since a square satisfies the definition of a rectangle, it also has four right angles. The additional property of a square is that all its sides are equal in length. Therefore, we simply need to recall the angle property of rectangles and squares. Step-by-Step Solution: Step 1: Recall that any quadrilateral has four interior angles. Step 2: By definition, a square is a quadrilateral where each interior angle is a right angle of 90 degrees. Step 3: Since there are four corners (vertices) in a square, and each corner contains a right angle, the total number of right angles is 4. Step 4: No additional interior angles exist beyond these four, so the answer is fixed and straightforward. Verification / Alternative Check: You can verify this by drawing a square on graph paper. Each corner clearly forms an L shape, which measures 90 degrees. Measuring with a protractor would confirm that each interior angle is a right angle. There is no possibility of more or fewer right angles in a standard square. Why Other Options Are Wrong: Option A (2) is wrong because it suggests that only two angles are right, which would not satisfy the definition of a square or a rectangle. Option B (6) and Option D (8) are not possible counts for interior angles of any quadrilateral, since a quadrilateral has only four vertices and thus four interior angles. Common Pitfalls: Sometimes learners confuse the total count of right angles with the total measure of all angles in the shape. While the sum of interior angles in any quadrilateral is 360 degrees, what matters here is the number of 90 degree angles, not their sum. All four angles in a square are right angles. Final Answer: A square has 4 right angles.

Question 14

In place value notation, what is the numerical value of 28 tens?

Accepted Answer

Introduction / Context: Place value is a core concept in the number system. It tells us how the position of a digit in a number determines its value in terms of ones, tens, hundreds and so on. Understanding statements such as 28 tens or 9 tens is important for interpreting numbers correctly in arithmetic reasoning questions. Given Data / Assumptions: - We are given the expression 28 tens. - One ten is equal to 10 units (or 10 ones). - We need to convert 28 tens into its equivalent numerical value in simple digits. Concept / Approach: The phrase 28 tens means 28 groups of 10. Therefore, the total value is 28 multiplied by 10. In place value terms, a digit in the tens place is worth 10 times its face value. Here we treat 28 as the count of tens and convert this into units by multiplication. Step-by-Step Solution: Step 1: Understand that 1 ten = 10 units. Step 2: Therefore, 28 tens = 28 * 10 units. Step 3: Multiply 28 by 10: 28 * 10 = 280. Step 4: Hence, the numerical value of 28 tens is 280. Verification / Alternative Check: You can think in terms of place value in a number like 280. The digit 8 is in the tens place and contributes 8 * 10 = 80, and the digit 2 in the hundreds place contributes 2 * 100 = 200, for a total of 280. Another way is to imagine writing 28 as 20 + 8 and then calculating 10 * 28 to verify that the answer is 280. Why Other Options Are Wrong: Option A (28) ignores the place value of tens and treats tens as ones. Option B (2) is far too small and does not correspond to 28 groups of anything. Option C (2800) comes from mistakenly multiplying by 100 instead of 10. Common Pitfalls: A common error is to confuse tens and hundreds, leading to multiplication by 100 instead of 10. Some learners also misread 28 tens as simply the number 28 rather than a count of tens. Carefully translating verbal place value statements into multiplication helps avoid these mistakes. Final Answer: The value of 28 tens is 280.

Question 15

In place value notation, 9 tens and 10 ones together are equal to which number?

Accepted Answer

Introduction / Context: Understanding expressions like 9 tens and 10 ones helps to build a strong foundation in place value. These expressions describe a number in terms of groups of tens and individual ones. Translating such descriptions into standard numerical form is a basic but important skill in arithmetic reasoning. Given Data / Assumptions: - We have 9 tens. - We also have 10 ones. - We must combine these to find the total number in standard numeric form. Concept / Approach: A ten represents 10 units, and a one represents 1 unit. So 9 tens means 9 * 10, and 10 ones means 10 * 1. Adding these values gives the total number. Sometimes, after addition, regrouping may show that the number of ones can be converted into more tens. Step-by-Step Solution: Step 1: Convert tens to units. 9 tens = 9 * 10 = 90 units. Step 2: Convert ones to units. 10 ones = 10 * 1 = 10 units. Step 3: Add the units: 90 + 10 = 100 units. Step 4: Therefore, 9 tens and 10 ones together equal 100. Verification / Alternative Check: You can also think of 9 tens and 10 ones as 9 tens plus 1 ten and 0 ones, because 10 ones equal 1 ten. Thus, you get 10 tens, which is 100. This confirms the result in another way and shows how regrouping works in the place value system. Why Other Options Are Wrong: Option A (910) comes from concatenating 9 and 10 as digits rather than adding their place values. Option B (90) counts only the tens and ignores the ones. Option C (91) would correspond to 9 tens and 1 one, not 9 tens and 10 ones. Common Pitfalls: Learners sometimes read tens and ones as separate digits and write them side by side, instead of converting them to total units. To avoid this, always translate phrases like 9 tens and 10 ones into multiplication and addition, then simplify. Final Answer: The expression 9 tens and 10 ones equals 100.

Question 16

What is the smallest prime number that is greater than 200?

Accepted Answer

Introduction / Context: Prime numbers are natural numbers greater than 1 that have no positive divisors other than 1 and themselves. Questions about primes often ask for the smallest or largest prime in a given range. Here, we need the smallest prime number strictly greater than 200. Given Data / Assumptions: - We are considering natural numbers greater than 200. - We must find the first number after 200 that is prime. - Prime numbers have exactly two distinct positive divisors: 1 and the number itself. Concept / Approach: We test successive integers starting from 201 and check whether each is composite or prime. A composite number has at least one divisor other than 1 and itself. We continue until we reach a number that has no such divisors. That number will be the smallest prime greater than 200. Step-by-Step Solution: Step 1: Check 201. It is divisible by 3 because the sum of its digits is 2 + 0 + 1 = 3, and 3 divides 201. So 201 is not prime. Step 2: Check 203. It is divisible by 7 because 7 * 29 = 203. So 203 is composite. Step 3: Check 205. It is divisible by 5 because it ends in 5. So 205 is composite. Step 4: Check 207 (not listed as an option but useful for verification). The sum of its digits is 2 + 0 + 7 = 9, and 3 divides 9, so 3 divides 207. Therefore, 207 is composite. Step 5: Check 209. It is 11 * 19, so it is composite. Step 6: Check 211. It does not have any divisors among small primes 2, 3, 5, 7, 11 and 13. None of these divide 211 evenly, so 211 is prime. Step 7: Since 211 is the first prime encountered after 200, it is the smallest prime greater than 200. Verification / Alternative Check: To reinforce the result, you may check divisibility of 211 up to the square root of 211, which is a bit more than 14. Prime candidates to test are 2, 3, 5, 7, 11 and 13. We already know 211 is not divisible by these, confirming its primality. Why Other Options Are Wrong: Option A (201) is divisible by 3, so it is composite. Option B (205) is divisible by 5, so it is composite. Option D (203) is divisible by 7, so it is composite. Common Pitfalls: A common mistake is to pick the first odd number greater than 200 without checking for divisors. Another error is to test only divisibility by 2 and 3, ignoring other small primes like 5, 7, 11 and 13. Systematic checking is essential for prime identification. Final Answer: The smallest prime number greater than 200 is 211.

Question 17

In basic statistics and data analysis, what does it mean to calculate frequencies within a dataset?

Accepted Answer

Introduction / Context: Frequencies are a core concept in descriptive statistics and data analysis. When you calculate frequencies, you are trying to understand how often certain values or categories appear in a dataset. This is often shown in frequency tables, bar charts or histograms and is usually one of the first steps in data exploration. Given Data / Assumptions: - We have a dataset consisting of multiple observations or cases. - Each observation belongs to some category or has a certain value on a variable (for example, age group, test score range, or survey response). - We want to understand what calculating frequencies actually means in this context. Concept / Approach: Calculating frequencies means counting the number of occurrences of each distinct value or category of a variable. For example, in a dataset of student grades, you might count how many students obtained grade A, how many obtained grade B and so on. These counts help summarize the distribution of data and are the basis for computing relative frequencies and percentages. Step-by-Step Solution: Step 1: Identify the variable of interest, such as gender, age group or response category. Step 2: List all distinct values or categories that occur for this variable. Step 3: For each category, count how many observations fall into that category. Step 4: Record these counts in a frequency table or chart. Step 5: Optionally, convert counts into relative frequencies or percentages by dividing each count by the total number of observations. Verification / Alternative Check: As an easy example, suppose you survey 10 people about their favorite fruit and get 4 responses for apple, 3 for banana and 3 for orange. The frequency of apple is 4, the frequency of banana is 3 and the frequency of orange is 3. These numbers are frequencies because they show how many cases fall into each category. Why Other Options Are Wrong: Option A describes calculating a product of variables, which is not what frequency means. Option C misinterprets frequency as the number of times a dataset is used by a student, which is unrelated to the internal distribution of data values. Option D describes calculating a sum over a column, which is an aggregate measure but not a frequency count of categories or values. Common Pitfalls: Learners sometimes confuse frequencies with measures like mean or sum. While mean and sum are numerical summaries, frequencies describe how often each category or value appears. Frequencies are particularly important in categorical data, where mean and standard deviation may not be meaningful. Final Answer: Calculating frequencies means counting the number of cases that fall into different subgroups within a dataset.

Question 18

In plane geometry, what is the term for the distance around a two dimensional figure such as a rectangle, triangle or polygon?

Accepted Answer

Introduction / Context: When studying plane figures such as rectangles, triangles and other polygons, we often measure two different things: how much space they cover and how long their boundary is. This question asks you to recall the specific geometric term used for the distance around a two dimensional figure. Given Data / Assumptions: - We are dealing with two dimensional shapes (figures in a plane). - We want the total length of the boundary that encloses the figure. - Common related terms include area and volume, which represent different concepts. Concept / Approach: The distance around a figure is called its perimeter. It is found by adding the lengths of all sides of the figure. For example, the perimeter of a rectangle is 2 times (length + breadth), while the perimeter of a triangle is the sum of its three side lengths. Area, by contrast, measures the amount of surface enclosed by the boundary, and volume applies to three dimensional solids. Step-by-Step Solution: Step 1: Recall that perimeter refers to the length of the outer boundary of a plane figure. Step 2: For a rectangle with sides length and breadth, perimeter = 2 * (length + breadth). Step 3: For a triangle, perimeter = side1 + side2 + side3. Step 4: The question simply asks for the term that describes this boundary length, not for a specific formula. Step 5: Therefore, the correct term is perimeter. Verification / Alternative Check: Consider a simple example of a rectangle that is 5 cm long and 3 cm wide. The perimeter is 2 * (5 + 3) = 16 cm. This 16 cm represents the total distance around the rectangle, which is exactly what is described in the question. The area, in contrast, would be 5 * 3 = 15 square centimeters, showing that area and perimeter are different measures. Why Other Options Are Wrong: Option A (Area) measures the amount of surface covered by a figure, not the distance around it. Option C (Volume) applies to three dimensional objects, not flat figures, and measures space enclosed in three dimensions. Option D (None of the above) is incorrect because perimeter is one of the given choices and it correctly describes the concept in the question. Common Pitfalls: Students sometimes confuse area and perimeter, especially when similar symbols or formulas are used. Always remember that area is measured in square units and refers to surface coverage, while perimeter is measured in linear units and refers to the boundary length. Final Answer: The distance around a two dimensional figure is called its Perimeter.

Question 19

In circle geometry, what do we call a line segment whose endpoints both lie on the circle?

Accepted Answer

Introduction / Context: Circles have several important associated lines and segments, including radii, diameters, chords, tangents and secants. Each term describes a specific geometric configuration. This question focuses on the definition of a chord in circle geometry. Given Data / Assumptions: - We are working with a circle in a plane. - We consider a line segment whose two endpoints lie on the circumference of the circle. - We must choose the correct term for such a segment from the given options. Concept / Approach: A chord is defined as a line segment with both endpoints on the circle. The diameter is a special type of chord that passes through the center of the circle. By contrast, a secant is a line (not just a segment) that intersects the circle at two points, and a tangent is a line that touches the circle at exactly one point. Recognizing these distinctions allows us to select the proper term. Step-by-Step Solution: Step 1: Identify the key condition in the question: both endpoints of the segment lie on the circle. Step 2: Recall that a chord is defined as a line segment joining two points on the circle. Step 3: A diameter is a chord that passes through the center, but the question does not require the center to be involved, so the more general term chord is appropriate. Step 4: A secant is an infinite line that cuts the circle at two points, not just a bounded segment. Step 5: A tangent touches the circle at exactly one point and does not have two endpoints on the circle. Step 6: Therefore, the correct term is chord. Verification / Alternative Check: Draw a circle and mark two distinct points on its circumference. Connect these points with a straight line segment. Geometry textbooks uniformly refer to this segment as a chord. You can further draw several such segments to observe that they all match the definition of a chord, whether or not they pass through the center. Why Other Options Are Wrong: Option A (secant) refers to the entire line that passes through the circle at two points, extending beyond the circle on both sides. It is not limited to the segment between the intersection points. Option C (tangent) refers to a line that touches the circle at exactly one point, not two. Option D (None of the above) is wrong because chord is a valid and correct choice among the options. Common Pitfalls: Students sometimes confuse chord and secant, or think of diameter as a separate concept rather than a special type of chord. Remember that chord always involves two endpoints on the circle and is a finite segment, whereas secant is an entire line extending beyond the circle. Final Answer: A line segment whose endpoints are both on the circle is called a chord.

Question 20

Five boys Prem, Raju, Sundar, Hari and Ompal own different shares of land. Raju has more land than Sundar but less than Prem. Ompal has the least share of land, and Hari has less land than Sundar. Who among them owns the highest share of land?

Accepted Answer

Introduction / Context: This is an ordering and comparison problem from verbal or arithmetic reasoning. You are given relational information about the relative amounts of land owned by five people and must deduce who has the largest share. No actual numerical values are provided, so you must rely entirely on inequalities and logical ordering. Given Data / Assumptions: - There are five boys: Prem, Raju, Sundar, Hari and Ompal. - Raju has more land than Sundar but less than Prem. - Ompal has the least share of land. - Hari has less land than Sundar. - All comparisons are consistent and there are no ties unless explicitly stated. Concept / Approach: We translate each verbal statement into an inequality and then combine them to form a complete ordering from highest to lowest share of land. The person at the top of this order will have the highest share. Careful reading is important to avoid misplacing people in the sequence. Step-by-Step Solution: Step 1: From the statement Raju has more land than Sundar but less than Prem, we write Prem > Raju > Sundar. Step 2: From Ompal has the least share, we conclude that Ompal is at the very bottom of the order. Step 3: From Hari has less land than Sundar, we write Sundar > Hari. Step 4: Combine all information: Prem is above Raju, Raju is above Sundar, Sundar is above Hari, and Ompal is below everyone. Step 5: The resulting order from highest to lowest is: Prem > Raju > Sundar > Hari > Ompal. Step 6: The person at the top of this chain, Prem, has the highest share of land. Verification / Alternative Check: You can assign hypothetical numerical values consistent with the inequalities, such as Prem = 50, Raju = 40, Sundar = 30, Hari = 20 and Ompal = 10. These numbers preserve all the given relations and clearly show Prem as having the largest share. The exact numbers do not matter, only the order does. Why Other Options Are Wrong: Option A (Sundar) is wrong because Raju has more land than Sundar, and Prem has more than Raju, so Sundar cannot be at the top. Option B (Raju) is wrong because the given condition states that Prem has more land than Raju, so Raju is not the highest. Option C (Hari) is wrong because Hari has less land than Sundar, who in turn has less than Raju and Prem. Common Pitfalls: A frequent mistake is to misread the comparative phrases or overlook the fact that Ompal has the least share. Another pitfall is jumping to conclusions without writing the inequalities explicitly. Drawing a simple order diagram helps prevent errors. Final Answer: The boy with the highest share of land is Prem.

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