Question 1

In this arithmetic reasoning puzzle, if you write down all the numbers from 1 to 100, how many times will the digit 3 be written?

Accepted Answer

Introduction / Context:This is a classic digit-counting problem from arithmetic reasoning, often asked in aptitude and placement exams. You are asked to write all the numbers from 1 to 100 and then count how many times the digit 3 appears. Such questions test your ability to systematically count occurrences of a specific digit in different place values without missing or double counting any cases. Given Data / Assumptions:We list all integers from 1 to 100 (both inclusive).We must count every occurrence of the digit 3 in the units and tens places.Numbers like 33 contain two occurrences of the digit 3.We assume standard decimal representation with no leading zeros for two-digit numbers. Concept / Approach:The key idea is to treat the tens place and the units place separately. First, we count how many numbers between 1 and 100 have 3 in the tens place. Then, we count how many numbers between 1 and 100 have 3 in the units place. After that, we add these two counts. For the number 33, which has a 3 in both places, we naturally count each occurrence separately, so it contributes two to the total count. Step-by-Step Solution:Step 1: Count numbers with 3 in the tens place. These are 30, 31, 32, 33, 34, 35, 36, 37, 38, and 39.Step 2: There are 10 such numbers (from 30 to 39), and each contributes one occurrence of 3 in the tens position, giving 10 occurrences so far.Step 3: Count numbers with 3 in the units place: 3, 13, 23, 33, 43, 53, 63, 73, 83, and 93.Step 4: Again, there are 10 such numbers, each contributing one occurrence of 3 in the units position, adding 10 more occurrences.Step 5: Note that 33 has already been handled correctly because it contains two 3s, one in the tens place and one in the units place, and we counted each occurrence separately.Step 6: Total occurrences of the digit 3 in all numbers from 1 to 100 = 10 (tens place) + 10 (units place) = 20. Verification / Alternative check:To verify, you can quickly list the key sets again. For tens place: 30 to 39 clearly gives 10 occurrences. For units place, the pattern 3, 13, 23, and so on up to 93 also clearly includes 10 terms, generated by adding 10 each time. The systematic separation of places ensures no number is missed. Since there is no additional place value for these two-digit numbers apart from tens and units, and 100 does not contain the digit 3, the total 20 is fully reliable. Why Other Options Are Wrong:Option A (11): This undercounts significantly and suggests an incomplete listing of cases where 3 appears.Option B (18): This misses two occurrences, likely due to incorrect handling of the overlap at 33.Option D (21): This overcounts, probably by double counting some numbers or misinterpreting the range up to 100. Common Pitfalls:Many students either forget that 33 contains two occurrences of the digit 3 or mishandle it by counting it incorrectly. Others miss numbers in the tens or units lists due to unsystematic enumeration. A structured method that separately counts tens and units positions, followed by careful addition, is the best way to avoid these errors in digit-counting problems. Final Answer:The digit 3 is written a total of 20 times when listing all numbers from 1 to 100.

Question 2

Two bus tickets from city A to city B and three tickets from city A to city C cost Rs 77, while three tickets from A to B and two tickets from A to C cost Rs 73. What are the individual fares from city A to B and from city A to C?

Accepted Answer

Introduction / Context:This arithmetic reasoning question involves forming and solving simultaneous linear equations based on cost information. We have bus ticket fares from city A to B and from city A to C. Different combinations of these tickets are purchased, and their total costs are given. The goal is to find the individual fares. This type of problem is very common in bank exams, aptitude tests, and school-level algebra practice. Given Data / Assumptions:Let the fare from city A to B be x rupees.Let the fare from city A to C be y rupees.Two tickets A to B plus three tickets A to C cost Rs 77, so 2x + 3y = 77.Three tickets A to B plus two tickets A to C cost Rs 73, so 3x + 2y = 73.We assume normal arithmetic with no discounts or hidden charges. Concept / Approach:We translate the verbal information into two linear equations in two variables. We then solve this system using either the elimination method or the substitution method. The elimination method is usually faster here: we can multiply equations to make the coefficients of one variable equal and then subtract to eliminate that variable. Once we find one fare, we substitute back to get the other. Step-by-Step Solution:Step 1: Write the equations clearly: 2x + 3y = 77 and 3x + 2y = 73.Step 2: Multiply the first equation by 3: 6x + 9y = 231.Step 3: Multiply the second equation by 2: 6x + 4y = 146.Step 4: Subtract the second new equation from the first new equation: (6x + 9y) - (6x + 4y) = 231 - 146.Step 5: This gives 5y = 85, so y = 17.Step 6: Substitute y = 17 into 2x + 3y = 77: 2x + 3*17 = 77, so 2x + 51 = 77.Step 7: Therefore 2x = 26 and x = 13.Step 8: The fares are Rs 13 from A to B and Rs 17 from A to C. Verification / Alternative check:We verify by substituting x = 13 and y = 17 back into the original statements. For two tickets to B and three to C: 2*13 + 3*17 = 26 + 51 = 77, which matches the first given total. For three tickets to B and two to C: 3*13 + 2*17 = 39 + 34 = 73, which matches the second total. Since both conditions are satisfied, our solution is fully consistent. Why Other Options Are Wrong:Option A (Rs. 4, Rs. 23): 2*4 + 3*23 = 8 + 69 = 77 works, but 3*4 + 2*23 = 12 + 46 = 58, not 73, so it fails the second equation.Option C (Rs. 15, Rs. 14): 2*15 + 3*14 = 30 + 42 = 72, not 77, so it is invalid.Option D (Rs. 17, Rs. 13): 2*17 + 3*13 = 34 + 39 = 73, which matches the second equation but not the first one. Common Pitfalls:Students sometimes mix up which equation corresponds to which combination of tickets, leading to wrong equations. Another common mistake is arithmetic error during elimination, especially when multiplying and subtracting equations. Writing each step clearly and double checking multiplication and subtraction prevents such errors. Final Answer:The fares are Rs 13 from city A to B and Rs 17 from city A to C, so the correct option is Rs. 13, Rs. 17.

Question 3

A tailor cuts shirt pieces from rolls of fabric. Each roll is cut into 10 equal pieces, and he makes cuts at the rate of 45 cuts per minute. How many rolls can he completely cut in 24 minutes?

Accepted Answer

Introduction / Context:This arithmetic reasoning question deals with rates of work and careful interpretation of what a “cut” means in the context of cutting fabric into pieces. A tailor has rolls of fabric, and each roll must be cut into equal shirt pieces. The cutting speed is given as a certain number of cuts per minute, and you must determine how many complete rolls can be processed in a given time. Such questions improve your understanding of time, work, and discrete operations. Given Data / Assumptions:Each roll is cut into 10 equal pieces.To get 10 pieces, only 9 cuts are required (each cut increases the number of pieces by one).The tailor makes 45 cuts per minute.Total time available is 24 minutes.We assume no time is wasted between cuts and cuts are uniform. Concept / Approach:The key idea is to recognise that the number of cuts is one less than the number of resulting pieces. Many students incorrectly assume that 10 pieces require 10 cuts, but actually, starting from a single roll, each cut divides one piece into two, so the number of pieces increases by one with every cut. Once we know how many cuts are needed per roll, we can compute the total number of cuts possible in 24 minutes and divide to find the number of rolls that can be fully cut. Step-by-Step Solution:Step 1: Each roll must be cut into 10 pieces.Step 2: To get 10 pieces from 1 roll, we need 10 - 1 = 9 cuts.Step 3: The tailor makes 45 cuts in one minute.Step 4: In 24 minutes, the total number of cuts he can make is 45 * 24 = 1080.Step 5: Number of complete rolls he can cut = total cuts / cuts per roll = 1080 / 9.Step 6: Compute 1080 / 9 = 120.Step 7: Therefore, he can completely cut 120 rolls. Verification / Alternative check:To verify, we can reverse check. If he cuts 120 rolls, and each roll needs 9 cuts, total cuts required are 120 * 9 = 1080. With 45 cuts per minute, the time needed is 1080 / 45 = 24 minutes, which matches the given time exactly. Since the numbers fit perfectly, there is no partial roll or leftover cutting capacity, confirming that 120 rolls is the correct count. Why Other Options Are Wrong:Option A (32 rolls): 32 * 9 = 288 cuts, which would require only 288 / 45 = 6.4 minutes, much less than 24 minutes.Option B (54 rolls): 54 * 9 = 486 cuts, which is still only 486 / 45 = 10.8 minutes.Option C (108 rolls): 108 * 9 = 972 cuts, requiring 972 / 45 = 21.6 minutes, which is less than the available time, so more rolls can be cut. Common Pitfalls:A very common error is to assume that 10 pieces require 10 cuts, which would incorrectly suggest that cuts per roll equal the number of pieces. Another mistake is to miscalculate the total number of cuts or to forget to divide correctly to get the number of rolls. Careful attention to the relationship between cuts and pieces helps avoid such errors. Final Answer:The tailor can completely cut 120 rolls of fabric in 24 minutes.

Question 4

Today is Varun's birthday. One year from today, he will be twice as old as he was 12 years ago. How old is Varun today?

Accepted Answer

Introduction / Context:This is a standard algebraic age problem often seen in quantitative aptitude exams. You are given information about Varun's age at different times relative to today: one year in the future and twelve years in the past. The relationship between these ages allows you to set up an equation and solve for Varun's present age. Such problems train you to translate verbal conditions into algebraic expressions. Given Data / Assumptions:Let Varun's present age be x years.One year from today, his age will be x + 1 years.Twelve years ago, his age was x - 12 years.We are told that one year from now, his age will be twice his age twelve years ago. Concept / Approach:The main idea is to convert the statement “one year from today he will be twice as old as he was 12 years ago” into an equation. This gives a linear relationship between x + 1 and x - 12. Solving this linear equation will directly yield the value of x, which is Varun's present age. Once we obtain a candidate value, we also verify it back in the verbal condition. Step-by-Step Solution:Step 1: Assume Varun's present age is x years.Step 2: According to the problem, one year from today his age will be x + 1.Step 3: Twelve years ago his age was x - 12.Step 4: The condition is x + 1 = 2 * (x - 12).Step 5: Expand the right-hand side: x + 1 = 2x - 24.Step 6: Rearrange the equation: bring x to one side, 1 + 24 = 2x - x, so 25 = x.Step 7: Therefore, Varun's present age is 25 years. Verification / Alternative check:We verify by substituting x = 25 into the original words. One year from today, Varun will be 26 years old. Twelve years ago, he was 25 - 12 = 13 years old. Twice 13 is 26, which matches his age one year from now. Thus, the condition is fully satisfied. None of the other options gives a present age that satisfies this relationship when substituted in the same way. Why Other Options Are Wrong:Option A (20 years): One year from now he would be 21, while 12 years ago he would have been 8. Twice 8 is 16, which does not equal 21.Option B (22 years): One year from now he would be 23, and 12 years ago he would have been 10. Twice 10 is 20, not 23.Option D (27 years): One year from now he would be 28, and 12 years ago he would have been 15. Twice 15 is 30, not 28. Common Pitfalls:A common mistake is to misread the phrase “one year from today” and to set x instead of x + 1 on the left-hand side of the equation. Another mistake is confusing “12 years ago” with “in 12 years”. Always carefully identify which time is in the past and which is in the future before forming the equation. Writing out each step clearly helps avoid sign and time-direction errors. Final Answer:Varun is currently 25 years old.

Question 5

Ayush was born two years after his father's marriage. His mother is five years younger than his father but 20 years older than Ayush, who is now 10 years old. At what age did the father get married?

Accepted Answer

Introduction / Context:This age puzzle involves three family members: Ayush, his father, and his mother. You are given information about age differences and the timing of Ayush's birth relative to his father's marriage. The goal is to compute the father's age at the time of marriage. Such questions test your ability to manipulate age differences and timelines logically. Given Data / Assumptions:Ayush is currently 10 years old.His mother is 20 years older than Ayush, so her present age is 10 + 20 = 30 years.His mother is five years younger than his father, so the father's present age is 30 + 5 = 35 years.Ayush was born two years after his father's marriage.All ages are in complete years, and we ignore months for this calculation. Concept / Approach:Once we determine the present ages of the parents, we can work backwards. If the father is 35 years old now and Ayush is 10, then the father's age when Ayush was born was 35 - 10 = 25 years. Since the marriage took place two years before Ayush was born, the father must have been two years younger at that time. Therefore, we subtract two more years from 25 to obtain the father's age at marriage. Step-by-Step Solution:Step 1: Ayush is 10 years old at present.Step 2: Mother is 20 years older than Ayush, so mother's present age = 10 + 20 = 30 years.Step 3: Mother is five years younger than father, so father's present age = 30 + 5 = 35 years.Step 4: When Ayush was born 10 years ago, father's age was 35 - 10 = 25 years.Step 5: Ayush was born two years after father's marriage, so at the time of marriage father was 25 - 2 = 23 years old.Step 6: Therefore, the father got married at the age of 23 years. Verification / Alternative check:We can check the overall consistency: if father married at 23 years, then two years later at age 25 he had Ayush. Ten years after that, he would be 35, matching the computed present age. At that same present time, mother is 5 years younger, so she is 30, which is also 20 years older than Ayush (10). All conditions in the problem are satisfied, confirming that the calculation is correct. Why Other Options Are Wrong:Option B (25 years): This would suggest father's age at marriage equals his age at Ayush's birth, which contradicts the condition that Ayush was born two years after the marriage.Option C (33 years) and Option D (35 years): These ages are too high and do not align with the given age differences and birth timelines when checked systematically. Common Pitfalls:A common mistake is to misinterpret the phrase “two years after his father's marriage” and either add or subtract incorrectly. Another frequent error is forgetting to use the mother's age relationship to verify or calculate the father's present age. Carefully drawing a time line for birth events and marriage often helps avoid such confusion and leads to a quick and correct solution. Final Answer:Ayush's father got married at the age of 23 years.

Question 6

In three coloured boxes (Red, Green and Blue) there are 108 balls in total. The combined number of balls in the green and red boxes is twice the number of balls in the blue box, and the number of balls in the blue box is twice the number of balls in…

Accepted Answer

Introduction / Context:This arithmetic reasoning problem deals with ratios and relationships between quantities in three different boxes: red, green, and blue. The question provides information about how the numbers of balls in the boxes relate to each other, and it asks specifically for the count in the green box. These types of questions are common in aptitude exams and test your ability to convert verbal relationships into algebraic equations. Given Data / Assumptions:Total number of balls in all three boxes = 108.Let the number of balls in the red box be R, in the green box be G, and in the blue box be B.There are twice as many balls in the green and red boxes combined as there are in the blue box: R + G = 2B.There are twice as many balls in the blue box as there are in the red box: B = 2R.We must find G (the number of balls in the green box). Concept / Approach:The key idea is to transform the relationships into equations and then solve them systematically. By substituting B = 2R into the other equations, we can express everything in terms of a single variable R. After finding R, we back-calculate B and G, and finally verify that their sum is 108. If the computed G does not match any of the numerical options given, then the correct choice is “None of these”. Step-by-Step Solution:Step 1: From the condition “twice as many in the blue box as in the red box”, we have B = 2R.Step 2: From “green and red combined are twice the blue”, we have R + G = 2B.Step 3: Substitute B = 2R into R + G = 2B, giving R + G = 2 * (2R) = 4R.Step 4: Simplify to get G = 4R - R = 3R.Step 5: Use the total balls condition: R + G + B = 108.Step 6: Substitute G = 3R and B = 2R into this equation: R + 3R + 2R = 108.Step 7: This simplifies to 6R = 108, so R = 18.Step 8: Then B = 2R = 36 and G = 3R = 54. Verification / Alternative check:Check the total: R + G + B = 18 + 54 + 36 = 108, which matches the given total. Check the conditions: blue is twice red (36 is twice 18), and green plus red equals 54 + 18 = 72, which is indeed twice blue (2 * 36 = 72). Thus, all relationships are satisfied, and the number of balls in the green box is 54. Since 54 is not listed among the numerical options given (18, 36, 45), we must select “None of these” as the correct answer. Why Other Options Are Wrong:Option A (18): If G were 18, you would not be able to satisfy all the relationships while keeping the total at 108.Option B (36): If G were 36, the equations would not hold consistently when you try to satisfy the given doubling relationships.Option C (45): Similarly, G = 45 does not lead to a consistent solution with the specified ratios and the total of 108 balls. Common Pitfalls:Some students incorrectly assume that one of the simple numeric options must be correct and do not consider “None of these” seriously. Others may make algebraic mistakes when substituting B = 2R into the equations. It is important to follow each substitution carefully and verify the final numbers against all given conditions, not just one of them. Final Answer:The number of balls in the green box is 54, which is not among the numeric options. Therefore, the correct option is None of these.

Question 7

In a class of 60 students, the ratio of boys to girls participating in the annual sports is 3 : 2. The number of girls not participating is 5 more than the number of boys not participating. If the number of boys participating is 15, how many girls a…

Accepted Answer

Introduction / Context:This question combines ratios, total counts, and participation information to test logical reasoning about a class of students. We know the total number of students, the ratio of participating boys to participating girls, and how many more girls than boys are not participating. We are also told the number of boys who do participate. The goal is to determine the total number of girls in the class. Such questions are standard in bank and SSC aptitude exams. Given Data / Assumptions:Total students in the class = 60.Ratio of participating boys to participating girls is 3 : 2.Number of boys participating = 15.Number of girls not participating is 5 more than number of boys not participating.We assume every student is either a boy or a girl and either participates or does not participate. Concept / Approach:We first use the participation ratio to determine how many girls participate. Then we set up variables for total boys and girls and express non-participating counts in terms of these. The relationship between non-participating girls and boys helps us form an equation, which we solve together with the total students equation to find the total number of girls in the class. Step-by-Step Solution:Step 1: Let total boys be B and total girls be G. So B + G = 60.Step 2: Participating boys to participating girls is 3 : 2, and participating boys are 15.Step 3: If 3 parts correspond to 15 boys, then 1 part = 5 and 2 parts = 10 girls participating.Step 4: Boys not participating = B - 15. Girls not participating = G - 10.Step 5: The number of girls not participating is 5 more than the number of boys not participating, so G - 10 = (B - 15) + 5.Step 6: Simplify the relation: G - 10 = B - 10, which gives G = B.Step 7: Substitute G = B into B + G = 60 to obtain 2B = 60, so B = 30 and G = 30.Step 8: Therefore, there are 30 girls in the class. Verification / Alternative check:With B = 30 and G = 30, check participation: boys participating = 15 (given), so boys not participating = 30 - 15 = 15. Girls participating = 10, so girls not participating = 30 - 10 = 20. The condition “girls not participating are 5 more than boys not participating” becomes 20 = 15 + 5, which is correct. The ratio of participating boys to participating girls is 15 : 10 = 3 : 2, matching the problem statement. All conditions are satisfied. Why Other Options Are Wrong:Option A (20) and Option B (25): These values for G, combined with B = 60 - G, fail to satisfy both the ratio and the non-participation difference simultaneously.Option D (Data inadequate): This is incorrect because the given data are sufficient to uniquely determine B and G through the equations. Common Pitfalls:Some students mistakenly apply the ratio 3 : 2 to total boys and girls instead of applying it only to participants. Others forget to link the non-participation counts correctly, or they mis-handle the equation G - 10 = B - 10. Writing clear expressions for participants and non-participants separately for boys and girls helps avoid such mistakes. Final Answer:There are 30 girls in the class.

Question 8

What is the product of all the numbers on a standard telephone dial?

Accepted Answer

Introduction / Context:This question tests your logical observation about numbers on a standard telephone dial. A normal dial contains the digits from 0 to 9. The problem asks for the product of all these numbers. It is a quick test of whether you recognise how multiplication behaves when any factor is zero, rather than trying to perform unnecessary large multiplications. Given Data / Assumptions:A standard telephone dial has digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.The product required is 0 * 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9.We assume the dial is standard and includes 0.We are not restricted to only non-zero digits unless explicitly stated, which is not the case here. Concept / Approach:The key concept is very simple: whenever you multiply any set of numbers and at least one of the factors is zero, the entire product becomes zero. Therefore, there is no need to calculate the factorial-like product of 1 through 9. Instead, you simply identify that 0 is included and conclude that the complete product equals zero. Then you check if zero is among the options; if not, the correct choice will be “None of these”. Step-by-Step Solution:Step 1: List the digits on a standard telephone dial: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.Step 2: The required product is 0 * 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9.Step 3: Recall that any number multiplied by 0 is 0.Step 4: Therefore, the entire product is 0, regardless of the values of the other digits.Step 5: Compare this result with the options and observe that none of the given numerical values equals 0.Step 6: Hence, the correct choice is “None of these”. Verification / Alternative check:If you ignore zero for a moment and compute the product of digits 1 through 9, you get 9 factorial, which is 362880. However, because 0 is part of the dial and is included in the product, the final product must be 0. None of the options matches 0 or 362880, which reinforces that the problem expects you to notice the presence of zero and to select “None of these”. Why Other Options Are Wrong:Option A (1,58,480), Option B (1,59,450), and Option C (1,59,480): These are arbitrary large numbers that might distract you into attempting long multiplications. None of them reflects the effect of multiplying by zero, and none equals the actual product. Common Pitfalls:A typical mistake is to start multiplying 1 * 2 * 3 and so on up to 9, completely overlooking the fact that 0 is included. This leads to wasted time and incorrect answers. Another pitfall is assuming the question might only involve digits 1 to 9 even though 0 is clearly part of a standard dial. Whenever you see a product including 0, immediately conclude that the result is 0 unless there is an explicit reason not to include that factor. Final Answer:The product of all the numbers on a standard telephone dial is 0, which is not listed among the given numbers, so the correct option is None of these.

Question 9

The sum of three consecutive multiples of 3 is 72. What is the second largest of these three numbers?

Accepted Answer

Introduction / Context:This question involves consecutive multiples of 3 and their sum. You are told that the sum of three such consecutive multiples equals 72 and are asked to find the second largest number. It is a straightforward algebra problem that helps you practise forming and solving simple equations with consecutive terms. Given Data / Assumptions:The three numbers are consecutive multiples of 3.Their total sum is 72.We must find the second largest number among these three.We assume positive integers and standard arithmetic operations. Concept / Approach:If the smallest of the three consecutive multiples is 3n, then the next two are 3n + 3 and 3n + 6. The sum of these three numbers is given, so we can express that sum in terms of n, set it equal to 72, and solve for n. Once we know n, we can write down each multiple and identify the second largest number among them. Step-by-Step Solution:Step 1: Let the three consecutive multiples of 3 be 3n, 3n + 3, and 3n + 6.Step 2: Their sum is 3n + (3n + 3) + (3n + 6).Step 3: Simplify this sum: 3n + 3n + 3 + 3n + 6 = 9n + 9.Step 4: According to the problem, 9n + 9 = 72.Step 5: Subtract 9 from both sides: 9n = 63.Step 6: Divide both sides by 9: n = 7.Step 7: The three numbers become 3*7 = 21, 3*7 + 3 = 24, and 3*7 + 6 = 27.Step 8: The second largest number is the middle one, which is 24. Verification / Alternative check:We can quickly verify the numbers: 21, 24, and 27 are indeed multiples of 3 and consecutive in that progression. Their sum is 21 + 24 + 27 = 72, which matches the problem statement. Among these, the largest number is 27, the second largest is 24, and the smallest is 21. Therefore, 24 is confirmed as the correct answer. Why Other Options Are Wrong:Option A (27): This is the largest number, not the second largest.Option C (21): This is the smallest of the three numbers.Option D (42): This number does not appear in the set of three consecutive multiples whose sum is 72. Common Pitfalls:One common mistake is to interpret “consecutive multiples of 3” as consecutive integers or to miswrite them as 3n, 3n + 1, 3n + 2. Another error is to forget that the second largest of three ordered numbers is simply the middle one. Carefully setting up the expression with 3n, 3n + 3, and 3n + 6 prevents such confusion. Final Answer:The three consecutive multiples of 3 are 21, 24, and 27, so the second largest number is 24.

Question 10

A, B, C, D and E play a game of cards. A tells B that if B gives A 3 cards, B will then have as many cards as A has now. A also says that if D takes 5 cards from B, D will then have as many cards as E has. A and C together have twice as many cards a…

Accepted Answer

Introduction / Context:This is a challenging logical and algebraic reasoning problem involving five players A, B, C, D and E sharing a total of 150 cards. Several conditions describe how the number of cards with one player changes when cards are given or taken, and how these counts compare to others. The goal is to determine how many cards C has. Problems like this are common in higher level aptitude tests and require careful equation setup and simultaneous solution. Given Data / Assumptions:Let the cards with A, B, C, D and E be a, b, c, d, and e respectively.When B gives A 3 cards, B will have as many cards as A has now: b - 3 = a.When D takes 5 cards from B, D will have as many cards as E: d + 5 = e.A and C together have twice as many cards as E: a + c = 2e.B and D together have the same number of cards as A and C together: b + d = a + c.Total cards: a + b + c + d + e = 150. Concept / Approach:We convert each statement into an algebraic equation in the variables a, b, c, d and e. This gives a system of linear equations. The strategy is to use substitution and elimination to reduce the system step by step until we find c, the number of cards with C. Because there are five variables and five independent equations, the system has a unique solution. Once we find values for all variables, we verify them against the original statements. Step-by-Step Solution:Step 1: From “B gives A 3 cards and then B has as many as A has now”, write b - 3 = a.Step 2: From “D takes 5 cards from B and then has as many as E”, write d + 5 = e.Step 3: From “A and C together have twice as many as E”, write a + c = 2e.Step 4: From “B and D together have the same as A and C together”, write b + d = a + c.Step 5: Total cards: a + b + c + d + e = 150.Step 6: Solve this system (by substitution or elimination). The consistent solution is a = 32, b = 35, c = 28, d = 25, e = 30.Step 7: Therefore C has c = 28 cards. Verification / Alternative check:Verify each condition using the solution a = 32, b = 35, c = 28, d = 25, e = 30. First, b - 3 = 35 - 3 = 32 which equals a, so the first statement is satisfied. Second, d + 5 = 25 + 5 = 30 which equals e, so the second statement is satisfied. Third, a + c = 32 + 28 = 60 and 2e = 2 * 30 = 60, so the third condition holds. Fourth, b + d = 35 + 25 = 60, which equals a + c = 60, so the fourth condition holds. Finally, total cards a + b + c + d + e = 32 + 35 + 28 + 25 + 30 = 150, matching the given total. All conditions are met, so the solution is fully consistent. Why Other Options Are Wrong:Option B (29), Option C (31) and Option D (35): Any choice other than 28 for C would break at least one of the relationships between A, B, C, D and E when you attempt to find consistent values for the other variables. No other value for c satisfies all the equations simultaneously. Common Pitfalls:This problem is error prone because of the number of variables and relationships. A common mistake is to misinterpret the phrase “B will have as many as A has at this moment” and write the equation incorrectly. Another pitfall is to forget to use the total number of cards to close the system. To handle such questions efficiently, it is helpful to label each equation clearly, avoid arithmetic slips, and consider solving them systematically, even with scratch algebra or a matrix approach if you are comfortable with it. Final Answer:C has 28 cards.

Question 11

Ravi's brother is 3 years older than Ravi. Ravi's father was 28 years old when Ravi's sister was born, and his mother was 26 years old when Ravi was born. If Ravi's sister was 4 years old when his brother was born, what were the ages of Ravi's fathe…

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Introduction / Context:This is a family age puzzle involving Ravi, his brother, his sister, and their parents. You are given information about the parents' ages at specific birth events and the age differences between the siblings. The question asks for the ages of the father and mother when Ravi's brother was born. This type of question tests your ability to manage time-line reasoning and age differences consistently. Given Data / Assumptions:Ravi's brother is 3 years older than Ravi.Ravi's father was 28 years old when the sister was born.Ravi's mother was 26 years old when Ravi was born.The sister was 4 years old when the brother was born.We assume ages are counted in whole years and ignore months. Concept / Approach:We can use age differences rather than absolute ages. The crucial piece is that the father is always 28 years older than the sister and the mother is always 26 years older than Ravi. We are told that the sister was 4 years old when the brother was born, so we can calculate the parents' ages at that time using their fixed differences with the sister and with Ravi, being careful with relative timing. A useful trick is to think of negative ages for children not yet born when going further back in time, simply to preserve the constant age differences. Step-by-Step Solution:Step 1: Since the father was 28 when the sister was born, the father is always 28 years older than the sister.Step 2: The mother was 26 when Ravi was born, so the mother is always 26 years older than Ravi.Step 3: At the time when the brother was born, the sister was 4 years old.Step 4: Therefore, at the brother's birth, father's age = sister's age + 28 = 4 + 28 = 32 years.Step 5: The brother is 3 years older than Ravi, so at the brother's birth, Ravi was not yet born and would be born 3 years later. This means at that moment, Ravi's age would be -3 if we extend the difference backwards.Step 6: Since the mother is always 26 years older than Ravi, at the brother's birth, mother's age = Ravi's age + 26 = -3 + 26 = 23 years.Step 7: Hence, when the brother was born, father was 32 years old and mother was 23 years old. Verification / Alternative check:We can move forward along the time line to double check. At the brother's birth, father is 32 and mother is 23. Four years later, when the sister is born, father would be 36 and mother would be 27, yet the problem states father was 28 at sister's birth, so this forward check appears inconsistent if we mistakenly treat the sister's age relation incorrectly. The key is that the sister being 4 years old at the time of the brother's birth means the sister was born 4 years before the brother, so earlier in the time line. If at the sister's birth father was 28, then 4 years later, at the brother's birth, father is 32, which fits perfectly. Similarly, Ravi is born 3 years after the brother, so at Ravi's birth father is 35 and mother is 26, matching the given mother's age at Ravi's birth and confirming the calculation. Why Other Options Are Wrong:Option B (32 years, 29 years): This implies the mother is only 3 years younger than the father, which does not match the ages at Ravi's birth.Option C (35 years, 29 years) and Option D (35 years, 33 years): These combinations do not maintain the correct age gaps with the sister and Ravi when checked against the original statements. Common Pitfalls:Many students misinterpret the direction of time when they read that the sister was 4 years old when the brother was born. It is easy to confuse which child is older. Another frequent mistake is to ignore that parent-child age differences remain constant through time. Drawing a simple time-line with labelled events for each birth helps to avoid such confusion and leads to accurate age calculations. Final Answer:Ravi's father and mother were respectively 32 years and 23 years old when the brother was born.

Question 12

In this number analogy, “7 is to 34” follows a certain rule. Using the same rule, “5 is to” which number?

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Introduction / Context:This is a simple number analogy question of the form “7 is to 34 as 5 is to what?”. You must identify the arithmetic rule that transforms 7 into 34 and then apply the same rule to 5 to find the corresponding number. Such problems are common in reasoning sections and usually involve basic algebraic expressions. Given Data / Assumptions:The given relation is 7 → 34.We must find the number N such that 5 → N under the same rule.We assume a simple linear relation, such as N = a * x + b, where x is the starting number.Exactly one option should satisfy the pattern in a reasonable way. Concept / Approach:We try to express 34 in terms of 7 using basic operations. One neat possibility is to consider a linear transformation of the form f(x) = 5x - 1. If x = 7, then f(7) = 5 * 7 - 1 = 35 - 1 = 34, which matches the given relation. Once we have identified such a pattern, we apply it to x = 5 and compare the result with the answer choices. Step-by-Step Solution:Step 1: Begin with the relation 7 → 34.Step 2: Try a simple function f(x) = 5x - 1.Step 3: Evaluate f(7) = 5 * 7 - 1 = 35 - 1 = 34, which matches the given number.Step 4: Assume the same rule holds for 5, so compute f(5) = 5 * 5 - 1.Step 5: f(5) = 25 - 1 = 24.Step 6: Therefore, 5 should correspond to 24. Verification / Alternative check:We can briefly test if any simpler pattern might exist, such as squaring or adding a fixed number, but none matches as neatly as 5x - 1. For example, 7 squared is 49, not 34. Adding a fixed number “n” directly to 7 to get 34 would require n = 27, but then 5 + 27 = 32, which is not in the options. The linear transformation 5x - 1 is therefore the simplest and most consistent pattern that explains the given pair. Why Other Options Are Wrong:Option A (33): This would require a different function or inconsistent pattern when applied to 7.Option C (54): There is no straightforward relationship from 5 that matches the rule producing 34 from 7.Option D (41): Again, this does not come from 5 using the rule that maps 7 to 34. Common Pitfalls:Some candidates try to force a complex rule involving squares, cubes, or multiple steps without checking for a simpler linear relation. Others may guess by proportion or random arithmetic, leading to incorrect answers. A good strategy is to test small linear forms like ax + b first, because exam setters usually prefer simple patterns for such single-pair analogies. Final Answer:Applying the rule f(x) = 5x - 1, we get 5 → 24, so the correct answer is 24.

Question 13

A cow is standing on a bridge, 5 m away from the middle of the bridge. A train is approaching the bridge from the nearer end. When the cow runs towards the train, it just manages to get off the bridge as the train is still 2 m away from the bridge. …

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Introduction / Context:This is a classic relative speed and distance puzzle involving a cow on a bridge and an approaching train. The cow can run either towards the train or away from it, and the problem describes exactly when the cow escapes or is hit under both choices. You are also told that the train moves four times as fast as the cow. The goal is to find the length of the bridge. Such puzzles appear in many aptitude exams and require careful equation setup using time and distance relationships. Given Data / Assumptions:Length of the bridge is L metres (unknown).The bridge runs from position 0 (near end) to L (far end).The cow initially stands 5 m away from the middle of the bridge, nearer to the approaching train.So the middle is at L/2 and the cow's starting position is x = L/2 - 5.The train approaches from the near end (position 0).The speed of the train is four times the speed of the cow.When the cow runs towards the train, it just leaves the bridge at the near end as the train is still 2 m away from the bridge.When the cow runs away from the train, it would be hit 2 m before the far end (position L - 2). Concept / Approach:We denote the cow's speed by v and the train's speed by 4v. We model two scenarios. In the first scenario, the cow runs from its starting point to the near end of the bridge, while the train runs from some starting point towards the bridge, remaining 2 m away when the cow leaves the bridge. In the second scenario, the cow runs towards the far end and is hit 2 m before that end, while the train runs the full distance from its starting point to that collision point. Using distance = speed * time, we set up time equality in both scenarios, express all distances in terms of L, and solve for L. Step-by-Step Solution:Step 1: Let the cow's starting position on the bridge be x = L/2 - 5.Step 2: Let the distance from the train's initial position to the near end of the bridge be D metres.Step 3: Scenario 1 (cow runs towards train): the cow runs distance x to reach the near end. Time taken by the cow is t1 = x / v.Step 4: In the same time, the train travels from its starting point to 2 m before the bridge, so its distance is D - 2. With speed 4v, time for the train is (D - 2) / (4v).Step 5: Since both events happen together, x / v = (D - 2) / (4v), giving 4x = D - 2, so D = 4x + 2.Step 6: Scenario 2 (cow runs away from train): the cow runs from x to position L - 2, so the distance is (L - 2) - x. Time t2 for the cow is (L - 2 - x) / v.Step 7: In this scenario, the train runs from distance D before the bridge to the collision point at L - 2, so its distance is D + (L - 2). Time for the train is (D + L - 2) / (4v).Step 8: Again, times are equal: (L - 2 - x) / v = (D + L - 2) / (4v).Step 9: Multiply by 4v to get 4(L - 2 - x) = D + L - 2.Step 10: Substitute D = 4x + 2 into this equation: 4(L - 2 - x) = (4x + 2) + L - 2.Step 11: Simplify left side: 4L - 8 - 4x. Right side simplifies to L + 4x.Step 12: So 4L - 8 - 4x = L + 4x.Step 13: Rearrange: 4L - L - 8 = 4x + 4x, giving 3L - 8 = 8x.Step 14: But x = L/2 - 5, so 8x = 8(L/2 - 5) = 4L - 40.Step 15: Substitute: 3L - 8 = 4L - 40. This gives 32 = L.Step 16: Therefore, the length of the bridge is 32 metres. Verification / Alternative check:With L = 32, the middle is at 16 m, so the cow starts at x = 16 - 5 = 11 m from the near end. From step 5, D = 4x + 2 = 4 * 11 + 2 = 46 m. In the first scenario, the cow runs 11 m, and the train runs 46 - 2 = 44 m. Time ratio is 11 / v for the cow and 44 / (4v) for the train, which are equal. In the second scenario, the cow runs from 11 m to 30 m (which is L - 2 = 30 m), a distance of 19 m, while the train runs 46 + 30 = 76 m. Time for the cow is 19 / v and for the train is 76 / (4v), again equal. Hence, L = 32 m satisfies all conditions. Why Other Options Are Wrong:Options 22, 28 and 34 metres: Substituting any of these values into the equations for the two scenarios will not satisfy both time equalities simultaneously. The distance and time relationships will not balance as required by the conditions in the problem. Common Pitfalls:Many students misinterpret the phrases “2 m away from the bridge” and “2 m before the end of the bridge” and place these positions incorrectly on the distance line. Another frequent mistake is to forget that the train's starting distance before the bridge is the same in both scenarios, leading to inconsistent equations. It is crucial to draw a clear diagram, assign positions on a number line, and carefully set up time equality equations for each scenario. Final Answer:The length of the bridge is 32 metres, which corresponds to option “32 mts”.

Question 14

In a certain office, one-third of the workers are women. One-half of these women are married and one-third of the married women have children. Of the men, three-fourths are married and two-thirds of the married men have children. What fraction of al…

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Introduction / Context: This arithmetic reasoning question is about fractions and population breakdown in an office. The workers are divided into women and men, and within each group, some are married and some of the married workers have children. The aim is to find what fraction of the total workers do not have children at all. Such questions test comfort with fractions, proportional reasoning, and careful reading of layered conditions. Given Data / Assumptions: One-third of the workers are women. One-half of the women are married. One-third of the married women have children. Three-fourths of the men are married. Two-thirds of the married men have children. All workers belong either to the group of men or to the group of women. Concept / Approach: To keep the calculation simple, treat the total number of workers as 1 unit (or 100 percent). Express all given pieces of information as fractions of this total. First, compute the fractions of women with children and women without children. Then do the same for men with children and men without children. Finally, add the fractions of women without children and men without children to obtain the fraction of all workers who are without children. Step-by-Step Solution: Step 1: Let the total number of workers be 1. Then women = 1/3 and men = 2/3 of the workers. Step 2: Married women = (1/2) * (1/3) = 1/6 of all workers. Step 3: Women with children = (1/3) * (1/6) = 1/18 of all workers. Step 4: Women without children = total women - women with children = 1/3 - 1/18 = 6/18 - 1/18 = 5/18. Step 5: Married men = (3/4) * (2/3) = 6/12 = 1/2 of all workers. Step 6: Men with children = (2/3) * (1/2) = 1/3 of all workers. Step 7: Men without children = total men - men with children = 2/3 - 1/3 = 1/3. Step 8: Total workers without children = women without children + men without children = 5/18 + 1/3 = 5/18 + 6/18 = 11/18. Verification / Alternative check: To verify, imagine a convenient total, for example 18 workers. Then women = 6 and men = 12. Married women = 3, women with children = 1, so 5 women have no children. Married men = 9, men with children = 6, so 6 men have no children. Total without children = 5 + 6 = 11 out of 18 workers, again 11/18, which confirms the fraction is correct. Why Other Options Are Wrong: Option A 5/18 counts only the women without children and ignores men. Option B 4/9 is less than 1/2 and does not match the computed combined fraction. Option D 17/36 is smaller than 11/18 and arises from incorrect subtraction or mixing of fractions. None of these match both the direct fraction calculation and the verification by assuming 18 workers. Common Pitfalls: A very common error is to treat percentages or fractions inside the women and men groups as if they were applied directly to the whole office, without first isolating the fraction of men and women. Another frequent mistake is to confuse fractions of married workers with fractions of all workers. Always anchor each fraction either to the whole office or to the subgroup, and convert all of them to fractions of the total before adding. Final Answer: The fraction of workers who are without children is 11/18.

Question 15

A man has a certain number of small boxes to pack into parcels. If he packs 3, 4, 5, or 6 boxes in each parcel, he is always left with one box over; but if he packs 7 boxes in each parcel, none of the boxes is left over. What is the possible total n…

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Introduction / Context: This arithmetic reasoning problem is based on remainders and divisibility, a typical application of the least common multiple and modular arithmetic. The man packs his boxes in parcels of different sizes. For 3, 4, 5, and 6 boxes per parcel he is left with one extra box, but for 7 boxes per parcel he has no remainder. The task is to determine a total number of boxes that fits all of these conditions using simple number theory ideas. Given Data / Assumptions: When boxes are packed 3 per parcel, the remainder is 1. When packed 4, 5, or 6 per parcel, the remainder is also 1 in each case. When packed 7 per parcel, there is no remainder, so the total is divisible by 7. The answer must be one of the given options and must satisfy all remainder conditions simultaneously. Concept / Approach: The statement that the number leaves remainder 1 when divided by 3, 4, 5, and 6 means the total is 1 more than a common multiple of all these divisors. The least common multiple (LCM) of 3, 4, 5, and 6 determines the basic repeating pattern for such numbers. Any valid total must be of the form 60k + 1, where 60 is the LCM. In addition, the number must be divisible by 7. Thus, among numbers of the form 60k + 1 we search for a value that is a multiple of 7 and also appears in the options. Step-by-Step Solution: Step 1: Compute the LCM of 3, 4, 5, and 6. The LCM is 60. Step 2: Numbers that leave remainder 1 when divided by 3, 4, 5, and 6 are of the form 60k + 1 for integer k. Step 3: List a few such numbers: 61, 121, 181, 241, 301, and so on. Step 4: Impose the extra condition that the total must be divisible by 7. Check 301: 301 / 7 = 43, so 301 is divisible by 7. Step 5: Check if 301 appears in the options. It is given as option B, so 301 is a valid candidate. Step 6: Other options such as 106, 309, and 400 either do not equal 60k + 1 or are not divisible by 7, so they fail the combined conditions. Verification / Alternative check: Verify all conditions for 301: 301 ÷ 3 gives remainder 1, 301 ÷ 4 gives remainder 1, 301 ÷ 5 gives remainder 1, and 301 ÷ 6 also gives remainder 1. Finally, 301 ÷ 7 gives 43 exactly with no remainder. This confirms that 301 satisfies every requirement in the question and is therefore a correct and complete answer. Why Other Options Are Wrong: Option A 106 is not 1 more than a multiple of 60, and 106 ÷ 5 does not leave remainder 1. Option C 309 also fails some remainder conditions and does not equal 60k + 1. Option D 400 is neither of the form 60k + 1 nor divisible by 7. Each alternative violates at least one of the divisibility constraints. Common Pitfalls: A frequent mistake is to test only one or two divisors, such as checking condition with 3 and 4 but not verifying 5, 6, and 7. Another pitfall is ignoring the power of the LCM and trying trial and error randomly. Using the LCM focuses attention on a small family of possible values and makes it much easier to find the number that meets all conditions. Final Answer: The total number of boxes the man may have is 301.

Question 16

In an examination, 10,500 candidates appeared. The statistics show: passed in all five subjects = 5,685; passed in three subjects only = 1,498; passed in two subjects only = 1,250; passed in one subject only = 835; failed in English only = 78; faile…

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Introduction / Context: This arithmetic reasoning problem uses examination statistics to test understanding of how passing and failing in individual subjects translate into passing in a certain number of subjects overall. The question asks for the number of candidates who passed in at least four subjects, which includes those who passed in all five as well as those who passed in exactly four subjects. Given Data / Assumptions: Total candidates who appeared: 10,500. Passed all five subjects: 5,685. Passed exactly three subjects: 1,498. Passed exactly two subjects: 1,250. Passed exactly one subject: 835. Failed in English only: 78. Failed in Maths only: 275. Failed in Physics only: 149. Failed in Chemistry only: 147. Failed in Biology only: 221. Concept / Approach: A candidate who fails in exactly one subject must have passed the remaining four subjects. Therefore, the counts “failed in English only,” “failed in Maths only,” and so on all represent candidates who passed in exactly four subjects. The number of candidates who passed in at least four subjects equals the number who passed all five plus the number who passed exactly four. The problem therefore reduces to adding the five failure-only counts and then adding the count of those who passed all five subjects. Step-by-Step Solution: Step 1: Recognize that failing in English only means passing the other four subjects, and similarly for each other subject. Step 2: Compute the number of candidates who passed exactly four subjects by summing the failure-only counts: 78 + 275 + 149 + 147 + 221. Step 3: 78 + 275 = 353; 353 + 149 = 502; 502 + 147 = 649; 649 + 221 = 870. So 870 candidates passed exactly four subjects. Step 4: Candidates who passed in at least four subjects = passed in all five + passed in exactly four = 5,685 + 870. Step 5: 5,685 + 870 = 6,555. Verification / Alternative check: As a check, verify that no candidate is double counted in the failure-only categories. Each person recorded as “failed in X only” cannot be in any other single-failure list, because that would mean failure in more than one subject. Hence these five groups are disjoint and can be safely added. The result 870 represents all four-subject passers. Adding 5,685 five-subject passers gives 6,555, which matches one of the options, confirming consistency. Why Other Options Are Wrong: Option B 5,685 counts only those who passed all five subjects and ignores the four-subject passers. Option C 1,705 could come from incorrect partial addition of the data. Option D 870 counts only those who failed in exactly one subject, missing all candidates who passed in all five subjects. Common Pitfalls: Some learners mistakenly interpret “failed in English only” as failing in all subjects except English, which reverses the meaning. Another mistake is to assume that the numbers provided for passing in one, two, or three subjects are needed to compute the four-subject passers, when in fact the four-subject group is directly given via the single-failure counts. Keeping track of the exact language “failed in X only” versus “passed in X only” is crucial. Final Answer: The number of candidates who passed in at least four subjects is 6,555.

Question 17

Ajay and Vijay each have some marbles. Ajay says to Vijay, "If you give me x marbles, then both of us will have an equal number of marbles." Vijay replies, "If you give me twice as many marbles, then I will have 30 more marbles than you will have." …

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Introduction / Context: This arithmetic reasoning question involves setting up and solving simple linear equations based on a word problem with marbles. Ajay and Vijay exchange marbles under two different hypothetical conditions, and both conditions must be satisfied for the same unknown x. By translating each statement carefully into an algebraic equation, you can solve for x and determine the number of marbles being discussed in each scenario. Given Data / Assumptions: Ajay initially has A marbles. Vijay initially has V marbles. If Vijay gives x marbles to Ajay, they both end up with the same number of marbles. If Ajay gives twice as many marbles, that is 2x marbles, to Vijay, then Vijay ends up with 30 more marbles than Ajay. The variable x is a positive integer and one of the given options. Concept / Approach: Two conditions are described, so we set up two equations involving A, V, and x. From the first condition we express the equality of marbles after Vijay gives x to Ajay. From the second condition we express that Vijay has 30 more marbles than Ajay after Ajay gives 2x to Vijay. Since both statements describe the same starting amounts A and V, the expressions for V - A derived from both situations must be equal. This leads to a single equation in x, which we solve using basic algebra. Step-by-Step Solution: Step 1: From the first statement, if Vijay gives x marbles to Ajay, Ajay has A + x and Vijay has V - x. These are equal, so A + x = V - x. Step 2: Rearranging A + x = V - x gives V - A = 2x. Step 3: From the second statement, if Ajay gives 2x marbles to Vijay, Ajay has A - 2x and Vijay has V + 2x. Vijay then has 30 more marbles than Ajay, so V + 2x = (A - 2x) + 30. Step 4: Simplify V + 2x = A - 2x + 30 to get V - A = -4x + 30. Step 5: Both expressions for V - A must be equal: 2x = -4x + 30. So 2x + 4x = 30 gives 6x = 30, hence x = 30 / 6 = 5. Verification / Alternative check: If x = 5, then V - A = 2x = 10. So Vijay always has 10 more marbles than Ajay initially. Check the second scenario: V + 2x - (A - 2x) = (V - A) + 4x = 10 + 4*5 = 10 + 20 = 30. This matches the condition that Vijay has 30 more marbles after Ajay gives him 2x marbles, confirming that x = 5 is consistent. Why Other Options Are Wrong: If x were 4, then V - A would be 8 from the first condition and 30 - 4*4 = 14 from the second, which is inconsistent. Similar contradictions appear for x = 8 or x = 10. Each of these values fails to satisfy both equations simultaneously, so they cannot be correct. Common Pitfalls: A common mistake is to misinterpret the second statement and mix up who is giving marbles to whom. Another error is to equate A and V incorrectly instead of equating the expressions for V - A. Carefully tracking who gains and who loses marbles in each scenario avoids these algebraic mistakes. Final Answer: The correct value of x is 5.

Question 18

In a note collection, there is one new Rs 2000 note for every three old Rs 500 notes. If 10 more new Rs 2000 notes are added to the collection, the ratio of new Rs 2000 notes to old Rs 500 notes becomes 1 : 2. Based on this information, what is the …

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Introduction / Context: This arithmetic reasoning question involves ratios and a simple linear relation between numbers of currency notes of two denominations. Initially the collection has a fixed ratio of new Rs 2000 notes to old Rs 500 notes. After adding some new notes, the ratio changes. Using this information, we determine the actual counts of each type of note and then calculate the total number of notes in the collection. Given Data / Assumptions: Initially, the ratio of new Rs 2000 notes to old Rs 500 notes is 1 : 3. Then 10 more Rs 2000 notes are added to the collection. After this addition, the new ratio of Rs 2000 notes to Rs 500 notes becomes 1 : 2. Old Rs 500 notes are not added or removed in this operation. We must find the total count of notes (both denominations) after adding the 10 new Rs 2000 notes. Concept / Approach: Let the initial number of new Rs 2000 notes be n. Then, by the given ratio 1 : 3, the initial number of old Rs 500 notes is 3n. After adding 10 new notes, the new count of Rs 2000 notes is n + 10, while old Rs 500 notes remain 3n. The new ratio condition states (n + 10) : (3n) = 1 : 2. Solving this proportion gives the value of n. Using n, we can compute both types of notes and hence the new total. Step-by-Step Solution: Step 1: Let the initial number of Rs 2000 notes be n, so initial Rs 500 notes = 3n. Step 2: After adding 10 new Rs 2000 notes, the new count of Rs 2000 notes = n + 10. The old Rs 500 notes remain 3n. Step 3: The new ratio is (n + 10) : 3n = 1 : 2. Step 4: Write the proportion as (n + 10) / (3n) = 1 / 2. Cross-multiplying gives 2(n + 10) = 3n. Step 5: Simplify: 2n + 20 = 3n, so n = 20. Step 6: Initial numbers: Rs 2000 notes = 20, Rs 500 notes = 3 * 20 = 60. Step 7: After adding 10 new Rs 2000 notes, total Rs 2000 notes = 20 + 10 = 30, Rs 500 notes remain 60. Step 8: Total number of notes in the collection after addition = 30 + 60 = 90. Verification / Alternative check: Check the final ratio: new Rs 2000 notes : old Rs 500 notes = 30 : 60 = 1 : 2, which matches the condition. Also verify the initial ratio: 20 : 60 = 1 : 3, which matches the original statement. Both ratio checks confirm that the counts are consistent and that the total 90 is correct. Why Other Options Are Wrong: Option A 60 would imply far fewer notes and cannot satisfy both ratio conditions with the addition of exactly 10 notes. Option B 30 and option C 70 also fail to match the new ratio 1 : 2 after adding 10 new Rs 2000 notes and the original ratio 1 : 3 simultaneously. Only 90 fits both stages of the problem. Common Pitfalls: A common mistake is to treat ratios as fixed differences instead of proportional relations, which leads to linear addition or subtraction errors. Another error is to forget that only the Rs 2000 notes increase while Rs 500 notes remain constant. Keeping the before and after stages clearly separated and writing equations from the ratio definitions prevents confusion. Final Answer: The total number of notes in the collection after adding the 10 new Rs 2000 notes is 90.

Question 19

The cost of 7 tables and 12 chairs together is Rs 48,250. Assuming that the cost of each table and each chair remains the same, what will be the total cost of buying 21 tables and 36 chairs?

Accepted Answer

Introduction / Context: This arithmetic reasoning question checks understanding of proportionality and linear scaling in cost problems. The combined cost of a certain number of tables and chairs is given, and you are asked to find the cost of a larger set of tables and chairs that has the same ratio between tables and chairs. Because the numbers of tables and chairs both increase by the same factor, the total cost scales by exactly the same factor. Given Data / Assumptions: The cost of 7 tables and 12 chairs together is Rs 48,250. The cost per table and the cost per chair remain constant. We want the cost of 21 tables and 36 chairs. Note that 21 = 3 * 7 and 36 = 3 * 12, so the quantities are scaled by a factor of 3. Concept / Approach: If both the number of tables and the number of chairs are multiplied by the same integer factor, and the unit costs stay the same, then the total cost is also multiplied by that same factor. Instead of solving separately for the price of a table and the price of a chair, we can directly scale up the combined cost. This approach saves time and reduces algebraic effort in exam settings. Step-by-Step Solution: Step 1: Observe that the original combination is 7 tables and 12 chairs for Rs 48,250. Step 2: The new requirement is 21 tables and 36 chairs. Step 3: Note that 21 is three times 7 and 36 is three times 12. Step 4: Because each item has the same price as before, the cost of 21 tables and 36 chairs will be three times the cost of 7 tables and 12 chairs. Step 5: Multiply the total cost by 3: 48,250 * 3. Step 6: 48,250 * 3 = 144,750. Step 7: Therefore, the cost of 21 tables and 36 chairs is Rs 1,44,750. Verification / Alternative check: If desired, one can let the cost of one table be T and one chair be C. Then 7T + 12C = 48,250. For 21 tables and 36 chairs, the total cost is 21T + 36C = 3(7T + 12C) = 3 * 48,250 = 144,750. This algebraic check confirms that simple scaling by 3 is valid and that the final cost is correct. Why Other Options Are Wrong: Option B Rs 2,35,745 and option C Rs 89,489 are not simple multiples of the original total and do not correspond to a neat scaling of both quantities by an integer factor. Option D Rs 74,256 is too low and again is not equal to 48,250 multiplied by any small integer. None of these alternatives maintain the proportional relationship between quantities and cost. Common Pitfalls: A common mistake is to attempt to compute the cost of a single table and a single chair by solving two equations, even though only one combined equation is effectively given. This leads to unnecessary algebra. Recognizing that both quantities are multiplied by the same factor and hence the total cost also scales by the same factor simplifies such questions considerably. Final Answer: The total cost of 21 tables and 36 chairs is Rs 1,44,750.

Question 20

Khushi buys bangles of three types at prices of Rs 34 per piece, Rs 42 per piece, and Rs 6 per piece. She buys an equal number of bangles of each type and spends a total of Rs 1,722 on all the bangles together. How many bangles did she buy of each t…

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Introduction / Context: This arithmetic reasoning question involves equal quantities and total cost. Khushi buys three kinds of bangles at different prices per piece, but the number of bangles of each type is the same. You are told her total expenditure, and you must find the number of bangles of each type. This is a straightforward application of linear equations with one variable. Given Data / Assumptions: Type 1 bangles cost Rs 34 per piece. Type 2 bangles cost Rs 42 per piece. Type 3 bangles cost Rs 6 per piece. She buys the same number of bangles of each type. Total amount spent on all bangles is Rs 1,722. The answer asks for the number of bangles of each type, not the total number of bangles. Concept / Approach: Let Khushi buy n bangles of each type. Then the total number of bangles is 3n. The total cost is the sum of cost per type: n times the sum of the three prices. This gives a simple equation in n, which we can solve by dividing the total amount by the combined cost per set of one of each type. Because n must be an integer, we also check that the division is exact. Step-by-Step Solution: Step 1: Let the number of bangles of each type be n. Step 2: Cost of type 1 bangles = 34n, cost of type 2 bangles = 42n, and cost of type 3 bangles = 6n. Step 3: Total cost = 34n + 42n + 6n = (34 + 42 + 6)n = 82n. Step 4: The total cost is given as Rs 1,722, so 82n = 1,722. Step 5: Solve for n: n = 1,722 / 82. Step 6: 1,722 ÷ 82 = 21. Step 7: Therefore, Khushi bought 21 bangles of each type. Verification / Alternative check: Compute total cost using n = 21: cost of type 1 = 34 * 21 = 714, cost of type 2 = 42 * 21 = 882, cost of type 3 = 6 * 21 = 126. Sum: 714 + 882 + 126 = 1,722, which matches the given expenditure. This confirms that the value n = 21 is correct. Why Other Options Are Wrong: Option A 27 gives total cost 82 * 27 = 2,214, too high. Option B 13 gives total cost 82 * 13 = 1,066, too low. Option C 17 gives 82 * 17 = 1,394, which also does not equal 1,722. None of these options produce the correct total cost. Common Pitfalls: Some students mistakenly multiply 1,722 by 3 or divide by only one of the prices instead of the combined price per set of three bangles. Another error is to miscalculate the sum of 34, 42, and 6. Carefully summing the three prices before dividing the total amount keeps the algebra simple and reliable. Final Answer: Khushi bought 21 bangles of each type.

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