Q: The number 54 is to be divided into two parts such that the sum of 10 times the first part and 22 times the second part is 780. What is the value of the bigger part?

Introduction:This is a standard linear equation word problem involving two parts of a given total. The problem asks you to split 54 into two numbers so that a weighted sum (10 times one part plus 22 times the other) equals 780. Such questions test your ability to convert a word statement into a pair of algebraic equations and solve them consistently. Given Data / Assumptions: The total of the two parts is 54. Let the first part be x and the second part be y. x + y = 54. 10x + 22y = 780. We must find the bigger of x and y. Concept / Approach:Use simultaneous equations. From the total equation, express one variable in terms of the other (for example, y = 54 - x). Substitute this into the second equation to obtain a single equation in x. Solve for x, compute y, and determine which is larger. This is a direct application of solving two linear equations in two unknowns. Step-by-Step Solution:Let first part = x and second part = y.Total: x + y = 54 => y = 54 - x.Condition: 10x + 22y = 780.Substitute y: 10x + 22(54 - x) = 780.10x + 1188 - 22x = 780.-12x + 1188 = 780.-12x = 780 - 1188 = -408.x = (-408)/(-12) = 34.Then y = 54 - 34 = 20.So the bigger part is 34. Verification / Alternative check:Check the condition: 10 * 34 + 22 * 20 = 340 + 440 = 780, which matches exactly. The total 34 + 20 = 54 is also satisfied. Thus the split (34, 20) is correct and 34 is indeed the larger part. Why Other Options Are Wrong:24, 44, 30, 20: none of these values, when paired with 54 minus that value, satisfy the equation 10x + 22y = 780. A quick substitution shows the weighted sum is different from 780. Common Pitfalls:Confusing which coefficient (10 or 22) multiplies which part when writing the equation.Arithmetic errors while simplifying 10x + 22(54 - x).Forgetting to verify both the total 54 and the weighted sum 780 after solving. Final Answer:34

Question 1

Three numbers are in the ratio 4:5:6 and their average is 25. What is the largest of these three numbers?

Accepted Answer

Introduction / Context:This question combines the concept of ratios with the concept of average. Three numbers are given in a fixed ratio, and their average is known. From this, we can find the actual numbers. We are asked specifically for the largest number among the three. This type of problem is very common in aptitude tests for reinforcing proportional reasoning. Given Data / Assumptions:- Three numbers are in the ratio 4:5:6.- Their average is 25.- All numbers are real and consistent with the ratio.- We need to find the largest number among them. Concept / Approach:When numbers are in a ratio, we can represent them using a common multiplier k. Thus, the three numbers can be written as 4k, 5k, and 6k. The average of three numbers is their sum divided by 3. Using the given average, we form an equation involving k and solve for it. Once we know k, we can compute each number and identify the largest one. Step-by-Step Solution:Step 1: Let the three numbers be 4k, 5k, and 6k.Step 2: Their sum is 4k + 5k + 6k = 15k.Step 3: The average of three numbers is total sum divided by 3, so average = 15k / 3.Step 4: Simplify 15k / 3 to get 5k.Step 5: The average is given as 25, thus 5k = 25.Step 6: Solve for k: k = 25 / 5 = 5.Step 7: Substitute k = 5 into the expressions for the numbers to get 4k = 4 * 5 = 20, 5k = 5 * 5 = 25, and 6k = 6 * 5 = 30.Step 8: The largest of these numbers is 30. Verification / Alternative check:We can verify by checking both the ratio and the average. The three numbers 20, 25, and 30 are indeed in the ratio 4:5:6 because dividing each by 5 yields 4, 5, and 6. Their sum is 20 + 25 + 30 = 75, and the average is 75 / 3 = 25, matching the given average. This confirms that the numbers are correctly found and that 30 is the largest number. Why Other Options Are Wrong:Values such as 40, 50, 60, or 24 do not fit as the largest number in a set that has ratio 4:5:6 with average 25. For example, if 40 were the largest number corresponding to 6k, then k would be 40 / 6, and the resulting numbers would not have an average of 25. Similar contradictions occur with the other options when checked carefully against the given average and ratio conditions. Common Pitfalls:Some learners mistakenly divide the average by each ratio part separately rather than expressing all numbers in terms of k. Others may incorrectly average the ratio values directly. The correct approach is to treat 4k, 5k, and 6k as the real numbers and use the given average to find k. This systematic method avoids confusion and errors. Final Answer:The largest of the three numbers is 30.

Question 2

The difference between two numbers is 1365. When the larger number is divided by the smaller number, the quotient is 6 and the remainder is 15. What is the value of the smaller number?

Accepted Answer

Introduction / Context:This is a number theory problem involving division with remainder and the difference between two numbers. We are told the difference between the larger and smaller numbers and given the quotient and remainder when the larger number is divided by the smaller. The goal is to find the smaller number. Such problems appear frequently in quantitative aptitude tests and require careful use of the division algorithm. Given Data / Assumptions:- There are two numbers, a larger one and a smaller one.- The difference between the larger and smaller numbers is 1365.- When the larger number is divided by the smaller number, the quotient is 6 and the remainder is 15.- We need to find the smaller number. Concept / Approach:The division algorithm states that if a larger number L is divided by a smaller number S, then L = S * quotient + remainder. Here, the quotient is 6 and the remainder is 15, so L = 6S + 15. We also know that the difference L - S equals 1365. Substituting the expression for L into the difference equation allows us to solve for S directly. Step-by-Step Solution:Step 1: Let the smaller number be S and the larger number be L.Step 2: From the division condition, L = 6S + 15.Step 3: The difference between the larger and smaller numbers is given as L - S = 1365.Step 4: Substitute L = 6S + 15 into the difference equation: (6S + 15) - S = 1365.Step 5: Simplify the left side: 6S + 15 - S = 5S + 15.Step 6: So we get 5S + 15 = 1365.Step 7: Subtract 15 from both sides to get 5S = 1350.Step 8: Divide both sides by 5: S = 1350 / 5 = 270.Step 9: Therefore, the smaller number is 270. Verification / Alternative check:Compute the larger number using L = 6S + 15. With S = 270, L = 6 * 270 + 15 = 1620 + 15 = 1635. The difference L - S is 1635 - 270 = 1365, which matches the given difference. Also, when 1635 is divided by 270, the quotient is 6 and the remainder is 15, since 270 * 6 = 1620 and 1635 - 1620 = 15. This confirms that the smaller number is indeed 270. Why Other Options Are Wrong:If S were 280, the larger number would be 6 * 280 + 15 = 1695, and the difference would be 1695 - 280 = 1415, not 1365. Similar contradictions arise with S = 370 or S = 170 or S = 260. None of those values give both the correct difference and the required quotient and remainder when the larger number is divided by the smaller. Common Pitfalls:Some learners misapply the division formula and write L = S / 6 + 15 or similar incorrect expressions. Others mistakenly set the difference as S - L instead of L - S. To avoid these errors, remember that L = S * quotient + remainder and that the difference is larger minus smaller. Writing these relationships clearly before solving makes the algebra straightforward. Final Answer:The smaller number is 270.

Question 3

The product of two numbers is 9375 and when the larger number is divided by the smaller number the quotient is 15. What is the sum of the two numbers?

Accepted Answer

Introduction / Context:This is a number theory question about two numbers whose product and quotient are known. We are told the product of the numbers and the quotient when the larger is divided by the smaller. From this information, we are required to find the sum of the two numbers. Questions like this are good practice for applying basic algebra and the relationships among product, quotient, and factors. Given Data / Assumptions:- There are two numbers, a larger one and a smaller one.- The product of the two numbers is 9375.- When the larger number is divided by the smaller number, the quotient is 15.- The remainder is zero because the problem does not mention any remainder.- We must find the sum of the two numbers. Concept / Approach:Let the smaller number be S and the larger number be L. The quotient condition tells us that L = 15S. The product condition tells us that L * S = 9375. Substituting L = 15S into the product equation allows us to solve for S. Once S is known, L can be found easily and their sum can be computed. Step-by-Step Solution:Step 1: Let the smaller number be S and the larger number be L.Step 2: From the quotient condition, L divided by S gives a quotient of 15, so L = 15S.Step 3: From the product condition, L * S = 9375.Step 4: Substitute L = 15S into L * S = 9375 to get 15S * S = 9375.Step 5: This simplifies to 15S^2 = 9375.Step 6: Divide both sides by 15 to obtain S^2 = 9375 / 15.Step 7: Compute 9375 / 15 = 625, so S^2 = 625.Step 8: Take the positive square root (since we consider positive numbers here) to get S = 25.Step 9: Then the larger number is L = 15 * 25 = 375.Step 10: The sum of the two numbers is S + L = 25 + 375 = 400. Verification / Alternative check:Check the product: 25 * 375 = 9375, which matches the given product. Check the quotient: when 375 is divided by 25, the quotient is 15 and the remainder is 0, which matches the given information. The sum 25 + 375 is 400, so all conditions are satisfied and the result is consistent. Why Other Options Are Wrong:If the sum of the numbers were 100, 200, 300, or 250, then the numbers would have to be very different, and it would be impossible for them to have both a product of 9375 and a quotient of 15. For example, a sum of 200 would imply average around 100, but 100 * 100 is only 10000, and making the product exactly 9375 with a quotient of 15 is not possible under that constraint. Only the sum 400 corresponds to the exact pair 25 and 375 that satisfies both product and quotient conditions. Common Pitfalls:Some learners may misinterpret the quotient and think that S = 15L instead of L = 15S, which reverses the relationship. Others may try to guess numbers randomly instead of using the algebraic relationship. Another mistake is to forget to take the positive square root when solving for S. Following the systematic approach of expressing one variable in terms of the other ensures a correct and efficient solution. Final Answer:The sum of the two numbers is 400.

Question 4

In a certain two digit number, the digit in the units place exceeds the digit in the tens place by 2. The product of the number and the sum of its digits is 144. What is the number?

Accepted Answer

Introduction / Context:This problem is a digit based number puzzle. It describes a two digit number with a specific relationship between its tens and units digits, and a product condition involving the number and the sum of its digits. Our goal is to translate the conditions into algebraic equations and determine the original number. Such problems are common in quantitative aptitude sections and help build skill in working with place value and simple equations. Given Data / Assumptions:- The number is a two digit number with tens digit x and units digit y.- The units digit y exceeds the tens digit x by 2, so y = x + 2.- The number is therefore 10x + y.- The sum of its digits is x + y.- The product of the number and the sum of its digits equals 144.- We must find the original two digit number. Concept / Approach:Using the digit relationship, we express y in terms of x. Then the number and the sum of its digits are both expressed as algebraic expressions in x. The given product condition yields an equation involving x only. Solving this equation gives the tens digit, and then the units digit follows from the relation y = x + 2. The actual two digit number is then formed as 10x + y. Step-by-Step Solution:Step 1: Let the tens digit be x and the units digit be y, so the number is 10x + y.Step 2: The units digit exceeds the tens digit by 2, so y = x + 2.Step 3: The sum of the digits is x + y. Using y = x + 2, this sum becomes x + (x + 2) = 2x + 2.Step 4: The product of the number and the sum of its digits is given as 144, so (10x + y) * (x + y) = 144.Step 5: Substitute y = x + 2 into the number: 10x + y becomes 10x + (x + 2) = 11x + 2.Step 6: The sum of the digits is 2x + 2, so the product condition becomes (11x + 2) * (2x + 2) = 144.Step 7: Expand the left hand side: (11x + 2)(2x + 2) = 22x^2 + 22x + 4x + 4 = 22x^2 + 26x + 4.Step 8: Set the expression equal to 144: 22x^2 + 26x + 4 = 144.Step 9: Subtract 144 from both sides to get 22x^2 + 26x - 140 = 0.Step 10: Divide the entire equation by 2 to simplify: 11x^2 + 13x - 70 = 0.Step 11: Solve this quadratic equation. One integer solution is x = 2.Step 12: With x = 2, y = x + 2 = 4, so the number is 10 * 2 + 4 = 24. Verification / Alternative check:Take the number 24 and check both conditions. The tens digit is 2 and the units digit is 4. The units digit exceeds the tens digit by 2, because 4 = 2 + 2. The sum of the digits is 2 + 4 = 6. The product of the number and the sum of its digits is 24 * 6 = 144, exactly matching the given condition. Therefore, 24 satisfies all requirements of the problem. Why Other Options Are Wrong:Consider 12, 36, 48, or 18. For 12, the units digit 2 does not exceed the tens digit 1 by 2. For 36, the units digit 6 exceeds the tens digit 3 by 3, not 2. For 48 and 18, the sum of digits multiplied by the number does not equal 144. For example, 36 * (3 + 6) = 36 * 9 = 324, which is not 144. None of these numbers satisfy both the digit difference and the product condition. Common Pitfalls:Common mistakes include misreading the phrase "exceeds by 2" and writing y = x - 2 instead of y = x + 2, or incorrectly forming the product equation. Some learners also try random numbers instead of forming and solving the equation, which can be time consuming and unreliable. Using a systematic algebraic approach is the most efficient and accurate method. Final Answer:The required two digit number is 24.

Question 5

In a fraction word problem, the denominator of a fraction is 3 more than its numerator. If both the numerator and the denominator are increased by 4, the value of the fraction becomes 4/5. What was the original fraction before this change?

Accepted Answer

Introduction:This question tests basic algebra with fractions and focuses on how changing numerator and denominator together affects the value of the fraction. You are given a relationship between the numerator and denominator, and then a new fraction formed after both parts are increased by the same amount. The aim is to form a simple equation in one variable and solve it to recover the original fraction. Given Data / Assumptions: Let the original numerator be n. The denominator is 3 more than the numerator, so denominator = n + 3. Both numerator and denominator are increased by 4. The new fraction after this increase is 4/5. We must find the original fraction n/(n + 3). Concept / Approach:Convert the description into an algebraic equation. After increasing numerator and denominator by the same number, we obtain a new fraction. Set this equal to 4/5 and solve for the unknown numerator. Once n is known, substitute back to get the denominator and hence the original fraction. Step-by-Step Solution:Let original fraction = n/(n + 3).After increasing both by 4, new fraction = (n + 4)/(n + 3 + 4) = (n + 4)/(n + 7).Given (n + 4)/(n + 7) = 4/5.Cross-multiply: 5(n + 4) = 4(n + 7).5n + 20 = 4n + 28.5n - 4n = 28 - 20 => n = 8.Denominator = n + 3 = 8 + 3 = 11.Original fraction = 8/11. Verification / Alternative check:If the original fraction is 8/11, then increasing numerator and denominator by 4 gives 12/15. Now 12/15 simplifies to 4/5, which matches the condition in the problem. So the solution is consistent. Why Other Options Are Wrong:7/11: After adding 4 to numerator and denominator we get 11/15, which is not 4/5.9/11: Leads to 13/15, still not equal to 4/5.5/8 and 3/7: Their transformed fractions do not simplify to 4/5 and do not satisfy the relation “denominator is 3 more than numerator”. Common Pitfalls:Forgetting that the denominator is 3 more than the numerator and instead assuming some random values.Adding 4 only to the numerator or only to the denominator, instead of both.Not simplifying or checking the resulting fraction after solving the equation. Final Answer:8/11

Question 6

What approximate value (correct to the nearest integer) will come in place of the question mark in the expression (47% of 1442 - 36% of 1412) + 63 = ?

Accepted Answer

Introduction:This is a percentage and approximation problem commonly seen in bank and aptitude exams. You are asked to evaluate an expression involving percentages of large numbers and then add a constant. The focus is on calculating accurately and then rounding sensibly to the nearest integer. Understanding how to convert percentages into simple multiplication is the key idea here. Given Data / Assumptions: We need to compute (47% of 1442 - 36% of 1412) + 63. 47% means 47/100 of the given number. 36% means 36/100 of the given number. We finally round the result to the nearest whole number for the approximate value. Concept / Approach:Convert each percentage to its decimal or fractional form and multiply by the number. Then subtract the two percentage values, and finally add 63. Since these are straightforward arithmetic operations, it is best to perform them step by step, and round only at the end to avoid compounding rounding errors. Step-by-Step Solution:47% of 1442 = (47/100) * 1442 = 677.74.36% of 1412 = (36/100) * 1412 = 508.32.Difference = 677.74 - 508.32 = 169.42.Now add 63: 169.42 + 63 = 232.42.Rounded to the nearest integer, 232.42 ≈ 232. Verification / Alternative check:You can approximate the calculations more roughly and still see that the final value is close to 232. For example, 47% of 1440 ≈ 0.47 * 1440 ≈ 677, and 36% of 1410 ≈ 0.36 * 1410 ≈ 508. The difference is around 169 and adding 63 gives roughly 232. This matches our precise calculation. Why Other Options Are Wrong:228 and 230: These are somewhat close but off by more than 2 units from the computed value 232.42.235 and 240: These are clearly larger than the accurate result and correspond to overestimation of the intermediate percentage calculations. Common Pitfalls:Rounding the percentage values too early, which can distort the final sum.Confusing percentage subtraction (47 - 36 = 11) and applying 11% directly without separating the numbers 1442 and 1412, which are different.Forgetting to add the constant 63 at the end. Final Answer:232

Question 7

There are two real numbers such that the sum of twice the first number and three times the second number is 39, while the sum of three times the first number and twice the second number is 36. What is the value of the larger of these two numbers?

Accepted Answer

Introduction:This is a classic pair of simultaneous linear equations in two variables. The question gives two different weighted sums of the same two unknown numbers and asks for the larger number. Such problems are very common in quantitative aptitude and test your ability to form equations, solve them systematically, and correctly identify which number is larger. Given Data / Assumptions: Let the first number be x. Let the second number be y. Equation 1: 2x + 3y = 39. Equation 2: 3x + 2y = 36. We must find the larger of x and y. Concept / Approach:We use standard methods for solving simultaneous equations, such as elimination or substitution. Elimination is usually faster here: multiply and subtract the equations to remove one variable, find the other, then back-substitute to find the remaining variable. After computing both values, compare them to identify the larger number. Step-by-Step Solution:Start with 2x + 3y = 39 … (1)And 3x + 2y = 36 … (2)Multiply (1) by 3: 6x + 9y = 117.Multiply (2) by 2: 6x + 4y = 72.Subtract second from first: (6x + 9y) - (6x + 4y) = 117 - 72.5y = 45 => y = 9.Substitute into (2): 3x + 2*9 = 36 => 3x + 18 = 36 => 3x = 18 => x = 6.Thus, the larger number is 9. Verification / Alternative check:Check equation (1): 2*6 + 3*9 = 12 + 27 = 39, correct. Check equation (2): 3*6 + 2*9 = 18 + 18 = 36, also correct. So (x, y) = (6, 9) satisfies both equations. Why Other Options Are Wrong:3 and 6: These are the individual numbers or less than them but do not represent the larger one.12 and 15: Substituting these into the equations fails to satisfy both equations simultaneously. Common Pitfalls:Making arithmetic mistakes while multiplying or subtracting equations.Stopping after finding only one variable and not checking the second equation.Accidentally reporting the smaller number instead of the larger one. Final Answer:9

Question 8

In a class IX General Knowledge quiz, 12 students took part. The number of boys is 4 more than the number of girls. How many boys and how many girls participated in the quiz?

Accepted Answer

Introduction:This is a straightforward numbers and equations problem involving a total and a difference. You are told how many students participated in a General Knowledge quiz and how many more boys there are than girls. Two simple equations in two variables are enough to find the exact number of boys and girls. Such problems are common in aptitude tests to check basic algebra and logical setup. Given Data / Assumptions: Total participants = 12 students. Number of boys is 4 more than the number of girls. All participants are either boys or girls, with no other category. Concept / Approach:Let the number of girls be G and the number of boys be B. Use the total equation B + G = 12 and the difference equation B = G + 4. Solve these simultaneously. This is a basic example of forming linear equations from a word problem. Step-by-Step Solution:Let G = number of girls and B = number of boys.From the statement: B + G = 12.Also, B = G + 4 (boys are 4 more than girls).Substitute B in the total: (G + 4) + G = 12.2G + 4 = 12 => 2G = 8 => G = 4.Then B = G + 4 = 4 + 4 = 8.So there are 8 boys and 4 girls. Verification / Alternative check:Check the total: 8 + 4 = 12. Check the difference: 8 is exactly 4 more than 4. Both conditions given in the question are satisfied, so the answer is correct. Why Other Options Are Wrong:7 boys and 5 girls: total 12 but the difference is only 2, not 4.9 boys and 3 girls: difference 6, not 4.6 boys and 6 girls: difference 0, not 4.10 boys and 2 girls: difference 8, not 4. Common Pitfalls:Reversing the difference and writing G = B + 4 instead of B = G + 4.Guessing numbers by trial and error instead of setting up simple equations.Forgetting that the sum must exactly match the total of 12 participants. Final Answer:8 boys and 4 girls

Question 9

The product of three consecutive integers, when divided by 8, is 720. If these three consecutive integers are all positive, what is the product of their square roots?

Accepted Answer

Introduction:This question mixes number theory with basic algebra and square roots. You are told that the product of three consecutive integers, when divided by 8, gives 720. From this, you must find the actual three integers, and then compute the product of their square roots. It tests your ability to form equations from word descriptions and to handle square roots of products correctly. Given Data / Assumptions: Three consecutive positive integers are involved. Their product divided by 8 equals 720. Let the integers be n, n + 1, and n + 2. The equation is n(n + 1)(n + 2) / 8 = 720. We must find √n * √(n + 1) * √(n + 2). Concept / Approach:First, use the condition on the product to determine the three consecutive integers. Multiply all three, equate the result to 720 * 8, and solve by trial or observation within a reasonable range. Once the integers are known, the product of their square roots can be simplified using the property √a * √b = √(ab). Step-by-Step Solution:Given: n(n + 1)(n + 2)/8 = 720.So n(n + 1)(n + 2) = 720 * 8 = 5760.Try n = 16: product = 16 * 17 * 18 = 4896 (too small).Try n = 18: product = 18 * 19 * 20 = 6840 (too large).Try n = 16 for even spacing in another way: check 16 * 18 * 20 = 5760.The three integers that satisfy the product condition are 16, 18, and 20.Now find product of square roots: √16 * √18 * √20.√16 = 4; √18 = √(9 * 2) = 3√2; √20 = √(4 * 5) = 2√5.So product = 4 * 3√2 * 2√5 = (4 * 3 * 2) * √(2 * 5) = 24 * √10. Verification / Alternative check:Check the product condition: 16 * 18 * 20 = 5760, and 5760 / 8 = 720, which exactly matches the statement. Then the square-root product simplification is purely algebraic and straightforward, confirming 24√10 is correct. Why Other Options Are Wrong:12√10 and 36√5: These are other square-root forms but do not come from the correct simplification of √16 * √18 * √20.120: Ignores the square roots and treats them as plain numbers.“None of these”: incorrect because 24√10 is available and valid. Common Pitfalls:Assuming the three numbers must be even only, instead of checking the product carefully.Forgetting to multiply the right-hand side by 8 when eliminating the division.Mistakes in simplifying square roots, such as mis-handling √18 or √20. Final Answer:24√10

Question 10

In the following number series, exactly one term is wrong. Identify the wrong number: 7.5, 47.5, 87.5, 157.5, 247.5, 357.5, 487.5

Accepted Answer

Introduction:This question is a number series puzzle where exactly one term is incorrect. The goal is to detect the underlying pattern that most of the numbers follow and then spot the term that breaks this pattern. Such series questions test observational skills, comfort with differences, and the ability to see second-level patterns in the sequence of differences between consecutive terms. Given Data / Assumptions: Series: 7.5, 47.5, 87.5, 157.5, 247.5, 357.5, 487.5. Only one term is wrong. All numbers end with .5, suggesting a consistent base increment. Concept / Approach:First, compute the differences between consecutive terms to look for a pattern in increments. If the first-level differences are not smooth, examine the second-level differences. Typically, you will find a simple arithmetic pattern (like adding 20 each time) that almost all terms follow, except one. That term is the wrong number. Step-by-Step Solution:Compute first-level differences:47.5 - 7.5 = 40.87.5 - 47.5 = 40.157.5 - 87.5 = 70.247.5 - 157.5 = 90.357.5 - 247.5 = 110.487.5 - 357.5 = 130.Now imagine a pattern where the increments increase by 20 each time, starting from 30: 30, 50, 70, 90, 110, 130.Using this ideal pattern from 7.5: 7.5 + 30 = 37.5, 37.5 + 50 = 87.5, 87.5 + 70 = 157.5, 157.5 + 90 = 247.5, 247.5 + 110 = 357.5, 357.5 + 130 = 487.5.So the correct series should be: 7.5, 37.5, 87.5, 157.5, 247.5, 357.5, 487.5.Comparing with the given series, only 47.5 is out of place; it should be 37.5. Verification / Alternative check:The corrected differences (30, 50, 70, 90, 110, 130) form a perfect arithmetic pattern with a constant increase of 20. Every term from the third term onwards fits this corrected pattern, confirming that only the second term is wrong. Why Other Options Are Wrong:87.5, 247.5, 357.5, 487.5: each of these fits the corrected pattern once the second term is fixed, so they are not wrong. Common Pitfalls:Stopping after looking at only first-level differences and not considering how they might ideally progress.Assuming the pattern must be very complex, when a simple increasing-difference pattern works after just one correction.Randomly guessing a term without verifying the full adjusted series. Final Answer:47.5

Question 11

A fraction becomes 2/3 when 1 is added to both its numerator and denominator, and it becomes 1/2 when 1 is subtracted from both the numerator and the denominator. What is the original fraction?

Accepted Answer

Introduction:This is a classic fraction puzzle where the same unknown fraction is modified in two different ways: adding 1 to both numerator and denominator gives one known value, and subtracting 1 from both gives another. The question tests algebraic setup using two equations in two variables (numerator and denominator) and solving them accurately without confusion between the different transformations. Given Data / Assumptions: Let the original fraction be n/d, with n and d positive integers and d ≠ 0. (n + 1)/(d + 1) = 2/3 (after adding 1). (n - 1)/(d - 1) = 1/2 (after subtracting 1). We must find n and d, hence the fraction n/d. Concept / Approach:Translate each condition into an equation and solve the resulting system. Both equations are rational, but cross-multiplication will reduce them to simple linear relationships between n and d. Solving the pair of equations simultaneously will give a unique solution for n and d, provided the fraction is well-defined. Step-by-Step Solution:From (n + 1)/(d + 1) = 2/3, cross-multiply: 3(n + 1) = 2(d + 1).3n + 3 = 2d + 2 => 3n + 1 = 2d … (1).From (n - 1)/(d - 1) = 1/2, cross-multiply: 2(n - 1) = d - 1.2n - 2 = d - 1 => d = 2n - 1 … (2).Substitute (2) into (1): 3n + 1 = 2(2n - 1) = 4n - 2.Rearrange: 4n - 2 - 3n - 1 = 0 => n - 3 = 0 => n = 3.Then d = 2n - 1 = 2*3 - 1 = 5.Original fraction = 3/5. Verification / Alternative check:Check adding 1: (3 + 1)/(5 + 1) = 4/6 = 2/3, correct. Check subtracting 1: (3 - 1)/(5 - 1) = 2/4 = 1/2, also correct. Thus 3/5 satisfies both transformed conditions perfectly. Why Other Options Are Wrong:1/3, 2/5, 4/7, 5/7: none of these fractions give 2/3 when you add 1 to numerator and denominator and 1/2 when you subtract 1 from both. At least one of the two conditions fails in each case. Common Pitfalls:Forgetting to cross-multiply correctly when forming the equations.Using the same modified fraction expression in both equations instead of the original n/d.Solving (1) and (2) with arithmetic mistakes while isolating n and d. Final Answer:3/5

Question 12

Four prime numbers are written in ascending order. The product of the first three primes is 385, and the product of the last three primes is 1001. What is the value of the fourth (largest) prime?

Accepted Answer

Introduction:This question combines prime factorization with logical ordering. You are told that four prime numbers are in ascending order, and you are given two overlapping products: the product of the first three and the product of the last three. The challenge is to factor each product into primes and then match the overlapping primes to identify the full set of four and the largest one. Given Data / Assumptions: The four primes are p1 < p2 < p3 < p4. p1 * p2 * p3 = 385. p2 * p3 * p4 = 1001. All numbers mentioned are positive primes. Concept / Approach:Factor 385 and 1001 into their prime factors. Since p2 and p3 are common to both products, they will appear in both factorizations. The extra prime in the first factorization will be p1, and the extra prime in the second factorization will be p4. This approach quickly reveals all four primes in ascending order. Step-by-Step Solution:Prime factorize 385: 385 = 5 * 7 * 11.So p1, p2, p3 must be 5, 7, 11 in ascending order.Prime factorize 1001: 1001 = 7 * 11 * 13.So p2, p3, p4 must be 7, 11, 13 in ascending order.Common primes in both products are 7 and 11, which must be p2 and p3.The remaining prime from 385 is 5 (p1), and the remaining prime from 1001 is 13 (p4).Thus the four primes in ascending order are 5, 7, 11, 13.The fourth (largest) prime is 13. Verification / Alternative check:Check product of first three: 5 * 7 * 11 = 385, correct. Check product of last three: 7 * 11 * 13 = 1001, also correct. The ordering 5 < 7 < 11 < 13 satisfies the condition of ascending primes. Why Other Options Are Wrong:11 and 17: 11 is one of the middle primes, not the largest, and 17 does not appear in the prime factorization of either 385 or 1001.19 and 23: these primes never feature in the factorization of 385 or 1001, so they cannot be any of p1, p2, p3, or p4. Common Pitfalls:Incorrect factorization of 385 or 1001, especially confusing 1001 with non-prime factors.Ignoring the ascending order condition and mis-assigning which factors correspond to which positions.Assuming the primes must be consecutive integers, which is not required by the question. Final Answer:13

Question 13

The number 54 is to be divided into two parts such that the sum of 10 times the first part and 22 times the second part is 780. What is the value of the bigger part?

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Introduction:This is a standard linear equation word problem involving two parts of a given total. The problem asks you to split 54 into two numbers so that a weighted sum (10 times one part plus 22 times the other) equals 780. Such questions test your ability to convert a word statement into a pair of algebraic equations and solve them consistently. Given Data / Assumptions: The total of the two parts is 54. Let the first part be x and the second part be y. x + y = 54. 10x + 22y = 780. We must find the bigger of x and y. Concept / Approach:Use simultaneous equations. From the total equation, express one variable in terms of the other (for example, y = 54 - x). Substitute this into the second equation to obtain a single equation in x. Solve for x, compute y, and determine which is larger. This is a direct application of solving two linear equations in two unknowns. Step-by-Step Solution:Let first part = x and second part = y.Total: x + y = 54 => y = 54 - x.Condition: 10x + 22y = 780.Substitute y: 10x + 22(54 - x) = 780.10x + 1188 - 22x = 780.-12x + 1188 = 780.-12x = 780 - 1188 = -408.x = (-408)/(-12) = 34.Then y = 54 - 34 = 20.So the bigger part is 34. Verification / Alternative check:Check the condition: 10 * 34 + 22 * 20 = 340 + 440 = 780, which matches exactly. The total 34 + 20 = 54 is also satisfied. Thus the split (34, 20) is correct and 34 is indeed the larger part. Why Other Options Are Wrong:24, 44, 30, 20: none of these values, when paired with 54 minus that value, satisfy the equation 10x + 22y = 780. A quick substitution shows the weighted sum is different from 780. Common Pitfalls:Confusing which coefficient (10 or 22) multiplies which part when writing the equation.Arithmetic errors while simplifying 10x + 22(54 - x).Forgetting to verify both the total 54 and the weighted sum 780 after solving. Final Answer:34

Question 14

Find the number which, when multiplied by 15, gives a result that is 196 greater than the original number itself.

Accepted Answer

Introduction:This is a simple linear equation problem involving one unknown number. The statement says that when the number is multiplied by 15, the resulting product is greater than the original number by 196. This kind of question is very common in aptitude tests and helps reinforce algebraic modeling of verbal descriptions using one equation in one variable. Given Data / Assumptions: Let the original number be x. When multiplied by 15, the result is 15x. This product is 196 more than the original number, so 15x = x + 196. We must solve for x. Concept / Approach:Translate the English sentence into an algebraic equation. The phrase “is increased by 196” means “is equal to the original plus 196.” Once the equation 15x = x + 196 is written, it is a straightforward linear equation that can be solved by isolating x on one side. Step-by-Step Solution:Let x be the original number.Given: 15x is 196 more than x.So, 15x = x + 196.Subtract x from both sides: 15x - x = 196.14x = 196.x = 196 / 14 = 14.Therefore, the required number is 14. Verification / Alternative check:Check the condition using x = 14: 15 * 14 = 210. The original number is 14. The increase is 210 - 14 = 196, which matches the problem statement exactly. Thus x = 14 is correct. Why Other Options Are Wrong:10: gives 15 * 10 = 150, increase of only 140.12: gives 15 * 12 = 180, increase of 168.16: gives 240, increase of 224.18: gives 270, increase of 252. None of these equal an increase of 196. Common Pitfalls:Misinterpreting the phrase “is increased by 196” as 15x = 196x or 15x + 196 incorrectly.Forgetting to subtract x from both sides when solving 15x = x + 196.Doing division 196/14 incorrectly due to a calculation slip. Final Answer:14

Question 15

The sum of the squares of two numbers is 3341, and the difference of their squares is 891. What are the two numbers?

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Introduction:This problem focuses on the algebraic properties of squares of numbers. You are given the sum of the squares of two numbers and the difference of their squares, and you must find the numbers themselves. The question is a good application of simultaneous equations and uses the identities involving a^2 + b^2 and a^2 - b^2 together. Given Data / Assumptions: Let the two numbers be a and b, with a > b. a^2 + b^2 = 3341. a^2 - b^2 = 891. Both a and b are real numbers; in this context they turn out to be positive integers. Concept / Approach:We have two equations in terms of a^2 and b^2. Adding the equations will eliminate b^2 and give a direct expression for a^2. Subtracting appropriately will then yield b^2. Once a^2 and b^2 are known, we take square roots to obtain a and b. This is a very standard approach when dealing with sums and differences of squares. Step-by-Step Solution:Given: a^2 + b^2 = 3341 … (1).And a^2 - b^2 = 891 … (2).Add (1) and (2): (a^2 + b^2) + (a^2 - b^2) = 3341 + 891.2a^2 = 4232.a^2 = 4232 / 2 = 2116.a = √2116 = 46 (taking the positive root for this context).Substitute a^2 back into (1): 2116 + b^2 = 3341.b^2 = 3341 - 2116 = 1225.b = √1225 = 35.Thus the two numbers are 35 and 46. Verification / Alternative check:Check the sum of squares: 35^2 + 46^2 = 1225 + 2116 = 3341, correct. Check the difference of squares: 46^2 - 35^2 = 2116 - 1225 = 891, also correct. Both given conditions are satisfied, confirming the result. Why Other Options Are Wrong:35 and 50; 40 and 55; 45 and 60; 38 and 45: none of these pairs simultaneously satisfy both the sum 3341 and the difference 891 when you compute their squares. Common Pitfalls:Trying to directly solve for a and b instead of working with a^2 and b^2 first.Making arithmetic mistakes while adding 3341 and 891 or subtracting 2116 from 3341.Forgetting to consider that the numbers must match both equations simultaneously, not just one. Final Answer:35 and 46

Question 16

Two numbers are related as follows: the greater number is 2 more than three times the smaller number, and 4 times the smaller number is 5 more than the greater number. What is the value of the greater number?

Accepted Answer

Introduction:This is a two-variable linear equation problem describing a relationship between a greater and a smaller number. Each sentence gives a different linear connection between the two numbers. The task is to convert both statements into equations and solve them together. This type of question is common in aptitude tests to check your ability to translate verbal relationships into algebra. Given Data / Assumptions: Let the greater number be G. Let the smaller number be S. G is 2 more than three times S. 4 times S is 5 more than G. We must find G. Concept / Approach:Convert each statement into an equation. “G is 2 more than three times S” becomes G = 3S + 2. “4 times the smaller is 5 more than the greater” becomes 4S = G + 5. Solving these two linear equations simultaneously will give unique values for S and G, after which we report the greater number G. Step-by-Step Solution:Let G = 3S + 2 … (1).Also, 4S = G + 5 … (2).Substitute G from (1) into (2): 4S = (3S + 2) + 5.4S = 3S + 7.4S - 3S = 7 => S = 7.Now use S in (1): G = 3*7 + 2 = 21 + 2 = 23.Thus the greater number is 23. Verification / Alternative check:Check the first condition: Three times the smaller is 3 * 7 = 21; G should be 2 more, so 23, correct. Check the second condition: 4 times the smaller is 4 * 7 = 28; G + 5 = 23 + 5 = 28, which also matches. Both conditions are satisfied by S = 7 and G = 23. Why Other Options Are Wrong:7 and 9: These are small numbers that do not satisfy both algebraic relations simultaneously.21: This is close to 3 * 7 but misses the extra 2 required by the first condition.25: If G were 25, there would be no integer S that satisfies both equations exactly. Common Pitfalls:Mixing up which number is greater and which is smaller when forming the equations.Interpreting “2 more than three times the second number” as 3S = G + 2 instead of G = 3S + 2.Solving only one equation and forgetting that both conditions must be satisfied at the same time. Final Answer:23

Question 17

On multiplying a certain positive integer by 7, the product is a number in which every digit is 3. What is the smallest such integer for which 7 times the integer consists only of the digit 3?

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Introduction:This question is about number patterns and divisibility. You are told that when a certain integer is multiplied by 7, the result is a number composed entirely of the digit 3 (for example, 33, 333, 3333, and so on). The task is to determine the smallest integer for which this happens. This involves understanding repdigit numbers (numbers with repeated identical digits) and checking divisibility by 7. Given Data / Assumptions: We need a number N such that 7N is a repdigit consisting only of 3s. The product 7N may be 33, 333, 3333, 33333, 333333, and so on. Among all such valid N, we need the smallest positive integer. Answer choices include several candidate integers like 47619 and 476190476. Concept / Approach:Repdigit numbers containing only 3s can be expressed as 3 * (111..., a string of 1s). One way to check candidates is to multiply them by 7 and see whether the product has only the digit 3. Since the options are specific and relatively few, direct checking is efficient and reveals the pattern. We must also confirm that there is no smaller candidate with fewer digits that satisfies the condition. Step-by-Step Solution:Check 47619: 7 * 47619 = 333333.This product, 333333, clearly consists only of the digit 3.Check 476190476: 7 * 476190476 = 3333333332, which contains a 2 and is therefore not valid.Check 48617: 7 * 48617 = 340319, which has digits other than 3.Check 4587962: 7 * 4587962 = 32115734, again not all 3s.Thus, 47619 is the only candidate among the given options that produces a product consisting entirely of the digit 3.To see that it is smallest, note that 7 * 333 = 2331, 7 * 3333 = 23331, 7 * 33333 = 233331, etc., and none of these produce an all-3 number until we reach 333333 = 7 * 47619. Verification / Alternative check:Compute 333333 / 7 = 47619 exactly, confirming no remainder. Also, check smaller repdigits (33, 333, 3333, 33333) to confirm that none are divisible by 7 without leaving other digits in the result. This confirms that 47619 is indeed the smallest integer with the required property. Why Other Options Are Wrong:476190476, 48617, and 4587962: their products with 7 contain digits other than 3.33333: this is not the integer N but rather a candidate product; it is not divisible by 7 into a whole repdigit of 3s as required by the problem description. Common Pitfalls:Confusing the integer N with the resulting number 7N and trying to match both with repeated 3s.Not actually checking the multiplication and assuming that a long number like 476190476 must be correct.Overlooking that the question asks for the smallest such integer, not any arbitrary integer that might work. Final Answer:47619

Question 18

How many integers between 101 and 300 (inclusive) are divisible by 11?

Accepted Answer

Introduction:This is a counting problem involving divisibility and a specified range. You are asked to find how many integers between 101 and 300 (inclusive) are divisible by 11. The method uses the idea of the first and last multiples of 11 within the given interval and then counts how many multiples lie between them using a simple formula. Given Data / Assumptions: We consider integers from 101 to 300 inclusive. We need numbers N in this range such that 11 divides N exactly (N is a multiple of 11). 11, 22, 33, … are the general multiples of 11. Concept / Approach:Find the smallest multiple of 11 that is at least 101, and the largest multiple of 11 that is at most 300. If those are 11 * a and 11 * b respectively, then every multiple of 11 in this interval can be written as 11 * k where k runs from a to b. The count is then simply b - a + 1. Step-by-Step Solution:Find the first multiple of 11 greater than or equal to 101.11 * 9 = 99 (too small), 11 * 10 = 110 (valid).So the smallest multiple in the range is 110.Find the last multiple of 11 less than or equal to 300.11 * 27 = 297 (valid), 11 * 28 = 308 (too large).So the largest multiple in the range is 297.Now multiples are 11 * k where k goes from 10 to 27 inclusive.Count = 27 - 10 + 1 = 18. Verification / Alternative check:You can list them quickly if needed: 110, 121, 132, 143, 154, 165, 176, 187, 198, 209, 220, 231, 242, 253, 264, 275, 286, 297. Counting them confirms there are 18 multiples of 11 in the interval from 101 to 300 inclusive. Why Other Options Are Wrong:17 or 19: These correspond to off-by-one counting errors, such as missing one endpoint or adding an extra multiple beyond 300.20 and 16: These would require either more or fewer multiples than actually exist, not matching the 10 to 27 factor range. Common Pitfalls:Not treating the interval as inclusive and accidentally excluding 110 or 297.Miscomputing 300 / 11 and rounding wrongly to find the last multiple.Confusing the count of multiples with the difference of the endpoints without adding 1. Final Answer:18

Question 19

The sum of two numbers is 44. One of the numbers is 3 times as large as the other. What are the two numbers?

Accepted Answer

Introduction:This is a very common linear equation word problem. You are told the sum of two numbers and that one number is three times the other. These conditions are enough to form and solve a simple algebraic equation. Problems like this strengthen basic algebra skills and help with quick mental modeling of relationships between numbers. Given Data / Assumptions: The sum of the two numbers is 44. One number is 3 times as large as the other. Let the smaller number be x and the larger number be 3x. Concept / Approach:Set up an equation using the sum condition: x + 3x = 44. Solve this simple equation for x, then multiply by 3 to get the larger number. Finally, present both numbers as a pair in ascending or logical order. Step-by-Step Solution:Let the smaller number be x.Then the larger number is 3x.Given: x + 3x = 44.4x = 44.x = 44 / 4 = 11.Larger number = 3 * 11 = 33.So the two numbers are 11 and 33. Verification / Alternative check:Check the sum: 11 + 33 = 44, correct. Check the relationship: 33 is indeed 3 times 11. Both conditions are satisfied exactly, so the solution (11, 33) is correct. Why Other Options Are Wrong:4 and 40: sum is 44 but 40 is 10 times 4, not 3 times.3 and 41: sum is 44 but 41 is not 3 times 3.10 and 34 or 12 and 32: both pairs have sum 44, but the larger numbers are not 3 times their smaller counterparts. Common Pitfalls:Confusing which number is three times the other and setting up reversed equations.Trying to guess pairs instead of writing the straightforward equation x + 3x = 44.Arithmetic mistakes when dividing 44 by 4. Final Answer:11 and 33

Question 20

If the product of 6 integers is negative, what is the maximum possible number of these integers that can be negative?

Accepted Answer

Introduction:This is a sign and parity reasoning question related to the product of integers. It tests your understanding of how a negative product arises from multiplying positive and negative numbers together. Specifically, for a product of 6 integers to be negative, the number of negative factors must be odd. The question asks for the maximum possible number of negative integers under this condition. Given Data / Assumptions: There are exactly 6 integers in the product. The overall product is negative. We can freely choose how many are negative and how many are positive, but the sign of the total product must remain negative. Zero is not explicitly discussed; we assume the product is truly negative (not zero), so no factor is zero. Concept / Approach:When multiplying real numbers, the sign of the product depends on how many factors are negative. Each pair of negative factors contributes a positive effect because (-1) * (-1) = +1. Therefore, the overall sign is negative if and only if the number of negative factors is odd. Since there are 6 integers in total, we need to find the largest odd number less than or equal to 6. Step-by-Step Solution:Let k be the number of negative integers among the 6.The sign of the product is (-1)^k times the product of absolute values.For the total product to be negative, (-1)^k must be -1.This happens exactly when k is odd (1, 3, 5, etc.).We must also have 0 ≤ k ≤ 6 because there are 6 integers in total.Among the allowed odd values 1, 3, 5, the maximum is 5.So at most 5 of the integers can be negative while still keeping the product negative. Verification / Alternative check:Example with 5 negatives: (-1) * (-2) * (-3) * (-4) * (-5) * (6). There are 5 negatives and 1 positive. Pair four of the negatives to get positive contributions and one leftover negative, so the overall sign is negative. With 6 negatives, such as (-1) * (-2) * (-3) * (-4) * (-5) * (-6), the product is positive because the exponent of -1 is 6, an even number. Why Other Options Are Wrong:1, 3: These are odd and would give a negative product, but they are not the maximum possible count of negative integers.2 and 4: These are even, and a product with an even number of negative factors is positive, not negative. Common Pitfalls:Thinking that any number of negatives will work as long as there is at least one negative, which is not true.Forgetting that six negatives produce a positive product.Confusing maximum possible count with the minimum or just any valid count. Final Answer:5

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