Question 1

The sum of two numbers is 25 and their difference is 13. What is their product?

Accepted Answer

Problem restatementGiven S = x + y = 25 and D = x − y = 13, find xy. Concept/ApproachUse the identity S2 − D2 = 4xy. Step-by-Step calculationxy = (S2 − D2) ÷ 4 = (252 − 132) ÷ 4= (625 − 169) ÷ 4 = 456 ÷ 4 = 114 Verification/AlternativeExplicit solution: x = (S + D)÷2 = 19, y = (S − D)÷2 = 6 ⇒ xy = 114. Common pitfallsComputing (S − D)2 instead of S2 − D2, or dividing by 2 instead of 4. Final Answer114

Question 2

A two-digit number has its unit's digit 2 more than its ten's digit. If the product of the number and the sum of its digits is 144, what is the number?

Accepted Answer

Problem restatementLet the number be 10x + y, with y = x + 2. Also, (10x + y) × (x + y) = 144. Find the number. Given data y = x + 2 Product with digit-sum equals 144 Concept/ApproachExpress everything in x, solve the quadratic, and retrieve the digits. Step-by-Step calculationNumber = 10x + (x + 2) = 11x + 2Digit-sum = x + (x + 2) = 2x + 2 = 2(x + 1)(11x + 2) × 2(x + 1) = 144 ⇒ (11x + 2)(x + 1) = 7211x2 + 13x + 2 = 72 ⇒ 11x2 + 13x − 70 = 0Discriminant = 132 + 4 × 11 × 70 = 3249 = 572x = (−13 + 57) ÷ 22 = 2 (valid digit); so y = 4Number = 24 Verification/AlternativeCheck: 24 × (2 + 4) = 24 × 6 = 144. Units digit (4) is 2 more than tens (2). Common pitfallsForgetting that digits must be integers in 0–9; discarding the negative/invalid root is essential. Final Answer24

Question 3

A 3-digit number has digit sum 10. The middle digit equals the sum of the other two, and reversing the digits increases the number by 99. What is the number?

Accepted Answer

Problem restatementLet the number be 100a + 10b + c with a + b + c = 10, b = a + c, and reverse − original = 99. Concept/ApproachTranslate the conditions into equations and solve for digits a, b, c. Step-by-Step calculationReverse − original = (100c + 10b + a) − (100a + 10b + c) = 99(c − a) = 99Therefore, c − a = 1 ⇒ c = a + 1b = a + c = a + (a + 1) = 2a + 1Sum: a + b + c = a + (2a + 1) + (a + 1) = 4a + 2 = 10 ⇒ a = 2Then c = 3, b = 5 ⇒ Number = 253 Verification/AlternativeReversal 352; 352 − 253 = 99. Digit sum 2 + 5 + 3 = 10; middle digit 5 equals 2 + 3. Common pitfallsSetting reverse − original = 0 or 990 by misplacing place values; remember only hundreds and units exchange roles. Final Answer253

Question 4

A number has two digits. If the digits are interchanged and the new number is added to the original, the result is divisible by what?

Accepted Answer

Problem restatementLet the original number be 10a + b. After swapping digits we get 10b + a. Show that their sum is always divisible by a fixed number. Concept/ApproachAlgebraic expression of the sum reveals a common factor. Step-by-Step calculation(10a + b) + (10b + a) = 11(a + b)Thus the sum is a multiple of 11 Common pitfallsGuessing 9 (digit-sum test) or 10 (last digit 0) without algebra; neither is generally true here. Final Answer11

Question 5

A two-digit number has digit product 8. Adding 18 to the number reverses its digits. What is the number?

Accepted Answer

Problem restatementLet the number be 10a + b with ab = 8 and 10a + b + 18 = 10b + a. Find the number. Concept/ApproachUse the reversal equation to relate a and b; combine with the product condition to solve integer digits. Step-by-Step calculation10a + b + 18 = 10b + a ⇒ 9a − 9b = −18 ⇒ a − b = −2 ⇒ b = a + 2ab = 8 ⇒ a(a + 2) = 8Try digit possibilities: a = 2 ⇒ b = 4 works (2 × 4 = 8)Number = 24 Verification/Alternative24 + 18 = 42, which is the reversal; product of digits = 8. Common pitfallsTaking a = 4, b = 2 (gives −2 = a − b = 2, contradiction). Final Answer24

Question 6

The product of two numbers is 9375, and the quotient (larger ÷ smaller) is 15. What is the sum of the two numbers?

Accepted Answer

Problem restatementLet the smaller number be x and the larger be 15x. Given their product equals 9375; find x and hence their sum. Concept/ApproachUse product and ratio to form an equation in one variable. Step-by-Step calculation(15x) × x = 9375 ⇒ 15x2 = 9375x2 = 9375 ÷ 15 = 625 ⇒ x = 25 (positive)Larger = 15x = 375Sum = 25 + 375 = 400 Verification/AlternativeProduct check: 25 × 375 = 9375. Quotient: 375 ÷ 25 = 15. Common pitfallsTaking square root as −25 (numbers are positive in this context) or mixing which is larger. Final Answer400

Question 7

The difference between a two-digit number and the number obtained by interchanging its digits is 36. What is the difference between its two digits?

Accepted Answer

Problem restatementA two-digit number (tens digit a, units digit b) exceeds its reverse by 36. Find |a − b|. Given data Original number = 10a + b Reversed number = 10b + a Difference = (10a + b) − (10b + a) = 36 Concept/ApproachSimplify the difference to get a linear equation in (a − b). Step-by-step calculation (10a + b) − (10b + a) = 9(a − b) = 36 a − b = 36 ÷ 9 = 4 Verification/AlternativeIf a − b = 4, then 9(a − b) = 36 exactly as given. Common pitfalls Using b − a instead of a − b; the absolute difference is 4 in any case. Final Answer4

Question 8

The sum of the digits of a two-digit number is 15 and their difference is 3. What is the two-digit number?

Accepted Answer

Problem restatementLet the tens digit be a and the units digit be b. Given a + b = 15 and a − b = 3, find the number 10a + b. Given data a + b = 15 a − b = 3 Concept/ApproachSolve the simultaneous linear equations. Step-by-step calculation Add the equations: (a + b) + (a − b) = 15 + 3 2a = 18 → a = 9 Then b = 15 − 9 = 6 Number = 10a + b = 10×9 + 6 = 96 Verification/AlternativeDigits 9 and 6 have sum 15 and difference 3; number 96 fits. Common pitfalls Taking 69 instead of 96; the equations make tens digit 9, units 6. Final Answer96

Question 9

If one-third of one-fourth of a number is 15, then three-tenths of that number is:

Accepted Answer

Problem restatementGiven (1/3) of (1/4) of N equals 15. Find (3/10) of N. Given data (1/12)N = 15 Concept/ApproachCompute N from the first relation, then take 3/10 of it. Step-by-step calculation N = 15 × 12 = 180 (3/10)N = (3/10) × 180 = 54 Verification/AlternativeCheck: (1/12) × 180 = 15, consistent. Final Answer54

Question 10

Three times the first of three consecutive odd integers is 3 more than twice the third. What is the third integer?

Accepted Answer

Problem restatementLet the consecutive odd integers be x, x + 2, x + 4. Use the given linear relation to find x and then the third integer. Given data 3x = 2(x + 4) + 3 Concept/ApproachTranslate the verbal condition into an equation and solve. Step-by-step calculation 3x = 2x + 8 + 3 = 2x + 11 x = 11 Third integer = x + 4 = 15 Verification/AlternativeCheck: 3×11 = 33 and 2×15 + 3 = 33, so the condition holds. Final Answer15

Question 11

The product of two numbers is 120 and the sum of their squares is 289. What is the sum of the numbers?

Accepted Answer

Problem restatementLet the numbers be a and b with ab = 120 and a2 + b2 = 289. Find a + b. Concept/ApproachUse the identity (a + b)2 = a2 + b2 + 2ab. Step-by-step calculation (a + b)2 = 289 + 2×120 = 289 + 240 = 529 a + b = √529 = 23 Verification/Alternative529 is 232. The sum could be −23 algebraically, but the context typically takes the positive sum. Final Answer23

Question 12

A two-digit number exceeds the number formed by interchanging its digits by 36. If the ratio between its digits is 1 : 2, what is the difference between the sum and the difference of its digits?

Accepted Answer

Problem restatementDigits (tens = a, units = b) satisfy 9(a − b) = 36 ⇒ a − b = 4. The digits are in the ratio 1 : 2. Find (a + b) − (a − b). Given data a − b = 4 Digits in ratio 1 : 2 ⇒ the two digits are proportional to 1 and 2 (smaller : larger). Concept/ApproachSince a − b > 0, the tens digit a is larger. Thus the pair must be (8, 4) up to the 1:2 ratio (4 : 8). Step-by-step calculation Take digits as 8 and 4 so that a − b = 4 and smaller : larger = 1 : 2 Sum = a + b = 12; Difference = a − b = 4 Required = (a + b) − (a − b) = 12 − 4 = 8 Verification/AlternativeCheck the interchange condition: 9(a − b) = 9×4 = 36 (satisfied). The ratio condition holds with digits {4, 8}. Common pitfalls Interpreting the ratio as a : b = 1 : 2 directly, which would contradict a − b = 4 (since that would make a < b). The ratio refers to values of the digits (smaller : larger). Final Answer8

Question 13

The sum of the squares of three numbers is 138, and the sum of their pairwise products is 131. What is their sum?

Accepted Answer

Problem restatementLet the numbers be x, y, z. Given x2 + y2 + z2 = 138 and xy + yz + zx = 131. Find S = x + y + z. Concept/ApproachUse the identity S2 = (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx). Step-by-step calculation S2 = 138 + 2×131 = 138 + 262 = 400 S = √400 = 20 (taking the principal value) Verification/AlternativeAlgebraically S could be −20 as well, but unless specified otherwise, the conventional answer reported is 20. Common pitfalls Forgetting the factor of 2 on the sum of pairwise products. Final Answer20

Question 14

Find the positive number which, when increased by 17, equals 60 times its reciprocal.

Accepted Answer

Problem restatementFind positive x such that x + 17 = 60 × (1/x). Concept/ApproachForm a quadratic equation and solve using the discriminant. Step-by-step calculation x + 17 = 60/x x2 + 17x − 60 = 0 Discriminant D = 172 + 4×60 = 289 + 240 = 529 = 232 x = (−17 ± 23)/2 ⇒ x = 3 or x = −20 Positive solution ⇒ x = 3 Verification/Alternative3 + 17 = 20 and 60 × (1/3) = 20; equality holds. Final Answer3

Question 15

What is the sum of two consecutive even numbers if the difference of their squares is 84?

Accepted Answer

Problem restatementLet the even integers be x and x + 2. Given (x + 2)2 − x2 = 84, find x + (x + 2). Concept/ApproachUse the identity (a + b)(a − b) = a2 − b2, or expand directly. Step-by-step calculation (x + 2)2 − x2 = 4x + 4 = 84 4(x + 1) = 84 → x + 1 = 21 → x = 20 Sum = x + (x + 2) = 20 + 22 = 42 Verification/AlternativeSquares: 222 − 202 = 484 − 400 = 84, consistent. Final Answer42

Question 16

Three numbers are in the ratio 1 : 2 : 3 and the sum of the cubes of these three numbers is 4500. What is the value of the smallest number?

Accepted Answer

Introduction: This question tests your understanding of ratios combined with powers of numbers, specifically cubes. It is a good example of how ratio-based expressions can be translated into algebraic equations and then solved using basic arithmetic and cube properties. Given Data / Assumptions: Three numbers are in the ratio 1 : 2 : 3. The sum of the cubes of these three numbers is 4500. We are asked to find the smallest of the three numbers. Concept / Approach: If three numbers are in the ratio 1 : 2 : 3, they can be represented as x, 2x, and 3x, where x is a positive real number. The cubes of these numbers are x^3, 8x^3, and 27x^3. Their sum gives an equation in x^3, which can be solved to find x, and hence the smallest number. Step-by-Step Solution: Step 1: Let the three numbers be x, 2x, and 3x. Step 2: Their cubes are x^3, (2x)^3 = 8x^3, and (3x)^3 = 27x^3. Step 3: Sum of cubes is given as 4500. Step 4: Form the equation: x^3 + 8x^3 + 27x^3 = 4500. Step 5: Simplify: (1 + 8 + 27)x^3 = 36x^3 = 4500. Step 6: So x^3 = 4500 / 36 = 125. Step 7: Take cube root: x = 5. Step 8: Therefore, the smallest number is x = 5. Verification / Alternative check: Compute the actual three numbers: 5, 10, and 15. Now check the cubes: 5^3 = 125, 10^3 = 1000, 15^3 = 3375. Sum of cubes = 125 + 1000 + 3375 = 4500, which matches the given condition exactly. Why Other Options Are Wrong: 6: If x = 6, the cubes would sum to 36 * 6^3 = 7776, not 4500. 10: This would be the second number, not the smallest. 4: With x = 4, the sum of cubes is 36 * 64 = 2304, which is too small. 8: With x = 8, the sum of cubes is 36 * 512 = 18432, far larger than 4500. Common Pitfalls: Some learners mistakenly cube the ratio numbers first and then try to match them directly to 4500 without introducing a common factor x. Others may miscalculate cube values or forget to divide correctly when solving for x^3. Careful algebra and arithmetic avoid these issues. Final Answer: The smallest of the three numbers is 5.

Question 17

What is the largest four-digit number that is exactly divisible by 88?

Accepted Answer

Introduction:This is a standard divisibility and multiples problem. You are asked to find the largest four-digit integer that is exactly divisible by 88. Such questions appear frequently in aptitude tests to check your comfort with division and multiples. Given Data / Assumptions: We consider only four-digit numbers, that is numbers from 1000 to 9999. We want the largest four-digit number N such that N is divisible by 88. Divisible by 88 means N = 88 × k for some integer k. Concept / Approach:To get the largest four-digit multiple of 88, we divide the largest four-digit number 9999 by 88 and take the integer part of the quotient. Then we multiply 88 by this quotient to obtain the required multiple. Step-by-Step Solution:Step 1: Start from the upper limit 9999.Step 2: Divide 9999 by 88 to get the approximate quotient.88 * 100 = 8800 (still four digits).88 * 110 = 9680.88 * 113 = 88 * (100 + 13) = 8800 + 1144 = 9944.88 * 114 = 88 * (100 + 14) = 8800 + 1232 = 10032 (which is five digits).Step 3: Conclude.The last product still under 10000 is 88 * 113 = 9944. Verification / Alternative check:Check that 9944 is divisible by 88 using quick division: 9944 ÷ 88 = 113 exactly, with no remainder. Also, the next multiple 9944 + 88 = 10032, which exceeds 9999, so 9944 is indeed the largest four-digit multiple. Why Other Options Are Wrong:8888 is divisible by 88 but is smaller than 9944. 9999 and 9988 are not exact multiples of 88. 9680 is also a multiple of 88 but smaller than 9944. So only 9944 satisfies both conditions: four-digit and largest possible. Common Pitfalls:Some learners try to test each option with long division, which is time-consuming. Others mistakenly choose 9999 or 9988 just because they look large. Always use multiplication or division systematically to identify the correct multiple. Final Answer:The largest four-digit number exactly divisible by 88 is 9944.

Question 18

In the two digit number 64, what is the place value of the digit 6?

Accepted Answer

Introduction / Context:This is a very basic problem from the topic of number systems. It tests understanding of the concept of place value in a two digit number. Even though the question looks simple, place value is a fundamental concept that underlies many more advanced numerical operations and decimal representations. Given Data / Assumptions:- The number under consideration is 64.- It is a two digit number with 6 in the tens place and 4 in the units place.- We are asked specifically for the place value of the digit 6 in this number. Concept / Approach:In a two digit number, the rightmost digit is in the units place and the left digit is in the tens place. The place value of a digit is the digit multiplied by the value of its position. In the tens place, the position value is 10, so the place value of the digit is that digit multiplied by 10. In the units place, the position value is 1, so the place value is the digit itself. Step-by-Step Solution:Step 1: Identify the digits and their positions in the number 64. The digit 6 is in the tens place and the digit 4 is in the units place.Step 2: The value of the tens place is 10. Therefore, the place value of a digit in the tens place is digit * 10.Step 3: For the digit 6 in the tens place, the place value is 6 * 10.Step 4: Compute the product: 6 * 10 = 60.Step 5: Thus, the place value of the digit 6 in the number 64 is 60. Verification / Alternative check:We can express 64 in expanded form based on place values. The number 64 can be written as 60 + 4, which is 6 * 10 + 4 * 1. Here, 6 * 10 represents the contribution of the digit 6 in the tens place, which is exactly 60. This confirms that the place value of 6 in 64 is 60. Why Other Options Are Wrong:The option 6 represents only the face value of the digit, not the place value. The option 64 represents the whole number, not the contribution from the tens place. The option 10 is just the place value of the position, not of the digit itself, and 600 is too large and would correspond to 6 in the hundreds place. Therefore, only 60 correctly represents the place value of 6 in 64. Common Pitfalls:Some learners confuse face value with place value and answer 6 instead of 60. Others may misinterpret the question and think it asks for the overall value of the number. Remember that face value is the digit itself, while place value is the digit multiplied by the value of its position in the number. Clear understanding of this distinction avoids such errors. Final Answer:The place value of the digit 6 in 64 is 60.

Question 19

In a two digit number, the digit in the units place is greater than twice the digit in the tens place by 1. If the digits in the units and tens places are interchanged, the difference between the new number and the original number is exactly 1 less …

Accepted Answer

Introduction / Context:This is a two digit number puzzle involving conditions on the digits and the difference between the original and reversed numbers. It uses basic algebra to represent the number and its digits, and then applies the given relationships to find the original number. Such problems are popular in aptitude tests to check number sense and equation forming skills. Given Data / Assumptions:- The number is a two digit number with tens digit x and units digit y.- The digit in the units place is greater than twice the tens digit by 1, so y is 2x plus 1.- When the digits are interchanged, a new two digit number is formed.- The difference between the new number and the original number is exactly 1 less than the original number.- We must find the original number. Concept / Approach:Let the tens digit be x and the units digit be y. This makes the original number 10x + y. The units digit condition allows us to express y in terms of x. When the digits are interchanged, the new number becomes 10y + x. We are told that the difference between the new number and the original number equals the original number minus 1. This information gives an equation that can be solved for x, and then y follows from the first condition. Step-by-Step Solution:Step 1: Let the tens digit be x and the units digit be y. Then the original number is 10x + y.Step 2: The digit in the units place is greater than twice the tens digit by 1, so y = 2x + 1.Step 3: When the digits are interchanged, the new number becomes 10y + x.Step 4: The problem states that the difference between the new number and the original number is 1 less than the original number, so (10y + x) - (10x + y) = (10x + y) - 1.Step 5: Simplify the left side: 10y + x - 10x - y = 9y - 9x.Step 6: The right side is 10x + y - 1. So we get 9y - 9x = 10x + y - 1.Step 7: Rearrange to 9y - 9x - 10x - y + 1 = 0, which simplifies to 8y - 19x + 1 = 0.Step 8: Substitute y = 2x + 1 into 8y - 19x + 1 = 0 to get 8(2x + 1) - 19x + 1 = 0.Step 9: Expand: 16x + 8 - 19x + 1 = 0, which gives -3x + 9 = 0, so 3x = 9 and x = 3.Step 10: Then y = 2x + 1 = 2 * 3 + 1 = 7. The original number is 10 * 3 + 7 = 37. Verification / Alternative check:With the original number 37, the digits are 3 and 7. Twice the tens digit is 6, and the units digit 7 is greater than 6 by 1, which satisfies the first condition. Reversing the digits gives the new number 73. The difference between the new and original numbers is 73 - 37 = 36. The original number minus 1 is 37 - 1 = 36. Thus, the second condition is also satisfied. This confirms that 37 is the correct original number. Why Other Options Are Wrong:Numbers such as 35, 36, 39, or 47 do not satisfy the conditions. For example, in 39, twice the tens digit is 6, but the units digit 9 is greater than 6 by 3, not 1. In 47, twice the tens digit is 8, but the units digit 7 is less than 8. Checking the reversed number differences also fails to match the requirement that the difference equals the original number minus 1. Common Pitfalls:Some learners may misinterpret the phrase "more than twice" and write y = 2x - 1 instead of y = 2x + 1. Others confuse which number is subtracted from which when forming the difference. It is important to read the problem carefully and translate each phrase correctly into algebraic equations before solving. Final Answer:The original two digit number is 37.

Question 20

Three times the first of three consecutive odd integers is 3 more than twice the third integer. What is the value of the third integer?

Accepted Answer

Introduction / Context:This is an algebraic puzzle involving three consecutive odd integers. The relationship between the first and third integers is given in terms of a linear equation. Our job is to translate the wording into algebra, solve for the first odd integer, and then find the third. This type of problem is common in aptitude exams and helps develop equation forming skills. Given Data / Assumptions:- There are three consecutive odd integers.- Let the first integer be some odd number n.- The second integer is then n + 2 and the third is n + 4.- Three times the first integer is 3 more than twice the third integer.- We need to find the third integer. Concept / Approach:Consecutive odd integers can be represented as n, n + 2, and n + 4. The condition "three times the first is 3 more than twice the third" becomes a linear equation connecting n and n + 4. Solving this equation gives n, and then the third integer is easily calculated as n + 4. Step-by-Step Solution:Step 1: Let the first odd integer be n.Step 2: Then the second odd integer is n + 2 and the third odd integer is n + 4.Step 3: The condition in the problem states that three times the first integer equals three more than twice the third integer.Step 4: Translate this into an equation: 3n = 2(n + 4) + 3.Step 5: Expand the right hand side: 2(n + 4) + 3 = 2n + 8 + 3 = 2n + 11.Step 6: The equation is now 3n = 2n + 11.Step 7: Subtract 2n from both sides to get n = 11.Step 8: Therefore, the first odd integer is 11 and the third odd integer is n + 4 = 11 + 4 = 15. Verification / Alternative check:By our solution, the three consecutive odd integers are 11, 13, and 15. Three times the first integer is 3 * 11 = 33. Twice the third integer is 2 * 15 = 30. The statement says that three times the first is 3 more than twice the third, which would require 33 = 30 + 3. This is true, so the integers satisfy the condition given in the problem. Therefore, 15 is confirmed as the correct third integer. Why Other Options Are Wrong:Options such as 12, 13, 14, or 17 are either not odd or do not arise as the third member of a set of three consecutive odd integers that satisfy the equation. For instance, if the third integer were 13, then the first would be 9, and three times 9 is 27 while twice 13 plus 3 is 29, which does not match. Similar checks show that these options fail the given condition. Common Pitfalls:One common error is to misrepresent consecutive odd integers as n, n + 1, n + 2 instead of n, n + 2, n + 4. Another is to misinterpret the phrase "3 more than twice the third" and write 3n = 3(2(n + 4)) or a similar incorrect expression. Carefully translating the language into an accurate equation is essential for solving such problems correctly. Final Answer:The third odd integer is 15.

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