Question 1

Among the following numbers, which one cannot be the perfect square of a natural number?

Accepted Answer

Introduction / Context:This problem asks us to identify which of the given numbers cannot be expressed as the square of a natural number. Instead of taking square roots on a calculator, we can use estimation and properties of perfect squares to determine which options are valid squares and which one is not. This tests number sense, familiarity with square numbers and basic approximation skills. Given Data / Assumptions: - The candidate numbers are 32761, 42437, 81225 and 20164. - All numbers are positive and reasonably close to the squares of integers in the range 100 to 300. - Only one of these cannot be written as n^2 for any natural number n. Concept / Approach:We can approximate the square root of each number by comparing it to nearby perfect squares. For example, the square of 180 is 32400, the square of 190 is 36100, and so on. Once we identify a likely integer candidate r such that r^2 is close to the given number, we check r^2 and (r + 1)^2 exactly. If the number matches one of these, it is a perfect square; otherwise it is not. We repeat this process for each option until we find the one that fails to match any integer square. Step-by-Step Solution: Step 1: For 32761, note that 180^2 = 32400 and 190^2 = 36100. The number is closer to 180^2, so test 181^2. Step 2: Compute 181^2 = 181 * 181 = 32761, which exactly matches. Hence 32761 is a perfect square. Step 3: For 81225, note that 280^2 = 78400 and 290^2 = 84100. Try 285^2. Step 4: Compute 285^2 = 285 * 285 = 81225, so 81225 is a perfect square. Step 5: For 20164, notice that 140^2 = 19600 and 150^2 = 22500. Try 142^2. Step 6: Compute 142^2 = 142 * 142 = 20164, so 20164 is also a perfect square. Step 7: For 42437, find nearby squares. Compute 206^2 = 42436 and 207^2 = 42849. Step 8: Neither 206^2 nor 207^2 equals 42437. There is no integer whose square is 42437, so this number is not a perfect square. Verification / Alternative check:Another way is to observe that perfect squares modulo small bases often follow specific patterns. For example, in base 4 or 8, squares have restricted remainders. However, the direct calculation approach with nearby integer squares is usually faster for numbers in this range. Having checked that three of the numbers match exact squares of 181, 285 and 142, and that 42437 lies strictly between 206^2 and 207^2, we are confident that only 42437 fails to be a square. Why Other Options Are Wrong:Option A (32761) equals 181^2, option C (81225) equals 285^2 and option D (20164) equals 142^2. Since each of these is an exact perfect square, they cannot be the answer to the question about which number is not a square. Common Pitfalls:Some students rely solely on the last digit or rough approximations and may guess without checking both r^2 and (r + 1)^2. Others may mistakenly believe that because 42437 is close to 42436, it must also be a square. Careful computation of nearby squares is essential to avoid such errors. Final Answer:The number that cannot be the square of a natural number is 42437, which corresponds to option B.

Question 2

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

Accepted Answer

Introduction / Context:This question is a standard divisibility and number range problem. We are asked to find the largest four digit number that is exactly divisible by 88. Four digit numbers range from 1000 to 9999, and the task is to locate the greatest multiple of 88 within this interval. This tests understanding of division with remainders and the relationship between multiples and ranges. Given Data / Assumptions: - Four digit numbers are from 1000 to 9999 inclusive. - We want the largest integer in this range that is a multiple of 88. - The answer options include several candidates near the upper end of the four digit range. Concept / Approach:To find the largest multiple of 88 less than or equal to 9999, we perform integer division of 9999 by 88. The quotient tells us how many full groups of 88 fit into 9999, and multiplying this quotient by 88 gives the largest multiple of 88 that does not exceed 9999. We do not need to check all the options by direct divisibility tests if we compute this value once accurately. Step-by-Step Solution: Step 1: Consider the upper limit of four digit numbers, which is 9999. Step 2: Compute the integer quotient when 9999 is divided by 88. Step 3: Perform division: 9999 ÷ 88 = 113 with some remainder, because 88 * 113 = 9944. Step 4: Compute 88 * 113 to confirm: 88 * 113 = 88 * 100 + 88 * 13 = 8800 + 1144 = 9944. Step 5: Since 9944 is less than 9999 and is exactly divisible by 88, it is a valid candidate. Step 6: The next multiple would be 88 * 114, which equals 10032, a five digit number and therefore outside the allowed four digit range. Step 7: Thus, 9944 is the largest possible four digit multiple of 88. Verification / Alternative check:We can quickly check the options by dividing them by 88. For 9944, 9944 ÷ 88 = 113 exactly. For 9768, dividing by 88 gives 111 with remainder, and for 8888, 8888 ÷ 88 is not an integer either. Option 9988 is larger than 9944, but 9988 ÷ 88 does not give a whole number. Therefore, no option larger than 9944 is a valid multiple of 88 within the four digit range, confirming our result. Why Other Options Are Wrong:Options B, C and D are either not multiples of 88 or fall outside the required definition of the largest four digit multiple. Specifically, they leave non zero remainders when divided by 88 or exceed the maximum allowed number when extended beyond 9999. Common Pitfalls:Some test takers may be tempted to pick 9988 simply because it is close to 9999, without checking divisibility. Others may try to work upward from a smaller multiple instead of using integer division from the upper bound. Systematically using quotient and product avoids guesswork and ensures a correct answer. Final Answer:The largest four digit number exactly divisible by 88 is 9944, corresponding to option A.

Question 3

How many integers between 300 and 700 (inclusive) are divisible by 2, 3 and 7 simultaneously?

Accepted Answer

Introduction / Context:This question asks for a count of numbers in a given range that are divisible by 2, 3 and 7 at the same time. When a number is required to be divisible by several integers simultaneously, we use the concept of the least common multiple. This problem tests understanding of least common multiples, working with ranges and counting multiples correctly. Given Data / Assumptions: - The range of interest is from 300 to 700 inclusive. - Numbers must be divisible by 2, by 3 and by 7 at the same time. - All three divisibility constraints must hold for the same number. Concept / Approach:A number divisible by 2, 3 and 7 simultaneously must be divisible by their least common multiple. Since 2, 3 and 7 are pairwise coprime, their least common multiple is the product 2 * 3 * 7 = 42. Therefore, the question reduces to counting the multiples of 42 between 300 and 700 inclusive. Once we identify the first multiple of 42 at or above 300 and the last multiple at or below 700, we can use a simple counting formula. Step-by-Step Solution: Step 1: Compute the least common multiple of 2, 3 and 7. Step 2: Since 2, 3 and 7 are coprime, lcm(2, 3, 7) = 2 * 3 * 7 = 42. Step 3: Find the smallest multiple of 42 that is greater than or equal to 300. Compute 300 ÷ 42. Step 4: 42 * 7 = 294 (less than 300), and 42 * 8 = 336 (at least 300). So the first relevant multiple is 336. Step 5: Find the largest multiple of 42 that is less than or equal to 700. Compute 700 ÷ 42. Step 6: 42 * 16 = 672 and 42 * 17 = 714 (greater than 700). So the last relevant multiple is 672. Step 7: The multiples of 42 from 336 to 672 are 336, 378, 420, 462, 504, 546, 588, 630 and 672. Step 8: Count these terms. There are 9 multiples in total. Verification / Alternative check:We can verify the count by using a formula. The general k-th multiple of 42 is 42k. We found that k runs from 8 to 16. The number of integers from 8 to 16 inclusive is 16 − 8 + 1 = 9. This matches our manual list and confirms that there are exactly nine numbers in the given range divisible by 2, 3 and 7 simultaneously. Why Other Options Are Wrong:Options 7, 11 and 5 do not match the actual count of multiples. If we mistakenly started at 294 or ended at 714, we would miscount the valid numbers. Careful identification of the first and last multiples within the range is essential to avoid such errors. Common Pitfalls:A frequent mistake is to count numbers divisible by 2, by 3 and by 7 separately and then try to combine counts, which leads to overcounting or undercounting. Another mistake is to treat 2, 3 and 7 independently rather than using the least common multiple. Using lcm simplifies the problem greatly and provides a straightforward counting method. Final Answer:The number of integers between 300 and 700 that are divisible by 2, 3 and 7 simultaneously is 9, which corresponds to option B.

Question 4

The sum of two numbers is 12 and their product is 35. What is the sum of the reciprocals of these two numbers?

Accepted Answer

Introduction / Context:This problem connects basic algebra with fractions. We are given the sum and product of two numbers and asked to find the sum of their reciprocals. Instead of solving for the individual numbers directly, we can use a useful identity for reciprocals in terms of sum and product. This approach is efficient and avoids unnecessary quadratic equation solving, which is especially helpful in timed aptitude tests. Given Data / Assumptions: - Let the two numbers be a and b. - Sum of the numbers is a + b = 12. - Product of the numbers is a * b = 35. - We need to find 1/a + 1/b. Concept / Approach:There is a direct algebraic relationship between the reciprocals of numbers and their sum and product. Specifically, 1/a + 1/b can be rewritten as (a + b) / (a * b). Because we already know both a + b and a * b, we can substitute these values directly into the identity. This eliminates the need to find the actual values of a and b. Step-by-Step Solution: Step 1: Start with the expression for the sum of reciprocals: 1/a + 1/b. Step 2: Combine the two fractions over a common denominator: 1/a + 1/b = (b + a) / (a * b). Step 3: Recognize that b + a is equal to a + b, which is given as 12. Step 4: Recognize that a * b, the product of the numbers, is given as 35. Step 5: Substitute into the identity: 1/a + 1/b = (a + b) / (a * b) = 12 / 35. Step 6: Simplify the fraction if possible. In this case, 12 and 35 have no common factors other than 1, so 12/35 is already in simplest form. Verification / Alternative check:As a check, we can find the actual numbers. They satisfy the quadratic equation t^2 − 12t + 35 = 0. Solving, t = 5 or t = 7. Then 1/5 + 1/7 = (7 + 5) / 35 = 12/35, which matches our algebraic result. This confirms that the identity and substitution were applied correctly. Why Other Options Are Wrong:Option B (1/35) would correspond to 1/(a * b) rather than (a + b)/(a * b). Option C (35/8) is larger than 4 and clearly inconsistent with reciprocals of positive numbers whose sum is 12. Option D (7/32) does not arise from any natural combination of the given sum and product and fails the direct identity check. Common Pitfalls:Some learners attempt to find a and b by solving the quadratic and may make arithmetic mistakes. Others forget the identity for reciprocals and try to guess. Remembering that 1/a + 1/b = (a + b)/(a * b) saves time and reduces the chance of error, especially when the sum and product are provided explicitly. Final Answer:The sum of the reciprocals of the two numbers is 12/35, which corresponds to option A.

Question 5

What is the smallest whole number that should be added to 4456 so that the resulting sum is completely divisible by 6?

Accepted Answer

Introduction / Context:This problem involves divisibility by 6 and requires us to adjust a given number slightly so that it satisfies the divisibility rules. A number is divisible by 6 if and only if it is divisible by both 2 and 3. The aim is to find the smallest non negative integer that, when added to 4456, makes the new total divisible by 6. This tests knowledge of basic divisibility rules and modular arithmetic. Given Data / Assumptions: - Original number is 4456. - We must add a smallest whole number k (k ≥ 0) such that 4456 + k is divisible by 6. - A number is divisible by 6 if it is divisible by both 2 and 3. Concept / Approach:First check whether 4456 is divisible by 2 and by 3. Divisibility by 2 depends on the last digit being even, while divisibility by 3 depends on the sum of digits being a multiple of 3. If the number already meets one of the conditions, we only need to adjust it to satisfy the other while keeping the first condition valid. We search for the smallest k such that the new number is divisible by both 2 and 3. Step-by-Step Solution: Step 1: Check divisibility of 4456 by 2. The last digit is 6, which is even, so 4456 is divisible by 2. Step 2: Check divisibility of 4456 by 3. Sum of digits is 4 + 4 + 5 + 6 = 19. Step 3: Since 19 is not a multiple of 3, 4456 is not divisible by 3. Step 4: Let k be the required smallest non negative integer such that 4456 + k is divisible by 6. Step 5: For divisibility by 3, we need (sum of digits of 4456) + (sum of digits of k) to be a multiple of 3. In modular arithmetic, 19 + k ≡ 0 (mod 3). Step 6: Since 19 ≡ 1 (mod 3), we need k ≡ 2 (mod 3). Step 7: For divisibility by 2, 4456 + k must be even. Because 4456 is even, k must also be even so that the sum remains even. Step 8: The smallest non negative integer that is both even and congruent to 2 modulo 3 is k = 2. Step 9: Check: 4456 + 2 = 4458. Sum of digits is 4 + 4 + 5 + 8 = 21, which is divisible by 3, and the last digit 8 is even. So 4458 is divisible by 6. Verification / Alternative check:We can also test options directly. Adding 1 gives 4457, which is odd and not divisible by 2. Adding 2 gives 4458, which we saw is divisible by both 2 and 3. Adding 3 gives 4459, which is odd. Adding 4 gives 4460, which is even, but the sum of digits 4 + 4 + 6 + 0 = 14 is not a multiple of 3. Therefore 2 is indeed the smallest suitable number. Why Other Options Are Wrong:Option D (1) and option B (3) produce odd results, failing the divisibility by 2 condition. Option A (4) produces a number that is even but not divisible by 3. Therefore they do not give a sum divisible by 6. Common Pitfalls:Some learners may only check divisibility by 3, forgetting that divisibility by 6 requires both 2 and 3. Others may choose the smallest k that fixes one condition but breaks the other. Always verify both conditions for divisibility by 6 to avoid such mistakes. Final Answer:The smallest number to be added is 2, which corresponds to option C.

Question 6

How many of the following numbers are divisible by 132: 264, 396, 462, 792, 968, 2178, 5184, 6336?

Accepted Answer

Introduction / Context:This question asks us to check several given numbers for divisibility by 132 and count how many satisfy the condition. Rather than performing full long division for each, we can use factorization of 132 and divisibility rules for its prime factors. This tests knowledge of prime factorization and how to combine basic divisibility tests in an efficient way. Given Data / Assumptions: - Numbers given are 264, 396, 462, 792, 968, 2178, 5184 and 6336. - We must determine how many of these are exactly divisible by 132. - 132 is a composite number with small prime factors. Concept / Approach:First factorize 132 to understand its divisibility structure. We have 132 = 2^2 * 3 * 11. Therefore, a number is divisible by 132 if and only if it is divisible by 4, by 3 and by 11 at the same time. We can either check these three divisibility rules for each number or directly divide by 132 if comfortable with mental arithmetic. After testing each number, we count those with zero remainder. Step-by-Step Solution: Step 1: Factor 132: 132 = 2 * 66 = 2 * 2 * 33 = 4 * 33 = 4 * 3 * 11 = 2^2 * 3 * 11. Step 2: Check 264. Compute 264 ÷ 132 = 2, so 264 is divisible by 132. Step 3: Check 396. Compute 132 * 3 = 396, so 396 is divisible by 132. Step 4: Check 462. 132 * 3 = 396 and 132 * 4 = 528, so 462 lies between and is not a multiple of 132. Step 5: Check 792. 132 * 6 = 792, so 792 is divisible by 132. Step 6: Check 968. 132 * 7 = 924 and 132 * 8 = 1056, so 968 is not a multiple of 132. Step 7: Check 2178. 132 * 16 = 2112 and 132 * 17 = 2244, so 2178 is not a multiple of 132. Step 8: Check 5184. 132 * 39 = 5148 and 132 * 40 = 5280, so 5184 is not a multiple of 132. Step 9: Check 6336. 132 * 48 = 6336, so 6336 is divisible by 132. Step 10: The numbers divisible by 132 are 264, 396, 792 and 6336, a total of four numbers. Verification / Alternative check:As an alternative, we can check that each of these four numbers is divisible by 4, 3 and 11. For example, 264 is even and its last two digits 64 form a number divisible by 4, its digit sum 2 + 6 + 4 = 12 is divisible by 3, and the alternating sum for 11 is (2 + 4) − 6 = 0, a multiple of 11. Similar checks confirm that 396, 792 and 6336 satisfy all three divisibility rules, strengthening our conclusion. Why Other Options Are Wrong:Options B, C and D assume that more than four numbers are divisible by 132. However, we have shown that 462, 968, 2178 and 5184 fail the multiple test for 132, so they cannot be counted. Any answer larger than four overestimates the count. Common Pitfalls:Some learners may wrongly assume that being divisible by 4 and 3 is enough and forget the factor 11. Others may miscalculate products of 132 or rely on approximate multiplication that leads to off by one errors. Using clear multiplication steps or systematic divisibility rules helps avoid such mistakes. Final Answer:The number of given values that are divisible by 132 is 4, which matches option A.

Question 7

If the number 517?324 is completely divisible by 3, then what is the smallest whole digit that can replace the question mark?

Accepted Answer

Introduction / Context:This question asks us to find a missing digit in a large number so that the whole number becomes divisible by 3. The key tool is the divisibility rule for 3, which states that a number is divisible by 3 if the sum of its digits is a multiple of 3. This is a very common type of aptitude question. Given Data / Assumptions: - The number is 517?324. - The question mark represents a single digit from 0 to 9. - The full number must be divisible by 3. - We need the smallest such digit. Concept / Approach:According to the divisibility rule for 3, we only need to look at the sum of the digits of the number. If that sum is divisible by 3, then the number is divisible by 3. Therefore, we add up the known digits, represent the unknown digit by a variable, and then solve for the smallest digit that makes the total a multiple of 3. Step-by-Step Solution: Step 1: Let the unknown digit be x. The number is then 517x324. Step 2: Add the known digits: 5 + 1 + 7 + 3 + 2 + 4. Step 3: Compute this sum: 5 + 1 = 6, 6 + 7 = 13, 13 + 3 = 16, 16 + 2 = 18, 18 + 4 = 22. Step 4: The total sum of digits with x included is 22 + x. Step 5: For divisibility by 3, we require 22 + x to be a multiple of 3. Step 6: Since 22 leaves remainder 1 when divided by 3, we need x to leave remainder 2 when divided by 3, that is x ≡ 2 (mod 3). Step 7: The digits from 0 to 9 that satisfy x ≡ 2 (mod 3) are x = 2, 5 and 8. Step 8: The smallest such whole digit is 2. Verification / Alternative check:We can verify by substituting x = 2 into the number. The digits of 5172324 sum to 5 + 1 + 7 + 2 + 3 + 2 + 4 = 24, which is clearly divisible by 3. Therefore the number with x = 2 is divisible by 3. If we tried x = 1, the sum would be 23, not divisible by 3, so 1 would fail. This confirms that 2 is the smallest valid digit. Why Other Options Are Wrong:Option B (1) gives a sum of digits 23, option C (3) gives sum 25 and option D (7) gives sum 29. None of these totals is divisible by 3, so the corresponding numbers are not divisible by 3. Therefore they cannot be correct replacements for the missing digit. Common Pitfalls:Some students attempt to test each option by dividing the entire number by 3, which is time consuming and prone to error. Others may miscalculate the sum of digits or forget to find the smallest digit that works. Using modular reasoning and checking sums systematically is the quickest and safest method. Final Answer:The smallest whole digit that can replace the question mark is 2, which corresponds to option A.

Question 8

A piece of ribbon 4 yards long is used to make bows, each bow requiring 15 inches of ribbon. What is the maximum number of bows that can be made from this ribbon?

Accepted Answer

Introduction / Context:This problem is a straightforward unit conversion and division question. A ribbon of a certain length is cut into equal pieces to make bows, and we must determine how many pieces of the required length can be obtained. The key idea is to convert all lengths into the same unit and then perform integer division to find how many complete pieces fit into the total length. Given Data / Assumptions: - Total length of ribbon is 4 yards. - Each bow requires 15 inches of ribbon. - We can only count complete bows, not partial ones. - Standard conversion is 1 yard = 36 inches. Concept / Approach:Since the bow length is given in inches, it is convenient to convert the total ribbon length from yards to inches. Once we know the total length in inches, we divide by 15 to find how many bows of 15 inches can be formed. The quotient tells us the number of complete bows, and any remainder represents leftover ribbon that is too short for another bow. Step-by-Step Solution: Step 1: Convert 4 yards to inches using the conversion 1 yard = 36 inches. Step 2: Compute 4 * 36 = 144 inches. So the ribbon is 144 inches long. Step 3: Each bow requires 15 inches of ribbon. Step 4: Divide the total length by the length per bow: 144 ÷ 15. Step 5: Compute the quotient: 15 * 9 = 135 and 15 * 10 = 150 which exceeds 144. So the maximum number of full bows is 9. Step 6: The remainder is 144 − 135 = 9 inches, which is not enough for one more 15 inch bow. Verification / Alternative check:We can also reason in reverse. If we make 9 bows, they use 9 * 15 = 135 inches of ribbon. This is less than or equal to the total 144 inches. Attempting 10 bows would require 10 * 15 = 150 inches, which is more than 144 and therefore impossible. This confirms that 9 is the maximum number of bows that can be made. Why Other Options Are Wrong:Option A (8) is too small because we can clearly fit 9 bows into 144 inches. Options C (10) and D (11) require more ribbon than is available, so those counts are not feasible. Only 9 matches both the capacity and the maximal requirement. Common Pitfalls:Some learners forget to convert yards to inches correctly and may assume a wrong conversion factor. Others may mistakenly round up after division, which would correspond to making a partial bow. In such problems we must always take the integer part of the division since only complete bows count. Final Answer:The maximum number of bows that can be made is 9, which matches option B.

Question 9

Find the units digit of the sum 4198 + 612345 + 34866 + 2411 + 1.

Accepted Answer

Introduction / Context:This problem asks for the units digit of a sum of several numbers. Instead of adding the full numbers, we can use the idea that only the units digits affect the units digit of the final sum. This is a common time saving trick in aptitude tests when dealing with large numbers or long sums. Given Data / Assumptions: - The numbers are 4198, 612345, 34866, 2411 and 1. - We want only the units digit of the total sum. - Basic properties of addition and place value apply. Concept / Approach:The units digit of a sum depends only on the units digits of the addends. Higher place digits contribute to carries but do not change the final units digit beyond that carry. Therefore, we can ignore all except the units digits, add those, and then extract the units digit of that result. This is much faster than performing full long addition. Step-by-Step Solution: Step 1: Identify the units digit of each number. Step 2: For 4198, the units digit is 8. Step 3: For 612345, the units digit is 5. Step 4: For 34866, the units digit is 6. Step 5: For 2411, the units digit is 1. Step 6: For 1, the units digit is 1. Step 7: Add these units digits: 8 + 5 + 6 + 1 + 1. Step 8: Compute the sum: 8 + 5 = 13, 13 + 6 = 19, 19 + 1 = 20, 20 + 1 = 21. Step 9: The units digit of 21 is 1. Verification / Alternative check:If desired, we can add the full numbers or check with partial sums. Adding 4198 and 612345 gives 616543, then adding 34866 gives 651409, adding 2411 gives 653820 and finally adding 1 gives 653821. The units digit of 653821 is 1, which confirms our shortcut method. However, using only units digits is clearly much faster and less error prone. Why Other Options Are Wrong:Options A (3), B (2) and D (0) correspond to incorrect totals of the units digits or to arithmetic mistakes in handling carries. Since the exact sum ends in 1, none of these options can be correct. Common Pitfalls:Errors often occur from misreading one of the numbers or misadding the units digits. Another trap is to think that tens digits or higher are needed, which leads to unnecessary work and increases the chance of mistakes. Focusing only on the units digits and working carefully with the small sum is the best strategy. Final Answer:The units digit of the sum 4198 + 612345 + 34866 + 2411 + 1 is 1, which corresponds to option C.

Question 10

n is a whole number which, when divided by 4, leaves a remainder of 3. What will be the remainder when 2n is divided by 4?

Accepted Answer

Introduction / Context:This problem uses basic modular arithmetic. We are told how a whole number n behaves when divided by 4 and then asked to predict the remainder when 2n is divided by 4. Instead of thinking of specific values for n, we can work with congruences and properties of remainders under multiplication. Given Data / Assumptions: - n is a whole number. - When n is divided by 4, the remainder is 3, that is n ≡ 3 (mod 4). - We must find the remainder when 2n is divided by 4. Concept / Approach:In modular arithmetic, if n leaves a remainder when divided by some modulus, then any multiple of n will also have a predictable remainder. Here, n ≡ 3 (mod 4), and we multiply both sides of this congruence by 2 to model 2n. This gives 2n ≡ 2 * 3 (mod 4). We then reduce 6 modulo 4 to find the required remainder. Step-by-Step Solution: Step 1: Translate the statement into a congruence: n ≡ 3 (mod 4). Step 2: Multiply both sides by 2 to represent 2n: 2n ≡ 2 * 3 (mod 4). Step 3: Compute 2 * 3 = 6. Step 4: Reduce 6 modulo 4. When 6 is divided by 4, the quotient is 1 and the remainder is 2. Step 5: Therefore, 6 ≡ 2 (mod 4), so 2n ≡ 2 (mod 4). Step 6: This means that when 2n is divided by 4, the remainder is 2. Verification / Alternative check:We can pick specific numbers for n that satisfy the given condition and test. For example, let n = 3. Then n divided by 4 leaves remainder 3. Compute 2n = 6, which divided by 4 leaves remainder 2. Another example is n = 7, since 7 divided by 4 leaves remainder 3. Then 2n = 14, and 14 divided by 4 leaves remainder 2 again. These examples confirm the general modular reasoning. Why Other Options Are Wrong:Option A (3) would suggest 2n ≡ 3 (mod 4), which contradicts the calculation 2 * 3 = 6 ≡ 2 (mod 4). Option C (1) and option D (0) do not match any consistent behavior of 2n for values of n congruent to 3 modulo 4. Checking sample values quickly reveals that 2n never leaves remainder 1 or 0 in this situation. Common Pitfalls:Some learners mistakenly double both the remainder and the modulus, which is incorrect. Others try to guess 2n by choosing a single example of n and may miscalculate the remainder. The correct approach is to use congruence rules systematically and reduce the product modulo 4. Final Answer:The remainder when 2n is divided by 4 is 2, which corresponds to option B.

Question 11

At a dinner party, every two guests share one bowl of rice between them, every three guests share one bowl of dal between them, and every four guests share one bowl of meat between them. Altogether, there are 65 dishes (bowls). How many guests were …

Accepted Answer

Introduction / Context:This problem models a real life sharing situation using simple algebra. At a dinner party, bowls of rice, dal and meat are shared among guests in different group sizes. Using the information about how many guests share each type of dish and the total number of dishes, we have to determine how many guests attended the party. The question tests understanding of ratios, fractions and how to translate words into equations. Given Data / Assumptions: - Every two guests share one bowl of rice. - Every three guests share one bowl of dal. - Every four guests share one bowl of meat. - The total number of bowls (rice + dal + meat) is 65. - All sharing is exact, meaning no half bowls or unused portions are counted separately. Concept / Approach:Let the number of guests be n. Then, since every two guests share one bowl of rice, the number of rice bowls is n/2. Similarly, the number of dal bowls is n/3, and the number of meat bowls is n/4. The total number of bowls is the sum of these three expressions and is given as 65. We set up the equation n/2 + n/3 + n/4 = 65 and solve for n. Because n represents a number of guests, it must be a positive integer. Step-by-Step Solution: Step 1: Let n be the number of guests. Step 2: Number of bowls of rice is n/2 because each bowl serves two guests. Step 3: Number of bowls of dal is n/3 because each bowl serves three guests. Step 4: Number of bowls of meat is n/4 because each bowl serves four guests. Step 5: According to the problem, total dishes are 65, so n/2 + n/3 + n/4 = 65. Step 6: Find a common denominator for 2, 3 and 4, which is 12. Step 7: Rewrite each term with denominator 12: n/2 = 6n/12, n/3 = 4n/12, n/4 = 3n/12. Step 8: Add them: 6n/12 + 4n/12 + 3n/12 = 13n/12. Step 9: So 13n/12 = 65. Step 10: Multiply both sides by 12 to get 13n = 65 * 12. Step 11: Compute 65 * 12 = 780, so 13n = 780. Step 12: Divide by 13 to get n = 780 / 13 = 60. Verification / Alternative check:We can verify by computing the number of bowls of each type for 60 guests. Rice bowls = 60/2 = 30. Dal bowls = 60/3 = 20. Meat bowls = 60/4 = 15. Total dishes = 30 + 20 + 15 = 65, matching the problem statement. Thus 60 guests satisfy all the conditions. Why Other Options Are Wrong:If we take 74 guests, then 74/2 + 74/3 + 74/4 is not an integer sum equal to 65. Similar checks show that 82 or 58 guests do not satisfy the equation with total dishes equal to 65. Hence those options are not consistent with the sharing rules. Common Pitfalls:Some learners incorrectly multiply rather than divide when converting guests to bowls, or they might forget to use a common denominator. Others may misinterpret the statement and assume that there are exactly two, three or four guests in total rather than per bowl. Translating carefully to algebra and using the least common multiple of denominators avoids these issues. Final Answer:The number of guests at the dinner party is 60, which corresponds to option A.

Question 12

Which natural number nearest to 8485 is completely divisible by 75?

Accepted Answer

Introduction / Context:This question asks for the natural number nearest to 8485 that is divisible by 75. To solve it, we must identify multiples of 75 close to 8485 and compare their distances from this target. This tests understanding of multiples, division with remainders and how to choose the nearest value. Given Data / Assumptions: - The reference number is 8485. - We seek a natural number that is a multiple of 75. - We want the multiple closest to 8485 in absolute difference. Concept / Approach:To find the nearest multiple of 75, we divide 8485 by 75 and look at the quotient and remainder. The quotient tells us the largest multiple of 75 less than or equal to 8485, and the next integer quotient gives the smallest multiple of 75 greater than 8485. By comparing the difference between 8485 and these two neighboring multiples, we can decide which is closer. Step-by-Step Solution: Step 1: Divide 8485 by 75 to find the nearest multiples. Step 2: Compute 75 * 100 = 7500 and 75 * 10 = 750, so 75 * 113 = 75 * (100 + 13) = 7500 + 975 = 8475. Step 3: The next multiple is 75 * 114. Compute 75 * 114 = 75 * (100 + 14) = 7500 + 1050 = 8550. Step 4: The two closest multiples around 8485 are 8475 and 8550. Step 5: Find the distance from 8485 to these multiples. For 8475, difference is 8485 − 8475 = 10. Step 6: For 8550, difference is 8550 − 8485 = 65. Step 7: Since 10 is less than 65, 8475 is closer to 8485 than 8550. Verification / Alternative check:We can also check other options quickly. Option B is 8500, and 8500 ÷ 75 = 113 with remainder 25, so it is not a multiple of 75. Option D is 8525, and 8525 ÷ 75 leaves a remainder of 50. Only 8475 and 8550 are exact multiples of 75 among the likely nearby numbers, and as shown, 8475 is closer. Why Other Options Are Wrong:Options B, C and D are either not divisible by 75 or not the nearest multiple. Specifically, 8500 and 8525 are not multiples of 75, and although 8550 is a multiple, it is farther away from 8485 than 8475 is. Common Pitfalls:Some learners may try to round 8485 to the nearest multiple of 100 or 50 instead of finding exact multiples of 75. Others may not check divisibility properly and choose a number that is close but not a correct multiple. Carefully computing 75 times nearby integers and comparing distances avoids such mistakes. Final Answer:The natural number nearest to 8485 that is completely divisible by 75 is 8475, which matches option A.

Question 13

Consider the sequence 3, 4, 5, 5, 12, 13, 7, 24, 25, 8, 15, ... which is built from Pythagorean triples. Which number should come next in place of the dots?

Accepted Answer

Introduction / Context:This sequence question hides a pattern based on Pythagorean triples, which are sets of three positive integers (a, b, c) that satisfy a^2 + b^2 = c^2. Recognizing that the numbers are grouped into such triples allows us to find the next term quickly. The problem tests pattern recognition and knowledge of common Pythagorean triples. Given Data / Assumptions: - The sequence given is 3, 4, 5, 5, 12, 13, 7, 24, 25, 8, 15, ... - We suspect the numbers may be grouped into meaningful sets of three. - We must determine the next single number after 8 and 15. Concept / Approach:Observe the sequence in groups of three terms. The first group is 3, 4, 5, which is the smallest Pythagorean triple. The second group is 5, 12, 13, another well known triple. The third group is 7, 24, 25, yet another triple. The pattern seems to be that each group of three numbers forms a Pythagorean triple, with the first two being the legs and the third being the hypotenuse. The next group starts with 8 and 15, suggesting we should look for the hypotenuse that completes the triple (8, 15, h). Step-by-Step Solution: Step 1: Group the terms as (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, ?). Step 2: Check each triple: 3^2 + 4^2 = 9 + 16 = 25 = 5^2, so (3, 4, 5) is a Pythagorean triple. Step 3: For (5, 12, 13), compute 5^2 + 12^2 = 25 + 144 = 169 = 13^2. Step 4: For (7, 24, 25), compute 7^2 + 24^2 = 49 + 576 = 625 = 25^2. Step 5: The pattern is clear: each group of three consecutive terms forms a Pythagorean triple. Step 6: Now consider (8, 15, ?). Compute 8^2 + 15^2 = 64 + 225 = 289. Step 7: Recognize that 289 is 17^2, since 17 * 17 = 289. Step 8: Therefore, the triple is (8, 15, 17), and the missing next term is 17. Verification / Alternative check:We can cross check by listing common Pythagorean triples from memory: (3, 4, 5), (5, 12, 13), (7, 24, 25) and (8, 15, 17). The sequence matches the first three triples exactly and starts the fourth triple with 8 and 15, confirming that 17 must follow. None of the other options fit neatly as a hypotenuse of these legs. Why Other Options Are Wrong:Option A (26) would not satisfy 8^2 + 15^2 = 26^2, because 64 + 225 = 289 while 26^2 = 676. Options B (16) and D (18) also do not yield the correct relation when squared and summed. Only 17 gives the correct Pythagorean relationship. Common Pitfalls:Some learners may treat the sequence as a simple numeric pattern without grouping the terms into triples, leading to incorrect differences or ratios. Others may not recall common Pythagorean triples and therefore miss the structure. Recognizing triples is a useful skill in many aptitude and geometry problems. Final Answer:The next number in the sequence is 17, which corresponds to option C.

Question 14

If x and y are the two digits of the number 653xy such that this number is divisible by 80, then what is the value of x + y?

Accepted Answer

Introduction / Context:This problem involves divisibility by 80 for a five digit number with two unknown digits. Divisibility by 80 can be analyzed using its prime factorization and standard divisibility rules for powers of 2 and for 5. The task is to find digits x and y in the number 653xy so that it becomes divisible by 80 and then compute x + y. This tests understanding of divisibility rules and handling of unknown digits in place value notation. Given Data / Assumptions: - The number has the form 653xy, where x is the tens digit and y is the units digit. - The number is divisible by 80. - x and y are digits from 0 to 9. - We must find x + y. Concept / Approach:First factorize 80. We have 80 = 2^4 * 5. A number divisible by 80 must be divisible by 16 and by 5 simultaneously. Divisibility by 5 requires the last digit to be 0 or 5, but because 80 is also a multiple of 10, any multiple of 80 must end in 0. Therefore, y must be 0. Then, for divisibility by 16, the last four digits of the number must form a number divisible by 16. Hence we examine the four digit number 53x0 and find which value of x makes it divisible by 16. Step-by-Step Solution: Step 1: Write the number as 653xy. For divisibility by 5 and 10, the last digit y must be 0 for a multiple of 80, so set y = 0. Step 2: The number is now 653x0. The last four digits are 53x0, which must be divisible by 16. Step 3: Consider possible values of x from 0 to 9 and test divisibility of 53x0 by 16. Step 4: Instead of testing blindly, note that 5300, 5310, 5320 and so on will vary in steps of 10. We seek 53x0 such that 53x0 ÷ 16 is an integer. Step 5: One reliable method is to test directly. For x = 6, the number is 5360. However our full number is 65360, whose last four digits are 5360, and 5360 ÷ 16 = 335 exactly. Step 6: Checking for other x values shows that none of them give a last four digit block divisible by 16. Therefore x = 6 is the unique solution. Step 7: We already have y = 0 from the divisibility by 80 condition. Step 8: Compute x + y = 6 + 0 = 6. Verification / Alternative check:Verify that 65360 is divisible by 80. Compute 65360 ÷ 80. This is the same as 65360 ÷ (16 * 5). First divide by 16 to get 4085, then divide 4085 by 5 to get 817. Since both divisions are exact, 65360 is indeed divisible by 80. Any change to x or y would break at least one of the divisibility conditions, confirming that x = 6 and y = 0 is the only suitable pair. Why Other Options Are Wrong:If x + y were 5, 4 or 3, then the sum of the digits would correspond to a different pair such as (5,0), (4,0) or (3,0) or similar combinations. None of these yield a number 653xy that is divisible by both 16 and 5 in the required way. Thus only the sum 6 is consistent with all the constraints. Common Pitfalls:Some students may incorrectly use divisibility by 8 and 10 instead of by 16 and 5, which guarantees divisibility by 40 rather than 80. Others may assume that a last digit of 5 is acceptable, forgetting that multiples of 80 must end with 0. Paying attention to prime factorization and correct divisibility rules avoids these mistakes. Final Answer:The value of x + y is 6, which corresponds to option A.

Question 15

How many terms are there in the geometric progression 3, 6, 12, 24, ..., 384?

Accepted Answer

Introduction / Context:This problem involves a geometric progression (G.P.), a sequence in which each term after the first is obtained by multiplying the previous term by a constant ratio. We are given the first few terms and the last term and asked how many terms are in the sequence. This tests understanding of the formula for the n-th term of a geometric progression and basic logarithmic or exponential reasoning. Given Data / Assumptions: - The G.P. starts with 3, followed by 6, 12, 24 and continues up to 384. - The first term a is 3. - The common ratio r is 6 ÷ 3 = 2. - The last term given is 384 and is part of the sequence. Concept / Approach:In a geometric progression, the n-th term T_n is given by T_n = a * r^(n − 1), where a is the first term and r is the common ratio. Here, we know a, r and T_n, and we need to find n. So we solve 3 * 2^(n − 1) = 384 for n. This involves dividing by 3 and then identifying a power of 2 that equals the result. Step-by-Step Solution: Step 1: Note that a = 3 and r = 2, because 6 ÷ 3 = 2, 12 ÷ 6 = 2 and 24 ÷ 12 = 2. Step 2: Let the number of terms be n. Then the n-th term is T_n = 3 * 2^(n − 1). Step 3: Given that T_n = 384, set up the equation 3 * 2^(n − 1) = 384. Step 4: Divide both sides by 3 to isolate the power of 2: 2^(n − 1) = 384 ÷ 3 = 128. Step 5: Recognize that 128 is a power of 2. Specifically, 2^7 = 128. Step 6: Therefore, 2^(n − 1) = 2^7, which implies n − 1 = 7. Step 7: Solve for n to get n = 8. Verification / Alternative check:We can list the terms: 3 (first), 6 (second), 12 (third), 24 (fourth), 48 (fifth), 96 (sixth), 192 (seventh) and 384 (eighth). Counting them confirms that 384 is the eighth term in the sequence, so there are eight terms in total. This matches the result from the formula. Why Other Options Are Wrong:Options B, C and D suggest 9, 10 or 11 terms. If there were 9 terms, then T_9 would be 3 * 2^8 = 3 * 256 = 768, which is larger than 384. Hence 384 would not be the last term. Similarly, for 10 or 11 terms, the last terms would be 1536 or 3072, which do not match the given last term. Common Pitfalls:One common mistake is to miscount the terms when listing the progression, accidentally including or skipping a term. Another is to confuse the formula for the n-th term with the formula for the sum of terms. Remember that T_n = a * r^(n − 1) is the correct formula for the n-th term in a geometric sequence. Final Answer:The geometric progression 3, 6, 12, 24, ..., 384 contains 8 terms, which corresponds to option A.

Question 16

In a certain class, about 4 out of every 5 students are left handed. If there are 40 students in the class, how many students are left handed?

Accepted Answer

Introduction / Context: This question tests basic percentage and ratio understanding in the context of a simple real life situation. The statement says that about 4 out of every 5 students are left handed and asks you to apply that ratio to a total of 40 students. Problems like this are very common in aptitude tests to check comfort with fractions, ratios and proportional reasoning. Given Data / Assumptions: In the class, 4 out of every 5 students are left handed. Total number of students in the class = 40. The ratio 4 : 5 applies uniformly across the whole class. We assume every student is either left handed or right handed, with no other category needed for the calculation. Concept / Approach: The phrase 4 out of 5 means that the fraction of left handed students is 4/5 of the entire group. To find how many students this represents in a class of 40, we multiply the total number of students by this fraction. This is the same as finding 80 percent of 40, since 4/5 is equal to 0.8 or 80 percent. Step-by-Step Solution: Ratio of left handed students = 4 out of 5 = 4/5. Total number of students in the class = 40. Number of left handed students = (4/5) * 40. Compute: (4/5) * 40 = 4 * (40/5) = 4 * 8 = 32. So there are 32 left handed students in the class. Verification / Alternative check: Another way is to first find how many students correspond to one part in the ratio 4 : 5. Each group of 5 students contains 4 left handed students. With 40 students, the number of groups of 5 is 40 / 5 = 8. Each group contributes 4 left handed students, so total left handed students = 8 * 4 = 32. This confirms the previous calculation. Why Other Options Are Wrong: 30: This corresponds to 30/40 = 3/4 = 75 percent, not 4/5. 28: This is 28/40 = 7/10 = 70 percent, again not equal to 4/5. 34 and 36: These are even larger proportions, more than the given 4 out of 5 ratio. Common Pitfalls: Learners sometimes mistakenly divide 40 by 4 instead of multiplying by 4/5, or they misread 4 out of 5 as 4 percent. Always convert the phrase into a clear fraction or percentage before computing. Writing the fraction 4/5 beside the total and then multiplying makes the process much easier to follow. Final Answer: The number of left handed students in the class is 32.

Question 17

What value should replace the question mark in the equation 22% of ? + 166.64 = 340?

Accepted Answer

Introduction / Context: This problem checks your ability to work with percentages and linear equations. You are given an equation where 22 percent of some unknown number plus a fixed amount equals 340. Your task is to isolate the unknown and solve for its value. Questions like this are typical in aptitude tests because they combine basic percentage manipulation with algebraic thinking. Given Data / Assumptions: The equation is 22% of X + 166.64 = 340, where X is the unknown number. Twenty two percent means 22/100 or 0.22. All values are real numbers, and standard arithmetic operations apply. We assume there is a unique solution for X. Concept / Approach: The key idea is to rewrite the percentage as a decimal multiplier and then solve a simple linear equation. First convert 22 percent of X into 0.22 * X. Then move the constant 166.64 to the other side of the equation by subtracting it from 340. Once you have an expression of the form 0.22 * X = some number, divide both sides by 0.22 to obtain X. Always keep track of decimal places carefully to avoid arithmetic errors. Step-by-Step Solution: Let the unknown number be X. Given: 22% of X + 166.64 = 340. Convert 22% to decimal: 22% = 22/100 = 0.22. So the equation becomes 0.22 * X + 166.64 = 340. Subtract 166.64 from both sides: 0.22 * X = 340 − 166.64 = 173.36. Now divide both sides by 0.22: X = 173.36 / 0.22. Compute: 0.22 = 22/100, so 173.36 / 0.22 = 17336 / 22 = 788. Therefore, X = 788. Verification / Alternative check: Check the result by substituting X = 788 back into the original equation. Twenty two percent of 788 is 0.22 * 788 = 173.36. Then 173.36 + 166.64 = 340 exactly. Because the left hand side equals the right hand side when X is 788, the solution is confirmed to be correct. Why Other Options Are Wrong: 786: 0.22 * 786 = 172.92; adding 166.64 gives 339.56, which is less than 340. 784: 0.22 * 784 = 172.48; adding 166.64 gives 339.12, again not equal to 340. 792: 0.22 * 792 = 174.24; adding 166.64 gives 340.88, which is too large. 780: 0.22 * 780 = 171.6; adding 166.64 gives 338.24, also incorrect. Common Pitfalls: Common mistakes include treating 22 percent as 22 instead of 0.22, or trying to compute the percentage of the wrong number. Some students also make rounding errors when working with decimals. A good tactic is to convert all decimals to fractions temporarily, perform the division exactly, and then convert back to a decimal only at the end if needed. Final Answer: The value that should replace the question mark is 788.

Question 18

The sum of three consecutive odd integers is 93. What is the value of the middle odd integer?

Accepted Answer

Introduction / Context: This question explores simple algebra using consecutive odd integers. Consecutive odd numbers follow a fixed pattern with a constant difference of 2 between them. By expressing these numbers in terms of a single variable, we can form an equation based on the given sum and solve it. This type of question is a staple in aptitude exams because it connects number patterns with algebraic expressions. Given Data / Assumptions: The three numbers are consecutive odd integers. Their total sum is 93. The difference between any two successive odd integers is 2. We are asked specifically for the middle integer. Concept / Approach: Let the middle odd integer be M. Then the previous odd integer is M − 2 and the next odd integer is M + 2. These three numbers form a simple arithmetic sequence. Their sum is (M − 2) + M + (M + 2). This expression simplifies neatly and allows us to solve for M directly. This approach avoids guessing and works for any similar problem involving sums of consecutive odd or even numbers. Step-by-Step Solution: Let the middle odd integer be M. Then the three consecutive odd integers are M − 2, M and M + 2. Their sum is (M − 2) + M + (M + 2). Simplify: (M − 2) + M + (M + 2) = 3M. Given that the sum is 93, we have 3M = 93. Divide both sides by 3: M = 93 / 3 = 31. Therefore, the middle odd integer is 31. Verification / Alternative check: Once we obtain M = 31, the three consecutive odd integers are 29, 31 and 33. Their sum is 29 + 31 + 33 = 93. This matches the sum given in the question, so the solution is correct. You can also reason that the average of three numbers is their sum divided by 3. Since they are equally spaced, the middle value must equal this average, and 93 / 3 = 31 gives the same result directly. Why Other Options Are Wrong: 33: If 33 were the middle number, the others would be 31 and 35, and the sum would be 99, not 93. 29: As the middle number, the sequence would be 27, 29 and 31, summing to 87. 27: This would give 25, 27 and 29, with a sum of only 81. 35: This would give 33, 35 and 37, which sum to 105. Common Pitfalls: Learners sometimes misidentify the pattern and use M, M + 1 and M + 2, which are consecutive integers, not consecutive odd integers. Another error is to forget that the sum of three numbers is three times their average, which is a quick shortcut here. Always check whether the difference between the numbers should be 1, 2, or some other fixed amount based on the wording of the question. Final Answer: The middle odd integer is 31.

Question 19

The angles of a quadrilateral are in the ratio 2 : 4 : 7 : 5. The smallest angle of this quadrilateral is equal to the smallest angle of a triangle. In that triangle, one angle is twice the smallest angle. What is the second largest angle of the tri…

Accepted Answer

Introduction / Context: This question links the internal angles of a quadrilateral with those of a triangle. It first gives the ratio of angles in a quadrilateral and then uses the smallest of these angles as the smallest angle of a triangle. In that triangle, another angle is twice the smallest angle. The task is to identify the second largest angle of the triangle. This combines knowledge about angle sums in polygons and basic algebraic manipulation of ratios. Given Data / Assumptions: The four angles of a quadrilateral are in the ratio 2 : 4 : 7 : 5. The sum of the interior angles of a quadrilateral is 360 degrees. The smallest angle of the quadrilateral equals the smallest angle of a triangle. One of the angles of that triangle is twice its smallest angle. The sum of interior angles of a triangle is 180 degrees. Concept / Approach: First, convert the ratio of the quadrilateral angles into actual angle measures. Let the common factor be k. Then the four angles are 2k, 4k, 7k and 5k. Their sum must equal 360 degrees, which allows us to find k and thus each angle. The smallest angle of the quadrilateral will be the smallest angle of the triangle. In the triangle, if the smallest angle is x degrees, then another angle is 2x degrees. The third angle can be found because the sum of angles in a triangle is 180 degrees. Finally, we order the three triangle angles and identify the second largest one. Step-by-Step Solution: Let the quadrilateral angles be 2k, 4k, 7k and 5k. Sum of angles of a quadrilateral = 360 degrees. So 2k + 4k + 7k + 5k = 18k = 360 ⇒ k = 20. Thus the four angles are 40°, 80°, 140° and 100°. The smallest angle is 40°, so the smallest angle of the triangle is also 40°. Let the triangle angles be 40°, 2 × 40° = 80°, and the third angle C. Sum of angles of triangle = 180°, so C = 180 − 40 − 80 = 60°. The three triangle angles are 40°, 60° and 80°; ordered from smallest to largest they are 40°, 60°, 80°. Therefore, the second largest angle is 60°. Verification / Alternative check: We can double check that 40° truly is the smallest quadrilateral angle, which it is compared to 80°, 100° and 140°. In the triangle, 40° and 80° satisfy the condition that one angle is twice the smallest. The remaining angle 60° completes the total of 180°. The largest triangle angle is 80°, the second largest is 60°, and the smallest is 40°, so the answer is consistent with all conditions. Why Other Options Are Wrong: 80 degrees: This is the largest triangle angle, not the second largest. 120 degrees: No triangle angle here is 120°, and such a large angle would violate the given conditions. 40 degrees: This is the smallest angle, not the second largest. 100 degrees: This comes from the quadrilateral but does not appear as a triangle angle. Common Pitfalls: A common mistake is to misidentify the smallest quadrilateral angle or to forget that triangle angles must sum to 180°. Another error is to assume that the second largest triangle angle must come directly from the quadrilateral without recalculating. Carefully computing all angles step by step avoids these errors and clarifies which angle is truly the second largest. Final Answer: The second largest angle of the triangle is 60 degrees.

Question 20

If V = 12R / (r + R), express R in terms of V and r.

Accepted Answer

Introduction / Context: This is a basic algebraic manipulation problem where you are asked to solve a formula for a different variable. Initially, the expression is given as V = 12R / (r + R), and you must rewrite it so that R is the subject. Such rearrangement tasks are fundamental in algebra and also appear in physics and engineering when you need to isolate a particular quantity in a formula. Given Data / Assumptions: Original formula: V = 12R / (r + R). V, r and R are real variables, with r + R not equal to zero. The goal is to express R explicitly in terms of V and r. We assume R is non zero so that division by R is valid when we factor. Concept / Approach: To solve for R, we treat the formula like an equation in R. The main steps are to clear the fraction by multiplying both sides by the denominator, then group all terms involving R on one side. After that, factor out R and finally divide to isolate R. Careful handling of each algebraic step ensures that we do not introduce or lose any valid solutions. Throughout, we keep track of conditions such as 12 − V not being zero in the final expression. Step-by-Step Solution: Given: V = 12R / (r + R). Multiply both sides by (r + R): V(r + R) = 12R. Expand the left side: Vr + VR = 12R. Move all R terms to one side: Vr = 12R − VR. Factor R on the right: Vr = R(12 − V). Assuming 12 − V is not zero, divide both sides by (12 − V): R = Vr / (12 − V). Thus R is expressed purely in terms of V and r. Verification / Alternative check: We can verify the rearranged formula by substituting R = Vr / (12 − V) back into the original expression for V. If we start with R = Vr / (12 − V) and compute 12R / (r + R), the algebra simplifies back to V, confirming that the manipulation is correct. This reverse substitution is a powerful way to check rearrangements of formulas and ensure no step was performed incorrectly. Why Other Options Are Wrong: (Vr + V) / 12: This treats 12 as if it were distributed differently and does not satisfy the original equation when substituted back. V: This would imply R is always equal to V, which clearly is not true for arbitrary r and V. V / (r - 12): This mixes constants and variables incorrectly and gives a dimensionally inconsistent result. 12r / (V - 12): This resembles the correct expression but uses r instead of R inside the factorised term, so it does not satisfy V = 12R / (r + R). Common Pitfalls: Students sometimes try to cross multiply too quickly and make sign errors when moving terms across the equals sign. Another frequent mistake is forgetting to factor R from all terms containing it, leading to incomplete or incorrect expressions. Always expand, collect like terms, factor the desired variable and only then divide to isolate it. This four step approach makes rearranging formulas more systematic and reliable. Final Answer: The correct expression for R in terms of V and r is R = Vr / (12 − V).

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