Math Problems and Logic Puzzles

If you take away 2 apples from a group, you are physically holding those 2 apples. The question asks how many you have, not how many are left in the original pile.

The statement 'all but 9 die' indicates that 9 pigs survived. The number of pigs left alive is exactly the number that did not die, which is 9.

This is a trick question testing careful reading. A dozen always means 12, regardless of what the item is. The phrase 'two cent stamps' is irrelevant - the answer is simply how many items are in a dozen, which is 12. Many students mistakenly calculate 12 divided by 2.

A man who has a widow is deceased. Since dead people cannot perform legal actions or marry, it is impossible for him to marry his widow's sister.

Every month of the year has at least 28 days. While February is the only month with exactly 28 days (usually), all 12 months contain a 28th day. This is a common lateral thinking puzzle.

Dividing by 1/2 is the same as multiplying by 2. So, 30 / 0.5 = 60. Adding 10 to 60 results in 70. Distractors like 25 result from incorrectly multiplying by 1/2 or subtracting.

The phrase 'all but 9 die' means that every pig died except for 9 of them. Therefore, 9 pigs remain alive. This is a classic linguistic trick question.

This is the same trick question as 82915. A dozen always means 12 items. The value of each stamp (2 cents) is irrelevant to counting how many stamps are in a dozen. The correct answer is 12.

This is a riddle. If a man has a 'widow', it means the man is dead. A dead man cannot marry anyone, regardless of the legality of marrying a sister-in-law. Therefore, the answer is 'No'.

Profit % = [(Items Bought - Items Sold) / Items Sold] * 100. Here, [(11 - 9) / 9] * 100 = 2/9 * 100 = 22.22%, which is expressed as the mixed fraction 22(2/9).

The last measure is A+B+C = 180. The 7 measures are A, B, C, A+B, B+C, C+A, and A+B+C. Summing these gives 4(A+B+C) = 4 * 180 = 720. The average of the 7 measures is therefore 720/7.

When two items are sold at the same price, one at a profit of x% and the other at a loss of x%, there is always a net loss given by (x/10)^2. Here, (10/10)^2 = 1% loss.

Let Eswar's salary = 100. Then Ram's salary = 105 (5% more). Eswar's salary relative to Ram = 100/105 ≈ 95.24%. Therefore Eswar's salary is 100 - 95.24 = 4.76% less than Ram's. Converting 0.76 to fraction: 0.76 = 76/100 = 19/25. Option C shows 4(16/21)% = 4.76%, which matches.

Working backwards: After giving 75% to charity, 261 remain. So 261 = 25% of good mangoes. Good mangoes = 261/0.25 = 1044. These 1044 represent the 87% that are good (100% - 13% bad). Total mangoes = 1044/0.87 = 1200. Verification: 13% of 1200 = 156 bad, 1044 good. 75% of 1044 = 783 to charity, leaving 261.

Original sum = 20 × 65 = 1300. But 96 was recorded instead of 69, so error = 96 - 69 = 27. Correct sum = 1300 - 27 = 1273. Correct mean = 1273/20 = 63.65. This matches option A. Option B (65.95) would result if we added 27 instead of subtracting.

Let one part = x, other = 1700 - x. Equation: 0.3x - 0.25(1700 - x) = 70. Simplifying: 0.3x - 425 + 0.25x = 70, so 0.55x = 495, giving x = 900. Other part = 1700 - 900 = 800. Verification: 30% of 900 = 270, 25% of 800 = 200, difference = 70.

Using the identity: if a + b = c, then a³ + b³ - c³ is divisible by a, b, and c. Here 21 + 33 = 54, so 54³ - 33³ - 21³ can be rearranged as 54³ - (33³ + 21³). However, note that for a + b - c = 0, the expression a³ + b³ + c³ - 3abc is divisible by (a + b + c). In this case: 54³ - 33³ - 21³ = 54³ + (-33)³ + (-21)³ - 3(54)(-33)(-21). Since 54 + (-33) + (-21) = 0, it's divisible by 0, meaning divisible by all three numbers individually.

n^200 < 5^300. Taking the 200th root of both sides: n < 5^(300/200) = 5^(3/2) = √125 ≈ 11.18. Since n must be an integer, the largest value is n = 11. Verification: 11^200 = (11^2)^100 = 121^100, while 5^300 = (5^3)^100 = 125^100. Since 121 < 125, 11^200 < 5^300 holds true.