Math and Logic Problem Solving

If n divided by 7 leaves remainder 3, then n = 7k + 3. Then (n-3) = 7k. Squaring both sides gives (n-3)^2 = (7k)^2 = 49k^2. Since 49k^2 is a multiple of 49, (n-3)^2 is exactly divisible by 49.

When 3t+4r > 0, comparing (3t+4r)/rt vs (3t+4r)/(r+t) depends on whether rt > r+t or rt < r+t. When 3t+4r < 0, the inequality direction reverses. Test cases: t=1, r=2 gives Col B greater; t=10, r=1 gives Col A greater; t=0.5, r=2 makes them equal. Since different values produce different results, the relationship cannot be uniquely determined.

Area of square = M^2. Area of circle = π(M/2)^2 = (π/4)M^2. Since π is approx 3.14, π/4 is approx 0.785. Because M^2 > 0.785M^2 for any positive M, the area of the square (Col A) is always greater.

Savings = x * n. Storage cost per day = y * n. Let d be the number of days. Storage cost = d * y * n. We want to find the maximum d such that d * y * n <= x * n. Dividing both sides by n*y gives d <= x/y.

Let x be gallons of water added. The salt amount stays constant: 0.15 gallons (15% of 1 gallon). Final concentration: 0.15/(1+x) = 0.10. Solving: 0.15 = 0.10 + 0.10x, so 0.05 = 0.10x, giving x = 0.5. The dilution formula is: C1V1 = C2V2 where C2 is the target concentration and V2 is the final total volume.

Sum of 7 numbers = 7 * 33 = 231. Sum of 8 numbers (including k) = 8 * 35 = 280. Therefore, k = 280 - 231 = 49. Since 49 (Col A) is greater than 47 (Col B), Column A is the correct answer.

Standard deviation measures spread from the mean, not the mean itself. Two datasets with different averages (40 vs 35) can have any standard deviation relationship depending on how their values are distributed around their respective means.

US:Indian = 5:2 and Indian:British = 5:1. To compare US to British, normalize the Indian count to 10 (LCM of 2 and 5). US:Indian = 25:10 and Indian:British = 10:2. Therefore, the ratio of US to British stamps is 25:2.

Total work needed: 20 large cakes (10 hours of work at 2/hr) and 700 small cakes (20 hours of work at 35/hr). Total work = 30 helper-hours. Since the kitchen is available for 3 hours, the number of helpers needed is 30 / 3 = 10.

When you double the length of a cube's sides, the volume increases by a factor of 8 (2³ = 8). Since weight is proportional to volume for the same material, if a 6-pound cube has its sides doubled, the new cube weighs 6 × 8 = 48 pounds. This demonstrates the cubic relationship between linear dimensions and volume.

The number of handshakes among n people is calculated using the formula n(n-1)/2. For 6 people, this is 6(5)/2 = 15. Each person shakes hands with 5 others, but we divide by 2 to avoid double-counting the same handshake.

Step 1: 2x + 5y. Step 2: 3(2x + 5y) = 6x + 15y. Step 3: (6x + 15y) - (x + y) = 5x + 14y. The calculation follows basic algebraic distribution and subtraction of like terms.

The wife died at 50. The husband was 30 when they married, and she was 25, so he was 5 years older. When she died at 50, he was 55. He cried for years, then stopped, and 10 years later he died. The question asks about widower years IF he lived to 80. From age 55 to 80 is 25 years of being a widower.

Let's work backwards: 2 hours ago it was 7 o'clock. At 7 PM, it is 6 hours after 1 PM and 6 hours before 1 AM, so the condition is satisfied. Therefore it is now 9 o'clock. The time symmetry works because 7 PM is exactly halfway between 1 PM and 1 AM.

This is a classic lateral thinking puzzle that relies on the gender assumption. The question states 'no man throws a punch' which is technically true if the boxers are women. Female boxers can and do knock each other out in professional boxing matches.

Let the son's age be AB (two digits), then father's age is BA. So (10A+B) + (10B+A) = 66, meaning 11(A+B) = 66, so A+B = 6. Valid pairs where father > son: (0,6), (1,5), (2,4). This gives ages 60 and 06, 51 and 15, 42 and 24. The option 82 and 28 sums to 110, not 66.

Since f(n+3) = f(n), the function has period 3. To find f(8), compute 8 - (-1) = 9, which is a multiple of 3 (9 = 3 × 3). Therefore f(8) = f(-1) = 6. The statement is true.

The first 100 even numbers are 2, 4, 6, ..., 200 and the first 100 odd numbers are 1, 3, 5, ..., 199. Each even number is exactly 1 greater than its corresponding odd number, so U − V = 100 × 1 = 100. Column A equals Column B.

Area R = πr², S = πs², T = πt². Given R=2S, then πr²=2πs² => r=s√2. Given S=2T, then πs²=2πt² => s=t√2. Substituting s: r = (t√2)√2 = 2t. Thus, r = 2t. The statement 'Col A is not equal to Col B' is False because they are equal.