Math & Logic Puzzles

Average × number of items = total sum. 250 × 4 = 1000 total salary. Subtract the three known salaries: 1000 - (120 + 50 + 100) = 730.

Probability of exactly 2 hits: sum of three cases where exactly one person misses. (6/7×4/5×1/4) + (6/7×1/5×3/4) + (1/7×4/5×3/4) = 54/140.

The pattern is multiply by 3 and subtract increasing numbers: 3×3-1=8, 8×3-2=22, 22×3-3=63, 63×3-4=185. Next: 185×3-5=550.

Let b be brothers and s be sisters. For Lata: s-1 = b. For Shyam: s = 2(b-1). Substituting s = b+1 into the second: b+1 = 2b - 2, so b = 3. Then s = 4. Total children = 3 + 4 = 7.

In the range 11-100, 6 appears in: 16, 26, 36, 46, 56, 60, 61, 62, 63, 64, 65, 66 (twice), 67, 68, 69, 76, 86, 96. Total count: 1 + 1 + 1 + 1 + 1 + (10 + 1 for the extra 6 in 66) + 1 + 1 + 1 = 19.

Time = distance ÷ speed. The bullet must travel 10 meters at 500 m/s, so t = 10/500 = 0.02 seconds. This is the reaction time available - extremely fast, showing why dodging bullets is unrealistic. Option A is a joke answer.

For free fall from height h: h = ½gt². With h = 100m and g = 9.825 m/s²: t = √(2h/g) = √(200/9.825) = √20.36 ≈ 4.51 seconds. Option B (4.4s) is the closest approximation - the slight difference comes from rounding in intermediate steps.

When hanging stationary, the rope exerts an upward force equal to the person's weight. Weight = mass × gravity = 8 kg × 10 m/s² = 80 N. This upward tension balances the downward gravitational force.

Let limit be L. A's speed = L + 10. B's speed = L + 20. Given B's speed is 35 km/hr, L + 20 = 35, so L = 15 km/hr.

Starting with 10 gallons at 10% water: 1 gallon water, 9 gallons paint. After adding x gallons water: total water = 1 + x, total volume = 10 + x. Need (1+x)/(10+x) = 0.20. Solve: 1+x = 2 + 0.2x → 0.8x = 1 → x = 1.25. Adding 1.25 gallons creates the 80/20 paint-to-water ratio.

Let x be the number of swans. x = (7/2)√x + 2. Let √x = y. y^2 - 3.5y - 2 = 0. Solving the quadratic gives y = 4 (since y must be positive). Thus x = y^2 = 16.

3 women = 2 men, so 21 women = 14 men. Using M1D1H1 = M2D2H2: 15 * 21 * 8 = 14 * Days * 6. 2520 = 84 * Days. Days = 2520 / 84 = 30 days.

Let son's age be x. Father's present age is x + 30. After 5 years: (x + 30) + 5 = 3(x + 5), so x + 35 = 3x + 15, giving x = 10. Father's present age = 10 + 30 = 40. Option A (30) would make father equal to son's age plus 30, not thrice after 5 years. C and D don't satisfy the condition.

The car saved 20 minutes total, meaning it saved 10 minutes of travel time to the office. Usually, it arrives at 7:00, so it met the man at 6:50. He walked from 5:30 to 6:50 (80 minutes).

To solve this question, the user needs to understand basic arithmetic, combinations, and the concept of counting. The boy starts with Rs 2 and can win or lose Re 1 at a time. He can lose only 5 times and is out of the game if he earns Rs 5. We need to find the number of ways in which this is possible.

Let's consider the possible outcomes for the boy's first 5 games:

• He can win all 5 games, which would result in him earning Rs 7.
• He can win 4 games and lose 1 game, which would result in him earning Rs 6.
• He can win 3 games and lose 2 games, which would result in him earning Rs 5.
• He can win 2 games and lose 3 games, which would result in him earning Rs 4.
• He can win 1 game and lose 4 games, which would result in him earning Rs 3.
• He can lose all 5 games, which would result in him earning Rs 2.

We can count the number of ways in which each outcome is possible using combinations. For example, the number of ways for the boy to win 4 games and lose 1 game can be found by calculating the number of ways to choose 4 games out of 5 to win, which is 5 choose 4, or ${5 \choose 4} = 5$. The number of ways for him to win all 5 games can be found by calculating the number of ways to choose 5 games out of 5 to win, which is 5 choose 5, or ${5 \choose 5} = 1$. We can use similar logic to count the number of ways for each outcome.

Let's list out the number of ways for each outcome:

• He can win all 5 games: 1 way
• He can win 4 games and lose 1 game: ${5 \choose 4} = 5$ ways
• He can win 3 games and lose 2 games: ${5 \choose 3} = 10$ ways
• He can win 2 games and lose 3 games: ${5 \choose 2} = 10$ ways
• He can win 1 game and lose 4 games: ${5 \choose 1} = 5$ ways
• He can lose all 5 games: 1 way

To find the total number of ways for the boy to earn Rs 5 or less, we just need to add up the number of ways for each outcome:
1 + 5 + 10 + 10 + 5 + 1 = 32

Therefore, the answer is:

The Answer is: C) 16

Let the ratio x:y = k. Then k = 25 × (y:x) = 25 × (1/k). This gives k = 25/k, so k² = 25, meaning k = 5 (positive ratio). Therefore x:y = 5:1.

Total ration = 100 soldiers × 10 days = 1000 soldier-days. After 2 days, 100 soldiers consumed 200 soldier-days, leaving 800 soldier-days. With 200 soldiers now, the ration lasts 800 ÷ 200 = 4 days. This is a standard inverse proportion problem.

Thicker candle burns in 6h, thinner in 4h. Let t be the time passed. (1 - t/6) = 2 * (1 - t/4). 1 - t/6 = 2 - t/2. t/2 - t/6 = 1. t/3 = 1. t = 3 hours.

Let dog leap be D and hare leap be H. 3D = 4H => D = 4/3 H. Speed ratio = (4 leaps of dog) / (5 leaps of hare) = (4 * 4/3 H) / 5H = 16/3 : 5 = 16:15.

Assume 30 boys and 70 girls. Non-scholarship boys = 75% of 30 = 22.5. Non-scholarship girls = 80% of 70 = 56. Total non-scholarship = 22.5 + 56 = 78.5 out of 100, which is 78.5%.