Math and Puzzles Challenge - 35

A gains 10 km/hr on B (25-15=10). In 20 hours, A gains 200 km, which is 2 full laps of the 100 km track. Starting from opposite points, they meet after 10 hours (A overtakes B), then at 20 hours (A's 2nd lap advantage means 3rd crossing). The key insight: relative gain of 100 km = 1 lap = 1 crossing. So 200 km gain = 2 laps = 3rd crossing.

When A beats B by 5 meters, B runs 95 meters when A finishes 100 meters, so B's speed is 95% of A's speed. When B beats C by 5 meters, C runs 95 meters when B finishes 100 meters, so C's speed is 95% of B's speed. Therefore, C's speed is 0.95 × 0.95 = 0.9025 times A's speed. In a 100m race, when A finishes, C will have run 100 × 0.9025 = 90.25 meters, so A beats C by 9.75 meters.

The ratio of speeds of two objects starting from A and B and meeting at a point, then reaching their destinations in times t1 and t2, is given by SpeedA / SpeedB = sqrt(t2 / t1). Here, sqrt(144 / 36) = sqrt(4) = 2.

If Chetan sold shares on three consecutive days, the price must have gone up on three consecutive days. Since Michael only sold shares once (which happens only when the price goes above 110, i.e., reaches 120), the price path must rise steadily. This sequence of consecutive up days means the price at the end of day 3 must be 110.

The price sequence that results in Chetan having exactly Rs.1300 more than Michael is: Day 1 (Rs.90), Day 2 (Rs.80), Day 3 (Rs.90), Day 4 (Rs.100), Day 5 (Rs.110). Chetan's strategy of buying when prices fall and selling when they rise gives him better returns compared to Michael who only trades at extreme prices. At day 4's Rs.100 price, both traders are in a neutral position relative to their strategies, setting up the final difference.

For Michael to end up with 20 more shares than Chetan, we analyze the possible price paths. Chetan always ends up with 10 fewer shares than his starting amount. Thus, Michael must end up with 10 more shares than his starting amount. This is only possible if Michael bought shares when the price went below 90 (i.e., reached 80). The only path satisfying this and ending at 110 on day 5 requires the price at the end of day 3 to be 90.

Let $U$ be the number of days the price went up and $D$ be the number of days it went down. Since the price went from 100 to 110 over 5 days with steps of $\pm 10$, we have $U + D = 5$ and $10(U - D) = 10$, which yields $U = 3$ and $D = 2$. Chetan sells 10 shares on up days and buys 10 on down days, ending with $30 - 20 = 10$ fewer shares. Michael only trades if the price goes above 110 or below 90. Analyzing the valid price paths shows that Michael's cash difference of -100 relative to Chetan forces both to end up with the same number of shares.

The maximum combined cash increase of Rs.5000 occurs with the price sequence: 100→90→80→90→100→110. Chetan's more active strategy (buying at Rs.80, Rs.90; selling at Rs.100, Rs.110) generates better returns than Michael's threshold-based approach. The key is maximizing price volatility while ensuring the sequence ends at Rs.110, allowing both traders to profit optimally from their respective strategies.

Let x = sqrt(2 + x). Squaring both sides gives x² = 2 + x, or x² - x - 2 = 0. Factoring gives (x - 2)(x + 1) = 0. Since the square root must be positive, x = 2.

Multiply the exponents: (2/3) * (2/3) * (3/4) = 1/3. The expression simplifies to 512^(1/3). Since 8^3 = 512, the cube root of 512 is 8.

The number of diagonals in an $n$-sided polygon is given by the formula $n(n - 3) / 2$. For an 11-sided polygon, this is $11 \times (11 - 3) / 2 = 11 \times 8 / 2 = 44$. Other options like 36, 48, or 55 represent calculation errors or incorrect formulas.

The largest square inscribed in a circle has a diagonal equal to the circle's diameter, which is 8 cm. The area of a square with diagonal $d$ is $d^2 / 2$. Thus, the area is $8^2 / 2 = 64 / 2 = 32$ square centimeters. Distractors represent incorrect formulas or area of the circumscribed square (64).

A square circumscribing a circle of diameter 8 cm has a side length equal to the diameter of the circle, which is 8 cm. The perimeter of this square is $4 \times 8 = 32$ cm. Distractors like 16, 24, and 48 represent incorrect side-to-perimeter scaling or area calculations.

Let the marked price be $M$ and cost price be $C$. The selling price with an 8% discount is $0.92M = 17940$, so $M = 19500$. The profit is 19.6%, meaning $1.196C = 17940$, so $C = 15000$. If no discount is offered, the selling price is $M = 19500$. The profit would be $(19500 - 15000)/15000 = 0.30$ or 30%. Since 30% is not listed among the numerical options, 'None of these' is correct.

From 8 a.m. on one day to 1 p.m. on the following day is 29 hours of indicated clock time. Since the clock gains 10 minutes every 24 hours, 24 hours and 10 minutes (1450 minutes) of indicated time corresponds to 24 hours (1440 minutes) of true time. Therefore, 29 hours of indicated time corresponds to $29 \times (1440 / 1450) = 28.8$ hours of true time, which is 28 hours and 48 minutes. Adding 28 hours and 48 minutes to 8 a.m. gives 12:48 p.m., or 48 minutes past 12.

This series follows an interleaved pattern. Odd positions (1st, 3rd, 5th, 7th, 9th): 8, 14, 11, 14, 18 follow the pattern: +6, -3, +3, +4. Even positions (2nd, 4th, 6th, 8th, 10th): 2, 6, ?, 6, 12 follow +4, +3, +3, +6. The missing number is 9, completing both patterns.

If 1200 km is 2/5 of the total trip, the total trip distance is $1200 \times (5/2) = 3000$ km. The distance traveled by car is one-third of the total trip, which is $3000 / 3 = 1000$ km. The remaining distance traveled by train is $3000 - 1200 - 1000 = 800$ km, making 800Km the correct choice.

Let original speed be $v$ and distance be $D$. The delay of 2 hours is caused by traveling the remaining distance at $3/5v$. If the fault occurs 50 miles later, the delay is reduced by 40 minutes (2/3 hours). This means traveling 50 miles at $3/5v$ instead of $v$ takes 2/3 hours longer. Thus, $50/(3v/5) - 50/v = 2/3$, which gives $v = 50$ mph. Substituting $v$ back into the first scenario yields a total distance of 300 miles.

Let the number be 10a + b (where a and b are digits, a ≠ 0). The reversed number is 10b + a. Average = (10a + b + 10b + a) / 2 = 11(a + b) / 2 = 9. This gives a + b = 18/11. However, a and b must be integers (0-9), and 18/11 is not an integer. Therefore, no such two-digit number exists - the problem has no valid solution with whole number digits.

Net rate per minute = 1/20 + 1/30 - 1/40. Common denominator is 120. Rate = (6 + 4 - 3) / 120 = 7 / 120. Time taken = 120 / 7 = 17 1/7 minutes.