Mathematics

When walking at 80% speed, the man takes longer to reach his office. Let the usual time be T minutes. At 80% speed, time taken is T/0.8 = 1.25T. The delay is 1.25T - T = 0.25T, which equals 10 minutes. Solving 0.25T = 10 gives T = 40 minutes. However, this contradicts the initial correction, so let's re-express carefully: If usual speed is S and usual time is T, distance D = S × T. At 80% speed, speed is 0.8S, time taken is D / (0.8S) = T / 0.8 = 1.25T. The delay is 1.25T - T = 0.25T = 10 minutes, so T = 40 minutes. Therefore, the usual time is 40 minutes, which corresponds to option C. The original correct answer C is correct.

Given a:b = 3:5, let a=3x and b=5x. After adding 2, (3x+2)/(5x+2) = 2/3. Cross-multiplying gives 9x + 6 = 10x + 4, so x = 2. Thus, b = 5x = 10.

Let the third number be x. The first number is 20% more than x, which is 1.2x. The second number is 50% more than x, which is 1.5x. The ratio of the two numbers is therefore 1.2x : 1.5x = 1.2 : 1.5 = 12 : 15 = 4 : 5. Hence, the correct ratio is 4 : 5, which corresponds to option C.

Given the equation XY = kZ, and the initial ratio X:Y:Z = 2:3:4 with X = 4, we find the values of Y and Z. Since X corresponds to 2 parts and equals 4, each part is 2. Therefore, Y = 3 parts = 6 and Z = 4 parts = 8. Using XY = kZ, we get 4 * 6 = k * 8, so k = 3. When Y changes to 12 and Z remains 8, the equation becomes X * 12 = 3 * 8 = 24. Solving for X gives X = 24 / 12 = 2. Thus, the new value of X is 2, which corresponds to option A.

Let the number of 25p, 10p, and 5p coins be x, 2x, and 3x respectively. The total value is 25x + 10(2x) + 5(3x) = 25x + 20x + 15x = 60x pence. Since Rs. 30 = 3000 pence, 60x = 3000, so x = 50. Therefore, the number of 5p coins is 3x = 150.

The fourth proportion to 3, 7, and 9 is found by solving the proportion 3 : 7 = 9 : x. Cross-multiplying gives 3x = 7 × 9, so x = 63.

The shares of A, B, C, and D are in the ratio 5:2:4:3. Let the common multiplier be x. Then C's share is 4x and D's share is 3x. Given C gets Rs. 1000 more than D, we have 4x - 3x = 1000, so x = 1000. Therefore, B's share is 2x = 2 * 1000 = Rs. 2000. Among the options, Rs. 2000 corresponds to option C, so the correct answer is C.

Let the speed of the boat in still water be b and the speed of the current be c. Let the distance between X and Y be d. Since Y is equidistant from X and Z, distance XZ = 2d. Time taken from X to Y and back is 10 hours: (d/(b-c)) + (d/(b+c)) = 10 Time taken from X to Z is 4 hours: 2d/(b-c) = 4 From the second equation, d = 2(b-c). Substitute d in the first equation: (2(b-c)/(b-c)) + (2(b-c)/(b+c)) = 10 => 2 + 2(b-c)/(b+c) = 10 => 2(b-c)/(b+c) = 8 => (b-c)/(b+c) = 4 This is not possible as the ratio must be less than 1. Re-examining the problem, the time from X to Z is 4 hours downstream, so speed downstream is (b+c), and time = distance/speed = 2d/(b+c) = 4. So, 2d/(b+c) = 4 => d = 2(b+c). Now substitute d in the first equation: (d/(b-c)) + (d/(b+c)) = 10 => (2(b+c)/(b-c)) + (2(b+c)/(b+c)) = 10 => 2(b+c)/(b-c) + 2 = 10 => 2(b+c)/(b-c) = 8 => (b+c)/(b-c) = 4 Let r = b/c. Then (r+1)/(r-1) = 4 => r+1 = 4r - 4 => 3r = 5 => r = 5/3 Therefore, the ratio of the speed of the boatman in still water to the speed of the current is 5:3.

Let the distance travelled on foot be x km. Then the distance travelled by bicycle is (61 - x) km. The time taken on foot is x/4 hours, and the time on bicycle is (61 - x)/9 hours. The total time is 9 hours, so x/4 + (61 - x)/9 = 9. Solving this equation gives x = 16 km, which matches option C.

A and B together complete 1/15 of the work per day, and B alone completes 1/20 per day. Therefore, A's work rate is 1/15 - 1/20 = 1/60 per day, so A alone can do the work in 60 days.