the graph shows that the equation has 2 positive roots and 1 negative root. So maximum positive roots for any given option should be two, ie no. of sign changes should be at least 2, its not in option 2 ,

hi, do you have any questions?

When the faster swimmer meets (overtakes) the slower swimmer for the first time, the faster swimmer would have covered 2 laps (one lap = 50m) more than the slower swimmer, and both are swimming in the same direction at this point of time. Let this time be t.
So, t/2 = t/3.25 + 2 => t = 10.4 minutes

So, the faster swimmer overtakes the slower swimmer for the first time at 10.4 minutes after the start of the race. Number of handshakes till then = 5 (they cross each other 4 times and meet one time)
So, number of handshakes by t = 31.2 minutes = 15
From this moment, the 5th handshake occurs at t = 41.6 minutes. But the faster swimmer would have stopped swimming by then.
So, number of handshakes from t = 31.2 to t = 40 is 4

Total number of handshakes = 19

Euler of 101 = 100

So, 7^100 mod 101 = 1
7 * 7^99 mod 101 = 1
Let 7^99 mod 101 = k
=> 7k mod 101 = 1

=> 7k = 101p + 1
p = 2 satisfies for k = 29
So, remaninder = 29

A ran 1000 m in the time that C ran 400 m. So, ratio of speeds of A and C is 5:2
Let the ratio of speeds of A, B and C be 5:x:2. So, speeds = 5k, xk and 2k

Time taken by B - Time taken by A = 1000/xk - 1000/5k = 60 seconds
Time taken by C - Time taken by B = 1000/2k - 1000/xk = 30 seconds
Equating for k, we get, 200*(5/x - 1)/60 = 500*(1-2/x)/30 => x = 2.5

So, speed of A : Speed of B = 5 : 2.5 = 2:1

In a kilometre race, P beats Q by 1 minute. Q beats R by 30 seconds. If Possible beats R by 200 Mt , find the time R takes to run the race?

Let the total distance be 2d.

Total time = d/10 + x/15 + (d-x)/20

x/15 = (d-x)/20 => x = 3d/7

Total time = d/10 + 3d/105 + 4d/140 = 11d/70

Average speed = Total distance/Total time = 2d/(11d/70) = 140/11 kmph

Hi, do you have any specific example in mind?

Since the decrease is uniform, the distance illuminated after 2 hours = 5 meters

Area initially illuminated = $$\pi * 9^2$$
Area illuminated after 2 hours = $$\pi * 5^2$$
Required area = $$81 \pi - 25 \pi = 56*22/7 = 176 m^2$$

Let the external angle X be 2a. So, angle XWY = a, angle YXZ = 180 - 2a
Let angle XYZ = 2b, so angle XYW = b.
Angle XZY = 180-(180-2a+2b) = 2(a-b) = 50 => a-b = 25
Angle XWY = 180-(180-2a+a+b) = a - b = 25 degrees
Required angle = 25 degrees

Can you explain the Angle XWY = 180-(180-2a+a+b) part?
Shouldn't it be equal to 180-(90-a+b) as external angle bisector of angle X should also bisect the internal angle X?

Finding a:

Case 1: All the boxes have 1 ball each => 7! ways
Case 2: One box has 0 balls and one box has 2 balls, all the other boxes have 1 ball each => 7C1 (for selecting the box with 0) * 6C1 (selecting the box with 1)* 7C2 (selecting 2 balls to be placed in the box) * 5! (distribution of the remaining balls in the 5 boxes) ways
Case 3: Case of (2 2 1 1 1 0 0) => 7C2 (for selecting 2 boxes that have 0 balls) * 5C2 (for selecting 2 boxes that will have 2 balls each)* 7C2 (balls to be placed in the first box)* 5C2 (balls to be placed in the second box) * 3! (distribution of the remaining 3 balls) ways
Case 4: Case of (2 2 2 1 0 0 0) => 7C3 (selecting 3 boxes that will have 0 balls each) * 4C3 (selecting the box with 1 ball) * 7C2 (selecting 2 balls) * 5C2 (selecting the next set of 2 balls) * 3C2 (selecting the third set of 2 balls) ways

Sum of these = 463680

Finding b: There in only 1 way in which 7 distinct balls can be put in 7 similar boxes such that each box has only 1 ball

So, a/b = 463680