Maths Riddles

526485
+197485
=723970

50%

The vents in this case are independent. If you form the possible outcomes, they will be as follows
HHH
HTT
HHT
HTH
TTT
TTH
THT
THH

Now among these outcomes, four will give the required results. Thus the probability is 50%.

4%

Let us assume that the initial volume of the water drop was k.

k - (1/4) k = (3/4) k

(3/4) k - (1/5) [(3/4) k] = (3/4) k * [1 - (1/5)] = (3/4) k * (4/5) = (3/5) k

(3/75) k = (1/25) k

1/25 = 4%

Thus, the answer is four percent.

7.5 Miles

(Speed of Bobby) = 60 MPH = 1.2 x (Speed of Wilbur) [MPH] = 1 M/ min

Thus (Speed of Wilbur) = 60/1.2 [MPH] = 50 [MPH] = 5/6 M/ min

(Time of Wilbur) = D [M] / (Speed of Wilbur) [M/min] = D/ (Speed of Wilbur) [min] = D/(5/6) min = 6D/5 min = 1.2 D min

(Time of Bobby) = D [M] / (Speed of Bobby) [M/min] = D/ (Speed of Bobby) [min] = D/(1) min = D min

Since (Time of Wilbur) - (Time of Bobby) = 1.5 min

Then (1.2 - 1) D min = 0.2D min = 1.5 min

Therefore D = 1.5/0.2 M = 1.5*5 = 7.5 Miles

(2^x)(30^3) = (2^3) (3^3) (4^3) (5^3)

(2^x)(30^3) = (120^3)

(2^x) = (120^3)/ (30^3)

(2^x) = (4^3)

(2^x) = 64

ln(2^x) = ln(64)

x ln(2) = ln(64)

x = ln(64)/ln(2)

x = 6

Thus x = 6.

infinite speed

Let us denote distance with d, time to get there by T, time to get back by t and the speed while travelling back with R.

Now by applying the simple time, distance and speed formula:
d = 20T
T = d/20

d = Rt
t = d/R

As you can see, we have the equations for both T and t now. We can now finally derive and equation for the round trip.

2d = 40(T + t)
2d = 40(d/20 + d/R)
2d = 40d(1/20 + 1/R)
1 = 20(R/20R + 20/20R)
20R = 20(R+20)
R = R + 20

You can clearly see that it is a paradox as to make the average speed 40 mph, you will have to travel back at an infinite speed. The faster you return, the quicker you can make it although your faster speed will offer a lesser impact on the average speed.

If you made the return trip instantly, it would be comparable to traveling double the distance in the same time that was taken on the one way trip.

Thus to reach the average speed of 40 mph, you must return with an infinite speed.

two hours and twelve minutes.

If you know about clocks, you must be knowing the fact that the hour hand is exactly on the minute mark five times each hour i.e. on the hour, twelve minutes past the hour, twenty four minutes past the hour, thirty six minutes past the hour and forty eight minutes past the hour.

Now let us suppose that X are the number of hours and Y are the number of minutes past the hour.
When the hour hand is on minute mark,
Position of hour hand:
5X + Y/12
Position of the minute hand:
Y

Now when the first observation is taken,
Y = Y = 5X + Y/12 + 6
This is also equivalent to 60X = 11Y - 72.

But from what we know, Y can only be equal to 0, 12, 24, 36 or 48. Thus keeping that in mind, the only value possible for X and Y are 1 and 12 respectively. This implies that the time is 1:12 on the first observation.

In the second observation, the equation will be:
60X = 11Y - 84
The possible values for X and Y here are 3 and 24 respectively. This implies that the time is 3:24.

Now time that has elapsed between 1:12 and 3:24 = two hours and twelve minutes.

72

The company owns seventy two buses at present.

Let us assume that they have B buses right now and S space before the expansion. Now the space is enough for
S = B - 12 ----- [1]

When they expand, they will have more space which will be enough for
S + 0.4 x S = B + 12 ------ [2]
=> 1.45 = B + 12
Also, B = 1.45 - 12 ----- [3]

Now we can rewrite the equation [1] as
B = S + 12 ----- [4]

Equating the third and fourth equation, we get
1.4 S - 12 = S + 12

Subtract S from both sides
0.4 S - 12 = 12

Now add 12 to both sides
0.4 S = 24

Multiply by ten and divide by four on both sides
S = 60

Put the value in fourth equation
B = 72

18

Assume that the total number of can before distribution were 3*C

After distribution, each one of them had C cans then.

Each one of them drank four cans and thus (C-4) cans remained.

Now according to the questions we can frame the following equation:

3 * (C-4) = C
(3 * C) - 12 = C
2* C = 12
C = 6

Thus, the total cans before distribution were 3 * C = 3 * 6 = 18.

119

Let the number of apples = X
X / 2 => Remainder = 1
X / 3 => Remainder = 2
X / 4 => Remainder = 3
X / 5 => Remainder = 4
X / 6 => Remainder = 5
X / 7 => Remainder = 0

Therefore, X is divisible by 7.

Whenever X is divided by any number less than 7, the remainder is 1 less than the divisor.
=> X + 1 is the LCM of 2, 3, 4, 5 and 6.

Now LCM of 2, 3, 4, 5 and 6 = 60
But 60 - 1 = 59 is not divisible by 7

60 * 2 = 120
120 - 1 = 119 which is divisible by 7

This means that X + 1 = 120 or X = 119

Number of minimum number of apples in the basket = 119