Latest Maths Riddles

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.

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

73rd

There is a simple logic behind solving this question. If the number of people are a power of 2, then the last person standing will be the person who started it.
Now, here we have 100 people. Thus, we will take the highest power of 2 which is less than the total number which is 64.
100 - 64 = 36.
This means that 36 people are killed.
If you go on with the series of even numbers, the 36 people will end at 72 (2, 4, 6, ..., 72). Thus the gun will be in the hand of 73rd person. Now the remaining people are 64 which is the power of 2. Thus the person who will survive in the end will be the 73rd person since he is starting it.

P = 20 units
Q = 12 units
R = 15 units

Explaining
If you have started looking for a complex formula, we inform you that this requires just the simple algebra equations.

Let the three sides of the box be P, Q and R.

Now we know the area of rectangle is l* b.

P * Q = 240
P * R = 300
Q * R = 180

Let us multiply these three equations:
(P*Q) (P*R) (R*S) = 240 * 300 * 180
P^2 * Q ^2 * R^2 = 12960000
(P*Q*R)^2 = 12960000
P*Q*R = 3600

Now we know that P*Q = 240

Thus R = 3600/240 = 15.

Similarly putting and solving for other variables, we get
P = 20 units
Q = 12 units
R = 15 units

23rd

The easiest way to find out without engaging in any formula would be to simply add them:
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 = 276

Thus, the birthday falls on the 23rd day of the month.

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.

1 green ball = $6
29 yellow balls = $87
70 black balls = $7

Detailed Explanation :
Let us denote the number of balls to be brought with a variable
p = green sponges
q = yellow sponges
r = black sponges

Now using a simple expression:
p + q + r = 100
6p + 3q + 0.1r = 100

Multiplying the first equation by 6 then 3 and subtracting the two equations, we get:

6p + 6q + 6r = 600
3p + 3q + 3r = 300

3q = 500 - 5.9r
3p = 2.9r - 200

Usually, we have three equations to solve three variables. However, we will manage with just two here because we know that p and q are non-negative integers.
If 3q >/= 0, then 500 - 5.9r >/= 0.
Using this, r Also, if 3p >/= 0, then 2.9r - 200 > 0. This implies that r >/= 68.97.

Now, the number of black balls must be an even dollar amount or we will land up with a fraction. Considering the expressions, you know that there can be just two integers that can satisfy the condition while lying in between 68.97 and 84.75.

If we take r as 80 and solve the equations (the last two equations), we will get p = 10.67 and q = 9.33. But p and q must be integer and not fraction. Thus you can cancel out 80.

If we take r as 70, we will get p = 1 and b = 29.

Therefore following are the number of balls that we must buy:
1 green ball = $6
29 yellow balls = $87
70 black balls = $7

No error is present in this proof and 10 is actually equal to 9.99999…. The proof is absolutely correct.

Let us find a bigger multiple first. Lets multiply the numerator and denominator by 9 in both the numbers.
Now, 1/3 = 9/27
And, 2/3 = 18/27

Clearly, 12/27 lies one third on the distance between the two numbers now.

As you can see, the person at D dies straight through the object directly. The object then travels further and the see saw thing kills the person at the C as it has spikes below that. Upon falling, over the see saw, the blue object lifts up and then kill the person at B.The solution ends here, but just for fun, The E stands for Ebola and thus the person at A dies as well.Thanks to 9gag for this gif.