#181 - Impossible Math Problem

Suppose you drive to a picnic spot at 20 mph, then how fast you must go while returning back home on the same route such that your average speed becomes 40 mph?

Impossible Math Problem
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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.

#182 - Interesting Math Problem

Alex and Martin are planning on a vacation. Alex says, "It will be better if we take the train to the hotel. We will reach faster."
To this, Martin says, "No. You are wrong. The train stops at halfway to the hotel and then we will have to walk the rest of the way. We should take the bikes to the hotel directly."
Alex says, "You are wrong."

Now, Alex takes the train while Martin takes the bike. The train travels with four times the speed of bike. The bike travels at twice the speed of walking.

Who do you think will reach the hotel first?

Interesting Math Problem
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Martin

S = total distance to the hotel.
V = walking speed.
2V = bike speed.
8V = train speed.
A = time taken by Alex
M = time taken by Martin

A = (S/2)/8V + (S/2)/V = S/16V + S/2V
M = S/2V

Now if you notice, A exceeds M by S/16V. Therefore, Martin will reach the hotel first.

#183 - Math Algebra Equation Problem

Can you find out the value of x in the following equation?

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

Math Algebra Equation Problem
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(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.

#184 - Interesting Math Question

Bobby and Wilbur decided to take their respective car out of the garage and race. None of them cheated and they both stood at the start time and decided to cover a distance in full throttle. The first to reach the mark was to be declared the winner.

Upon reaching the finishing mark, they found out that Bobby car is 1.2 times faster than Wilbur. Now, Wilbur had reached the mark about 1 minute and 30 seconds later than Bobby. Bobby car reached the mark at 60 MPH on average.

Can you calculate the distance between the starting mark and the final mark with the help of the given data?

Interesting Math Question
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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

#185 - Hard Guess My Number Puzzle

One night, I thought of ways that can be used for creating a palindrome. So I decided that I will turn into a larger number by adding the reversed digits to the original number and keep doing it till I finally obtained a palindrome.

I am not sure if this process will always result in a palindrome eventually but I was able to produce a four digit palindrome. Can you guess my starting number?

Hard Guess My Number Puzzle
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My starting number was 192.
192 + 291 = 483
483 + 384 = 867
867 + 768 = 1635
1635 + 5361 = 6996

#186 - Can You Solve This Math Problem

A rain drop fell from one leaf to another leaf and lost 1/4th of its volume. It then fell to another leaf and lost 1/5th of the volume. It again fell on another leaf and lost 1/5th of the volume.

This process kept repeating till it fell on the last leaf losing 1/75th of its volume.

Can you calculate the total percentage of loss from the initial volume when the drop has fallen to the last leaf accurate up to two decimal places?

Can You Solve This Math Problem
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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.

#187 - Coin Probability Problem

Three fair coins are tossed in the air and they land with heads up. Can you calculate the chances that when they are tossed again, two coins will again land with heads up?

Coin Probability Problem
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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%.

#188 - Math Algebra Problem

You are given with the following sum. Each of the letters can be decoded as a digit. If we tell you that D = 5, then can you solve it entirely?

DONALD
+GERALD
=ROBERT

Math Algebra Problem
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526485
+197485
=723970

#189 - Sports Logic Puzzle

A game is being played where eight players can last for seventy minutes. Six substitutes alternate with each player in this game. Thus, all players are on the pitch for the same amount of time including the substitutes.

For how long is each player on the pitch?

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40 minutes

Total players = 8 + 6 = 14
Now, the question has told us that only eight players are present on the field at one time.
70 * 8 = 560 min in total.
Now dividing that with 14, we can conclude that each player was there on the pitch for 40 minutes (560/14)

#190 - Find The Number Riddle

I have thought of a number that is made up by using all the ten digits just once. Here are a few clues for you to guess my number:

First digits is divisible by 1.
First two digits are divisible by 2.
First three digits are divisible by 3.
First four digits are divisible by 4.
First five digits are divisible by 5.
First six digits are divisible by 6.
First seven digits are divisible by 7.
First eight digits are divisible by 8.
First nine digits are divisible by 9.
The number is divisible by 10.

Can you find out the number ?

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The number is 3816547290