#151 - Simple Maths Age Problem

Peter"s adolescence lasted for 1/6 of his life. He grew facial hair after 1/12 more. He married Susanne after 1/7 more of his life. She gave birth to a beautiful daughter after 5 years. The daughter lived 1/2 of what Peter lived. Peter died four years after his daughter.

Can you find out how long did Peter live?

84

Suppose that Peter lived for x years.

According to the question:
x/6 + x/12 + x/7 + 5 + x/2 + 4 = x
=> x = 84.

Thus, Peter lived for 84 years.

#152 - Distance Math Problem

There is a straight highway. Four different villages lie on that highway. The distance between them is different. The third village is 60km away from the first village; the fourth is 40 km away from the second; the third is 10 km near to the fourth that it is to the second.

Can you calculate the distance between the fourth and the first village ?

75

First Village = A
Second Village = B
Third Village = C
Fourth Village = D

Distance between B and D = 40km
=> Distance between B and C + Distance between C and D = 40km
=> Distance between B and C + (Distance between B and C - 10) = 40 km
=> 2 * Distance between B and C = 50km
=> Distance between B and C = 25km

Therefore distance between C and D = 15km

Thus, distance between A and D = 60 + 15 = 75km

#153 - Cost Pen Cap Puzzle

A pen along with its cap costs Rs. 5.50. If the cost of pen is Rs. 5, more than the cap, can you calculate the cost of the cap ?

Let the cost of cap = c
Cost of pen = (5 + c)

As per the condition

5.50 = 5 + c + c
2c = 0.5
C = 0.25

Thus the cost of the cap is 25 Paise

#154 - Smallest Number Problem

There is a number which when you multiply by 3 and subtract 2 from the result, then the resulting number is the reverse of the actual number.

What is the smallest number that stands true on the statement?

The number is 28.

28 * 3 - 2 = 82.

#155 - Find The Number Puzzle

There is a square of a particular number which when doubled, becomes 7 more than its quarter.

Can you find the number?

16

Let the number be m, then the square will be m^2
According to question:
2m = (m/4) + 7
=> 8m = m + 28
=> m = 4

m^2 = 4^2 = 16.

The required square is 16.

#156 - Distance Tyre Maths Problem

A man always keeps a spare tyre in his car. To make full use of all the five tyres, he changes the tyres in a manner that for a distance of 1, 00,000 km, each of them runs the same distance.

Can you calculate the distance traveled by each tyre in that journey ?

80,000.

We need to calculate the 4/5th of the total distance to solve the problem.
4/5 * 1, 00,000 km = 80, 000 km

Thus, each tyre traveled 80, 000 km.

#157 - Confusing Lost Chocolates Puzzle

Alex bought 150 chocolates but he misplaced some of them. His mother asked him how many chocolates were misplaced and he gave the following answer to her:
If you count in pairs, one remain
If you count in threes, two remain
If you count in fours, three remain
If you count in fives, four remain
If you count in sixes, five remain
If you count in sevens, no chocolate remain.

Can you analyze the statements and tell us how many chocolates were lost?

119 chocolates were lost.

#158 - Trick Time Riddle

Doctor advised his patient to take a tablet after every fifteen minutes. He gave him five tablets in total.

How much time will he take to have all the five tablets ?

60

The patient will be able to take all the tablets in an hour. This is because the first tablet will start the time.
1st tablet = 0 min
2nd tablet = 15 min
3rd tablet = 30 min
4th tablet = 45 min
5th tablet = 60 min

#159 - Find Ages Puzzle

Aditya is 45 years older than his son Atul. If you find similarities between their ages, both of their ages contains prime numbers as the digits. Also, Aditya age is the reverse of Atul age.

Can you find their ages ?

72,27

2, 3, 5 and 7 are the only single digit prime numbers.
Now we can list the possible combinations of age:
Aditya = 32, Atul = 23
Aditya = 52, Atul = 25
Aditya = 72, Atul = 27

Now, the third one meets our requirement.
72 - 27 = 45

Thus Aditya is 72 years old and his son is 27 years old.

#160 - Six Digit Number Puzzle

Calculate a six digit number that is smaller than 500000 such that the sum of its digits is 43 exactly.

499849

Let us begin with the maximum sum of a digit under 500000. Of course it will be the sum of 499999.
4 + 9 + 9 + 9 + 9 + 9 = 49.

Now, the required number must be close to it.
Rooting 499999, we get 707.106
Taking the square of 707, we get 499849.
4 + 9 + 9 + 8 + 4 + 9 = 43

Thus, the required number is 499849.