Some Interesting Math Problems Part Two

Some Interesting Math Problems Part Two

This hub is a sequel to my previously published hub "Some Interesting Math Problems". Here is another five interesting Math problems. Hope you enjoy it.

Problem Number One :

It costs $4.00 to have a shave at Arthur's Barber Shop and it cost $7.00 to have a haircut . One Sunday 14 people went to the shop and three of them had both a haircut and a shave. If Arthur earned $98.00 that day, how many did he shave?

Solution : Let X = number of people who had a haircut only

Y = number of people who had both a haircut and a shave

Z = number of people who had shave only

Since a haircut costs $7, a shave cost $4 and both $11 we have

98 = 7X + 11Y + 4Z equation (1) But Y = 3 and since X + Y + Z = 14, we get X + Z = 11 or X = 11 - Z We plug in the values of Y and X in equation (1) 98 = 7(11 - Z) + 11 (3) + 4Z so 98 = 77 - 7Z + 33 + 4Z Then 98 - 77 - 33 = - 3 Z, we get Z = 4 , the number had a shave only But 3 persons who had a haircut also had a shave so the total number of person who had a shave is 4 + 3 = 7

Problem Number Two :

2 dozens of apples and 4 dozens of mangoes cost 52 dollars. 3 dozens of mangoes and 4 dozens of bananas cost 59 dollars. If apples and bananas cost the same how much does each mango cost ?

Solution :

Since apples and bananas cost the same let their cost denoted b X and the cost of mangoes be Y, So we have (1) 2X + 4Y = 52

(2) 3Y + 4X = 59

From equation (1) we obtain X = (52 - 4Y)/ 2 and substitute this for X in (2) we get 3Y + 4(52 -4Y) (1/2) = 59 therefore 5Y = 45 or Y = 9

A dozen of mangoes cost 9 dollars therefore a piece of mango costs 9/12 = .75 dollars.

Problem Number Three :

Eight teams entered in the first conference of the NBA. If one team plays the other teams once and only once, how many games will be played in the first conference ?

Solution :

Let us say Team 1 is one of those who entered. Team 1 will play the 7 other teams once, so there are seven games, at least. Now team 2 already played Team 1, so it will only have 6 games remaining. Team 3 already played Teams 1 and 2, so it will have 5 more games. Team 4 already played Teams 1, 2 and 3 so it will play 4 more games. By the same reasoning , team 5 will have 3, Team 6 will have 2 and Team 7 will meet Team 8 for the last game. Therefore there are : 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 games in all.

Problem Number Four :

The Incredible Hulk can dig a tunnel 4 days longer than Spiderman can dig one. If they both work togerher , they can dig a tunnel in 4 4/5 days. How fast can Spiderman dig a tunnel ?

Solution : This is a Work Problem

Let X = number of days can spiderman dig a tunnel

X + 4 = number of hays can Incredible Hul dig a tunnel

Working equation : ( 1/X + 1/X + 4 ) 24/5 = 1

( 24/5X + 24/5X + 20 = 1 ) 5X(5X + 20 )

24(5X + 20 ) + 24 (5X ) = 25 X^2 + 100X

120X + 480 + 120X = 25X^2 + 100X

240X + 480 = 25X^2 + 100X

( 25X^2 - 140X - 480 = 0) 1/5 =

5X^2 28X -96 = 0 By facroring we get (5X + 12 ) ( X - 8 ) = 0

=== X - 8 = 0 = X = 8 Answer : 8 days

Problem Number Five :

A deadly epidemic has swept the small planet ZNNX - 8. On the first day 10 died, on the second day 20 died and on the third 40 died. If the progression of mortality continues this way , in how many days will the planet population be wiped out if the population of ZNNX -8 has 20,470 inhabitants. This is a problem involving geometric sequences.

(20,470 = (10 ) 2 ^(n-1) ) 1/10

2047 = 2^(n-1)

Log 2047 = log 2^(n-1)

(Log 2047 = (n - 1) log 2 ) 1/log 2

Log 2047/log 2 = n - 1

3.31/.3 = n - 1

11 = n - 1

N = 12 days.