Rectangular Parallelepiped Problem, 4

Category: Solid Geometry

"Published in Vacaville, California, USA"

Compute the cost of the lumber necessary to resurface a foot-bridge 16 ft. wide, 150 ft. long with 2-in. planks, if lumber is $40 per 1000 board feet. Neglect waste. (One board foot = 1 ft. by 1 ft. by 1 in.)

Solution:

To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life

The volume of a rectangular parallelepiped is given by the formula

Substitute the values of length, width, and height of a lumber to resurface a foot-bridge, we have 

Therefore, the cost of the lumber necessary to resurface a foot-bridge is

 

Rectangular Parallelepiped Problem, 3

Category: Solid Geometry, Physics

"Published in Vacaville, California, USA"

Counting 38 cu. ft. of coal to a ton, how many tons will a coal bin 19 ft. long, 6 ft. wide, and 9 ft. deep contain, when level full?

Solution:

To illustrate the problem, it is better to draw the figure as follows 

Photo by Math Principles in Everyday Life

The volume of a rectangular parallelepiped is given by the formula

Substitute the values of length, width, and height of a bin, we have

The density of a substance is given by the formula

where ρ is the density, W is the weight, and V is the volume of a substance respectively. 

Therefore, the weight of a coal in a bin is

Finding Missing Digit - Divisibility Rule, 14

Category: Arithmetic

"Published in Vacaville, California, USA"

Find the missing digit so that it becomes divisible by 15 for

a. 13?2
b. 337?

Solution:

a. Consider the given number

A number is divisible by 15 if it is both divisible by 3 and 5. The given number is not divisible by 5 because the last digit is not 5 or 0. Because of this, the given number is not divisible by 15. The multiples of 15 ends with 5 or 0 always.There's nothing that we can do in order to become divisible by 15 since the last digit of a given number is not 5 or 0. You can assign any number to the missing digit but still, the given number will never become divisible by 15.

b. Consider the given number

A number is divisible by 15 if it is both divisible by 3 and 5. Since the last digit of a given number is missing, then we can assign 5 or 0. A number is divisible by 5 if the last digit ends with 5 or 0 and a number is divisible by 3 if the sum of the digits is a multiple of 3. Let's do the test for 5 and 0 as their last digit.

If the last digit is 0, then the sum of the digits is

Since 4 is not a multiple of 3, then the given number is not divisible by 3. 0 is not the last digit of a given number. Let's do the test again for the other last digit which is 5.

If the last digit is 5, then the sum of the digits is

Since 9 is a multiple of 3, then the given number is divisible by 3. 5 is the last digit of a given number. Because of this, the given number is divisible by 15. Therefore, the possible number is only 3375.