SOLUTION: A number, one fifth of that number, and one fourth of that number are added. The result is 58. What is the original number?
Algebra.Com
Question 1145211: A number, one fifth of that number, and one fourth of that number are added. The result is 58. What is the original number?
Found 2 solutions by Alan3354, greenestamps:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A number, one fifth of that number, and one fourth of that number are added. The result is 58. What is the original number?
---------
n*(1 + 0.2 + 0.25) = 58
Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!
The sum is a whole number, so 1/5 of the number and 1/4 of the number are almost certain to be both whole numbers.
That means the number is divisible by both 4 and 5; and that means it is divisible by 4*5=20.
Then since the sum is 58, the number is probably the largest multiple of 20 less than 58.
So try 40 as the number:
40+40/5+40/4 = 40+8+10 = 58
RELATED QUESTIONS
a number, one-fifth of that number, and one-sixth of that number are added. the result... (answered by josgarithmetic)
a number, one third of that number, and one fourth of that number are added. the result... (answered by mananth)
a number one third of that number and one fourth of that number are added. the result is... (answered by mananth)
a number, one fourth of that number, and one third of that number are added. the result... (answered by ankor@dixie-net.com)
A number, twice that number, and one-fourth of that number are added. The result is... (answered by checkley77)
A number, half of that number, and one-third of that number are added. The result is 22.... (answered by checkley75)
a number, half of that number, and one-third of that number are added. the result is 22.... (answered by edjones)
a number, half of that number and one third of that number are added the result is 22.... (answered by checkley77)
A number, half of that number, and one-third of that number are added. the result is 22.... (answered by Fombitz)