Mr. Oliver has three pieces of string of length 50 cm, 90 cm and 100 cm. He wishes to cut the three pieces of string into smaller pieces of equal length with no remainder.
- What is the greatest possible length of each of the smaller pieces of string?
- How many of the smaller pieces of string of equal length can he get?
(a)
Factors of 50: 1, 2, 5, 10, 25, 50
Factors of 90:1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
Factors of 100:1, 2, 4, 5, 10, 20, 25, 50, 100
The highest common factor of 50, 90 and 100 is 10.
Greatest possible length of each smaller piece of string = 10 cm
(b)
Total length of string
= 90 + 100 + 50
= 240 cm
Number of smaller pieces
= 240 ÷ 10
= 24
Answer(s): 10 cm; (b) 24