wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Jenna, Carl, and Peter are organizing a surprise party for Martha. They are using ribbons to decorate the table on which the cake will be placed.
They take ribbons of length 8 m, 16 m, and 12 m, respectively. Now, they have to cut the ribbons into pieces of equal length.

Find the maximum possible length of each piece.

A
1 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
8 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 4 m
Length of the ribbons:
Jenna = 8 m
Carl = 16 m
Peter = 12 m

To find the maximum possible length of each piece, we will find the highest common factor (H.C.F.) of 8 m, 16 m, and 12 m.

Factors of 8 = 1, 2, 4, 8

Factors of 16 = 1, 2, 4, 8, 16

Factors of 12 = 1, 2, 3, 4, 6, 12

The common factors of 8, 16, and 12 are 1, 2, and 4.

The maximum possible length of each piece of ribbon of the same length = 4 m

Thus, the maximum possible length of each piece of ribbon is 4 m.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
HCF
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon