Match the equivalent fractions and write two more for each.
(i)
(a) ![]()
(ii)
(b) ![]()
(iii)
(c) ![]()
(iv)
(d) ![]()
(v)
(e) ![]()
i. We have,
![]()
![]()
Hence,
(i) and (d) are pairs
Two more fractions are: ![]()
ii. We have,
![]()
![]()
Hence,
(ii) and (e) are pairs
Two more fractions are: ![]()
iii. We have,
![]()
![]()
Hence,
(iii) and (a) are pairs.
Two more fractions are: ![]()
iv. We have,
![]()
![]()
Hence,
(iv) and (c) are pairs.
Two more fractions are: ![]()
v. We have,
![]()
![]()
Hence,
(v) and (b) are pairs.
Two more fractions are: ![]()