Find:
(a) 1/2 of (i)
(ii) ![]()
(b) 5/8 of (i)
(ii) ![]()
(a) We have,
(i)
of 2![]()
=
× 2![]()
=
× ![]()
= ![]()
= 1![]()
Also,
(ii)
of 4![]()
=
× 4![]()
=
× ![]()
= ![]()
= 2![]()
(b) We have,
(i)
of 3![]()
=
× 3![]()
=
× ![]()
= ![]()
= 2![]()
Also,
(ii)
of 9![]()
=
× 9![]()
=
× ![]()
= ![]()
= 6![]()
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