Find the sum: [3(4/5)] + [2(3/10)] + [1(1/15)]

1.	Find the sum: [3(4/5)] + [2(3/10)] + [1(1/15)]
Answer» [3(4/5)] + [2(3/10)] + [1(1/15)] First convert each mixed fraction into improper fraction. We get, = [3(4/5)] = (18/5) = [2(3/10)] = (23/10) = [1(1/15)] = (16/15) Then, (19/5) + (23/10) + (16/15) LCM of 5, 10, 15 = 30 Now, let us change each of the given fraction into an equivalent fraction having 30 as the denominator. = [(19/5) × (6/6)] = (114/30) = [(23/10) × (3/3)] = (69/ 30) = [(16/15) × (2/2)] = (32/30) Now, = (114/30) + (69/30) + (32/30) = [(114+69+32)/30] = (215/30) = [7 (5/30)] = [7 (1/5)]

Answer»

[3(4/5)] + [2(3/10)] + [1(1/15)]

First convert each mixed fraction into improper fraction.

We get,

= [3(4/5)] = (18/5)

= [2(3/10)] = (23/10)

= [1(1/15)] = (16/15)

Then,

(19/5) + (23/10) + (16/15)

LCM of 5, 10, 15 = 30

Now, let us change each of the given fraction into an equivalent fraction having 30 as the denominator.

= [(19/5) × (6/6)] = (114/30)

= [(23/10) × (3/3)] = (69/ 30)

= [(16/15) × (2/2)] = (32/30)

Now,

= (114/30) + (69/30) + (32/30)

= [(114+69+32)/30]

= (215/30)

= [7 (5/30)]

= [7 (1/5)]

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