# Mixed Word Problem5

In this page mixed word problem5 you can find the question as well as solution for this question with detailed explanation.Like this you can see every problems with detailed solution and explanation.

Question:

The total weight of 4 girls is 155 ⅕ . If three of them weighs 20 ⅕  kg,44 ½ kg, and 30 ⅓ kg.Find the weight o the remaining girl.

Basic idea:

In this problem we have total weight of 4 girls that is 155 ⅕ .If the weight of three girls is 20 ⅕  kg,44 ½ kg, and 30 ⅓ kg.Now we need to find the weight o the remaining girl.So let u consider the weight of the remaining girl is x.If we add the given three weights with x we will get  155 ⅕ .Then we can easily solve for x.

Total weight of 4 girls = 155 ¼

Sum of the weight of three girls   = (20 ⅕ + 44 ½ + 30 ⅓ )

=  (100+1)/5 + (88+1) /2 + (90+1)/3.

=  101/5 + 89/2 + 91/3

The denominators of the three fractions are not same.So we have to take L.C.M

Here L.C.M is 5 x 2 x 3 = 30

=  (101/5) x (6/6) +  (89/2) x (15/15) + (91/3) x(10/10)

=  606/30 + 1335/30 + 910/30

= (505 + 1335 + 910) / 30

=  2851 /30

Weight of remaining girl  = 155 ⅕ - 2851 /30

=  (620+1) /5 - 2851/30

=  621/5 - 2851/30

L.C.M = 30

= (606/5) x (6/6) -(2851/30)

= (3636 -2851)/30

= 785/30