Mixed Word Problem5

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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


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