Mathematics
Grade6
Easy

Question

Robert, a milkman, has 7 fraction numerator 2 over denominator 3 end fraction liters of milk. He wants to pour the milk into 5 bottles equally, then find the quantity of milk each bottle contains.

  1. 1 fraction numerator 8 over denominator 15 end fraction liters    
  2. 1 fraction numerator 7 over denominator 15 end fraction liters    
  3. 2 fraction numerator 8 over denominator 15 end fraction liters    
  4. 2 fraction numerator 7 over denominator 15 end fraction    

Hint:

To find the quantity of milk for each bottle , we divide the total quantity of milk by number of bottles to be poured.

The correct answer is: 1 fraction numerator 8 over denominator 15 end fraction liters


    Total quantity of milk Robert has = 7 fraction numerator 2 over denominator 3 end fraction liters ; Number of bottles = 5
    Quantity of milk in each bottle =7 2 over 3 space divided by space 5

    We can multiply the reciprocal of the denominator with the numerator to find the value of the problem.
    7 2 over 3 divided by 5 equals fraction numerator 3 cross times 7 plus 2 over denominator 3 end fraction cross times 1 fifth liters
    equals space 23 over 15 space equals 1 8 over 15 liters per bottle.
    Therefore, Quantity of milk in each bottle is 1 8 over 15 spaceliters.

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