Mathematics
Grade5
Easy

Question

Brian's team practiced soccer for 2 over 3 of an hour on Wednesday, and for 5 over 6 of an hour on Friday. His team practiced _________hours for soccer altogether.

  1. 1 half
  2. 7 over 9
  3. 7 over 6
  4. 3 over 2

hintHint:

We are given that Brian's team practiced soccer for 2 over 3 of an hour on Wednesday, and for 5 over 6 of an hour on Friday and we need to find how much in total they have practiced and this can be found by adding the fraction and for that we have to make denominator same this can be done by taking LCM. LCM is the least common multiple of two integers. We will use the prime factorization method to find the LCM. In prime factorization method the simply put the factor of each number and then for LCM we take product of factors by considering the repeated factors only one time with highest exponent.

The correct answer is: 3 over 2


    Step1: First we will write all the fractions and then we will write the denominator and then we will use prime factorization method to add the fraction.
    Fractions are:2 over 3 and 5 over 6.
    In this question the denominators are 3 and 6.
    Factors of 3 are:3 to the power of 1
    Factors of 6 are:2 to the power of 1 cross times 3 to the power of 1
    Step2:  For calculating the LCM we will multiply the factors by considering the common factors only one time with the highest exponent.
    In this case the factor 3 is common in both 3 and 6 so we will multiply it only once with the highest exponent
    LCM (3,6) = 2 to the power of 1 cross times 3 to the power of 1
    LCM (3,6) = 6
    Step3: Now what we do is divide the LCM by each denominator and multiply the numerator by quotient and then add them
    So, 6 over 3 equals 2
    and 6 over 6 equals 1
    So, we will multiply the numerator of 3 by 2 and numerator of 6 by 1.
    fraction numerator 2 cross times 2 plus 5 cross times 1 over denominator 6 end fraction equals 9 over 6
    So, Brian's team practiced a total of 9 over 6  which is equivalent to 2 over 3 of hours as both numerator and denominator are divisible by 3.

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