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Question

Hi, 

 

How do I draw the model for this question?  Appreciate very much.

 

At first  2/3 of Roy’s marbles is equal to 3/4 of Tom’s marbles. After Roy gives 48 marbles to Tom, both boys have an equal number of marbles. How many more  marbles does Roy have than  Tom at first?

Answer

will be better answered in ratio. 

make the both the numerator of the fractions the same, the denominator will represent the number of units they had at first. The ratio of R:T will be 9:8. Because of internal transfer,  make the before and after ratio the same. It will result in a single unit transfer. Hence, 1 unit = 48. 48 x 2= 96(ANS)

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At first  2/3 of Roy’s marbles is equal to 3/4 of Tom’s marbles. After Roy gives 48 marbles to Tom, both boys have an equal number of marbles. How many more  marbles does Roy have than  Tom at first?

2 Units Roy = 3 Units Tom

3 Units Roy – 48 = 4 Units Tom + 48

3 Units Roy – 96 = 4 Units Tom

Ans = 96 

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This is a Equal Fraction question. 

LCM

2/3 = 6/9

3/4 = 6/8

  Roy  Tom   Total   
Before    9u  8u    
 After  9u – 48  8u + 48    
         

9u – 48 = 8u + 48

9u – 8u = 48 + 48

1u = 96

 

 

 

 

 

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Not model but fractions method. 

2/3 R = 3/4 T

make numerator same:

6/9 R = 6/8 T

R has 9u and T has 8u. 

To be equal: R must give 1/2 u to T. 

So 1/2 u is 48

1 u is 96. 

It is likely a common MCQ-type question, the shorter working can save some time if kids can understand. 

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This would be what is taught in Primary school. Strongly advise students to get familiar recognising a common numerator problem such as this one.

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post-19795

This is quite a long answer …. as try to be as detailed as possible in the logic thinking.

 

Otherwise, alternatively……  you can also

After Step 1 , deduce that  ½u = 48  

and finally ……..  1 u = 2 x 48 = 96

 

Using model drawing will likely be more time consuming but is more easily understand for those strong in visual. For those students who can accept more abstract-type solution, using fraction or other methods to solve will probably be slightly faster.

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