Q:

# Help with distance math problem

Nancy lives 12 3/5 km from school.she walks 3 1/3 km to the taxi rank and then takes the taxi to school.how far does Nancy travel by taxi?
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> 60 days ago

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Hi Annelisa.....

This problem may looks complicated at first glance but it's really just a subtraction equation.  The problem is subtracting fractions with different denominators.  The one rule you need to solve this problem is converting all equations to have the same common denominators before doing subtraction or addition.

Here goes:

12 3/5 km is equal to (12/1) + (3/5) = (60/5) + (3/5) = 63/5

3 1/3  km is equal to (3/1) + (1/3) = (9/3) + (1/3) = 10/3

Since Nancy lives 12 3/5 km from school and she walked 3 1/3 so the distance she travel by taxi would be the difference of these two distance.

So to subtract: (63/5) - (10/3)
You need to convert these two fractions to have the same denominators:

(63/5)(3/3) - (10/3)(5/5) =
(189/15) - (50/15) = 139/15
139/15 = 9 4/15

9 4/15 km is the distance Nancy traveled by taxi.

Hope this helps!

Vicki
> 60 days ago

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First you have to find the lowest common denominator for 3/5 and 1/3. In other words, the lowest number both 5 and 3 will divide into evenly which is 15. So you have to multiply the 5 in 3/5 by 3 to make it 15. Since you multiplied the 5 by 3 you have to multiply the 3 by 3 also. You turned 3/5 into 9/15. To change the 3 in 1/3 to 15 you multiply it by 5 so you also multiply the 1 by 5. You turned 1/3 into 5/15. Once you have fractions in both numbers with the same denominator you can then subtract the 3 5/15 that she walked from the 12 9/15 total distance she lives from school. When you subtract 3 5/15 from 12 9/15 you are left with 9 4/15km.
> 60 days ago

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The far does Nancy travel by taxi is:123/5-31/3=(123.3)-(5.31)/5.3
=214/15
> 60 days ago

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