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Math Ratio (1 Viewer)

Lukybear

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In one of the papers 2007 i think theres this question

Given a:b = 2:3
and b:c = 4:5

what is a:c

a 1:2
b 2:5
c 3:5
d 8:15

the answer is d but how??
isnt is 18:25...
 
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Lukybear said:
In one of the papers 2007 i think theres this question

Given a:b = 2:3
and b:c = 4:5

what is a:c

a 1:2
b 2:5
c 3:5
d 8:15

the answer is d but how??
isnt is 18:25...
lukybear, since you know what a and c is, you want to find the LCM of b (common value that they both divide in) - so it is 12. You times the 1st fraction by 4 then to get 8:12, and then times the second one by 3 to get 12:15.
Now that you've got a common value for b (12), the answer is 8:12. d.
 

lyounamu

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lookoutastroboy said:
lukybear, since you know what a and c is, you want to find the LCM of b (common value that they both divide in) - so it is 12. You times the 1st fraction by 4 then to get 8:12, and then times the second one by 3 to get 12:15.
Now that you've got a common value for b (12), the answer is 8:12. d.
you mean, 8:15
 

Lukybear

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but isnt what they asking 2/5 : 5/9

where did i read it wrong?
 

lyounamu

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Lukybear said:
but isnt what they asking 2/5 : 5/9

where did i read it wrong?
Just look at this and see where you got it wrong:

a:b = 2 and 3

a/b = 2/3

and b/c = 4/5

b = 3a/2 in other words.

Then b:c = 4 and 5

which means b/c = 4/5

sub b = 3a/2 here

3a/2c = 4/5
15a = 8c

So a =8 when c = 15
 

Lukybear

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aaa i get it now

was think of it differently

thxs lyounamu
 
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bored of sc

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Lukybear said:
In one of the papers 2007 i think theres this question

Given a:b = 2:3
and b:c = 4:5

what is a:c
This might be easier to understand:

Get b to the same value in each ratio.

Easiest way to do this is do b in the first ratio multiplied by b in the second ratio = 3 x 4 = 12. Now change each ratio appropriately.

a:b = 2:3 = 8:12 <--- x4 to each part of ratio.
b:c = 4:5 = 12:15 <--- x3 to each part of ratio.

Since b is the same in both ratios you can create one ratio a:b:c = 8:12:15.

Therefore a:c = 8:15.

Whoops did exactly the same method as lookoutastroboy.
 
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troongy

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^ Lol. That could work. But the other options are safer.
 

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