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# Help!

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Suppose a is directly proportional to b, but inversely proportional to c. If a = 2 when b = 5 and c = 9, then what is c  when b =3?

Aug 23, 2018

#1
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You can first find the second $$a$$, by setting up an equation and solving it. We can make the equation $$\frac{2}{5} = \frac{a}{3}$$, which we can solve using cross-multiplication. Then, we get $$6 = 5a$$, which gives us $$a = \frac{6}{5}$$. So if $$a = \frac{6}{5}$$, and $$a$$ is inversely proportional to $$c$$, than we can see that $$\frac{6}{5} \times c$$ must equal $$18$$, because $$2 \times 9 = 18$$$$\frac{18}{\frac{6}{5}} = 18 \times \frac{5}{6} = 15$$. So $$c = 15$$.

- Daisy

Aug 23, 2018
#2
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2 = 5k
k=2/5 constant of direct proportionality. But:............................(1)
2 =k/9
k=18 constant of inverse proportionality. But when:..................(2)
b =3, from (1) above:
a =2/5 x 3=6/5 = 1.2.  Using (2) above for the new value of a:
1.2 =18 / c
c = 18 / 1.2
c =15 when b = 3

Aug 24, 2018
edited by Guest  Aug 24, 2018
#3
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Look at this problem in a much simpler and intuitive way as follows:

When "b" = 5 goes to "b"= 3, that is a direct decrease of 3/5 =0.6, when c=9

And since "b" is directly proportional to "a" and inversely proportional to "c", it follows that:

New "c" = 9 x 1/0.6 =15

Aug 24, 2018