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Re: If p/3 = r/2 and r = 0.75s, what is the ratio of p to r to s? [#permalink]
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Arvind42 wrote:
Bunuel wrote:
If p/3 = r/2 and r = 0.75s, what is the ratio of p to r to s?

(A) 2 : 3 : 4
(B) 8 : 12 : 9
(C) 9 : 6 : 8
(D) 6 : 4 : 3
(E) 12 : 8 : 3


r=.75s multiply by 4 to obtain 4r=3s; p/3=r/2 to be multiplied by 6 to get 2p=3r

Multiply 4r=3s and 2p=3r by 3 and 4 respectively to get 8p=12r=9s.

IMO B


Eg: If 2a = 3b. a:b will be 3:2 NOT 2:3.
That’s why we can’t take ratio of above as 8:12:9.

Assume 8p=12r=9s=k (k is any constant)

—> p=k/8, r=k/12 and s=k/9
—> p:r:s = k/8:k/12:k/9 = 1/8:1/12/1/9
Multiply with constant 72 to the above ratio
—> p:r:s = 72*8:72*12:72*9 = 9:6:8.

Hope it’s clear.

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Re: If p/3 = r/2 and r = 0.75s, what is the ratio of p to r to s? [#permalink]
Bunuel wrote:
If p/3 = r/2 and r = 0.75s, what is the ratio of p to r to s?

(A) 2 : 3 : 4
(B) 8 : 12 : 9
(C) 9 : 6 : 8
(D) 6 : 4 : 3
(E) 12 : 8 : 3


given
p:r=3:2
and r=3:4
so
p:r:s
3:2:
:3:4
9:6:8
IMO C
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Re: If p/3 = r/2 and r = 0.75s, what is the ratio of p to r to s? [#permalink]
1
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Expert Reply
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Given that p/3 = r/2 and r = 0.75s and we need to find the value of p : r : s

To solve this problem we need to express all the variables in terms of one single variable first. We are doing this as when we take the ratio of p : r : s then we can cancel out the common variable

So, lets express all variables in terms of s.

r = 0.75s = \(\frac{3}{4}\)*s

\(\frac{p}{3}\) = \(\frac{r}{2}\)
=> p = \(\frac{3*r}{2}\) = \(\frac{3r}{2}\) = \(\frac{3}{2}\) * r = \(\frac{3}{2}\) * \(\frac{3}{4}\)*s = \(\frac{9}{8}\)s

=> p : r : s = \(\frac{9}{8}\)s : \(\frac{3}{4}\)*s : s = \(\frac{9}{8}\) : \(\frac{3}{4}\) : 1
= \(\frac{9}{8}\) : \(\frac{3*2}{8}\) : \(\frac{8}{8}\)
= 9 : 6 : 8

So, Answer will be C
Hope it helps!
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Re: If p/3 = r/2 and r = 0.75s, what is the ratio of p to r to s? [#permalink]
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