Home » Past Questions » Mathematics » If p : q = \(\frac{2}{3}\) : \(\frac{5}{6}\) and q : r = \(\frac{3}{4}\) : \(\frac{1}{2}\), find p : q : r...

If p : q = \(\frac{2}{3}\) : \(\frac{5}{6}\) and q : r = \(\frac{3}{4}\) : \(\frac{1}{2}\), find p : q : r...


Question

If p : q = \(\frac{2}{3}\) : \(\frac{5}{6}\) and q : r = \(\frac{3}{4}\) : \(\frac{1}{2}\), find p : q : r

Options

A) 12 : 15 : 10

B) 12 : 15 : 16

C) 10 : 15 : 24

D) 9 : 10 : 15

The correct answer is A.

Explanation:

If p : q = \(\frac{2}{3}\) : \(\frac{5}{6}\), then the sum S1 of ratio = \(\frac{2}{3}\) + \(\frac{5}{6}\) = \(\frac{9}{6}\)
If q : r = \(\frac{3}{4}\) : \(\frac{1}{2}\), then the sum S2 of ratio = \(\frac{3}{4}\) + \(\frac{1}{2}\) = \(\frac{5}{4}\)
Let p + q = T1, then
q = (\(\frac{5}{6} \div \frac{9}{6}\))T1 = (\(\frac{5}{6} \times \frac{6}{9}\))T1 = \(\frac{5}{9}\)T1
Again, let q + r = T2, then
q = (\(\frac{3}{4} \div \frac{5}{4}\))T2 = (\(\frac{3}{4} \times \frac{4}{5}\))T2 = \(\frac{3}{5}\)T2
Using q = q
\(\frac{5}{9}\)T1 = \(\frac{3}{5}\)T2
5 x 5T1 = 9 x 3T2
\(\frac{T_1}{T_2}\) = \(\frac{9 \times 3}{5 x 5}\) = \(\frac{27}{5}\)
Giving that, T1 = 27 and T2 = 25
P = (\(\frac{2}{3} \div S_1\))T1 = (\(\frac{2}{3} \div \frac{9}{6}\))T1
= (\(\frac{2}{3} \times \frac{6}{9}\))27 = 12
q = (\(\frac{5}{6} \div S_1\))T1 = (\(\frac{5}{6} \div \frac{9}{6}\))T1
= (\(\frac{5}{6} \times \frac{6}{9}\))27 = 15
and r = (\(\frac{1}{2} \div S_2\))T2 = (\(\frac{1}{2} \div \frac{5}{4}\))T2
= (\(\frac{1}{2} \times \frac{4}{5}\))25 = 10
Hence p : q : r = 12: 15 : 10

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Dicussion (1)

  • If p : q = \(\frac{2}{3}\) : \(\frac{5}{6}\), then the sum S1 of ratio = \(\frac{2}{3}\) + \(\frac{5}{6}\) = \(\frac{9}{6}\)
    If q : r = \(\frac{3}{4}\) : \(\frac{1}{2}\), then the sum S2 of ratio = \(\frac{3}{4}\) + \(\frac{1}{2}\) = \(\frac{5}{4}\)
    Let p + q = T1, then
    q = (\(\frac{5}{6} \div \frac{9}{6}\))T1 = (\(\frac{5}{6} \times \frac{6}{9}\))T1 = \(\frac{5}{9}\)T1
    Again, let q + r = T2, then
    q = (\(\frac{3}{4} \div \frac{5}{4}\))T2 = (\(\frac{3}{4} \times \frac{4}{5}\))T2 = \(\frac{3}{5}\)T2
    Using q = q
    \(\frac{5}{9}\)T1 = \(\frac{3}{5}\)T2
    5 x 5T1 = 9 x 3T2
    \(\frac{T_1}{T_2}\) = \(\frac{9 \times 3}{5 x 5}\) = \(\frac{27}{5}\)
    Giving that, T1 = 27 and T2 = 25
    P = (\(\frac{2}{3} \div S_1\))T1 = (\(\frac{2}{3} \div \frac{9}{6}\))T1
    = (\(\frac{2}{3} \times \frac{6}{9}\))27 = 12
    q = (\(\frac{5}{6} \div S_1\))T1 = (\(\frac{5}{6} \div \frac{9}{6}\))T1
    = (\(\frac{5}{6} \times \frac{6}{9}\))27 = 15
    and r = (\(\frac{1}{2} \div S_2\))T2 = (\(\frac{1}{2} \div \frac{5}{4}\))T2
    = (\(\frac{1}{2} \times \frac{4}{5}\))25 = 10
    Hence p : q : r = 12: 15 : 10

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