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Pre-RMO 2014/14

One morning, each member of Manjul's family drank an \(8\)- ounce mixture of coffee and milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Manjul drank \(\frac{1}{7}\) -th of the total amount of milk and \(\frac{2}{17}\) -th of the total amount of coffee. How many members are there in Manjul's family?


This note is part of the set Pre-RMO 2014

Note by Pranshu Gaba
3 years, 2 months ago

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Let the total number of members be \(k\), total amount of milk \(= x\) ounce, and total amount of coffee \(= y\) ounce.

We know that \(\dfrac{x}{7} + \dfrac{2y}{17} = 8\)

Therefore, \(17x + 14y = 8 \times 7 \times 17\)

Also, \(x + y = 8k \)

So, \( 3x + (14\times 8)k = 8 \times 7 \times 17\)

\(3x = 56( 17 - 2k)\)

\(\dfrac{x}{7} = \dfrac{8}{3} (17 - 2k)\)

It is given \( 0 < \dfrac{x}{7} < 8\)

\(0 < \dfrac{8}{3} (17 -2k) < 8\)

\( 0 < 17 - 2k < 3\)

\(0 > k - \dfrac{17}{2} > - \dfrac{3}{2}\)

\(\dfrac{17}{2} > k > 7\)

\(7 < k < 8.5\)

The only integral value of \(k\) satisfying this is \(k= 8\)

Therefore, there are \(\boxed{8}\) members in Manjul's family.

Pranshu Gaba - 3 years, 2 months ago

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did the same thing..

Abhishek Bakshi - 3 years, 2 months ago

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Approx 8 members

Mayyank Garg - 3 years, 2 months ago

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We can't say apprx 8 members as that might mean there are 7.99 or 8.01 members.

Siddhesh Naik - 3 years, 1 month ago

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