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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
2 years, 5 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. · 2 years, 5 months ago

did the same thing.. · 2 years, 5 months ago

Approx 8 members · 2 years, 5 months ago