Question #8d811

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Question #8d811

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Answer 1 · 2016-07-01T03:33:07

The right answer is (b) $69$ $69$

Explanation:

Let the height of five men be $H_{1}$ $H_1$ , $H_{2}$ $H_2$ , $H_{3}$ $H_3$ , $H_{4}$ $H_4$ and $H_{5}$ $H_5$ .

We know that

(1) $\frac{H_{1} + H_{2} + H_{3} + H_{4} + H_{5}}{5} = 68$ $(H_1+H_2+H_3+H_4+H_5)/5 = 68$

(2) $H_{1} = 70$ $H_1 = 70$

(3) $\frac{H_{2} + H_{3} + H_{4}}{3} = 67$ $(H_2+H_3+H_4)/3 = 67$

Our task is to determine $H_{5}$ $H_5$

SOLUTION

From (1):
(4) $H_{1} + H_{2} + H_{3} + H_{4} + H_{5} = 5 \cdot 68 = 340$ $H_1+H_2+H_3+H_4+H_5 = 5 * 68 = 340$

From (3):
(5) $H_{2} + H_{3} + H_{4} = 3 \cdot 67 = 201$ $H_2+H_3+H_4 = 3*67 = 201$

Substitute (2) and (5) into (4) getting
$70 + 201 + H_{5} = 340$ $70+201+H_5 = 340$
from which follows
$H_{5} = 340 - 70 - 201 = 69$ $H_5 = 340-70-201 = 69$

CHECK the average height of all five men:
$\frac{H_{1} + H_{2} + H_{3} + H_{4} + H_{5}}{5} = \frac{70 + 201 + 69}{5} = \frac{340}{5} = 68$ $(H_1+H_2+H_3+H_4+H_5)/5 = (70+201+69)/5 = 340/5=68$

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Question #8d811

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