Question

Between the three persons, the weight of B and are 160% and 70% of the weight of A. If the difference between the weights of B and C is 36, then find the average weight of A,B and C?​

Answer 1

Answer:

Therefore,

B - C = 160 - 70 = 36 = 90 % of A

$\frac{90}{100} a = 36 \\ a = 36 \times \frac{10}{9} \\ a = 40$

Therefore,

A = 40

B = 160 % of 40

$b = \frac{160}{100} \times 40 \\ b = \frac{8}{5} \times 40 \\ b = 64$

C = 70 % of a

$c = \frac{70}{100} \times 40 \\ c = \frac{70}{5} \times 2 \\ c = 14 \times 2 \\ c = 28$

Therefore Average will be :-

=>( 28 + 64 + 40 ) / 3

=> 132 / 3

=> 44

ANSWER

44

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Answer 2

Given:

Between the three persons, the weight of B and C are 160% and 70% of the weight of A. If the difference between the weights of B and C is 36,

Solution:

Use given information and form the following equation,

$\[B = A \times \frac{{160}}{{100}}\]$

$\[\therefore 5B = 8A\]$                 -------(1)

And,

$\[C = A \times \frac{{70}}{{100}}\]$

$\[\therefore 10C = 7A\]$                 -------(2)

Perform, $\[\left( 1 \right) \times 2 - \left( 2 \right)\]$

$\[\begin{array}{l}10B - 10C = 16A - 7A\\ \Rightarrow 10\left( {B - C} \right) = 9A\end{array}\]$

Know that  the difference between the weights of B and C is 36,

$\[\begin{array}{l}\therefore 10 \times 36 = 9A\\ \Rightarrow A = 40\end{array}\]$

Put $\[A = 40\]$ in equation (1),

$\[\begin{array}{l} \Rightarrow 5B = 8 \times 40\\ \Rightarrow B = 64\end{array}\]$

Put $\[A = 40\]$ in equation (2),

$\[\begin{array}{l} \Rightarrow 10C = 7 \times 40\\ \Rightarrow C = 28\end{array}\]$

Calculate the average age of A, B and C,

$\[ = \frac{{40 + 64 + 28}}{3}\]$

$\[ = \frac{{132}}{3}\]$

$\[ = 44\]$

Hence the average age of A, B and C is 44.