Question

A factory packs tea in three containers of different weights. Container A weighs 3/2 kg more than container B.
and container weighs 5/2 kg less than container B. The total weight of the three containers is 79/2
kg. What
is the weight of container A, B and C?​

Answer 1

Given:

A factory packs tea in three containers of different weights.

Container A weighs 3/2 kg more than container B.

Container C weighs 5/2 kg less than container B.

The total weight of the three containers is 79/2  kg.

To find:

What  is the weight of container A, B and C?​

Solution:

Let's assume,

"A kg" → represents the weight of the container A.

"B kg" → represents the weight of the container B.

"C kg" → represents the weight of the container C.

Since Container A weighs 3/2 kg more than container B, so we get the equation as:

$A = \frac{3}{2} + B$ ...... (i)

Since Container C weighs 5/2 kg less than container B, so we get the equation as:

$C = B - \frac{5}{2}$ ...... (ii)

The total weight of the 3 containers = $7\frac{9}{2} \:kg = \frac{23}{2} \:kg$

∴ $A + B + C = \frac{23}{2}$

substituting from (i) & (ii), we get

$\implies \frac{3}{2} + B + B + B - \frac{5}{2} = \frac{23}{2}$

$\implies \frac{3}{2} + 3B - \frac{5}{2} = \frac{23}{2}$

$\implies 3B - 1 = \frac{23}{2}$

$\implies 3B = \frac{23}{2} + 1$

$\implies 3B = \frac{23 + 2}{2}$

$\implies 3B = \frac{25}{2}$

$\implies \bold{B = \frac{25}{6}\:kg}$

On substituting the value of B in eq. (i), we get

$\bold{A} = \frac{3 }{2} + \frac{25}{6} = \frac{9 + 25}{6} = \frac{34}{6} = \bold{\frac{17}{3}\:kg }$

On substituting the value of C in eq. (ii), we get

$\bold{C} = \frac{25}{6} - \frac{5 }{2} = \frac{25 - 15}{6} = \frac{10}{6} = \bold{\frac{5}{3}\:kg }$

Thus, the weight of the containers A, B & C are:

$\boxed{\bold{A = \underline{\frac{17}{3}\: kg }}}\\\\\boxed{\bold{B = \underline{\frac{25}{6}\: kg }}}\\\\\boxed{\bold{C = \underline{\frac{5}{3}\: kg }}}$

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Answer 2

hi friend I am Rohan Kumar Singh