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Question

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(A) $\dfrac{3}{{17}}$

(B) $\dfrac{4}{{17}}$

(C) $\dfrac{6}{{19}}$

(D) $\dfrac{3}{{19}}$

Answer

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Now we assume that a full container without the handle weighed 100g to find the weight fraction.

So total weight of container without handle is 100g

Now according to the question the weight of an empty container is 15% of the total weight of the container.

Weight of empty container $ = \dfrac{{15 \times 100}}{{100}}$

So weight of empty container is $15g$

Now we have to find weight of liquid filled in container

Weight of liquid = total weight of container – weight of empty container

Weight of liquid =$100g - 15g$

Weight of liquid = $85g$

Now, adding a handle adds 5% or 5 grams,

So a full container with a handle would weight = $100g + 5g$

Full weight of container with handle = $105g$

Given, the weight of a partly filled container is 1/3 of the completely filled container with handle

So weight of partly filled container = $\dfrac{1}{3} \times 105g$

Weight of partly filled container = $35g$

Now to find fraction of container utilize is find by subtracting the weights of container and the handle

So \[35g - 5g - 15g = 15g\]

Now calculating our fraction we have,

Ratio of liquid present in partly filled container and fully filled container

So $\dfrac{{15g}}{{85g}}$

fraction of container utilize = $\dfrac{3}{{17}}$

So option A is the correct answer.

So total weight of container without handle$ = x$

Weight of container= $15\% $of $x$

$ \Rightarrow \dfrac{{15 \times x}}{{100}}$

$ \Rightarrow .15x$

Now weight of liquid filled in container

$x - .15x = .85x$

Now weight of container with handle is $x + 5\% $of $x$

$x + 0.05x$

$1.05x$

the weight of a partly filled container is 1/3 of the completely filled container with handle

So weight of partly filled container=$\dfrac{1}{3} \times 1.05x$

$ \Rightarrow .35x$

Now fraction of container utilize is find by subtracting the weights of container and the handle

$.35x - .05x - .15x$

$ \Rightarrow .15x$

Now fraction of container =$\dfrac{{.15x}}{{.85x}}$

So fraction of container = $\dfrac{3}{{17}}$