Questions & Answers

Question

Answers

(a) 4.6kg

(b) 6 kg

(c) 9.6 kg

(d) 11.4 kg

Answer
Verified

Hint: Divide the amount of apples in two parts, one for gain and another at loss. Now calculate the selling price of each of them and the selling price of total fruits as well. Now, get the amount of apples at loss by solving the equations formed by the above information (selling prices).

Complete step by step solution:

Total amount of fruits (apples) given is 24kg.

Let us suppose the fruit seller sold â€˜xâ€™ kg of apples at gain of 20% and (24-x) kg of apples at loss of 5%.

Now, we need to use the price of fruits/apples as well .So suppose the rate of apples be Rs. y/ kg i.e., the cost of 1 kg apple is Rs.y (cost price).

Let us calculate the cos of x kg apples and (24-x) kg of apples in terms of y.

So, the cost price of 1 kg apple is Rs.y.

Hence, the cost price of x kg apples be Rs.xy.

And the cost price of (24-x) kg apples be Rs.y(24-x).

Now, fruit seller received a gain of 20% at x kg of apples and loss of 5% at (24-x) kg of apples.

So, selling price of x kg apples can be calculated as

$s{{p}_{1}}$ = cost price + gain at cost price.

$s{{p}_{1}}=xy+20%$ of cost of kg apples

$\begin{align}

& s{{p}_{1}}=xy+\dfrac{20}{100}\times xy \\

& s{{p}_{1}}=xy\left[ 1+\dfrac{20}{100} \right] \\

& \Rightarrow s{{p}_{1}}=xy\dfrac{120}{100}=\dfrac{12}{10}xy...........(i) \\

\end{align}$

Where $s{{p}_{1}}$ is the selling price of x kg apples

Similarly, selling price of (24-x)kg apples can be given as

$s{{p}_{2}}$= cost price â€“ loss amount at cost price

$\begin{align}

& s{{p}_{2}}=\left( 24-x \right)y-\dfrac{\left( 24-x \right)y\times 5}{100} \\

& s{{p}_{2}}=\left( 24-x \right)y\left[ 1-\dfrac{5}{100} \right] \\

& \Rightarrow s{{p}_{2}}=\dfrac{95}{100}\left( 24-x \right)y............(ii) \\

\end{align}$

Where $s{{p}_{2}}$ is the selling price of (24-x) kg apples.

Now, we also know that fruit seller earn a profit of 10% on the whole.

So, selling price of whole amount fruits (apples) i.e. 24kg can be given as

Sp = cost price of 24 kg apples + gain at this cost price.

Hence,

$\begin{align}

& sp=24y+\dfrac{10}{100}\times 24y \\

& sp=24y\left[ 1+\dfrac{10}{100} \right] \\

& sp=24y\times \dfrac{110}{100}=\dfrac{11}{10}\times 24y \\

& \Rightarrow sp=\dfrac{264}{10}y............(iii) \\

\end{align}$

Here sp is the selling price of 24kg apples.

Now, we know the total selling price and selling price of x kg apples and selling price of (24-x) kg apples.

So, the sum of sp of x kg and sp of (24-x) kg should be equal to 24kg apples as well.

Hence, from equation (i), (ii) and (iii) we get

$\begin{align}

& s{{p}_{1}}+s{{p}_{2}}=sp \\

& \dfrac{12}{10}xy+\dfrac{95}{100}\left( 24-x \right)y=\dfrac{264}{10}y \\

& \Rightarrow \dfrac{12}{10}x+\dfrac{95}{100}\left( 24-x \right)=\dfrac{264}{10} \\

& \Rightarrow \dfrac{12}{10}x+\dfrac{95}{100}\times 24-\dfrac{95}{100}x=\dfrac{264}{10} \\

& \dfrac{12}{10}x-\dfrac{95}{100}x=\dfrac{264}{10}-\dfrac{95}{100}\times 24 \\

\end{align}$

Now, taking L.C.M to both sides, we get

$\begin{align}

& \dfrac{120x-95x}{100}=\dfrac{2640-95\times 24}{100} \\

& 25x=2640-2280 \\

& x=\dfrac{360}{25}=\dfrac{72}{5}=14.4kg \\

& x=14.4kg \\

\end{align}$

Hence, apples sold at gain are 14.4kg.

So, amount of apples sold at a loss = 24-14.4 = 9.6 kg

Hence, option (c) is the correct answer.

Note: One can suppose the price of 1 kg apples as 100 rs, or 1 rs or any other variables as well. It will not affect the answer. As we can observe that â€˜yâ€™ from the equation of $sp=s{{p}_{1}}+s{{p}_{2}}$ is cancelled out from each side. So, donâ€™t get confused with the price of apples.

Applying the relation $sp=s{{p}_{1}}+s{{p}_{2}}$ is the key point of the question.

One can go wrong with the relations s.p = c.p + gain and sp = cp â€“ loss

We donâ€™t need to remember any formula for these kinds of questions. We can go with the basic terminology as well. As gain will be added and loss should be subtracted from the original cost price to get the selling price of any object. So, donâ€™t confuse the plus and minus signs of the above relations.

Complete step by step solution:

Total amount of fruits (apples) given is 24kg.

Let us suppose the fruit seller sold â€˜xâ€™ kg of apples at gain of 20% and (24-x) kg of apples at loss of 5%.

Now, we need to use the price of fruits/apples as well .So suppose the rate of apples be Rs. y/ kg i.e., the cost of 1 kg apple is Rs.y (cost price).

Let us calculate the cos of x kg apples and (24-x) kg of apples in terms of y.

So, the cost price of 1 kg apple is Rs.y.

Hence, the cost price of x kg apples be Rs.xy.

And the cost price of (24-x) kg apples be Rs.y(24-x).

Now, fruit seller received a gain of 20% at x kg of apples and loss of 5% at (24-x) kg of apples.

So, selling price of x kg apples can be calculated as

$s{{p}_{1}}$ = cost price + gain at cost price.

$s{{p}_{1}}=xy+20%$ of cost of kg apples

$\begin{align}

& s{{p}_{1}}=xy+\dfrac{20}{100}\times xy \\

& s{{p}_{1}}=xy\left[ 1+\dfrac{20}{100} \right] \\

& \Rightarrow s{{p}_{1}}=xy\dfrac{120}{100}=\dfrac{12}{10}xy...........(i) \\

\end{align}$

Where $s{{p}_{1}}$ is the selling price of x kg apples

Similarly, selling price of (24-x)kg apples can be given as

$s{{p}_{2}}$= cost price â€“ loss amount at cost price

$\begin{align}

& s{{p}_{2}}=\left( 24-x \right)y-\dfrac{\left( 24-x \right)y\times 5}{100} \\

& s{{p}_{2}}=\left( 24-x \right)y\left[ 1-\dfrac{5}{100} \right] \\

& \Rightarrow s{{p}_{2}}=\dfrac{95}{100}\left( 24-x \right)y............(ii) \\

\end{align}$

Where $s{{p}_{2}}$ is the selling price of (24-x) kg apples.

Now, we also know that fruit seller earn a profit of 10% on the whole.

So, selling price of whole amount fruits (apples) i.e. 24kg can be given as

Sp = cost price of 24 kg apples + gain at this cost price.

Hence,

$\begin{align}

& sp=24y+\dfrac{10}{100}\times 24y \\

& sp=24y\left[ 1+\dfrac{10}{100} \right] \\

& sp=24y\times \dfrac{110}{100}=\dfrac{11}{10}\times 24y \\

& \Rightarrow sp=\dfrac{264}{10}y............(iii) \\

\end{align}$

Here sp is the selling price of 24kg apples.

Now, we know the total selling price and selling price of x kg apples and selling price of (24-x) kg apples.

So, the sum of sp of x kg and sp of (24-x) kg should be equal to 24kg apples as well.

Hence, from equation (i), (ii) and (iii) we get

$\begin{align}

& s{{p}_{1}}+s{{p}_{2}}=sp \\

& \dfrac{12}{10}xy+\dfrac{95}{100}\left( 24-x \right)y=\dfrac{264}{10}y \\

& \Rightarrow \dfrac{12}{10}x+\dfrac{95}{100}\left( 24-x \right)=\dfrac{264}{10} \\

& \Rightarrow \dfrac{12}{10}x+\dfrac{95}{100}\times 24-\dfrac{95}{100}x=\dfrac{264}{10} \\

& \dfrac{12}{10}x-\dfrac{95}{100}x=\dfrac{264}{10}-\dfrac{95}{100}\times 24 \\

\end{align}$

Now, taking L.C.M to both sides, we get

$\begin{align}

& \dfrac{120x-95x}{100}=\dfrac{2640-95\times 24}{100} \\

& 25x=2640-2280 \\

& x=\dfrac{360}{25}=\dfrac{72}{5}=14.4kg \\

& x=14.4kg \\

\end{align}$

Hence, apples sold at gain are 14.4kg.

So, amount of apples sold at a loss = 24-14.4 = 9.6 kg

Hence, option (c) is the correct answer.

Note: One can suppose the price of 1 kg apples as 100 rs, or 1 rs or any other variables as well. It will not affect the answer. As we can observe that â€˜yâ€™ from the equation of $sp=s{{p}_{1}}+s{{p}_{2}}$ is cancelled out from each side. So, donâ€™t get confused with the price of apples.

Applying the relation $sp=s{{p}_{1}}+s{{p}_{2}}$ is the key point of the question.

One can go wrong with the relations s.p = c.p + gain and sp = cp â€“ loss

We donâ€™t need to remember any formula for these kinds of questions. We can go with the basic terminology as well. As gain will be added and loss should be subtracted from the original cost price to get the selling price of any object. So, donâ€™t confuse the plus and minus signs of the above relations.

×

Sorry!, This page is not available for now to bookmark.