
A woman buys toffees at \[Rs\text{ }2.50\] a dozen and an equal number at \[Rs.3\] a score. She sells them at \[Rs\text{ 3}.60\]a score and thus makes a profit of \[Rs\text{ }10\]. How many toffees did she buy?
A $10000$
B $12000$
C $5000$
D $6000$
Answer
592.5k+ views
Hint:
Here we need to apply the concept of conversions from one unit to another unit and Profit formula.
Conversions Required: One score equals $20$ objects.
One dozen equals $20$ objects.
Formula Required: $Profit=Selling\text{ }\Pr ice\text{-}\operatorname{Cos}t\text{ }\Pr ice$
Profit incurs when selling Price is more than cost price.
We need to find how many toffees she buys.
Complete step by step solution:
Let the number of toffee bought in each case is\[x\] .
Total toffees bought is \[2x\].
Cost of \[12\] toffees is \[Rs\text{ }2.50\]
$\Rightarrow $ Cost of \[1\] toffees is \[Rs\text{ }\dfrac{2.50}{12}\]
Cost of \[x\] toffees bought in first case is \[Rs\text{ }x\times \left( \dfrac{2.50}{12} \right)\]
Cost of $20$ toffees bought in first case is \[Rs.3\]
$\Rightarrow $Cost of \[1\] toffees is \[Rs\text{ }\dfrac{3}{20}\]
Cost of \[x\] toffees bought in second case is \[Rs\text{ }x\times \left( \dfrac{3}{20} \right)\]
Total cost price of \[2x\] toffees is \[Rs\text{ }x\times \left( \dfrac{2.50}{12} \right)+Rs\text{ }x\times \left( \dfrac{3}{20} \right)\]
According to the question,
Selling Price of $20$ toffees is \[Rs\text{ 3}.60\]
Selling Price of \[1\] toffees is \[Rs\text{ }\dfrac{\text{3}.60}{20}\]
Total Selling price of \[2x\] toffees is \[Rs\text{ 2}x\times \left( \dfrac{3.60}{20} \right)\]
Profit is \[Rs\text{ }10\]
$\Rightarrow $ Total Selling price of \[2x\] toffees- Total cost price of \[2x\] toffees is \[Rs\text{ }10\]
$\begin{align}
& \Rightarrow \text{2}x\times \left( \dfrac{3.60}{20} \right)-\left( x\times \left( \dfrac{2.50}{12} \right)+\text{ }x\times \left( \dfrac{3}{20} \right) \right)=10 \\
& \Rightarrow \text{2}x\times \left( \dfrac{3.60}{20} \right)-x\left( \dfrac{25}{120}+\dfrac{3}{20} \right)=10 \\
& \Rightarrow \dfrac{9x}{25}-\dfrac{43x}{120}=10 \\
& \Rightarrow \dfrac{216x-215x}{600}=10 \\
& \Rightarrow \dfrac{x}{600}=10 \\
& \Rightarrow x=6000 \\
\end{align}$
Total toffees bought is \[2x\],
\[\begin{align}
& \Rightarrow 2x=2\times 6000 \\
& =12000 \\
\end{align}\]
Therefore, the total toffees she bought is $12000$ .
Hence, Option choice B is the correct answer.
Note:
In such types of questions the concept of conversions from one unit to another unit and Profit formula is needed. Assigning the variable to the unknown and equations are framed as per the relation in the question, then solved to get the required value.
Here we need to apply the concept of conversions from one unit to another unit and Profit formula.
Conversions Required: One score equals $20$ objects.
One dozen equals $20$ objects.
Formula Required: $Profit=Selling\text{ }\Pr ice\text{-}\operatorname{Cos}t\text{ }\Pr ice$
Profit incurs when selling Price is more than cost price.
We need to find how many toffees she buys.
Complete step by step solution:
Let the number of toffee bought in each case is\[x\] .
Total toffees bought is \[2x\].
Cost of \[12\] toffees is \[Rs\text{ }2.50\]
$\Rightarrow $ Cost of \[1\] toffees is \[Rs\text{ }\dfrac{2.50}{12}\]
Cost of \[x\] toffees bought in first case is \[Rs\text{ }x\times \left( \dfrac{2.50}{12} \right)\]
Cost of $20$ toffees bought in first case is \[Rs.3\]
$\Rightarrow $Cost of \[1\] toffees is \[Rs\text{ }\dfrac{3}{20}\]
Cost of \[x\] toffees bought in second case is \[Rs\text{ }x\times \left( \dfrac{3}{20} \right)\]
Total cost price of \[2x\] toffees is \[Rs\text{ }x\times \left( \dfrac{2.50}{12} \right)+Rs\text{ }x\times \left( \dfrac{3}{20} \right)\]
According to the question,
Selling Price of $20$ toffees is \[Rs\text{ 3}.60\]
Selling Price of \[1\] toffees is \[Rs\text{ }\dfrac{\text{3}.60}{20}\]
Total Selling price of \[2x\] toffees is \[Rs\text{ 2}x\times \left( \dfrac{3.60}{20} \right)\]
Profit is \[Rs\text{ }10\]
$\Rightarrow $ Total Selling price of \[2x\] toffees- Total cost price of \[2x\] toffees is \[Rs\text{ }10\]
$\begin{align}
& \Rightarrow \text{2}x\times \left( \dfrac{3.60}{20} \right)-\left( x\times \left( \dfrac{2.50}{12} \right)+\text{ }x\times \left( \dfrac{3}{20} \right) \right)=10 \\
& \Rightarrow \text{2}x\times \left( \dfrac{3.60}{20} \right)-x\left( \dfrac{25}{120}+\dfrac{3}{20} \right)=10 \\
& \Rightarrow \dfrac{9x}{25}-\dfrac{43x}{120}=10 \\
& \Rightarrow \dfrac{216x-215x}{600}=10 \\
& \Rightarrow \dfrac{x}{600}=10 \\
& \Rightarrow x=6000 \\
\end{align}$
Total toffees bought is \[2x\],
\[\begin{align}
& \Rightarrow 2x=2\times 6000 \\
& =12000 \\
\end{align}\]
Therefore, the total toffees she bought is $12000$ .
Hence, Option choice B is the correct answer.
Note:
In such types of questions the concept of conversions from one unit to another unit and Profit formula is needed. Assigning the variable to the unknown and equations are framed as per the relation in the question, then solved to get the required value.
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