
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
493.2k+ 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.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
Explain the system of Dual Government class 8 social science CBSE

What is Kayal in Geography class 8 social science CBSE

Who is the author of Kadambari AKalidas B Panini C class 8 social science CBSE

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

Advantages and disadvantages of science

Write the smallest number divisible by both 306 and class 8 maths CBSE
