Question

The class XI students of a school wanted to give a farewell party to the outgoing students of class XII. They decided to purchase two kinds of sweets, one costing Rs. 250 per kg and the other costing Rs. 350 per kg. They estimated that 40 kg of sweets were needed. If the total budget for the sweets was Rs. 11,800, find the sweets of each kind were brought?
(a) 18 kg and 14 kg
(b) 22 kg and 18 kg
(c) 12 kg and 23 kg
(d) 19 kg and 22 kg

Answer 1

Hint: To solve the question given above, we will first assume that the amount of sweets costing Rs. 250 per kg bought is x kg and the amount of sweets costing Rs. 350 kg bought is y kg. Then we will form a pair of linear equations according to the information given in the question. The first condition is given as the estimated total amount of sweets is 40 kg. So, we get x + y = 40. Similarly, other equations can be formed. Then we will solve these equations with the help of the substitution method and we will find x and y.

Complete step by step solution:
To start with we will assume that the sweets costing Rs. 250 per kg has the name ‘A’ and the sweets costing Rs. 350 per kg has the name ‘B’. Now, we will assume that the amount of sweets ‘A’ bought is x kg and the amount of sweets ‘B’ bought is y kg. Now, it is given in the question that the total amount of sweets bought is 40 kg. Thus, we will get the following equation.
\[x\text{ }kg+y\text{ }kg=40\text{ }kg\]
\[\Rightarrow x+y=40......\left( i \right)\]
Now, the cost of 1 kg of sweets ‘A’ = Rs. 250.
So, the cost of x kg of sweets ‘A’ \[=Rs.250\times x=Rs.250x\]
The cost of 1 kg of sweets ‘B’ = Rs. 350
So, the cost of y kg of sweets ‘B’ \[=Rs.350\times y=Rs.350y\]
Now, the total cost of sweets is Rs. 11800. Thus, we will get the following equation.
\[Rs.250x+Rs.350y=Rs.11800\]
\[\Rightarrow 250x+350y=11800\]
Now, we will divide the whole equation by 50. After doing this, we will get,
\[\Rightarrow \dfrac{250x}{50}+\dfrac{350y}{50}=\dfrac{11800}{50}\]
\[\Rightarrow 5x+7y=236.....\left( ii \right)\]
Now, we have a pair of linear equations in two variables. We will solve them with the help of the method of substitution. Now, from (i), we can say that,
\[x+y=40\]
\[\Rightarrow y=40-x....\left( iii \right)\]
Now, we will put the value of y from (iii) to (ii). Thus, we will get the following equation:
\[\Rightarrow 5x+7\left( 40-x \right)=236\]
\[\Rightarrow 5x+280-7x=236\]
\[\Rightarrow 5x-7x=236-280\]
\[\Rightarrow -2x=-44\]
\[\Rightarrow 2x=44\]
\[\Rightarrow x=22\]
Now, we will put the value of x in (iii). Thus, we will get,
\[\Rightarrow y=40-22\]
\[\Rightarrow y=18\]
Therefore, the students bought 22 kg of ‘A’ and 18 kg of ‘B’.
Hence, option (b) is the right answer.

Note: The linear equations obtained from the equations (i) and (ii) can also be solved by the method of elimination. For this, we will multiply the equation (i) with 5. Thus, we will get,
\[5x+5y=200.....\left( iv \right)\]
Now, we will subtract (iv) from (ii). Thus, we will get,
\[\left( 5x+7y \right)-\left( 5x+5y \right)=236-200\]
\[\Rightarrow 7y-5y=36\]
\[\Rightarrow 2y=36\]
\[\Rightarrow y=18\]
On putting y = 18 in (i), we will get, x = 22.