Question

400 kg of wheat and 600 kg of flour costs Rs 6500.Due to shortage in supply a month later ,the price was increased by 15% and that of flour by 10%.As a result,the same quantities of these two commodities would now cost Rs 7250.Calculate the original price of one kg each commodity

Answer 1

The original price of 1 kg of wheat is Rs 5, and the original price of 1 kg of flour is Rs 7.5.

Let's assume the original price of 1 kg of wheat is Rs "x" and the original price of 1 kg of flour is Rs "y."

According to the given information, we have two equations:

400 kg of wheat and 600 kg of flour cost Rs 6500:

$400x + 600y = 6500$

After a month, the price increased by 15% for wheat and 10% for flour, and the new cost of the same quantities is Rs 7250:

$400(1.15x) + 600(1.10y) = 7250$

Now, we need to solve these two equations to find the values of "x" and "y."

Let's start with the first equation:

$400x + 600y = 6500$

Now, let's divide the whole equation by 100 to make the calculations easier:

$4x + 6y = 65$ ...........(Equation 1)

Now, let's move on to the second equation:

$400(1.15x) + 600(1.10y) = 7250\\460x + 660y = 7250$

Divide the whole equation by 10:

$46x + 66y = 725$ ...........(Equation 2)

Now, we have a system of equations:

$4x + 6y = 65$ ...........(Equation 1)

$46x + 66y = 725$...........(Equation 2)

Let's solve this system of equations to find the values of "x" and "y." We can use the method of substitution or elimination.

Using elimination:

Multiply Equation 1 by 11 to make the coefficient of "y" the same in both equations:

$44x + 66y = 715$...........(Equation 3)

Now, subtract Equation 3 from Equation 2 to eliminate "y":

$(46x + 66y) - (44x + 66y) = 725 - 715\\2x = 10\\x = 5$

Now that we have the value of "x," we can find the value of "y" using Equation 1:

$4x + 6y = 65\\4(5) + 6y = 65\\20 + 6y = 65\\6y = 65 - 20\\6y = 45\\y = 45 \div 6\\y = 7.5$

So, the original price of 1 kg of wheat is Rs 5, and the original price of 1 kg of flour is Rs 7.5.

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