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The cost of 4kg of radishes and 1.5kg of carrots is £14.80
The cost of 3kg of radishes and 2kg of carrots is £12.50
work out the costs of
a) 1kg of carrots
b) 1kg of radishes


Sagot :

Answer:

1kg of carrots = 1.60

1kg of radishes = 3.10

Step-by-step explanation:

To solve the given problem, we can set up a system of linear equations based on the information provided. Let's denote the cost per kilogram of radishes as \( r \) and the cost per kilogram of carrots as \( c \).

From the first statement, we have the equation:

\[ 4r + 1.5c = 14.80 \]

This equation represents the total cost of 4 kg of radishes and 1.5 kg of carrots.

From the second statement, we have the equation:

\[ 3r + 2c = 12.50 \]

This equation represents the total cost of 3 kg of radishes and 2 kg of carrots.

Now, we have a system of two equations with two unknowns:

\[ \begin{cases} 4r + 1.5c = 14.80 \n 3r + 2c = 12.50 \end{cases}\]

To solve for \( c \) and \( r \), we can use the method of substitution or elimination. Let's use the elimination method by multiplying the second equation by 2 and the first equation by 3 to make the coefficients of \( c \) the same:

\[ \begin{cases} 12r + 4.5c = 44.40 \n 6r + 4c = 25 \end{cases} \]

Now, subtract the second equation from the first:

\[ (12r + 4.5c) - (6r + 4c) = 44.40 - 25 \]

\[ 6r + 0.5c = 19.40 \]

Now, we can solve for \( r \) by multiplying the entire equation by 2 to eliminate the decimal:

\[ 12r + c = 38.80 \]

\[ c = 38.80 - 12r \]

We can now substitute \( c \) back into one of the original equations to find \( r \). Let's use the second original equation:

\[ 3r + 2(38.80 - 12r) = 12.50 \]

\[ 3r + 77.60 - 24r = 12.50 \]

\[ -21r = 12.50 - 77.60 \]

\[ -21r = -65.10 \]

\[ r = (-65.10)/(-21) \]

\[ r = 3.10 \]

Now that we have the value of \( r \), we can find \( c \) by substituting \( r \) back into the equation we derived for \( c \):

\[ c = 38.80 - 12(3.10) \]

\[ c = 38.80 - 37.20 \]

\[ c = 1.60 \]