High School

14.

1. 5 kg of rice and 3 kg of flour cost Rs. 145; and 3 kg of rice and 5 kg of flour cost Rs. 151. Find the cost of each item.

2. (ii) 5 kg of apples and 15 kg of mangoes cost Rs. 2350 and 3 kg of apples and 7 kg of mangoes cost Rs. 1190.
- (a) Find the cost of each item.
- (b) What is the increase or decrease?

Answer :

(i) The cost per kg of rice [tex]x[/tex] is Rs. 4.28, and the cost per kg of flour [tex]y[/tex] is Rs. 27.65.

(ii) The cost per kg of apples [tex]a[/tex] is Rs. 140, and the cost per kg of mangoes [tex]m[/tex] is Rs. 110.

(i) We have two sets of equations based on the given information:

  1. [tex]14.5x + 3y = 145[/tex]
  2. [tex]3x + 5y = 151[/tex]

where, [tex]x[/tex] is the cost per kg of rice, and [tex]y[/tex] is the cost per kg of flour.

Solve Equation (1) for one variable:

[tex]14.5x + 3y = 145[/tex]

[tex]3y = 145 - 14.5x[/tex]

[tex]y = \frac{145 - 14.5x}{3}[/tex] (Equation 3)

Substitute Equation (3) into Equation (2):

[tex]3x + 5\left(\frac{145 - 14.5x}{3}\right) = 151[/tex]

Multiply through by 3 to clear the fraction:

[tex]9x + 5(145 - 14.5x) = 453[/tex]

[tex]9x + 725 - 72.5x = 453[/tex]

Combine like terms:

[tex]-63.5x + 725 = 453[/tex]

Subtract 725 from both sides:

[tex]-63.5x = -272[/tex]

Divide both sides by -63.5:

[tex]x = \frac{272}{63.5}[/tex]

[tex]x = 4.28[/tex]

Now, substitute [tex]x[/tex] back into Equation (3):

[tex]y = \frac{145 - 14.5(4.28)}{3}[/tex]

[tex]y = \frac{145 - 62.06}{3}[/tex]

[tex]y = \frac{82.94}{3}[/tex]

[tex]y = 27.65[/tex]

(ii) We have two sets of equations based on the given information:

  1. [tex]5a + 15m = 2350[/tex]
  2. [tex]3a + 7m = 1190[/tex]

where, [tex]a[/tex] is the cost per kg of apples, and [tex]m[/tex] is the cost per kg of mangoes.

Solve Equation (1) for one variable:

[tex]5a + 15m = 2350[/tex]

[tex]a + 3m = \frac{2350}{5}[/tex]

[tex]a + 3m = 470[/tex] (Equation 3)

Solve Equation (2) for one variable:

[tex]3a + 7m = 1190[/tex]

Solve using substitution method:

Subtract Equation (3) from (2) to eliminate one variable:

[tex]3a + 7m - (a + 3m) = 1190 - 470[/tex]

[tex]3a + 7m - a - 3m = 720[/tex]

[tex]2a + 4m = 720[/tex]

Divide by 2:

[tex]a + 2m = 360[/tex]

Now we have:

[tex]a + 3m = 470[/tex] (Equation 3)

[tex]a + 2m = 360[/tex] (Equation 4)

Subtract Equation (4) from (3):

[tex]a + 3m - (a + 2m) = 470 - 360[/tex]

[tex]m = 110[/tex]

Substitute [tex]m[/tex] back into Equation (4):

[tex]a + 2(110) = 360[/tex]

[tex]a + 220 = 360[/tex]

[tex]a = 140[/tex]

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