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Joe and Ann buy some fruit from the same shop. Joe buys 4 apples and 3 bananas for £2.50 Ann buys 3 apples and 4 bananas for £2.40 Work out the cost of
(i) one apple
(ii) one banana​

Sagot :

Answer:

apple costs $0.40 , banana costs $0.30

Step-by-step explanation:

let a represent apples and b represent bananas , then

4a + 3b = 2.5 → (1)

3a + 4b = 2.4 → (2)

multiplying (1) by 4 and (2) by - 3 and adding will eliminate b

16a + 12b = 10 → (3)

- 9a - 12b = - 7.2 → (4)

add (3) and (4) term by term to eliminate b

7a + 0 = 2.8

7a = 2.8 ( divide both sides by 7 )

a = 0.4

substitute a = 0.4 into either of the 2 equations and solve for b

substituting into (1)

4(0.4) + 3b = 2.5

1.6 + 3b = 2.5 ( subtract 1.6 from both sides )

3b = 0.9 ( divide both sides by 3 )

b = 0.3

(i) one apple cost $0.40

(ii) one banana costs $0.30

semsee45

Answer:

(i) One apple = $0.40

(ii) One banana = $0.30

Step-by-step explanation:

Given information:

  • Cost of 4 apples and 3 bananas = £2.50
  • Cost of 3 apples and 4 bananas = £2.40

Define the variables:

  • Let x = The cost of one apple.
  • Let y = The cost of one banana.

Create two equations with the given information and defined variables:

  [tex]\textsf{Equation 1}: \quad 4x+3y=2.50[/tex]

  [tex]\textsf{Equation 2}: \quad 3x+4y=2.40[/tex]

Multiply the first equation by 4:

[tex]\implies 4 \cdot 4x+4 \cdot 3y=4 \cdot 2.50[/tex]

[tex]\implies 16x+12y=10.00[/tex]

Multiply the second equation by 3:

[tex]\implies 3 \cdot 3x + 3 \cdot 4y = 3 \cdot 2.40[/tex]

[tex]\implies 9x + 12y = 7.20[/tex]

Subtract the equations to eliminate the term in y:

[tex]\begin{array}{crcccc}& 16x & + & 12y & = & 10.00\\-& (9x & + & 12y & = & 7.20)\\\cline{2-6} & 7x & & &=&2.80\\\end{array}[/tex]

Solve for x:

[tex]\implies 7x=2.80[/tex]

[tex]\implies \dfrac{7x}{7}=\dfrac{2.80}{7}[/tex]

[tex]\implies x=0.40[/tex]

Therefore, the cost of one apple is $0.40.

Substitute the found value of x into one of the equations and solve for y:

[tex]\implies 4x+3y=2.50[/tex]

[tex]\implies 4(0.40)+3y=2.50[/tex]

[tex]\implies 1.60+3y=2.50[/tex]

[tex]\implies 1.60+3y-1.60=2.50-1.60[/tex]

[tex]\implies 3y=0.90[/tex]

[tex]\implies \dfrac{3y}{3}=\dfrac{0.90}{3}[/tex]

[tex]\implies y=0.30[/tex]

Therefore, the cost of one banana is $0.30.

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