Question

1) If Agnes buys six apples, five bananas, and two oranges, the cost will be $2.85. If she

buys three apples, seven bananas, and four oranges, the cost will be $2.70. If apples are 1

cent less than twice as expensive as oranges, what is the cost of each?

Answer 1

One apple cost 27 cent or $0.27

One banana cost 19 cent or $0.19

One orange cost 14 cent or $0.14

Explanation:

Let x represent the cost of one apple

Let y represent the cost of one banana

Let z represent the cost of one orange

Since the monetary unit is both in dollars and cents, we use the cent (¢) unit.

1 dollar ($) = 100 cent (¢)

Therefore, $2.85 and $2.70 will be 285¢ and 270¢ respectively.

According to the question, Agnes buys six apples, five bananas, and two oranges with a total cost of 285¢. Mathematically, this means:

6x + 5y + 2z = 285¢ ......... eqn1

Agnes buys three apples, seven bananas, and four oranges, which cost 270¢ in total. Mathematically,

3x + 7y + 4z = 270¢ ........ eqn2

The question further states that apples are 1

cent less than twice the cost of oranges. This mathematically means:

x = 2z - 1 ........... eqn3

We use simultaneous equation to solve this, but we have to reduce the variables in each equation 1&2 to 2 by substituting eqn3 into both equations 1&2 i.e.

6x + 5y + 2z = 285 .......... eqn1

6(2z-1) + 5y + 2z = 285

12z - 6 + 5y + 2z = 285

14z + 5y = 285 + 6

14z + 5y = 291 .......... eqn4

3x + 7y + 4z = 270 ........ eqn2

3(2z-1) + 7y + 4z = 270

6z - 3 + 7y + 4z = 270

10z + 7y = 270 + 3

10z + 7y = 273 ............. eqn5

Now we can simultaneously solve equations 4&5 using elimination method.

14z + 5y = 291

10z + 7y = 273

We multiply eqns 4 and 5 by 7 and 5 respectively in order to eliminate variable y

7 × 14z + 5y = 291

5 × 10z + 7y = 273

98z + 35y = 2037

50z + 35y = 1365

We subtract to have

48z = 672

z = 672/48

z = 14

If z is 14, we substitute the value of z into eqn3;

x = 2z - 1

x = 2(14) - 1

x = 28 - 1

x = 27

If z= 14 and x=27, we substitute both values into eqn 2

3x + 7y + 4z = 270

3(27) + 7y + 4(14) = 270

81 + 7y + 56 = 270

137 + 7y = 270

Collect like terms

7y = 270-137

7y = 133

y= 133/7

y = 19

Hence, z= 14, y=19 and x=27.

This means that the cost of one apple, one banana and one orange is 27¢, 19¢ and 14¢ respectively.

Answer 2

An apple costs $0.27 or ¢27

A banana costs $0.19 or ¢19

And an orange costs $0.14 or ¢14

Explanation:

Lets assume that the cost of each apple is x,that of a banana is y and the cost of each orange is z

And since the prices of apples are a cent less than twice the price of oranges,we have the equation like this: x = 2z - 1

And since ¢100 = $1

We can therefore change the total amount of money paid for the fruits to cents

6x + 5y +2z = 285_____ equation 1

3x + 7y + 4z = 270____ equation 2

Remember that x = 2z - 1,so we substitute into both equation

6(2z - 1) + 5y + 2z = 285

12z - 6 + 5y + 2z = 285

14z + 5y = 291____ equation 3

3(2z - 1) + 7y + 4z = 270

6z - 3 + 7y + 4z = 270

10z + 7y = 273____ equation 4

Now we want to make the coefficient of y from both equation 3 and 4 to be same by multiplying them by 7 and 5 respectively

98z + 35y = 2037____ equation 5

50z + 35y = 1365____ equation 6

Subtract equation 6 from 5

48z = 672

Z = 14

Remember that x = 2z - 1

(14 × 2) - 1 = 27

X = 27

Finally substitute x = 27 and z = 14 in equation 1

6(27) + 5y + 2(14) = 285

162 + 5y + 28 = 285

5y = 285 - 162 - 28

5y = 95

Y = 19

Here the cost of apple is 27 cents($0.27), banana is 19 cents($0.19] and oranges is 14 cents ($0.14)