Answer:
An apple costs $0.27 or ¢27
A banana costs $0.19 or ¢19
And an orange costs $0.14 or ¢14
Step-by-step 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)
Answer:
One apple cost 27 cent or $0.27
One banana cost 19 cent or $0.19
One orange cost 14 cent or $0.14
Step-by-step explanation:
Let x represent the cost of one apple
Let y represent the cost of one banana
Let z represent the cost of one orange
Since the monetary unit is both in dollars and cents, we use the cent (¢) unit.
1 dollar ($) = 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.
