Question

Isabella bought apples, oranges, and pineapples. The cost of the oranges is $2 less than thrice the cost of the apples and the cost of the pineapples is $2 more than twice of the cost of the apples. The total cost of the fruits is $18.95. Find the cost of each fruit. (Estimate to nearest tenths place).
The cost of the oranges is $4.8, the cost of the apples is $8.6 and the cost of the pineapples is $9.6
The cost of the oranges is $7.6, the cost of the apples is $3.2 and the cost of the pineapples is $8.4
The cost of the oranges is $5.3, the cost of the apples is $8.2 and the cost of the pineapples is $8.8
The cost of the oranges is $3.8, the cost of the apples is $8.8 and the cost of the pineapples is $9.8

Answer 1

Let's denote the cost of the apples as \( a \), the cost of the oranges as \( o \), and the cost of the pineapples as \( p \). According to the problem, we have the following relationships:

1. The cost of the oranges is $2 less than thrice the cost of the apples:
\[
o = 3a - 2
\]

2. The cost of the pineapples is $2 more than twice the cost of the apples:
\[
p = 2a + 2
\]

3. The total cost of the fruits is $18.95:
\[
a + o + p = 18.95
\]

Now we can substitute the expressions for \( o \) and \( p \) in terms of \( a \) into the total cost equation:

\[
a + (3a - 2) + (2a + 2) = 18.95
\]

Combine like terms:

\[
a + 3a + 2a - 2 + 2 = 18.95
\]
\[
6a = 18.95
\]

Now, solve for \( a \):

\[
6a = 18.95 \implies a = \frac{18.95}{6} \approx 3.1583 \implies a \approx 3.2
\]

Now we can substitute \( a \) back to find \( o \) and \( p \):

1. For \( o \):
\[
o = 3(3.2) - 2 \approx 9.6 - 2 = 7.6
\]

2. For \( p \):
\[
p = 2(3.2) + 2 \approx 6.4 + 2 = 8.4
\]

So we can summarize the estimated costs of the fruits as:
- Cost of Apples, \( a \approx 3.2 \)
- Cost of Oranges, \( o \approx 7.6 \)
- Cost of Pineapples, \( p \approx 8.4 \)

Thus, the correct choice from the provided options is:
- The cost of the oranges is $7.6, the cost of the apples is $3.2, and the cost of the pineapples is $8.4.