Let the cost of an apple be \(a\), an orange be \(o\), and a mango be \(m\).
The given basket costs are:
1. Basket 1: \(2a + 4o + 6m\)
2. Basket 2: \(a + 4o + 8m\)
3. Basket 3: \(8o + 7m\)
Since all baskets have the same cost, we can set up equations:
\(2a + 4o + 6m = a + 4o + 8m\)
Simplifying this equation yields \(a = 2m\) (Equation i).
Equating the costs of the second and third baskets:
\(a + 4o + 8m = 8o + 7m\)
This simplifies to \(a + m = 4o\) (Equation ii).
Substitute \(a = 2m\) from Equation (i) into Equation (ii):
\(2m + m = 4o\), which simplifies to \(3m = 4o\), or \(o = 0.75m\) (Equation iii).
To determine the number of mangoes equivalent to the cost of the other baskets, we use the cost from the first basket:
\(2a + 4o + 6m = 2(2m) + 4(0.75m) + 6m\)
\(= 4m + 3m + 6m\)
\(= 13m\)
Therefore, a basket containing 13 mangoes has the same cost as the other baskets.