Question

A basket of 2 apples, 4 oranges and 6 mangoes cost the same as a basket of 1 apple, 4 oranges and 8 mangoes, or a basket of 8 oranges and 7 mangoes. Then the number of mangoes in a basket of mangoes that has the same cost as the other baskets is

• A. 12
• B. 13
• C. 11
• D. 10

Answer 1

The Correct Option is B

Let's assume the cost of an apple is \(a\), an orange is \(o\), and a mango is \(m\)

From the given information:

1. For the first basket: \(2a + 4o + 6m\)

2. For the second basket: \(a + 4o + 8m\)

3. For the third basket: \(8o + 7m\)

Given that all the baskets cost the same, we can equate the cost expressions:

\(2a + 4o + 6m = a + 4o + 8m\)

From the above equation, \(\left( a = 2m \right) ...(i) \)

Similarly, from the second and third baskets: 

\(a + 4o + 8m = 8o + 7m\)

Which gives, \(\left( a + m = 4o \right) ...(ii)\)

Substituting \(a\) from equation (i) in equation (ii): 

\(( 2m + m = 4o ) ( 3m = 4o ) ( o = 0.75m ) ...(iii) \)

Now, let's find the number of mangoes in a basket that costs the same as the other baskets, using only mangoes. 

From the first basket: 

\((2a + 4o + 6m ) = ( 2(2m) + 4(0.75m) + 6m)\)

\(= \left( 4m + 3m + 6m \right) \)

\(= ( 13m )\)

So, a basket of 13 mangoes has the same cost as the other baskets.

Answer 2

Let the costs of an apple, an orange, and a mango be \(a, o\) and \(m\) respectively.
According to the question,
\(2a+4o+6m=a+4o+8m\)
\(⇒2a=8m-6m\)
\(⇒ a=2m\)
Also, \(a+4o+8m=8o+7m\)
\(10m−7m=4o\)
\(3m=4o\)
Now, we can express the cost of a basket solely in terms of mangoes:
\(2a+4o+6m\)
\(=4m+3m+6m\)
\(=13m\)

So, the correct option is (B): \(13\)