CAT 2021 Set-1 | Quantitative Aptitude | Question: 2

Question

A basket of $2$ apples, $4$ oranges and $6$ mangoes costs 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. $10$
• C. $11$
• D. $13$

Answer 1

Let the cost of $1$ apple, $1$ orange, $1$ mango be $x, y, $ and $z$ respectively.

Now, $2x + 4y + 6z = x + 4y + 8z = 8y + 7z \quad \longrightarrow (1)$

Let's take the first two relations.

$\Rightarrow 2x + 4y + 6z = x + 4y + 8z $

$ \Rightarrow \boxed{x = 2z}$

$ \Rightarrow \frac{x}{z} = \frac{2}{1}$

$ \Rightarrow \boxed{x : z = 2 : 1} \quad \longrightarrow (2)$

Take the second and third relation.

$ \Rightarrow x + 4y + 8z = 8y + 7z$

$ \Rightarrow 2z + 4y + 8z = 8y + 7z$

$ \Rightarrow \boxed{3z = 4y}$

$ \Rightarrow \frac{z}{y} = \frac{4}{3}$

$ \Rightarrow \boxed{ z : y = 4 : 3} \quad \longrightarrow (3)$

Now, from equation $(2),$ and $(3)$

• $x : z = (2 : 1) \times 4 = 8 : 4$
• $z : y = (4 : 3) \times 1 = 4 : 3$

$ \Rightarrow \boxed{x : y : z = 8 : 3 : 4}$

Let

• $x = 8k$
• $y = 3k$
• $z = 4k,$  where $k$ is constant

The cost of third basket $ = 8y + 7z = 8 (3k) + 7 (4k)  = 24k + 28k = 52k $

The cost of mangoes $ = z = 4k $

$\therefore$ The number of mangoes $ = \frac{52k}{4k} = 13.$

Correct Answer $:\text{D}$