Question

A man buys a certain quantity of apples, mangoes and bananas. If the mangoes were to cost the same as apples, he would have to forego the bananas to buy the same number of mangoes as he had bought earlier (for the same total amount). The amount spent by him on mangoes and bananas together is 50\% more than the amount spent on apples. The total amount spent is ₹ 140. The number of mangoes bought is the same as the number of bananas. If he wishes to buy the same number of apples as well, how much additional amount would he have to spend?

• A. 56
• B. 140
• C. 28
• D. 42

Hint:

When multiple categories share a simple price relation, first convert the statements into linear equations in "amount spent". Frequently, equal–number conditions ($M=B$) collapse the algebra via identities like $m+b=a$.

Answer 1

The Correct Option is C

Step 1: Set variables and translate the statements.
Let the unit prices be: apple $=a$, mango $=m$, banana $=b$.
Let the numbers purchased be: apples $=A$, mangoes $=M$, bananas $=B$.
Given $M=B$ (same number of mangoes and bananas). Let $M=B=n$. Step 2: "If mangoes cost as apples, bananas must be foregone."
With the hypothetical $m=a$, and keeping the same total spend (₹ 140), the cost of \emph{only} apples and mangoes uses the full amount: \[ Aa + Ma = 140. \] But in reality, \[ Aa + Mm + Bb = 140. \] Equating the two totals gives \[ Ma = Mm + Bb \;\Rightarrow\; M(a-m)=Bb. \] Since $M=B=n$, this simplifies to \[ a - m = b. \qquad (1) \] Step 3: "Mangoes + Bananas cost is 50\% more than Apples."
\[ Mm + Bb = 1.5\,(Aa). \] Using $M=B=n$ and \((1)\Rightarrow m+b=a\), we obtain \[ n(m+b) = n a = 1.5\,Aa \;\Rightarrow\; n = 1.5\,A = \frac{3A}{2}. \qquad (2) \] Step 4: Use the total bill ₹ 140 to find $Aa$.
\[ 140 = Aa + n(m+b) = Aa + n a \quad (\text{since } m+b=a). \] Thus \[ 140 = (A+n)a = \Big(A + \tfrac{3A}{2}\Big)a = \tfrac{5A}{2}\,a \;\Rightarrow\; Aa = 140\cdot \tfrac{2}{5} = 56. \qquad (3) \] Step 5: Extra money to make apples as many as mangoes/bananas.
Target apples $=n=\tfrac{3A}{2}$, so extra apples needed $= n-A = \tfrac{A}{2}$.
Additional cost $= \tfrac{A}{2}\cdot a = \tfrac{1}{2}(Aa) = \tfrac{1}{2}\times 56 = \boxed{28}. $