Math Problem Statement

Harsh receives an inheritance of a certain amount from his grandfather. Of this he loses 32.5% in his effort to produce a film. From the balance, a taxi driver stole the sum of 1,00,000 that he used to keep in his pocket. Of the rest, he donated 20% to a charity. Further he purchases a flat in Ganga Apartment for 7.5 lakh. He then realises that he is left with only ` 2.5 lakh cash of his inheritance. What was the value of his inheritance?

Solution

Let the total value of Harsh's inheritance be denoted by xx.

Step 1: Loss due to film production

Harsh loses 32.5% of his inheritance in his effort to produce a film, which leaves him with: Remaining after film loss=x0.325x=0.675x\text{Remaining after film loss} = x - 0.325x = 0.675x

Step 2: Loss due to theft

After that, a taxi driver stole ₹1,00,000. This leaves him with: Remaining after theft=0.675x1,00,000\text{Remaining after theft} = 0.675x - 1,00,000

Step 3: Donation to charity

Harsh then donates 20% of the remaining amount to charity. The remaining amount before donation is 0.675x1,00,0000.675x - 1,00,000, and 20% of this amount is: Donation=0.2×(0.675x1,00,000)\text{Donation} = 0.2 \times (0.675x - 1,00,000) The remaining amount after donation is: Remaining after donation=(0.675x1,00,000)0.2×(0.675x1,00,000)=0.8×(0.675x1,00,000)\text{Remaining after donation} = (0.675x - 1,00,000) - 0.2 \times (0.675x - 1,00,000) = 0.8 \times (0.675x - 1,00,000)

Step 4: Purchase of the flat

He then purchases a flat for ₹7.5 lakh. After this, he is left with ₹2.5 lakh cash. So the remaining amount after the flat purchase is: 0.8×(0.675x1,00,000)7,50,000=2,50,0000.8 \times (0.675x - 1,00,000) - 7,50,000 = 2,50,000

Step 5: Solve the equation

Now, we solve the equation: [ 0.8 \times (0

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Math Problem Analysis

Mathematical Concepts

Percentage
Subtraction
Algebra

Formulas

Remaining after percentage loss: Remaining = x - p%x
Donation calculation: Donation = 0.2 × Remaining
Final equation: 0.8 × (Remaining after film loss - 1,00,000) - 7,50,000 = 2,50,000

Theorems

Basic Percentage Theorem
Linear Equations

Suitable Grade Level

Grades 9-12